Instructor’s ManualFor Principles of Econometrics, Fourth Edition Instructor’s Manual For Principles of Econometrics, Fourth Edition WILLIAM E. GRIFFITHS University of Melbourne R. CARTER HILL Louisiana State University GUAY C. LIM University of Melbourne SIMON YUNHO CHO University of Melbourne SIMONE SI-YIN WONG University of Melbourne JOHN WILEY & SONS, INC New York / Chichester / Weinheim / Brisbane / Singapore / Toronto PREFACE This Instructor’s Manual contains solutions to the Exercises in the Probability Primer, Chapters 216 and Appendices A, B and C in Principles of Econometrics, 4th edition, by R. Carter Hill, William E. Griffiths and Guay C. Lim (John Wiley & Sons, 2011). There are several other resources available for both students and instructors. Full details can be found on the Web page http://principlesofeconometrics.com/poe4/poe4.htm. These resources include: Answers to Selected Exercises. These answers are available to both students and instructors. They are shortened versions of the solutions in this Manual for exercises that are marked in POE4 with a *. Supplementary computer handbooks designed for students to learn software at the same time as they are using Principles of Econometrics to learn econometrics. These handbooks are available for the following software packages: EViews Stata GRETL Excel SAS Data files for all text examples and exercises. The following types of files are available: Data definition files (*.def) are text files containing variable names, definitions and summary statistics. Text files (*.dat) containing only data. Variable names are in *.def files. EViews workfiles (*.wf1) compatible with EViews Versions 6 or 7. Stata data sets (*.dta) readable using Stata Version 9 or later. Excel spreadheets (*.xlsx) for Excel 2007 or 2010. GRETL data sets (*.gdt). SAS data sets (*.sas7bdat) compatible with SAS Version 7 or later. We welcome any comments on this manual. Please feel free to contact us if you discover errors or have suggestions for improvements. William E. Griffiths
[email protected] Simon Yunho Cho
[email protected] R. Carter Hill
[email protected] Simone Si-Yin Wong
[email protected] Guay C. Lim
[email protected] October 1, 2011 CONTENTS Solutions to Exercises in: Probability Primer 1 Chapter 2 The Simple Linear Regression Model 21 Chapter 3 Interval Estimation and Hypothesis Testing 54 Chapter 4 Prediction, Goodness of Fit and Modeling Issues 97 Chapter 5 The Multiple Regression Model 132 Chapter 6 Further Inference in the Multiple Regression Model 178 Chapter 7 Using Indicator Variables 225 Chapter 8 Heteroskedasticity 271 Chapter 9 Regression with Time Series Data: Stationary Variables 308 Chapter 10 Random Regressors and Moment Based Estimation 360 Chapter 11 Simultaneous Equations Models 387 Chapter 12 Regression with Time Series Data: Non-Stationary Variables 424 Chapter 13 Vector Error Correction and Vector Autoregressive Models 448 Chapter 14 Time-Varying Volatility and ARCH Models 472 Chapter 15 Panel Data Models 489 Chapter 16 Qualitative and Limited Dependent Variable Models 527 Appendix A Mathematical Tools 576 Appendix B Probability Concepts 586 Appendix C Review of Statistical Inference 604 PROBABILITY PRIMER Exercise Solutions 1 Probability Primer, Exercise Solutions, Principles of Econometrics, 4e 2 EXERCISE P.1 (a) X is a random variable because attendance is not known prior to the outdoor concert. Before the concert, attendance is uncertain because the weather is uncertain. (b) Expected attendance is given by E( X ) x f ( x ) 500 0.2 1000 0.6 2000 0.2 1100 x (c) Expected profit is given by E (Y ) (d) E (5 X 2000) 5E ( X ) 2000 5 1100 2000 3500 The variance of profit is given by var(Y ) var(5 X 2000) 52 var( X ) 25 240,000 6,000,000 Probability Primer, Exercise Solutions, Principles of Econometrics, 4e EXERCISE P.2 (a) The completed table is Y X (b) E( X ) x f ( x) f ( x, y ) 0 1 f ( x) 10 0 10 0.18 0.00 0.07 0.00 0.30 0.45 0.18 0.30 0.52 f ( y) 0.25 0.75 10 0.18 0 0.3 10 0.52 3.4 x You should take the bet because the expected value of your winnings is positive. (c) The probability distribution of your winnings if you know she did not study is f x| y 1 f ( x,1) fY (1) for x 10, 0,10 It is given in the following table f ( x,1) fY (1) f x| y 1 10 0.00 0.75 0.0 0 0.30 0.75 0.4 10 0.45 0.75 0.6 X (d) Given that she did not study, your expected winnings are E X |Y 1 x f x| y 1 x 10 0.0 0 0.4 10 0.6 6 3 Probability Primer, Exercise Solutions, Principles of Econometrics, 4e EXERCISE P.3 Assume that total sales X are measured in millions of dollars. Then, X P X 3 P Z 3 2.5 0.3 P Z 1.6667 1 P Z 1.6667 1 0.9522 0.0478 N 2.5,0.32 , and 4 Probability Primer, Exercise Solutions, Principles of Econometrics, 4e 5 EXERCISE P.4 Extending the table to include the marginal distributions for political affiliation (PA) and CITY yields Political Affiliation (PA) CITY I D f (CITY ) Southern 0.24 0.04 0.12 0.4 Northern 0.18 0.12 0.30 0.6 f ( PA) 0.42 0.16 0.42 P R | CITY (a) (b) R Northern f ( R, Northern) fCITY ( Northern) 0.18 0.6 0.3 Political affiliation and region of residence are not independent because, for example, (c) f ( R, Northern) 0.18 f PA ( R) fCITY ( Northern) 0.42 0.6 0.252 E ( PA) R f PA ( R) I f PA ( I ) D f PA ( D) 0 0.42 2 0.16 5 0.42 2.42 E( X ) (d) E 2 PA 2 PA2 2 E ( PA) 2 E PA2 where E PA2 R2 f PA ( R) I 2 f PA ( I ) D 2 f PA ( D ) 02 0.42 22 0.16 52 0.42 11.14 Thus, E ( X ) 2 E ( PA) 2 E PA2 2 2.42 2 11.14 27.12 Probability Primer, Exercise Solutions, Principles of Econometrics, 4e 6 EXERCISE P.5 (a) The probability that the NFC wins the 12th flip, given they have won the previous 11 flips is 0.5. Each flip is independent; so the probability of winning any flip is 0.5 irrespective of the outcomes of previous flips. (b) Because the outcomes of previous flips are independent and independent of the outcomes of future flips, the probability that the NFC will win the next two consecutive flips is 0.5 multiplied by 0.5. That is, 0.52 0.25 . Go Saints! Probability Primer, Exercise Solutions, Principles of Econometrics, 4e EXERCISE P.6 (a) E ( SALES ) (b) var( SALES ) var(40710 430PRICE ) (c) E (40710 430 PRICE ) 40710 430 E ( PRICE ) 40710 430 75 8460 P SALES 6300 P Z 6300 8460 4622500 P Z 1.00465 P Z 1.00465 0.8425 2 430 var( PRICE ) 4302 25 4,622,500 7 Probability Primer, Exercise Solutions, Principles of Econometrics, 4e 8 EXERCISE P.7 After including the marginal probability distributions for both C and B, the table becomes B 0 1 2 f (c ) 0 0.05 0.05 0.05 0.15 1 0.05 0.20 0.15 0.40 2 0.05 0.25 0.15 0.45 f (b) 0.15 0.50 0.35 C (a) (b) The marginal probability distribution for C is given in the last column of the above table. E (C ) c f (c ) 0 0.15 1 0.40 2 0.45 1.3 c (c) c 2 f (c ) var(C ) E (C ) 2 02 0.15 12 0.40 2 2 0.45 (1.3) 2 0.51 c (d) For the two companies’ advertising strategies to be independent, the condition f (c , b ) fC (c) f B (b) must hold for all c and b. We find that f (0,0) 0.05 fC (0) f B (0) 0.15 0.15 0.0225 Thus, the two companies’ advertising strategies are not independent. (e) (f) Values for A are given by the equation A 5000 1000 B . Its probability distribution is obtained by matching values obtained from this equation with corresponding probabilities for B. A f (a) 5000 6000 7000 0.15 0.50 0.35 Since the relationship between A and B is an exact linear one, they are perfectly correlated. The correlation between them is 1. .5 as the number of rolls increases. 4e 9 EXERCISE P.2.5 and are expected to become closer as the number of rolls increases. Exercise Solutions. (d) E X2 x 2 f ( x) 12 x (e) (f) var( X ) E X2 E( X ) 1 6 22 2 1 6 32 1 6 42 15.52 1 6 52 1 6 62 1 6 15.4.16667 3.4.200 X 20 3.6.3.91667 The results for this part will depend on the rolls obtained by the student.1. the average of all the values shown will be 3. Principles of Econometrics.4.5 means that if a die is rolled a very large number of times.2.3 X10 3.5.5.2. it will approach 3. Let X n denote the average value after n rolls.5.000 2.200 These values are relatively close to the mean of 3.4. The values obtained by one of us and their averages are: 20 values of X X5 3.3.Probability Primer.2.1.4.5 The result E ( X ) 3.5.16667 2.8 (a) (b) P X (c) E( X ) X f ( x) 1 16 2 16 3 16 4 16 5 16 6 16 1 6 4 P X x f ( x) 1 x 1 6 2 1 6 3 4 or X 1 6 1 6 4 1 6 5 5 1 6 1 6 6 1 3 1 6 3.2. 4 . Recalling that the formula for the area of a triangle is half the base multiplied by the height. The latter probability is X 5 2 5 9 25 36 0.27778 1 2 3 4 2 9 4 3 X 1 4 1 2 f 3 4 . When x 1 2 . Principles of Econometrics.69444 Therefore. Exercise Solutions. The probability is given by the area under the triangle between 0 and 1/2.Probability Primer.30555 To compute this probability we can subtract the area under the triangle between 3/4 to 3 from the area under the triangle from 1/4 to 3. Doing so yields P 14 X 34 P 1 4 1 2 2 1 2 121 144 5 18 X 3 4 f 3 P 1 4 1 2 11 11 4 18 9 16 0. it is given by (b) 1 2 3 2 3 1.6 .9 (a) 0 . This can be calculated as 1 P 1 / 2 P 1/ 2 X 3 1 bh 2 1 2 3 .2/9x 0 1 2 x 3 The area under the curve is equal to one.8 f(x)=2/3 . f ( x) 5 9 . 4e 10 EXERCISE P. P (0 (c) X 1 / 2) 1 P 1 / 2 X 3 25 36 1 11 36 0.2 fx . 5 var( X ) var(Y ) 2cov( X .5 var( Z ) 2 Y 2 2 1 2 2 1 2 2 var( X ) var(Y ) 1 ( 4 2 2 2 ) . var( Z ) (c) 1 2 2 X var 2 Assuming that cov( X . Y ) 2 2 3 4 2 . Principles of Econometrics. 4e 11 EXERCISE P.10 (a) (b) E (Z ) E X Y 1 ( 2 E ( X ) E (Y ) ) Assuming X and Y are independent.Y ) 0. Exercise Solutions.Probability Primer. X var 1 4 Y 2 2 0. 645) 0.6 1.11 Let X denote the length of life of a personal computer selected at random.319 (which is approximately 16 months) .4743 0.645 X 0 3. The fraction of computers that fail within a given time interval is equal to the probability that X lies in that interval.7115 0.645 1.6824 0.64 Solving for X 0 yields X0 3.05 . P( Z 1.2384 0.8658 P 0.6 P Z Z 0.4 1. 1 3.4743 0.5 X 4 P P Z 2.Probability Primer. Principles of Econometrics.6 P Z 0.7115 Z 0. Now.0289 0.6 (a) P X 1 P Z (b) P X 4 P Z 4 3.4 1.3176 1.4 1.5 3.4 1.1068 4 3.05 .4 1.4 1.8974 0.4743 1.6 P Z (d) P 2. 4e 12 EXERCISE P. and thus a suitable X 0 is such that 1. Exercise Solutions.444 (e) We want X 0 such that P X X0 0.6 P Z (c) P X 2 P Z 2 3.4 1. either 2.22 x 0 5 0.20 . 4e EXERCISE P.00 0 1 2 3 4 5 6 7 x (b) The probability that.04 0.22 0. or 4 students will be absent is 4 f ( x) f (2) f (3) f (4) 0.12 (a) The probability function of X is shown below.26 32 0. Principles of Econometrics.30 .5944 1. f ( x) 0 0.04 7 0.02 12 0.01 0.Probability Primer. on a given Monday. Exercise Solutions.03 2 0.08 6 0.16 3 25.2627 E (7 X 3) 7 E ( X ) 3 7 3. more than 3 students are absent is 7 f ( x) f (4) f (5) f (6) f (7) 0. the average number of students absent on Mondays is 3.02 1 0.01 = 11. or 3. .35 x 4 (d) 7 E( X ) x.15 .16.01 3.82 x 2 (c) The probability that.04 7 2 0.08 0.26 0.26 3 0.5944 78.22 52 0.12 var(Y ) var(7 X 3) 72 var( X ) 49 1.58 2 var( X ) 2 (f) E (Y ) 11.22 0.03 22 0.40 f(x) .58 (3.10 .34 x 0 42 0.08 62 0.1256 13 . (e) E X2 var( X ) 7 E X2 [ E ( X )]2 x 2 f ( x) 02 0.35 .34 0.16 Based on information over many Mondays. on a given Monday.05 .34 4 0.25 .16)2 1. 04 2 .25) 0.07 0.87%.62% to 12. P(Y 0) P Z P( X 0.1265 The calculations show that the probability of a negative return has increased from 10. Whether fund managers should or should not change their portfolios depends on their risk preferences.05 0.15) 0 0. 0.0062 Let Y be the return from the alternative portfolio.15 0.04 P Z P( Z 0.5) 0. Then.13 Let X be the annual return from the mutual fund. . 4e 14 EXERCISE P.65%.15 0.04 1.56% to 15.15) (c) P Z 0 0.05.05 0.1587 P( Z 1. X ~ N 0. Then.07. (a) P( X 0) (b) P( X 0. 0. Principles of Econometrics. while the probability of a return greater than 15% has increased from 0. Y ~ N 0.1429) 0.07 0.07 P Z P( Z 0.Probability Primer.1056 P( Z 2.07 1) 0. Exercise Solutions.072 . 14 Expressing the returns in terms of percentages. we have RA E ( P) (a) E 0.252 82 var( P ) (c) A 0.75 96 121 121 11 P When cov RA . RB var( P ) 0.252 var RA 0.25RA 0. RB ) 8 12 96 0. RB 1 var RA var RB Hence.252 82 0 . and the variance and standard deviation of the portfolio are 0.5 8 12 0.75E RB 0.75 cov RA . cov RA . 0.25RA 0. cov RA .25 4 0.752 12 2 48 2 0. Exercise Solutions.252 82 85 9. RB Now.25E RA 4.752 12 2 85 .752 var RB 2 0.5 0. 122 . 82 and RB 8. Principles of Econometrics.25 0. RB When var( P ) P var RB A B 0.25 0.75RB 0.75RB 0. 4e 15 EXERCISE P.Probability Primer.75 8 7 2 P var P (b) var 0. cov( R A .75 48 103 103 10.15 P (d) B 0.5 var RA cov RA .22 0 . RB 0.752 12 2 2 0.25 0. 5)2 (2 1.5 5.5) (2 1.75 1.5) (12 5.5) (2 5.5 ( 3.5) ( 7 1.5) 0.25 66 109 0. 4e EXERCISE P.5 (7 1.52 2.5 x (7 1.6055 16 .5) 2.5) (4 1.5)2 5.5) (3 5. Principles of Econometrics.5) ( 8. Exercise Solutions.15 (a) 2 xi i 1 (b) x1 4 x i 1 (c) 4 7 2 9 1 x1 4 xi 4 xi i 1 x2 x x1 x2 x3 x x2 1 7 2 4 7 4 x4 x x3 x x4 1.75 6.5) (5 5.5) 6.5 49 4 16 49 4 2.Probability Primer.5) (4 1.5) 5.5)2 (e) 4 y i 1 4 i 1 xi 1 y1 4 yi 4 x yi y y2 y3 109 1 5 2 3 12 4 y4 5.5 ( 2.52 ( 8.52 0.5) 5.5) ( 7 1.5 ( 0.25 55.5 0 (d) 4 xi i 1 2 x x1 x 2 x2 x 2 x3 x 2 x4 x 2 (7 1.5 2.25 66 4 (f) i 1 xi yi 4 i 1 2 i x 4 x 4 x y 2 x1 y1 x12 x2 y 2 x22 x3 y3 x4 y4 4 xy x32 x42 4 x 2 7 5 2 2 4 3 ( 7) 12 4 1.5 2.5)2 ( 7 1.5 0.5 8.5)2 (4 1.5) (2 1. Probability Primer.16 (a) (b) (c) (d) (e) (f) x1 x2 x3 x2 x3 3 i 2 4 x4 i 1 xi xi x1 y1 x2 y2 x3 y3 x4 y4 x1 y3 x2 y 4 x3 y5 x4 y 6 x3 y32 x4 y42 ( x1 y1 ) ( x2 4 i 3 4 i 1 4 i 1 xi yi xi yi 2 xi yi2 y2 ) ( x3 y3 ) 3 i 1 ( xi yi ) 17 . Exercise Solutions. Principles of Econometrics. 4e EXERCISE P. 4e EXERCISE P.Probability Primer. y ) f (6) f (2. Exercise Solutions. y ) f (0. y ) f (5) f (1. y ) x 0 (f) 4 2 x 2y 1 ( x 2 y) 4 x 2 ( x 2 1) ( x 2 2) 4 (2 x 6) x 2 (2 2 6) (2 3 6) (2 4 6) 10 12 14 36 2 3 2) 18 .17 (a) 4 i 1 (a bxi ) (a bx1 ) (a bx2 ) ( a bx3 ) (a bx4 ) 4a b( x1 x2 (b) 3 i 2 12 22 x3 x4 ) 32 1 4 9 14 i 1 (c) 3 ( x2 2 x 2) (02 2 0 2) (12 2 1 2) (22 2 2 2) (32 x 0 = 2 + 5 + 10 + 17 = 34 (d) 4 f ( x 2) f (2 2) f (3 2) f (4 2) x 2 f (4) (e) 2 f ( x. Principles of Econometrics. Probability Primer. 4e EXERCISE P. Principles of Econometrics.18 (a) x 4 i 1 (b) 4 i 1 xi xi 4 x ( x1 x1 x2 x 1 3 (c) 4 i 1 xi x 2 x1 x 1 3 (d) 4 i 1 (e) n i 1 xi 2 4 x 2 xi x 2 x12 n i 1 n i 1 xi 2 x4 ) 4 x2 x (1 3 5 3) 4 x3 3 3 5 3 2 x x2 2 3 3 x22 xi 2 x3 x32 2 n i 1 x 3 3 2 2 x3 5 3 x42 4 x 2 xi x 2nx 2 n x2 i 1 nx 2 x4 x 2 i 1 x 0 2 3 3 x4 x 2 8 2 1 9 25 9 4 32 n i 1 n 3 xi 2 xi 2 nx 2 2 xn 1 n xi ni1 8 nx 2 19 . Exercise Solutions. 19 n i 1 xi x yi y n i 1 n i 1 n i 1 n i 1 xi yi n i 1 xi y n i 1 1 n xi ni1 xi yi yn xi yi 2nxy xi yi nxy nxy n xyi xy i 1 xn 1 n yi ni1 nxy 20 .Probability Primer. Exercise Solutions. 4e EXERCISE P. Principles of Econometrics. CHAPTER 2 Exercise Solutions 21 . 4 2. Exercise Solutions.4 1.6 0. (c) 5 i 1 5 i 1 5 i 1 5 i 1 xi2 xi yi xi2 xi yi 02 12 22 32 42 30 0 6 1 2 2 3 3 1 4 0 11 N x2 30 5 22 10 5 i 1 N x y 11 5 2 2.2 0.4 0.22 Chapter 2.01 0.7 1.6 1. (b) y xi 12 x xi 0 2 y y x x 3.7 0. Principles of Econometrics.3 b2 is the estimated slope of the fitted line.1 0.1 (a) x y x x x x 0 1 2 3 4 6 2 3 1 0 -2 -1 0 1 2 4 1 0 1 4 xi = yi 10 x 2.4 x 2 y 10 y 7.3 0.4 0 1.8 y x x 0 y 13 2.2 eˆi = 0 eˆi2 1 2.8 xi eˆi = 0 y .4 xi 13 x 5 i 1 2 xi x yi y (d) yˆ i xi yi 0 1 2 3 4 xi = 6 2 3 1 0 yi = 5 3.1 0.2 0.3 y xi eˆi 0 1.89 0.4 4.04 eˆi2 = 4.2 yˆi = 10 12 12 eˆi 1 1. 4e EXERCISE 2.4 x b2 x x y x y 13 10 2 1. b1 y b2 x 2.3 2 5 b1 is the estimated value of E ( y ) when x 0 .6 0.4 1. it is the intercept of the fitted line.7 2.36 0. x (g) Given b1 5.4 b1 b2 x 2. we have y 12 5 2. 2.14333 2. 5 1.4 . Principles of Econometrics.4333 10 0. Exercise Solutions.1 Observations and fitted line 0 1 2 x y 3 4 Fitted values (f) See figure above.1 (continued) (e) 0 2 4 6 Figure xr2.3 2 y 2.3 1. The fitted line passes through the point of the means.4 y 4. 4e Exercise 2.23 Chapter 2.4333 3 ˆ2 xi x 2 1. b2 (h) yˆ yˆ i N (i) ˆ2 (j) var b2 eˆi2 N 2 1.4 .3 and y b1 b2 x . 3 .2 190 5 .8413 .2 .1 P X 0 (b) 1.Chapter 2. 4e EXERCISE 2.4 Figure xr2-2a -2 -5 P X y|x $2000 190 2 y| x $2000 P Z y| x $2000 2 y| x $2000 190 200 100 1 P Z 1 0.9104 0 . Principles of Econometrics.2 (a) P 180 X 215 180 P X y|x $2000 y| x $2000 2 y| x $2000 180 200 100 P P 215 2 y| x $2000 215 200 100 Z 2 Z 1. Exercise Solutions.1 f(z) .3 .5 0.5 0 z -5 -1 0 z y| x $2000 2 y|x $2000 5 24 .4 Figure xr2-2b f(z) . 8667 1.9391 (d) P X 190 P X P Z y|x $2000 2 y| x $2000 190 200 81 1 P Z 0. Principles of Econometrics. Exercise Solutions.2222 Z 1. 4e Exercise 2.2 (continued) (c) P 180 X 215 P 180 y|x $2000 2 y| x $2000 180 200 81 P P X Z y| x $2000 2 y| x $2000 215 200 81 2.6666 0.1111 190 y| x $2000 2 y| x $2000 215 y| x $2000 2 y|x $2000 25 .Chapter 2. Exercise Solutions.5 3. it has been omitted.5 1.5 The least squares estimates are: x x b2 b1 y x x y b2 x y 26.514286 3. This implies that the least-squares estimated line passes through the point ( x .5 5. The line drawn for part (a) will depend on each student’s subjective choice about the position of the line.5 17.8 1.3 (a) The observations on y and x and the estimated least-squares line are graphed in part (b). Principles of Econometrics.26 Chapter 2.5 x xi N 21 6 3.5 x 33 xi x yi y 26. 4e EXERCISE 2. That is. the predicted value at the sample mean x is the sample mean of the dependent variable y . (b) Preliminary calculations yield: xi y 21 yi 5.514286 3.5 The predicted value for y at x yˆ x 4 5 6 Fitted values x is b1 b2 x 10. This point is at the intersection of the two dashed lines plotted on the graph in part (b) .5 2 4 6 8 10 Figure xr2.3 Observations and fitted line 1 2 3 y (c) y yi N 33 6 5.5 10. y ) . For this reason.5 We observe that yˆ b1 b2 x y .5 xi x 2 17.8 2 5. .514286 1. 714286 0.257143 6 1.27 Chapter 2.3 (Continued) (d) The values of the least squares residuals. xi eˆi 1 0.285714 4 0.257143 0.228571 1.228571 1.257143 b1 b2 xi .228571 3 5 =0 yi 1.228571 1.257143 1. Exercise Solutions. Principles of Econometrics.714286 2 0. 4e Exercise 2. computed from eˆi Their sum is (e) eˆi xi yi 1 2 3 4 5 6 10 8 5 5 2 3 yˆ i yi eˆi 0. are: .285714 0. 4 (a) If 1 0. 0).2 2. The sum of squares function becomes 2 N ( yi2 i 1 91 2 2 2 xi yi xi2 ) 352 352 91 yi2 2 xi yi 2 xi2 2 40 35 SUM_SQ 30 25 20 15 10 1. we have b xi2 xi yi or b xi yi xi2 Thus. (d) that minimizes S ( ) we obtain To find the value of dS d 2 xi yi 2 xi2 Setting this derivative equal to zero.6 1. 4e EXERCISE 2.8 2. the least-squares estimate is b2 176 1. setting 1 0 implies the mean of the simple linear regression model E ( yi ) 2 xi passes through the origin (0.0 2. Exercise Solutions.4 BETA Figure xr2.95 that we obtained geometrically. 2 1.9341 91 which agrees with the approximate value of 1.95. the simple linear regression model becomes yi x ei 2 i (b) Graphically. Principles of Econometrics. (c) To save on subscript notation we set S( ) N i 1 ( yi xi )2 352 2 176 .4(a) Sum of squares for 2 The minimum of this function is approximately 12 and occurs at approximately The significance of this value is that it is the least-squares estimate.28 Chapter 2. . 7.3846. Principles of Econometrics.6044 3.7363 4 0.6703 b2 xi are: eˆ3 eˆ6 1.4 (Continued) (e) 12 10 Y1 8 * (3.1978 0.333) 6 4 2 0 0 1 2 3 4 5 6 X1 Figure xr2.1978 3 0. (f) The least squares residuals.1319 2 1.5.7363 eˆi eˆ2 eˆ5 yi 2.6044 6 0 1 0.4 (b). Exercise Solutions. 4e Exercise 2.4(b) Fitted regression line and mean The fitted regression line is plotted in Figure xr2.6703 5 0.0659 1 2. . y ) does not lie on the fitted line in this instance.0659 0. Note this value is not equal to zero as it was for 2. Note that the point ( x . obtained from eˆi eˆ1 eˆ4 Their sum is (g) xi eˆi 2.29 Chapter 2.1319 0. 4e 30 EXERCISE 2.5 The consultant’s report implies that the least squares estimates satisfy the following two equations b1 500b2 10000 b1 750b2 12000 Solving these two equations yields 250b2 2000 b2 2000 250 8 b1 6000 Therefore. Principles of Econometrics. Exercise Solutions. the estimated regression used by the consultant is: SALES 6000 8 ADVERT 8000 sales 10000 12000 14000 Figure xr2.5 Regression line 6000 (a) 0 200 400 advert 600 800 1000 .Chapter 2. Clearly. it is impossible to sell 240 sodas and so this estimate should not be accepted as a sensible one. (b) If temperature is 80 F.6 (a) The intercept estimate b1 240 is an estimate of the number of sodas sold when the temperature is 0 degrees Fahrenheit. This estimate does make sense. the predicted number of sodas sold is yˆ (c) 240 8 80 400 If no sodas are sold.6 Regression line -200 (d) 0 20 40 x 60 80 100 . The slope estimate b2 8 is an estimate of the increase in sodas sold when temperature increases by 1 Fahrenheit degree. then the estimated relationship may not be a good approximation to reality in that region.Chapter 2. If we have no observations in the region where temperature is 0. One would expect the number of sodas sold to increase as temperature increases. Exercise Solutions. 4e 31 EXERCISE 2. A graph of the estimated regression line: 0 y 200 400 600 Figure xr2. A common problem when interpreting the estimated intercept is that we often do not have any data points near x 0 . and 0 240 8x or x 30 Thus. y 0. Principles of Econometrics. she predicts no sodas will be sold below 30 F. 879 For Arkansas eˆi yi yˆ i yi b1 b2 xi 12.742 N x 2 .04672 0.139 2 xi2 xi2 (f) ˆ2 2 2088.04672 ( N 2) 2.3 0.04672 49 100.031305 Also. Exercise Solutions.04672 it follows that eˆi2 (b) 2.00098 0.274 2.32 Chapter 2.18 suggests that a 1% increase in the percentage of males 18 years or older who are high school graduates will lead to an increase of $180 in the mean income of males who are 18 years or older.1392 = 245. we have xi x 2 N x2 2088. (d) (e) x 2. ˆ2 ( xi x ) 2 var(b2 ) Thus. 4e EXERCISE 2.7 (a) Since ˆ eˆi2 N 2 2 2.5 The value b2 0.18 58. xi (c) b1 Since y b2 x var b2 2.29 The standard error for b2 is se(b2 ) var(b2 ) 0.742 0.962 .18 69.187 0.00098 xi x 15.5 51 69. Principles of Econometrics. 2 x2 x2 2 x1 2 2 x1 2 x2 2 x1 2 33 . then bEZ ~ N ki2 var ei ki yi ) 1 2 x1 2. = kN = 0 Thus.Chapter 2.8 (a) The EZ estimator can be written as y2 x2 bEZ y1 x1 1 x2 1 y2 x1 x2 x1 y1 ki yi where k1 1 . 4e EXERCISE 2. Exercise Solutions. Principles of Econometrics.. 2 .. (c) The variance is given by var bEZ var( 1 2 x2 (d) If ei ~ N 0. (b) Taking expectations yields E bEZ y2 x2 E y1 x1 1 x2 1 x2 x1 x x2 x2 x 2 1 x2 2 x1 x2 E y1 x1 1 x1 x 2 1 x1 x2 1 2 2 2 2 x2 x1 x 1 1 E y2 x1 x1 x2 x1 2 ki2 2 Thus. and k3 = k4 = . k2 x2 x1 1 x2 x1 . bEZ is an unbiased estimator. bEZ is a linear estimator. This last inequality clearly holds. Stuff to the Gauss Markov theorem.34 Chapter 2.Z. Principles of Econometrics. we need to show that 2 2 x2 2 2 x1 xi x 2 or that x1 x xi x x2 2 x1 2 2 Consider x2 2 x1 x2 x 2 2 x2 x 2 x1 x 2 2 2 x2 x x1 x 2 Thus. Thus. we need to show that 2 N i 1 xi x 2 x2 x 2 x1 2 x 2 x2 x x1 x or that x1 x 2 x2 x x2 x 2 2 x2 x x1 x 2 N i 3 xi x 2 0 or that x1 x 2 2 N i 3 xi x 2 0. as we have done above. Stuff that var(b2) < var(bEZ).Z. . we could also refer Professor E.8 (continued) (e) To convince E. bEZ is not as good as the least squares estimator. Rather than prove the result directly. 4e Exercise 2. Exercise Solutions. .9 (a) Plots of the occupancy rates for the motel and its competitors for the 25-month period are given in the following figure. 4e 35 EXERCISE 2. 1=march 2003. competitor occupancy rates reflect overall demand in the market for motel rooms. A plot of MOTEL_PCT against COMP_PCT yields: Figure xr2. 25=march 2005 percentage motel occupancy percentage competitors occupancy The repair period comprises those months between the two vertical lines. the damaged motel and the competitors had similar occupancy rates..9b Observations on occupancy 100 90 percentage motel occupancy (b) 80 70 60 50 40 40 50 60 70 80 percentage competitors occupancy There appears to be a positive relationship the two variables. Exercise Solutions. Principles of Econometrics.Chapter 2. The graphical evidence suggests that the damaged motel had the higher occupancy rate before and after the repair period.9a Occupancy Rates 100 90 80 70 60 50 40 30 0 2 4 6 8 10 12 14 16 18 20 22 24 26 month. That is. Such a relationship may exist as both the damaged motel and the competitor(s) face the same demand for motel rooms. . During the repair period. Figure xr2.. 25=march 2005 Figure xr2.12 RELPRICE The sign of the estimated slope is negative. (d) 30 Repair period 20 residuals 10 0 -10 -20 -30 0 4 8 12 16 20 24 28 month. indicating that the model has over-predicted the motel’s occupancy rate during the repair period. we have that: E MOTEL _ PCT 1 2 REPAIR 1 1 2 if REPAIR 1 if REPAIR 0 . Exercise Solutions. All except one are negative.. (e) We would expect the slope coefficient of a linear regression of MOTEL_PCT on RELPRICE to be negative. holding other factors constant.8646 percentage points.8646 COMP _ PCT .40 0..Chapter 2. 1=march 2003. the lower the demand will be for those rooms. as expected. (f) The linear regression with an indicator variable is: MOTEL _ PCT 1 2 REPAIR e From this equation.9(d) Plot of residuals against time The residuals during the occupancy period are those between the two vertical lines. The estimated regression is: MOTEL _ PCT 166. as expected. the damaged motel’s occupancy rate is expected to increase by 0. 4e 36 Exercise 2. as the higher the relative price of the damaged motel’s rooms. The regression indicates that for a one percentage point increase in competitor occupancy rate. Principles of Econometrics.66 122.9 (continued) (c) The estimated regression is MOTEL _ PCT 21.. The competitors’ occupancy rates are positively related to motel occupancy rates. 74% This comparison supports the motel’s claim for lost profits during the repair period.37% During the non-repair period.2357 REPAIR In the non-repair period.1183 REPAIR The intercept estimate in this equation (16. the damaged motel had an estimated occupancy rate of 79. Exercise Solutions. we have MOTEL 0 b1 79. their occupancy rate was 16. MOTEL1 COMP1 . the estimated occupancy rate was 79.Chapter 2.35 13.12) 2. MOTEL0 COMP 0 . When there were no repairs. the difference between the average occupancies was: MOTEL0 COMP 0 79. the estimated regression is: COMP _ PCT 62. during the repairs it was only 2.3500 13. Principles of Econometrics.9(f) (continued) The expected occupancy rate for the damaged motel is 1 2 during the repair period. (h) The estimated regression is: MOTEL _ PCT COMP _ PCT 16. COMP 0 b1 62. . 4e 37 Exercise 2.35 13.35% MOTEL1 b1 b2 79.11% For competitors. During the repair period. (g) From the earlier regression.8825 REPAIR Thus.37 2.11%.8611 14.86% higher than that of their competitors.86% During the repair period it was MOTEL1 COMP1 66.35 62. The estimated regression is: MOTEL _ PCT 79. it appears the motel did suffer a loss of occupancy and profits during the repair period.35%. Thus 2 is the difference between the expected occupancy rates for the damaged motel during the repair and non-repair periods.74% higher. The sum of the two coefficient estimates 16.49% COMP1 b1 b2 62.49 16. it is 1 outside of the repair period.11 63.24 66.86 ( 14.74 is equal to the difference in average occupancies during the repair period.24 = 66.86) is equal to the difference in average occupancies during the non-repair period. Thus.49 0.4889 0.88 63. This relationship exists because averaging the difference between two series is the same as taking the difference between the averages of the two series. 38 Chapter 2.4140 .4140 j The stocks Microsoft. .2614 1. The fitted regression line and data scatter for Microsoft are plotted in Figure xr2. General Electric.4 -. Principles of Econometrics.20 -.8993 1.10 Scatter plot of Microsoft and market rate (d) The estimates for j given j 0 are as follows.1878 0.05 .2622 1.1 -. 4e EXERCISE 2.15 -. (c) Firm b1 = ˆ Microsoft General Electric General Motors IBM Disney ExxonMobil 0. it has not changed the aggressive or defensive nature of the stock.0011 0.5 . .0079 j All estimates of the j are close to zero and are therefore consistent with finance theory.0012 0.8979 0.3189 0. x rm rf .3189 .1 .0059 0.1882 0.8993 1.10 -.10.2 -.3185 0.0116 0. xr2.05 .3 -.2 .10 (a) The model is a simple regression model because it can be written as y where y rj rf .4134 j The restriction j = 0 has led to small changes in the ˆ j .00 . 1 2 x e (b) Firm ˆ b2 Microsoft General Electric General Motors IBM Disney ExxonMobil 1.10 MKT-RISKFREE Fig. 1 j and 2 j.4 MSFT-RISKFREE . Exercise Solutions. General Motors and IBM are aggressive with Microsoft being the most aggressive with a beta value of b2 1.0061 0. Firm Microsoft General Electric General Motors IBM Disney ExxonMobil ˆ 1.0 -.8978 0.3 . Disney and ExxonMobil are defensive with Exxon-Mobil being the most defensive with a beta value of b2 0. 772 suggests that expected house price increases by approximately $73. Figure xr2. Exercise Solutions.11b Observations and fitted line 1000000 800000 600000 400000 200000 0 0 2000 4000 total square feet sale price.772 SQFT The slope of 73.77 for each additional square foot of house size. 4e 39 EXERCISE 2. Once again. this estimate should not be accepted as a serious one.11 (a) Figure xr2. Principles of Econometrics.Chapter 2. The intercept term is 28. dollars 6000 Fitted values 8000 .11(a) Price against square feet for houses of traditional style (b) The estimated equation for traditional style houses is: PRICE 28408 73.408 which would be interpreted as the dollar price of a traditional house of zero square feet. A negative value is meaningless and there is no data in the region of zero square feet. 11c Observations and quadratic fitted line 1000000 800000 600000 400000 200000 0 0 2000 4000 total square feet sale price.825 That is. Figure xr2. the elasticity of price with respect to living space for a traditional home with 2000 square feet of living space is: ˆ slope SQFT PRICE d PRICE dSQFT SQFT PRICE 2 0.825%.40 Chapter 2.012063 SQFT 2 PRICE The marginal effect on price of an additional square foot is: d PRICE slope 2 0. Exercise Solutions.25 That is.012063 SQFT dSQFT For a home with 2000 square feet of living space. we first need to compute an estimate of the expected price when SQFT 2000 : PRICE 68710 0.0120632 2000 2000 116963 0. dollars tangent 6000 Fitted values 8000 .012063 2000 48.11 (continued) (c) The estimated equation for traditional style houses is: 68710 0. for a 2000 square foot house. Principles of Econometrics. To obtain the elasticity.0120632 2000 2 116963 Then.25. the marginal effect is: d PRICE dSQFT 2 0. 4e Exercise 2. an additional square foot of living space for a traditional home of 2000 square feet is expected to increase its price by $48. we estimate that a 1% increase in house size will increase price by 0. 11d Residuals from quadratic relation 400000 Residuals 200000 0 -200000 -400000 0 2000 4000 total square feet 6000 8000 The magnitude of the residuals tends to increase as housing size increases suggesting that 2 [homoskedasticity] could be violated. The larger residuals for larger SR3 var e | x houses imply the spread or variance of the errors is larger as SQFT increases. 4e 41 Exercise 2.11d Residuals from linear relation 600000 Residuals 400000 200000 0 -200000 0 2000 4000 total square feet 6000 8000 Figure xr2.Chapter 2. .11 (continued) (d) Residual plots: Figure xr2. indicates a lower value for the squared distance between a regression line and data points. or sum of squared residuals. Principles of Econometrics. Exercise Solutions. A lower SSE. Or. (e) SSE of linear model.23 1012 The quadratic model has a lower SSE. (c): SSE eˆi2 1. in other words. (b): SSE eˆi2 1.37 1012 SSE of quadratic model. there is not a constant variance of the error term for all house sizes. indicating a line that better fits the data. then take the antilogarithm to make predictions in terms of PRICE. dollars tangentl (g) 6000 8000 pricel The SSE from the log-linear model is based on how well the model fits ln PRICE .11f Observations and log-linear fitted line 1000000 800000 600000 400000 200000 0 0 2000 4000 total square feet sale price. Exercise Solutions. the SSE from this specification is not comparable to the SSE from the models with PRICE as the dependent variable. This is smaller than the SSE from the fitted linear relationship. with a tangent line included.11 (continued) (f) The estimated equation for traditional style houses is: ln PRICE 10. . Since the log scale is compressed.Chapter 2. 4e 42 Exercise 2. Principles of Econometrics.000413235 SQFT The fitted line. One way to correct this problem is to obtain the predicted values from the log-linear model. is Figure xr2. but not as small as the SSE from the fitted quadratic model.31 1012 .79894 0. Then a residual can be computed as eˆ PRICE exp ln PRICE Using this approach the SSE from log-linear model is 1. hundreds of square feet selling price of home.7 LIVAREA The coefficient 9181.12(a) Scatter plot of selling price and living area (b) The estimated equation for all houses in the sample is SPRICE 30069 9181. Exercise Solutions.Chapter 2. a meaningless value.12b Observations and fitted line 800000 600000 400000 200000 0 10 20 30 40 living area. The intercept. Principles of Econometrics. Figure xr2. The model should not be accepted as a serious one in the region of zero square feet. Figure xr2. dollars Fitted values 50 .7 suggests that selling price increases by approximately $9182 for each additional 100 square foot in living area.069.12 (a) The scatter plot in the figure below shows a positive relationship between selling price and house size. 4e 43 EXERCISE 2. if taken literally. suggests a house with zero square feet would cost $30. it is better at capturing the proportionally higher prices for large houses. (d) Figure xr2.33 That is.12 (continued) (c) The estimated quadratic equation for all houses in the sample is 57728 212.83LIVAREA2 The estimated equation for houses that are on small lots in the sample is: SPRICE 62172 186. dollars Fitted values 50 Fitted values The quadratic model appears to fit the data better.611 LIVAREA2 SPRICE The marginal effect of an additional 100 square feet is: d SPRICE slope dLIVAREA 2 212. Exercise Solutions. adding 100 square feet of living space to a house of 1500 square feet is estimated to increase its expected price by approximately $6378. SSE of linear model.12d Linear and quadratic fitted lines 800000 600000 400000 200000 0 10 20 30 40 living area.23 1012 SSE of quadratic model. Principles of Econometrics. hundreds of square feet selling price of home. 4e 44 Exercise 2.86 LIVAREA2 . (c): SSE eˆi2 2. indicating that it is a better fit. the marginal effect is: d SPRICE dLIVAREA 2 212.Chapter 2. (e) The estimated equation for houses that are on large lots in the sample is: SPRICE 113279 193. (b): SSE eˆi2 2.611 15 6378.03 1012 The SSE of the quadratic model is smaller.611 LIVAREA For a home with 1500 square feet of living space. Figure xr2.66 LIVAREA for houses on large lots and $373. with some very large positive values. The estimated intercept.e..12f sprice vs age regression line 800000 600000 400000 200000 0 0 20 40 60 age of home at time of sale. The following figure contains the scatter diagram of PRICE and AGE as well as the estimated equation which is SPRICE 137404 627. years selling price of home. 4e 45 Exercise 2.12f residuals from linear model 600000 400000 Residuals (f) 200000 0 -200000 0 20 40 60 80 age of home at time of sale. a new house) would cost $137. years 100 . Exercise Solutions. dollars 80 100 Fitted values We estimate that the expected selling price is $627 less for each additional year of age. suggests a house with zero age (i. The marginal effect of living area on price is $387. if taken literally. The model residuals plotted below show an asymmetric pattern. For these observations the linear fitted model under predicts the selling price. which is 2 2 LIVAREA . we estimate that the mean price of large lots is $113.279 and the mean price of small lots is $62. Thus. Principles of Econometrics.12(e) (continued) The intercept can be interpreted as the expected price of the land – the selling price for a house with no living area.Chapter 2. The coefficient of LIVAREA has to be interpreted in the context of the marginal effect of an extra 100 square feet of living area.72 LIVAREA for houses on small lots.16 AGE Figure xr2.404.172. 48%. and find an estimate of the price of a new home as exp ln SPRICE exp(11. To find an interpretation from the intercept.12f log(sprice) vs age regression line 14 13 12 11 10 0 20 40 60 age of home at time of sale. the log-linear model is preferred. we set AGE 0 . Figure xr2. Principles of Econometrics. The estimated expected selling price for a house not on a large lot (LGELOT = 0) is $115220.00476 AGE Figure xr2. years 80 100 The estimated equation for all houses is: SPRICE 115220 133797 LGELOT The estimated expected selling price for a house on a large lot (LGELOT = 1) is 115220+133797 = $249017. Exercise Solutions.746 0. based on the plots and visual fit of the estimated regression lines. the residuals are distributed more symmetrically around zero.74597) $126.244 The following residuals from the fitted regression of ln(SPRICE) on AGE show much less of problem with under-prediction. Thus.12(f) (continued) The following figure contains the scatter diagram of ln(PRICE) and AGE as well as the estimated equation which is ln SPRICE 11.12f transformed residuals from loglinear model 2 Residuals 1 0 -1 -2 0 (g) 20 40 60 age of home at time of sale. 4e 46 Exercise 2. each extra year of age reduces the selling price by 0. . years lsprice 80 100 Fitted values In this estimated model.Chapter 2. (c) The estimated equation using a sample of regular classes and regular classes with a fulltime teacher aide is: TOTALSCORE 918. as measured by higher totals of all achievement test scores.13 (a) The estimated equation using a sample of small and regular classes is: TOTALSCORE 918.31 = 918. This is a 1.0 while students in regular classes with a teacher aide achieve an average total score of 918. (d) The estimated equations using a sample of regular classes and regular classes with a fulltime teacher aide are: READSCORE 434.08 = 491.310 0. .31 + 8. This result suggests that small classes have a positive impact on learning.7 while students in small classes achieve an average of 434.04 + 0.899SMALL Comparing a sample of small and regular classes. Exercise Solutions.4. (b) The estimated equations using a sample of small and regular classes are: READSCORE 434. is negligible. In math students in regular classes achieve an average score of 483.0 while students in small classes achieve an average of 918. This result does not differ from the case using total scores.819SMALL MATHSCORE 483.4.9 = 931.043 0. This is a 1. we find students in regular classes achieve an average total score of 918. This is a 1.0 + 13.82 = 440.6.705 AIDE MATHSCORE 483.310 8.50% increase.67% increase.043 13.Chapter 2. These results suggests that small class sizes also have a positive impact on learning math and reading. These results suggest that having a full-time teacher aide has little impact on learning outcomes as measured by totals of all achievement test scores.391AIDE The effect of having a teacher aide on learning.733 5. 4e 47 EXERCISE 2.733 0.31 while students in small classes achieve an average of 483.314 AIDE Students in regular classes without a teacher aide achieve an average total score of 918.73 + 5.080 SMALL Students in regular classes achieve an average reading score of 434. Principles of Econometrics. as measured by reading and math scores.34% increase.9. 88595 suggests that for a 1 percentage point increase in the growth rate of GDP in the 3 quarters before the election there is an estimated increase in the share of votes of the incumbent party of 0. We estimate. A graph of the fitted line and data is shown in the following figure. Principles of Econometrics.220. we obtain the predicted vote share for the incumbent party: . 4e 48 EXERCISE 2. that that the incumbent party’s expected vote is 50.877982GROWTH The actual 2008 value for growth is 0. (b) The estimated equation for 1916 to 2008 is VOTE 50.2004 is VOTE 51.88595GROWTH The coefficient 0. based on the fitted regression intercept.848% when the growth rate in GDP is zero. 30 Incumbent vote 40 50 60 xr2-14 Vote versus Growth with fitted regression -15 -10 -5 0 Growth rate before election Incumbent share of the two-party presidential vote (c) 5 10 Fitted values The estimated equation for 1916 .848 0.Chapter 2.14 (a) 30 Incumbent vote 40 50 60 xr2-14 Vote versus Growth -15 -10 -5 0 Growth rate before election 5 10 There appears to be a positive association between VOTE and GROWTH.053 0. Putting this into the estimated equation. the incumbent party will still maintain the majority vote. This suggests that when there is no real GDP growth.88595 percentage points. Exercise Solutions. 4%. The figure below shows a plot of VOTE against INFLATION.408 0.246 This suggests that the incumbent party will maintain the majority vote in 2008.14(c) (continued) VOTE 2008 51. the incumbent party did not maintain the majority vote.60. Exercise Solutions. 30 Incumbent vote 40 50 60 xr2-14 Vote versus Inflation 0 2 4 Inflation rate before election 6 8 The estimated equation (plotted in the figure below) is: VOTE = 53. There appears to be a negative association between the two variables.053 0.Chapter 2. Principles of Econometrics.877982GROWTH 2008 51. the expected share of votes won by the incumbent party is 53.053 0. However.444312 INFLATION We estimate that a 1 percentage point increase in inflation during the incumbent party’s first 15 quarters reduces the share of incumbent party’s vote by 0. Incumbent vote 40 50 60 xr2-14 Vote versus Inflation 30 (d) 0 2 4 Inflation rate before election Incumbent share of the two-party presidential vote 6 8 Fitted values . the actual vote share for the incumbent party for 2008 was 46.444 percentage points. which is a long way short of the prediction. 4e 49 Exercise 2. The estimated intercept suggests that when inflation is at 0% for that party’s first 15 quarters. the incumbent party is predicted to maintain the majority vote when inflation.220 51. is at 0%.877982 0. during its first 15 quarters. 140 Series: WAGE Sample 1 1000 Observations 1000 120 100 80 60 40 20 Mean Median Maximum Minimum Std. Principles of Econometrics.7103 represents the estimated wage rate of a worker with no years of education. respectively. The coefficient 6.30 dollars per hour. (b) The estimated equation is WAGE 6. .30000 76. Dev. 4e 50 EXERCISE 2. representing those who did not complete high school.7103 1. There are a few observations at less than 12. The maximum earned in this sample is 76. Half of the sample earns an hourly wage of more than 17.7260 0.97 dollars per hour.39 dollars per hour and the least earned in this sample is 1. Spikes at 13 and 14 years are people who had one or two years at college.15(a) Histogram and statistics for WAGE The observations for WAGE are skewed to the right indicating that most of the observations lie between the hourly wages of 5 to 40.921362 Jarque-Bera Probability 773.9803 represents the estimated increase in the expected hourly wage rate for an extra year of education. implying that they finished their education at the end of high school. The spike at 16 years describes those who completed a 4year college degree.583909 5.39000 1.000000 0 0 10 20 30 40 50 60 70 Figure xr2.61566 17. Skewness Kurtosis 20. It should not be considered meaningful as it is not possible to have a negative hourly wage rate. and that there is a smaller proportion of observations with an hourly wage greater than 40.970000 12. and further education such as a PhD. while those at 18 and 21 years represent a master’s degree.15 (a) Figure xr2. Exercise Solutions. with the average being 20.15(a) Histogram and statistics for EDUC Most people had 12 years of education.62 dollars per hour.83472 1.Chapter 2.9803EDUC The coefficient 1. 5507 1.9919 EDUC The white equation is obtained from those workers who are neither black nor Asian. There is a pattern evident.1681 2.3575 EDUC If male: WAGE 3. And an extra year of education increases the wage rate of a female worker more than it does for a male worker.0859 2. . Exercise Solutions. From the results we can see that an extra year of education increases the wage rate of a black worker more than it does for a white worker. suggesting that the error variance is larger for larger values of EDUC – a violation of assumption SR3.15(c) Residuals against education (d) The estimated equations are If female: WAGE 14.76.4491EDUC If white: WAGE 6. 4e 51 Exercise 2.08283 0. Principles of Econometrics. the estimated marginal effect of an additional year of education on expected wage is: slope d WAGE dEDUC 2 0. an additional year of education for a person with 12 years of education is expected to increase wage by $1.073489 12 1.073489 EDUC For a person with 12 years of education.7637 That is. as EDUC increases.073489 EDUC 2 The marginal effect is therefore: slope d WAGE dEDUC 2 0. 60 50 40 RESID 30 20 10 0 -10 -20 -30 0 4 8 12 16 20 24 EDUC Figure xr2. there should not be any patterns evident in the residuals.0544 1. the magnitude of the residuals also increases.8753EDUC If black: WAGE 15.Chapter 2. If the assumptions SR1-SR5 hold.15 (continued) (c) The residuals are plotted against education in Figure xr2. (e) The estimated quadratic equation is WAGE 6.15(c). the marginal effect of an additional year of education is: d WAGE slope 2 0.4 .6 . The linear model in (b) suggested that an additional year of education is expected to increase wage by $1.15(e) (continued) For a person with 14 years of education. .8 The histogram of ln(WAGE) in the figure below is more symmetrical and bell-shaped than the histogram of WAGE given in part (a). 4e 52 Exercise 2. That is.98 regardless of the number of years of education attained.073489 14 dEDUC 2. Principles of Econometrics. 0 (g) 1 2 3 lwage 4 Figure xr2.06.Chapter 2.15(f) Quadratic and linear equations for wage on education The quadratic model appears to fit the data slightly better than the linear equation. The quadratic model suggests that the effect of an additional year of education on wage increases with the level of education already attained.2 Density . (f) Figure xr2.15(g) Histogram for ln(WAGE) 5 .0577 An additional year of education for a person with 14 years of education is expected to increase wage by $2. Exercise Solutions. the rate of change is constant. 60944 0.090408EDUC We estimate that each additional year of education increases expected wage by approximately 9.76 and $2.090408 14 17. For workers with 12 and 14 years of education we predict the wage rates to be WAGE WAGE EDUC 12 EDUC 14 exp 1. For the quadratic relationship the corresponding marginal effect estimates are $1.04%. Principles of Econometrics. The estimated marginal effect of education on WAGE is dWAGE dEDUC 2 WAGE This marginal effect depends on the wage rate.6028 EDUC 14 For the linear relationship the marginal effect of education was estimated to be $1.728 Evaluating the marginal effects at these values we have dWAGE dEDUC b2 WAGE 1.06. Exercise Solutions. Figure xr12.15 (continued) (h) The estimated log-linear model is ln WAGE 1. The marginal effects from the log-linear model are lower. 4e 53 Exercise 2.60944 0.3377 EDUC 12 1.15(h) Observations with linear and log-linear fitted lines .98. A comparison of the fitted lines for the linear and log-linear model appears in the figure below.Chapter 2.60944 0.796 exp 1.090408 12 14. CHAPTER 3 Exercise Solutions 54 . 30. Exercise Solutions.416 2. The level of significance.410 Since the t = 1. tc t(0. (b) To test H 0 : 1 0 against H1 : t1 b1 1 se(b1 ) 1 0 we compute the t-value 83.1(c) we can see that having a p-value > 0.1(c) Critical and observed t values .416 0 1. 4e 55 EXERCISE 3.92 and to the right of t = 1.46.0311.975.30) We estimate that 1 lies between 4. so the area under the curve for t tc is / 2 . When b1 83. From Figure xr3. 95% of similarly constructed intervals would contain the true 1 .92.410.975. is given by the sum of the areas under the PDF for | t | | tc |. . we get the interval estimate: 83.410 = ( 4.05 is equivalent to having tc t tc . each of the tail areas that make up the p-value are p / 2 0.1 (a) The required interval estimator is b1 tc se(b1 ) . or tc t tc .46 and 171. 171. is the same as not rejecting the null hypothesis because because / 2 p / 2. Since the t distribution is symmetric.0622 2 0. In repeated samples.92 value does not exceed the 5% critical value tc t(0.024 43.416.0622 represents the sum of the areas under the t distribution to the left of t = 1.025 and likewise for t tc . The data does not reject the zero-intercept hypothesis. we do not reject H 0 . Figure xr3.024 . (c) The p-value 0. Principles of Econometrics.38) 2.024 and se(b1 ) 43.38) 2.Chapter 3.92 43. Therefore not rejecting the null hypothesis p. Exercise Solutions.0311. the alternative is accepted. the critical value is chosen such that there is a probability of 0.3 PDF 0.1 (continued) (d) Testing H 0 : 1 0 against H1 : 1 0. and we conclude that the intercept is positive. not hypothesis tests. It is careless and dangerous to equate 5% level of significance with 95% confidence.05 is equivalent to having t > 1.2 0.Chapter 3. When carrying out a two-tailed test of the form H 0 : k c versus H1 : k c.0 -3 -2 -1 0 1 tc 2 3 T Figure xr3.1 1. 4e 56 Exercise 3. 0. Because it is a one-tailed test. t = 1.92 0. The 5% significance is a probability statement about a procedure not a probability statement about 1 . providing the level of significance is equal to one minus the level of confidence. tc t(0. . which relates to interval estimation procedures.686 .92 > tc = 1. Principles of Econometrics.4 Rejection Region 0. in repeated samples we have a 5% chance of rejecting the null hypothesis when it is true. That is. (f) False.1(d) Rejection region and observed t value (e) The term "level of significance" is used to describe the probability of rejecting a true null hypothesis when carrying out a hypothesis test.92. non-rejection of H 0 implies c lies within the confidence interval.38) 1.1(d) that having the p-value < 0. That is. uses the same t-value as in part (b).95.92) = 0. The test in (d) uses the level of significance 5%. We see from Figure xr3. and vice versa.05 in the right tail. Since t = 1. H 0 is rejected. The term "level of confidence" refers to the probability of an interval estimator yielding an interval that includes the true parameter. In this case p-value = P(t > 1. which is the probability of a Type I error.69.69. Experience has a positive effect on quality rating. Since 2. 3.Chapter 3.074 0.2 3.074 . Principles of Econometrics.5 3.074 < 1. on average.6 3.044 = 1.074 we fail to reject the null hypothesis.9 3.717 .727 1. a technical artist's quality rating goes up by 0.167) We are 95% confident that the procedure we have used for constructing a confidence interval will yield an interval that includes the true parameter 2 . . 22) 1.076 2.1 0 1 2 3 4 5 6 7 8 EXPER Figure xr3.717 .727 < 2. 22) 2.95.727.8 3. The t critical value for a two tail test with N 2 = 22 degrees of freedom is 2.2(a) Estimated regression function (b) Using the value tc b2 t(0. namely t 1. 4e 57 EXERCISE 3.076 for every additional year of experience.074. we use the test statistic t = b2/se(b2) = 0.4 3.076/0.727 . Exercise Solutions. we use the t-value from part (c).3 3. 0.975.7 RATING 3. we reject H 0 and is positive. (d) To test H 0 : 2 0 against H1 : but the right-tail critical value tc conclude that 2 2 0. the 95% confidence interval for 2 is given by tc se(b2 ) 0.2 (a) The coefficient of EXPER indicates that. t(0.015. (c) To test H 0 : 2 0 against H1 : 2 0.044 ( 0. Since 1. 4e 58 Exercise 3. p-value > 0. Principles of Econometrics.727 and to the right of 1.2 (continued) (e) The p-value of 0. or fail to reject. the null hypothesis just based on an inspection of the p-value. We do not reject H 0 because.0982 is given as the sum of the areas under the t-distribution to the left of 1.727.Chapter 3. Having the p-value > is equivalent to having t tc 2. for 0.074 . We can reject.05. Exercise Solutions.2(e) p-value diagram . Figure xr3.05. 3 One tail rejection region (d) (e) Hypotheses: Calculated t-value: Critical t-value: H 0 : 2 0.508.310 2.079 and 0.05. 0.99.717 Decision: Do not reject H 0 because t 2 0 3.78 tc t(0.78 tc t(0.541 using a procedure that works 99% of the time in repeated samples.310 0.78 tc 1.082 2. Exercise Solutions. 22) 2.78 tc 2 0 3.78 tc t(0. Principles of Econometrics. A 99% interval estimate of the slope is given by b2 tc se(b2 ) = 0.310 0.717.Chapter 3. 22) 2.32 tc t(0. 4e 59 EXERCISE 3.32 tc 2. 22) 1.310 0.082 3. 2.541) We estimate 2 to lie between 0. .975.819 0.074.3 (a) (b) (c) Hypotheses: Calculated t-value: Critical t-value: H 0 : 2 0 against H1 : t 0.079.819.310 0.082 3. 0 3.78 tc 2 2.5) 0. Figure xr3.082 = (0. 22) 2.082 3.5 against H1 : 2 0.995.819 Decision: Reject H 0 because t Hypotheses: Calculated t-value: Critical t-value: H 0 : 2 0 against H1 : t 0.074 Decision: Reject H 0 because t 2.508 Decision: Reject H 0 because t Hypotheses: Calculated t-value: Critical t-value: H 0 : 2 0 against H1 : t 0.5 t (0. The positive sign is as expected.180 5.257 2. Exercise Solutions.68 0.264) . 4e 60 EXERCISE 3.995.4 (a) b1 t se(b1 ) = 1. a 99% confidence interval for the slope is given by tc se(b2 ) = 0. (e) Using tc t(0.18 indicates that a 1% increase in males 18 and older. who are high school graduates.180 2.0313 = (0.0313 (c) p-value = 2 1 P(t 1. Principles of Econometrics.4(c) p-value diagram (d) The estimated slope b2 0.257) = 2 (1 0. increases average income of those males by $180. 0.4(a) Estimated regression function (b) se(b2 ) b2 t 0.68 .8926) = 0.49) b2 2.733 24 20 MIM 16 12 8 4 0 0 10 20 30 40 50 60 70 80 90 100 PMHS Figure xr3.174 = 2.754 0. more education should lead to higher salaries.096.2147 Figure xr3.Chapter 3. 01. Principles of Econometrics. Non-rejection of H 0 means that this claim is compatible with the sample of data. 2. .0313 2 0. 4e 61 Exercise 3. we calculate 0. Since t = 0. Exercise Solutions.01.Chapter 3. we do not freedom are tc reject H 0 .639 The critical values for a two-tailed test with a 5% significance level and 49 degrees of 2.634 lies in the interval ( 2.2 against H1 : 0. leads to an increase in average income for those males of $200.2 0.2.4 (continued) (f) For testing H 0 : t 2 0.180 0. The null hypothesis suggests that a 1% increase in males 18 or older. who are high school graduates.01). 1121. Exercise Solutions.101.5 Fitted regression line and mean (b) The relationship in part (a) indicates that. (i) If income increases by $1000. . If taken literally. (ii) The standard error of b2 is 0. this value is not reliable.8802 5 0. The rejection region (18 degrees of freedom) is | t | > 2. then they would purchase $6855 worth of insurance. where tc is the critical level of significance. then an estimate of the resulting change in the amount of life insurance is $3880.3835 0.101 .1121 9.5 (a) The linear relationship between life insurance and income is estimated as INSURANCE (se) 6. However. before we sample. To test the claim.99 2. we reject the null hypothesis and conclude that the estimated relationship does not support the claim. The value of the test statistic is t b2 5 se b2 3. we have no reason to suspect 2 > 5 or 2 2 < 5. the value of b1 = 6. 4e 62 EXERCISE 3.99 As t = 9.Chapter 3.20. the amount of life insurance increases.8802 INCOME 7.8550 3. given the lack of data in the region where INCOME 0 . Principles of Econometrics. The test statistic is that given in part (b) (ii) with 2 set equal to 5. b2 2) degrees of freedom at the tc se b2 .8550 implies that if a family has no income. The alternative 5 has been chosen because.1121 Figure xr3. To test a hypothesis about b2 2 ~tN se b2 (c) the test statistic is 2 An interval estimator for value for t with ( N 2 2 is b2 tc se b2 . the relevant hypotheses are H0: 2 = 5 versus H1: 2 5. as income increases. as is expected. Thus. we do not give this value a direct economic interpretation. and to quantify its effect on insurance. One likely important determinant of life insurance cover is household income. INCOMEi is household income. To estimate our hypothesized relationship. The estimated equation could be used to assess likely requests for life insurance and what changes may occur as a result of income changes. 1 and 2 are unknown parameters that describe the relationship. Principles of Econometrics.Chapter 3. we take a random sample of 20 households.1121 The point estimate for the response of life-insurance coverage to an income increase of $1000 (the slope) is $3880 and a 95% interval estimate for this quantity is ($3645. we set up the model INSURANCEi 1 2 INCOMEi ei where INSURANCEi is life insurance cover by the i-th household.7 0.101. suggesting we have reliable information about the response. 4e 63 Exercise 3.1121 Since t 25. as mentioned in part (b). The value of the test statistic is t 3. The intercept estimate is not significantly different from zero. we proceed as we did in part (c). The estimated equation.3835 0. our hypotheses are H0: 2 = 1 versus H1: 2 1.8550 3.8802 1 25. (e) Life insurance companies are interested in household characteristics that influence the amount of life insurance cover that is purchased by different households.8802 INCOME 7. The rejection region is | t | > 2. but this fact by itself is not a matter for concern. . collect observations on INSURANCE and INCOME and apply the least-squares estimation procedure. To see if income is important. We conclude that the amount of life insurance does not increase at the same rate as income increases. is INSURANCE se 6. Exercise Solutions. and ei is a random uncorrelated error that is assumed to have zero mean and constant variance 2 . we reject the null hypothesis. using 1 instead of 5. with standard errors in parentheses. This interval is a relatively narrow one.5 (continued) (d) To test the hypothesis that the slope of the relationship is one.101. $4116).7 tc 2. One would expect both occupancy rates to be high in periods of high demand and low in periods of low demand.500 .6 (a) The estimated model is MOTEL _ PCT 21.40 0. the motel’s occupancy rate is also high.500.86464 0. The p-value is 0.Chapter 3. and vice versa.91) (0.99. 4e 64 EXERCISE 3. The calculated value of the t-statistic is t b2 se(b2 ) 0.2027) The null and alternative hypotheses are H0 : 2 0 H1 : 2 0 The test statistic and its distribution assuming the null hypothesis is true at the point 0 are 2 t b2 se(b2 ) t(23) At a 1% significance level.265 > 2. we reject H 0 when t t(0. Principles of Econometrics.6(a) Rejection region and p-value . we reject H 0 and conclude that when the competitors’ occupancy rate is high.23) 2.000145. Exercise Solutions. Figure xr3.8646COMP _ PCT (se) (12.20271 4.265 Since 4. 35 2.12 RELPRICE (se) (43. we would expect the occupancy rate to be negatively related to relative price. 4e 65 Exercise 3.01. Exercise Solutions.23) 2. There is insufficient evidence to conclude that there is an inverse relationship between MOTEL_PCT and RELPRICE.093 2.0238.57) (58.6 (continued) (b) The model is MOTEL _ PCT 1 2 RELPRICE e The null and alternative hypotheses are H0 : 2 0 H1 : 2 0 The test statistic and its distribution assuming the null hypothesis is true are t b2 se(b2 ) t(23) At a 1% significance level.500 .500 .093 Since 2. Figure xr3. at a 5% significance level we would have rejected H 0 . we reject H 0 when t t(0.Chapter 3. the data are not sufficiently informative to do so at a 1% level. From demand theory. The estimated regression is MOTEL _ PCT 166.12 58.35) The calculated value of the t-statistic is t b2 se(b2 ) 122. However. we do not reject H 0 at a 1% significance level. The p-value is 0.656 122. Principles of Econometrics. This result is a surprising one.6(b) Rejection region and p-value . and that the motel suffered a loss in occupancy during the repair period.154) (5.9606 ( 25. . The data suggest that the motel’s occupancy rate is significantly lower during the repair period.91) With 95% confidence we estimate that the effect of the repair period is to reduce the motel’s occupancy rate by a percentage between 0.2357 5. The estimated regression model is MOTEL _ PCT 79.2357 2. Exercise Solutions. Our confidence is in the procedure: 95% of intervals constructed in this way with new samples of data would yield an interval that contains 2 . (d) A 95% interval estimate for ˆ 2 2 is given by t(0. Principles of Econometrics.2357 REPAIR (se) (3.35 13.9606 2. 4e 66 Exercise 3. The effect of repairs on the occupancy rate has not been estimated precisely. If we are able to reject the null hypothesis that the difference in occupancy rates between the repair and non-repair period is zero.23)se ˆ 2 13. we reject H 0 at a 5% significance level.57.57.714 . or positive.Chapter 3. The test statistic and its distribution assuming H 0 is true at the point ˆ 2 t t(23) se ˆ 2 where ˆ 2 is the least squares estimator of 2 .714 .23) 1.975. Our interval suggests it could be anywhere from almost no effect to a 25% effect.221 1. We reject H 0 when t 2 0 are t(0. which implies that 2 0 .05.0687 5.91 and 25. 0.6 (continued) (c) The model is MOTEL _ PCT 1 2 REPAIR e The expected occupancy rate in the repair and non-repair periods is E MOTEL _ PCT 1 2 REPAIR 1 1 2 REPAIR 0 REPAIR 1 The null and alternative hypotheses are H0 : 0 2 H1 : 0 2 We wish to show that the motel occupancy rate is less during the repair period. we will then conclude “beyond reasonable doubt” that this difference is negative.9606) The calculated value of the t-statistic is ˆ t 2 se ˆ 2 13.221 Since 2. During the repair period this advantage fell by 14.87) With 95% confidence we estimate that the effect of the repair period is to reduce the difference between the motel’s occupancy rate and the competitors’ occupancy rate by a percentage between 5. This test overcomes one of the potential problems of the test in part (c). we have not estimated the effect precisely. Our test shows that this decline is statistically significant at the 0. (f) A 95% interval estimate for 2 is given by ˆ 2 t(0.36. then ignoring the competitor’s occupancy rate could have led the low demand to be incorrectly attributable to the repairs.1183REPAIR (2. The estimated regression model is MOTEL _ PCT (se) COMP _ PCT 16.23) 2. 4e 67 Exercise 3. The regression estimates show that during the non-repair period the motel enjoyed an occupancy rate 16.8611 14.118 2.1183 3.12%.36.01. we reject H 0 when t t(0.0687 3.9863) The calculated value of the t-statistic is t Since 3.542 2. we reject H 0 at a 5% significance level.1092) (3.975. .Chapter 3.542 ˆ2 se ˆ 2 14. namely.86% higher than its competitors’ rate.9863 ( 22.9863 3.87 and 22.01 level of significance.23)se ˆ 2 14. if the repair period was a period in which demand was normally low.500 . 5. but there does appear to have been a reduction in the motel’s occupancy rate.6 (continued) (e) The model is MOTEL _ PCT COMP _ PCT 1 2 REPAIR e The difference in the expected occupancy rates in the repair and non-repair periods is E MOTEL _ PCT COMP _ PCT 1 2 REPAIR 1 1 2 REPAIR 0 REPAIR 1 The null and alternative hypotheses are H0 : 2 0 H1 : 2 0 The test statistic and its distribution assuming the null hypothesis is true are t ˆ2 se ˆ 2 t(23) At a 5% significance level.500 . This interval is a relatively wide one. Exercise Solutions. Including the competitor’s occupancy rate controls for normal fluctuations in demand. Principles of Econometrics. indicating that the sample data are consistent with the conjecture that the Disney.89794 1 0. fail to reject H 0 GE t 0. is t bj 1 se b j The rejection region is t ~ t 130 1.978 and t 1.18821 1 1. reject H 0 Exxon-Mobil t 0.41397 1 0.978 . where t(0. Principles of Econometrics.89926 1 0.978 .089713 Since t 6.293 0.978 .16079 Since t 1.126433 Since 1.826 Since 1. and conclude that these stocks do not have the same volatility as the market portfolio.12363 0.130) 1. GE.20222 Since 1.26141 1 1. The results for each company are given in the following table: Stock t-value Decision rule Disney t 0. GE.532 1. Each beta measures the volatility of the stock relative to the market portfolio and volatility is often used to measure risk.978 t 1.489 0. 4e 68 EXERCISE 3.978 .978 t 1. For Microsoft and Exxon-Mobil. GM.984 0.978 .975.978 .978 t 1. Exercise Solutions. fail to reject H 0 GM t 1.31895 1 1. GM and IBM we fail to reject the null hypothesis. .978 . The economic relevance of this test is to test whether the return on the firm’s stock is risky relative to the market portfolio.978 . and IBM stocks have the same volatility as the market portfolio. fail to reject H 0 Microsoft t 1. The test statistic given H0 is true.978 t 1. A beta value of one indicates that the stock’s volatility is the same as that of the market portfolio. we reject the null hypothesis.020 Since 1.Chapter 3. reject H 0 For Disney.098782 1.7 (a) We set up the hypotheses H 0 : j 1 versus H1 : j 1 . fail to reject H 0 IBM t 1. The value of the test statistic is t 0.532 < tc = 1. A beta equal to 1 suggests a stock's variation is the same as the market variation. we reject H0 and conclude that Mobil-Exxon’s beta is less than 1. Exercise Solutions. but we would be surprised if it did.658 where tc t(0. we reject H0 and conclude that Microsoft’s beta is greater than 1. A beta less than 1 implies the stock is less volatile than the market.9836 > tc = 1.975.657.05. 1.532 Since t = 6. (d) A 95% interval estimator for Microsoft’s beta is b j t(0.Chapter 3. The j relevant test statistic. it is an aggressive stock. A beta greater than 1 implies the stock is more volatile than the market.7 (continued) (b) We set up the hypotheses H 0 : j 1 versus H1 : 1 where j = Mobil-Exxon.637. that Microsoft’s beta falls in the interval 1. The relevant test statistic. (c) We set up the hypotheses H 0 : j 1 versus H1 : j 1 where j = Microsoft.9836 0.95.6567 where t(0. with 95% confidence.6567 . A beta equal to 1 suggests a stock's variation is the same as the market variation. This result appears in line with our conclusion in both parts (a) and (c). .089713 6. is t bj 1 se b j ~ t 130 The rejection region is t > 1. is t bj 1 se b j ~ t 130 The rejection region is t < 1.16079 Since t = 1.657 .16079 = (1.001. given H0 is true.6567.3190 1.130) 1. it is a defensive stock.41397 1 0. given H0 is true.31895 1 1.130) 1. Principles of Econometrics.637) Thus we estimate.130) se(b j ) . Using our sample of data the corresponding interval estimate is 1. 4e 69 Exercise 3.001 to 1. It is possible that Microsoft’s beta falls outside this interval. because the procedure we used to create the interval works 95% of the time.978 0. The value of the test statistic is t 1. fail to reject H 0 Mobil-Exxon t 0.978 .978 . This result indicates that the sample data is consistent with the conjecture from economic theory that the intercept term equals 0.978 .978 .978 t 1.01155 0. fail to reject H 0 0.823 0.978 t 1.245 1.005956 GE t 0.978 t 1.007747 0.193 0.004759 GM t 0.961 Since 1.009743 IBM t 0.006098 0.978 and t 1. The test statistic.00788 1.00115 0.978 .130) 1. fail to reject H 0 We do not reject the null hypothesis for any of the stocks.185 Since 1.978 t 1.006091 0.978 . fail to reject H 0 Disney t 0.975.005851 0.978 t 1. is ~ t 130 1.978 . .001167 0.978 . where t(0. fail to reject H 0 Microsoft t 0. 4e 70 Exercise 3.7 (continued) (e) The two hypotheses are H0: t aj se a j The rejection region is t j = 0 versus H1: j 0. Principles of Econometrics.978 t 1. fail to reject H 0 Since 1. The results for each company are given in the following table: Stock t-value Decision rule Since 1. Exercise Solutions. given H0 is true.004322 Since 1.787 Since 1.Chapter 3. We conclude that the slope is not zero and that there is a statistically significant relationship between house size in square feet and house sale price.333 t(0. Principles of Econometrics.000. we first calculate the estimated variance: var b1 2000b2 var b1 (2000) 2 var b2 2 2000 cov b1 . There is not enough evidence to suggest that the expected price of a house of 2000 square feet is greater than $120.01 . The value of the test statistic is t 73. given H0 is true. .99.333 t(0. we do not reject the null hypothesis. we reject the null hypothesis that 2 = 0 and accept the alternative that 2 0 .772SQFT (5728) (2.400 2.99.Chapter 3. is b2 ~ t(580) se(b2 ) 0.580) .772 2.333. The test statistic. Exercise Solutions.580) . The value of the test statistic is 28407.34 4648311 The corresponding standard error is: se b1 2000b2 The rejection region is t t var b1 2000b2 4648311 2156 2.333 . we set up the hypotheses H0 : 1 2000 2 120000 H1 : 1 2000 2 120000 The test statistic. 4e 71 EXERCISE 3.8 (a) The estimated linear regression is: PRICE (se) 28408 73. b2 32811823 4000000 5. the rejection region is t 2.301 32.06 > 2.77195 2156 120000 0.56 2000 73. given H0 is true.294462 4000 12335. is t b1 2000b2 se(b1 120000 2000b2 ) ~ t(580) To obtain the standard error se(b1 2000b2 ) .301) The hypotheses are H0: t With 2 0 versus H1: 2 0 .06 Since t = 32.000.401 Since 0. (b) For testing H 0 : E PRICE | SQFT 2000 120000 against an alternative 1 2 SQFT that the expected price is greater than $120. 4 114902.975.253 SQFT 2000 2 0.580 se b1 2000b2 28407.012063 4000 96. (d) The estimated quadratic regression is: PRICE (se) 68710 0. Principles of Econometrics.371.000346) The marginal effect of an additional square foot of living area is dPRICE dSQFT 2 2 SQFT Its estimates for houses of 2000 and 4000 square feet are dPRICE dSQFT dPRICE dSQFT 2 0.8(b) p-value (c) A 95% interval estimate for the expected price of a house of 2000 square feet is b1 2000b2 t 0.902 and $123.656 Figure xr3. Exercise Solutions.3 4234.506 SQFT 4000 .012063 2000 48. 4e 72 Exercise 3.Chapter 3. 123371 We estimate with 95% confidence that the expected house price of a 2000 square foot house lies between $114.012063SQFT 2 (2873) (0.77195 1.56 2000 73.401 0.8(b) (continued) The p-value of the test is p P t 580 0.964 2156 119136. we wish to test the hypotheses H 0 : 4000 against the alternative H1 : 4000 2 75 . is t 75 8000a2 75 ~ t(580) se(8000 a2 ) The rejection region is t t 2 2. we wish to test the hypotheses H 0 :8000 against the alternative H1 : 8000 2 75 . we do not reject the null hypothesis 8000 2 75 in favor of the alternative that 8000 2 75 .00034626 2. The value of the test statistic is 26.000013927) The marginal effect of an additional square foot of living area is dPRICE dSQFT 2 PRICE The estimated value of PRICE when SQFT PRICE exp ˆ 1 ˆ 2 SQFT 2000 is exp(10. (e) The estimated log-linear model is ln PRICE 10. 4e 73 Exercise 3.580) . we reject the null hypothesis that 4000 2 75 and accept the Since t alternative that 4000 2 75 .31 19.000413235 2000) 111905. Principles of Econometrics.Chapter 3.03467) (0. For the case of a 4000 square foot house. given H0 is true. We conclude that the marginal effect of an additional square foot of living area in a home with 2000 square feet is less than $75.01 .506 75 8000 0. The test statistic.8(d) (continued) For the case of a 2000 square foot house. given H0 is true.580) .506 2. Exercise Solutions.79894 0. The test statistic.747 1.333 .01.31 2.253 75 4000 0.333 t(0. The two different hypothesis test outcomes occur because the marginal effect of an additional square foot is increasing as the house size gets larger.5 .01.333 t(0.000413235 SQFT (se) (0.333 .76 Since t 7.79894 0. the rejection region is t t 2 48. The value of the test statistic is 96. is t For 75 4000a2 75 ~ t(580) se(4000a2 ) 0.385 19.76 2.00034626 21.770 7. There is no evidence to suggest that the marginal effect of an additional square foot of living area in a home with 4000 square feet is less than $75. 01. Exercise Solutions.333 . Note: The above solution to part (e) assumes that the predicted values of price for SQFT = 2000 and SQFT 4000 are known with certainty. we wish to test the hypotheses H 0 : 255731 H1 : 255731 2 75 . The value of the test statistic is 255731 0. we do not reject H 0 : 255731 2 75 in favor of the alternative that 255731 2 75 .Chapter 3. the rejection region is t t 2 2.757 1.5 0. we reject the null hypothesis that 111905.562 Since t 8.333 t(0.6. in part (e) we need to test the hypothesis H0 : 2 exp 1 2 SQFT 75 Techniques for testing nonlinear functions of parameters such as this one are considered in Chapter 5. the two different hypothesis test outcomes occur because the marginal effect of an additional square foot is increasing as the house size gets larger.45 18. is t With 75 against 111905. The value of the test statistic is 111905.5 2 75 .000413235 75 111905. we wish to test the hypotheses H 0 :111905. given H0 is true. given H0 is true.01 . Like in part (d).79894 0.333 t(0.613 3.000413235 4000) 255731 Thus.01.5 ˆ 2 75 ~ t(580) se(111905. There is no evidence to suggest that the marginal effect of an additional square foot of living area in a home with 4000 square feet is less than $75. is t For 75 against the alternative 255731ˆ 2 75 ~ t(580) se(255731ˆ 2 ) 0.5 2 75 .5 2 75 and accept Since t the alternative that 111905.8(e) (continued) For a 2000 square foot house. Principles of Econometrics. The test statistic.5 0.000413235 30. Because PRICE exp ˆ 1 ˆ 2 SQFT and ˆ 1 and ˆ 2 contain sampling error. 4e 74 Exercise 3. We conclude that the marginal effect of an additional square foot of living area in a home with 2000 square feet is less than $75.677 8.580) .01 . For the case of a 4000 square foot house. PRICE will also be subject to sampling error.5 ˆ 2 ) 0.580) .3.45 2. . the rejection region is t t 2 2.613 2. the estimated price is exp ˆ 1 PRICE ˆ 2 SQFT exp(10.333 .5 the alternative H1 :111905.000413235 28. it assumes there is no sampling error associated with these predictions.559 18. The test statistic. To accommodate this sampling error.000413235 75 255731 0. given H0 is true. (b) A 95% interval estimate for b2 2 from the regression in part (a) is: t(0.4443 0. There is not enough evidence to suggest inflation has a negative effect on the vote.975.263) This interval estimate suggests that.717.5999 0. Since 2 represents the change in percentage vote due to economic growth. The alternative 2 0 is chosen because we assume that inflation.8859 0.8859GROWTH (1. if it does influence the vote. Exercise Solutions. will do so in a negative way. 1.Chapter 3.8859 2. The test statistic. (c) We set up the hypotheses H0: 2 0 versus H1: 2 0 .509.870 > 1.1819 = (0. .22) .870 Since t = 4. if it does influence the vote.1819) The value of the test statistic is t 0.074 0. The estimated regression model is VOTE 53.509 and 1. with 95% confidence. We conclude that economic growth has a positive effect on the percentage vote earned by the incumbent party.263.4443INFLATION (se) (2.4077 0.741 Since 0.9 (a) We set up the hypotheses H0: 2 0 versus H1: 2 0 . we reject the null hypothesis that 2 = 0 and accept the alternative that 2 0 . Principles of Econometrics. The test statistic. The alternative 2 0 is chosen because we assume that growth. The estimated regression model is VOTE (se) 50.741 1. is t b2 ~ t(22) se(b2 ) Selecting a 5% significance level.717 t(0. will do so in a positive way.717 t(0.2500) (0. 4e 75 EXERCISE 3.05.509 and 1.95.263.22) .5999) The value of the test statistic is t 0.22) se(b2 ) 0. the rejection region is t 1.1819 4. we expect that an increase in the growth rate of 1% will increase the percentage vote by an amount between 0.0125) (0.717 .8484 0. we do not reject the null hypothesis. given H0 is true. is b2 t ~ t(22) se(b2 ) The rejection region is t 1. the true value of 2 is between 0. 4077 50 1. Principles of Econometrics.22) se(b2 ) 0. b2 4 1.975.5191 2. In repeated samples of elections with inflation at 2%. the expected vote in favor of the incumbent party is E VOTE | INFLATION Thus. 55. The test statistic.4077 2 ( 0. we do not reject the null hypothesis.688 and 0. There is no evidence to suggest that the expected vote in favor of the incumbent party is less than 50% when there is no inflation. we wish to test H 0 : 1 0 1 0 2 1 50 against the alternative H1 : assuming H0 is true at the point 1 1 50 . with 95% confidence.2653 The 95% interval estimate is therefore: b1 2b2 t 0.800) 2.Chapter 3.1216 49.3599 2 2 cov b1 .9 (continued) (d) A 95% interval estimate for b2 from the regression in part (c) is: 2 t(0. It suggests that an increase in the inflation rate of 1% could increase or decrease or have no effect on the percentage vote earned by the incumbent party.44431) 52.800. 0. the true value of 2 is between 1.22) .074 2. The value of the test statistic is The rejection region is t 53. we expect the mean vote to lie within 95% of the interval estimates constructed from the repeated samples.975.515 2.074 This interval estimate suggests that.2500 t Since 1. Exercise Solutions.515 1.0625 4 0.5191 3. (e) When INFLATION 0 .64 We estimate with 95% confidence that the expected vote in favor of the incumbent party when inflation is at 2% is between 49.717 t(0.22 se b1 2b2 52.5191 The standard error of this estimate is the square root of var b1 2b2 var b1 22 var b2 5.5999 = ( 1.05. (f) A point estimate of the expected vote in favor of the incumbent party when INFLATION 2 is E VOTE b1 2b2 53. 4e 76 Exercise 3.0592 2.688.4443 0.2653 52. is b1 50 ~ t(22) se(b1 ) t 1.40% and 55.40. .64%.717 . 50 . 95.043 13. The mean score of students in small classes is significantly greater than that of students in regular-sized classes.Chapter 3.899 2.667) (2.681 7.681 > 1.4466 (9. 4e 77 EXERCISE 3.3741) se(b2 ) 13.681 Since 5. Figure xr3.4466 5. .645 . we estimate that the average score for students in small classes is between 9. It suggests that governments should invest in more teachers and classrooms so that class sizes can be smaller.10(a) Illustration of p-value (b) A 95% interval estimate for 2 is b2 t(0. The calculated value of the test statistic is 13. we reject H 0 .9606 2. 18.3741) 1.10 and 18.21 10 9 .975. The p-value of the test is p P t(3741) 5. Principles of Econometrics.645.70) With 95% confidence.447) The null and alternative hypotheses are H0 : 2 0 H1 : 2 0 The test statistic and its distribution when the null hypothesis is true are t b2 se(b2 ) We reject H 0 when t t t(3741) t(0. Exercise Solutions.70 points higher than the average score for students in regular-sized classes.10 (a) The estimated equation using a sample of small and regular-sized classes (without aide) is: TOTALSCORE 918.10.899SMALL (se) (1.899 1. No differences are uncovered if scores in math and reading tests are considered separately. test statistic and rejection region as in part (a).5865 5.314 AIDE (se) (1.95.605 > 1. the estimated equation is MATHSCORE 483. the value of the test statistic is t b2 se(b2 ) 8.10 (continued) (c) For READSCORE.586) Using the same hypotheses.081) (1. we reject H 0 : 2 0 in favor of H1 : 2 0 .310 8.10(a).8191 5. Principles of Econometrics. The p-value diagram is similar to that given in Figure xr3. the estimated equation is READSCORE 434.093 > 1.270) The null and alternative hypotheses are H0 : 0 2 H1 : 2 0 The test statistic and its distribution when the null hypothesis is true are t ˆ2 se ˆ 2 We reject H 0 when t t(3741) t(0.Chapter 3.3741) 1.038) Using the same hypotheses.10(a).093 Because 5.645.707) (1. Having a smaller class has a positive effect on the learning of both math and reading. The mean reading score of students in small classes is significantly greater than that of students in regularsized classes. 4e 78 Exercise 3. The mean math score of students in small classes is significantly greater than that of students in regular-sized classes.605 1.645 . test statistic and rejection region as in part (a). The calculated value of the test statistic is .0382 Because 5.0799 1.819SMALL (se) (0.080SMALL (se) (1.043 0. For MATHSCORE.613) (2.645.733 5. The p-value diagram is similar to that given in Figure xr3. we reject H 0 : 2 0 in favor of H1 : 2 0 . Exercise Solutions. the value of the test statistic is t b2 se(b2 ) 5. (d) The estimated equation using regular-sized classes with and without a teacher aide is: TOTALSCORE 918. 7054 0.2704 0.733 0.705 AIDE (0. test statistic and rejection region as in part (d).14.10(d) (continued) t 0.138 Since 0.645.697) (0. . Principles of Econometrics. The mean math score of students in classes with a teacher aide is not significantly greater than that of students in classes without a teacher aide.310 0. or it may have no effect.719 < 1. For MATHSCORE.645.975.267 < 1.038 < 1. In other words having an aide may improve scores. The mean score of students in classes with a teacher aide is not significantly greater than that of students in classes without a teacher aide.4687 0.719 Because 0. It suggests that governments should not invest in providing more teacher aides in classrooms. 4. No differences are uncovered. (f) For READSCORE. (e) A 95% interval estimate for 2 is ˆ 2 t(0. the estimated equation is MATHSCORE (se) 483.14 and 4.9606 2. the value of the test statistic is t ˆ2 se ˆ 2 0. we do not reject H 0 : 2 0 in favor of H1 : 2 0 .3741)se ˆ 2 0.Chapter 3. 4e 79 Exercise 3.469) Using the same hypotheses. the value of the test statistic is t ˆ2 se ˆ 2 0.982) Using the same hypotheses. The mean reading score of students in classes with a teacher aide is not significantly greater than that of students in classes without a teacher aide.77) With 95% confidence.3915 1.645. it may lead to scores that are worse.3139 2.043) (1. we do not reject H 0 .2704 ( 4. Exercise Solutions.391AIDE (1.9817 0. test statistic and rejection region as in part (d). we estimate that the difference in average scores for students from classes with and without a teacher aide lies between 4. Having a teacher aide improves neither the average reading score nor the average math score.77.267 Because 0. we do not reject H 0 : 2 0 in favor of H1 : 2 0 .3139 1. the estimated equation is READSCORE (se) 434. 26.646 t(0. The value of the test statistic is t 0. it tells us that the average wage for those without experience is $18. There is a positive relationship between the hourly wage and a worker’s experience.998) . 80 70 60 WAGE 50 40 30 20 10 0 0 10 20 30 40 50 60 70 EXPER Figure xr3.2577 0.Chapter 3.826 Decision: Reject H 0 because 2.826 1. 4e 80 EXERCISE 3.646 . The relatively large t-values suggest that the least squares estimates are statistically significant at a 5% level of significance. . Furthermore. given H0 is true.6885 2. We conclude that the estimated slope of the relationship b2 is statistically significant. Exercise Solutions.95. Principles of Econometrics.0315 2. is t b2 ~ t(998) se(b2 ) The rejection region is t 1.9273 (t ) 0.0890.0890 0. 2 0 because we expect experience to have a The test statistic.11(a) Fitted regression line and observations (b) We set up the following hypothesis test: H0 : 2 0 H1 : 2 0 The alternative hypothesis is set up as positive effect on wages.11 (a) The estimated equation is: WAGE 18.8257 The estimated equation tells us that with every additional year of experience.0890 EXPER (se) 0. the associated increase in hourly wage is $0.0315 19. 8222 3.1407 EXPER (se) 1.11 (continued) (c)(i) For females. the associated increase in average hourly wage for males is $0. however.3349 0.0096 1. This estimate is not significantly different from zero.11(c)(ii) Fitted regression line and observations for males .0460 (t ) 13.1407.11(c)(i) Fitted regression line and observations for females (c)(ii) For males.0427 (t ) 14.84. 80 70 60 WAGE 50 40 30 20 10 0 0 10 20 30 40 50 60 70 EXPER Figure xr3. the estimated equation is: WAGE 18. Principles of Econometrics.45. The average wage for females without experience is $17. 4e 81 Exercise 3. the estimated equation is: WAGE 17.4511 0.8413 0. 80 70 60 WAGE 50 40 30 20 10 0 0 10 20 30 40 50 60 70 EXPER Figure xr3. Exercise Solutions.1650 With every extra year of experience the associated increase in average hourly wage for females is $0.0497 EXPER (se) 1.2735 0.0497. The average wage for males without experience is $18.Chapter 3.0619 With every extra year of experience. however.66.9146 With every extra year of experience the associated increase in average hourly wage for white males is $0. Principles of Econometrics. Exercise Solutions.8858 With every extra year of experience.5319 0.2362 0.4607 0. the estimated equation is: WAGE 18. the associated increase in average hourly wage for blacks is $0.0834 t 6.11(c)(iv) Fitted regression line and observations for white males .Chapter 3. The average wage for white males without experience is $18.0738 EXPER se 2.7893 0.1455 EXPER (se) (t ) 1. The average wage for blacks without experience is $15.1455.6556 0.11(c)(iii) Fitted regression line and observations for blacks (c)(iv) For white males. 80 70 60 WAGE 50 40 30 20 10 0 0 10 20 30 40 50 60 70 EXPER Figure xr3.79. 4e 82 Exercise 3.0499 12. the estimated equation is: WAGE 15. This estimate is not significantly different from zero.0738.7715 2.11(c) (continued) (c)(iii) For blacks. 80 70 60 WAGE 50 40 30 20 10 0 0 10 20 30 40 50 60 70 EXPER Figure xr3. For those with no experience the wage ranking is white males. There are very few negative residuals with an absolute magnitude larger than 20. The effect is only slightly less for males in general. It is approximlately halved for blacks and is less still for females.Chapter 3.11(c) (continued) Comparing the estimated wage equations for the four categories. Exercise Solutions. 4e 83 Exercise 3. for white males. Principles of Econometrics. males. or leads to the largest increase in wages. The residual plots appear in the figures below. females.11(d)(i) Plotted residuals for female regression . These characteristics suggest a distribution of the errors that is not normally distributed. but skewed to the right. whereas the magnitude of the positive residuals cover a much greater range. 60 50 40 RESID 30 20 10 0 -10 -20 -30 0 10 20 30 40 50 60 70 EXPER Figure xr3.11(d) Plotted residuals for full sample regression 60 50 40 30 RESID (d) 20 10 0 -10 -20 0 10 20 30 40 50 60 70 EXPER Figure xr3. and blacks. we find that experience counts the most. The main observation that can be made from all the residual plots is that the pattern of positive residuals is quite different from the pattern of negative residuals. Chapter 3.11(d) (continued) 60 50 40 RESID 30 20 10 0 -10 -20 -30 0 10 20 30 40 50 60 70 EXPER Figure xr3.11(d)(iv) Plotted residuals for white male regression 84 .11(d)(iii) Plotted residuals for black regression 60 50 40 RESID 30 20 10 0 -10 -20 -30 0 10 20 30 40 50 60 70 EXPER Figure xr3.11(d)(ii) Plotted residuals for male regression 60 50 40 RESID 30 20 10 0 -10 -20 0 10 20 30 40 50 60 70 EXPER Figure xr3. Principles of Econometrics. Exercise Solutions. 4e Exercise 3. 013828 0. 2 0 . The distribution of wages is skewed to the right with the majority of people having a wage less than $30. 998) 1. Exercise Solutions.013828EXPER30 2 WAGE (se) (0. we reject H 0 : 2 0 .527) (0. The calculated value of the test statistic is 0. 4e 85 EXERCISE 3. but it is hard to discern a strong relationship.001956 7. To test H 0 : 0 against the alternative H 1 : 2 t ˆ2 se ˆ 2 The rejection region is t t ˆ2 se ˆ 2 2 0 .068) The t-values for both coefficient estimates are greater than 2.068 implies a significant quadratic relationship in the shape of an inverted “U”. Principles of Econometrics. indicating that they are significantly different from zero at a 5% significance level.646 . and with a small number having wages more than double this amount.Chapter 3. 80 70 60 WAGE 50 40 30 20 10 0 -30 -20 -10 0 10 20 30 40 EXPER30 Figure xr3.12 (a) The required scatter diagram is displayed in Figure xr3.068 1.646 . The few observations with the largest experience have a low wage. Accepting the alternative H 1 : Because 7.12(a) Scatter diagram for WAGE and EXPER30 (b) The estimated equation is 23. There are no distinct patterns evident.80) ( 7.12(a). and those with the highest wages tend to be those where EXPER30 lies between –10 and 15.001956) (t ) (43.05. we use the test statistic t(998) t(0.067 0. 962 se me50 0.962 se me30 0.5531 0.0019564 0.40. (d) The 95% confidence intervals for the slopes are as follows me10 1.07826 The marginal effect at 30 years of experience is not significantly different from zero since it is zero for all possible values of 2 .0138283 60 60 0. Both me10 and me 50 are significantly different from zero at a 5% significance level because the values t 0.962 and t(0. The marginal effects for EXPER 10 and 50 are relatively precise. 30.975. 0.07826 (0.5531 0.998) +1.0) me50 1.068 do not lie between t(0.400. Principles of Econometrics. No estimation was necessary.0138283 20 60 0.962.400) The marginal effect for EXPER 30 is exact.5531 1.07826 7.707. 4e 86 Exercise 3. the estimated marginal effects for persons with 10. Exercise Solutions.962 se me10 0.Chapter 3. They suggest an extra year of experience will change the wage by an amount between $0.5531 1. and 50 years experience are me10 me 30 me50 d E (WAGE ) d EXPER 0.962 0.0138283 100 60 0.0 EXPER 10 d E (WAGE ) d EXPER d E (WAGE ) d EXPER EXPER 30 0.962 0.962 0.07826 ( 0.998) 1.0138283 .0 (0.0.0 1.71 and $0. 0.5531 EXPER 50 Their standard errors are se me10 se me50 se me30 0 40 se ˆ 2 40 0. 0.025. .12 (continued) (c) Noting that E (WAGE ) 1 EXPER 2 60 EXPER 900 2 the marginal effect of experience on wage is given by d E (WAGE ) d EXPER 2 2 EXPER 60 Using ˆ 2 0.707) me 30 1. 9263 0. The estimates in part (c) 20 . b2 0.088953EXPER30 (0.927 .Chapter 3.17567076 900 0. The slope is positive when EXPER30 zero when EXPER30 0 .088953EXPER (se) (0. it is are consistent with the fitted values. Exercise Solutions.12(e) Plot of fitted and actual values of WAGE (f) The two estimated regressions are WAGE (se) 20.031480) The two equations have the same slope coefficient but different intercepts.0889534EXPER 18. Principles of Econometrics.258 0.12(e).927) (0.088953EXPER which agrees with the second equation. and negative when EXPER30 20 .0889534( EXPER 30) 20.0889534 30 0. 80 70 60 50 WAGE Fitted WAGE 40 30 20 10 0 -30 -20 -10 0 10 20 30 40 EXPER30 Figure xr3. we note that ˆ 1 b1 30b2 and hence that se ˆ 1 var b1 302 var b2 2 30 cov b1 . To reconcile the two intercepts we note that the right-hand side of the first equation can be written as 20.00099100088 60 0.419) (0.9263 0. 4e 87 Exercise 3.258 0.12 (continued) (e) A plot of the actual and fitted WAGE appears in Figure xr3.926 0. To derive the standard error of ˆ 1 from the covariance matrix of the estimates from the first equation.00346057507 0.031480) WAGE 18. we use the test statistic 2 t(4836) t(0. To test H 0 : 0 against the alternative H 1 : 2 t ˆ2 se ˆ 2 The rejection region is t t ˆ2 se ˆ 2 0 . the estimated marginal effects for persons with 10.08895. The assumption of a constant slope does not appear to be a good one.4957 .000879) (t ) (94.098 1. (g) Using the larger data set in cps4.645 .098) The t-values for both coefficient estimates are very large. 30. The marginal effect of experience is greatest when a worker has little or no experience. 4836) 1. The results from parts (b) to (d) suggest the slope will decline with experience and eventually become negative.05.237) (0.01239307 60 60 0.355 0. we reject H 0 : 2 0 . Accepting the alternative H 1 : implies a significant quadratic relationship in the shape of an inverted “U”.dat. The calculated value of the test statistic is 0.645 .Chapter 3.00087909 14.012393EXPER30 2 WAGE (se) (0. 4e 88 Exercise 3.01239307 20 60 0. 2 0 The marginal effect of experience on wage is given by d E (WAGE ) d EXPER 2 2 EXPER 60 Using ˆ 2 0.12(f) (continued) The estimated marginal effect of experience on wage from the two regressions is 0.0 EXPER 10 EXPER 30 0.01239307 0. we obtain the following results The estimated equation is 22.01239307 100 60 EXPER 50 0.48) ( 14.4957 0. Exercise Solutions. Principles of Econometrics.098 Because 14.01239307 . and 50 years experience are me10 me30 me50 d E (WAGE ) d EXPER d E (WAGE ) d EXPER d E (WAGE ) d EXPER 0. indicating that they are significantly different from zero at a 5% significance level. 565.138. do not lie between t(0.025.4957 0.00087909 0. 0.4957 1.962 se me50 0.96 0.0 1.307 to 0. Principles of Econometrics. 0.975.962 se me30 0.4957 1.960 and t(0.0) me50 1.12(g) (continued) Their standard errors are se me10 se me50 se me30 0 40 se ˆ 2 40 0.962 se me10 0.098 1.96 0.0 (0.565) me30 1.0. Both me10 and me 50 are significantly different from zero at a 5% significance level because the values t 0.4836) +1.03516 14.Chapter 3.427) The larger sample has increased the precision of estimation by reducing the width of the confidence intervals by more than half: from 0.03516 (0.03516 ( 0. 4e 89 Exercise 3.427. .4836) The 95% confidence intervals for the slopes are as follows me10 1.96 0. Exercise Solutions.960. 0.03516 The marginal effect at 30 years of experience is not significantly different from zero since it is zero for all possible values of 2 . 5 LNWAGE 3. indicating that they are significantly different from zero at a 5% significance level.646 . The calculated value of the test statistic is 0. To test H 0 : 0 against the alternative H 1 : 2 t ˆ2 se ˆ 2 The rejection region is t t ˆ2 se ˆ 2 2 0 .1) ( 8.5 1. 4.646 .13 (a) The scatter diagram appears below.07) The t-values for both coefficient estimates are greater than 2.13(a) Scatter plot of ln(WAGE) against EXPER30 (b) The estimated log-polynomial model is: ln WAGE 2. Accepting the alternative H 1 : 2 0 Because 8. we reject H 0 : 2 0 . after which wages decrease with experience.9826 0.067 1.067 implies a significant quadratic relationship in the shape of an inverted “U”. .237) (0.5 4. 4e 90 EXERCISE 3.05.0007088 EXPER30 2 (se) (0.0 2.0 3.0000879) (t ) (126.5 -30 -20 -10 0 10 20 30 40 EXPER30 Figure xr3. 998) 1. Wages increase with experience until a turning point is reached. we use the test statistic t(998) t(0.Chapter 3.000087872 8.0 1. Principles of Econometrics. Exercise Solutions.00070882 0. It is difficult to discern any strong pattern.5 2.0 0. 000708822) 50 30 14. 30 and 50 are WAGE10 exp 2.4215 EXPER 50 A plot of the actual and fitted WAGE appears in Figure xr3.000708822 (30 30)2 19.Chapter 3.8665 Using these values and ˆ 2 the marginal effects me10 me30 me50 (d) d WAGE d EXPER d WAGE d EXPER d WAGE d EXPER 0.7398 WAGE50 exp 2.982638 0.982638 0. 4e 91 Exercise 3.000708822 .13 (continued) (c) Using the hint.982638 0.0 2 ( 0.000708822 (10 30)2 14. 80 70 60 50 WAGE fitted WAGE 40 30 20 10 0 -30 -20 -10 0 10 20 30 40 EXPER30 Figure xr3.7398 0. and negative when EXPER30 20 . it is are consistent with the fitted values. we can compute the following estimates for 2 ( 0.8665 0.000708822) 10 30 14. The estimates in part (c) 20 . The slope is positive when EXPER30 zero when EXPER30 0 .13(d). we have d WAGE d EXPER 2 2 EXPER 30 WAGE The predicted values for WAGE when EXPER 10.4215 2 ( 0.13(d) Plot of fitted and actual values of WAGE . Exercise Solutions. Principles of Econometrics.8665 EXPER 10 EXPER 30 0.000708822 (50 30)2 14.8665 WAGE30 exp 2.000708822) 30 30 19. 403 t(0.85807 2.8581 0. Instead.616 . Since brands 2 and 3 are substitutes for brand 1.8581RPRICE2 (se) (0. when the two prices are the same.2758 1.90%. a 10% increase in the relative price of brand 1 tuna to brand 2 tuna will decrease sales of brand 1 by between 8. it is not so meaningful to talk about a 1-unit increase in RPRICE2. 4e 92 EXERCISE 3. Demand is elastic. and considering a price change of a realistic magnitude: If the prices of brands 1 and 2 are the same.5139 2.890. we consider the elasticity d ( SAL1) RPRICE2 d ( RPRICE2) SAL1 2 RPRICE2 We can interpret 2 as the percentage change in sales from a 1% increase in the relative price when the prices of the two brands are identical ( RPRICE2 1) . In terms of our estimate. and the relative price of brand 1 to brand 2 increases by 10%. A 95% interval estimate for 2 from the regression is: b2 t(0. In this particular case where RPRICE2 is a unit-free relative price variable.5139) The typical interpretation of 2 in a log-linear model is that 1-unit increase in x will lead to a 100 2 % increase in y. Principles of Econometrics. is t b2 ~ t(50) se(b2 ) The rejection region is t t 1.Chapter 3. will lead to a decline in the sales of brand 1. Exercise Solutions.50)se(b2 ) 1.58%. the sales of brand 1 will decline by 18.975. (c) We set up the following hypothesis test: H0 : 2 0 H1 : 2 0 The test statistic. 0. or relative to the price of brand 3.26% and 28. an increase in the price of brand 1 relative to the price of brand 2.826 This interval estimate suggests that.50) . The value of the test statistic is 3.01.5139 2.5185) (0. given H0 is true.14 (a) The relationship between sales (SAL1) and the relative price variables is expected to be a negative one. (b) The estimated log-linear regression is: ln( SAL1) 10.009 0. with 95% confidence. as it is expected that demand for a good should be inversely related to the relative price of that good to a substitute good. the sales of brand 1 will decline by 30. Exercise Solutions.403 .009 0.14(c).14(c) Rejection region for hypothesis test.975. 1. a 10% increase in the relative price of brand 1 tuna to brand 3 tuna will decrease sales of brand 1 by between 19. with 95% confidence. If the prices of brands 1 and 3 are the same. and the relative price of brand 1 to brand 3 increases by 10%.0543 2. when the two prices are the same. This result is consistent with economic theory.991 This interval estimate suggests that. Figure xr3.17%.5291 4. 4e 93 Exercise 3.54%.5291) The estimate of 2 can be interpreted as follows.Chapter 3. A sketch of the rejection region is We conclude that there is a statistically significant inverse relationship between the unit sales of brand 1 tuna and the relative price of brand 1 tuna to brand 2 tuna.50)se( ˆ 2 ) 3.5347) (0.91% and 41.14(c) (continued) 3.616 Decision: Reject H 0 because displayed in Figure xr3.4810 3. Principles of Econometrics.0543RPRICE3 (se) (0. . Demand is elastic. (d) The estimated log-linear regression is: ln( SAL1) 11. A 95% interval estimate for 2 from the regression is: ˆ 2 t(0.117. 2. 14(e).52913 Decision: Reject H 0 because displayed in Figure xr3.772 2. 4e 94 Exercise 3. A sketch of the rejection region is We conclude that there is a statistically significant inverse relationship between the unit sales of brand 1 tuna and the relative price of brand 1 tuna to brand 3 tuna. given H0 is true.Chapter 3.50) .403 . . as it is expected that demand for a good should be inversely related to the relative price of that good to a substitute good. is t ˆ2 ~ t(50) se( ˆ 2 ) The rejection region is t t 2. This result is consistent with economic theory.14(e) Rejection region for hypothesis test.05425 0.01. Principles of Econometrics.403 t(0. The value of the test statistic is 3. Figure xr3.772 5. 5. Exercise Solutions.14 (continued) (e) We set up the following hypothesis test: H0 : 2 0 H1 : 2 0 The test statistic. 8844 PRBARR (0.0966) (0.69224 1.9873 0.14775 ( 0.369 t(0.Chapter 3. A 95% interval estimate for b2 2 from the regression is: t(0. Principles of Econometrics. a 95% interval estimate for the percentage change in the crime rate after an increase in the probability of conviction of 0.86.28.1 is ( 26. and that this relationship is an inverse relationship. A 95% interval estimate for b2 from the regression is: 2 t(0.1).15 (a) The estimated log-linear regression using data from 1987 is: LCRMRTE (se) 2.3744 5. Exercise Solutions.88)se(b2 ) 0.84% . (b) We set up the following hypothesis test: H0 : 2 0 H1 : 2 0 The test statistic.01.9859.1). 3.88) .975. the crime rate will decrease by 0. We conclude that there is a statistically significant relationship between the crime rate and the probability of arrest.8844 0.88)se(b2 ) 1.6922 PRBCONV (0.1478) If the probability of conviction increases by 10% (or 0. 1. 11. is t b2 ~ t(88) se(b2 ) The rejection region is t t 2.033 Decision: Reject H 0 because 5.1 is ( 9. the crime rate will decrease by 0.033 2.3744 ( 2.99) .1218) (0.8844 1. given H0 is true. 4e 95 EXERCISE 3.884 100% 18.1 0. 0. The value of the test statistic is 1.1604 0.140) Thus.1 1.40) . a 95% interval estimate for the percentage change in the crime rate after an increase in the probability of arrest of 0.3744) If the probability of arrest increases by 10% (or 0. (c) The estimated log-linear regression using data from 1987 is: LCRMRTE (se) 3. .9854 1.975.374 .3986) Thus.9873 0.628.692 100% 6.92% . 685 Decision: Reject H 0 because 4.69224 0.374 .14775 4.369 t(0.15(c) (continued) To test the relationship between crime rate and the probability of conviction at the 1% significance level. . and that this relationship is an inverse relationship. given H0 is true. we set up the following hypothesis test: H0 : 2 0 H1 : 2 0 The test statistic. The value of the test statistic is 0. Exercise Solutions. We conclude that there is a statistically significant relationship between the crime rate and the probability of conviction.685 2. 4e 96 Exercise 3.01.Chapter 3. is t b2 ~ t(88) se(b2 ) The rejection region is t t 2. Principles of Econometrics.88) . CHAPTER 4 Exercise Solutions 97 . 71051 .5155 0.36 (1 0. N y2 5930.Chapter 4.85 631.1 eˆi2 (a) R2 1 (b) To calculate R 2 we need yi yi y y 182.5155 98 .7911) (20 2) 6. ˆ2 SST (1 R 2 ) N K 552. 4e EXERCISE 4. R2 (c) SSR SST 666. Exercise Solutions.4104 788.94 20 16. Principles of Econometrics.0352 Therefore.63 1 2 yi 2 y yi2 2 0.72 788.8455 From R2 1 eˆi2 SST 1 (N K)ˆ 2 SST we have. 38 x (1.00234) where yˆ yˆ 50 yˆ 0.117) where yˆ yˆ and x 20 The values of R 2 remain the same in all cases.83 17. x 20 99 .2 (a) (b) (c) yˆ 5.0246) (0. Exercise Solutions. Principles of Econometrics.01738 x (0.0615) (0.2915 0.869 x (0.Chapter 4.23) (2.34) where x x 20 yˆ 0.1166 0. 4e EXERCISE 4. 5144 ( 5.1824 1. and 1.72 1.293333 3.4 3. 4.019. Predictions are more precise when made for x values close to the mean. 4e 100 EXERCISE 4.1824 .574 1.3) 1 5 (4 2) 2 10 2.619 5.5144 (c) b1 b2 x0 Using x x0 0.638 Width in part (d) 6.574) Width in part (c) 4. Principles of Econometrics.2 3. as expected.774 8. the prediction is yˆ 0 1 N 5 1.Chapter 4.774.3115 ( 1.2 ( x0 x ) 2 ( xi x ) 2 1 N Using se( f ) from part (b) and tc yˆ 0 (d) 5 1.975.1824 1.3 (a) yˆ 0 (b) var( f ) ˆ2 1 se( f ) 2.6. 0.019 9.3115 yˆ 0 1.43333 1 2.348 1.619) 2 .3 2 ( x0 x ) 2 ( xi x ) 2 var( f ) ˆ2 1 se( f ) 1.4 .72 The width in part (d) is smaller than the width in part (c).43333 1 1 5 (2 2) 2 10 tc se( f ) 2. .3 4 tc se( f ) t(0. Exercise Solutions.29333 1. 05343 The 95% interval estimates for the marginal effect from each model are Model 1: me t(0. 2.4 3.4276 0.5343 1 EXPER 0.07295 Model 2: (d) d RATING dEXPER 0.5345 Estimates of the marginal effects at EXPER 10 are Model 1: d RATING dEXPER 0.4 (a) Graphs for each of the models are given below.47 se me 0. (b) (c) 20 EXPER 30 40 Model 2: the linear-log model.001459(2 EXPER 70) 0.8 RATING 3 RATING 3 3. 4e 101 EXERCISE 4.05343 2.0433 10 (0.975.001459(10 35)2 Model 2: RATING 1.4(a) 10 20 EXPER 30 40 10 Model 1: the quadratic model.5 2.0000786 50 (0.0650.0809) Model 2: me t 0.0117 0.0621) .Chapter 4.48)se me 0.07295 2.4464 0. The predicted ratings for a worker with 10 years of experience are Model 1: RATING 3.975. Exercise Solutions.6579 2.5343 1 10 0.001459(2 10 70) 0.5343ln(10) 2.2 3.0106 0.0.6 2.0447.5 Figure xr4.0. Principles of Econometrics. . e e 50 The estimated equation becomes yˆ yˆ 50 b1 50 b2 50 x Thus. e where * 2 2 20 and x x 20 The estimated equation becomes b2 20 yˆ b1 Thus. the variance of the error term var(e) (b) 2 is unaffected. Exercise Solutions. Multiplying all the y values by 50 in the simple linear regression model y gives the new model y 50 1 50 * 2 x e 50 x 2 1 2 x e e 50 or y * 1 where y * 1 y 50. Principles of Econometrics. Since e does not change.5 (a) If we multiply the x values in the simple linear regression model y the new model becomes y 2 1 20 * 2 1 x x 20 e 1 2 x e by 20. * 2 2 50.Chapter 4. 1 x 20 and b1 do not change and and b2 become 20 times smaller than their original 2 values. the variance of the error term is 2500 times larger than its original value. 1 50. The variance of the new error term is var(e ) var e 50 2500 var(e) 2500 2 Thus. both 1 and 2 are affected. They are 50 times larger than their original values. 4e 102 EXERCISE 4. Similarly. b1 and b2 are 50 times larger than their original values. Thus. x lies on the fitted line. (b) Consider the fitted line yˆ i b1 yˆi N 1 N yˆ = From part (a). y b1 b2 x . Averaging over N. we obtain b1 xi b2 1 b1 N N b1 b2 x .6 (a) The least squares estimator for 1 is b1 y b2 x . we also have y xi b2 . b2 xi b1 b2 xi N b1 b2 x .Chapter 4. y yˆ . and hence y . Exercise Solutions. Principles of Econometrics. 4e 103 EXERCISE 4. Thus. 333 . 4e 104 EXERCISE 4. 2.370 11.333 and SSR 67.06592 b2 x0 .6703 yi2 42 Ru2 1 (c) 62 72 11.6044 78.7 (a) The least squares predictor in this model is yˆ 0 (b) Using the solution from Exercise 2.6044 2 2 11.4 part (f) eˆi2 SSE (2. (d) Calculations reveal SST SSR SSE yi y 2 29. Thus.967 The squared correlation between the predicted and observed values for y is ryy2ˆ yˆ i ˆ 2yyˆ ˆ 2y ˆ 2yˆ yi yˆ y 2 yi 2 y yˆi yˆ 2 (42.974 The decomposition does not hold. yˆi SST y 2 67.370 .943 The two alternative goodness of fit measures Ru2 and ryy2ˆ are not equal.7363 0.333 0.461 29. Exercise Solutions. 29.6044 352 72 92 112 2 0.Chapter 4.6044 352 0.19782 0.13192 1. Principles of Econometrics.549) 2 65. 4 -0. 4e EXERCISE 4.2 .0719 0.2 1.000048 *** (i) (ii) 2.2254 R2 0.4 0.8 5 10 15 20 Residual 25 Actual 30 35 40 45 Fitted Figure xr4.7994 0.0150 t *** (se) 0.8(b) Fitted line and residuals for the simple linear regression 2.8 5 10 15 20 Residual 25 30 Actual 35 40 45 Fitted Figure xr4.8(b) Fitted line and residuals for the linear-log regression 105 . Exercise Solutions.8 0.0 -.8 .8 0.8 (a) Linear regression results: yˆ t 0.0485 (b) *** 0.6 1.4 0.4 2.0026 R2 0.2 0.1425 *** 0.000338 t 2 yˆ t (se) 0.6954 0.8 0.Chapter 4.0469 *** Quadratic regression results: 0. Principles of Econometrics.4 .5623 0.4245 R2 0.0 1.1696 ln t (se) 0.6 1.4 -.4 0.5252 *** Linear-log regression results: yˆ t 0.0 -0.4 2.0 1. 0 -.6 0.5 1 2 Figure xr4-8 Residual histogram: quadratic relation 1. Exercise Solutions.5 0 Residuals -.8 .5 Quadratic: -.645 Density 1 Linear log: JB = 2.5 2 Figure xr4-8 Residual histogram: linear relation -.2 -.0 1.416 p-value = 0.4 .2 .6 5 10 15 20 Residual 25 Actual 30 35 40 45 Fitted Figure xr4.5 .8(b) continued (b) (i) (ii) 2.5 0 Residuals .5 1.5 (iv) 0 Residuals .249 0 .5 Values of R 2 are given in part (a) JB = 0.878 p-value = 0.4 -.8 1.5 Linear: JB = 0.5 Density 1 Density 1 1.4 0.778 p-value = 0. Principles of Econometrics.6 .5 .2 .4 2. 4e 106 Exercise 4.Chapter 4.812 .8(b) Fitted line and residuals for the quadratic regression (iii) Residual histograms and Jarque-Bera error normality tests: Figure xr4-8 Residual histogram:linear-log relation 0 0 . 4e 107 Exercise 4. Exercise Solutions. 4.8(b) continued To choose the preferred equation we consider the following. Principles of Econometrics. Considering the plots for the fitted equations.5685. The signs and significance of the estimates of the response parameters 2 . the one obtained from the third equation seems to fit the observations best. The third equation provides a more balanced distribution of positive and negative residuals throughout the sample. 3. Considering all these factors. The plots of the fitted equations and their residuals: The upper parts of the figures display the fitted equation while the lower parts display the residuals. namely 0. the third equation is preferable. visual inspection of the histograms suggests those from the linear and quadratic equations more closely resemble a normal distribution. R 2 : The value of R 2 for the third equation is the highest. 2 and 2 : We expect them to be positive because we expect yield to increase over time as technology improves. The residual histograms and Jarque-Bera tests: Normality of the residuals is not rejected in any of the cases. In terms of the residuals. 2. the first two equations have concentrations of positive residuals at each end of the sample. . However. 1.Chapter 4. All estimates have the expected signs and are significantly different from zero at a 1% significance level. 28787 (0.577 The standard error of the forecast error is se( f ) 0.23454 .0141 0.219 48 1. but outside the interval estimate from the linear-log model.45) se( f ) 1.0150 2 0.45) se( f ) 1.975. The 95% prediction interval is given by yˆ 0 t(0.907.417 Using computer software.0003375 482 1. respectively.01502 0.139 2 0.4166 0.28787 . .Chapter 4. The 95% prediction interval is given by yˆ 0 t(0.45) se( f ) 1. we have 0. 2.1696 48 0.25293 .69538 0. that can be expected from technological change in the next year.79945 0.0324 Evaluating the elasticities at t Equation 1: dyt t dt yt ˆ t 2 yˆ 0 Equation 2: dyt t dt yt ˆ2 yˆ 0 Equation 3: dyt t dt yt 2 ˆ 2 t2 yˆ 0 48 and the relevant value for yˆ 0 .219 The standard error of the forecast error is se( f ) 0.0035 2 0.577145 2. the 95% prediction interval is given by yˆ 0 t(0. (b) (c) (d) Equation 1: dyt dt ˆ Equation 2: dyt dt ˆ2 t Equation 3: dyt dt 2ˆ2t 0.4166 2.639. 1. Then.015025 48 1. The results show that the predicted effect of technological change is very sensitive to the choice of functional form.234544 (1.0141 0.0141 0.2189 2.56231 0.000337543 (48) 2 1.25293 (0. we find the standard error of the forecast error is se( f ) 0. which lies within the interval estimates from the linear and quadratic models.577 0.1696 1. Exercise Solutions.0003375 48 0.16961 ln(48) 1.975.9 (a) Equation 1: yˆ 0 0.105. Principles of Econometrics.844.509 0. 4e 108 EXERCISE 4.926) Equation 2: yˆ 0 0.1.799) Equation 3: yˆ 0 0.986 The slopes dy dt and the elasticities dy dt t y give the marginal change in yield and the percentage change in yield.050) The actual yield in Chapman was 1.975. 12944 ln(101.4 0.0 -0.4 0.6363 0.848) 1 1.2 0.10) ( 16.2 RESID WFOOD1 Figure xr4.0 3 4 5 X1 Fitted equation 6 3 4 5 X1 Residual plot 6 . (c) 0.4 0.5461 For households with 2 children. therefore.1495 ln(94.0099 0.3203 ( 16. Principles of Econometrics.0099 0.19) R2 0. the average total expenditure is 94.6 0.10(c) Plots for 1-child households 0.9535 0.12944 ln(101.70) For households with 2 children: WFOOD 0.2206 (t ) (26.1495ln(TOTEXP ) (se) (0.9535 0. 4e 109 EXERCISE 4.168 and b1 b2 ln TOTEXP ˆ 1 0. The standard errors for b1 and b2 from both estimations are relatively small resulting in high values of t ratios and significant estimates.0365) (0. Both estimations give the expected sign.1294ln(TOTEXP ) (se) R2 (0.0401) (0.0090) (t ) (25.0099 0.8 0.2 -0.9535 0. food is a necessity.848) b1 b2 ln TOTEXP 0. the average total expenditure is 101.10 (a) For households with 1 child WFOOD 1.0080) 0.168) b1 b2 ln TOTEXP Both of the elasticities are less than one.168) 1 0.Chapter 4. (b) For households with 1 child.16) For 2 we would expect a negative value because as the total expenditure increases the food share should decrease with higher proportions of expenditure devoted to less essential items.848 and b1 b2 ln TOTEXP ˆ 1 1.1495 ln(94. Exercise Solutions. 0 -0. 5.991 .1105 The estimated slope coefficient is negative.5 0.8 0.5 6.10(c) Plots for 2-child households 0. 0.2 0.2 4.0 5.2 RESID WFOOD2 Figure xr4. However.0 4.4 0. respectively.5 5.0 X2 Fitted equation (d) 5.0198 ) (0.3009 0. We conclude that for both cases the errors are not normally distributed.0045 and 0. Exercise Solutions. and statistically significant at the 5% level.4 3. the absolute magnitude of the residuals appears to decline as ln(TOTEXP) increases.7941 and 6. In Chapter 8 we discover that such behavior suggests the existence of heteroskedasticity.71) R2 0.Chapter 4. The p-values obtained are 2) 0.0 -0. The negative sign suggests that as total expenditure increases the share devoted to fuel will decrease. Also.6 0.0 4.0464ln(TOTEXP ) (se) (0. The values of JB for testing H 0 : the errors are normally distributed are 10.0412.0 Residual plot The estimated equation for the fuel budget share is WFUEL 0. we reject H 0 . respectively.5 6.0043) (t ) (15. . consistent with the low R 2 value. 4e 110 Exercise 4.5 X2 5.0 3. The plots of the fitted equation and the residuals for households with 2 children lead to similar conclusions. confirming that H 0 is rejected. Principles of Econometrics.95.4 0.10(c) (continued) (c) The fitted curve and the residual plot for households with 1 child suggest that the function linear in WFOOD and ln(TOTEXP) seems to be an appropriate one.3794 for households with 1 child and 2 children.22) ( 10. the observations vary considerably around the fitted line. Since both values are 2 greater than the critical value (0.5 4. 046409 ln(90) 1 0.300873 0. but there is not a big difference in their magnitudes at 90 and 180 pounds. Fuel need to heat houses would be considered essential.300873 0.046409 ln(180) 0. Classifying transportation as a luxury is consistent with households moving to more expensive and quicker modes of transportation as their incomes increase. 4e 111 Exercise 4.2640. (e) The elasticity for quantity of fuel with respect to total expenditure. . we find that the elasticity for transportation at median total expenditure is 1. evaluated at median total expenditure is ˆ 0. Principles of Econometrics. At the higher expenditure level the elasticity is smaller. indicating that fuel is a necessity. The positive sign suggests that as total expenditure increases the share devoted to transportation will increase.046409 ln(90) 0.0091) (t ) ( 1.3232.51 ) The estimated slope coefficient is positive. Exercise Solutions. indicating that transportation is a luxury. and statistically significant at the 5% level.300873 0. and at the 95th percentile of total expenditure it is 1. These results for fuel are consistent with economic reasoning. indicating that for these households additional percentage increases in total expenditure lead to smaller percentage increases in the quantity of fuel used.2249 These elasticities are less than one. These elasticities are greater than one.0410ln(TOTEXP ) (se) (0.Chapter 4. Using similar calculations.39) R2 0.10(d) (continued) The estimated equation for the transportation budget share is WTRANS 0.0576 + 0.4958 and at the 95th percentile of total expenditure it is ˆ 0.0216 ( 4.046409 ln(180) 1 0.300873 0. indicating that for these households additional percentage increases in total expenditure lead to smaller percentage increases in the quantity of transportation used. The share devoted to transportation increases as total expenditure increases. At the higher expenditure level the elasticity is slightly smaller. One might expect the elasticity to be higher for the higher level of total expenditure.0414) (0. The share devoted to fuel declines as total expenditure increases. and those households with higher incomes (higher total expenditures) are likely to make a smaller adjustment because they would be using an amount closer to what they consider necessary. 1% of the vote.018.87798 0.11 (a) The estimated regression model for the years 1916 to 2008 is: VOTE 50.Chapter 4.05325 0.5189 0. Exercise Solutions.0796 4.475) The actual 2008 outcome VOTE2008 (d) 46.246 The prediction error is: f VOTE2008 VOTE 2008 46.6 falls within this prediction interval.443 The estimated regression model for the years 1916 to 2004 is: VOTE 51.05325 0.61.1% of the vote is that value of GROWTH for which 50.877982 1.043 The least squares residual is: VOTE2008 VOTE 2008 (b) 46.877982 GROWTH Solving for GROWTH yields GROWTH 50.600 51. (c) The 95% prediction interval is: VOTE 2008 t(0.9185 (41.220 51.086 We estimate that real per capita GDP would have had to decrease by 1.8484 0.600 51. 4e 112 EXERCISE 4.8780GROWTH (se) (1.1825) R2 0.05325 0. This result is expected because the estimated regression in part (b) does not contain information about VOTE in the year 2008.21) se( f ) 51. . The estimated value of GROWTH that would have given the incumbent party 50.246 4.0533 0.1819 The predicted value of VOTE in 2008 is: VOTE 2008 50.1 51.043 4.1 51.0379) (0.8859 0.0125 R2 0.5243 The predicted value of VOTE in 2008 is: VOTE 2008 51.975.2464 2.8484 0.8859GROWTH (se) 1.22 51. Principles of Econometrics.646 This prediction error is larger in magnitude than the least squares residual.086% in the first three quarters of the election year for the incumbent party to win 50. 895 1.5 2 Price of Chicken Fitted values 2.0244 48. a similar. 1 1.31 and 48.3650 1 P 0.365 PQˆ When P 1.5 3 Quantity of Chicken Figure xr4.12(a) Scatter of data points and fitted reciprocal model (b) The derivative of the reciprocal model is dQˆ dP 48. The reciprocal model fits the data relatively well.365 1 30. 1.365 1 P2 Thus.0592 (2. but slightly .895 1.31 .12 (a) The estimated reciprocal model is: Qˆ R2 6.5612) 10 20 30 40 50 A plot of this equation appears below.Chapter 4.31 30. Qˆ 6. There is some tendency to underestimate quantity in the middle range of prices and overestimate quantity at the low and high extreme prices. Principles of Econometrics. the elasticity is given by dQˆ P dP Qˆ 48.0244 48. Exercise Solutions.8770 (se) 2. 4e 113 EXERCISE 4.195 The elasticity found using the log-log model was smaller absolute value than that for the reciprocal model.365 1.121 . Chapter 4.1584) R2 0.8138 .9898) (2. Principles of Econometrics.2111 31.9078 Qˆ 31.8817 2 Rreciprocal 0. Thus.195 . respectively.8138 10 20 30 40 50 A plot of this equation appears below. Exercise Solutions. its fit appears slightly worse than that of the reciprocal model.9078ln( P) (se) (0.5 2 Price of Chicken Fitted values 2. 4e 114 Exercise 4.979 The elasticities for the log-log and reciprocal models were 1. As shown in the plots.9078 41. it exhibits the least variation between the actual data and its fitted values. (e) After considering the data plots in parts (a) and (c) and Figure 4.31) 0.31 is given by dQˆ P dP Qˆ 31. Like the reciprocal model.2111 31.12(c) Scatter of data points and fitted reciprocal model (d) The derivative of the linear-log model is dQˆ dP 31.9078ln(1.121 and 1. Rg2 log log 0. 1 1.12 (continued) (c) The estimated linear-log model is: Qˆ 41. elasticity when P 1. we can conclude that the log-log model fits the data best.16 in the text.9078 1 P Thus. the linear-log model yields a lower elasticity (in absolute value) than the other models.8770 2 Rlinear-log 0.5 3 Quantity of Chicken Figure xr4. this log-linear model tends to over predict for low and high prices and under predict for mid-range prices. Also. This is confirmed by comparing the R 2 values for each model. at the mean.23 The value 67. elasticity = 2 SQFT 0.06% increase in the price of the house.00059596 1611.23 is interpreted as the increase in price associated with a 1 square foot increase in living area at the mean. This is an unrealistic and unreliable value since there are no prices for houses of zero area. Exercise Solutions.81 67. .5938 0.9682 0.00059596 112810. Principles of Econometrics.0219 t 484.000596 suggests an increase of one square foot is associated with a 0.9607 This result tells us that. 4e 115 EXERCISE 4. To find the slope d PRICE d SQFT we note that d ln( PRICE ) dSQFT d ln( PRICE ) dPRICE dPRICE dSQFT dPRICE dSQFT 2 PRICE 2 PRICE 1 PRICE dPRICE dSQFT 2 Therefore At the mean dPRICE dSQFT 0. a 1% increase in area is associated with an approximate 1% increase in the price of the house. The coefficient 0. The elasticity is calculated as: 2 SQFT 1 PRICE dPRICE SQFT dSQFT dPRICE PRICE dSQFT SQFT % PRICE % SQFT At the mean.84 0.30 The intercept 10.13 (a) The regression results are: ln( PRICE ) 10.000013 46.Chapter 4.000596 SQFT se 0.5938 is the value of ln(PRICE) when the area of the house is zero. Chapter 4.1707 is the value of ln(PRICE) when the area of the house is 1 square foot. 2 g R cov( y .1707 1.672 . yˆ ) 2 [corr( y . From the log-linear function in part (a). Principles of Econometrics. In this sense the log-linear model fits the data best. Exercise Solutions.20 44.99573 109 2 2. 2 At the means.715 From the log-log function in part (b).0066 is the elasticity since it is a constant elasticity functional form. R 2 0.1655 0.0066 112810.81 1611.444 The value 70. (c) From the linear function.673 The highest R 2 value is that of the log-linear functional form. This is an unrealistic and unreliable value since there are no prices for houses of 1 square foot in area. In other words.65 The intercept 4.0066 says that an increase in living area of 1% is associated with a 1% increase in house price.13 (continued) (b) The regression results are: ln( PRICE ) 4. To find the slope d PRICE d SQFT note that d ln( PRICE ) d ln( SQFT ) SQFT dPRICE PRICE dSQFT dPRICE dSQFT PRICE SQFT 2 Therefore. 4e 116 Exercise 4. yˆ )] 2 var( y ) var( yˆ ) 1. yˆ )] cov( y .57631 109 2 2. .0225 t 25.99996 109 0. The coefficient 1. yˆ ) 2 var( y ) var( yˆ ) 1.444 is interpreted as the increase in price associated with a 1 square foot increase in living area at the mean. The coefficient 1.9682 70. dPRICE dSQFT 2 PRICE SQFT 1. the linear association between the data and the fitted line is highest for the log-linear functional form.32604 109 0.78614 109 1. 2 g R 2 [corr( y .0066ln( SQFT ) se 0.78614 109 1. 85 p -value = 0.13(d) Histogram of residuals for simple linear model All Jarque-Bera values are significantly different from 0 at the 1% level of significance.50 0.50 -0.13 (continued) (d) 120 100 80 Jarque-Bera = 78. particularly in the simple linear model.50 0.75 Figure xr4.25 0.0000 60 40 20 0 -0. We can conclude that the residuals are not compatible with an assumption of normality.0000 120 80 40 0 -100000 0 100000 200000 Figure xr4.50 -0.00 0.25 0.25 0. 4e 117 Exercise 4.Chapter 4. .75 -0.74 p -value = 0.75 -0.0000 80 60 40 20 0 -0. Exercise Solutions.25 0.13(d) Histogram of residuals for log-linear model 120 100 Jarque-Bera = 52.13(d) Histogram of residuals for log-log model 200 160 Jarque-Bera = 2456 p -value = 0.75 Figure xr4. Principles of Econometrics.00 0. 8 0.4 0.4 -0.2 residual 0. Principles of Econometrics. 4e 118 Exercise 4.8 0 1000 2000 3000 4000 5000 SQFT Figure xr4.8 0 1000 2000 3000 4000 5000 SQFT Figure xr4. This is most evident in the residuals of the simple linear functional form.Chapter 4.2 residual 0.13 (continued) (e) 1.8 0. the residuals for the simple linear model in the area less than 1000 square feet are all positive indicating that perhaps the functional form does not fit well in this region. .4 0.0 -0.13(e) Residuals of log-linear model 1.0 -0.13(e) Residuals of simple linear model The residuals appear to increase in magnitude as SQFT increases.4 -0. Furthermore. Exercise Solutions.13(e) Residuals of log-log model 250000 200000 150000 residaul 100000 50000 0 -50000 -100000 -150000 0 1000 2000 3000 4000 5000 SQFT Figure xr4. 59379+0.3355 85.20876 The 95% confidence interval for the prediction from the log-log model is exp ln( y ) t(0.203034 1 x0 xi 1 880 2 x x 2 2700 1611. 4e Exercise 4.1 The 95% confidence interval for the prediction from the log-linear model is: exp ln( y ) t(0.878) se f exp 4.683.878) se f exp 10.65 81. 221 Prediction for simple linear model: PRICE (g) 18385.13 (continued) (f) Prediction for log-linear model: PRICE exp b1 b2 SQFT ˆ2 2 exp 10.316 The standard error of forecast for the log-log model is 1 se( f ) 0.000595963 2700+ 0. Exercise Solutions.59379+0.975.203032 2 203.3890 2700 201.968 2 0.2082512 2 188.96267 0.96267 0.006582 log(2700) 1.975.170677 + 1. 297.170677 + 1.006582 log(2700)+ 0.000595963 2700 1.20363 133.208251 1 880 7.Chapter 4.20363 248768933.20876 122.90101 7.516 Prediction for log-log model: PRICE exp 4.34453 2 0. Principles of Econometrics. 277. 267. 454 119 .365 The standard error of forecast for the log-linear model is se( f ) ˆ 2 1 1 N 0. 878) se f 201.364.96267 30.26 The 95% confidence interval for the prediction from the simple linear model is yˆ 0 t(0. characteristics that suggest it is the model that best fits the data. a 1% change in square feet leads to a 1% change in price. 4e 120 Exercise 4.968 248768933.Chapter 4. namely.2 1 1 880 2700 1611.26 141. Their residuals have similar patterns and they both lead to a plausible elasticity of price with respect to changes in square feet.801. Principles of Econometrics. 260. The loglinear model is favored on the basis of its higher Rg2 value.348. .13(g) (continued) The standard error of forecast for the simple linear model is se( f ) 30259.62 1.1 2 30348. and its smaller standard deviation of the error.928 (h) The simple linear model is not a good choice because the residuals are heavily skewed to the right and hence far from being normally distributed.975. Exercise Solutions. It is difficult to choose between the other two models – the log-linear and log-log models. 0000) for the residuals from the linear model and JB 27.1361 The estimated return to education at mean wage b2 100 WAGE 1. The Jarque-Bera test results are JB 839. However.6349 (p-value = 0. that for ln(WAGE) exhibits a more symmetric normal shape. 4e 121 EXERCISE 4.6157 The results for the log-linear model are ln WAGE se 1.0 1. ln(WAGE) more closely resembles a normal distribution.14(a) 140 80 120 70 100 60 50 80 40 60 30 40 20 20 10 0 0 0 10 20 30 40 50 60 70 1.82 (p-value = 0.5 3. Exercise Solutions.9142 0.6094 0. in both cases.73 (p-value = 0. a null hypothesis of normality is rejected at a 1% level of significance.0 2.0 Histogram of ln(WAGE) Neither WAGE nor ln(WAGE) appear normally distributed. . Both the histograms and the Jarque-Bera test results suggest the residuals from the log-linear model are more compatible with normality.04%. (c) The histograms of residuals are displayed in Figure xr4.0 3. (b) The regression results for the linear model are WAGE se R2 6.1782 0.0000) for WAGE and JB 0.1980 EDUC 1. While the distribution for WAGE is positively skewed.53 (p-value = 0.7103 1.14(c).5 Histogram of WAGE 2. This conclusion is confirmed by the Jarque-Bera test results which are JB 773.9803 100 9.7280) for ln(WAGE).0061 The estimated return to education b2 100 9.61% 20.14 (a) Figure xr4.0904 EDUC 0. However. Principles of Econometrics.5 4.Chapter 4.0000) for the residuals from the log-linear model.0864 R2 0.1750 0. suggesting heteroskedasticity which is covered in Chapter 8.5 50 -2. (e) Figure xr4.Chapter 4.0 1. It is also apparent.1859 where WAGE Since.WAGE 2 0. the positive residuals tend to be greater in absolute magnitude than the negative residuals.5 0. Exercise Solutions. that there are only positive residuals in the early range of EDUC. . Rg2 0.14(e) Residuals plotted against EDUC 60 2 50 1 40 20 residual residual 30 10 0 -1 0 -10 -2 -20 -30 -3 0 4 8 12 16 educ Simple linear model 20 24 0 4 8 12 16 20 24 educ Log-linear model The absolute value of the residuals increases in magnitude as EDUC increases.14(c) (continued) Figure xr4.1750 corr WAGE .5 log-linear regression 0.14(c) Histograms of residuals 140 200 120 160 100 120 80 60 80 40 40 20 0 0 -30 -20 -10 0 10 20 30 40 -2.5 1.0 -1. This suggests that there might be a threshold effect – education has an impact only after a minimum number of years of education. R 2 we conclude that the log-linear model fits the data better. We also observe the non-normality of the residuals in the linear model.0 -0. Principles of Econometrics.5 Simple linear regression (d) Linear model: R2 Log-linear model: Rg2 -1.0 exp(b1 b2 EDUC ). 4e 122 Exercise 4. for both models. 60944 0. when predicting the average age of workers with 16 years of education.98029 16 24. .5266112 ) / 2 24.71028 1.501 (g) The log-linear function is preferred because it has a higher goodness-of-fit value and its residuals are more consistent with normality. However.Chapter 4.401 Actual average wage of all workers with 16 years of education = 25. Principles of Econometrics.090408 16 (0.974 Prediction for log-linear model: WAGE c exp 1. the linear model had a smaller prediction error. Exercise Solutions.14 (continued) (f) Prediction for the simple linear model: WAGE 0 6. 4e 123 Exercise 4. 368 13.206 17.500 19.30 1.142 19.1095 (0.0067) 9.39 76.4244) 0.4 57.13 57.97 1.7 These results show that.Chapter 4.1175 (0.3712 white males (se) The return to education is highest for black females (15.50 72.7 69.5 74.60%) and lowest for black males (6.8778 (0.1147 (0.0904 (0.9345 (0.50 72.6 65.0325) 6.218 12.13 76.39 72.0254) 15.6 61. Principles of Econometrics.1092) 0.1560 (0.6250 (0.1612 (vi) white females (se) 1.0079) 7.0941) 0.50 2.04 0. Black females have the highest coefficient of variation and all males and white males have the lowest.2437 (iii) all whites (se) 1.0100) 12.5% for all other sub-samples. (c) Regression results Sub-sample Constant EDUC % return R2 (i) all males (se) 1. 4e 124 EXERCISE 4.606 2.780 22.0933 (viii) black females (se) 0.0197 (0.172 20.965 12.1243 (0.2716) 0.2656 (vii) black males (se) 1.1693 (0.3552) 0. on average.765 12.39 52.75 0.1716 (ii) all females (se) 1. Exercise Solutions.92 0. The wage of white females is approximately the same as that of all females. white males have the highest wages and black males the lowest. .6 66.dat (a).47 0.60 0. It varies approximately from 8 to 12.539 10.97 6.45 6.30 1.13 76.13 72.0770 (0.3 60.5610 (0.0086) 7.150 18.744 12. (b) Summary statistics for WAGE Sub-sample (i) all males (ii) all females (iii) all whites (iv) all blacks (v) white males (vi) white females (vii) black males (viii) black females Mean Std Dev Min Max CV 22.0200) 11.0796 (0.839 17.8068 (0.1770 (iv) all blacks (se) 1.1314) 0.1439) 0.92%).339 12.0092) 11.96 0.97 7.1176) 0.43 0.70 0.851 12.15 Results using cps4_small.2310 (v) 1.0692 (0. 965 0.155 Fail to reject H 0 (iv) all blacks 0. For males and white males.056 Fail to reject H 0 (iii) all whites –1.016 Reject H 0 (vii) black males 0.463 Fail to reject H 0 (v) white males 2.207 64 1.. .10 against H1 : 0. the wage return to an extra year of education is estimated as less than 10%. Test results for H 0 : 2 0.349 Fail to reject H 0 (viii) black females 2.963 0.736 110 1.982 0.008 Reject H 0 (vi) white females 2. Those for white males and black males are particularly poor. In all other sub-samples.425 843 1.679 417 1.10 Sub-sample t-value df tc p-value Decision (i) –2.947 44 2.15 (continued) Results using cps4_small. Exercise Solutions.1 se(b2 ) tc or t tc where tc t(0.dat (d) The model does not fit the data equally well for each sub-sample.015 0.965 0. the data do not contradict the assertion that the wage return is 10%.569 484 1.011 Reject H 0 (ii) all females 1.917 512 1.10 versus H1 : 2 0.966 0. 4e 125 Exercise 4.998 0. (e) The t-value for testing H 0 : t We reject H 0 if t 2 0. Principles of Econometrics. The results are given in the following table.975.031 Reject H 0 all males The null hypothesis is rejected for males. white males. The best fits are for black females and white females.966 0. white females and black females. df ) .10 is given by 2 b2 0.420 424 1. suggesting that there is statistical evidence that the rate of return is different to 10%. while it is greater than 10% for the other two sub-samples where H 0 was rejected.Chapter 4. 7909 (0.0861 (0.0115) 12.2105) 0.62%).0911 (0. (b) Summary statistics for WAGE Sub-sample (i) all males (ii) all females (iii) all whites (iv) all blacks (v) white males (vi) white females (vii) black males (viii) black females Mean Std Dev Min Max CV 22. For all other sub-samples.1057 (0.5924 (0.013 9.2541 (0.13 72. 4e 126 EXERCISE 4.62 0.834 18. .119 16.0617) 0.9 These results show that.61 0.444 22.64 0.1592) 0.671 11.13 173.0036) 8.0559) 0. males have higher average wages than females and whites have higher average wages than blacks.00 96.1052 (0.2033 (v) 1.75 173.Chapter 4.2043 (ii) all females (se) 1.1064 (0.0522) 0.8 58.00 1.62 0.11 0.17 173.054 20.5 %.9 60.136 13.14 1. on average.2456 (0.6 59.52 0. Black females have the highest coefficient of variation and black males have the lowest.3024 white males (se) The return to education is highest for black females (12.6 63.14 1.213 16.2264 (vii) black males (se) 1.50 1.dat (a).00 72.0762 (0.0499) 0.0094) 10.493 10.0158) 7.14 1. (c) Regression results Sub-sample Constant EDUC % return R2 (i) all males (se) 1.1278) 0.157 12.0983 (viii) black females (se) 0.638 10.2427 (0.13 60.84 0.2312 (iii) all whites (se) 1.621 13.5 to 10.17 72. Overall.57 0. white males have the highest wages and black males the lowest.9395 (0.0039) 10.2059 (vi) white females (se) 1.485 16.00 3.473 11.0029) 9. Exercise Solutions.15 Results using cps4. Principles of Econometrics.62%) and lowest for black males (7. it varies from approximately 8.1923 (iv) all blacks (se) 1.0037) 8.258 18.8 61.7 61.0043) 10.6521 (0.5 61.00 1.7326 (0.1262 (0.0411) 0.616 1.00 96. The highest wage earner is a white male.0884 (0. 629 2441 1.961 0. In all other sub-samples.0021 Reject H 0 (iv) all blacks 0.961 0.326 2049 1. That for black males is particularly poor.1 se(b2 ) tc or t tc where tc t(0. .971 0.504 212 1.273 277 1.5816 Fail to reject H 0 –3.10 against H1 : 0.969 0.dat (d) The model does not fit the data equally well for each sub-sample. suggesting that there is statistical evidence that the rate of return is different to 10%. while it is greater than 10% for the other two sub-samples where H 0 was rejected.10 t-value df tc p-value Decision 3. Test results for H 0 : 2 0.1851 Fail to reject H 0 (vii) black males 1.0238 Reject H 0 Sub-sample (i) (v) all males white males The null hypothesis is rejected for males.551 491 1. white males and black females. Exercise Solutions. all whites.10 versus H1 : 2 0. the wage return to an extra year of education is estimated as less than 10%.975.15 (continued) Results using cps4. (e) The t-value for testing H 0 : t We reject H 0 if t 2 0.961 0.961 0. For males and all whites.0002 Reject H 0 (vi) white females 1. the data do not contradict the assertion that the wage return is 10%. The results are given in the following table.263 2393 1.965 0.720 2063 1.10 is given by 2 b2 0. df ) .1034 Fail to reject H 0 (iii) all whites –3.Chapter 4.1341 Fail to reject H 0 (viii) black females 2.075 4114 1.0011 Reject H 0 (ii) all females 1.961 0. 4e 127 Exercise 4. Principles of Econometrics. The best fits are for all females and black females. 2 2.4 1940 1950 1960 1970 1980 YEAR Greenough .16(b) Plots of the reciprocal of yield against time 3. the inverse of yield is 1 AREA hectares / tonne YIELD PRODUCTION Thus.8 0.8 0.16 (a) By definition.0 1. RYIELD can be interpreted as the number of hectares needed to produce one tonne of wheat RYIELD (b) Figure xr4.0 1. Exercise Solutions.0 RMULLEWA 1960 YEAR 2.Chapter 4.6 1.2 0.2 0.8 1950 1960 1970 1980 1990 0.6 2.6 1.4 1940 1950 1950 1960 1970 YEAR Mullewa 1980 1990 2000 0.4 2.4 RNORTHAMPTON 2. Principles of Econometrics.4 1940 0.2 2.8 3.8 0.4 2.0 RGREENOUGH 2.6 1.4 1940 2000 1970 1980 YEAR Chapman Northampton 1990 2000 1990 2000 2.4 2.8 RCHAPMAN 2. yield is given as PRODUCTION AREA YIELD tonnes / hectare So.2 1.8 1. 4e 128 EXERCISE 4.6 1. 3934 0.Chapter 4.0024) R2 0.0686) (0. . Exercise Solutions.2950 (0. 4e 129 Exercise 4. A check of rainfall data at http://www. There were similar but less extreme outliers in Mullewa in 1976.4552 0.0132TIME (se) (0. Thus. (c) The estimated equations are Northampton RYIELD 1. Wheat production in Western Australia is highly dependent on rainfall. The test statistic assuming the null hypothesis is true is: t ˆ2 se( ˆ 2 ) t(46) We reject H 0 if t t(0.1300) (0. since technological change will lead to a reduction in the number of hectares needed to produce one tonne of wheat. and in Chapman in 1976 and 1977.4954 Mullewa Greenough In each case the estimate of 2 is an estimate of the average annual change in the number of hectares needed to produce one tonne of wheat.0039) Chapman RYIELD 1. A one-tail test is used because. The test results for testing H 0 : 2 0 against the alternative H 1 : 2 0 are given in the table below.46) 1.16(b) (continued) There is an outlier in 1963 across all four shires.1087) (0.678 or p-value 0.gov. excess rainfall may have caused rust or other disease problems during the growing season. and so one would suspect that rainfall was low in the above years.3594 0. or rain at harvest time may have led to a deterioration in wheat quality.au/climate/data/ reveals that rainfall was lower than usual in 1976 and 1977. implying that a greater number of hectares was needed to produce one tonne of wheat than in any other year. but higher than normal in 1963. if 2 is not zero. Principles of Econometrics. for Greenough.0121TIME (se) (0. we expect it to be negative. we estimate that the number of hectares needed declines by 0. 1977 and 1979.05.0164 per year.3485 0.2869 RYIELD 1. For example.0164TIME (se) (0.bom. In all four cases the null hypothesis is rejected indicating that the required number of hectares is decreasing over time.1306 RYIELD 1.05 .0862) (0.0046) R2 0. it is difficult to assess why 1963 was a bad year.0031) R2 0.0169TIME (se) R2 0. 25 -0. Principles of Econometrics.0 -0.2 RES_RGREENOUGH RES_RMULLEWA 0.0000 Reject H 0 (iii) Mullewa 2.679 0.5 1940 2000 1950 1960 YEAR 1970 1980 1990 2000 1990 2000 YEAR Chapman Northampton 1.75 0.0000 Reject H 0 (d) 2. the residuals for 1976.6 0.0 1. Exercise Solutions.0 -0.00 2.5 1950 1960 1970 1980 1990 -0. in addition. confirming that this observation is an outlier.50 1940 1. .6 1.679 0.0000 Reject H 0 (ii) Chapman 4. this observation is also an outlier but.0 RES_RNORTHAMPTON RES_RCHAPMAN Figure 4.0 0.00 1.0 -0.679 0.629 1.721 1.2 1.16(c) Residual plots from estimated equations 1.Chapter 4. In Mullewa.2 0.8 0.50 0.25 0.4 -0.8 0. 1978 and 1980 are relatively large.0058 Reject H 0 (iv) Greenough 6.4 0. 4e 130 Exercise 4.4 0.387 1.5 0.16(c) (continued) Test results for H 0 : Shire 2 0 versus H 1 : tc t-value 2 0 p-value Decision (i) Northampton 4.8 1940 0.2 1950 1960 1970 1980 1990 2000 -0.302 1.679 0.5 1.25 0.4 1940 1950 1960 1970 1980 YEAR YEAR Mullewa Greenough The residual for 1963 is clearly much larger than all others for all shires except Mullewa. 5515 RYIELD 1.0017) R2 0. The standard errors for the coefficient of interest ˆ 2 decrease for all four shires.0472) (0.2850 0.0024) R2 0.0019) R2 0.6448 Chapman Mullewa Greenough When we re-estimate the reciprocal model without data for the year 1963.0117TIME (se) (0.1211) (0. 4e 131 Exercise 4.0107TIME (se) (0.16 (continued) (e) The estimated equations with the observation for 1963 omitted are Northampton RYIELD 1. Principles of Econometrics.0150TIME (se) (0. Also the value of R 2 increases considerably for Northampton.0686) (0. Exercise Solutions.3429 RYIELD 1. but that for Mullewa shire decreases slightly.0549) (0. in all cases the coefficient of time declines slightly in absolute value.3010 0. suggesting that the earlier estimates may have exaggerated the effect of technological change. Chapman and Greenough.2862 0.Chapter 4.1222 RYIELD 1. .0043) R2 0.3929 0.0144TIME (se) (0. CHAPTER 5 Exercise Solutions 132 . (b) (c) 0. x3 xi*2 xi*3 xi*2 xi*3 2 13 10 4 0 16 10 02 0.4. Exercise Solutions.8125 4 16 13 0 16 10 02 0.Chapter 5.1625 xi 2 xi 3 x3 ) 2 xi 2 2 xi 3 2 ˆ2 x2 ) 2 (1 r232 ) SST ( yi R2 0 0. x2 ) var(b2 ) 1.6396 x2 )( xi 3 2 0.8375 SSE 12. 0. Principles of Econometrics.8375 9 3 ( xi 2 ( xi 2 0.1 (a) y 1. 0.4 y b2 x2 b3 x3 1 (d) eˆ 0. yi* xi*2 13.4125. xi*2 xi*3 xi*2 xi*3 xi*22 yi* xi*3 b3 b1 xi*3 xi*32 yi* xi*2 b2 xi*2 xi*22 16. 4e EXERCISE 5.375. 0.1999 y )2 16.025.1875 0.1625 16 0. x3 ) ( xi 3 ( xi 2 3.6.025. SSR SST 12.4125. (e) ˆ2 eˆi2 N K (f) r23 (g) se(b2 ) (h) SSE eˆi2 SSR SST 3. 0. x2 0 yi* 0 1 2 2 1 2 0 1 1 1 2 1 0 1 1 1 1 0 0 1 2 2 1 2 1 0 1 yi* xi*3 yi* xi*3 *2 i2 *2 i3 x x 2 yi* xi*2 *2 i2 *2 i3 x x xi*32 10 4. 0.7602 133 .9875.6396 16 0. Principles of Econometrics.8125 1 0. 6) 2. .975. we reject H 0 if t we do not reject H 0 .8125 2. t(0. Since 0. 4e 134 EXERCISE 5.2 (a) A 95% confidence interval for 2 is b2 t(0.975.447 0.447 . Exercise Solutions.9377 At a 5% significance level.1999 (0.9377 2.3017) (b) The null and alternative hypotheses are H0 : 2 1.3233. H1 : 2 1 The calculated t-value is t b2 1 se(b2 ) 0.1999 0. 1.447 .Chapter 5.6) se(b2 ) 0. To find SSE (N K ) 5.054 (v) The estimated error standard deviation is ˆ (b) 5.0133.061622 The value b2 0. (c) A 95% confidence interval for 3 b3 t0.08246 .975.00418 .3 (a) (i) The t-statistic for b1 is b1 se(b1 ) 0. From the output.9624) 0.752896 .08246 1 6.0014. 0. 0.6086 0.0002 ( 0. the share of the alcohol expenditure is estimated to decrease by an amount between 0. (iv) To compute R 2 .0014 . SSE SST.96 0. .0276 implies that if ln(TOTEXP) increases by 1 unit the alcohol share will increase by 0.0133 suggests that if the household has one more child the share of the The value b4 alcohol expenditure decreases by 0.Chapter 5.001.75290 6. we use the result ˆy SST N 1 0.0276 TOTEXP .0276 6.0018 and 0. The change in the alcohol share from a 1-unit change in total expenditure depends on the level of total expenditure.0633) 2 R2 1 SSE SST 5.1515) se(b3 ) is 0.0010) This interval tells us that. 0. Specifically. Thus.0091 0. Principles of Econometrics.476 .0191 (ii) The standard error for b2 is se(b2 ) (iii) The estimate for 3 is b3 0. d (WALC ) d (TOTEXP) 0. Exercise Solutions.0018.0002 ( 6. 0.0014 suggests that if the age of the household head increases by 1 year The value b3 the share of alcohol expenditure of that household decreases by 0. 0. 4e 135 EXERCISE 5. we need SSE and SST.000276 increase in the alcohol share of expenditure. if the age of the household head increases by 1 year.0276.0014 1.0633 which gives SST 1518 (0. A 1% increase in total expenditure leads to a 0. 0.752896 1519 4 0. we reject H 0 if t t(0. 1515) 1. H1 : 4 0. we reject H 0 and conclude that the number of children in the household influences the budget proportion on alcohol. .96 .3 (Continued) (d) The null and alternative hypotheses are H 0 : The calculated t-value is t b4 se(b4 ) 4 0.075 At a 5% significance level.Chapter 5. Having an additional child is likely to lead to a smaller budget share for alcohol because of the non-alcohol expenditure demands of that child. Also. Exercise Solutions.975. 4e 136 Exercise 5. Principles of Econometrics. 4. Since 4. believing that drinking may be a bad example for their children. perhaps households with more children prefer to drink less.075 1.96 . 013.7) 0.0001 implies that as the age of the head of the household increases by 1 The value b3 year the budget share for transport decreases by 0. (d) The proportion of variation in the budget proportion allocated to transport explained by this equation is 0.0414ln(TOTEXP0 ) 0.004. Similar tests for the coefficients of the other two variables yield p-values less than 0. Alternatively. suggesting that AGE could be excluded from the equation. Alternatively. Principles of Econometrics.3.0247 (0.0004) R2 0.4 (a) The regression results are: WTRANS se (b) 0.0055) The value b2 0.0315 0.0414ln TOTEXP (0.0315 0.0315 0.869.0414. The expected sign for b3 is not clear.0322) (0.0130 NK (0. (e) For a one-child household: WTRANS 0 0.0414 ln(98.013 NK 0 0.0315 0. one can say that a 10% increase in total expenditure will increase the budget proportion for transportation by 0.05.013 NK 0 0.0001 AGE0 0. The negative sign means that adding children to a household increases expenditure on other items (such as food and clothing) more than it does on transportation. having more children may lead a household to turn to cheaper forms of transport. it is difficult to predict the effect of age on transport share. For a given level of total expenditure and a given number of children.0001.0414ln(TOTEXP0 ) 0.0001 36 0.013 2 0.) The positive sign of b2 is according to our expectation because as households become richer they tend to use more luxurious forms of transport and the proportion of the budget for transport increases.0130 implies that an additional child decreases the budget share for The value b4 transport by 0.0001 AGE 0.013 1 0.0247.3.0315 0.0001 AGE0 0.0414 suggests that as ln TOTEXP increases by 1 unit the budget proportion for transport increases by 0.0001 36 0. 0.0414 ln(98.Chapter 5. 0. 4e 137 EXERCISE 5. Exercise Solutions.0071) 0. (c) The p-value for testing H 0 : 3 0 against the alternative H1 : 3 0 where 3 is the coefficient of AGE is 0.1420 For a two-child household: WTRANS 0 0.1290 . (See Chapter 4.7) 0. 3924) (0. An increase in the proportion of owner-occupied units built prior to 1940 leads to a decline in the home value.0478 AGE (0. The data is consistent with the hypothesis that increasing the number of rooms by one increases the value of a house by $7000.811. A 95% confidence interval for the coefficient of ACCESS is b7 t(0. we reject H 0 if the absolute calculated t is greater than 1.0723 (0.0365) (4. The higher the tax rate per $10.05 .2001) 6.8109 NITOX (se) (5. we do not reject H 0 .1768 PTRATIO (0.1834CRIME 22.05 . We cannot value t(0. Since 1.497) conclude that reducing the pupil-teacher ratio by 10 will increase the value of a house by more than $10.130.1607) 1.2723 ACCESS 0.965 0.1834 1.965 .000 the lower the home value.0365 ( 0.4.497)se(b2 ) 0.497)se(b7 ) 0.335 for every unit of weighted distance.965.414) (c) We want to test H 0 : t At room 7 against H1 : brooms 7 se(brooms ) room 6.112) .6017 1. (d) We want to test H 0 : t ptratio 1 against H1 : 1.3353DIST (0. The further the weighted distances to the five Boston employment centers the lower the home value by $1. Since 1. Exercise Solutions.3659) (0.0723) (0. we reject H 0 if the calculated t is less than the critical 1. Finally.000.255.965 0.0038) (0.05.2723 1.975. a one unit increase in the nitric oxide concentration leads to a decline in value of $22. an increase in one room leads to an increase of $6.372. The value of the t statistic is 1.2683 1.0141) 0.648.648 . .4067 0.Chapter 5.0126TAX 1. the lower the home value.975. the higher the pupil-teacher ratio.5 (a) The estimated equation is VALUE 28. Increasing the average number of rooms leads to an increase in the home value.6017 0.3924 7 . The value of the t statistic is 1. The higher the level of air pollution the lower the value of the home. (b) A 95% confidence interval for the coefficient of CRIME is b2 t(0.3715 7 0. we do not reject H 0 .1768 1 0.1394) The estimated equation suggests that as the per capita crime rate increases by 1 unit the home value decreases by $183.3715ROOMS 0. Principles of Econometrics.1394 ptratio 1 . 0. 4e 138 EXERCISE 5. 0.2683 At a significance level of 0. Exercise Solutions. 4e 139 EXERCISE 5.5 se(b2 ) 4 0 against the alternative 2 Since 2 1.60) 2. (a) The value of the t statistic for testing the null hypothesis H 0 : H1 : 2 0 is b2 3 t 1. There is no sample evidence to suggest that 2 2 5.975.60) 2.000 and t(0.000 .5 2 . (b) For testing H0: 1 +2 2 = 5 against the alternative H1: b1 2b2 t 1 +2 2 5. we use the statistic 5 se b1 2b2 For the numerator of the t-value. b3 ) 3 4 3 2 2 2 1 0 Thus. we fail to reject H0 and conclude that there is no sample evidence to suggest that 2 0.5 2 .025. 1 (c) For testing H 0 : t Now.3166 0. we use (b1 b2 b3 ) 4 se(b1 b2 b3 ) b3 ) 4 2 3 1 4 se(b1 b2 b3 ) 6 . and var(b1 b2 b3 ) var(b1 ) var(b2 ) var(b3 ) 2cov(b1 . we fail to reject H0 and conclude that there is insufficient sample evidence to suggest that 1 2 + 3 = 4 is incorrect.Chapter 5. t 6 4 4 1. . The critical values are given by t(0.9045 Since 2 0. t 3 3.3166 var(b1 ) 4 var(b2 ) 4 cov(b1 . we have b1 2b2 5 2 2 3 5 3 The denominator is given by se(b1 b2 ) var(b1 2b2 ) 3 4 4 4 2 Therefore.9045 2 . b2 ) 11 3. b2 ) 2cov(b1 . The rejection region is t 2 or t 2 . we fail to reject H0. Principles of Econometrics. b3 ) 2cov(b2 . (b1 b2 1 2 3 4 against the alternative H1 : 1 2 3 4 .5 Since 2 1.6 In each case we use a two-tail test with a 5% significance level. 00684 0.Chapter 5.05590 1 ( 0. Exercise Solutions.05590 Thus.57 0. Principles of Econometrics. 4e 140 EXERCISE 5. b2 and b3 are : se b2 var(b2 ) se b3 var(b3 ) ˆ2 2 23 1 r ( xi 2 x2 ) ˆ2 2 23 1 r ( xi 3 x3 ) 2 2 0. the standard errors of the least square estimates.114255) 2 1210.05590 1 ( 0.00137 .178 0.12389 202 3 0.7 The variance of the error term is given by: ˆ2 SSE N K 11.114255)2 30307. (c) To test these hypotheses.800 1. the marginal effect of one fertilizer is smaller.778 PHOS 0. with that for nitrogen becoming negative.233 7.944 NITRO 0.888 2 0. 4e 141 EXERCISE 5. we test H 0 : 2 2 4 0 against the alternative if 2 2 4 6 6 H1 : 2 2 4 0.567 2.567 PHOS E YIELD PHOS 4. Principles of Econometrics.888 NITRO 0.800 2 0.567 2 2 are 0.567 NITRO 4.011 3.556 4.800 1.554 When NITRO 1 and PHOS 1 .011 3. Increasing the fertilizer applications to NITRO 2 and PHOS 2 reduces the marginal effects of both fertilizers. Exercise Solutions. the larger is the amount of the other fertilizer that is applied.011 2 1.567 PHOS 8.556 0. Thus.899 4. (b) (i) The marginal effects when NITRO 1 and PHOS 1 are E YIELD NITRO E YIELD PHOS (ii) 8.944 0. The value of the test statistic is 6 t b2 2b4 b6 se b2 2b4 b6 8.567 3.Chapter 5.556 PHOS 0.556 2 0. Also.011 2 1.011 3.567 2 0.367 .567 NITRO These equations indicate that the marginal effect of both fertilizers declines – we have diminishing marginal products – and these marginal effects eventually become negative.677 The marginal effects when NITRO 2 and PHOS E YIELD NITRO E YIELD PHOS 8. the coefficients are defined according to the following equation YIELD 1 2 NITRO 3 PHOS 4 NITRO 2 5 PHOS 2 6 NITRO PHOS e (i) The settings NITRO 1 and PHOS 1 will yield a zero marginal effect for nitrogen 0 .800 1.8 (a) Equations describing the marginal effects of nitrogen and phosphorus on yield are E YIELD NITRO 8.888 0.567 0. the marginal products of both fertilizers are positive. 567 0.778) 4.975.701 ( 0.567) 2 4 ( 1. A zero marginal yield Since |t| < 2. 21) .567 0. Principles of Econometrics.011 ( 0. we reject the null hypothesis and conclude that the marginal effect of nitrogen on yield is not zero when NITRO = 1 and PHOS = 1.944 0.567) 2 4 ( 1.040 1. the optimal levels are those for which E YIELD PHOS PRICEPHOS PRICEPEANUTS and E YIELD NITRO PRICENITRO PRICEPEANUTS . we reject the null hypothesis and conclude that the Since |t| > 2.944)( 0. we test H 0 : 2 4 4 0 against H1 : 2 4 4 0 . (iii) To test whether the marginal effect of nitrogen is zero when NITRO 3 and PHOS 1 . 4e 142 Exercise 5. 21) 2. The value of 6 6 the test statistic is t b2 4b4 b6 se b2 4b4 b6 8.778) 2 4. we do not reject the null hypothesis.778) 2.660 t(0.8(c)(i) (Continued) Since t > tc t(0.800 ( 0.567) 1.742 t(0. we test H 0 : 2 6 4 0 against the alternative H1 : 2 6 4 0.Chapter 5. (ii) To test whether the marginal effect of nitrogen is zero when NITRO 2 and PHOS 1 .800 ( 1.080 with respect to nitrogen cannot be rejected when NITRO = 1 and PHOS = 2.975.944 0.011 6 1.080 .975. E YIELD PHOS 3 2 5 PHOS 2 2 4 NITRO E YIELD NITRO 6 NITRO 0 PHOS 0 6 The solutions and their estimates are NITRO PHOS 2 2 2 2 6 5 3 6 4 3 4 2 6 4 4 5 2 6 4 5 2 8.080 marginal product of yield to nitrogen is not zero when NITRO = 3 and PHOS = 1. 21) . Exercise Solutions.011 ( 0. The optimal levels are those where the marginal cost of the inputs is equal to the marginal value product of those inputs.567) ( 0. Thus. (d) The maximizing levels NITRO and PHOS are those values for NITRO and PHOS such that the first-order partial derivatives are equal to zero.944) 8.465 The yield maximizing levels of fertilizer are not necessarily the optimal levels.944)( 0. 6 6 The value of the test statistic is t b2 6b4 b6 se b2 6b4 b6 8.011 4 1.233 8. Chapter 5.998)se b3 8b4 0.0101 0.998)se b2 (ii) 2. Also.975.9 (a) The marginal effect of experience on wages is WAGE EXPER 2 4 EXPER 3 (b) We expect 2 to be positive as workers with a higher level of education should receive higher wages.4238.2.7788) . A negative 3 and a positive 4 gives a quadratic function with these properties. additional experience leads to a larger increase in their wages than it does after they become relatively experienced. Also. EXPER .2774 A 95% interval estimate is given by b2 t(0. When workers are relatively inexperienced. we expect 3 and 4 to be positive and negative.6013 1. the number of years of experience at which wages start to decline. b4 0.2774 1.6013 To compute an interval estimate. The maximum is reached when the first derivative is zero.1394 (2. 4e 143 EXERCISE 5.6821 8 0.975. (c) Wages start to decline at the point where the quadratic curve reaches a maximum. eventually we expect wages to decline with experience as a worker gets older and their productivity declines.0.5509) A point estimate of the marginal effect of experience on wages when EXPER WAGE EXPER b3 4 is 2b4 (4) 0.000189259 0.09045 (0. Exercise Solutions. we need the standard error of this quantity which is given by se b3 8b4 var b3 82 var b4 2 8 cov b3 .09045 A 95% interval estimate is given by b3 8b4 t(0.962 0. respectively.000003476 16 0. Thus. is such that 3 2 4 EXPER 3 EXPER (d) (i) 0 2 4 A point estimate of the marginal effect of education on wages is WAGE EDUC b2 2.010987185 64 0. Principles of Econometrics.962 0.0039. 0101 0.6821 33. b4 0. we need the standard error of this quantity which is given by se b3 50b4 50 2 var b4 var b3 2 50 cov b3 . EXPER To obtain an interval estimate for EXPER .000003476 100 0.02741 A 95% interval estimate is given by b3 50b4 t(0.0.962 0.9(d) (continued) (iii) A point estimate of the marginal effect of experience on wages when EXPER is WAGE EXPER b3 25 2b4 (25) 0.975.010987185 2500 0.Chapter 5.77 2b4 2 0. Exercise Solutions. EXPER var EXPER 2 2 EXPER var b3 3 var b4 4 2 EXPER EXPER 3 cov b3 .02741 (0.0101 We estimate that wages will decline after approximately 34 years of experience. we require se requires the derivatives EXPER 3 EXPER 1 2 4 2 4 b3 2b4 which in turn 3 2 4 Then. b4 4 and var EXPER 1 2b4 2 var b3 Substituting into this expression yields b3 2b42 2 var b4 2 1 2b4 b3 cov b3 .000189259 0.998)se b3 50b4 0. b4 2b42 . Principles of Econometrics.2309) (iv) Using the equation derived in part (c).1233. 4e 144 Exercise 5.1771 To compute an interval estimate. we find: b3 0.6821 50 0.1771 1. 975.1237.975.3) .000003476 0.962 0.Chapter 5.010987185 2 1 2 0.4239.7762 (30.000189259 3.998)se EXPER 33.01012 2 0.3.0. If the computations are made using software and the file cps4c_small.9(d)(iv) (continued) 1 2 0.77 (30.77 1.0.9.2314) (iv) EXPER t(0.9 of the text.3.01012 0.798 1.998)se b3 8b4 0.131785 se EXPER 3.770 A 95% interval estimate for EXPER is EXPER t(0.6821 2 0.60137 1.0101 var EXPER 2 0. Principles of Econometrics. Exercise Solutions. 37.(iii). 37.998)se EXPER 33.17756 1. 4e 145 Exercise 5.6821 2 0. slightly different results are obtained.962 1. The answers obtained using software for parts (d)(ii) .975.0101 0. These results do not suffer from the rounding error caused by truncating the number of digits reported in Table 5. and (iv) are: (d) (ii) b3 8b4 t(0.dat.090418 (0.962 0.131785 1.2) Note: The above answers to part (d) are based on hand calculations using the estimates and covariance matrix values reported in Table 5.962 1.027425 (0.998)se b3 50b4 0.975.7789) (iii) b3 50b4 t(0. 000000 1.4000 (e) 0.000000 0. Method: Least Squares Dependent Variable: Y Method: Least Squares Included observations: 9 Coefficient Std.717882 1. Exercise Solutions.680208 0.D. (c) The least squares estimates can be read directly from the table.1648 X1 X2 X3 R-squared Adjusted R-squared S.000000 0.728217 0.000000 0.0066 0.799742 0.063823 1.934631 Mean dependent var S.000000 0.652140 2.E.1625 0.581654 0. 4e 146 EXERCISE 5.751221 4.0250 –0.000000 1.316228 0. (d) The residuals from the estimated equation are: –0.799740 3. That is. Then.316228 1. Error t-Statistic Prob. ˆ2 (f) 0.10 The EViews output for verifying the answers to Exercise 5. (g) The standard error for b2 can be read directly from the EViews output.4142142 SST SSR ( yi SST y )2 2 .760156 .4125 0. dependent var Akaike info criterion Schwarz criterion Hannan-Quinn criter.3750 –1.812500 0.E. 1. Principles of Econometrics.0250 1.8375 12.266580 0. SSE = “Sum squared resid” = 3.9875 –0.252900 3. (h) From the EViews output.414214 2.812500 0.400000 0. (n 1) s y2 8 2 16 SSE 16 3.000000 The correlation between x2 and x3 is zero.0095 0.837500 –8.1 is given in the following table. .639584 The correlation matrix for the three variables is X2 X3 Y X2 X3 Y 1.Chapter 5.760156 0. and R 2 To obtain SST note that s y2 1.8375.4125 0.6000 0.199935 0.1875 The estimate ˆ 2 is given by the square of “S. of regression Sum squared resid Log likelihood 0.812500 0. of regression”. To summarize.0044 0.00227 0.0929 0.00055 0. alcohol.00029 0.0881 0.00062 0.0459 0. but the effect of AGE and NK is only significant in the food and alcohol equations.2167 0. 4e 147 EXERCISE 5.0327 0.0000 0.1477 0.1587 0.0397 0.3179 0.0321 0.6878 0.0560 0.11 (a) Estimates. alcohol and other.0001 0.2816 0. . Exercise Solutions. At a 5% significance level.0118 0.00077 0.0058 0.2242 NK Estimate Std Error p-value 0.00040 0. standard errors and p-values for each of the coefficients in each of the estimated share equations are given in the following table. transport and other.0126 0.1157 An increase in total expenditure leads to decreases in the budget shares allocated to food and fuel and increases in the budget shares of the commodity groups clothing.0123 0.0000 0. ln(TOTEXP ) has a clear impact in all equations. NK has an impact on the food and alcohol shares. age has a significant effect on the shares of food and alcohol.00056 0.0510 0.0084 0.0088 0. Principles of Econometrics.0061 0.0112 0.0048 0. but its impact on the other budget shares is measured less precisely.0113 0.00071 0.7382 0. Dependent Variable Explanatory Variables Food Fuel Clothing Estimate Std Error p-value 0.1245 0.0082 0.0536 0.0094 0. but we can be less certain about the effect on the other groups.5658 0.0000 0. The coefficients of ln(TOTEXP ) are significantly different from zero for all commodity groups.0000 0. Households with an older household head devote a higher proportion of their budget to food.Chapter 5. fuel and transport and a lower proportion to clothing.0000 0. Having more children means a higher proportion spent on food and fuel and lower proportions spent on the other commodities.0149 0.0152 0.8798 0.0265 0. Significance tests for the coefficients of the number of children yield a similar result.0000 0.1921 0.0000 0.00220 0.3062 0.0111 0.0572 0.0000 0.0191 0.0001 Constant Alcohol Transport Other AGE Estimate Std Error p-value 0.0062 0.0148 0.00044 0.1006 ln(TOTEXP) Estimate Std Error p-value 0.0000 0.0512 0.0084 0.0139 0.00055 0.0370 0.00058 0. 012 2. Perhaps a finer classification is necessary to distinguish between basic and luxury clothing.496) 1. The tests suggest the rest are luxuries.0000 0.0000 0.Chapter 5. transportation and other might be luxuries. it is difficult to see clothing categorized as a luxury. While alcohol. Using a 5% level of significance. Exercise Solutions.648 .0000 1.11 (continued) (b) The t-values and p-values for testing H 0 : 2 0 against H1 : 2 0 are reported in the table below.884 0.95.0001 Reject H 0 Those commodities which are regarded as necessities ( b2 0 ) are food and fuel. .083 9. 4e 148 Exercise 5. t-value p-value decision WFOOD WFUEL WCLOTH WALC WTRANS 13.0056 Do not reject H 0 Do not reject H 0 Reject H 0 Reject H 0 Reject H 0 WOTHER 3. the critical value for each test is t(0.548 1.0000 0.569 8.266 4. Principles of Econometrics. 892 . implying a discount for larger sales. . (c) The proportion of variation in cocaine price explained by the variation in quantity.1162 QUAL 2. Principles of Econometrics.3546 . we do not reject H 0 .3546 .0600. Exercise Solutions. A possible reason for a decreasing price is the development of improved technology for producing cocaine. We expect 3 to be positive. At 0.0600 QUANT 0. We cannot conclude that a premium is paid for better quality cocaine.5717.0102) (0. All the signs turn out according to our expectations.5097. the price will go down by 0. respectively. quality and time is 0. the higher the price. The sign for 4 will depend on how demand and supply are changing over time.Chapter 5. a fixed demand and an increasing supply will lead to a fall in price. The calculated t-value is 1.675.5097 The estimated values for 2 . Since 5.12 (a) The expected sign for 2 is negative because. the price decreases by 2.5717) ( 1. Since for this case. (f) 2.6987) R2 0. It The average annual change in the cocaine price is given by the value of b4 has a negative sign suggesting that the price decreases over time. A fixed supply and increased demand would lead to a rise in price. (b) The estimated equation is: PRICE 90.5803) (0. 3 and 4 are 0. As time increases by 1 year.588) ( 5. such that suppliers can produce more at the same cost.05 we reject H 0 if the calculated t is greater than 1.52 the calculated t is less than the critical t value. the price per gram should decrease.675 . (d) For this hypothesis we test H 0 : 2 0 against H1 : 2 0 . Also. The calculated t-value is 0. as the number of grams in a given sale increases.3861) (t ) (10. we reject H 0 and conclude that sellers are willing to accept a lower price if they can make sales in larger quantities.0600 . 0.1162.95. the purer the cocaine. as the quality increases by 1 unit the price goes up by 0. (e) We want to test H 0 : 3 0 against H1 : 3 0 . the calculated t is not greater than the critical t. We reject H 0 if the calculated t is less than the critical t 0.892) (0. For example.8467 0.3546 TREND (se) (8.3546. with 4 implying supply has been increasing faster than demand. 4e 149 EXERCISE 5. They imply that as quantity (number of grams in one sale) increases by 1 unit.1162 and 2.2033) (1. The critical value for a 5% 1. we refer to these estimates as b1 .646 .0414 ( 1000) 1.8936 The corresponding p-value is P t(1077) 1.403) (140.160 AGE 2 (se) (10432) (6.389) (0.95. with corresponding parameters 1 . We conclude that the estimated equation is compatible with the hypothesis that an extra year of age decreases the price by $1000 or less. 2 . b2 . a one square foot increase in the size of the house increases the selling price by 90. (i) The marginal effect of SQFT on PRICE is given by PRICE SQFT 2 2 3 SQFT .8 AGE 30. Principles of Econometrics.9698 1. 3 .89) The estimate b2 90.Chapter 5.04 dollars. 5 .04 implies that holding SQFT constant.4031 (86.962 2. on average.97 implies that holding age constant.9698 t(0. given that tc b2 90.97 dollars. we fail to reject the null hypothesis. 4e 150 EXERCISE 5.25. Since the t-value is greater than the critical value and the p-value is greater than 0.69) (iii) The t-value for testing H 0 : t 2 PRICE SQFT b3 ( 1000) se(b3 ) 1000 against H1 : 3 3 1000 is 755.646 .1077) 1.05.1077) 0. we view the equation as a poor model for data values in the vicinity of SQFT 0 and AGE 0 . (b) The estimated regression is: PRICE 170150 55.023153SQFT 2 2797.1) (5. The estimate b1 could be interpreted as the average price of land if its value was meaningful. Since a negative price is unrealistic.04 AGE (6990) (2. 4 .784SQFT 0. b4 . The rejection region is t 1. on average.000964) (305. b3 .975. (ii) A point estimate for the price increase is A 95% interval estimate for b2 tc se(b2 ) .05. 755.7386 significance level is t(0.970SQFT 755.7386 140. b5 in the same order as they appear in the equation. an The estimate b3 increase in the age of the house by one year decreases the selling price by 755. Exercise Solutions.962 is 90.071) For the remainder of part (b).13 (a) The estimated regression is PRICE (se) (i) 41948 90.959 . The quadratic function has a minimum at an earlier age than is desirable.16033 80 2027.023153 662 25. and that.89 These values suggest that as the size of the house gets larger the price or cost for extra square feet gets larger.788 2 30.13(b)(i) (continued) The estimated marginal effect of SQFT on PRICE for the smallest house where SQFT 662 is PRICE SQFT 55.47 When a house is new.7842 2 0.38 The estimated marginal effect of AGE on PRICE for the newest house ( AGE 1) is PRICE AGE 2797. it is unrealistic for the oldest houses to increase in price as they continue to age. However. Aging has a smaller and smaller negative effect as the house gets older. The result for small houses is unrealistic. . extra years of age have the greatest negative effect on price. Exercise Solutions.788 2 30.7842 2 0.Chapter 5. extra space leads to a decline in price.7842 2 0. for small houses. as is suggested by the marginal effect for AGE 80 .788 2 30.13 The estimated marginal effect of SQFT on PRICE for a house with SQFT = 2300 is PRICE SQFT 55.16033 20 20 is 1591. Principles of Econometrics. it is possible that additional square feet leads to a higher price increase in larger houses than it does in smaller houses. (ii) The marginal effect of AGE on PRICE is given by PRICE AGE 4 2 5 AGE The estimated marginal effect of AGE on PRICE for the oldest house ( AGE 80) is PRICE AGE 2797. unless a house has some kind of heritage value. However.023153 7897 309.86 The estimated marginal effect of AGE on PRICE for a house when AGE PRICE AGE 2797.023153 2300 50. This result is as expected.72 The estimated marginal effect of SQFT on PRICE for the largest house where SQFT 7897 is PRICE SQFT 55. 4e 151 Exercise 5.16033 1 2737. 72. for a 20-year old house.13(b) (continued) (iii) A 95% interval for the marginal effect of SQFT on PRICE when SQFT using tc t(0.5472 (iv) The null and alternative hypotheses are H0 : 4 40 5 H1 : 1000 40 4 5 1000 The t-value for the test is t b4 40 b5 ( 1000) se b4 40 b5 591. with corresponding parameters 1 .022185SQFT 2 (0. b3 .93062 SQFT (410.005870334) 2.05.561) 139.03 AGE 26.729 SQFT (se) (12143) (6. 6 . We conclude that.5472 2300 .238 0.939) AGE (0. 4 . 2 . The standard error se b4 se b4 40 b5 40 b5 can be found using software or from var b4 402 var b5 2 40cov b4 .554 The corresponding p-value is P t(1075) 5% significance level is t(0. Since the t-value is less than the critical value and the p-value is less than 0. b3 40.375 139.719 1. The critical value for a 1. b5 . 5 .1075) = 1.238 4. 3 .82499 46002 9.000943) 442.Chapter 5. Exercise Solutions.1075) 4. an extra year of age decreases the price by more than $1000.296015 10 7 9200 ( 0.975.72) The standard error for me can be found using software or from se me var b2 46002 var b3 2 4600cov b2 . we reject the null hypothesis.646 .55 (c) The estimated regression is: PRICE 114597 30. 4e 152 Exercise 5.962.519 AGE 2 0.55.48 1600 25.898) 0.71554 80 ( 1434. and (45. b4 .962 2. b2 .11244) For the remainder of part (c). b5 93095.646 . we refer to these estimates as b1 . The rejection region is t 1. Principles of Econometrics. b6 in the same order as they appear in the equation.61) (4. .0000 . is: me tc se me 50.05. 97 When AGE 20 the estimated marginal effect of SQFT on PRICE for a house with SQFT = 2300 is PRICE SQFT 30. extra space appears to lead to a decline in price. For small houses. the estimated marginal effect of SQFT on PRICE for the largest house where SQFT 7897 is PRICE SQFT 30.6 2300 . Exercise Solutions.7 .Chapter 5.93062 2300 1660.7289 2 0.7289 2 0. 4e 153 Exercise 5.0336 2 26. It would be more realistic if the quadratic reached a minimum before the smallest house in the sample. the estimated marginal effect of AGE on PRICE for a house of PRICE AGE 442.93062 20 52.04 These values lead to similar conclusions to those obtained in part (b). the estimated marginal effect of AGE on PRICE for the oldest house ( AGE 80) is PRICE AGE When SQFT AGE 20 is 442.71 When AGE 20 . This result for small houses is unrealistic.519 20 0.0221846 7897 0.022185 662 0.93062 2300 1521.93062 20 19.0336 2 26. (ii) The marginal effect of AGE on PRICE is given by PRICE AGE 4 2 5 AGE 6 SQFT When SQFT 2300 . As the size of the house gets larger the price or cost for extra square feet gets larger. Principles of Econometrics.930621 20 301.022185 2300 0.13(c) (continued) (i) The marginal effect of SQFT on PRICE is given by PRICE SQFT 2 2 3 SQFT 6 AGE When AGE 20 . the estimated marginal effect of SQFT on PRICE for the smallest house where SQFT 662 is PRICE SQFT 30.7289 2 0.519 80 0. with a similar pattern in (b) and (c). and hence has the potential to be interpreted as the average price of the land.646 . .When a house is new.4 These results lead to similar conclusions to those reached in part (b). These differences carry over to the interval estimates for the marginal effect of SQFT and to the hypothesis tests on the marginal effect of AGE. is: me tc se me 52. 4e 154 Exercise 5. The marginal effect of AGE is constant at 755 in part (a) but varies from approximately 2600 to +1800 in parts (b) and (c). (iv) The null and alternative hypotheses are 1000 H1 : b4 40 b5 2300b6 ( 1000) se b4 40 b5 2300b6 521. This result is as expected.84. but some noticeable differences in magnitudes. Principles of Econometrics. The rejection region is t 1. we reject the null hypothesis. whereas in parts (b) and (c).4825 (47. but they are vastly different from those from the linear model in part (a).847 0. and using tc t(0.57. the positive marginal effect for AGE 80 is unrealistic.847 3.Chapter 5.708 1. the estimated marginal effect of AGE on PRICE for the newest house ( AGE 1) is PRICE AGE 442. it varies from approximately 20 to +300. The critical value for a 1. However.1074) 3. We do not expect the oldest houses to increase in price as they continue to age.1074) = 1.0336 2 26.13(c)(ii) (continued) When SQFT 2300 . for a 20-year old house with SQFT 2300 . The marginal effects are clearly not constant and so the linear function is inadequate. 2300 and (iii) A 95% interval for the marginal effect of SQFT on PRICE when SQFT AGE 20 . an extra year of age decreases the price by more than $1000.701 135. extra years of age have the greatest negative effect on price.05. Aging has a smaller and smaller negative effect as the house gets older.93062 2300 2529. We conclude that.05.962. relative to house prices. It is interesting that the intercept is positive in the quadratic equations. unless a house has some kind of heritage value. Since the t-value is less than the critical value and the p-value is less than 0.630 H0 : 4 40 5 2300 6 4 40 5 2300 6 1000 The t-value for the test is t The corresponding p-value is P t(1074) 5% significance level is t(0. In part (a) the marginal effect of SQFT is constant at 91. Exercise Solutions. Both quadratic functions are an improvement.975. (d) The results from the two quadratic specifications in parts (c) and (d) are similar. Both estimates seem large however.646 .519 1 0.58) The standard error for me was found using software.0001 .962 2. but they do give some counterintuitive results for old houses and small houses. 0387624 23 0. ln( PRICE ) AGE 0. holding SQFT constant.06% per year.58% per year.00002266) (b) The estimate ˆ 2 0.1196 0.Chapter 5.03876 suggests that. Principles of Econometrics.017555 20 0.00017336 5 0.038762 SQFT100 0. (d) The required derivatives are given by PRICE AGE 3 2 4 AGE PRICE 3 2 4 AGE exp PRICE SQFT100 2 PRICE 2 exp 1 2 1 2 SQFT100 SQFT100 3 AGE 3 4 AGE 4 AGE 2 AGE 2 where exp x is notation for the exponential function e x . the price of a 5-year old house will decrease at a rate of 1. holding SQFT constant.0274) (0. 2 4 AGE 0.11959 0.00017336 AGE 2 ln( PRICE ) 11.00017336 20 0.001356) (0.88% on average.14 (a) The estimated regression is: 0. an increase in the size of the house by one hundred square feet increases the price by 3. . holding age constant. When AGE 20 .000869) (0. (e) To estimate these marginal effects we first find PRICE 0 exp ˆ 1 ˆ 2 SQFT100 ˆ 3 AGE ˆ 4 AGE 2 exp 11. 4e 155 EXERCISE 5.017555 2 0.01062 This estimate implies that. Exercise Solutions.017555 2 0.01582 This estimate implies that. (c) The required derivative is given by ln( PRICE ) AGE When AGE 3 ln( PRICE ) AGE 5.017555 AGE (se) (0.00017336 202 124165 Then. the price of a 20-year old house will decrease at a rate of 1. 14(e) (continued) (f) PRICE AGE 0.386 4.386 0. we find the standard errors are se (g) PRICE AGE 72. .00017336 20 PRICE SQFT100 0.9 1.5052) The null and alternative hypotheses are H0 : 3 40 4 exp 1 23 2 H1 : 3 40 4 exp 1 23 2 20 20 3 3 400 400 4 4 1000 1000 The calculated value of the t-statistic is t 1318.13(c). Principles of Econometrics.646 .637 A 95% interval estimate for the marginal effect of SQFT100 is me t(0.671 121.13(c) shows that similar conclusions are reached.962 121.05.Chapter 5. we need the delta method introduced on pages 193-4 of the text. The rejection region is t 1.017555 2 0. and the estimated marginal effect is smaller. Since the t- value is less than the critical value and the p-value is less than 0. for a 20-year old house with SQFT 2300 . To compute their standard errors.1076) se me (h) PRICE SQFT100 se 4812.0387624 124165 4813 124165 1318. Using computer software. an extra year of age decreases the price by more than $1000.671 The corresponding p-value is P t(1076) significance level is t(0. The critical value for a 5% 1. we reject the null hypothesis. 4e 156 Exercise 5. Remark: A comparison of the results in parts (g) and (h) with those from the quadratic function with the interaction term in Exercise 5.646 . the marginal effect in (h) is smaller (in absolute value) and estimated more precisely than its counterpart in Exercise 5.05.7 We require the standard errors of PRICE AGE ˆ3 PRICE SQFT100 40 ˆ 4 exp ˆ 1 23 ˆ 2 20 ˆ 3 400 ˆ 4 ˆ 2 exp ˆ 1 23 ˆ 2 20 ˆ 3 400 ˆ 4 These expressions are nonlinear functions of the least squares estimators for the ’s. Exercise Solutions.1076) 4.7 ( 1000) 72.975.0000 .637 (4574. although the interval estimate in (g) is narrower. We conclude that. Similarly. Chapter 5.16 0.4290) The hypothesis test results on the significance of the coefficients are: H0 : 2 0 H1 : 2 0 p-value = 0. for re-election of the incumbent party.1656) (0.3456 not significant at 10% level One-tail tests were used because more growth is considered favorable. (b) (i) For INFLATION VOTE 0 (ii) 4 and GROWTH 52.46) (0.1565 0. b3 2 3cov b1 . the incumbent party will get the majority of the vote when 1 2 When INFLATION 1 (i) GROWTH 4 GROWTH When GROWTH 1 Given that t(0.0003 significant at 10% level H0 : 3 0 H1 : 3 0 p-value = 0.184003 6 0.172076 4 49.30) t 50 INFLATION 4 .99. b3 2.048748 8 0.172076 4 53.498011 24 0.127815 9 0. Exercise Solutions.1721 INFLATION (se) (1. the hypotheses are 3 2 4 3 50 H1 : 1 3 2 4 3 50 2.457 .54 For INFLATION VOTE 0 4 and GROWTH (c) 0 .15 (a) The estimated regression model is: VOTE 52. 4e 157 EXERCISE 5. we reject H 0 when b1 3b2 4b3 50 se b1 3b2 4b3 2. b2 2 3 4cov b2 . the predicted percentage vote is 52. this requirement becomes 2 H0 : 3 3 50 3 . the predicted percentage vote is 52.027433 16 0. var b1 3b2 4b3 var b1 32 var b2 42 var b3 2 4cov b1 .64342 (0) 0.47 (iii) For INFLATION VOTE 0 3 .64342 ( 3) 0.6434 GROWTH 0.011860 1.34252 .172076 4 51. Principles of Econometrics.1565 0.1565 0. the predicted percentage vote is 4 and GROWTH 3 .64342 3 0.457 Now. and more inflation is considered not favorable.40 Ignoring the error term. There is insufficient evidence to suggest that the incumbent part will get the majority of the vote when INFLATION 4 and GROWTH 0 . 4e 158 Exercise 5.399 Since 0.4682 50 1. you would not want to conclude that you would be reelected without strong evidence to support such a conclusion. the hypotheses are 3 2 We reject H 0 when t 4 3 50 H1 : b1 3b2 4b3 50 se b1 3b2 4b3 1 3 2 4 3 50 2.15(c)(i) (continued) The calculated t-value is t b1 3b2 4b3 50 se b1 3b2 4b3 49.34252 0. the hypotheses are 4 50 3 We reject H 0 when t H1 : b1 4b3 50 se b1 4b3 1 4 50 3 2.Chapter 5. we reject H 0 .04296 .408 1. (ii) When GROWTH H0 : 1 0 . Using computer software.3985 50 1.457 . Setting up re-election as the alternative hypothesis with a 1% significance level reflects this scenario.457 . . The calculated t-value is t b1 3b2 4b3 50 se b1 3b2 4b3 53. we do not reject H 0 .15188 .04296 Since 1. we find se b1 3b2 4b3 1.950 Since 2.538 50 1. Exercise Solutions. Using computer software. we do not reject H 0 . As a president seeking re-election. We conclude that the incumbent part will get the majority of the vote when INFLATION 4 and GROWTH 3 . There is no evidence to suggest that the incumbent part will get the majority of the vote when INFLATION 4 and GROWTH 3. The standard error can be calculated from a similar expression to that given in (c)(i). The calculated t-value is t b1 4b3 50 se b1 4b3 51. Principles of Econometrics.457 .457 .399 2.457 .950 2. (iii) When GROWTH H0 : 1 3 .408 2. we find se b1 4b3 1.15188 2. The standard error can be calculated from a similar expression to that given in (c)(i). Principles of Econometrics. (c) (d) The hypothesis test results on the significance of the coefficients are: (i) H0 : 2 0 H1 : 2 0 p-value = 0. holding PR2 and PR3 constant. The t-value is t b2 300 se b2 470.011 .013) R2 0.990 PR2 165.845 PR1 92.147 2.011 .113PR3 (se) (b) (9806) (79.443 (93.0422 significant at 5% level The hypotheses are H0 : Since t(0.16 (a) The estimated regression is: SAL1 22963 470. The estimate b3 92. we reject H 0 if t b2 300 se b2 2. while the signs of the other two coefficients are positive.0952 not significant at 5% level H0 : 4 0 H1 : 4 0 p-value = 0.845 suggests that. reflecting the fact that quantity demanded will fall as price rises. The estimate b4 165. 4e 159 EXERCISE 5.011 .975. a one cent increase in the price of brand 3 leads to an increase in the sales of brand 1 by 165 units.670) 470.113 suggests that. .147 price of brand 1 does not reduce its sales by 300 cans.Chapter 5. The estimates of 2 . a one cent increase in the price of brand 2 leads to an increase in the sales of brand 1 by 93 units.48) 2 300 H1 : 2 300 2.011 or t 2. Increases in their prices will increase the demand for brand 1. The sign of 2 is negative.990 suggests that. reflecting the fact that brands 2 and 3 are substitutes.578 2. 3 and 4 have the expected signs.0000 significant at 5% level H0 : 3 0 H1 : 3 0 p-value = 0. holding PR1 and PR 2 constant. holding PR1 and PR3 constant. we reject H 0 and conclude that a 1-cent increase in the Since 2.845 300 79.578) (70. a one cent The estimate b2 increase in the price of brand 1 leads to a decrease in the sales of brand 1 by 471 units. Exercise Solutions. 011 . There is no evidence to suggest that the increase in sales of brand 1 from a 1-cent increase in the price of brand 3 is different from 300 cans.763 8774. we reject H 0 if t b3 300 se b3 2. Thus we test H 0 : 3 4 against the alternative H1 : 3 4 and we reject H 0 if t 2.113 300 93.011 .011 or t 2.440 2.975.670 1.011 .011 or t 2.990 165.011 .Chapter 5. The t-value is t b3 300 se b3 92. b4 4901.957 2.113 123. There is no evidence to suggest that price changes in brands 2 and 3 have different effects on sales of brand 1.118 Since 2.48) 3 300 H1 : 3 300 2.990 300 70.48) 4 300 H1 : 4 300 2.011 .586 The standard error se(b3 b4 ) 123.011 . (iv) Price changes in brands 2 and 3 will have the same effect on sales of brand 1 if 3 4 .586 2.011 or t 2.011 .16(d) (continued) (ii) The hypotheses are H0 : Since t(0. The t-value is t b4 300 se b4 165. we reject H 0 and conclude that a 1-cent increase in the Since 2.118 can be calculated using computer software or from the coefficient covariance matrix as follows se b3 b4 var b3 var b4 2cov b3 . Exercise Solutions.011 1.118 0.957 price of brand 2 does not increase sales of brand 1 by 300 cans. 4e 160 Exercise 5. we fail to reject H 0 .011 0. Principles of Econometrics.013 2.048) 123. . The t-statistic is calculated as follows: t b3 b4 se(b3 b4 ) 92.975.440 Since 2. (iii) The hypotheses are H0 : Since t(0.011 .127 2 ( 741. we reject H 0 if t b4 300 se b4 2. we do not reject H 0 . And in part (iv) it could be true that 3 4 . (v) Suppose that prices are set at PR10 . SAL10 1 2 PR10 3 PR20 4 PR30 (Strickly speaking. we can use computer software or se b2 b3 b4 var b2 var b3 var b4 2cov b2 . we fail to reject H 0 . in part (iii).845 92. And in part (iv) we concluded that the effect of increases in prices for brands 2 and 3 could be equal. 4e 161 Exercise 5.) Now suppose that all prices go up by 1 cent and that average sales do not change. That is. That is. On the surface.724 2.127 2 1642. only that we have insufficient evidence to reject them.763 8774. PR20 and PR30 and that average sales are SAL10 .598 2 4. 0 The t-value is calculated as follows: t b2 b3 b4 se(b2 b3 b4 ) 470. Thus. but it also could be something else.048 123. b3 2cov b2 .16(d)(iv) (continued) In part (ii) we concluded that the effect of a price increase in brand 2 was not 300 cans.Chapter 5. b4 2cov b3 .416 .724 Since 2.990 165. SAL10 1 2 1 2 PR10 1 PR10 3 3 PR20 PR20 1 4 4 PR30 PR30 1 2 3 4 For SAL10 to be the same in these two equations we require we test H0 : 2 3 4 0 H1 : 2 3 4 2 3 4 0 . Thus.815 2 741. In part (iii) we concluded that the effect of a price increase in brand 3 could be 300 cans. but it could also be true that they are not equal.416 .113 123. For calculation of se(b2 b3 b4 ) 123.416 1. The results are compatible with the hypothesis that sales remain unchanged if all 3 prices go up by 1 cent. the effect of a price increase in brand 3 could be 300 cans.011 1. To appreciate that the hypothesis-test conclusions are indeed compatible. this may seem like a contradiction: the results from parts (ii) and (iii) suggest the effects are different and the part (iv) result suggests they are the same.011 . Principles of Econometrics. it must be appreciated that we never conclude null hypotheses are true. b4 6332. Exercise Solutions.635 4901. we are looking at no change in average sales so we can ignore the error term. .17 (a) The estimated linear regression from Exercise 5.145589 (7.48) 2. (b) The estimated log-linear regression is ln( SAL1) 10.53356 Using tc t(0.113PR3 (se) (9806) (79.009843) A point estimate for expected log-sales when PR1 90. the linear model is not a good one for forecasting.45595 0. by multiplying the coefficients by 100. The values PR1 90.845 PR1 92.53356 2. Principles of Econometrics.008362) (0. exp(7.8447 (90) 92. PR2 75 and PR3 75 is ln( SAL1) 10.007357) (0. (c) When SAL1 is the dependent variable the coefficients show the change in number of cans sold from a 1-cent change in price. The interval estimate contains a wide range of negative values which are clearly infeasible. 4e 162 EXERCISE 5.021472 75 7. 2506) Comparing this interval with the one obtained from the linear function.88 2.523 found using computer software.975.062176 PR1 0.24083.45595 0.021472 PR3 (se) (1. Sales cannot be negative.011 1385.013) (93. we have exp(7. PR2 75 and PR3 75 are unfavorable ones for sales of brand 1. but they are nevertheless within the ranges of the sample data.014174 75 0.443 SAL1 22963.975.43 470.011 .24083).7.88 SAL1 22963 470. the width of the interval from the log-linear model is much narrower.010635 . suggesting more accurate estimation of expected sales.990 PR2 165.9900 (75) 165.48) 2. but the lower bound for the interval from the log-linear model is positive and much larger than that from the linear model.82629) Converting this interval into one for sales using the exponential function. a 95% interval estimate is given by SAL1 tc se( SAL1) 54. PR2 75 and PR3 75 is Using tc t(0.82629) (1395. Exercise Solutions.062176 90 0. we find that the two upper bounds of the intervals are of similar magnitude.523 ( 2841.010635 0. When ln( SAL1) is the dependent variable.1129 (75) 54.16 is R2 0.014174 PR2 0.670) A point estimate for expected sales when PR1 90. Also. we get the the percentage change in number of cans sold from a 1-cent change in price. 2731) with se( SAL1) se b1 90b2 75b3 75b4 1385.03046) (0. Thus. a 95% interval estimate for expected log-sales is given by ln( SAL1) tc se ln( SAL1) 7.Chapter 5.578) (70. which is contrary to our expectations. Thus.1110) 0.433PRBARR 0. the probability of a prison sentence. Principles of Econometrics.320) R2 0. . but the other two. It is less clear why WCON should have a positive sign.000834) All five variables are expected to have negative effects on the crime rate.002187WCON (0. high crime rates may be more likely to exist in counties with greater numbers of police because more police are employed to counter high crime rates. Exercise Solutions. Of these three variables.Chapter 5.18 The estimated regression is LCRMRTE (se) 3. are significantly different from zero. POLPC and WCON. it appears that the variables.4700) (43. the number of police and the weekly wage in construction have positive signs. the coefficients of the other three variables. In the context of this example. We expect each of them to act as a deterrent to crime. PRBARR and PRBCONV are the most important for crime deterrence.3338PRBPRIS 200.6 POLPC 0. the coefficient of PRBARR is not significantly different from zero. 4e 163 EXERCISE 5. The positive sign for the coefficient of POLPC may have been caused by endogeneity. In the estimated equation the probability of an arrest and the probability of conviction have negative signs as expected. Perhaps construction companies have to pay higher wages to attract workers to counties with higher crime rates.0000.601 (0.6) (0. and have unexpected positive signs. a concept considered in Chapter 10.8077 PRBCONV (0.482 2. On the other hand.351) (0. and both coefficients are significantly less than zero with p-values of 0. 005776 implies that holding other variables constant. The coefficient estimates for variables EDUC 2 . The coefficient of the remaining variable EDUC is not significantly different from zero.03% on average.646 .1 se(b2 ) 1.005776 EXPER 0.008941HRSWK (se) (0.1581 (0.1 se(b2 ) 0.634.Chapter 5.00608 1. All coefficient estimates are significantly different from zero. an additional year of education increases wage by 9. The value of the t-statistic is t b2 0.001581) R2 0. Principles of Econometrics.996) se 100 b4 0.1005 0.1 0.00608) (0. The estimate b3 0. EXPER.565 is not sufficient evidence to show that the return to another year of education is less than 10%. That for EDUC EXPER is significant at a 10% level. Exercise Solutions.008941 implies that holding other variables constant. Since 1. (d) The estimates with quadratic terms and interaction term for EDUC and EXPER are given in the table on page 165. we do not reject H 0 .0903 implies that holding other variables constant.001275) (0.595 1.0555.646 0.1 The critical value for a 5% significance level is t(0. an extra year of related work experience increases wage on average by 0.646.154) We estimate with 90% confidence that the wage return to working an extra hour per week lies between 0.8941 1.58%. The estimate b4 0.2197 The estimate b2 0.63% and 1.09031EDUC 0. . We reject H 0 when 1.996) t b2 0.19 (a) The estimated regression is: ln(WAGE ) 1. 4e 164 EXERCISE 5.1 2 H1 : 2 0. (b) The null and alternative hypotheses are H0 : 0.09031 0.0000. with p-values of 0.15%. There The corresponding p-value is 0. (c) A 90% confidence interval for 100 100 b4 4 is given by t(0. working an extra hour per week increases wage by 0. EXPER 2 and HRSWK are significantly different from zero at a 5% level of significance.1095) (0.646 .05.1.95.89% on average. ln(WAGE ) EDUC b2 24b3 10b6 0.722 0.410 0.268 0.Chapter 5.0005054 –1. 4e 165 Exercise 5.0009238 0. . As education increases.080 0.0000 EDUC EXPER 6 –0.0006287 0.3404072 2.19(d) (continued) Estimates of wage equation with quadratic and interaction terms included Variable Coefficient Estimate Std.0066930 0.0066 2 0.0000888 –7. ln(WAGE ) EDUC 0.0009238 0. 3 From the above table. the marginal effect of education increases. the marginal effects on ln(WAGE ) are ln(WAGE ) EDUC 2 2 3 EDUC 4 2 5 EXPER b2 32b3 10b6 ln(WAGE ) EXPER (f) t-value p-value 6 EXPER 6 EDUC For Jill.0009238 0. Now the inequality 32 3 24 3 holds if 0 and does not hold if 3 0 . Thus.0527446 0.1810 3 0.339 0.0325 2 0.115 For Wendy. Error 1 0. Exercise Solutions. There are “increasing returns” to education.141 0.0023649 0.0000 C EDUC EDUC 2 EXPER EXPER (e) 2 Defining the coefficients as they appear in the above table.0490281 0.9266081 0.0163 . (g) Jill’s marginal effect of education will be greater than that of Wendy if 2 32 3 10 6 2 24 3 10 6 which will be true if and only if 32 3 24 3 .0366258 1. Principles of Econometrics.049028 32 0. we reject H 0 and conclude that Jill’s marginal effect of education is greater than that of Wendy.0011048 2. Thus a suitable test is H 0 : 3 0 against H1 : 3 0 .0325 4 0.097 We estimate that Jill has a greater marginal effect of education than Wendy. the p-value for this test is 0.0015681 4.0097493 5.049028 24 0.0000 5 –0.0023649 10 0.828 0.0023649 10 0.0679 HRSWK 7 0. 993)se EXPER* 30.2 years. Exercise Solutions. 4e 166 Exercise 5. (i) For someone with 16 years of education.962 1.22.0006287 30.0128 For Dave. ln(WAGE ) EXPER b4 40b5 16b6 0. the marginal effect of experience will be negative when EXPER 16 4 2 6 EXPER 5 A point estimate for EXPER * is EXPER b4 16b6 2b5 0.Chapter 5.052745 40 0. .0009238 0.0002 We estimate that Chris has a greater marginal effect of exerience than Dave. 0 .19 The delta method is required to get the standard error se EXPER se b4 16b6 2b5 1.052745 60 0.0009238 2 0.0006287 16 0. the marginal effect of experience decreases.0009238 0.5163 (27.0006287 16 0.191 1.19 (continued) (h) For Chris.0527446 16 0. ln(WAGE ) EXPER b4 60b5 16b6 0.17) We estimate with 95% confidence that the number of years of experience after which the marginal return to experience becomes negative is between 27. There are “decreasing returns” to experience.975. the marginal effect of experience is ln(WAGE ) EXPER Assuming 5 4 2 5 EXPER 16 6 . Principles of Econometrics.5163 A 95% interval estimate is given by EXPER* t(0.2 and 33. 33. As experience increases. 6 3 2.1512 3.5 4 1 . In this case the t value is .75 is not optimal.76796) 1 2.8b4 1 se b3 3. This outcome illustrates why we never accept null hypotheses as the truth. Thus we fail to reject H 0 and conclude that ADVERT0 2.1381.05435 1. note that for part (c) we could also have set up the hypothesis H 0 : ADVERT0 which is identical algebraically to H 0 : 1 2 4 3 4.9 hypotheses are H 0 : 3 3.3 hypotheses are H 0 : 3 4.75 4 1 . Note that we have found that both 1.6 ( 2.8 4 1 and H 1 : 3 3.6 4 1 and H 1 : 3 4.149 0. Thus the null and alternative hypotheses are H 0 : 3 3. To appreciate how the difference can arise.8b4 4 4 1 .968 0.8 t b3 3.3365.5 4 1 and H 1 : 3 3. Thus the null and alternative 1 . The best we can do is to say there is insufficient evidence to conclude a null hypothesis is not true.6b4 1 se b3 4.Chapter 5.3 could be optimal values for advertising expenditure. The t-value is 12.9 will be optimal if 3 2 1.1512 4.3 will be optimal if 3 2 2. Principles of Econometrics.3 4 1 .75 will be optimal if 3 2 1. A null hypothesis that used any value for ADVERT0 in between these two values would also not be rejected.20 (a) ADVERT0 1.3 lies outside the 95% interval estimate for ADVERT0 found on page 195 of the text. (b) ADVERT0 1.76796) 1 1.5b4 12.76796) 1 0. The t-value is t b3 3.65419 and the corresponding p-value is 0.5 ( 2. Thus the null and alternative 1 .8 ( 2.5b4 1 se b3 3.6b4 4 4 1 . 4e 167 EXERCISE 5.6 t b3 4.0350.500 and the corresponding p-value is 0.3 could be optimal. Thus we fail to reject H 0 and conclude that ADVERT0 1. (c) ADVERT0 2.68085 and the corresponding p-value is 0. Thus we reject H 0 and conclude that ADVERT0 1.9 and 2. The t-value is 12. You might be surprised by the fact that 2. Exercise Solutions.1512 3.9 could be optimal. 3 2.6b4 involves getting the standard error for a linear function of the b’s.20 (continued) t 1 b3 2b4 1 b3 se 2b4 2.219 The p-value is 0. whereas se b3 4. Exercise Solutions. and H 0 is rejected.Chapter 5.1512 2 ( 2. The different outcome arises because the delta method used to find se 1 b3 2b4 is a large sample approximation needed for nonlinear functions of the b’s. Principles of Econometrics.12872 2.3 1 12.0297. 4e 168 Exercise 5.76796) 0. something we can do exactly without a large sample approximation. . 652 . This t-value is t 1.652 . (d) The hypotheses are H 0 : and t(0.970 0. The data are consistent with the hypothesis that each train delays Bill by 3 minutes.975.227)se(b2 ) 0. The 95% confidence intervals for the coefficients are: 1 : b1 t(0.975. we do not reject H 0 .78 1. 3 : Each red light increases traveling time by 1.3038 0.44. We have obtained precise estimates of each of the coefficients. his traveling time increases by 3.3038) Interpretations of each of the coefficients are: 1 : The estimated time it takes Bill to get to work when he leaves Carnegie at 6:30AM and encounters no red lights and no trains is 19.9166 0.227) 4 3 and H 1 : 4 3 .3353 2 0.652 .975.05.7 minutes for every 10 minutes that his departure time is later than 6:30AM (assuming the number of red lights and trains are constant).227)se(b1 ) 19.652 1.339.9166 1.652 .39) 2 : b2 3 : b3 t(0. 4 (b) : Each train increases traveling time by 2.227) 1.227) 1.400) t(0.227) se(b3 ) 1.16.1390) (0.3038 (2. 3.34 minutes.652 0.807 Since 1.652 .21 (a) The estimated equation is TIME 19.970 0. (c) The hypotheses are H 0 : 3 2 and H1 : 3 2 . This t-value is t 2. 0.227)se(b4 ) 2. We reject H 0 when the calculated t-value is such that t 1.61) 4 : b4 t(0.807 1.970 0.05.1.970 1.7548TRAINS (se) (1. We reject H 0 when the calculated t-value is less than 1.06.3353REDS 2.3353 1.95.36923 1. The critical values are t(0. 22.2548) (0.2548 (17. We conclude that the delay from each red light is less than 2 minutes. .1390 4. 4e 169 EXERCISE 5.652 or t 1.75 minutes.1390 (1.Chapter 5.01553) (0.7548 1.78 Since 4. these intervals are relatively narrow ones.975.36923DEPART 1. Principles of Econometrics.92 minutes.35) In the context of driving time.7548 3 0. The critical value is t(0. we reject H 0 . Exercise Solutions.01553 (0.652 . 2 : If Bill leaves later than 6:30AM. 33333 0.36923 0.31 1. where the calculated t-value is t 0. then the minimum time it takes Bill to travel to work is 1 .05.066 Since 0. The data are consistent with the hypothesis that delaying departure time by 30 minutes increases travel time by at least 10 minutes.652 .21 (continued) (e) Delaying the departure time by 30 minutes. where the calculated t-value is t 19. Exercise Solutions. Thus. We reject H 0 if t t(0.01553 2. otherwise increasing one of the other variables will lower the travel time and the hypothesis would need to be framed in terms of more coefficients than 1 .31 Since 2.066 1. .95. Thus. we do not reject H 0 . The data support the null hypothesis that the minimum travel time is less than or equal to 20 minutes. increases travel time by 30 2 .9166 20 1. We reject H 0 if t t(0.652 . It was necessary to assume that 2 .652 . 4e 170 Exercise 5. and the alternative is H1 : 2 1 3 . (f) If we assume that 2 .227) 1.2548 0. or H 0 : 2 1 3 .227) 1. 3 and 4 are all positive or zero. the hypotheses are H 0 : 1 20 and H1 : 1 20 .652 . the null hypothesis is H 0 : 30 2 10 . Principles of Econometrics. 3 and 4 are all non-negative. we do not reject H 0 .Chapter 5. 4e 171 EXERCISE 5.227) t b4 3b3 se(b4 3b3 ) 4 3 3 1. Exercise Solutions. where the t-value is calculated as 3 2.9174 .05.3353 2.7548 5 0.36923DEPART 1.970 . Thus.95.7548TRAINS (se) (a) (1.227) 1. the hypotheses are H0 : 3 4 5 5 and 2 The rejection region for the t-test is t t 3b4 5b2 5 se(3b4 5b2 ) H0 : 3 4 5 5 2 t(0. The hypotheses are H0 : 3 4 H1 : and 3 The rejection region for the t-test is t t(0.5205 The null hypothesis is rejected because 2.7548 3 1.227) region is t (0. The delay from a train is not equal to three times the delay from a red light.22 The estimated equation is TIME 19.2548) (0.1390) and the delay from a red light is H1 : 3 and 4 The critical values for the t-test are t(0.3353REDS 2.01553) The delay from a train is alternative hypotheses are H0 : 3 3 4 (0.652 .404 1.652 . and the calculated t-value is 2.5205 2. (b) This test is similar to that in part (a).404 where the standard error is computed from se(3b3 b4 ) 9 var(b3 ) var(b4 ) 2 3 cov(b2 .404 1. The rejection 1.546 0.00081 0.975.970 or t 1. we reject H 0 . (c) The delay from 3 trains is 3 4 .3353 0. The p-value is 0. the null and 4 1. b3 ) 9 0.970 and t(0.3038) 3 3 . The calculated value of the t-test statistic is t 3b3 b4 se(3b3 b4 ) 3 1.36923 5 1.227) 1.Chapter 5.975.970 .017.092298 6 0.5205 2.019311 0.970 .9166 0. Thus. The extra time gained by leaving 5 minutes earlier is 5 5 2 . Principles of Econometrics.652 .404 Since 2.7548 0. The delay from a train is less than three times the delay from a red light. but it is a one-tail test rather than a two-tail test. Chapter 5, Exercise Solutions, Principles of Econometrics, 4e 172 Exercise 5.22(c) (continued) and the standard error is computed from se(3b4 5b2 ) 9 var(b4 ) 25 var(b2 ) 30 cov(b2 , b4 ) 9 0.092298 25 0.000241 30 0.000165 0.9174 Since 1.546 1.652 , we do not reject H 0 at a 5% significance level. Alternatively, we do not reject H 0 because the p-value = 0.0617, which is greater than 0.05. There is insufficient evidence to conclude that leaving 5 minutes earlier is not enough time. (d) The expected time taken when the departure time is 7:15AM, and no red lights or trains are encountered, is 1 45 2 . Thus, the null and alternative hypotheses are H0 : 1 45 2 45 and The rejection region for the t-test is t t b1 45b2 45 se(b1 45b2 ) H1 : 1 45 2 45 t(0.95,227) 1.652 , where the t-value is calculated as 19.9166 45 0.36923 45 1.1377 7.44 and the standard error is computed from se(b1 45b2 ) var(b1 ) 452 var(b2 ) 90 cov(b1 , b2 ) 1.574617 2025 0.00024121 90 0.00854061 1.1377 Since 7.44 1.652 , we do not reject H 0 at a 5% significance level. Alternatively, we do not reject H 0 because the p-value = 1.000, which is greater than 0.05. There is insufficient evidence to conclude that Bill will not get to the University before 8:00AM. Chapter 5, Exercise Solutions, Principles of Econometrics, 4e 173 EXERCISE 5.23 The estimated model is 39.594 47.024 AGE 20.222 AGE 2 SCORE (se) (28.153) (27.810) (8.901) 2.749 AGE 3 (0.925) The within sample predictions, with age expressed in terms of years (not units of 10 years) are graphed in the following figure. They are also given in a table on page 176. 15 10 5 0 SCORE SCOREHAT -5 -10 -15 20 24 28 32 36 40 44 AGE_UNITS Figure xr5.23 Fitted line and observations (a) To test the hypothesis that a quadratic function is adequate we test H 0 : 4 0. The t-value is 2.972, with corresponding p-value 0.0035. We therefore reject H 0 and conclude that the quadratic function is not adequate. For suitable values of 2 , 3 and 4 , the cubic function can decrease at an increasing rate, then go past a point of inflection after which it decreases at a decreasing rate, and then it can reach a minimum and increase. These are characteristics worth considering for a golfer. That is, the golfer improves at an increasing rate, then at a decreasing rate, and then declines in ability. These characteristics are displayed in Figure xr5.23. (b) (i) Using the predictions in the table on page 176, we find the predicted score is lowest ( 6.29) at the age of 30. Thus, we predict that Lion was at the peak of his career at age 30. Mathematically, we can find the value for AGE at which E ( SCORE ) is a minimum by considering the derivative dE ( SCORE ) dAGE 2 3 AGE 3 4 AGE 2 2 Setting this derivative equal to zero and solving for age yields AGE 2 3 4 6 2 3 4 12 2 4 Chapter 5, Exercise Solutions, Principles of Econometrics, 4e 174 Exercise 5.23(b)(i) (continued) Replacing 2 , 3 AGE1* AGE *2 , 4 by their estimates b2 , b3 , b4 gives the two solutions 2 ( 20.2222) 4 ( 20.2222)2 12 47.02386 2.74934 6 2.74934 3.008 2 ( 20.2222) 4 ( 20.2222)2 12 47.02386 2.74934 6 2.74934 1.895 The second derivative d 2 E ( SCORE ) dAGE 2 is positive when AGE 2b3 AGE 6b4 AGE AGE1* and negative when AGE expected score E ( SCORE ) is a minimum when AGE to 30.08 years. (ii) AGE *2 . Thus, the 3.008 , which is equivalent Lion’s game is improving at an increasing rate between the ages of 20 and 25, where the differences between the predictions are increasing. (iii) Lion’s game is improving at a decreasing rate between the ages of 25 and 30, where the differences between the predictions are declining. We can consider (ii) and (iii) mathematically in the following way. When Lion’s game is improving the first derivative will be negative. It can be verified that the estimated first derivative will be negative for values of AGE between 2 and 3. If Lion’s game is improving at an increasing rate, the second derivative will also be negative; it will be positive when Lion’s game is improving at a decreasing rate. Thus, to find the age at which Lion’s improvement changes from an increasing rate to a decreasing rate we find that AGE for which the second derivative is zero, namely AGE *3 2b3 6b4 2 ( 20.2222) 6 2.74934 2.452 which is equivalent to 24.52 years. (iv) At the age of 20, Lion’s predicted score is –4.4403. His predicted score then declines and rises again, reaching –4.1145 at age 36. Thus, our estimates suggest that, when he reaches the age of 36, Lion will play worse than he did at age 20. (v) (c) At the age of 40 Lion’s predicted score becomes positive implying that he can no longer score less than par. At the age of 70, the predicted score (relative to par) for Lion Forrest is 241.71. To break 100 it would need to be less than 28 ( 100 72) . Thus, he will not be able to break 100 when he is 70. Chapter 5, Exercise Solutions, Principles of Econometrics, 4e Exercise 5.23 (continued) Predicted scores at different ages age predicted scores 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 4.4403 4.5621 4.7420 4.9633 5.2097 5.4646 5.7116 5.9341 6.1157 6.2398 6.2900 6.2497 6.1025 5.8319 5.4213 4.8544 4.1145 3.1852 2.0500 0.6923 0.9042 2.7561 4.8799 7.2921 10.0092 175 Chapter 5, Exercise Solutions, Principles of Econometrics, 4e 176 EXERCISE 5.24 (a) The coefficient estimates, standard errors, t-values and p-values are in the following table. Dependent Variable: ln(PROD) Coeff Std. Error t-value p-value C -1.5468 0.2557 -6.0503 0.0000 ln(AREA) 0.3617 0.0640 5.6550 0.0000 ln(LABOR) 0.4328 0.0669 6.4718 0.0000 ln(FERT) 0.2095 0.0383 5.4750 0.0000 All estimates have elasticity interpretations. For example, a 1% increase in labor will lead to a 0.4328% increase in rice output. A 1% increase in fertilizer will lead to a 0.2095% increase in rice output. All p-values are less than 0.0001 implying all estimates are significantly different from zero at conventional significance levels. (b) The null and alternative hypotheses are H 0 : 2 0.5 and H1 : 2 0.5 . The 1% critical 2.59 . Thus, the rejection region is t 2.59 or values are t(0.995,348) 2.59 and t(0.005,348) t 2.59 . The calculated value of the test statistic is t 0.3617 0.5 0.064 2.16 Since 2.59 2.16 2.59 , we do not reject H 0 . The data are compatible with the hypothesis that the elasticity of production with respect to land is 0.5. (c) A 95% interval estimate of the elasticity of production with respect to fertilizer is given by b4 t(0.975,348) se(b4 ) 0.2095 1.967 0.03826 (0.134, 0.285) This relatively narrow interval implies the fertilizer elasticity has been precisely measured. (d) This hypothesis test is a test of H 0 : 3 0.3 against H1 : 3 0.3 . The rejection region is t t(0.95,348) 1.649 . The calculated value of the test statistic is t 0.433 0.3 1.99 0.067 We reject H 0 because 1.99 1.649 . There is evidence to conclude that the elasticity of production with respect to labor is greater than 0.3. Reversing the hypotheses and testing H 0 : 3 0.3 against H1 : 3 0.3 , leads to a rejection region of t 1.649 . The calculated t-value is t 1.99 . The null hypothesis is not rejected because 1.99 1.649 . Chapter 5, Exercise Solutions, Principles of Econometrics, 4e 177 EXERCISE 5.25 (a) Taking logarithms yields the equation ln(Y ) 1 2 ln( K ) 3 ln( L) 4 ln( E ) 5 ln( M ) e where 1 ln( ) . This form of the production function is linear in the coefficients 3, 4 and 5 , and hence is suitable for least squares estimation. (b) , 2 , Coefficient estimates and their standard errors are given in the following table. 2 3 4 5 (c) 1 Estimated coefficient Standard error 0.05607 0.22631 0.04358 0.66962 0.25927 0.44269 0.38989 0.36106 The estimated coefficients show the proportional change in output that results from proportional changes in K, L, E and M. All these estimated coefficients have positive signs, and lie between zero and one, as is required for profit maximization to be realistic. Furthermore, they sum to approximately one, indicating that the production function has constant returns to scale. However, from a statistical point of view, all the estimated coefficients are not significantly different from zero; the large standard errors suggest the estimates are not reliable. CHAPTER 6 Exercise Solutions 178 Chapter 6, Exercise Solutions, Principles of Econometrics, 4e 179 EXERCISE 6.1 (a) To compute R 2 , we need SSE and SST. We are given SSE. We can find SST from the equation ( yi y ) 2 N 1 ˆy SST N 1 13.45222 Solving this equation for SST yields ˆ 2y ( N 1) (13.45222) 2 39 SST 7057.5267 Thus, R2 (b) 1 SSE SST The F-statistic for testing H 0 : F At 1 2 979.830 7057.5267 3 0.8612 0 is defined as ( SST SSE ) ( K 1) SSE ( N K ) 0.05 , the critical value is F(0.95, 2, 37) (7057.5267 979.830) / 2 979.830 / (40 3) 114.75 3.25 . Since the calculated F is greater than the critical F, we reject H 0 . There is evidence from the data to suggest that 0. 3 2 0 and/or Chapter 6, Exercise Solutions, Principles of Econometrics, 4e 180 EXERCISE 6.2 The model from Exercise 6.1 is y 1 e . The SSE from estimating this 2x 3z model is 979.830. The model after augmenting with the squares and the cubes of 2 3 2 ˆ 3 e . The SSE from estimating predictions yˆ and yˆ is y 1 2 x 3 z 1 yˆ 2y this model is 696.5375. To use the RESET, we set the null hypothesis H 0 : 1 0. 2 The F-value for testing this hypothesis is F ( SSER SSEU ) J SSEU ( N K ) (979.830 696.5375) 2 696.5373 (40 5) The critical value for significance level 7.1175 0.05 is F(0.95,2,35) 3.267 . Since the calculated F is greater than the critical F we reject H 0 and conclude that the model is misspecified. Chapter 6, Exercise Solutions, Principles of Econometrics, 4e 181 EXERCISE 6.3 (a) Let the total variation, unexplained variation and explained variation be denoted by SST, SSE and SSR, respectively. Then, we have eˆi2 SSE N K ˆ2 (20 3) 2.5193 42.8281 Also, R2 1 SSE SST 0.9466 and hence the total variation is SST SSE 1 R2 42.8281 1 0.9466 802.0243 and the explained variation is SSR (b) SST SSE A 95% confidence interval for 2 802.0243 42.8281 759.1962 is b2 t(0.975,17)se(b2 ) 0.69914 2.110 A 95% confidence interval for 3 To test H0: 1 against the alternative H1: 2 t b2 2 se b2 0.69914 1 0.048526 2 0.037120 < 1, we calculate t(0.05,17) 2 3 F 0 against the alternative H 1 : explained variation K 1 unexplained variation N K 2 2 = 3 2 0 and/or 759.1962 / 2 42.8281 / 17 The critical value for a 5% level of significance is F(0.95,2,17) reject H0 and conclude that the hypothesis 1.3658 1.740 . Since we fail to reject H 0 . There is insufficient evidence to conclude To test H 0 : (1.3704, 2.1834) 1.3658 At a 5% significance level, we reject H0 if t (d) (0.2343, 1.1639) is b2 t(0.975,17)se(b3 ) 1.7769 2.110 (c) 0.048526 1.740 , 1. 3 0 , we calculate 151 3.59 . Since 151 3.59 , we = 0 is not compatible with the data. Chapter 6, Exercise Solutions, Principles of Econometrics, 4e 182 Exercise 6.3 (continued) (e) The t-statistic for testing H 0 : 2 2b2 t 2 3 against the alternative H1 : 2 2 3 is b3 se 2b2 b3 For a 5% significance level we reject H 0 if t t(0.025,17) 2.11 or t t(0.975,17) 2.11 . The standard error is given by se 2b2 b3 2 2 var(b2 ) var(b3 ) 2 2 cov(b2 , b3 ) 4 0.048526 0.03712 2 2 0.031223 0.59675 The numerator of the t-statistic is 2b2 b3 2 0.69914 1.7769 0.37862 leading to a t-value of t Since 2 2 2.11 3 . 0.37862 0.59675 0.634 0.634 2.11 , we do not reject H 0 . There is no evidence to suggest that Chapter 6, Exercise Solutions, Principles of Econometrics, 4e 183 EXERCISE 6.4 (a) The value of the t statistic for the significance tests is calculated from: t bk se(bk ) tc 2 . The t-values for each of the coefficients are given in the following table. Those which are significantly different from zero at an approximate 5% level are marked *. When EDUC and EDUC 2 both appear in an equation, their coefficients are not significantly different from zero, with the exception of eqn (B), where EDUC 2 is significant. In addition, the interaction term between EXPER and EDUC is not significant in eqn (A). We reject the null hypothesis H 0 : k 0 if t t-valuesa Variable Eqn (A) Eqn (B) Eqn (C) Eqn (D) Eqn (E) 1 3.97* 6.59* 8.38* 23.82* 9.42* 2 1.26 0.84 1.04 3 1.89 2.12* 1.73 4 4.58* 6.28* 5.17* 6.11* EXPER 5 –5.38* –5.31* –4.90* –5.13* EXPER*EDUC 6 –1.06 HRSWK 7 10.11* 8.71* C EDUC 2 EDUC EXPER 2 8.34* 8.43* 9.87* 15.90* a Note: These t-values were obtained from the computer output. Some of them do not agree exactly with the t ratios obtained using the coefficients and standard errors in Table 6.4. Rounding error discrepancies arise because of rounding in the reporting of values in Table 6.4. (b) Using the labeling of coefficients in the above table, we see that the restriction imposed on eqn (A) that gives eqn (B) is 6 0 . The F-test value for testing H 0 : 6 0 against H1 : 6 0 can be calculated from restricted and unrestricted sums of squared errors as follows: ( SSER SSEU ) J (222.6674 222.4166) 1 F 1.120 SSEU ( N K ) 222.4166 993 The corresponding p-value is 0.290. The critical value at the 5% significance level is F(0.95,1,993) 3.851 . Since the F-value is less than the critical value (or the p-value is greater than 0.05), we fail to reject the null hypothesis and conclude that the interaction term, EDUC EXPER is not significant in determining the wage. The t-value for testing H 0 : 6 0 against H1 : 6 0 is –1.058. At the 5% level, its absolute value is less than the critical value, t(0.975,993) 1.962 . Thus, the t-test gives the same result. The two tests are equivalent because 1.120 1.058 and 3.851 1.962 . 994) 3.993) 2.05). All three coefficients relate to variables that include EXPER.Chapter 6. The critical value at a 5% significance level is F(0. Thus. The test outcome suggests that education is indeed a relevant variable.1 222.3. 4 or 5 or 6 is By performing this test.0000. Since the F-value is greater than the critical value (or the p-value is less than 0.6674) 2 129. Both coefficients relate to variables that include EDUC. Since the F-value is greater than the critical value (or the p-value is less than 0.05). By performing this test. 5 0 and H1 : At least one of 4 6 or 4 0. we reject the null hypothesis and conclude at least one of 2 or 3 is nonzero. The F-value is calculated from: F ( SSER SSEU ) J SSEU ( N K ) (233.005 .5061 222. . 5 0 and 0 .8317 222.614 . (d) The restrictions imposed on eqn (B) that give eqn (D) are H0 : 2 0.95. we are asking whether experience is relevant for determining the wage level.4166 993 The corresponding p-value is 0. Thus. we test 3 0 H1 : At least one of 2 or 3 is nonzero .4166) 3 16.6674 994 The corresponding p-value is 0. we reject the null hypothesis and conclude at least one of nonzero. 3 2 0 and 0 . we are asking whether education is relevant for determining the wage level. The F-value is calculated from: F ( SSER SSEU ) J SSEU ( N K ) (280.4 (continued) (c) The restrictions imposed on eqn (A) that give eqn (C) are we test H0 : 4 0.0000.2.988 222. The test outcome suggests that experience is indeed a relevant variable. The critical value at a 5% significance level is F(0.95. Principles of Econometrics. 6 0 5 or 6 is nonzero . 4e 184 Exercise 6. Exercise Solutions. we do not reject the null hypothesis. 1. Exercise Solutions. Principles of Econometrics. 6 0 is By performing this test. (g) The AIC for eqn (D): AIC D ln SSE N 2K N ln 280. All its estimated coefficients are significantly different from zero. It includes both EXPER and EXPER2 which were shown to be jointly significant.4 (continued) (e) The restrictions imposed on eqn (A) that give eqn (E) are H0 : 3 0.263 The SC for eqn (A): SC A ln SSE N K ln( N ) N ln 222. 3 0.6716 222. The test outcome suggests that including it as a linear term is adequate.455 .Chapter 6. The critical value at a 5% significance level is F(0. we test 0 H1 : At least one of 3 or 6 is nonzero .05). The F-value is calculated from: F ( SSER SSEU ) J SSEU ( N K ) (223. Eqn (E) is favored by the SC criterion. 6 3 0 and 6 0 . 4e 185 Exercise 6.0612.95. Thus.993) 3. were not significant.005 . Since the F-value is less than the critical value (or the p-value is greater than 0. jointly.2.4166 1000 7 ln(1000) 1000 Eqn (B) is favored by the AIC criterion.802 The corresponding p-value is 0.4166) 2 222.4166 993 2. we are asking whether it is sufficient to include education as a linear term or whether we should also include it as a quadratic and/or interaction term.5061 1000 8 1000 1. The assumption compatible with the data. and it excludes the interaction term and EDUC 2 which. (f) Eqn (E) is the preferred model. 005 .6674 994 70. The null and alternative hypotheses are: H0 : 2 4 and H1 : 2 4 or 3 4 and 5 5 or both The restricted model assuming the null hypothesis is true is ln(WAGE ) (c) 3 2 1 4 ( EDUC EXPER) 5 ( EDUC 2 EXPER2 ) 6 HRSWK e The F-value is calculated from: F ( SSER SSEU ) J SSEU ( N K ) (254.994) 3.5 (a) (b) Education and experience will have the same effects on ln(WAGE ) if 3 5 .05).6674) 2 222. Since the F-value is greater than the critical value (or the p-value is less than 0. the critical value at a 5% significance level is F(0. Principles of Econometrics.32 The corresponding p-value is 0. Also. Exercise Solutions. 4e 186 EXERCISE 6.1726 222.0000.2. we reject the null hypothesis and conclude that education and experience have different effects on ln(WAGE ) . .Chapter 6.95. for example. . yˆ b1 b2 x b3 z is perfectly collinear with x and z . the model y 1 2 x 3 z e If we augment the model with the predictions yˆ the model becomes y 1 2 x 3 z yˆ e However. Exercise Solutions. This perfect collinearity means that least-squares estimation of the augmented model will fail.6 Consider. Principles of Econometrics. 4e 187 EXERCISE 6.Chapter 6. Thus.7 (a) Least squares estimation of y e gives b3 0.95.1072 when w is omitted from the equation. se(b3 ) 0.4979 0.95.1174 1 2x 3w and t 0. Thus. Exercise Solutions.1.4979 .Chapter 6. This result is consistent with what we observe. . respectively. the RESET based on the equation y e gives F-values of 17. 4e 188 EXERCISE 6. the model omitting w is inadequate.31) 3.1174 4.30 .15 and F(0.72 which are much higher 1 2x than the 5% critical values of F(0. Principles of Econometrics. (b) Let b2 be the least squares estimator for variable bias is given by E (b2* ) 2 in the model that omits w . The omitted- cov(x. (c) The high correlation between x and w suggests the existence of collinearity. The estimated coefficient for 2 changes from 0. cov( x .98 and 8. This result suggests that b3 is significantly different from zero and therefore w should be included in the model.9985 to 4. w) 2 3 var( x) Now. Additionally. w ) 0 because rxw 0 .2. The observed outcomes that are likely to be a consequence of the collinearity are the sensitivity of the estimates to omitting w (the large omitted variable bias) and the insignificance of b2 when both variables are included in the equation.24 .32) 4. the omitted variable bias will be positive. 8 ADVERT 4 4 into the equation ADVERT 2 e yields 2 1.9 3 3.61 3.8 4 ADVERT 4 4 2 PRICE ADVERT 2 e Rearranging this equation into a form suitable for estimation yields SALES ADVERT 78. Principles of Econometrics.1 2 PRICE 6 4 3.Chapter 6.61 4 Substituting the first restriction into the second yields 1 80 6 2 1. With this in mind.9(1 3.8 ADVERT ADVERT 2 e .8 There are a number of ways in which the restrictions can be substituted into the model. with each one resulting in a different restricted model.8 4 1 80 6 2 1.9(1 3. Exercise Solutions.8 4 ) 3.61 1 3. We have chosen to substitute out 1 and 3 .61 Substituting this restriction and the first one SALES 1 SALES 80 6 2 PRICE 3 3 4 1 3.8 4 ) 3. 4e 189 EXERCISE 6. we rewrite the restrictions as 3 1 3. 20) (2. Correlation with Variables Variable Auxiliary R 2 ln(L) ln(E) ln(M) ln(K) ln(L) ln(E) ln(M) 0.800 0. (f) Part H0 F-value df Fc (5%) p-value (a) (b) (c) (d) (e) =0 2= 3=0 2= 4=0 2= 3= 4=0 + 2 3+ 4+ 5=1 0.980 2 The auxiliary R 2 s and the explanatory-variable correlations that are exhibited in the following table suggest a high degree of collinearity in the model.670 ln M 0.112 1. there is insufficient evidence to reject the null hypothesis at the 5% level of significance.959 0.952 The very small t-values for all variables except ln(M ) .20) (2.056 ln K 0.226 ln L 0.181 0.947 0. 4e 190 EXERCISE 6. we have not been able to estimate the effects of the individual explanatory variables with any reasonable degree of precision.Chapter 6. However.127 0. ln Y (t ) 0.511 0. and the high R 2 .35 3. all the variables produce a model with a high level of explanation and a good predictive ability. .216 R2 0.972 0.044 ln E 0.49 3. our inability to reject any of the null hypotheses in parts (a) through (e).035 0. Note that.973 0.9 The results of the tests in parts (a) to (e) appear in the following table.987 0. Principles of Econometrics.047 0. in all cases.983 To examine the effect of collinearity on the reliability of estimation.35 0.10 4.969 0.20) (3.862 0.984 0. we examine the estimated equation.831 0.855 0. Furthermore.150 0. Collectively.49 3.001 (1. Exercise Solutions.881 0. our economic theory tells us that all the variables are important ones in a production function. are indicative of high collinearity. with t values in parentheses.984 0.908 0.986 0.20) (1.20) 4. 25) 4.127.809 0.955 0.020 (0.955 0.583 (0.694 0.243 (3.50. Most standard errors for the restricted estimates are less than their counterparts for the unrestricted estimates. The F-value can be calculated from restricted and unrestricted sums of squared errors as follows ( SSER SSEU ) J (0. The two sets of estimates are similar except for the noticeable difference in sign for ln(PL).10 (a) (b) The restricted and unrestricted least squares estimates and their standard errors appear in the following table.187 is more in line with our a priori views about the cross-price elasticity with respect to liquor than the negative estimate 0.24 .416) 0. we do not reject H 0 . Exercise Solutions. The relatively large standard error and the wrong sign for ln( PL) are a likely consequence of this correlation.971 0. This idea is consistent with economic theory.798 (3.Chapter 6.284) 0. CONST ln(PB) ln(PL) ln(PR) ln(I ) Unrestricted 3. 4e 191 EXERCISE 6.08992) 1 F 2.50 SSEU ( N K ) 0.95. Sample Correlation With (c) Variable Auxiliary R2 ln(PL) ln(PR) ln(I) ln(PB) ln(PL) ln(PR) ln(I) 0.743) 1. with a p3 4 5 value of 0.50 4. The evidence from the data is consistent with the notion that if prices and income go up in the same proportion. Since 2. Principles of Econometrics.239) Restricted 4.560) 0. The critical value is F(0.1.077) 0.964 0.080) 0.427) The high auxiliary R2s and sample correlations between the explanatory variables that appear in the following table suggest that collinearity could be a problem.167 (0.714) 1.923 (0. The value of the test statistic is F = 2.821 We use the F-test to test the restriction H 0 : 2 0 against the alternative 3 4 5 hypothesis H 1 : 2 0 .967 0.187 (0.583.774 0.08992 25 .210 (0. supporting the theoretical result that restricted least squares estimates have lower variances.299 (0.946 (0. The positive restricted estimate 0.098901 0.24 .166) 0.971 0. demand will not change. 7 . The two 95% prediction intervals are (70.7). The effect of the nonsample restriction has been to increase both endpoints of the interval by approximately 10 litres.088 4.6.060 4. Principles of Econometrics.975. 127. ln(Q) (d) (e) Restricted Unrestricted Q ln(Q ) se( f ) tc lower upper lower upper 4.4239 0.6.Chapter 6. Exercise Solutions.9) and (59. 116.851 4.16285 2.257 4.975. The t-values used to construct the interval estimates are t(0.14446 0.9 116.5541 4.10 (continued) (d)(e) The results for parts (d) and (e) appear in the following table.056 2. 26) 2.6 59.056 for the restricted model.6 127.060 for the unrestricted model and t(0. 25) 2. 4e 192 Exercise 6.759 70. 020 0.559ln L 0. implying that estimation is still imprecise.704 0. and the standard errors have declined.602 ln K (se) 0. However. the standard errors are still relatively large.688 The magnitudes of the elasticities of production (coefficients of ln(L) and ln(K)) seem reasonable.816 R2 0. Principles of Econometrics.986.11 (a) The estimated Cobb-Douglas production function with standard errors in parentheses is ln Q (se) 0. Exercise Solutions. implying the estimates are unreliable.559 We note that the relative magnitude of the elasticities of production with respect to capital and labor has changed. but their standard errors are very large.Chapter 6. leading to a collinearity problem which has produced the unreliable estimates. (b) After imposing constant returns to scale the estimated function is ln Q 0.053 0. .398ln L 0. It seems that labor and capital are used in a relatively fixed proportion.129 0. 4e 193 EXERCISE 6.559 0.488ln K 0. The sample correlation between ln(L) and ln(K) is 0.546 0. 0186 for one and two terms respectively).12 The RESET results for the log-log and the linear demand function are reported in the table below. .8377 2 terms 4. Test F-value df 5% Critical F p-value Log-log 1 term 0.7028 Linear (1.7618 Because the RESET returns p-values less than 0.260 3.23) 4. Principles of Econometrics.23) 4.422 0. On the other hand.0066 and 0. the log-log model appears to suit the data well with relatively high p-values of 0.260 3.Chapter 6.24) (2.05 (0.0186 1 term 8. based on the RESET we conclude that the log-log model better reflects the demand for beer. Thus.0066 0.9319 0.24) (2. Exercise Solutions. 4e 194 EXERCISE 6.7028 for one and two terms respectively.3581 (1.9319 and 0.0075 2 terms 0. at a 5% level of significance we conclude that the linear model is not an adequate functional form for the beer data.422 0. 303504 (48 5) The corresponding p-value is 0.0918) ( 0.6254 0.0567) (t ) (2.2443) With the restrictions imposed the signs of all the estimates are as expected.0302 t 0.972) (0.2582) (0.0005 RD 0.047) We expect the signs for 2 .2443) (0.0138 RG 0. That is. 4 and 5 to be all positive. although b3 and b4 are negative. positive values of 3 and 4 are not in conflict with the data. we do not reject H 0 .05). the response estimates for rainfall in all periods are not significantly different from zero. These terms could capture a declining marginal effect of rainfall.0567) (0. RD and RF could be included in the model.0035) (0. Also. the t -statistics for testing significance of b3 and b4 are very small.2443) (0.95.0567) (0. Interval estimates for 3 and 4 would include positive ranges.303504) 2 2.863664 4.432) (8.7985 4.2679) (0.0314 t 0.Chapter 6. However. but those for b3 and b4 are not.0138 RD 0.0034) (0.89) (0. The signs of b2 and b5 are as expected. indicating that both of them are not significantly different from zero. 3 . . (b) We want to test H 0 : 3 4. The value of the F test statistic is F (SSER SSEU ) J SSEU (T K ) 3 . development or flowering. Exercise Solutions. Thus.6889 (0. 4e 195 EXERCISE 6. Principles of Econometrics.43) 3.2.005) 0.0817) (t ) (2.3387 RF (se) (0.6515 0. 4 and 5 are not (4.1654) (2.785) ( 0.214 . One possibility for improving the model is the inclusion of quadratic effects of rainfall in each 2 2 2 period. However. the critical value for a 5% significance level is F(0. Since the F-value is less than the critical value (and the p-value is greater than 0. the squared terms RG . (c) The estimated model under the restriction is Yˆ 0.0794 RG 0.422) (8. We expect the wheat yield to increase as technology improves and additional rainfall in each period should increase yield.072. 3 5 against the alternative H1 : all equal. The data do not reject the notion that the response of yield is the same irrespective of whether the rain falls during germination.13 (a) The estimated model is Yˆ R2 0.0138 RF (se) (0. 0301HA2 0.0081 3.06022 48 years.1997 HA 4.1655 2.8895 0. . The effects of changes in education and in age on wages are given by the partial derivatives HW HE 1.528 with p-value = 0.4580 0.458 0. (c) The estimated equation is: HW (se) (t ) 45.7329 0.1236 2.0021. Also.1801 0.8895 0. Both p-values are smaller than a significance level of 0.5436 1.0602 HA The first of these two derivatives suggests that the wage rate declines with education up to an education level of HEmin 1.Chapter 6.954 12. older husbands earn 20 cents more on average per year of age. Thus.298 3. but at a decreasing rate.0086. (d) A RESET with one term yields F 0.958 An increase of one year of a husband’s education leads to a $2. so the estimated relationship might not be reliable in this region. and with two terms F 4.8 years. A negative value of HW HE for low values of HE is not realistic.703 Wages are now quadratic functions of age and education. and with two terms F 0.597 1.788 and p-value = 0. there is no evidence from the RESET test to suggest the model in part (c) is inadequate.882 and p-value = 0.8895HA 0. Exercise Solutions. The derivative with respect to age suggests the wage rate increases with age. leading us to conclude that the linear model suggested in part (a) is not adequate.5675 1.3022HE HW HA 2.1511HE 2 17.414.19 increase in wages. 4e 196 EXERCISE 6.182 R2 0.8.326 with p-value = 0.05. Only 7 of the 753 observations have education levels less than 4.298 2.943 R2 0.1918 3.14 (a) The estimated model is HW (se) 8. and then increases at an increasing rate. other things equal.1228 0.0458 2.568.0675 (t ) 1.05.4580 HE 0. (b) A RESET with one term yields F 9.30522 4. reaching a maximum at the age HAmax 2. Both p-values are much larger than a significance level of 0. Principles of Econometrics.1583 0.1933HE 0. 182 15 . Principles of Econometrics.8895 0.0602 HA HW HE 2. . is 0.3376HE HW HA 2.855 HA 35 and HA HW HA 35 HA 0.800 3.691 0.356 0. (f) The p-value for b6 .2443 The wage rate in large cities is. Note that when CIT was excluded from the equation in part (c).1688 HE 2 2.178 2.123 0.0676 .6213HA 17.0540 2. (g) From part (c). $7. HE 0.0160 1.2076 0.023 3. and the marginal effects presented in the above table are similar for both parts with the exception of HW HE for HE 6 where the sign has changed.0079 1.1012 3. its omission was not picked up by RESET.6213 0.678 50 leads to the 50 0.0444 0. The remaining coefficients have similar signs and magnitudes for both parts (c) and (e).208 R2 0. on average.3022HE HW HA 2. The likely reason for the absence of strong omitted variable bias is the low correlations between CIT and the included variables HE and HA.94 higher than it is outside those cities. The RESET test does not always pick up misspecifications. This suggests that b6 is significantly different from zero and CIT should be included in the equation. Exercise Solutions. HE Evaluating these expressions for HE following results. These correlations are given by corr CIT . HA) 0.2333 and corr(CIT .0278 HA2 7.525 7. the coefficient associated with CIT. 4e 197 Exercise 6.156 The omitted variable bias from omission of CIT does not appear to be severe.7101 (t ) 2.4580 0. HW HE Part (c) Part (e) 6 0.0000.0556 HA and from part (f) 6 .9379CIT 0.2076 HE 0.0914 0. we have HW HE 1.14 (continued) (e) The estimated model is: HW (se) 37.076 2.Chapter 6. HA HE HE 15 3.781 0. 15 (a) The estimated model is: SPRICE 11154. (c) Given that the living area is measured in hundreds of square feet. on average the price of a 2-year old house is 90.67 dollars more than the price of a 10-year old house. that we write alternatively as H 0 : 2 10000 and H1 : 2 10000 . Taking space from elsewhere to add bedrooms or bathrooms might reduce the price. Exercise Solutions.1495) se( 8b3 ) 90.1) (273.962 8 80.672 1. This interval is a relatively narrow one. ) . that their coefficients measure the effects on price of adding more bedrooms or more bathrooms. Recall. A 95% interval is given by: SPRICE AGE SPRICE AGE 2 10 t(0. however.30 BATHS (se) (6555. we estimate that the average price difference between houses that are 2 and 10 years old lies between –$1173 and $1354. The null and alternative hypotheses are H 0 : 2 2 20000 and H1 : 2 2 20000 .975.0) (2903. we estimate that extending the living area by 200 square feet will increase the price of the house by $21360. The negative signs on BEDS and BATHS might be puzzling.0 LIVAREA 11. the wording of the question suggested the alternative hypothesis should be H1 : 2 10000 .502 ( 1173.34) 90. (Note: In the first printing of the text.334 AGE 15552. Since a null hypothesis should always include an equality.82) All coefficients are significantly different from zero with the exception of that for AGE.Chapter 6.3 10680. the expected increase in price is estimated as: SPRICE LIVAREA 22 SPRICE LIVAREA 20 b2 22 b2 20 10680 2 21360 Holding other variables constant.4 BEDS 7019.1354) With 95% confidence.502) (1970. we have change the hypotheses accordingly. 4e 198 EXERCISE 6. (b) An estimate of the expected difference in prices is: SPRICE AGE SPRICE AGE 2 10 b3 2 b3 10 22. Principles of Econometrics. but it is uninformative in the sense that the difference could be negative or positive.668 ( 113. while keeping LIVAREA constant.672 Holding other variables constant.1) (80. 646 (or because the p-value is less than 0.0065.43 3880922 4 170680. 2 10680 15552. 4e 199 Exercise 6. The standard error can be found from computer software or from se 2b2 b4 22 var b2 var b4 2 2cov b2 .489 The corresponding p-value is 0.2 1869.9 (e) A RESET with one term yields F 117. and with two terms F 73.15(c) (continued) At a 5% significance level we reject H 0 if t t b2 10000 se(b2 ) t(0.646 . (d) Adding a bedroom of size 200 square feet will change the expected price by 2 Thus. we estimate the price increase will be between $2139 and $9476. Both p-values are smaller than a significance level of 0.95.9476) With 95% confidence.05).1495)se 2b2 b4 5807.4 5808 A 95% interval estimate of the price change is 2b2 b4 t(0.80 with p-value = 0. we reject the null hypothesis and conclude that an increase in the price of the house is more than 20000 dollars.975. Since the t-value is greater than the critical value of 1. b4 4 74610.05. The calculated t-value is 2.9 (2139. .1495) 1. Principles of Econometrics.0000. Exercise Solutions.985 and p-value = 0. leading us to conclude that the linear model suggested in part (a) is not reasonable.Chapter 6.0000.962 1869. an estimate of the price change is 2b2 b4 2 4 .6 1. we estimate that the average price difference between houses that are 2 and 10 years old lies between $2742 and $7812. but a more realistic one than that obtained using the specification in Exercise 6.95 (2741.13) (b) (3.7 2994.09LIVAREA2 14.78) (1972. (c) (b) An estimate of the expected difference in prices is: SPRICE AGE SPRICE AGE 2 10 b3 2 b7 22 b3 10 b7 102 8b3 96b7 8 ( 830.1495) se 8b3 96b7 5276.9434999 1012 (1500 7) 64.7 1.9434999 1012 2 1. we reject the null hypothesis and conclude that including LIVAREA2 and AGE 2 has improved the model.1111419 1012 .1) (2797. This interval is a relatively wide one. Since the F-value is larger than the critical value (or because the p-value is smaller than 0.06 BATHS (8744. Exercise Solutions.3559) To see if LIVAREA2 and AGE 2 are relevant variables.37) +169.4 The corresponding p-value is 0.2326 AGE 2 (16.05). we test the hypotheses H0 : 6 0.3785) 96 14.15.962 1291.65LIVAREA 830.00.15(a): SSER unrestricted SSE is that from part (a). The critical value at a 5% significance level is 3.38 AGE 11921. Principles of Econometrics. Using se 8b3 96b7 SPRICE AGE 1291. 4e 200 EXERCISE 6. a 95% interval is: 2 SPRICE AGE 10 t(0.3) (772.30) (197. .9.975. The F-value is calculated as follow: F ( SSER SSEU ) J SSEU ( N K ) 2.5) With 95% confidence.9 BEDS 4971.95 from computer software. H1 : 6 0 and/or 7 0 7 0 2. with LIVAREA2 and AGE 2 included. 7811.7 Holding other variables constant.1111419 1012 1. The The restricted SSE is that from Exercise 6.23261 5276.Chapter 6.16 (a) The estimated regression is: SPRICE (se) 79755. we estimate that the average price difference between a 2-year old house and a 10-year old house is $5277.0000. ) At a 5% significance level we reject H 0 if t t(0. Exercise Solutions. we have change the hypotheses accordingly.10455) With 95% confidence. (c) (d) Adding a bedroom of size 200 square feet will change the expected price by 20 2 202 6 4 ( BEDS 1) 18 2 182 6 4 BEDS 2 2 76 6 Thus.646 (or because the p-value is greater than 0.646 . Since a null hypothesis should always include an equality. the estimated price increase is between $3382 and $10.Chapter 6. It shows that test conclusions can be sensitive to the model specification.00 534.3 1.1493) 1. The null and alternative hypotheses are H0 : 2 2 84 6 20000 H1 : 2 2 84 6 20000 (Note: In the first printing of the text.193. Principles of Econometrics. 4e 201 Exercise 6.16 (continued) (c) (c) An estimate of the expected increase in price is SPRICE LIVAREA SPRICE LIVAREA 22 222 b6 22b2 20 20b2 202 b6 2b2 84b6 2 2994.1493)se 2b2 76b6 b4 6918.3591.55 0.000 dollars. The calculated t- value is (2b2 84b4 ) 20000 se(2b2 84b4 ) t 193.652 84 169.0916 11921.468 (3382. 4 . we fail to reject the null hypothesis and conclude that there is not sufficient evidence to show that the increase in the price of the house will be more than 20.652 76 169.361 The corresponding p-value is 0. This test outcome is opposite to the conclusion reached in Exercise 6.962 1802.455.95.92 6918. we estimate that extending the living area by 200 square feet will increase the price of the house by $20.3 A 95% interval estimate of the price change is 2b2 76b6 b4 t(0. the wording of the question suggested the alternative hypothesis should be H1 : 2 2 84 6 20000 .975. Since the t-value is less than the critical value of 1.15.0916 20193 Holding other variables constant.05). an estimate of the price change is 2b2 76b6 b4 2 2994. . despite being an improvement over the model in Exercise 6. leading us to conclude that the model with LIVAREA2 and AGE 2 included is not adequate. Exercise Solutions. 4e 202 Exercise 6. Principles of Econometrics.Chapter 6. with two terms it yields F 32.90 with p-value = 0.0017.16 (continued) (c) (e) A RESET with one term yields F 9.0000. Both p-values are smaller than a significance level of 0.15.56 with p-value = 0.05. 0079785 AGE (se) (0.4625 1494 25.2.04 The corresponding p-value is 0.05). 4e 203 EXERCISE 6.004477) (0. we reject the null hypothesis and conclude that age of house helps explain the selling price.7453 0. .0011799) 0.00014110 AGE 0.00002001) (b) (0.00050364 LIVAREA2 0. Also.0505) (0.0000. H1 : 2 0 or 0 3 0 or both are nonzero 3 The F-value can be calculated as: F ( SSER SSEU ) J SSEU ( N K ) (177. The F-value can be calculated as: F ( SSER SSEU ) J SSEU ( N K ) (71.4625) 2 1166.05).00009629) (0.075423BEDS (0.96 69.17 (a) The estimated regression is ln( SPRICE ) 10.002 . H1 : 4 0 or 0 5 5 0 or both are nonzero . the critical value is F(0.002.0000.4625 1494 The corresponding p-value is 0.Chapter 6.011316) The null and alternative hypotheses are H0 : 2 0. we reject the null hypothesis and conclude that living area helps explain the selling price. (c) The null and alternative hypotheses are H0 : 4 0. Since the Fvalue is greater than the critical value (or because the p-value is less than 0.9768 69.4625) 2 69. Exercise Solutions. The relevant critical value is 3.7908 69. Principles of Econometrics. Since the F-value is greater than the critical value (or because the p-value is less than 0.1494) 3.95.082609 LIVAREA 0. 075423BEDS exp 10.851.480484).00014110 10 2 0.74528 0. Principles of Econometrics.0079785 10 0.215938 .082609 20 0.480484.0079785 AGE 0.74528 0. se( f ) 0.0079785 10 0.327630) (96808.00014110 102 0. we first find such an interval for ln( SPRICE ) ln( SPRICE ) t(0. the estimated price of Wanling’s house after the extension is SPRICE n exp 10.082609 LIVAREA 0.975.327630) which yields the following prediction interval for SPRICE exp(11.075423 3 147865 The predicted price using the corrected predictor is: SPRICE c (e) SPRICE n exp ˆ 2 2 147865 exp 0.000503644 202 0.075423 3 167204 . Exercise Solutions. 12. 4e 204 Exercise 6. was found using computer software.74528 0.904057 1.808 and $225. 225851) With 95% confidence.1494) se( f ) 11. (f) Using the natural predictor.Chapter 6. we predict that the selling price of a house with the specified characteristics will lie between $96.96155 0.000503644 LIVAREA2 0.082609 22 0. The standard error of the forecast error for ln( SPRICE ) .17 (continued) (d) The predicted price using the natural predictor is: SPRICE n exp 10. exp(12.0464941 2 151343 To find a 95% prediction interval for SPRICE.00014110 AGE 0.215938 (11.000503644 222 0. 15 and 6. Then. Since a null hypothesis should always include an equality.17 (continued) (g) Ignoring the error term. A comparison of this test result to that from similar tests in Exercises 6.15 and 6. the null and alternative hypotheses are H0 : g( ) 20000 H1 : g ( ) 20000 (Note: In the first printing of the text. Principles of Econometrics. Since the t-value is less than the critical value of 1. we have change the hypotheses accordingly.951 1.282 (or because the p-value is greater than 0.139 The corresponding p-value is 0.6094. The standard error se g (b) 580.16. In Exercises 6. the t-values were 2.16 illustrates the sensitivity of test results to model specification.282 .464 580. respectively. we fail to reject the null hypothesis and conclude that there is not sufficient evidence to show that the increase in the price of the house will be more than $20.3254.951 was found using computer software that utilized the delta method since g (b) is a nonlinear function.361. Exercise Solutions.495 with p-value = 0.90.Chapter 6.8725.968 with p-value = 0. Both p-values are larger than a significance level of 0.05).05. (h) A RESET with one term yields F 0.1494) 1. The calculated t-value is t g (b) 20000 se g (b) 661.16.000. This conclusion is in contrast to those from similar tests in Exercises 6.15 and 6. It appears that the log specification is a better model than the linear and quadratic ones considered earlier. the wording of the question suggested the alternative hypothesis should be H1 : g ( ) 20000 .489 and 0. using two terms yields F 0. . 4e 205 Exercise 6. leading us to conclude that the model suggested in part (a) is a reasonable specification. ) At a 10% significance level we reject H 0 if t t(0. the increase in price of the house is given by SPRICE LIVAREA SPRICELIVAREA 22 exp 22 1 exp exp Let g ( ) exp 1 10 1 4 2 20 1 10 100 222 4 5 10 202 2 100 3 3 20 6 5 3 102 4 3 10 102 4 exp 22 6 exp 22 2 3 5 5 3 484 2 484 6 3 6 3 exp 20 exp 20 2 400 400 2 3 3 . 071877) (0.174 The corresponding p-value is 0. Since the F-value is less than the critical value (or because the p-value is greater than 0.0116321 0.00020366 3 10.013812( LIVAREA BEDS ) (0.02920) 4 69.085363 0. .02125) (0. 9 0 and 10 is nonzero The value of F-statistic is F ( SSER SSEU ) J SSEU ( N K ) (69. 9 and 10 are jointly not significantly different from zero.00030730 0.5033 0.12680 LIVAREA 0. 9 6 .00054740 0.062799 BEDS 0.00015854 BEDS 4 10.0005148) 0.4.0026419( AGE BEDS ) (0.00013163) (0.4405 0.1490) 2.005844) 0.18 (a) The estimated regression is: ln( SPRICE ) 10.05).Chapter 6.00024011(LIVAREA2 BEDS ) 0. 8 10 .007373) 0.00029391 AGE 2 (0. This results suggests that the number of bedrooms effects the price only through its interaction with the living area.0021610) 0.0012677 LIVAREA2 0.0063483 0. 8 . the critical value is F(0.000036997) The estimated relationships for 2. 8 0. H1 : At least one of 0.24671 69.95. Exercise Solutions.02920 1490 1.000045123( AGE 2 BEDS ) (0.099175 0.016916 AGE (se) (0.071550 0.2408) (0. we do not reject the null hypothesis at the 5% level.3205. 4e 206 EXERCISE 6.00078751 0. Principles of Econometrics. Also.00012498) (0.3149 0. 3 and 4 bedroom houses are as follows: BEDS C LIVAREA LIVAREA2 AGE AGE2 (b) 2 BEDS 10.5661 0.00011342 The null and alternative hypotheses are H0 : 6 0.0089902 0. and conclude that 6 .378 . 090116LIVAREA 0.0080479 0. .0080479 0.00034819 0.074133 0.00014243 AGE 2 0.078129 0.00001998) (0. 4e 207 Exercise 6. (d) The AIC and SC values for the two models are: Model in part (a): AIC 3.00014243 In this case only the coefficient of LIVAREA changes with the number of bedrooms.00009426) (0.00034819 0.030 Model in part (c) AIC 3. Exercise Solutions.0005695) The estimated relationships for 2.00034819 LIVAREA2 0.00034819 0.0039957( LVAREA BEDS ) (0.0011784) 0.0080479 0. Principles of Econometrics.5518 10.Chapter 6.065 SC 3.004903) (0.0080479 AGE (se) (0.068 SC 3.00014243 0. the model in part (c) is favored by both the AIC and the SC. 3 and 4 bedroom houses are as follows: BEDS C LIVAREA LIVAREA2 AGE AGE2 2 BEDS 3 BEDS 4 10.0479) (0.18(continued) (c) The estimated regression is: ln( SPRICE ) 10.5518 10.046 Thus.5518 0.082125 0.5518 0.00014243 0. a 95% interval estimate for Bill’s arrival time is from 8:49AM to 9:05AM.0704 .Chapter 6. (b) The predicted time it takes Bill to reach the University if he leaves at 7:45AM is TIME b1 b2 75 b3 10 b4 4 19. 227)se( f ) 71. a 95% interval estimate for the travel time is TIME t(0.9166 0. Exercise Solutions. Thus.9166 0.33532 10 2.369227 30 1.63. a 95% interval estimate for the travel time is TIME t(0.970 4.49.981 1. the standard error of the forecast error can be calculated as se( f ) 4.2396 (63.970 4.760 1.975. 4e 208 EXERCISE 6.78) Rounding this interval to 34 – 50 minutes. a 95% interval estimate for Bill’s arrival time is from 7:34AM to 7:50AM.760 Using suitable computer software. the standard error of the forecast error can be calculated as se( f ) 4. .74.33) Rounding this interval to 64 – 80 minutes.75483 41.33532 6 2.75483 4 71.981 Using suitable computer software. Principles of Econometrics. Thus.80.19 (a) The predicted time it takes Bill to reach the University if he leaves at 7:00AM is TIME b1 b2 30 b3 6 b4 1 19.0704 (33.369227 75 1.227)se( f ) 41.2396 .975. (c) In this case the null and alternative hypotheses are H0 : 2 3 0 2 3 4 We reject H 0 when F 1 F(0. The calculated F-value is 0.12165 0. Principles of Econometrics. Thus we do not reject H 0 because 0. where F(0. we reject H 0 when t and t(0.348) . . Using a t-test.95. Using a t-test. the p-value of the test is 0.649 .05.348) 2.348) or t t(0.90. Also.025. The test can be performed with an F 3 4 or a t statistic.348) 3. The calculated t-value is t b2 b3 se(b2 b3 ) 0.649 0. the p-value of the test is 0.36174 0. we do not reject H 0 because 0.585 In this case H 0 is not rejected because 1.2.172 0. The calculated F-value is 0. The p-value of the test is 0.348) where t(0. we do not reject H 0 because 0.864. Using an F-test.348) H1 : 2 3 2 3 0 and/or 1 4 3.Chapter 6.967 . we reject H 0 when F F(0.1.0295.348) where t(0. Also. The hypothesis that the land and labor elasticities are equal cannot be rejected at a 5% significance level.02 .025. we reject H 0 when F F(0.95.0295 2.348) 1.023797 In this case H 0 is not rejected because 1. The test can be performed with an F or a t statistic. confirming non-rejection of H 0 .36174 0.02 . Thus. Using an F-test.868 .72 .348) t(0.975.1.559.864. Thus.348) 1.833.95. The joint null hypothesis of constant returns to scale and equality of land and labor elasticities cannot be rejected at a 5% significance level.1.348) or t t(0.868 . Exercise Solutions.95.20 (a) We are testing the null hypothesis H 0 : 2 3 against the alternative H 1 : 2 3 .342 3.649 . The hypothesis of constant returns to scale cannot be rejected at a 10% significance level.72 . The calculated t-value is t b2 b3 b4 1 se(b2 b3 b4 ) 0.975. 4e 209 EXERCISE 6. using a 10% significance level.559.967 .209502 1 0.342.348) t(0.585 1. we reject H 0 when t t(0. the p-value of the test is 0.43285 0.967 0. The calculated F-value is 0.649 and 1.43285 0. The hypothesis of constant returns to scale is not rejected at a 10% significance level. (b) We are testing the null hypothesis H 0 : 2 1 against the alternative 3 4 H1 : 2 1 .183 3. confirming non-rejection of H 0 .183. confirming nonrejection of H 0 . The p-value of the test is 0. The hypothesis that the land and labor elasticities are equal cannot be rejected at a 5% significance level.95.05.172 1.967 1. Also. 0188) ln( LABOR ) 0.4030 (0.5381 (0.20 (continued) (d) The restricted model for part (a) where ln( PROD) 1 2 1 2 3 is ln( AREA LABOR) The restricted model for part (b) where ln( PROD) 2 2 3 ln( AREA) ln( FERT ) e 1 is 4 1 4 2 ln( LABOR) 4 4 ln( FERT ) e or.3595 (0.2557) –1. imposing this restriction changes the unrestricted estimates the restriction 2 3 4 and their standard errors very little.3964 (0.4299 (0. ln PROD LABOR 1 2 ln The restricted model for part (c) where ln PROD FERT 1 2 ln AREA LABOR 2 3 4 and FERT LABOR ln 2 3 AREA LABOR FERT 2 4 e 1 is e The estimates and (standard errors) from these restricted models. Adding 2 3 4 further. leaving the coefficient estimates essentially unchanged. and the unrestricted model.2106 (0.3964 (0.0646) 0. Imposing the restriction 2 3 has an impact. changing the estimates for both 2 and 3 .5468 (0.0382) 0.2502) –1.0241) 0.0913) ln( AREA) 0. Because the unrestricted estimates almost satisfy 1 .6079 1 .3617 (0.0377) 0. Principles of Econometrics.0383) 0.1011) –1.2095 (0.0376) SSE 40.4328 (0.Chapter 6.0625) 0. Exercise Solutions.4095 (0.0241) 0.3941 (0.0188) ln( FERT ) 0. and reducing their standard errors 1 to this restriction reduces the standard errors even considerably. 4e 210 Exercise 6.0669) 0. Unrestricted 2 3 2 3 4 2 1 2 3 3 4 C –1.2118 (0.5654 40.3941 (0.2109 (0.6052 40.5688 40.0640) 0. are given in the following table. 3015.863 (0. but again the omitted variable is not picked up by the RESET.21 The results are summarized in the following table.2095 to 0.3617 0.4567 0. The RESET F-values (p-values) for 1 and 2 extra terms are 0. Principles of Econometrics.428) and 0.511 (0.4328 to 0.572).5689 0.0083 (i) With FERT omitted the elasticity for AREA changes from 0.3015 0.3617 to 0.559 (0. respectively.779 (0.2682 0.2095 0.4328 0. respectively.6633 RESET(1) p-value RESET(2) p-value 0. 4e 211 EXERCISE 6.629 (0.1140 0. Omitting LABOR also appears to bias the other elasticities upwards. The RESET F-values (p-values) for 1 and 2 extra terms are 0. and the elasticity for FERT changes from 0. Full model FERT omitted LABOR omitted b2 ( AREA) b3 ( LABOR) b4 ( FERT ) 0. (iii) With AREA omitted the elasticity for FERT changes from 0. Omitting AREA appears to bias the other elasticities upwards. particularly that for LABOR.5721 0.3617 to 0.024 (0.877) and 0.6633. but the omitted variable is not picked up by the RESET. The RESET F-values (p-values) for 1 and 2 extra terms are 2. Exercise Solutions.Chapter 6.114) and 4.2682. respectively. (ii) With LABOR omitted the elasticity for AREA changes from 0. .5688 0.4567.2095 to 0.8771 0.4328 to 0.460). and the elasticity for LABOR changes from 0. Omitting FERT appears to bias the other elasticities upwards.5689.7084. In this case the omitted variable misspecification has been picked up by the RESET with two extra terms.008). and the elasticity for LABOR changes from 0.4281 0.7084 0.4598 AREA omitted 0.2761 0. However.905 0.731 2.36) 7.432 1. are given in the following table.818 0. all the confidence intervals do overlap.283 2.993 2. we reject H0 and conclude that age does affect pizza expenditure. Principles of Econometrics. for different ages.991 The interval estimates were calculated using tc Confidence Interval Lower Upper 1. 36) degrees of freedom is Fc = 3. Thus. standard errors and 95% interval estimates for this quantity. Exercise Solutions.K 819286 580609 2 580609 36 7.515 3. .002.622 1.202 1.710 0.050 0.808 t(0. 4e 212 EXERCISE 6. Age Point Estimate Standard Error 20 30 40 50 55 4. the confidence intervals are relatively wide indicating that our information on the marginal propensities is not very reliable.107 0.448 1. as we would expect.Chapter 6.40 The 5% critical value for (2. (b) The marginal propensity to spend on pizza is given by E PIZZA INCOME 3 4 AGE Point estimates. The point estimates for the marginal propensity to spend on pizza decline as age increases.22 The model for parts (a) and (b) is PIZZA (a) 1 2 AGE 3 INCOME 4 ( AGE INCOME ) e The hypotheses are H0: 2 = 4 =0 and H1: 2 0 and/or 4 0 The value of the F statistic under the assumption that H0 is true is F SSER SSEU J SSEU N .26 and the p-value of the test is 0.598 4.465 0.212 2.258 2.0281 .975.520 0. Indeed. 949 2.4139 0.136 0.056 0. in this case.496 1.962 2.675 3.4012 The marginal propensity to spend on pizza. .1198 0.634 2.004205 0.664 1.945 2.72 135. Principles of Econometrics.575 1. The results are given in the table below.401.595 –1.0383 3. up to a point.35) 11.57 –2.Chapter 6.769 2.325 The interval estimates were calculated using tc Confidence Interval Lower Upper 0.0301 .090 0. is given by E PIZZA INCOME 3 4 AGE + 5 AGE 2 Point estimates.781 1. are given in the following table.4238 0.469 0. and then decline. Error 109.22 (continued) (c) This model is given by PIZZA 1 + 2 AGE 3 INC 4 AGE INC 5 AGE 2 INC e The marginal effect of income is now given by E PIZZA INCOME 3 4 AGE + 5 AGE 2 If this marginal effect is to increase with age. but. The sign of the estimated coefficient b5 = 0.975.8399 –0.779 5. with a p-value of 0. Exercise Solutions.963 1.5419 14.074 0. Variable C AGE INCOME AGE INCOME AGE2 INCOME (d) Coefficient Std. 4e 213 Exercise 6.2635 0.589 1.371 3.5687 0. for different ages. Age Point Estimate Standard Error 20 30 40 50 55 6.850 p-value 0.004948 t-value 0.009 1.809 –0. standard errors and 95% interval estimates for this quantity. it was not significantly different from zero. then 5 < 0.744 t(0.0962 8.0042 did not agree with our expectation.4704 0. Also. we reject the null hypothesis and conclude at least one of 2 . including squared and cubed variables can lead to collinearity if there is inadequate variation in a variable.8 568869. the critical value at the 5% significance level is F(0. respectively. 0. (f) Two ways to check for collinearity are (i) to examine the simple correlations between each pair of variables in the regression. Thus. There is no “life-cycle effect” where the marginal propensity increases up to a point and then declines. Again. collinearity is potentially a problem.874 . In the tables below there are 3 simple correlations greater than 0.94 for the regression in part (c) and 5 when AGE 3 INC is included.2635 and 0. the confidence intervals are relatively wide indicating that our information on the marginal propensities is not very reliable. Thus.05). the relevant section of the quadratic function is that where the marginal propensity to spend on pizza is declining. 4e 214 Exercise 6. for the range of ages in the sample.4704 2 0. and (ii) to examine the R2 values from auxiliary regressions where each explanatory variable is regressed on all other explanatory variables in the equation. Exercise Solutions. and AGE 2 INCOME are 0.95. we set up the hypotheses H0 : 2 4 5 0 H1 : At least one of 2 .2 35 The corresponding p-value is 0. despite the fact each of the three coefficients was insignificant when considered separately. To perform a joint test of the significance of all three coefficients. In general.004205 AGEmin 55. each of these coefficients is not significantly different from zero. the point estimates for the marginal propensity to spend on pizza decline as age increases. It is decreasing at a decreasing rate. The number of auxiliary regressions with R2s greater than 0. (e) The p-values for separate t tests of significance for the coefficients of AGE. .99 is 3 for the regression in part (c) and 4 when AGE3 INC is included. 4 and 5 is nonzero The F-value is calculated as follows: F ( SSER SSEU ) J SSEU ( N K ) (819285.93 Thus. Examining the estimates and their standard errors confirms this fact. None of the coefficients are reliably estimated. The range of ages in the sample is 18-55.Chapter 6.2) 3 5.3. In both cases there are no t-values which are greater than 2 and hence no coefficients are significantly different from zero. Since the F-value is greater than the critical value (or because the pvalue is less than 0. This result suggests that age is indeed an important variable for explaining pizza consumption.35) 2.136 568869.5687.4012.0048. The quadratic function reaches a minimum at 0. AGE INCOME . 4 and 5 is nonzero.22(d) (continued) As in part (b). Collinearity is the likely reason for this outcome. We investigate it in part (f). Principles of Econometrics. Exercise Solutions.9436 0.Chapter 6.99994 .9636 0.6887 0.99796 0.68400 0.9812 0.9893 0.99956 0. Principles of Econometrics.99983 0.99859 0. 4e 215 Exercise 6.99999 0.99999 0.22(f) (continued) Simple Correlations INC AGE AGE INC AGE 2 INC AGE AGE INC AGE 2 INC AGE3 INC 0.5862 0.9921 R2 Values from Auxiliary Regressions LHS variable R2 in part (c) R2 in part (f) INC AGE AGE INC AGE 2 INC AGE3 INC 0.8975 0.4685 0.6504 0.82598 0. 08918 ln(WAGE ) EDUC 0. 4. and p-values obtained for this model are given in the following table.0009 0.4685 (6.0011122 0.383 5.875 –2.04.046418 2 0.036936 0.0026509 10 0.0010256 10 0. standard errors.014 0.23 Coefficient estimates.9624 0.363 1.0005092 3. ln(WAGE ) EDUC se 0.11.0000882 0.4262 (2.0010256 10 0.2092 0.057775 –0.13408 0.9624 1.918 1.0006946 10 0. a 95% interval estimate for 100 ln(WAGE ) EDUC is 8.975.257 2.0442 The percentage change in WAGE from an extra year of education is calculated from: ln(WAGE ) 100 EDUC 2 2 3 EDUC + 6 EXPER 100 The percentage change in WAGE from an extra year of experience is calculated from: ln(WAGE ) 100 EXPER (i) 4 2 5 EXPER + 6 EDUC 100 When EDUC 10 and EXPER 10 .0026509 0.004262 A 95% interval estimate for 100 ln(WAGE ) EXPER is 3.0173 0.014685 Using t(0.009761 0.03363 0.33982 0.0006946 –0.20) . (a) Variable Coefficient Std.80) (ii) When EDUC 10 and EXPER 10 . t-values.057775 2 ln(WAGE ) EXPER 0. 994) 1.53.0000 0.9624 . Exercise Solutions.919 –7. ln(WAGE ) EXPER se 0. 4e 216 EXERCISE 6.337 1.0010256 0.046418 0. Error t-value p-value C EDUC EDUC2 EXPER EXPER2 EDUC EXPER 1.Chapter 6. Principles of Econometrics.0000 0. 1 and/or 20 4 20 4 10 5 0. a 95% interval estimate for 100 ln(WAGE ) EDUC is 13.0010256 20 0. the F-value and the p-value are computed as 1.04 6 Using econometric software.9478 1.12 and/or 4 40 4 5 40 20 5 0. the return to education increases with further education whereas the return to experience decreases with further experience. Exercise Solutions.975.9624 0.3273. 4e 217 Exercise 6.7154.13194 ln(WAGE ) EDUC 0.29. Since the p-value is larger than 0. the data are compatible with the hypothesis that the return to an extra year of education is 10% and the return to an extra year of experience is 4%. we do not reject the null hypothesis. for 20 years of experience and 20 years of education.118 and 0.003324 A 95% interval estimate for 100 ln(WAGE ) EXPER is 0.9624 1.05. We conclude that. 0.335 and 0. 0.057775 2 ln(WAGE ) EXPER 0. 994) 1.16. (c) The null and alternative hypotheses are: H0 : 2 40 3 20 6 0.60) These results suggest that the return to an extra year of education is greater than the return to an extra year of experience. Furthermore. .0010256 20 0.04 6 10 5 0.23(a) (continued) (iii) When EDUC 20 and EXPER ln(WAGE ) EDUC se 20 .0026509 20 0.9624 .014807 Using t(0.Chapter 6.1 and H1 : 2 20 3 10 6 0.05. We conclude that. we do not reject the null hypothesis. (b) The null and alternative hypotheses are: H0 : 2 20 3 10 6 0.0006946 20 0. Principles of Econometrics. respectively.194 1.046418 2 0. the F-value and the p-value are computed as 0. Since the p-value is larger than 0.01 6 20 6 0. respectively.10) (iv) When EDUC 20 and EXPER ln(WAGE ) EXPER se 20 .4807 (10.12 and H1 : 2 40 3 20 6 0.01 Using econometric software. for 10 years of experience and 10 years of education.1.009478 0. the data are compatible with the hypothesis that the return to an extra year of education is 12% and the return to an extra year of experience is 1%.30.3324 (0. 0570590 –0.9759 2 252. We conclude that the data are compatible with the hypothesis that. we do not reject the null hypothesis. the F-value and the p-value are computed as 0.335 2 . for 10 years of experience and 10 years of education.01EXPER 6 EDUC 6 EXPER 3 EDUC 2 EXPER 2 EDUC EXPER e 5 EXPER 2 6 EDUC EXPER 20 EDUC 20 EXPER e 1 3 5 EDUC 2 6 40 EDUC 40 EXPER Estimating the above model.0004659 0.1.12 40 0.553 0.24712 4.0000879 0.021249 0. we have 2 0. 4e 218 Exercise 6.230 0. we can manually calculate the F-value. and for 20 years of experience and 20 years of education. and substituting into the restrictions to find estimates for and 4 yields Variable C EDUC EDUC2 EXPER EXPER2 EDUC EXPER Coefficient Std.05.058 6. F SSER SSEU J SSEU ( N K ) 253. (e) From the joint hypotheses in part (c). the return to an extra year of education is 10% and the return to an extra year of experience is 4%.04522 0.04.842 –7.0000 0.934 –3.01 6 H1 : At lease one of the above equations does not hold Using econometric software.12 40 3 20 6 4 0. Since the p-value is larger than 0.0004 To confirm the result in (c).01 40 5 20 6 Substituting these expressions into the original equation yields ln(WAGE ) 1 0. 20 2 40 3 20 6 0. the return to an extra year of education is 12% and the return to an extra year of experience is 1%.0006974 –0.0002697 2.01 40 5 20 3 20 ln(WAGE ) 0.0000 0. respectively.0000 0.0001 0.0083390 0.12 and 4 10 5 4 40 6 5 0.Chapter 6. Exercise Solutions. Principles of Econometrics. Error t-value p-value 1.9759 994 0.063536 0.0029 0.12 EDUC 0.990 4.1464 252.0009582 0.23 (continued) (d) The null and alternative hypotheses are: H0 : 2 20 3 10 6 0.0018907 0. 20 0.7695 and 0.5452. 0000 0. The expected sign of 3 is positive. Holding other variables constant. a 1% increase in price per can of brand 1 changes brand 1’s sales by 2 % .6871 Std. we can rewrite the regression equation as: ln( SAL1) ( 1 (d) 1 1 . The data do not support the marketing manager’s claim. 4 )ln( APR1) 3 ln( APR2) ln( APR2) ln( APR1) 1 3 1 3 ln APR2 APR1 1 3 ln APR1 APR2 1 where we have set 3 2 ln 2 3 APR1 APR2 and ln APR3 APR1 e 4 ln APR1 APR3 e 4 APR1 APR3 ln ln( APR3) e ln( APR3) ln( APR1) 4 3 3 4 4 e e .376 –7.0697 0.Chapter 6. 2 is the cross price elasticity of sales of brand 1 with respect to changes in the price of brand 2. 3 is the cross price elasticity of sales of brand 1 with respect to changes in the price of brand 3.10.841. 4 (b) The regression results are Variable C ln(APR1) ln(APR2) ln(APR3) Coefficient 7.0588 < 0. Principles of Econometrics.262 p-value 0. Error 0. Since 0. The expected sign of 4 is positive. we find the F-value for this hypothesis to be 3.0588. The expected sign of 2 is negative.245 1. a 1% increase in price per can of brand 3 changes brand 1’s sales by 4 % .9904 1. we reject H 0 at a 10% significance level.6383 0. with corresponding p-value of 0.855 2.24 (a) is the direct price elasticity of sales of brand 1 with respect to changes in the price of brand 1.7460 t-value 31.5338 0. Exercise Solutions. .0283 All coefficients have the expected signs and all are significantly different from zero at a 5% level of significance with the exception of b3 which is the coefficient of ln( APR 2) . (c) If 2 3 4 0 . a 1% increase in price per can of brand 2 changes brand 1’s sales by 3 % .8894 –4. The null and alternative hypotheses are: H0 : 2 3 4 0 H1 : 2 3 4 0 Using econometric software.6246 0. Holding other variables constant.0000 0. Holding other variables constant. 4e 219 EXERCISE 6.2514 0. The standard error can be calculated as follows se b4 b3 var b4 var b3 2 cov b4 .677 . The F-test result in part (d) can be confirmed using the sums of squared errors from the restricted and unrestricted models SSER SSEU J SSEU N . the evidence is not sufficiently strong to confirm that brand 3 is a stronger competitor than brand 2. Principles of Econometrics. holding other variables constant.70%. 4e 220 Exercise 6.556547 0. Exercise Solutions. Also. b3 0.841 15.234. a 1% increase in the price ratio of brand 1 to brand 3 tuna decreases the sales of brand 1 tuna by 2.733 The corresponding p-value is 0. the critical value at a 5% level of significance is t(0.7001ln APR1 APR3 (0. Since t 1. The t-values for a2 and a3 are 2.95. a3 2.6956 15.K F 16.4585 48 (f) Both estimated models in parts (b) and (e) suggest that brand 3 is the stronger competitor to brand 1 because b4 > b3 and a3 a2 .5534) a2 1. For the model in part (a).527 and 4.9507 0.0820) (0.6871 0.318 implies that.70 implies that. respectively.9904 0. At a 5% level of significance.5215) 2. we do not reject the null hypothesis.Chapter 6.24 (continued) (e) The estimated regression is: ln( SAL1) 8.4585 1 3.9507 . we set up an alternative hypothesis that brand 3 is a stronger competitor than brand 2. A price change in brand 3 has a greater effect on sales of brand 1 than a price change in brand 2.879 . indicating that both these estimated coefficients are significantly different from zero.3567 1. holding other variables constant.284986 2 ( 0.677 .031110) 0. H0 : 4 3 against H1 : 4 3 The value of the t-statistic is t b4 b3 se b4 b3 1. a 1% increase in the price ratio of brand 1 to brand 2 tuna decreases the sales of brand 1 tuna by 1. (g) To confirm that brand 3 is the stronger competitor.318%.48) 1.3177ln (se) APR1 APR2 (0. 9092 .677 .271995 2 ( 0.520 The corresponding p-value is 0. At a 5% level of significance. a2 0. t(0. H0 : 3 2 against H1 : 3 2 The value of the t-statistic is t a3 a2 se a3 a2 2.677 . The opposite conclusion is reached if we use a 10% significance level.7001 ( 1.9092 1. Since t 1.Chapter 6.299 1.124246) 0. In this case.0674. The standard error can be calculated as follows se a3 a2 var a3 var a2 2 cov a3 . the critical value at a 5% level of significance is t(0.49) 1.3177) 0.49) 1. we do not reject the null hypothesis. 4e 221 Exercise 6. and the evidence is sufficiently strong to confirm that brand 3 is a stronger competitor. Principles of Econometrics. Exercise Solutions.306213 0. the evidence is not sufficiently strong to confirm that brand 3 is a stronger competitor than brand 2.10. Also.24(g) (continued) For the model in part (c).520 .05. 4 4 100 .43 470.47084PR1 0. we begin by rewriting the first equation as follows SAL1 1 2 1 2 1 APR1 2 3 APR2 PR1 100 PR1 3 3 4 APR3 e PR2 100 PR2 4 4 PR3 100 e PR3 e where 1 1 .1129 PR3 SALES 22. 4e 222 EXERCISE 6. we write SAL1 1 2 PR1 1000 SALES SALES 3 1 1 2 2 1000 1000 1 2 PR1 PR2 4 PR1 PR1 3 3 PR3 e PR2 3 1000 PR2 4 PR2 4 PR3 e 4 1000 PR3 e 1000 PR3 e where 1 1 1000. will be 1000 times smaller than those in the second equation.8447 PR1 92.00 PR2 16511. 2 2 100. Thus.09299PR2 0. 2 2 1000. . Principles of Econometrics.963 0. 4 4 1000 . the coefficients of PR1. For the third equation.43 47084. 3 3 100. The estimated regressions are: SAL1 22963. APR2. including the intercept. PR2. Exercise Solutions. and APR3 in the first equation. 3 3 1000.25 (a) To appreciate the relationship between the 3 equations. The intercept coefficient remains unchanged.Chapter 6.9900 PR2 165. Thus.16511PR3 The relationships between the estimated coefficients in these three equations agree with the conclusions we reached by algebraically manipulating the equations.29 PR3 SAL1 22963.47 APR1 9299. and PR3 in the second equation will be 100 times smaller than the coefficients of APR1. all coefficients in the third equation. we write ln( SAL1) 1 2 PR1 ln( SALES 1000) ln( SALES ) 1 1 1 3 2 PR2 PR1 ln(1000) 2 PR1 2 3 4 3 PR3 e PR2 PR1 PR2 3 4 4 PR2 PR3 e 4 PR3 e PR3 e ln(1000).54819 0. The coefficients of the third equation are identical to those of the second equation. The estimated regressions are: ln( SAL1) 10. The intercept coefficient remains unchanged.907755 . 2 2 100. we write ln( SAL1) 1 2 1 2 1 APR1 3 APR2 PR1 100 2 PR1 APR3 e PR2 100 3 3 4 PR2 4 4 PR3 100 e PR3 e where 1 1 .45595 6.014174 PR2 0. 4 4 . with the exception of the intercept which differs by the amount ln(1000) 6. 4 4 100 . 4e 223 Exercise 6. and PR3 in the second equation will be 100 times smaller than the coefficients of APR1.45595 6. APR2. 3 3 100.4174 APR2 2.062176 PR1 0. and APR3 in the first equation.021472 PR3 These estimates agree with the relationships established algebraically.45595 0.062176 PR1 0.014174 PR2 0.1472 APR3 ln( SAL1) 10. The coefficients of PR1. Principles of Econometrics. To obtain the third equation from the second.2176 APR1 1.Chapter 6. The relationships between the coefficients are the same as those in part (a).90776 3. PR2. 3 3. Note that a1 ln(1000) 10.021472 PR3 ln( SALES ) 3.25 (continued) (b) To obtain the relationship between the coefficients of the first two equations. 2 where 1 1 2. Exercise Solutions.54819 ˆ1 . we write ln( SAL1) 1 2 ln( APR1) 1 2 ln 1 2 ln( PR1) 1 2 PR1 100 3 ln( PR1) 3 ln( APR2) 3 ln PR2 100 ln( PR2) 3 4 ln( PR2) 4 ln( APR3) e 4 ln PR3 100 e ln( PR3) 4 2 3 4 ln(100) e ln( PR3) e where 1 1 2 3 4 ln(100).85591 a1 .9904ln( PR2) 1. 2 2.85591 4. 3 3. we note that b1 b2 b3 b4 ln(100) 7. The coefficients of the third equation are identical to those of the second equation. Thus.6871ln( PR3) As expected.9904ln( PR2) 1.88938 4. 4e 224 Exercise 6.624576 0. Principles of Econometrics. the elasticity estimates are the same in all three equations. 4 4 . 3 3. To reconcile the three different intercepts.687140) 4. Exercise Solutions.948158 ˆ 1 Comparing equations 1 and 2. we write ln( SAL1) 1 2 ln( PR1) ln( SALES 1000) ln( SALES ) 1 1 1 2 3 ln( PR1) ln(1000) 2 ln( PR2) ln( PR1) 2 3 4 ln( PR2) ln( PR1) 3 ln( PR3) e ln( PR2) 3 4 ln( PR2) 4 ln( PR3) e 4 ln( PR3) e ln( PR3) e ln(1000). To obtain the third equation from the second.889381 ( 4.855913 6.9904ln( APR2) 1.907755 .6246ln( PR1) 0. 4 4 .6871ln( PR3) ln( SALES ) 9.25 (continued) (c) To obtain the relationship between the coefficients of the first two equations. all coefficients of the second equation are identical to those of the first equation with the exception of the intercept which differs by the amount 2 3 4 ln(100) .6246ln( APR1) 0. This is a general result.6871ln( APR3) ln( SAL1) 16.Chapter 6.990379 1.907755 9. The estimated regressions are: ln( SAL1) 7. with the exception of the intercept which differs by the amount ln(1000) 6.6246ln( PR1) 0. first note that a1 ln(1000) 16.94816 4.60517 16. This result is the same as that obtained in part (b). 2 where 1 1 2. In all three cases only the intercept changes. Changing the units of measurement of variables in a log-log model does not change the values of the coefficients which are elasticities. CHAPTER 7 Exercise Solutions 225 . relative to males.03 . The intercept suggests the starting salary for someone with a zero GPA and who did not take econometrics is $24. Principles of Econometrics.Chapter 7. and other variables are held constant.1 (a) When a GPA is increased by one unit. Students who take econometrics are estimated to have a starting salary which is $5033 higher.66 . The R2 = 0.001). we estimate that the average starting salary is estimated to increase by the amount $1643 ( t 4. E SAL 1 2 1 GPA 4 3 2 METRICS GPA 3 if FEMALE = 0 5 METRICS if FEMALE = 1 . all else held constant. this figure is likely to be unreliable since there would be no one with a zero GPA. than the starting salary of those who did not take econometrics ( t 11. all else held constant.74 implies 74% of the variation of starting salary is explained by GPA and METRICS (b) A suitably modified equation is SAL 1 2 GPA 3 METRICS 4 FEMALE e The parameter 4 is an intercept indicator variable that captures the effect of gender on starting salary. Exercise Solutions. 4e 226 EXERCISE 7. we change the model to SAL 1 2GPA 3 METRICS 4 FEMALE 5 METRICS FEMALE e The parameter 4 is an intercept indicator variable that captures the effect of gender on starting salary. and the coefficient is significant at = 0. on average. The parameter 5 is a slope-indicator variable that captures any change in the slope for females. However. and the coefficient is significant at = 0. E SAL 1 2 1 (c) GPA 4 3 2 METRICS GPA 3 if FEMALE = 0 METRICS if FEMALE = 1 To see if the value of econometrics is the same for men and women.001).200. 003 < 0. Intercept: At the beginning of the time period over which observations were taken. However. However.05 level of significance. on a day which is not Friday.01 level.06 and a pvalue = 0. The “holiday effect” is significant at the 0. Saturday or a holiday. NEWMOON: The average number of emergency room cases is estimated to go up by 6. Principles of Econometrics.6 on Fridays and Saturdays. Therefore.2 (a) Considering each of the coefficients in turn. and their standard errors.69. a null hypothesis stating that a new moon has no influence on the number of emergency room cases would not be rejected at the usual 10% level. and also should expect a steady increase over time. These estimated coefficients are both significant at the 0.45 on days when there is a full moon. T: We estimate that the average number of emergency room cases has been increasing by 0. or smaller. other factors held constant.01. as expected when variables are omitted. when FULLMOON and NEWMOON are omitted. FRI and SAT: The average number of emergency room cases is estimated to go up by 6.86 on holidays. holding all else constant. Based on these casual observations the consequences of omitting FULLMOON and NEWMOON are negligible.9 and 10. respectively. .Chapter 7. holding all else constant. and a day which has neither a full moon nor a half moon. This time trend has a t-value of 3.0338 per day. all else constant. Exercise Solutions.4 on days when there is a new moon. (b) There are very small changes in the remaining coefficients. all else held constant. Fridays and Saturdays. we have the following interpretations. The equation goodness-of-fit statistic decreases slightly. HOLIDAY: The average number of emergency room cases is estimated to go up by 13. the estimated average number of emergency room cases was 93. hospitals should expect more calls on holidays. a null hypothesis stating that a full moon has no influence on the number of emergency room cases would not be rejected at any reasonable level of significance. 4e 227 EXERCISE 7. FULLMOON: The average number of emergency room cases is estimated to go up by 2. 277.Chapter 7. Exercise Solutions. 2.05 critical value is F(0. The . .2 (continued) (c) The null and alternative hypotheses are H0 : 6 7 0 H1 : 6 or 7 is nonzero. The test statistic is F ( SSER SSEU ) 2 SSEU (229 7) where SSER = 27424. Thus. Principles of Econometrics. 222) 3. we do not reject the null hypothesis that new and full moons have no impact on the number of emergency room cases.82 is the sum of squared errors from the estimated equation with these variables included.307 .29.95. The calculated value of the F statistic is 1. and corresponding p-value is 0.19 is the sum of squared errors from the estimated equation with FULLMOON and NEWMOON omitted and SSEU = 27108. 4e 228 Exercise 7. Chapter 7.035 0.564 less days. 4e 229 EXERCISE 7. Keeping in mind that the critical t-value is 1. (e) The effect of GENDER suggests that. It is given by q 4. Those in the 12-20 age range consume alcohol on 1. if the price of pure alcohol goes up by $1 per liter. On average. respectively. the average number of days (out of 31) that alcohol is consumed will fall by 0.78 0.045 24.807 0.960.320 We estimate that a 1% increase in the price of alcohol will reduce the number of days of alcohol usage by 0. (b) The price elasticity at the means is given by qp pq 0. males consume alcohol on 1. Exercise Solutions.580 and 0. This last estimate is not significantly different from zero.099 0.035 more days than those who are over 30. the price elasticity is qp pq 0.580 3.000057 12425 1. than individuals from other races.045 24. however. holding all else fixed. two age ranges instead of three (12-20 and an omitted category of more than 20).32%. Principles of Econometrics. which would be a sensible thing to do. The magnitude of the estimated effect is small. married people consume alcohol on 0.045.637 0. (c) To compute this elasticity.3 (a) The estimated coefficient of the price of alcohol suggests that.637 more days than women.01 level. Those in the 21-30 age range consume alcohol on 0. but based on the t-statistic the estimate is statistically significant at the 0. the estimated coefficient would change to 0.531 less days than those who are over 30. . are likely to be adequate.807 less days than single people.057.78 3.78 3.280 We estimate that a 1% increase in the price of alcohol will reduce the number of days of alcohol usage by a married black male by 0. except that for the indicator variable for the 21-30 age range. all coefficients are significantly different from zero.97713 Thus. on average. we need q for married black males in the 21-30 age range.49 0. Black and Hispanic individuals consume alcohol on 0. holding all else fixed.97713 0.28%.000057. (d) The coefficient of income suggests that a $1 increase in income will increase the average number of days on which alcohol is consumed by 0.045 24. Thus. If income was measured in terms of thousand-dollar units. 79. holding constant both the size and age of the house. Exercise Solutions. The estimated coefficient for AGE implies the house price is $179 less for each year the house is older. other things being equal. Thus. The coefficient is significantly different from zero. the value of the constant term.Chapter 7. . Exact collinearity would cause least squares estimation to fail. holding all other factors fixed.4 (a) The estimated coefficient for SQFT suggests that an additional square foot of floor space will increase the price of the house by $72. The collinearity arises between the dummy variables and the constant term because the sum of the dummy variables equals 1. The positive sign is as expected. The negative sign implies older houses cost less. and the estimated coefficient is significantly different from zero. (b) The estimated coefficients for the indicator variables are all negative and they become increasingly negative as we move from D92 to D96. house prices have been steadily declining in Stockton over the period 1991-96. (c) Including a indicator variable for 1991 would have introduced exact collinearity unless the intercept was omitted. Principles of Econometrics. 4e 230 EXERCISE 7. we have 2 dx Thus. a 1 unit change in x leads to approximately a percentage change in y equal to 100 2 .00104 0.01899 POOL 0. 4e 231 EXERCISE 7.0264 0.0359 0.8619 0.00510 In the log-linear functional form ln( y ) dy 1 dx y 2 1 dy y or R2 0. and the slope-indicator variable SQFT_UTOWN are significantly different from zero at the 5% level of significance. A house which is a year older is estimated to sell for 0.5 (a) The model to estimate is ln PRICE + 1 3 UTOWN 2 SQFT POOL 3 FPLACE e 1 AGE 2 SQFT UTOWN The estimated equation.0904% less.000904 AGE 0.6% for a house not in University town and 3.4638 0.03596 SQFT 0.000218 (b) 0. Principles of Econometrics. with standard errors in parentheses.006556 FPLACE 0.003428 SQFT UTOWN 0. AGE.3334UTOWN 0. Exercise Solutions. we estimate that an additional 100 square feet of floor space is estimated to increase price by 3.25% for a house in University town. is ln PRICE (se) 4. The estimated coefficients of UTOWN.001414 0. holding all else constant. holding all else fixed.Chapter 7. In this case PRICE 1 SQFT PRICE PRICE 1 AGE PRICE 2 UTOWN 3 Using this result for the coefficients of SQFT and AGE. .004140 2 x e. ln( PRICE fireplace ) ln( PRICEnofireplace ) 100 3 100 % PRICE An approximation of the percentage change in price due to the presence of a fireplace is 0. (e) In this case the difference in log-prices is given by ln PRICEutown SQFT 25 ln PRICEnoutown SQFT 25 0. ln( PRICE pool ) ln( PRICEnopool ) 100 2 100 % PRICE An approximation of the percentage change in price due to the presence of a pool is 1. (d) From Section 7.90%. Exercise Solutions.3334 0.003428 25 UTOWN 0. is e0.3.5 (continued) (c) Using the results in Section 7.11% .1.2. 4e 232 Exercise 7. Using the results in Section 7. Principles of Econometrics.66%.Chapter 7.3.3334UTOWN 0.003428 25 0.66%.3. PRICE pool PRICEnopool PRICEnopool 100 e 2 1 100 The exact percentage change in price due to the presence of a pool is estimated to be 1.1. for a 2500 square-feet home.2.92%. PRICE fireplace PRICEnofireplace PRICEnofireplace 100 e 3 1 100 The exact percentage change in price due to the presence of a fireplace is also 0. From Section 7.3.2477 1 100 28.2477 and the percentage change in price attributable to being near the university. Furthermore.4313DISPAD 0.Chapter 7. a one-unit increase in the price of Brand 1 is estimated to lead to a 375% decline in sales.8428 0.7463 APR1 1. the coefficients are interpreted as proportional changes in quantity from a 1-unit change in price. The sign of 2 is negative. For example.15%.5765 0.6053 R2 1. Exercise Solutions.1495 APR2 1. These signs imply that Brands 2 and 3 are substitutes for Brand 1. (i) No display and no advertisement SAL11 exp 1 2 APR1 3 APR2 4 APR3 Q APR1 3 APR2 4 APR3 5 Q exp 5 APR1 3 APR2 4 APR3 6 Q exp 6 (ii) A display but no advertisement SAL12 exp 1 2 (iii) A display and an advertisement SAL13 exp 1 2 The estimated percentage increase in sales from a display but no advertisement is Q exp{b5 } Q 100 (e 0.8% Q SAL11 The estimated percentage increase in sales from a display and an advertisement is SAL12 SAL11 SAL13 SAL11 SAL11 100 100 Q exp{b6 } Q 100 (e1.6464 0. 4e 233 EXERCISE 7.75% and 1.288 APR3 0. If the price of Brand 1 rises.4313 1) 100 318% Q The signs and relative magnitudes of b5 and b6 lead to results consistent with economic logic. (c) There are three situations that are of interest. while the signs of the other two coefficients are positive. If we wish to consider a 1 cent change in price – a change more realistic than a 1-dollar change – then the percentages 375 and 115 become 3. .1052 0. with the log-linear function. but a price rise for Brand 2 or 3 will increase sales of Brand 1.4237 1) 100 52. a one-unit increase in the price of Brand 2 is estimated to lead to a 115% increase in sales. 3 and 4 are all significant and have the expected signs.9848 3.4237 DISP (se) 0. holding all else fixed. A display increases sales.6 (a) The estimated equation is ln SAL1 8. These percentages are large because prices are measured in dollar units. a display and an advertisement increase sales by an even larger amount. respectively. then sales of Brand 1 will fall.1562 (b) The estimates of 2 .4486 0. Principles of Econometrics. Part (i) (ii) (iii) (iv) (e) H0 =0 6=0 5 5 6 = 6 5 =0 Test Value Degrees of Freedom 5% Critical Value Decision t = 4.03 t = 9.20 1.6 (continued) (d) The results of these tests appear in the table below. .Chapter 7.01 Reject H0 Reject H0 F = 42.68 Reject H0 Reject H0 The test results suggest that both a store display and a newspaper advertisement will increase sales.86 (2. 4e 234 Exercise 7.46) 46 3.0 t = 6.17 46 46 2. Exercise Solutions. and that both forms of advertising will increase sales by more than a store display by itself. Principles of Econometrics.01 2. 0127) (0. it is more likely that the borrower will face delinquency.02364) (0.00078) (0.7 (a) The estimated regression is E ( DELINQUENT ) 0. .Chapter 7. RATE. RATE: A higher interest rate of the mortgage will result in a higher probability of delinquency. Principles of Econometrics.0319) The explanatory variables with the positive signs are LVR. On the other hand.00020) (0.00354) (0. holding all else fixed. which leads to a higher probability of delinquency. CREDIT: A borrower with a higher credit rate will have a lower probability of delinquency. 4e 235 EXERCISE 7. reducing the probability of delinquency. perhaps indicating financial stress. AMOUNT: As the amount of mortgage gets larger.0238) (0.1283 ARM (0. it is less likely that the borrower faces delinquency.00162 LVR 0. and these signs are as expected because: LVR: A higher ratio of the amount of loan to the value of the property will lead to a higher probability of delinquency.0344 RATE (se) (0. the interest rate may rise above what the borrower is able to repay. the higher credit rate is earned by borrowers who have a good track record of paying pack loans and debts in a timely fashion. A longer term means lower monthly payments which are easier to fit into a budget. the magnitude of the estimated coefficient is unreasonably large. However. INSUR: Taking insurance is an indication that borrower is more reliable. CREDIT and TERM. the explanatory variables with the negative signs are REF. and these signs are also as expected because: REF: Refinancing the loan is usually done to make repayments easier to manage. TERM: As the term of the mortgage gets longer.01262TERM 0.0238 AMOUNT 0. The higher the ratio the less the borrower has put as a down payment. After all.2115) (0.6885 0.0593REF 0. which has a negative impacts upon the loan delinquency. Exercise Solutions. Lenders target higher risk borrowers and charge a higher rate as a risk premium. INSUR.00044CREDIT 0.4816 INSUR 0.0086) 0. ARM: With the adjustable rate. AMOUNT and ARM. 01262 30 0. This is an extremely large effect. if CREDIT increases by 50 points. We wonder if INSUR has captured some omitted explanatory variable and thus has an inflated coefficient. (c) The predicted value of DELINQUENT at the 1000th observation is E ( DELINQUENT ) 0. the predicted values of 135 observations were less than zero but none of the observations had its predicted value greater than 1.6885 0.2 0.4816 0 0. the 1000th borrower was not in fact delinquent. (d) Out of the 1000 observations.00162 88. Thus.Chapter 7.00044.5785. If a borrower is insured.4816.022.00044 suggesting an increase in the credit score by one point decreases the probability of missing at least three payments by 0.5785 the exact calculation using software This suggests that the probability that the last observation (an individual) misses at least three payments is 0.0238 2.910 0.5.00044 624 0. . we estimate that the probability of their having a delinquent payment falls by 0. Despite the fact that this predicted probability is greater than 0. Principles of Econometrics.1283 1 = 0. The estimated coefficient of CREDIT is 0. This is problematic because we cannot have a negative probability.4816.0593 1 0. the estimated probability of delinquency decreases by 0. Exercise Solutions.650 0.7 (continued) (b) The coefficient estimate for INSUR is 0.0344 7. 4e 236 Exercise 7. 37 16..12 $56. based on lost revenue from 14. 1=march 2003. the damaged motel and the competitor had similar occupancy rates.49 16.12 Therefore. 40 60 80 100 (a) 0 5 10 15 20 month.11 63.37 The difference is MOTEL1 COMP1 66. 25=march 2005 percentage motel occupancy 25 percentage competitors occupancy The graphical evidence suggests that the damaged motel had the higher occupancy rate before the repair period.23 * MOTEL1 MOTEL1 80.35 62. The average occupancy rates during the repair period: MOTEL1 COMP1 66.86 .Chapter 7.86 80.35 62.39 .37 2. (b) The average occupancy rates during the non-repair period: MOTEL 0 COMP 0 79.49 The difference is MOTEL1 COMP1 79.11 63. During the repair period.. Principles of Econometrics. 215 14.61 $171.8 The line plots of variables against TIME.12 rooms.835. Exercise Solutions. 4e 237 EXERCISE 7.. The reference lines are a TIME = 17 and TIME = 23.12% × 100 = 14.23 66.74 The estimate of lost occupancy is computed as follows: * MOTEL1 63. the estimated amount of revenue lost is.11 14. 97 . D and E.12%. Point C is inferred under the “common trend” assumption. on average. the estimated reduction in the occupancy rate is 10. Principles of Econometrics.6326COMP _ PCT 106. The significance test suggests that the estimate is significant both at the one and five percent levels.7561 0.23% 66. a one unit increase in the relative price of the damaged motel and its competitor decreases the occupancy rate of the damaged motel by 107%.192) b2 0.35% .37% . D = MOTEL1 66. 4e 238 Exercise 7.6326 . . Exercise Solutions. * Point A = COMP 0 62.23% is an estimate of what occupancy rate would have been in the absence of the damage. (d) The estimated model is MOTEL _ PCT (se) 120. which is too large to be relevant.194) (49. C = MOTEL1 80.378) (4.11% . E = COMP1 63. A one-unit change is a change in relative price of 100%.7%. B = MOTEL0 79. on average.9659 RELPRICE 18.11% = 14. Holding other variables constant. The significance test suggests that the estimate is significant at the five percent level but not at the one percent level.Chapter 7.49% . Loss = 80. This implies that holding other variables constant.8 (continued) (c) In the figure below we observe Points A and B. If the relative price increases by only 10%. b3 106.735) (0. a one percentage increase in the competitor’s occupancy rate is estimated to increase the damaged motel’s occupancy rate by 0.63 percent.1441REPAIR (45. 61 b4 ) 220784. 114722) The simple estimate from part (b) is within this interval estimate.Chapter 7. with a p-value of 0.08 51025.834. The significance test suggests that the estimate is significant at the one percent level. (g) The graph below depicts the least square residuals over time.54. The simple estimate of the revenue loss calculated in part (b) is $171. or 18.61 18.21) se(215 56. the occupancy rate of the damaged motel when it is under repair is 18. The 95% interval estimate for the estimated loss is calculated as follows: 215 56. (e) The expected revenue loss is computed as 215 $56. Exercise Solutions.. Holding other variables constant.14 percent less than when it is not under repair.14 $220. 1=march 2003.04 ( 326947 ..835. Testing for serial correlation is delayed until Chapter 9.61 b2 t(0.8(d) (continued) b4 18.. . There is no evidence from this RESET to suggest the model in part (c) is misspecified.39 . The RESET value with three terms is 0. 4e 239 Exercise 7.144 .4 . -10 0 Residuals 10 20 (f) 0 5 10 15 20 month. Principles of Econometrics.6601. 25=march 2005 25 The residuals trend down a little over time.975. on average.14 rooms per day. This calculation is based on the 18.66 2.14% decline in the occupancy of a 100 unit motel. 315*** (3.280 BIC 66169.233*** (2.314 30777099.668*** (2.837*** (2.807 65407.750** (2.0429 regular sized class with aide = 918.9 (a) The estimated average test scores are regular sized class with no aide = 918.002 1.970 SSE 31232400.409) (2.782 (2.211) TCHWHITE 16.006*** (2.114*** 1. .896*** 13.121) (3.746*** (2.940) (4.720*** (0.101 0.780) -7.395) (2.167) 25.287 28203498.310) (2.510) -1.469*** (0. ** p<0.250*** 931.209) (2.721*** 923. Exercise Solutions. *** p<0. The results of the estimated models for parts (b)-(g) are summarized in the following table.05.01 . the average scores are higher with the small class than the regular class.167) BOY FREELUNCH WHITE_ASIAN 1. -------------------------------------------------------------------------------------------N adj.126) (2.3568 small class = 931.532*** -34.104 0.858) SCHRURAL -7.559) (0. Principles of Econometrics.006*** 13.980*** 15.641) (2.481*** (2.166) -14.899*** 14.538 (3.Chapter 7.846) (1.019) SCHURBAN -5.272 65418.755*** 918.294) (2.096) 0. R-sq 5786 5766 5766 5766 5766 0.965 28089837.184) .314 -0. The effect of having a teacher aide is negligible. 4e 240 EXERCISE 7.947 22271314.843) (1.601 0. Exercise 7-9 -------------------------------------------------------------------------------------------(1) (2) (3) (4) (5) (b) (c) (d) (e) (g) -------------------------------------------------------------------------------------------C SMALL AIDE 918.9419 From the above figures.217) (2. .626 64062.302) 0.121*** (1.698 1.842) TCHMASTERS -3.161) (0.228) (3.025) TCHEXPER 1.156*** (2.284) -2.500 65884.272*** (1.955 -------------------------------------------------------------------------------------------Standard errors in parentheses * p<0.007 0.045*** -14.043*** 904.621 (2. .306) (2.020 0.117*** -32.011) 11.662) -34.008*** -12.064) (2.357) 13.10.560* (2. boys score 14.01 level of significance.84 points higher. (c) The estimated regression after including TCHEXPER is in column (2) above. The estimates of the school effects themselves are suppressed.11 points lower than those who do not.15. (d) and (e) suggest that while some additional variables were found to have a significant impact on total scores. FREELUNCH and WHITE_ASIAN is in column (3) of the Table above. The coefficient of SMALL is the difference between the average of the scores in the regular sized classes (918. The inclusion of these variables has little impact on the coefficients of SMALL and AIDE. However AIDE’s coefficient is not significantly different from zero and this change is of negligible magnitude. Similarly the coefficient of AIDE is the difference between the average score in classes with an aide and regular classes. The t-test of significance of 3 is t b3 se(b3 ) 0.94). so the sign change is not important. that students receiving a free lunch score 34. The inclusion of these variables has only a very small and negligible effect on the estimated coefficients of AIDE and SMALL. the estimated advantage of being in small classes. The fact that the estimates of the key coefficients did not change is support for the randomization of student assignments to the different class sizes. . We estimate that. and that white and/or Asian students score 11. Principles of Econometrics.136 The critical value at the 5% significance level is 1.05 points lower than girls.9 (continued) (b) The estimated regression results are in column (1) of the Table above.0429 = 13. The regression result suggests that TCHWHITE. (d) The estimated regression after including BOY.9419 918.310 0.28. SCHRURAL and SCHURBAN are significant at the 5% level and TCHMASTERS is significant at the 10% level. holding all of the factors constant. The variables themselves are statistically significant at the 0. (e) The estimated regression after including the additional four variables is in column (4) of the Table above. We cannot conclude that there is a significant difference between test scores in a regular class and a class with an aide. We find that inclusion of the school effects increases the estimates of the benefits of small classes and the presence of a teacher aide. and the insignificance of the presence of a teacher aide. The addition or deletion of uncorrelated factors does not affect the estimated effect of the key variables. Exercise Solutions. thus we reject the null hypothesis that all the school effects are zero. The variables SCHURBAN and SCHRURAL drop out of this model because they are exactly collinear with the included 78 indicator variables. The F-test of the joint significance of the school indicators is 19. The tstatistic for its significance is 8. That is b2 = 931. 4e 241 Exercise 7.899. although the latter effect is still insignificant statistically.Chapter 7. and the coefficient of AIDE has gone from positive to negative. The inclusion of this variable has a small impact on the coefficient of SMALL. (g) The estimated model including school fixed effects is in column (5) of the Table above. and conclude that at least some are not zero.96.78 and we reject the null hypothesis that a teacher’s experience has no effect on total test scores. is unaffected. (f) The results found in parts (c).314 2. The 5% F-critical value for 78 numerator and 5679 denominator degrees of freedom is 1.36) and the average of the scores in small classes (931. There are. assuming that the common trend assumption is valid.8914 12. We see that in the absence of the law. The line segment BE represents what would have happened to LNUNITS in the absence of the law. For LNPRICE the line segment AD represents what happens in cities without the law. For LNUNITS the diagram follows. 4e 242 EXERCISE 7.8992 9.9950 9. Exercise Solutions.4176 0 The diagrams for LNUNITS and LNPRICE are on the following page.01 The approximate percentage differences in the price and units for cities with and without the law are 60.01% respectively. IZLAW 1 IZLAW Pct.63% and 45. The line segment BE represents what would have happened to LNPRICE in the absence of the law. assuming that the common trend assumption is valid. 0 LNPRICE 12.63 LNUNITS 9.2851 60. as well as the percentage differences using only the data for 2000.5449 45. we estimate that the average price of units would have been smaller. The line segment BC represents what happened in cities with the law. it suggests that the law failed to achieve its objective of making housing more affordable.3383 12. Principles of Econometrics. The line segment AD represents what happens in cities without the law. (b) The sample means of LNPRICE and LNUNITS before the year 1990 are IZLAW 1 IZLAW LNPRICE 12.0646 LNUNITS 9. using the approximation 100 ln y1 ln y0 % y . . more units available in cities with the law.10 (a) The table below displays the sample means of LNPRICE and LNUNITS. however. Diff. We see that in the absence of the law. Since the average price is higher under the law. The line segment BC represents what happened in cities with the law. we estimate that the number of units would have actually been larger.Chapter 7. 4e Exercise 7.10(b) (continued) 243 . Principles of Econometrics. Exercise Solutions.Chapter 7. 141) (0.506 176.790) D 0.039*** (0.109 0.296) LPOP 0.515* (0.333** 0.065*** -1.300*** 0.150*** -0. 4e Exercise 7.124 367.059) (0.194*** (0.398) (0.589*** (0.498 -----------------------------------------------------------Standard errors in parentheses * p<0.038) (0. Principles of Econometrics.050) IZLAW_D 0. Discussion follows Exercise 7-10 LNPRICE -----------------------------------------------------------(1) (2) (3) (c) (d) (e) -----------------------------------------------------------C 12. *** p<0.694 0. ** p<0.781 BIC 1026.103 SSE 181.100) (0.610*** 5.221*** -0.940*** (0.070) LMEDHHINC 1.029) (0.Chapter 7.182*** 0.439 44.083) (0.518*** (0.238*** 0.033) (0.126) PROPPOVERTY -0.058 (0.011) -----------------------------------------------------------N 622 622 622 adj.046) (0.147*** (0.10.032) IZLAW 0.074) EDUCATTAIN 1. Exercise Solutions.274*** 0.05.10 (continued) The regressions for parts (c)-(e) are summarized in the following tables.01 244 .891 62. R-sq 0. Principles of Econometrics.087) (0.021 0.020 0.980 BIC 1732.176) (0. Exercise Solutions. For the LNUNITS equation the effect carries a negative sign.176) (0.404) D 0.846 565.039 1738.249) (0. which is represented in the table as IZLAW_D.006) -----------------------------------------------------------N 622 622 622 adj. ** p<0.343*** (0.007 (0.116 0.038) EDUCATTAIN 1.023*** (0.114) (0.479*** 0. The treatment effect is estimated by the coefficient of D IZLAW . To summarize. which is opposite the direction we expect.034 -0.10.077*** (0.199) (0.057) (1. contrary to the intention of the policy.249) (0.482*** 0.081) (0. . so that its sign should not be interpreted (t = 0.151) LPOP 0.2] and this effect is statistically significant at the 5% level (t = 2.418*** 9.998*** (0.764*** (0.036) LMEDHHINC 0. R-sq 0.13).016) IZLAW 0.01 (c) See column (1) in each of the above tables.026) IZLAW_D -0.5% using the exact calculation of Chapter 7. In the LNPRICE equation we estimate that the result of the law was to increase prices by about 33.Chapter 7.559 SSE 565. *** p<0.352 -658.3% [39.10 (continued) Exercise 7-10 LNUNITS -----------------------------------------------------------(1) (2) (3) (c) (d) (e) -----------------------------------------------------------C 9.620*** (0.027 (0.031 -0. but the coefficient is not statistically different from zero.737 11. 4e 245 Exercise 7.3. these models suggest that the policy effect is to increase prices but not to increase the number of housing units.05.064) PROPPOVERTY -2.630 -----------------------------------------------------------Standard errors in parentheses * p<0.35).005*** 14.039 -0.127 0. 10 (continued) (d) See column (2) in each of the above tables. an increase in the percent of the population with a college degree increases by 0.515%. Exercise Solutions. holding other variables constant.998% (or about 1%) increase in housing units. A one-unit change of a proportion is very large. we estimate that an increase in the proportion of the population holding a college degree will increase prices by a statistically significant amount. . Principles of Econometrics. Again this effect is strongly significant. an increase in the proportion of the population in poverty decreases house prices by a statistically significant amount.3%.343 percent. an increase of the proportion living in poverty of 0.01. There is no evidence that the policy increased the number of housing units. In the LNPRICE equation the effects are: EDUCATTAIN: Holding all else constant.4%. or 1%. The treatment effect is statistically significant at the 1% level.3 percent. The inclusion of this control variable reduces the magnitude of the estimated treatment effect to approximately 28. LPOP: Holding all else constant. we estimate that house prices will fall by 0. with a t-value of 2. The inclusion of these control variables does not alter the insignificance of the treatment effect.94% PROPOVERTY: Holding all else constant. The addition of these additional controls slightly reduces the estimated treatment effect to 19. In the LNUNITS equation the effects are: EDUCATTAIN: We estimate. and this effect is significant at the 1% level. This effect is statistically significant with a t-value of 34.Chapter 7. that holding other factors fixed. (e) See column (3) in the above tables. we estimate that a 1% increase in population is associated with a 0. which is significant at the 1% level. PROPOVERTY: We estimate that holding other factors constant. LPOP: Holding all else constant. an increase in the population of 1% is estimated to increase house prices by 0. or 1%.039 percent.01. 4e 246 Exercise 7. If there is an increase in the proportion by 0. is associated with a decrease of housing units of 2. If the poverty rate increases by 0.87.01. In the LNPRICE equation. In the LNUNITS equation the median income variable is not statistically significant and the estimate of the treatment effect remains statistically insignificant. or 1%. This effect is statistically significant at the 1% level. the number of housing units will increase by 1. we estimate that a one percent increase in the households’ median income increases the price of housing by 1.01. the estimated increase in house prices is 1. The treatment remains statistically significant at the 1% level.62%.36. or 1%. One conjecture is that the law reduces the profitability of builders and thus actually may reduce the supply of homes.000. the developer is required to provide some at a much lower price.4% (the low estimate). to housing in cities before the policy change. the percent of the population living in poverty. The policy. there does in fact appear that there has been an increase in average price resulting from the policy change. the level of educational attainment. and the population size. if the average price of homes in a development is $900. Exercise Solutions. which control for median income. from the model providing the low estimate.3% (the high estimate) and 19. A 95% interval estimate of the effect on prices. but it has failed based on an analysis of the data. That is.6% to 33. is 5. Indeed. Using an array of models. after the policy change was implemented in some cities.10 (continued) (f) California’s Inclusionary Zoning policies are designed to increase the supply of affordable housing. we find that there has been no significant increase in the number of housing units attributable to the policy change.Chapter 7. requires developers to provide a percentage of homes in new developments at below market price. the data show that the number of housing units in cities in which the policy was implemented has increased less than in cities in which the policy was not implemented. Principles of Econometrics. However. 4e 247 Exercise 7. The policy has a noble intention. which is implemented in some California cities. we estimate the increase in average house price due to the law change to be between 33. .2%. Comparing housing in cities across California in 2000. 4e 248 EXERCISE 7. equation (A.1762) 0.4815IZLAW (0. we 100 ln y percentage change in y . approximately.2205 0.0314075 IZLAW (se) (0.0812) 0. .1273 0. approximately.0466) The estimated differences-in-differences regression is LNPRICE 12.11 Note: In the following question the interpretation of coefficient estimates is based on the characteristics of changes in logarithms of variables. If there is an indicator variable D on the right-hand side. In Appendix A.3326323IZLAW (se) (0.0152) (0.2205D (se) (0.2737 IZLAW (0.0646 + 0.3326323 IZLAW D (0. though standard errors are different due to estimation with different numbers of observations.Chapter 7. in a regression note that 100 ln y1 ln y0 equation ln y 1 2 ln x 100 ln y 100 1 2 100 ln x A percentage change in x is associated with a 2 percent change in y.0314075 IZLAW D (0. Thus.4176 + 0.3).0999) 0.0574) (0.1413) Note that the estimate of the treatment effect is the same in both equations.4602) 0. (a) The estimated regression for price is DLNPRICE 0. The estimated regression for changes in LNUNITS is DLNUNITS 0.0325) (0. Exercise Solutions. Principles of Econometrics. then ln y 1 D 100 ln y 100 1 100 D The effect of the indicator variable is 100 % change in y.2492) The estimate of treatment effects are the same as the treatment effects from the differences-in-differences regression though the standard errors are different.1273D (0.0366) And for LNUNITS LNUNITS (se) 9.0119) (0. The least squares estimator of is N ˆ i 1 yi y di N 2 di i 1 where y 1 N N d d yi . abbreviating Treatment. . The second term in the last line of (1) is N1 y N1 N N1 N1 yta N N i 1 yia N1 ytb yib N 0 yca y N i 1 i d N1 N N i 1 N 0 ycb N1 y and di yia yib N1 N1 yta N 1 di ytb yia N 0 yca yib ycb . i 1 N0 N1 N . 4e 249 Exercise 7. and with di being the treatment variable. for example. with a denoting After and b denoting Before.24) is yi 3 where yi yia yib . Exercise Solutions. Working then with the numerator of the expression we have N yi i 1 y di N yi i 1 N y di N N d y N i 1 N y yi i 1 yi y d y di (1) di i 1 N where we have used the fact that yi i 1 N y di ydi i 1 yi di i 1 yi i 1 N yi di i 1 N d i 1 yi yia yib d i 0 . N i 1 yia di is the sum of the outcome variable only for the treated group. effect is ˆ di error .Chapter 7. Using the differenced data. Principles of Econometrics. We can simplify the first term in the y last line of (1) as N i 1 N N1 N1 yta N yi d i i 1 i 1 yia d i N1 N1 ytb N N1 i 1 N i 1 N yia d i i 1 yib d i yib d i (2) N1 N1 yta ytb The last line arises from the fact that. i 1. Control. where N1 is the number receiving From Appendix 7B the denominator is treatment and N 0 is the number in the control group.18) we see that the differences-in-differences estimator of the treatment yta yca ytb ycb . Before and After. where di = 1.11 (continued) (b) From equation (7. the regression (7. N . The estimate of the treatment effect falls from 33.6157 DLMEDHHINC (0. the presence of the law is associated with 7.Chapter 7.2397 IZLAW 1. (c) The estimated regression for price is DLNPRICE (se) 0.2801DLMEDHHINC (0. We estimate that.77. holding all else constant. 4e 250 Exercise 7.6% fewer housing units being available. Exercise Solutions.0480 0. This estimate is statistically very significant with a t-value of 10.26% to 23. The denominator is. N0 N1 N .0384) (0.97%. we estimate that one percent growth in the median household income between 1990 and 2000 increases housing price by 1.0415) (0. so that ˆ yta ytb yca ycb This is exactly the differences-in-differences estimator.1439 0. one percent growth in median household income between 1990 and 2000 is associated with an increase of 0.09.1268) The interpretation of the coefficient estimate for DLMEDHHINC is: Holding other factors constant.0761IZLAW 0. The estimated regression for units is DLNUNITS (se) 0.0331) (0. but the estimate remains statistically significant with a t-value of 5.28 percent. The coefficient of IZLAW is negative and now statistically significant at the 5% level. The last line of (3) is the numerator of ˆ . Principles of Econometrics.11(b) (continued) Then expression (1) becomes N i 1 yi di N1 N yta N y ytb N di N1 yta N12 yta N ytb i 1 N12 N ytb N1 N N yta ytb N1 N N N1 ytb yca N1 N 0 N yta N1 N 0 yca N N1 N 0 yca N yta N1 N1 yta N ytb ytb N 0 yca ycb (3) ycb N1 N 0 yca N ycb ycb ycb where in the last line we have used the fact that N N1 N0 .62 percent increase in the number of housing units. already noted.1094) The interpretation of the coefficient estimate for DLMEDHHINC is: Holding other factors constant. .0358) (0. 0481) (0.63. a 1% increase in the proportion of people below the poverty level between 1990 and 2000 is associated with a decrease in housing prices by 0. We now estimate that a 1% increase in median income is associated with a 0.0148) (0. DLPOP: Holding other variables constant. This estimate is statistically significant with a t-value of 4.Chapter 7. This estimate is not statistically significant from zero. we estimate that a 1% increase in the size of the population between 1990 and 2000 is associated with an increase in the housing supply by 0.24%. DPROPPOVERTY and DLPOP: DEDUCATION: Holding other factors constant.8489 DLPOP (0. DEDUCATTAIN.2448 DLPOP (0.85%. This estimate is not statistically significant.1896 IZLAW 1. DEDUCATTAIN. This estimate is significantly different from zero at the 1% level. DPROPPOVERTY: Holding other factors constant.0456) +0.1494 0. a 1% increase in the size of population between 1990 and 2000 is associated with a decrease in housing prices by 0.325%. 4e 251 Exercise 7.48.05. with a t-value of 52. but the sign is difficult to rationalize.0372 DLMEDHHINC (0.5609) (0.0115) (0. Principles of Econometrics.3238 DPROPPOVERTY 0. This estimate is significant at the 1% level.0223IZLAW 0.0163) First note that the effect of the law passage is associated with a numerically smaller fall in the number of housing units available of 2. with a t-value of 6.0371) (0. we estimate that a 1% increase in the proportion of people below the poverty level between 1990 and 2000 is associated with a decrease in the housing supply by 0.0528) Interpretation of new variables. The estimated regression for units is DLNUNITS (se) 0. DLPOP: Holding other factors constant. This estimate is very significant.0424% increase in the number of housing units. DPROPPOVERTY: Holding other factors constant.32%.187%. but this estimate is not statistically significant.0564) (0.2%.11 (continued) (d) The estimated regression for price is DLNPRICE (se) 0.0640 0. but the effect is still statistically significant at close to the 5% level.3251DEDUCATTAIN 0. DPROPPOVERTY and DLPOP: DEDUCATION: Holding other factors constant.0424 DLMEDHHINC (0.1841DEDUCATTAIN 0.18%. Interpretation of new variables.1478) 1. a 1% increase in the proportion of people with a college education between 1990 and 2000 is associated with an increase in the housing price by 1. we estimate that a 1% increase in the proportion of people with a college education between 1990 and 2000 is associated with an increase in the housing supply by 0. .1731) (0.1873DPROPPOVERTY 0.1828) (0. Exercise Solutions. 0331 2 0. 2.2044 FULLTIME 0.14% less than their male counterparts. FEMALE – We estimate that. or.0059)*** 0.0000835)*** (0. These estimated changes are all statistically different from zero. 5. holding all else constant. That more educated workers earn significantly higher salaries may occur because of their accumulated human capital. and it is the base from which all our indicator variables are measured.0346) 0. compared to a worker with long years of experience whose productivity changes little as experience is accumulated. Exercise Solutions.0331) (0.05% increase in hourly wages.3 The “life-cycle” effect of experience on earnings reflects the additional productivity that less experienced workers receive from additional experience.12 (a) The estimated regression is ln WAGE (se) 0.2014 FEMALE (0. EDUC – We estimate that an increase in education by one year is associated with an approximate 9. because smarter people stay in school longer.83% and 1.24%. This estimate is significantly different from zero at a 1% level of significance. Its presence facilitates predictions and is present for mathematical completeness.Chapter 7. 4e 252 EXERCISE 7.03315 2 0. Principles of Econometrics.975. This estimate is statistically different from 0 at the 1% level. 3.7% respectively.8%. For workers with 1. approximately.0004973 EXPER . holding all else constant. The turning point in the relationship occurs at EXPER * bEXPER 2bEXPER 2 0.0512) (0.0460)*** (0.0318)*** 0. Discrimination in the workplace is reflected in these lower wages. and smarter workers earn higher salaries. Using the exact calculation.0301MARRIED 0. Considering the variables individually: The intercept estimate cannot be reliably interpreted in this equation. .1039) (0.0158SOUTH (0.0905 EDUC 0.0048)*** (0.000497 32.990) 1.1713METRO (0. EXPER and EXPER2 – The marginal effect of another year of experience is estimated to be 0. females earn approximately 20.0377)*** The 5% critical t-value for testing the significance of the coefficients and for other hypothesis tests is tc t(0.000497 EXPER 2 0.962 . 0. the difference is 18.9561 0. 25 and 50 years of experience these marginal effects are estimated to be.1191BLACK ** (0.2%. perhaps.0331EXPER 0. so using these data there is no significant evidence that married workers earn more.01% higher than those who are not married. Discrimination in the workplace is reflected in these lower wages. Full time workers tend to have more specialized training and more education as well.15). MARRIED – We estimate that wages for married workers are 3. This estimate is significant at the 1% level. holding all else constant. so the effect may not be estimated precisely.44% (22. holding all else equal.58% less than their non-southern counterparts. SOUTH – We estimate that wages for southerners are 1. These data are from the 2008 CPS (see Exercise 2. That wages are higher for full-time workers than part-time workers is not surprising. This estimate is statistically different from 0 at the 5% level. This estimate is not statistically different from zero.12(a) (continued) BLACK – We estimate that wages for black workers are approximately 11.Chapter 7. we cannot reject the hypothesis that southern workers do not earn less than non-southern workers. METRO – We estimate that the hourly wage for someone who lives in a metropolitan area is approximately 17. Workers in metropolitan areas have a wider variety of work opportunities resulting in higher average wages. The estimate is statistically different from zero at the 1% level. . The current sample is only 1000 observations.13% higher (18. 4e 253 Exercise 7.68% using the exact calculation) higher than it is for those who do not work full time.69% using the exact calculation) than non-metro workers.9% lower than they are for non-black workers. Exercise Solutions. Principles of Econometrics. FULLTIME – We estimate that the hourly wage for full time workers is approximately 20. This outcome is different from results in many model estimations using data from earlier periods. This estimate is not statistically significant. dat relative to the 1000 observations in cps5. 4e 254 Exercise 7.dat.033) (0. .092*** (0.032) (0.666 1057.014) BLACK -0.006) (0.190*** (0.201*** -0. ** p<0.002) EXPER^2 -0.956*** 0.091*** 0.015) SOUTH -0.030 0.10.12 (continued) (b) To facilitate comparison from using the alternative data sets we have tabled them.042*** (0.015) FULLTIME 0.000) (0. This reflects the use of a larger sample size of 4838 observations in cps4.Chapter 7.430E-3*** (0. Using a larger sample size improves the reliability of our estimated coefficients because we have more information about our regression function.000) FEMALE -0.05.336 SSE 231. The larger t-values also mean that the estimates have smaller p-values and will therefore be significantly different from zero at a smaller level of significance. *** p<0.204*** 0.047) EDUC 0.values are all much larger in magnitude for estimation from the cps4.306 0.146*** (0. Exercise Solutions.051) (0.266*** (0.119** -0.171*** 0. that the effects of being married and being a southern worker are statistically significant using cps4. Exercise 7-12 -------------------------------------------(1) (2) CPS5 CPS4 -------------------------------------------C 0.dat.029*** (0.016 -0.01 There are only slight differences in the estimated coefficient values.033*** 0. What is evident is that the t.723 -------------------------------------------Standard errors in parentheses * p<0.003) EXPER 0.020) METRO 0.906*** (0. 497E-3*** -0.038) (0.104) (0.dat data.083*** (0.023) MARRIED 0.145*** (0. We now find.dat.005) (0. whereas they were not using cps5. for example.017) -------------------------------------------N 1000 4838 adj. Principles of Econometrics.046) (0.035) (0. R-sq 0. and the signs of the coefficients are the same. 003) BLACK -5.236 0.6517 0.239 0.034) BLACK_FEMALE_SOUTH -2.4824*** -0.153) (0.021) EDUC_SOUTH -0.6400*** (0.9091 36969.9787 579789.003) (0.0977*** (0.4128 1189.5892*** 0.010) (0.1615 0.9040*** -0.555) (0.249 BIC 36931.8271 1195.0315*** 0.01 .05.3000*** (0.0950*** 2.6721*** -0.0075 (0.2764 0.5658 0. 4e 255 EXERCISE 7.1386*** -0.2589*** -5.006) BLACK_SOUTH 1.2299 7011.1053*** 0.8406 -0.071) (0.1203 (2.0254 (0.6046 SSE 577188.768) (0.071) SOUTH -0.018) BLACK_FEM 4.893) (0.465) (0.388) (0.058) (0.3191*** -0.9545 7049.8691*** 1.253 0.2147*** 6.021) WEST 0. Exercise 7-13 ---------------------------------------------------------------------------(1) (2) (3) (4) (a) (b) (c) (d) ---------------------------------------------------------------------------C -4.123) (0.2463*** -5.755) (0.015) (0.5431*** 1.790) (0. Principles of Econometrics.0878 ---------------------------------------------------------------------------Standard errors in parentheses * p<0.041) (1.0460** 2.333) (0.Chapter 7.0212 (0.073) FEMALE_SOUTH 0.0934 (1.2642*** (0.597) (0.145) (0.080) MIDWEST -1.048) (1.0724*** (0. R-sq 0.1055*** 0.2077* -0.2800*** (1.048) (0.052) FEMALE -5.020) (1. *** p<0. ** p<0.0612 (0.036) (1.097) ---------------------------------------------------------------------------N 4838 4838 4838 4838 adj.13 The regressions for parts (a) – (d) are summarized in the following table.451) (0.046) EDUC 2.8266* -0.6894*** -5.10. Exercise Solutions.465) (0. we reject the null hypothesis at the 5% level and conclude the regional effect is significant in determining the wage level.2404 (0.0578 0.3191FEMALE (se) 0.4653) 0.8266SOUTH 1.607.6721MIDWEST 1. Principles of Econometrics. F SSE R SSEU J SSEU N K 580544.5658WEST (0.4648) (i) To test whether there is interaction between BLACK and FEMALE. Since this value is less than 0.1386 BLACK 5. against the alternative that it is not zero.38 with a p-value of 0. MIDWEST and WEST are jointly zero. Also. The t-statistic given by the computer output is 4. the critical value at the 5% significance level is 2.05).01. against the alternative that at least one of the indicator variables’ coefficients is not zero. we test the null hypothesis that the coefficient of BLACK FEMALE is zero. Exercise Solutions. .4 3 577188.3325 4. we test that the coefficients of SOUTH.5 577188. Since the F-value is larger than the critical value (or the p-value is less than 0. (ii) To test the hypothesis that there is no regional effect.13 (continued) (a) The estimated regression with standard errors in parentheses is WAGE 4. 4e 256 Exercise 7.4510) R2 (0.0315EDUC 5.5892 BLACK FEMALE 0.5431 2. we reject the null at a 1% level of significance and we conclude that there is a significant interaction between BLACK and FEMALE. The F-value can be calculated from the restricted (regression without regional variables) and the unrestricted models.7903 0.0475 +0.000.4 (4838 8) 9.8925 0.3615 The corresponding p-value is 0.000.Chapter 7. in model (a) we estimate that each additional year of education. we test the null hypothesis that the coefficient of BLACK FEMALE is zero. against the alternative that it is not zero. The F-value can be calculated from the restricted (regression without regional variables) and the unrestricted models. (ii) To test whether there is interaction between BLACK and FEMALE.01.05).0950 EDUC 0.2540 (0. Since the F-value is larger than the critical value (or the p-value is less than 0. In part (b) we estimate that the effect of an extra year of education. is associated with an increase in the hourly wage of $2. Principles of Econometrics.03. MIDWEST and WEST are jointly zero. 4e 257 Exercise 7. The log-linear model suggests that the variable SOUTH is significant at the 5% level while in the linear model in part (a) it is significant at only the 10% level. we reject the null at a 1% level of significance and we conclude that there is a significant interaction between BLACK and FEMALE. Since this value is less than 0.0460 SOUTH 0.2147 BLACK FEMALE 0.5% increase in the hourly wage. holding all else constant. The major difference lies in the value of coefficient estimates and their respective standard errors.0405 0. In part (b) they measure an impact on ln(WAGE).0476 +0.6894 0.979 3 1189. the critical value at the 5% significance level is 2. The t-statistic given by the computer output is 4.0026 0. (iii) To test the hypothesis that there is no regional effect.000. we test that the coefficients of SOUTH.0151 0.979 (4838 8) 9.302 The corresponding p-value is 0.0724 MIDWEST (0. holding all else constant.51 with a p-value of 0.2463BLACK 0.0211) 0. is associated with approximately a 9.000. we reject the null hypothesis at the 5% level and conclude the regional effect is significant in determining the ln(WAGE) level. This is due to the nature of the linear versus the log-linear model. against the alternative that at least one of the indicator variable’s coefficients’ is not zero. Exercise Solutions.2589 FEMALE (se) 0.Chapter 7. In part (a) the estimated coefficients measure an impact on WAGE.0254WEST (0.854 1189.607.0211) (i) Comparing the results with the estimated equation in part (a). .0359 0. we find the signs of all the coefficient estimates are exactly the same. Also.13 (continued) (b) The estimated regression using ln(WAGE) as a dependent variable: ln(WAGE ) 1.0204) R2 0. F SSE R SSEU J SSEU N K 1196. For example. The F-statistic is calculated from the sum of squared residuals of restricted and unrestricted models.Chapter 7.2764 BLACK SOUTH 0.7554) 2. the critical value at the 5% significant level is 2.1053EDUC 5.05).6517 FEMALE SOUTH (1.13 (continued) (c) The estimated regression is WAGE (se) 5.3885) 6.4824 FEMALE (1.1535) (0. . Also.1450) To test the null hypothesis that the wage equation in the south is the same as the wage equation for non-southerners. and is given by F SSE R SSEU J SSEU N K 580544.1615SOUTH 0.8 5 579789.8691 2.216.8406 BLACK FEMALE SOUTH (2.2077 EDUC SOUTH (1. The alternative is that at least one these coefficients is not zero. we do not reject the null hypothesis at the 5% level and conclude that there is no significant difference between wage equations for southern and non-southern workers.5969) (0.0708) (1.5 579789.7682) (0.1535) (1.9040 BLACK 5. Exercise Solutions.0099) (0. we test the joint hypothesis that the coefficients of SOUTH and all the interaction variables with SOUTH are zero.257 The corresponding p-value is 0.1055BLACK FEMALE 2. Since the F-statistic is less than the critical value (or the p-value is greater than 0.1229) 1. 4e 258 Exercise 7.8 (4838 10) 1. Principles of Econometrics.2798. which would indicate a difference between south and nonsouth wage equations. 0212 FEMALE SOUTH (0. and is given by F SSE R SSEU J SSEU N K 1196.77% increase in the hourly wage.0934 BLACK SOUTH 0. we find the signs of all the coefficient estimates are exactly the same. Otherwise. . Principles of Econometrics. The alternative is that at least one these coefficients is not zero.0459) (0. In part (a) the estimated coefficients measure an impact on WAGE.0177) 0.2800 BLACK FEMALE 0. (ii) To test the null hypothesis that the wage equation in the south is the same as the wage equation in the non-south.13 (continued) (d) The estimated regression for the log-linear model is ln(WAGE ) 1. SOUTH and its interactions are not significantly different from zero in both models. which would indicate a difference between south and non-south wage equations. holding all else constant. holding all else constant. we do reject the null hypothesis at the 5% level and conclude that there is no significant difference between wage equations for southern and nonsouthern workers.0725) (0. the critical value at the 5% significance level is 2.05).3000 BLACK 0. The major difference lies in the value of the coefficient estimates and their respective standard errors.1203BLACK FEMALE SOUTH (0. in model (a) we estimate that each additional year of education. while in the linear relationship it is. For example. Exercise Solutions.216.0706) (0.0977 EDUC 0. In part (b) they measure an impact on ln(WAGE).0803) (0.0974) (i) Comparing the results with the estimated equation in part (a).11.0056) 0. This is due to the nature of the linear versus the log-linear model.088 5 1195.0343) 0.854 1195.088 (4838 10) 1. we test the joint hypothesis that the coefficients of SOUTH and all the interaction variables with SOUTH are zero. 4e 259 Exercise 7.0075EDUC SOUTH (0.2110.Chapter 7.0032) (0. is associated with an increase in the hourly wage of $2. The F-statistic is calculated from the sum of squared residuals of restricted and unrestricted models.6400 0.427 The corresponding p-value is 0. In the log-linear model the interaction between EDUC and SOUTH is not significant at even the 10% level.0524) (0. is associated with approximately a 9. Since the F-value is less than the critical value (or the p-value is greater than 0. In part (b) we estimate that an extra year of education.2642 FEMALE (se) (0.0612 SOUTH 0. Also. Exercise Solutions. We expect the parameter for WAR to be positive because voters were more likely to stay with the incumbent party during the World Wars.6879 The signs are as expected.2628 0.Chapter 7.3300 DURATION 2. We expect the coefficient of GROWTH to be positive because society rewards good economic growth. We expect the coefficient for PERSON to be positive because a party is usually in power for more than one term. For the same reason we expect the coefficient of GOODNEWS to be positive.9052 suggests that the model fits the data very well.6763PARTY 1.6572 INFLATION 1.4081 1. we expect the incumbent to get the majority vote for most of the elections. The sign for PARTY is as expected if one knows that the Democratic Party was in power for most of the period 1916-2004. Thus. The coefficient of WAR is statistically insignificant at a level of 5%.0749GOODNEWS 0. 4e 260 EXERCISE 7. on average.2493 3. The coefficients of INFLATION.14 (a) We expect the parameter estimate for the dummy variable PERSON to be positive because of reputation and knowledge of the incumbent. therefore we expect the parameter for DURATION to be negative. We expect the parameter estimate for WAR to be positive reflecting national feeling during and immediately after first and second world wars. (c) The estimated regression using observations for 1916-2004 is VOTE (se) 47. however. (b) The regression functions for each value of PARTY are: E VOTE | PARTY 1 1 7 5 E VOTE | PARTY 1 1 GROWTH PERSON 7 5 2 2 6 6 INFLATION DURATION GROWTH PERSON 3 3 GOODNEWS 4 GOODNEWS WAR 8 INFLATION DURATION 4 WAR 8 The intercept when there is a Democrat incumbent is 1 7 .5384 0. PERSON. However.2124 0. an R2 of 0. unpopular and/or ineffective.6149WAR 2.2914 0. All the estimates are statistically significant at a 1% level of significance except for INFLATION. Lastly.2983PERSON 3.6797GROWTH 2.1107 0.6264 5. DURATION and PERSON are statistically significant at a 5% level of significance. the effect of PARTY on the vote is 2 7 with the sign of 7 indicating whether incumbency favors Democrats ( 7 0) or Republicans ( 7 0) . When there is a Republican incumbent it is 1 7 . We expect that for each subsequent term it is more likely that the presidency will change hands. . DURATION and WAR. We expect a negative sign for the coefficient of INFLATION because increased prices impact negatively on society. Principles of Econometrics. it could be negative if the incumbent was. meaning that anything could happen.4 35.econ.815 (42. (e) A 95% confidence interval for the vote in the 2008 election is VOTE 2012 (f) t(0.6 40.5 48.9% of the popular vote to 45. as the incumbent party.091 2. and based on the estimates from part (c). We use the value of inflation 3.yale.14 (continued) (d) Using the data for 2008. Barack Obama.6 growth inflation . If there is poor economic performance.5 46.3 52. is running for election and so we set PERSON = 1.edu/vote2012/index2. then President Obama can expect to be re-elected. along with a listing of the values of the explanatory variables.22 2. 4e 261 Exercise 7.Chapter 7. Principles of Econometrics.15) se( f ) 48.0 We see that if there is good economic performance. We consider 3 scenarios for GROWTH and GOODNEWS representing good economic outcomes.5 1 -3 INFLATION 3 3 3 GOODNEWS 6 3 1 lb 45. we predict that the Republicans. though the interval estimate upper bound is greater than 50%.0% anticipating higher rates of inflation after the policy stimulus.091% of the vote.0 vote 51. In the intermediate case. The values and the prediction intervals based on regression estimates with data from 1916-2008.88 goodnews 3 person 0 duration 1 party -1 war 0 votehat 48. Exercise Solutions.09079 Thus.5 41. This prediction was correct.7%. with only modest growth and less good news. will lose the 2008 election with 48. WAR = 0. we summarize the actual and predicted vote as follows. then we predict he will lose the election. then we predict he will lose the election with the upper bound of the 95% prediction interval for a vote in his favor being only 48%. are GROWTH 3. Readers can keep up with Professor Fair’s http://fairmodel.1315 2.htm model and predictions at . with Democrat Barack Obama defeating Republican John McCain with 52.09) For the 2012 election the Democratic party will have been in power for one term and so we set DURATION = 1 and PARTY = 1. moderate and poor. the incumbent.5 ub 57.975. if there is a “double-dip” recession. Also.09. vote 46. 54. Exercise Solutions.938 0.000 is very different from the maximum price of $1. Principles of Econometrics.15 A table of selected summary statistics: Mean Median Std.098 0. 4e 262 EXERCISE 7.292.538889 18 2 3 1 0 0 130000 2186.19425 0.044455 3.000. We can see that the median price $130.8 1008.562963 0.2 2325.561539 6.912199 0.49625 0.25387 0.488889 0.15603 3.500108 0.57407 1.55344 5.93851 0.64466 60.291909 1.001976 10.537512 -0.542671 1.17963 0.599577 -0. dollars 1500000 Figure xr7.751031 1.5 1 17.07963 154863.15 Histogram of PRICE We can see from Figure xr7.024345 10 Percent 20 30 40 Variable 0 (a) 0 500000 1000000 sale price.498716 0. the measure of skewness is 6. Dev. Skewness Kurtosis AGE BATHS BEDROOMS FIREPLACE OWNER POOL PRICE SQFT TRADITIONAL 19. .Chapter 7.15 that the distribution of PRICE is positively skewed.94976 7.973148 3.064451 1.580.105585 6.709496 0. In fact.270844 122912.612067 0. Homes that are occupied rather than vacant are estimated to sell for 6.058 -.30 0.0469639 .0014059 21.00 0.019015 4.980833 .0042748 .001 .0674655 . though we should recall that the dependent variable is logarithmic.8092 . Since style is a matter of taste.070886 SQFTS | .0062145 .62 % per year.000 . The estimated effect of an increase in the number of BATHS is positive and significant. This is consistent with the notion that more bedrooms. Empty houses may also indicate sellers are more anxious for a sale because they have moved on. Err.0005179 -12.000 3.001 -.14 0.1215857 WATERFRONT | . The estimated coefficient of AGE suggests that depreciation reduces the value of the home by 0.0271425 . Again this estimate is significant at the 1% level. One would think that an amenity such as a pool would carry a positive value.80 0. The estimated effect of an increase in the number of BEDROOMS is to reduce the house price by 3.74 0.737 .0560925 .033355 3.0315812 -0.0170267 -3.0072308 -. with R 2 0. The generalized R2 value. It is reasonable that a lived-in looking home is more attractive than a vacant one. with additional baths increasing the value of the house by approximately 19%.10997 . . so this result is somewhat surprising. calculated as the squared correlation between price and its predictor.6% less.29 0.Chapter 7. Exercise Solutions.000 .890779 4.000 -.0895021 -. This estimate is significant at the 1% level.0576933 TRADITIONAL | -.1754186 ------------------------------------------------------------------------------ The estimated model fits the data well. results in smaller bedrooms which is less desirable.190119 .017746 3. This estimate is significant at the 10% level. TRADITIONAL style homes are estimated to sell for 5. holding all else fixed.000 . Principles of Econometrics.892 -.90 0. PRICE )]2 0.000 .0010875 BATHS | . 4e 263 Exercise 7.0458947 86.27 0. However the presence of a pool does increase maintenance costs and thus it is not a totally positive factor. other things being equal.022683 FIREPLACE | .0662429 .15%. indicating that an additional 100 square feet of living space.031506 . The estimated coefficient of SQFT is positive and significant.0051982 OWNER | . Interval] -------------+---------------------------------------------------------------C | 3. t P>|t| [95% Conf.0640996 .2304573 AGE | -. holding all else constant.0326445 .25 0. Std. will increase the price of the house by approximately 3%.1022864 POOL | -.0445213 .0842748 . The presence of a POOL is statistically insignificant.0299011 .0205579 9. is [corr( PRICE . holding all else fixed.1497807 .0166109 -1. it is difficult to form an a priori expectation about the sign of this factor.15 (continued) (b) The results from estimating the regression model are below: -----------------------------------------------------------------------------| Coef.0326597 BEDROOMS | -.43 0. holding all else constant.7% more. 9711*** (0. and the positive and significant estimate is as we would expect.5256 -------------------------------------------- .069) -------------------------------------------N 1080 1080 adj.0684*** (0.046) SQFTS 0.001) (0.040) WF_TRAD -0.0024 (0.0561*** -0.0043 -0.735 0.1901*** 0.4% increase in the house value is perhaps a bit high.1883*** (0. On average.001) (0.0299*** 0. The estimated 8. (c) After including the variable TRADITIONAL WATERFRONT .1654*** (0.1722** (0.018) (0.046) (0.1100 1 11. Exercise Solutions. the results from estimating the two regression models are summarized below: -------------------------------------------(1) (2) (b) (c) -------------------------------------------C 3.018) FIREPLACE 0.0873*** (0.033) (0.021) (0.017) (0. a waterfront house sells for 100 exp 0.0315* -0.0675*** 0.0313* (0.0300*** (0.Chapter 7.001) BEDROOMS -0.032) TRADITIONAL -0. 4e 264 Exercise 7.021) AGE -0. The coefficient of WATERFRONT can be used to tell us the percentage increase or decrease associated with a waterfront house.0843*** 0.018) POOL -0.9809 77.019) WATERFRONT 0. Principles of Econometrics.032) (0.9808*** 3.0062*** -0.0449** (0.736 SSE 77.15(b) (continued) A FIREPLACE is a nice amenity for a home.001) OWNER 0.017) (0.62% higher than a house that is not waterfront. R-sq 0.0061*** (0.019) (0.1100*** 0.017) BATHS 0. traditional houses on the waterfront sell for less than traditional houses elsewhere.15(c) (continued) Let ln( P0 ) be the mean log-price for a non-traditional house that is not on the waterfront. The price advantage from being on the waterfront is lost if the house is a traditional style. For traditional houses not on the waterfront: 100 exp 0. Exercise Solutions. Its estimate is significant at a 5% level of significance.98% For traditional houses on the waterfront: 100 exp 0.1654 1 17. respectively.39% For non-traditional houses on the waterfront: 100 exp 0.5% 10 The approximate percentage difference in price for traditional houses on the waterfront is [ln( PTW ) ln( P0 )] 100% ( 9 10 11 ) 100% 5. and let 9 .1722 1 5.0449 0. 4e 265 Exercise 7. The corresponding exact percentage price differences are as follows. Then the mean log-price for a traditional house not on the waterfront is ln( PT ) ln( P0 ) 9 The mean log-price for a non-traditional house on the waterfront is ln( PW ) ln( P0 ) 10 The mean log-price for a traditional house on the waterfront is ln( PTW ) ln( P0 ) 9 10 11 The approximate percentage difference in price for traditional houses not on the waterfront is [ln( PT ) ln( P0 )] 100% 100% 9 4. 10 and 11 be the coefficients of TRADITIONAL. Principles of Econometrics. The approximate proportional difference in price for houses which are both traditional and on the waterfront cannot be obtained by simply summing the traditional and waterfront effects 9 and 10 .Chapter 7.1654 0. must also be added. 11 .5% The approximate percentage difference in price for non-traditional houses on the waterfront is [ln( PW ) ln( P0 )] 100% 100% 16.17% Thus. The extra effect from both characteristics. WATERFRONT and TRADITIONAL WATERFRONT .0449 1 4.04% . the null hypothesis is rejected at a 5% level of significance.889 . and the last model is the fully interacted model. The “natural predictor” is PRICE n exp ln PRICE 1000 1000 exp(4.1062) 1. Rejecting the null indicates that the equations for traditional and non-traditional home prices are not the same.992.Chapter 7. We want to test the joint null hypotheses that the coefficients of TRADITIONAL and all its interactions are zero. We conclude that there are different regression functions for traditional and non-traditional styles.7413). On the following page four models are summarized. The restricted model is the one in which it is assumed that there is no difference between TRADITIONAL and non-traditional houses (Rest). Exercise Solutions. The F-value for this test is F Since 4.627 SSE R SSEU J SSEU ( N K ) 78.0725 2 152.9.7995 1080 18 4.6272 F(0.992) 1000 147.7719-75. 265 The “corrected predictor” is PRICEc PRICE n exp ˆ 2 / 2 147. Two models are for the subsets of the data for which the variable TRADITIONAL is 1 or 0. Principles of Econometrics.7995 is equal to the sum of the SSE from traditional houses (31.703 . 265 0. (e) Using the model from part (c) we find that the prediction for ln PRICE 1000 is 4. against the alternative that at least one is not zero.15 (continued) (d) The Chow test requires the original model plus an interaction variable of TRADITIONAL with every other variable. 4e 266 Exercise 7.95. Note that SSEU 75.0582) and the SSE from non-traditional houses (44.7995 9 75. 035) pool_tr -0.034) bath_tr 0. ** p<0.0068*** -0.034) (0. *** p<0.063) fp_tr 0.0989*** (0.752 0.0053* (0.1226*** -0.0271*** 0.033) (0.021) (0.0062*** -0.002) (0.046) (0.001) (0.019) (0. R-sq 0.0302*** 0.0216 0.1730*** (0.029) age -0.033) (0.032) (0.026) pool 0.0650*** 0.030) waterfront 0.018) (0.041) age_tr -0.022) (0.733 0.15(d) (continued) ---------------------------------------------------------------------------Rest Trad=1 Trad=0 Unrest ---------------------------------------------------------------------------sqfts 0.0388 0.0714*** (0.2070*** (0.026) (0.094) sqft_tr -0.041) (0.05.002) (0.Chapter 7.7413 75.1228*** 0.041) traditional -0.0975*** 0.0055*** (0.001) (0.0912*** 0.0021 0.0340 0.0405** 0.1831*** (0.7719 31.0388 (0.0714*** -0.0650* (0.0673*** 4.0311 (0.0673*** (0. 4e 267 Exercise 7.001) (0.0055*** -0.051) (0.065) (0.001) own_tr 0.741 SSE 78.0578* (0.01 .730 0. Exercise Solutions.002) bedrooms -0.1831*** 0.7322*** 4.7995 ---------------------------------------------------------------------------Standard errors in parentheses * p<0.0324*** 0.1894*** 0.10.065) (0.0582 44.0238 (0.003) beds_tr 0.047) (0.001) owner 0.027) (0.0275 -0.0021 (0.021) (0.001) (0.046) (0. Principles of Econometrics.0587* (0.016) (0.3351*** (0.0008 -0.071) _cons 3.042) fireplace 0.0578* 0.0324*** (0.039) wf_tr -0.021) (0.024) baths 0.058) ---------------------------------------------------------------------------N 1080 582 498 1080 adj.0013 (0.9701*** 3.1730*** 0.029) (0.2142*** 0. 0 200000 400000 600000 selling price of home. 4e 268 EXERCISE 7. Exercise Solutions.16(b) Histogram of ln(PRICE) 14 . the logarithm of PRICE is much less skewed and is more symmetrical. Principles of Econometrics.16 0 10 Percent 20 30 The histogram for PRICE is positively skewed. Thus. dollars 800000 5 Percent 10 15 Figure xr7. the histogram of the logarithm of PRICE is closer in shape to a normal distribution than the histogram of PRICE.Chapter 7.16(a) Histogram of PRICE 0 (a) 10 11 12 log(selling price) 13 Figure xr7. On the other hand. 0161LGELOT LIVAREA (0.9860 0.0480 BEDS 0.0255) (0. However.0114 (0.0231) All coefficients are significant with the exception of that for BATHS.0017 0. SSER 72. All signs are reasonable: increases in living area.5 acres is approximately 100 exp( 0. (e) To carry out a Chow test. Principles of Econometrics.0113 (0. holding all else fixed. larger lot sizes and the presence of a pool are associated with higher selling prices.0013 AGE 0.0382 BEDS 0.0005) (0.0633 and the sum of squared errors from the unrestricted model. The number of baths is statistically insignificant. note that by adding this interaction variable into the model.0164) 0.6134 LGELOT 0.0005) (0.0632) 0. implies smaller bedrooms which are less valued by the market.2531LGELOT 0. Exercise Solutions.8% larger than the price of houses on lot sizes less than 0. (c) The price of houses on lot sizes greater than 0.4712 Then the value of the F-statistic is . we use the sum of squared errors from the restricted model that does not distinguish between houses on large lots and houses that are not on large lots. 4e 269 Exercise 7.61% less.89%.0373 0.0016 AGE 0.27%.0228) (0.0026) Interpretation of the coefficient of LGELOT LIVAREA: The estimated marginal effect of an increase in living area of 100 square feet in a house on a lot of less than 0.0165) 0.5 acres is 5. The inclusion of the interaction variable separates the effect of the larger lot from the fact that larger lots usually contain larger homes.0539 LIVAREA 0. Older homes depreciate and have lower prices.0370 0. Increases in the number of bedrooms.2531) 1 28.0201BATHS 0. so its negative sign cannot be reliably interpreted.Chapter 7. The same increase for a house on a large lot is estimated to increase the house selling price by 1.0853POOL (0.0787 POOL (0.0019 0. that includes LGELOT and its interactions with the other variables. (d) The estimated regression after including the interaction term is: ln PRICE 1000 (se) 3.16 (continued) (b) The estimated equation is ln PRICE 1000 (se) 3.5 acres. holding other factors constant.9649 0.0589 LIVAREA 0. which is SSEU 65. or 4. the coefficient of LGELOT increases dramatically.0103BATHS 0. 0633 65.0604*** (0.037) (0.97 2.004) LOT_BEDS 0.0989*** 0.038) LIVAREA 0. R-sq 0.0522*** (0.0337*** 0.0088 -0.005) (0.9794*** 3.696 BIC 50. Thus.0018 -0.002) (0. 4e 270 Exercise 7.060) ---------------------------------------------------------------------------N 95 1405 1500 1500 adj.048) (0.024) (0. A summary of the alternative model estimations follows.017) (0.8699 -439.0002 (0.0016*** (0.012) (0.000) (0.4712 6 65.4121*** 3.0266*** (0. Exercise Solutions.0434 (0.017) AGE -0.0594*** -0. Principles of Econometrics.183) (0.8402 SSE 7.16 (continued) F SSER SSEU SSEU ( N J K) / The 5% critical F value is F(0.0262 -0.4293*** (0.074) (0.141) LOT_AREA -0.676 0.0016*** -0.037) LOT_BATHS 0.039) (0.0697*** 0.1259* 0.10. Exercise 7-16 ---------------------------------------------------------------------------(1) (2) (3) (4) LGELOT=1 LGELOT=0 Rest Unrest ---------------------------------------------------------------------------C 4.012) BATHS 0.2028 -252.000) (0. we conclude that the pricing structure for houses on large lots is not the same as that on smaller lots.0633-65.0334** -0.0008* -0.002) (0.8181 -352.1488) 72.4712 1488 24.0607*** 0.0604*** 0.1161** (0.0697*** (0.01 ** LOT_X indicates interaction between LGELOT and X .052) LOT_AGE -0.05.066) (0.95.0522*** -0.10 .024) (0.3445 72.002) BEDS -0.667 0.012) (0.000) POOL 0.4712 ---------------------------------------------------------------------------Standard errors in parentheses * p<0.002) (0.0827 -0.9828*** (0. ** p<0.0334* (0.025) LGELOT 0.017) (0.0562 (0.6.608 0. *** p<0.Chapter 7.9828*** 3.1268 58.001) LOT_POOL 0. CHAPTER 8 Exercise Solutions 271 . 4e 272 EXERCISE 8. Exercise Solutions. Principles of Econometrics.Chapter 8.1 When 2 i 2 N xi i 1 N i 1 xi x x 2 2 i 2 2 N xi i 1 N i 1 xi x 2 2 N 2 i 1 x 2 2 N i 1 xi xi x x 2 2 2 2 N i 1 xi x 2 . Exercise Solutions.2 (a) 1 Multiplying the first normal equation by 1 i ˆ 2 xi 2 1 i 1 1 i xi ˆ 1 i i i i xi 2 xi and the second one by 2 ˆ 1 2 i 1 xi i xi 2 ˆ 2 2 2 i yields yi xi yi* i Subtracting the first of these two equations from the second yields 2 xi 2 i 1 i 2 xi ˆ 2 2 xi yi* i 1 i 1 xi i yi Thus. 2 i xi 2 xi . 2 ˆ xi yi* i 2 1 i 2 xi 2 i 2 2 yi xi i i 2 1 xi i 1 i i 2 i i xi 2 i 2 2 xi 2 yi yi i 2 i 2 x i 2 xi 2 2 i i In this last expression. See pages 52 and 83-84 of the text. yields 2 i 2 for all i. . i yi 2 yi . . the second line is obtained from the first by making the 1 1 substitutions yi i yi and xi i xi . and by dividing numerator and denominator by 2 ˆ 1 1 2 2 x ˆ y for ˆ . Principles of Econometrics.Chapter 8. Making these substitutions into the expression for ˆ 2 yields 2 yi xi N 2 2 2 2 N 2 N 2 yi xi2 N 2 2 xi xi 2 2 2 N yi xi N xi2 N yx x2 and that for ˆ 1 becomes ˆ 2 1 N yi 2 2 N xi 2 ˆ 2 y x ˆ2 These formulas are equal to those for the least squares estimators b1 and b2 . 2 xi 2 i (b) yi and xi 2 yi 1 i 2 yi xi 2 yi xi . Solving the first normal equation i i 1 and making the substitutions yi ˆ 2 i 1 i i 2 When and 2 i 2 2 i N ˆ ˆ i 1 i i 2 i i 1 xi . 4e 273 EXERCISE 8. 2 (continued) (c) The least squares estimators b1 and b2 are functions of the following averages x 1 N xi y 1 N yi 1 N xi yi 1 N xi2 For the generalized least squares estimator for ˆ 1 and ˆ 2 .Chapter 8. Exercise Solutions. 4e 274 Exercise 8. Principles of Econometrics. . Reliable observations with small error variances are weighted more heavily than those with higher error variances that make them more unreliable. these unweighted averages are replaced by the weighted averages 2 i xi 2 i 2 i yi 2 i yi xi 2 i 2 i 2 i xi2 2 i In these weighted averages each observation is weighted by the inverse of the error variance. Exercise Solutions. the generalized least squares estimates are ˆ 1 5(47 8) (37 12)(7) 5(349 144) (37 12) 2 2. Principles of Econometrics. 5 . Thus.984 and ˆ 2 y* ˆ x* 1 (7 5) 2. and ei* ei xi .Chapter 8. xi* yi* 47 8 . This model can be estimated by least squares with the usual simple regression formulas.984 (37 12) 5 0. we find 7. 4e 275 EXERCISE 8. the transformed model that gives a ei* where yi* yi xi . but with 1 and 2 reversed.3 yi* 1 xi* 2 2 i ei where var(ei ) For the model yi 1 2 xi constant error variance is 2 x . yi* With N xi* 37 12 .44 ( xi* ) 2 349 144 . xi* 1 xi . the generalized least squares estimators for 1 and 2 are ˆ N 1 N xi* yi* ( xi* ) 2 xi* yi* xi* 2 and ˆ y* 2 ˆ x* 1 Using observations on the transformed variables. (ii) The White standard errors are slightly larger but very similar in magnitude to the conventional ones from least squares.96. using White’s standard errors leads one to conclude estimation is less precise. Thus.8124 The 5% critical value for (96. (b) Since the residual plot shows that the error variance may increase when income increases.95. namely 62. but not age. there is not one unique ordering. and this is a reasonable outcome since greater income implies greater flexibility in travel.4 (a) In the plot of the residuals against income the absolute value of the residuals increases as income increases. Exercise Solutions. Principles of Econometrics.dat after the observations have been ordered according to INCOME reveals 7 middle observations with the same value for INCOME.96) 1. but the same effect is not apparent in the plot of the residuals against age. Remark: An inspection of the file vacation.0479 107 ) (100 4) 2. and the values for SSE1 and SSE2 will depend on the ordering chosen. Thus. Those specified in the question were obtained by ordering first by INCOME and then by AGE. 2 we set up the null and alternative hypotheses as the one tail test H 0 : 12 2 versus 2 2 2 H1 : 12 2 . but it does not have a big impact on assessment of the precision of estimation. (iii) The generalized least squares standard errors are less than the White standard errors for least squares. where 1 and 2 are artificial variance parameters for high and low income households. 4e 276 EXERCISE 8. The value of the test statistic is F ˆ 12 ˆ 22 (2. Thus. the graphs suggest that the error variance depends on income. 96) degrees of freedom is F(0. . Thus.401 .9471 107 ) (100 4) (1. when the data are ordered only on the basis of INCOME. suggesting that generalized least squares is a better estimation technique. In this latter case there is no apparent relationship between the magnitude of the residuals and age. (c) (i) All three sets of estimates suggest that vacation miles travelled are directly related to household income and average age of all adults members but inversely related to the number of kids in the household.Chapter 8. we reject H 0 and conclude that the error variance depends on income. AGE. when they occur.020 Upper 0. (c) As mentioned in parts (a) and (b).600 0.2). but it could lead to slightly misleading information on the reliability of the estimates for ROOMS.Chapter 8. and PTRATIO.252 5.255 5. are no greater than 0.007 (b) Most of the standard errors did not change dramatically when White’s procedure was used.7). In particular. there is very little difference. after recognizing heteroskedasticity and using White’s standard errors. we can say that. the inferences drawn from use of the two sets of standard errors are likely to be similar. 95% confidence intervals Least squares standard errors Lower CRIME ROOMS AGE TAX 0.076 0.005 White’s standard errors Lower 0. AGE. . 4e 277 EXERCISE 8. AGE and TAX are narrower while the confidence interval for ROOMS is wider. However. the standard errors for CRIME.001. Those which changed the most were for the variables ROOMS. TAX. but only 3 in this exercise (Table 8. Principles of Econometrics. heteroskedasticity does not appear to present major problems. Thus. DIST.5 (a) The table below displays the 95% confidence intervals obtained using the critical t-value t(0.965 and both the least squares standard errors and the White’s standard errors. Exercise Solutions.497) 1.143 0.679 0. keeping in mind that the differences are not great. After recognizing heteroskedasticity and using White’s standard errors. TAX and PTRATIO and overstates the reliability of the estimates for the other variables. TAX and PTRATIO. the confidence intervals for CRIME. there will be some rounding error differences in the interval estimates in the above table. in terms of the magnitudes of the intervals.5 (Table 5.070 0.065 0. However.019 Upper 0. TAX and PTRATIO decrease while the others increase. DIST. Remark: Because the estimates and standard errors are reported to 4 decimal places in Exercise 5.975. none of the intervals contain zero and so all of the variables have coefficients that would be judged to be significant no matter what procedure is used.114 7. These differences.020 0.112 7. using incorrect standard errors (least squares) understates the reliability of the estimates for CRIME.026 0. and the inferences that would be drawn from each case are similar. Therefore. suggesting that the further the house is from the employment centre. Because the coefficient of ROOMS 2 is positive.822 6.84 11.5) test value is 2 N R2 506 0. The variance of house prices is negatively related to DIST.3 Thus.95. 11. The coefficients for ROOMS and ROOMS 2 are both significantly different from zero at a 1% level of significance.07 . generally.95.07 .4 Thus. this number of rooms is ˆ2 2ˆ3 ROOMS min 305. We reject H 0 if 2 2 (0. The crime rate for which it is a maximum is ˆ4 2ˆ5 CRIMEmax 2. The variance of house prices is also a quadratic function of CRIME. the variance of house prices increases with the crime rate up to crime rates of around 30 and then declines.08467 42. 4e 278 EXERCISE 8.039 29. but at a decreasing rate. but this time the quadratic function has a maximum. for houses of 6 rooms or less the variance of house prices decreases as the number of rooms increases and for houses of 7 rooms or more the variance of house prices increases as the number of rooms increases. (b) We can test for heteroskedasticity using the White test. The .285 2 0. Exercise Solutions. and so we can say that.Chapter 8. the smaller the variation in house prices. we reject H 0 and conclude that heteroskedasticity exists.84 Since 42. Principles of Econometrics. the variance increases as the crime rate increases.311 2 23. There are very few observations for which CRIME 30 .5) where 2 (0. The null and alternative hypotheses are H0 : 2 3 H1 : not all The test statistic is 2 6 s 0 in H 0 are zero N R 2 . the quadratic function has a minimum which occurs at the number of rooms for which eˆ 2 2 3 ROOMS 0 2 ROOMS Using the estimated equation.6 (a) ROOMS significantly effects the variance of house prices through a relationship that is quadratic in nature. 4.8875 y b2 x The least squares residuals eˆi yi eˆ observation 1 2 3 4 5 6 7 8 (c) To estimate ln( 2 i ) yˆi and other information useful for part (c) follow 1. The least squares estimate for is 8 ˆ i 1 zi ln(eˆi2 ) 8 i 1 zi2 86.382787 2.618977 4.591219 5. Exercise Solutions.465187 5. we begin by taking logs of both sides of zi . 4e 279 EXERCISE 8.4853 .8875 The least squares estimates are given by b2 N xi yi N xi xi 2 yi xi 8 89. See.527295 18.742369 16.714707 3.513031 1. Exercise 2.549756 1.34 3. Then.733822 9.8875 1.7071 and b1 (b) 3.35 52. that yields with eˆi2 to give the estimating equation vi from this model is equivalent to a simple linear Using least squares to estimate regression without a constant term.484558 3.291665 3.068875 4.887376 3.7 (a) Hand calculations yield xi x 0 yi 0 y 31.641419 4. for example.185693 31.353113 0.319125 0.933946 0.1 2 8 52.746079 ln(eˆ 2 ) z ln(eˆ2 ) 1. we replace the unknown ln(eˆi2 ) zi 2 i 2 i exp( zi ) .715469 2.35 0 31.7071 0 3.34 0 2 =1.17 0.Chapter 8.1 xi 2 xi yi 89.078484 2.905082 .4674 178.496682 2. Principles of Econometrics. 383851)2 2. These xi ˆi xi* are given in the following table.373502 0.575401 2.2.042323 0.036003 0.193812 ( 0.442137 ( 0.008623 8.457624 0.672141 1.960560 1.575316 0.383851) 1.345673 2.7 (continued) (d) Variance estimates are given by the predictions ˆ i2 values and those for the transformed variables yi* yi . the generalized least squares estimate for 2 yi xi ˆ 2 2 yi i i 2 i xi is xi 2 i 2 2 i 2 2 i 2 2 xi 2 i i 15.33594 2.193812 ( 0.585418 0.879147 9.053260 22.789371 0. Exercise Solutions.Chapter 8.1242 The generalized least squares estimate for ˆ 2 i 1 yi 2 i 2 i xi 2 i 1 ˆ 2 is 2. 4e 280 Exercise 8.224494 2.042323 From Exercise 8.1242 2.493887 0. (e) observation ˆ i2 yi* xi* 1 2 3 4 5 6 7 8 4.785981 2.383851) 2.156725 29.477148 7. Principles of Econometrics.115325 3.287700 4.6253 .008623 15.464895 3.4853 zi ) .330994 0. ˆi exp( ˆ zi ) exp(0.540580 1.514531 27.144126 0. which is 20 years old. 4e 281 EXERCISE 8.Chapter 8.8433 20 96. Also. (c) For the White test we estimate the equation eˆi2 1 2 SQFT and test the null hypothesis H 0 : 2 Since 2 2 (0. (d) 6 Estimating the regression log(eˆi2 ) ˆ 1 16.95.940.583.8433 AGE 35. .001414SQFT ) .15 68.0976 These results tell us that an increase in the house size by one square foot leads to an increase in house price of $63.3907 1800 217.001414 2 i as exp(16.25 11. Thus.95.5) 2 (0. For SQFT = 1800 and AGE = 20 PRICE 5193.3907 SQFT (se) 3586. ˆ2 With these results we can estimate ˆ i2 1 2 SQFT vi gives the results 0.3786. Principles of Econometrics.1687 217.3907 1400 217.5) N R2 3 2 AGE 3 4 SQFT 2 5 AGE 2 SQFT AGE vi 0 .15 68. relative to new houses of the same size. is $96.3786 0.8 (a) The regression results with standard errors in parenthesis are PRICE 5193.583 The estimated price for a 1400 square foot house. the calculated value is larger than the critical value.940 The estimated price for a 1800 square foot house. is $123.0375 35.07 .15 68.8433 20 123. The value of the test statistic is 6 940 0.39.84. we reject the null hypothesis and conclude that heteroskedasticity exists. That is. Exercise Solutions. (b) For SQFT = 1400 and AGE = 20 PRICE 5193. which is 20 years old. each year of age of a house reduces its price by $217.64 2. 326 The estimated price for a 1800 square foot house. is $96.6587 20 122.196. we replace the unknown i with the estimated standard deviations ˆ i . each year of age of a house reduces its price by $187.33. relative to new houses of the same size. with standard errors in parenthesis. which is 20 years old. is $122.43 2. The regression results.3269 1800 187.14 65.2844 These results tell us that an increase in the house size by one square foot leads to an increase in house price of $65.14 65.66.14 65. Exercise Solutions.6587 AGE (se) 3109.326.3269SQFT 187.0825 29. which is 20 years old.6587 20 96. Also. are PRICE 8491.Chapter 8. (f) For SQFT = 1400 and AGE = 20 PRICE 8491.8 (continued) (e) Generalized least squares requires us to estimate the equation PRICEi SQFTi 1 1 i 2 i AGEi ei 2 i i i When estimating this model. For SQFT = 1800 and AGE = 20 PRICE 8491. Principles of Econometrics. .196 The estimated price for a 1400 square foot house.3269 1400 187. 4e 282 Exercise 8. .9 (a) (i) Under the assumptions of Exercise 8.8(a) are 20 .000 is 0. P PRICE 115000 P Z 115000 96583 22539.207.Chapter 8. The probability that your 1400 square feet house sells for more than $115. 4e 283 EXERCISE 8.6 0.1) .207 where Z is the standard normal random variable Z N (0.8171 0. (ii) For houses of size SQFT 1800 and AGE prices from Exercise 8.6 P Z 0.6185 0. the mean and variance of house var( PRICE ) 22539. The probability is depicted as an area under the standard normal density in the following diagram.632 E ( PRICE ) 123940 The required probability is P PRICE 110000 P Z P Z 110000 123940 22539.8 part (a). Principles of Econometrics. the mean and variance of house prices for houses of size SQFT 1400 and AGE 20 are E ( PRICE ) 1 1400 2 20 var( PRICE ) 3 2 Replacing the parameters with their estimates gives var( PRICE ) 22539.268 The probability that your 1800 square feet house sells for less than $110. Exercise Solutions.000 is 0.632 E ( PRICE ) 96583 Assuming the errors are normally distributed.268. 152. . (c) In part (a) where the heteroskedastic nature of the error term was not recognized.8(d). In part (b) recognition of the heteroskedasticity has led to a standard deviation of prices that is smaller than that in part (a) for the case of the smaller house.893127 108 (24275.2704 2 1800) exp(16.3 P Z 1. from Exercise 8.0278 0. These differences have in turn led to a smaller probability for part (i) where the distribution is less spread out and a larger probability for part (ii) where the distribution has more spread. 4e 284 Exercise 8.347172 108 (18295.378549 1.000 is 0.5077 0. 2 and 3 . the variance of these house types is var( PRICE ) exp( 1 1. Principles of Econometrics.00141417691 1800) 5. and larger than that in part (a) for the case of the larger house. E ( PRICE ) 96196 .2704 0.152 The probability that your 1400 square feet house sells for more than $115.3) 2 Thus.306.000 is 0.8) 2 Thus.2704 2 1400) exp(16.2704 0.378549 1.8(f). Using estimates of 1 and 2 from Exercise 8. we find that E ( PRICE ) 122326 and 1.306 The probability that your 1800 square feet house sells for less than $110.Chapter 8.8 0. the mean of house prices for houses of size SQFT 1400 and AGE 20 is.00141417691 1400) 3.9 (continued) (b) (i) Using the generalized least squares estimates as the values for 1 . (ii) For your larger house where SQFT var( PRICE ) exp( 1 1800 . Exercise Solutions. P PRICE 115000 P Z 115000 96196 18295. the same standard deviation of prices was used to compute the probabilities for both house types. P PRICE 110000 P Z P Z 110000 122326 24275. Chapter 8.79 23. Thus. and regress the squares on xi to obtain 2 eˆ 123.05896 x R2 0. the variance assumption i eliminate heteroskedasticity. Principles of Econometrics.95.59 A null hypothesis of no heteroskedasticity is rejected because 5. Exercise Solutions.13977 5. we compute a value of the 2 N R2 2 test statistic as 40 0.10 (a) The transformed model corresponding to the variance assumption yi 1 1 xi 2 xi xi ei 2 i 2 xi is ei where ei xi We obtain the residuals from this model. square them.09 A null hypothesis of no heteroskedasticity is not rejected because 1.84 . square them. 4e 285 EXERCISE 8.02724 To test for heteroskedasticity.32923872 ln( xi ) 1 ˆi 2 xi ˆi ei where ei ei ˆi and We obtain the residuals from this model. we compute a value of the 2 N R2 2 test statistic as 40 0.1) 3.09 is less than the 5% 2 2 2 xi is adequate to critical value (0. .117 0.1) 3.13977 To test for heteroskedasticity.02724 1.27) is yi ˆi 1 ˆi exp(0.84 .35 x R2 0. and regress the squares on xi to obtain 2 eˆ 1.93779596 2. the variance assumption i eliminate heteroskedasticity.59 is greater than the 5% 2 2 2 xi was not adequate to critical value (0. (b) The transformed model used to obtain the estimates in (8.95. Thus. 2) 5. (a) (b) (c) yi xi 1 1 0. and.270 81.Chapter 8. ˆ 1 se( ˆ 1 ) ˆ 2 part (a) part (b) part (c) 81.000 76.665 6. importantly. The results are substantially different for part (b).822 12. The critical chi-squared value for the White test at a 2 5% level of significance is (0. we reject the null hypothesis for parts (a) and (c) because. 2 2 2 2 ln( xi ) do not eliminate xi and var(ei ) (0.95. Heteroskedasticity has been eliminated with the assumption that var(ei ) 2 2 i x .004 33.11 The results are summarized in the following table and discussed below.25 i ei xi 1 ln( xi ) 1 2 ln( xi ) ln( xi ) where ei ei xi0. The two transformations have similar effects.612 10. These equations were of the form eˆ 2 1 2 x 3 x2 v 0 against the alternative For the White test we are testing the hypothesis H 0 : 2 3 hypothesis H1 : 2 0 and/or 3 0. particularly the standard errors.024 1. Principles of Econometrics. Exercise Solutions. we do not reject the 2 null hypothesis in part (b) because 2 (0.991 .955 se( ˆ 2 ) 2 N R 2 The transformed models used to obtain the generalized estimates are as follows. the coefficient estimates and their standard errors are almost identical.2) . the results can be sensitive to the assumption made about the heteroskedasticity.95.009 32.95.25 i x 1 xi yi x xi xi 2 ei 0. .323 1.641 2.806 10. After comparing the critical value with our test statistic values. In the two cases where heteroskedasticity has not been eliminated (parts (a) and (c)). whether that assumption is adequate to eliminate heteroskedasticity.706 1.328 10.2) . On the other hand.733 6. 4e 286 EXERCISE 8.25 where ei ei xi ei where ei ei ln( xi ) In each case the residuals from the transformed model were squared and regressed on income and income squared to obtain the R 2 values used to compute the 2 values. The assumptions var( ei ) heteroskedasticity in the food expenditure model. Thus.25 i x yi xi 1 2 0. in these cases. however. 29298 9. we reject the null hypothesis and (0. The plotted values are more dispersed about the fitted regression line for larger values of GDP per capita.8 0. Since 9. and thus they have greater flexibility in terms of how much they can spend on education. countries with a higher GDP per capita. have more money to distribute.0 -0.961 The critical chi-squared value for the White test at a 5% level of significance is 2 5.991 .991. 1.0052 The fitted regression line and data points appear in the following figure. For the White test we are testing the 0 against the alternative hypothesis H1 : 2 0 and/or 3 0.2 0 2 4 6 8 10 12 14 16 18 GDP per capita (c) For the White test we estimate the equation 2 i eˆ 1 2 GDPi Pi 3 GDPi Pi 2 vi This regression returns an R2 value of 0.6 0.12 (a) This suspicion might be reasonable because richer countries.4 0.29298.2) conclude that heteroskedasticity exists.961 is greater than 5.0732 (se) 0.95.0485 0.4 1. with the standard errors in parentheses are EEi Pi GDPi Pi 0. Exercise Solutions. There is evidence of heteroskedasticity. hypothesis H 0 : 2 3 The White test statistic is 2 N R2 34 0. and therefore the amount they spend on education is likely to vary less. a country with a smaller GDP will have fewer budget options. Principles of Econometrics.2 1. (b) The regression results. In comparison. 4e 287 EXERCISE 8.0 0.2 0.Chapter 8. . This suggests that heteroskedasticity exists and that the variance of the error terms is increasing with GDP per capita.1246 0. 975.0044 Using these estimates. the 95% confidence interval for b2 2 2 is t(0.0.073173 2.073173 2. Principles of Econometrics. reflecting that generalized least squares is more precise than least squares when heteroskedasticity is present.Chapter 8.0603.0.0693 0. we expect the generalized least part (d).12 (continued) (d) Using White’s formula: se b1 0.975. The 95% confidence interval for b2 2 0. Exercise Solutions.32) se(b2 ) 0.0837) using White’s standard errors is t(0.00621162 (0.0605. we obtain GDPi Pi 0.040414.0289 xi .0369 0. however.006212 using the conventional least squares standard errors is t(0.0369 0.32) se(b2 ) 0. The confidence interval from White’s standard errors is wider.0783) The width of this confidence interval is less than both confidence intervals calculated in 2 xi is true.00441171 (0.0858) In this case.00517947 The 95% confidence interval for b2 se b2 2 (0.069321 2. A direct comparison of the generalized least squares interval with that obtained using the conventional least squares standard errors is not meaningful. because the least squares standard errors are biased in the presence of heteroskedasticity. Given the assumption var(ei ) squares confidence interval to be narrower than that obtained from White’s standard errors.0369 0. ignoring heteroskedasticity tends to overstate the precision of least squares estimation. . 4e 288 Exercise 8.0626.32) se(b2 ) 0.0929 0.975. (e) Re-estimating the equation under the assumption that var(ei ) EEi Pi (se) 0.0. 24) distribution is 3.4 The average cost function is given by C1t Q1t Thus.744 1.65 6111. Principles of Econometrics.40.Chapter 8.134 24 108. Exercise Solutions.595 68. . Since the calculated F is greater than the critical F. Q12t 3 4 and 93. the generalized The calculated F value for testing the hypothesis that 1 = 4 = 0 is 108. average cost is a linear function of output.4. where var e1t estimated coefficient 1 (b) 4 Q13t 1 4 1 1 Q1t 2 3 Q1t 4 Q12t et Q1t 0 .484 10.134 2 24 6111. 3 2. The F value can be calculated from SSER F SSEU SSEU 2 61317. if (d) 2 are: standard error 2 (c) e1t . The 5% critical value from the F(2. The average cost function is an appropriate transformed model for estimation when 2 2 heteroskedasticity is of the form var e1t Q1t .422 17. 4e 289 EXERCISE 8.592 23. we reject the null hypothesis that 1 = 4 = 0.774 0.2425 3 4 Q1t .0086 3.13 (a) For the model C1t 1 2 least squares estimates of Q1t 1. 3026 Q 3 (se) 16. Principles of Econometrics.933 ˆ 22 847.984 . we expect the quadratic MC function to have a minimum.24) 0.415 0. and positive for large Q (increasing MC). (b) 2 2 2 Testing H 0 : 12 2 against H1 : 1 2 is a two-tail test. thus.05. for which we require 4 > 0.24. but the estimated error variances are quite different. .2721 ˆ 12 324. The value of the F statistic is ˆ 22 ˆ 12 F Since F 2. 4e 290 EXERCISE 8. Comparing the two estimated equations.270 89.655 4.Chapter 8.66 SSE2 20343.49 Cˆ 2 51.796 Q12 1. we apply generalized least squares to the combined set of data. The estimated equation is Cˆ 67.24.85 SSE1 7796.24. we require 3 < 0.156 0. the standard errors of all the coefficients except the constants are quite small indicating reliable estimates.14 (a) The least squares estimated equations are Cˆ1 72.015 Q22 1. Both our least-squares estimated equations have these expected signs.185 108.659 Q1 13. 2 Since the test outcome in (b) suggests 12 2 . For this slope to be negative for small Q (decreasing MC).3802 To see whether the estimated coefficients have the expected signs consider the marginal cost function MC dC dQ 2 2 3Q 3 4 Q 2 We expect MC > 0 when Q = 0.61 F(0. with the observations transformed using ˆ 1 and ˆ 2 .95.66 324.83 20.920 Q 15.1911Q13 (se) 23. Furthermore.408 Q 2 1.597 0.973 3. The slope of the MC function is d ( MC ) dQ 2 3 6 4Q .29 Q2 (se) 28.0504 and F(0.24) . but we are assuming both firms have the same coefficients. Exercise Solutions. we expect 2 > 0. we reject H0 and conclude that the data do not support the proposition that (c) 847.85 2 1 2 2 . Also.2065 Remark: Some automatic software commands will produce slightly different results if the transformed error variance is restricted to be unity or if the variables are transformed using variance estimates from a pooled regression instead of those from part (a).6131Q23 6.24) 1.774 83.95. The critical values for performing a two-tail test at the 10% significance level are F(0. we see that the estimated coefficients and their standard errors are of similar magnitudes. Principles of Econometrics. These values are both much less than 2 their respective 5% critical values F(0. we have the interesting result SSE1 ˆ 12 SSEU* SSE2 ˆ 22 ( N1 K1 ) ˆ 12 ˆ 12 (N2 K 2 ) ˆ 22 ˆ 22 N1 K1 N2 K2 48 Thus. There is no evidence to suggest that the firms have different coefficients.14 (continued) (d) 2 1 Although we have established that H0 : 1 1 .95. it is instructive to first carry out the test for 3 .32) 4 28140.2412 48) 4 48 48 0.252 .49 20343. note that the number of observations N is the total number from both firms. It could give misleading 2 results.488 . For unrestricted estimation. Doing so by combining all 56 observations yields SSER 28874. we follow the same steps. To perform the test under the assumption that 12 2 . F (49. 4. but we use values for SSE computed from transformed residuals.3103 and 2 1. The test statistic is F SSER SSEU SSEU N J K where SSEU is the sum of squared errors from the full dummy variable model. The F-value is given by F SSER SSEU SSEU N J K (28874.565 and (0. 2 Assuming that 12 2 . In the formula for F. . We can also calculate SSEU as the sum of the SSE from separate least squares estimation of each equation.83 28140.34 .32 2 The restricted model has not yet been estimated under the assumption that 12 2 . In this case SSEU SSE1 SSE2 7796. and then under the assumption that 2 1 2 2 . The above test is not valid in the presence of heteroskedasticity. however.4) 9. The 2 and F test values can also be conveniently calculated by performing a Wald test on the coefficients after running weighted least squares on a pooled model that includes dummy variables to accommodate the different coefficients.2412 . 4e 291 Exercise 8. the test is equivalent to the Chow test discussed on pages 268-270 of the text.313 The corresponding 2 -value is 2 4 F 1.32 (56 8) 0.95. Exercise Solutions.Chapter 8.34 28140. 2 2 3 . 2 2 1 under the assumption that 2 2 2 . The dummy variable model does not have to be estimated. 4 4 . For restricted estimation from part (c) the result is SSER 49. There is no evidence to suggest that the firms have different coefficients. and K is the number of coefficients from both firms.241 The same conclusion is reached. 48) 2. 1332EDUCi 1.508448 and ˆ 2 Using these estimates.2185EDUCi 0. .243 ).2704) yields ˆ 22. we use the estimated variances above to transform the model in the same way as in (8.508448 0. these standard errors are larger than those in part (b).35).Chapter 8. although the variance estimates can be sensitive to the estimation technique.338041 .338041 1) 6. After doing so the regression results are. reflecting the lower precision of least squares estimation. 4e 292 EXERCISE 8.2124 0.824 and ˆ 2R 15.3375 and 2 R ˆ 4.1328EDUCi 1. These values are closer.508448 0. Multiplying ˆ 2M 2 R 2 M 6. The differences occur because the residuals from the combined model are different from those from the separate sub-samples.2340EDUCi 0. Principles of Econometrics.9140 1.7052 1.5301METROi 1.519711 exp(1.15 (a) To estimate the two variances using the variance model specified.0158 0. the resulting generalized least squares estimates of the mean function are much less sensitive.0835 0.5241METROi 1. ˆ 2M 0.337529 These error variance estimates are much smaller than those obtained from separate subsamples ( ˆ 2M 31. we have METRO = 0 ˆ 2R METRO = 1.0150 0. (c) The regression output using White standard errors is WAGEi (se) 9. One reason is the bias factor from the exponential function – see page 317 of the text.36).5197 by the bias factor exp(1. Thus. we first estimate the equation WAGEi 1 2 EDUCi 3 EXPERi 4 METROi ei From this equation we use the squared residuals to estimate the equation ln(eˆi2 ) 1 2 METROi vi The estimated parameters from this regression are ˆ 1 1.3445 With the exception of that for METRO.576 and ˆ 16.338041 0) 4.0694 0. exp(1.100 . Exercise Solutions. (b) To use generalized least squares.0485 0. with standard errors in parentheses WAGEi (se) 9. but still different from those obtained using separate sub-samples.3858 The magnitudes of these estimates and their standard errors are almost identical to those in equation (8. 34) 1.349 and F(0.298 > 2. i.899215 10-4 Since 5.4466 0.298 ˆ G2 2.1138 Testing the null hypothesis that demand is price inelastic. are not the same.1262 1 against the 5.2954 ( 1) 0.1838 0.0268 0.691 . The value of the F-statistic is ˆ 2A 15. .05. Since 5.1262 0.Chapter 8.1039 0.15. The value of our test statistic is alternative H1 : 3 t 0. H 0 : 3 1 .2954 0.15. is a one-tail t test.36132 10-4 . we do not reject the null hypothesis and conclude that there is not enough evidence to suggest that demand is elastic. Principles of Econometrics. Exercise Solutions.16 (a) Separate least squares estimation gives the error variance estimates ˆ G2 2.025.862. Austria and Germany.36132 10-4 5.58 The critical t value for a one-tail test and 34 degrees of freedom is t(0. 4e 293 EXERCISE 8.691 .58 1.4005 0.15) 2. (b) 2 G The critical values for testing the hypothesis H 0 : H1 : 2 G 2 A at a 5% level of significance are F(0. we reject the null hypothesis and conclude that the error variances of the two countries..e.899215 10 4 and ˆ 2A 15. F (c) The estimates of the coefficients using generalized least squares are 1 2 3 4 (d) [const] [ln(INC)] [ln(PRICE)] [ln(CARS)] estimated coefficient standard error 2.862 .15) 2 A against the alternative 0.975. 00087) (0.5 0.274) (0. This conclusion agrees with our speculation from inspecting the figures in part (b). although in part (b) we did suggest the sign of SQFT100 might be negative. We can conclude that AGE has a significant effect on variance while SQFT100 is not significant. the absolute magnitude of the residuals appears to decrease as SQFT100 increases. 40 50 60 70 80 SQFT100 Plot of residuals against AGE (c) 30 h( 3 1 0 Plot of residuals against SQFT100 2 AGE H1 : 3 2 SQFT 100) and test the hypotheses: 0 and/or 3 0 The test statistic value is 2 N R2 1080 0.7139 0.01756 AGE 0.5 -1.1196 0.17(b) 1.0 -1.006945. Exercise Solutions. 4e 294 EXERCISE 8.210. Principles of Econometrics. respectively.5 0.00404 and 0. 1.5 RESID RESID Figure xr8.99.0000227) The residual plots are given in the figures below.210 . The absolute magnitude of the residuals increases as AGE increases.1082 116.2) 9.03876SQFT100 0. with the variance dependent on the age of the house.5 -0. The variance might decrease as the house size increases.0 -0. but we cannot be certain.3589.17 (a) The estimated regression is ln( PRICE ) 11.02177 AGEi 0. Since 116.0 1.0001734 AGE 2 (se) (b) (0.00136) (0.0 0.Chapter 8. we reject the null hypothesis and conclude that heteroskedasticity exists. although this pattern is less pronounced. (d) The estimated variance function is given as ˆ i2 exp( 4. .0 0. Conversely. Corresponding p-values are 0.5 -1.006377 SQFT 100i ) The robust standard errors for AGE and SQFT100 are 0.88 is greater than 9.0000 and 0.5 1. suggesting heteroskedasticity.0 -1.5 0 10 20 30 40 50 60 70 80 90 0 10 20 AGE We set up the model var( e ) H0 : 2 0.876 The critical chi-squared value at a 1% level of significance is 2 (0. suggesting that GLS has improved the efficiency of estimation.024) 11.033) 0.00136) 0.00175) 0.01540 AGE 0.95.027) 0. Principles of Econometrics.01540 (0.0000272) (f) b1 b2 b3 b4 11.105 0.03881SQFT100 0. 4e 295 Exercise 8.0000227) (ii) with HC standard errors 11.028) 0.17 (continued) (e) The estimated generalized least squares model is ln( PRICE ) 11.00082) 0.Chapter 8.01756 (0.0000272) 0.00136) –0.00136) (0.99 and the p-value of the test is 0.03881 (0.120 (0.0001734 (0.024) (0.105 (0.00144) 0. The GLS HC standard errors are slightly larger than the conventional GLS ones.0000314) (i) Least Squares (iv) with HC standard errors The coefficient estimates from least squares and GLS are similar.01540 (0. with the greatest differences being those for AGE and AGE2.120 (0.105 (0.2) 5.62 The 5% critical value is 2 (0. The variance function that we used does not appear to have been adequate to eliminate the heteroskedasticity.01756 (0.03881 (0.00123) –0. Exercise Solutions. The conventional GLS standard errors are smaller than the least squares HC standard errors.00087) –0.0001297 (0. this could be indicative of some remaining heteroskedasticity.0001734 (0.018169 19. and for all coefficients.03876 (0. Thus we reject a null hypothesis of homoskedastic errors. .0001297 AGE 2 (se) (0.0001.0000372) (iii) GLS 11.00082) (0. (g) The Breusch-Pagan test statistic obtained by regressing the squares of the transformed residuals on AGE and SQFT100 is 2 N R 2 1080 0.00105) –0.0001297 (0. The heteroskedasticity-consistent (HC) standard errors are higher than the conventional standard errors for both least squares and GLS.03876 (0. Therefore. which is COKE i .06594 PRATIOi (0.3207) (0.08571DISP _ COKEi 0.1097 DISP _ PEPSI i (0. Both DISP_PEPSI and DISP_COKE have significant coefficients if one-tail tests and a 5% significance level are used. or the probability that a customer in store i will purchase Coke.0469) The results suggest that PRATIO and DISP_PEPSI have negative impacts on the probability of purchasing coke. We can think of it as the proportion evaluated for a large number of customers in store i. although the coefficient of the price ratio is not significantly different from zero at a 5% significance level. DISP_COKE has a positive impact on the probability of purchasing coke.31199) 0.Chapter 8.04671) (0. Exercise Solutions. 4e 296 EXERCISE 8. We can write pi (d) 1 2 PRATIOi 3 DISP _ COKEi 4 DISP _ PEPSI i The estimated regression is: COKE i (se) 0. the total number of shoppers who buy coke in store i is given by 1 Ni and the proportion will be given by (b) E COKE i 1 E Ni 1 Ni Ni j 1 Ni j 1 1 N i2 COKEij . Principles of Econometrics. .18 (a) COKEij is a binary variable which assigns 1 if the shopper buys coke and zero otherwise. pi Ni j 1 Ni j 1 Ni j 1 COKEij var COKEij zero covariance terms pi (1 pi ) Ni pi (1 pi ) N i2 (c) COKEij E COKEij 1 var N i2 1 N i2 j 1 j 1 COKEij 1 N i pi Ni var COKE i Ni Ni pi (1 pi ) Ni pi is the population proportion of customers in store i who purchase Coke.5196 0. 07831DISP _ COKE 0.4485 0. 4e 297 Exercise 8.30205) (0. Since 7.067 .887 < 14. or the sample size is too small.15774 7. both in terms of coefficient magnitudes and significance.18 (continued) (e) The null and alternative hypotheses are H 0 : errors are homoskedastic H1 : errors are heteroskedastic The test statistic is 2 N R2 50 0. but remains insignificant. Given the relative importance of PRATIO.04568) (0. The variance of the error term is var COKE i pi (1 pi ) Ni 1 2 PRATIOi 1 2 3 DISP _ COKEi PRATIOi 3 4 DISP _ PEPSI i DISP _ COKEi 4 DISP _ PEPSI i Ni The product in the above equation means that the variance will depend on each of the variables and their cross products.7) insufficient evidence to conclude that the errors are heteroskedastic. for the test to be conclusive. Principles of Econometrics.3099) (0.04135 0. Thus. (f) The estimated results are reported in the table below: pˆ (g) Mean Standard Deviation Maximum Minimum 0.5503 0. we do not reject the null hypothesis. it makes sense to include the cross-product terms when carrying out the White test. .5459 0.Chapter 8.0449) The results are very similar to those obtained in part (d). Exercise Solutions. There is (0. The p-value of the test is 0. It is surprising that the White test did not pick up any heteroskedasticity. this insignificance is puzzling.343.887 The critical chi-squared value for the White test at a 5% level of significance is 2 14.09673PRATIO 0. It could be attributable to the small variation in PRATIO. The coefficient of PRATIO is a mild exception.067.1009 DISP _ PEPSI (0. Or the omission of Ni could be masking the effect of the variables. Perhaps the variation in pi is not sufficient.95.3385 The estimated GLS regression is: COKE i (se) 0. it is larger in absolute value than its least squares counterpart. 0000920) 0.0006933EXPER 2 (se) (0.01138) (0. 4e 298 EXERCISE 8.5297 0.03392) The null and alternative hypotheses for testing whether married workers get higher wages are given by H0 : 6 0 H1 : 0 6 The test value is: t b6 se(b6 ) 0.1272 EDUC 0.1261EDUC 0.06298 EXPER 0. Also.000638) (0. We conclude that there is insufficient evidence to show that wages of married workers are greater than those of unmarried workers.0403MARRIED (0.188 0.001309 EXPER EDUC 0.00339 The corresponding p-value is 0. we do not reject the null hypothesis at the 5% level.0000956) 0.05).2528) (0. The residual plot 2 1 0 EHAT (c) -1 -2 -3 -1 0 1 2 MARRIED Figure xr8.001322 EXPER EDUC (0.19 (a) The estimated least square regression with heteroskedasticity-robust standard errors is ln(WAGE ) 0. Exercise Solutions.01159) (0.5411 0.0170) (0.06137 EXPER 0.0171) (0.0007139 EXPER 2 (se) (0.2542) (0.1176.646. Principles of Econometrics. Since the test value is less than the critical value (or because the pvalue is less than 0.04029 1.000637) (b) Adding marriage to the equation yields ln(WAGE ) 0. there is the evidence of heteroskedasticity.19(c) Plot of least squares residuals against marriage The residual plot suggests the variance of wages for married workers is greater than that for unmarried workers.Chapter 8. Thus. the critical value at the 5% level of significance is 1. . 0144) (0.Chapter 8.3558) (0.0007128EXPER 2 (se) (0.0007478) The estimated regression when MARRIED 0 is ln(WAGE ) 0.000654) The Goldfeld-Quandt test The null and alternative hypotheses are: 2 M H0 : 2 U against H1 : 2 M 2 U The value of the F statistic is F ˆ U2 ˆ 2M 0.1513EDUC 0. 4e 299 Exercise 8.0194) (0. the standard errors in the GLS-estimated equation are all less than their counterparts in the least squares-estimated equation.9197 0.06452 EXPER 0. However.0222) (0. reflecting the increased efficiency of least squares estimation.19 (continued) (d) The estimated regression when MARRIED 1 is ln(WAGE ) 0.0001193) 0.0007014 EXPER 2 (se) (0. Because 0. (e) The generalized least squares estimated regression is ln(WAGE ) 0.835 .00932) (0.194 .743 F(0.21285 0.0007088 EXPER 2 (se) (0. we reject H 0 and conclude that the error variances for married and unmarried women are different.000484) There are no major changes in the values of the coefficient estimates.4780 0. .07284 EXPER 0.0004620 EXPER EDUC (0. 414.835 and FUc F(0.05069 EXPER 0.2212) (0.414.025.2945) (0.1008 EDUC 0.001443EXPER EDUC (0.1309 EDUC 0.01271) (0.0000862) 0.01493) (0.743 F FLc 0.576) 0.0001379) 0. Exercise Solutions. Principles of Econometrics.28658 The critical values are FLc 0.002145EXPER EDUC (0.576) 1.1975 0.975. 1364) The interval estimate from the GLS equation is slightly narrower than its least squares counterpart. 0.962 0. the two interval estimates are as follows. Exercise Solutions.010193 (0.011335 (0.19 (continued) (f) The marginal effect for a worker with 10 years of experience is given by E ln(WAGE ) EDUC 2 5 EXPER 2 10 5 The estimate for the marginal effect calculated using the regression in part (a) is E ln(WAGE ) EDUC 0.11643 Its standard error is se ˆ 2 10 ˆ 5 0. .0013224 10 0.11643 1.010193 . Principles of Econometrics. 0. The estimate for the marginal effect calculated using the regression in part (e) is E ln(WAGE ) EDUC 0. but overall.Chapter 8. Thus.0014426 10 0.995) 1.478006 0.11397 Its standard error is se b2 10 b5 0.962 .962 0.11397 1.1362) From the GLS-estimated equation in part (e): me tc se ˆ 1 10 ˆ 5 0.127195 0.0964. From the least squares-estimated equation in part (a): me tc se b1 10b5 0.011335 . The t-value for computing the interval estimates is tc t(0. there is very little difference.975. 4e 300 Exercise 8.0917. 20 Residual polots against EDUC and EXPER Both residual plots exhibit a pattern in which the absolute magnitudes of the residuals tend to increase as the values of EDUC and EXPER increase.65 The critical chi-squared value at a 5% level of significance is 2 (0.815. EDUC or MARRIED.01391EDUCi 0.0021.0255 0.815 .Chapter 8. 4e 301 EXERCISE 8. (b) The null and alternative hypotheses are H 0 : errors are homoskedastic H1 : errors are heteroskedastic with H1 implying the error variance depends on one or more of EXPER. Exercise Solutions.65 is greater than 7. we reject the null hypothesis and conclude that heteroskedasticity exists. Principles of Econometrics. . the plots suggest there is heteroskedasticity with the variance dependent on EDUC and possibly EXPER.00516 EXPERi 0.20 The residual plots against EDUC and EXPER are as follows 2 2 1 1 0 0 RESID RESID (a) -1 -2 -1 -2 -3 0 4 8 12 16 20 24 -3 0 10 20 30 EDUC 40 50 60 70 EXPER Figure xr8.95. The value of the test statistic is 2 N R 2 1000 0. Thus.04547 MARRIEDi ) The standard deviations for each observation are calculated by getting the square roots of the forecast values from the above equation. The p-value of the test is 0.3) 7.01465 14. (c) The estimated variance function is ˆ i2 exp( 3. although for EXPER the increase is not very pronounced. Since 14. The first ten estimates are presented in the following table. Chapter 8.5297 0.26123 1. 2.00944) (0. 8. the standard errors in the GLS-estimated equation are all less than their counterparts in the least squaresestimated equation.0000920) 0. Principles of Econometrics.1274 EDUC 0.27944 0. 4.001322 EXPER EDUC (0.0144) (0. 4e 302 Exercise 8. (d) The generalized least squares estimated regression is ln(WAGE ) 0. reflecting the increased efficiency of least squares estimation.2528) (0.0007151EXPER 2 (se) (0.1272 EDUC 0.5265 0.26470 0.0170) (0.20(c) (continued) Observation Standard deviation 0.27287 0. 9.06298 EXPER 0.000637) The coefficient estimates in both equations are very similar. Exercise Solutions.2203) (0. . 5. 7.06365EXPER 0.000492) The least squares estimated equation with heteroskedasticity-robust standard errors is ln(WAGE ) 0. However.01138) (0.24982 0.27856 0.26049 0.0000887) 0.0007139 EXPER 2 (se) (0.24957 0. 6. 10.001369 EXPER EDUC (0. 3.27217 0.26745 0. there is very little difference.01723) From the GLS-estimated equation in part (d): me tc se b3 40 b4 16 b5 0.013265 1.00136903 0.975.002020 .002020 (0.962 . 4e 303 Exercise 8. From the least squares-estimated equation: me tc se b3 40 b4 16 b5 0.00941.995) 1.063646 40 0.962 0.00930.Chapter 8. Principles of Econometrics.01686) The interval estimate from the GLS equation is slightly narrower than its least squares counterpart. 0.0007139386 16 0. Thus.001322388 0.013136 1.013136 Its standard error is se b3 40 b4 16 b5 0.062981 40 0.013265 Its standard error is se b3 40 b4 16 b5 0. The t-value for computing the interval estimates is tc t(0.962 0.001898 .20 (continued) (e) The marginal effect for a worker with 16 years of education and 20 years of experience is given by E ln(WAGE ) EXPER 3 2 4 EXPER 5 EDUC 3 40 4 16 5 The least squares estimate for the marginal effect is E ln(WAGE ) EDUC 0. . the two interval estimates are as follows. The generalized least squares estimate for the marginal effect is E ln(WAGE ) EDUC 0. but overall.001898 (0. 0. Exercise Solutions.0007151398 16 0. 00136903402 16 18) 26.072 To compute the forecast using the corrected predictor. the forecast wage for a married worker with 18 years of education and 16 years of experience is WAGE n exp(0.0454734) 0. Principles of Econometrics.01391 18 0.0708577 2 27.43.0051605 16 0.0636458 16 0.00071513983 162 0. This estimate is given by ˆ2 exp( 3. we first need to estimate the variance for a married worker with 18 years of education and 16 years of experience.958) 0.260868 1.526482 0.072 exp 0. 4e 304 EXERCISE 8.025504 0.464.Chapter 8. Exercise Solutions.21 (a) Using the natural predictor.962 (15.0708577 .012 (b) The 95% forecast interval is given by exp ln WAGE n tc se( f ) exp 3.0708577 Then the forecast from the corrected predictor is WAGE c WAGE n exp ˆ 2 2 26.127412 18 0. 01262TERM (0. The cross product terms are included because in the linear probability model var( DELINQUENT ) E ( DELINQUENT ) 1 E ( DELINQUENT ) where E ( DELINQUENT ) is a linear function of all the explanatory variables.2112) (0.22 (a) The estimated linear probability model is DELINQUENT (se) 0.4816 INSUR 0.000785) (0. 4e 305 EXERCISE 8. Principles of Econometrics.40) conclude that heteroskedasticity exists.02383) 0.21997 219.0002018) (0. .01267) (0. as expressed in the estimated equation. The number of observations where var( DELINQUENT ) 0.00860) 0.02377 AMOUNT (0. The value of the test statistic is 2 N R 2 1000 0. we reject the null hypothesis and (0. (b) The error variances are estimated using var( DELINQUENT ) DELINQUENT 1 DELINQUENT The number of observations where var( DELINQUENT ) 1 is zero.0319) The White test The null and alternative hypotheses are H 0 : errors are homoskedastic H1 : errors are heteroskedastic Under H1 we are assuming that the error variance depends on one or more of the explanatory variables.03438RATE (0. Exercise Solutions.05932 REF 0.974 is greater than 55.01 is 158.0004419CREDIT (0.6885 0.758 .758. The number of observations where var( DELINQUENT ) 0 is 135.974 The critical chi-squared value for the White test at a 5% level of significance is 2 55.001624 LVR 0.1283 ARM (0. their squares and their cross products.95.00354) 0.Chapter 8.0236) 0. Since 219. 1283 (0.0419 (0.0277) (iii) <0.00068) –0.0041) 0.0002 (0. Exercise Solutions.000162 (0.0292) 0.0238) –0.0109) For most of the coefficients the least squares and generalized least squares estimates are similar. Moreover.00086.0126 (0.22 (continued) (c) LVR REF INSUR RATE AMOUNT CREDIT TERM ARM (i) LS 0.0413 (0.00159 (0.0045 (0. if the loan was for refinancing.4816 (0.00001.0099) –0.0121) –0.0319) (ii) LS-HC 0. an exception is AMOUNT whose coefficient is not significantly different from zero in the least squares estimations.0327 (0. (d) LVR: The estimated coefficient is 0.0082) 0. The positive sign is reasonable as a higher ratio of the amount of loan to the value of the property will lead to a higher probability of delinquency.0048) -0.0187 (0.000442 (0.0327.0018) 0.0105) –0.00038) –0. In the other cases almost all of the 8 coefficients were significant.000118) –0.4086) 0. The coefficient of LVR is significantly different from zero at the 5% level.4770 (0.000024 (0. holding other variables constant.00086.0238 (0. .0126 (0.000207) –0.0297) 0.0098) 0.00024) –0.00078) –0.5016 (0.00162 (0. a one unit increase in the ratio of the loan amount to the value of property increases the probability of delinquency by 0.000184) –0.0236) 0.01 0. and the estimates are very sensitive to the threshold which is chosen.0057) 0.0327.0593 (0. which has a negative impact upon the loan delinquency.Chapter 8.0258 (0.5127 (0.0035) 0. Principles of Econometrics.0188 (0.000382 (0.0571 (0.0190 (0.000442 (0.0211) –0.4816 (0.0146) –0.0140) (v) =0.0021) 0.0036) 0. In the extreme case where variances less than 0.0086) 0. This suggests that.00001. holding other variables constant.0204 (0.01.0018 (0. or less than 0.0267 (0. The coefficient of REF is significantly different from zero at the 5% level.0593 (0. This suggests that. Setting small and negative variances equal to a small number seems to be a practice fraught with danger.0240) –0.01.000202) –0.000085) –0. are set equal to one of these threshold values.0065 (0.0127) –0.0407) (iv) =0.00162 (0. the probability of delinquency decreases by 0.00054 (0.00086 (0.0304) 0. The negative sign is reasonable as refinancing the loan is usually done to make repayments easier to manage. providing the GLS estimates are obtained by discarding observations with variances less than 0.1283 (0.0344 (0.2089 (0.0089) –0. REF: The estimated coefficient is –0.00001 are set equal to 0.0238 (0. only two of the estimated coefficients are significantly different from zero.00001 0.01 0. the standard errors from the first three sets of estimates are sufficiently similar for the same conclusions to be reached about the significance of estimated coefficients. 4e 306 Exercise 8.0145) –0. It places very heavy weights on a relatively few number of observations.00081) –0. The magnitudes of the coefficients change considerably when variances less than 0.0344 (0. holding other variables constant.000162. holding other variables constant. taking insurance is an indication that a borrower is more reliable. if a mortgage carries mortgage insurance. .000162. The positive sign is reasonable as a higher interest rate will result in a higher probability of delinquency.0065. the probability of delinquency decreases by 0. the less likely it is that the borrower will face delinquency. holding other variables constant. a one unit increase in the initial interest rate of the mortgage increases the probability of delinquency by 0. The coefficient not significantly different from zero at the 5% level.0419. the borrower is more likely to face delinquency. CREDIT: The estimated coefficient is –0.4770. Exercise Solutions. holding other probability of with a higher of CREDIT is TERM: The estimated coefficient is –0. with the adjustable rate.22(d) (continued) INSUR: The estimated coefficient is –0.0187. The coefficient of TERM is significantly different from zero at the 5% level. The coefficient of RATE is significantly different from zero at the 5% level. the longer the term of the loan. This suggests that.0419. This suggests that. This suggests that. This suggests that.0187. The coefficient of INSUR is significantly different from zero at the 5% level.0204.0204. This suggests that. The coefficient of ARM is significantly different from zero at the 5% level. decreases the probability of delinquency by 0. 4e 307 Exercise 8.0065. The positive sign is reasonable because. The negative sign is reasonable as a borrower credit rating will have a lower probability of delinquency. the probability of delinquency increases by 0. holding other variables constant. This suggests that. a one unit increase in the credit score decreases the delinquency by 0.Chapter 8. The coefficient of AMOUNT is not significantly different from zero at the 5% level. a one unit increase in the amount of the mortgage increases the probability of delinquency by 0.4770. holding other variables constant. which leads to a higher probability of delinquency. The positive sign is reasonable because. Principles of Econometrics. The negative sign is reasonable. variables constant. The negative sign is reasonable because. the interest rate may rise above what the borrower is able to repay. if the mortgage interest rate is adjustable. given AMOUNT is constant. a one year-increase in the term between disbursement of the loan. RATE: The estimated coefficient is 0. ARM: The estimated coefficient is 0. reducing the probability of delinquency. and the date it is expected to be fully repaid. AMOUNT: The estimated coefficient is 0. as the amount of the mortgage gets larger. CHAPTER 9 Exercise Solutions 308 . 4 1 0.8FFRATE8 0.2 1 2 (b) If FFRATEt 1.2 FFRATE4 4 0. 0. INVGWTH 6 4 0.1 .2 .6 1 0. 2. 0.6 FFRATE3 0.8 1 0.5 for t 5 and FFRATEt 1 for t 6.8 1. 6. 7 and 8 is given by 0.5 0 .2 FFRATE1 4 0.8FFRATE3 0. in periods 5.2 FFRATE6 4 0.5 0. INVGWTH 7 4 0.6 FFRATE4 0. then: For t 5.8 1 0.3 and 0.4 FFRATE4 0.2 FFRATE3 4 0.8 FFRATE5 0.8 1 0.4 1 0.2 1 1.9 For t 9. INVGWTH 5 4 0.5 0. respectively.8 1 0. 0. 6. and noting that the increase was 0.4 FFRATE5 0.2. Examining the forecasts given above.6 For t 7. INVGWTH 9 4 0.8. 0.6 1. Using the notation 0 .7.4.6 1 0. and 0.6 1 0. we use the impact and delay multipliers to examine the effect of the increase.7 For t 8. 2 and 3 for the impact and delay multipliers. 4e 309 EXERCISE 9.5. we find that. INVGWTH has declined by 0.5 1 . The estimates of these values are 0. then INVGWTH 4 4 0.2 FFRATE2 4 0.8 FFRATE7 0.5 3 .2 1 2 Since FFRATE was increased from 1% to 1.6 FFRATE5 0.8 FFRATE6 0.9 . Since the delay multiplier for period 4 is zero ( 4 0) . the effect of the increase in periods 5.8 1 0. INVGWTH returns to its original level of 2% in period 9.4 1 0.3. respectively. 0.6 1 0.5 0.8 FFRATE4 0.2 1 1.4 FFRATE6 0.4 . 0. .4 1. Thus.5 2 and 0. INVGWTH 8 4 0.2 FFRATE5 4 0.4 1 0.3.5 1. our forecasts agree with the estimates we get from using the impact and delay multipliers.6 FFRATE6 0.1.4 1 0. Exercise Solutions.5% in period 5 and then returned to its original level. relative to the initial value of INVGWTH of 2% (when t 4) . 4 .4 FFRATE8 0. 1 .Chapter 9.6 FFRATE2 0.4 FFRATE9 0.2 1.8 For t 6. 7 and 8.1 (a) If FFRATEt 1 for t 1. Principles of Econometrics.4 FFRATE7 0.6 FFRATE7 0.2 1 1.6 1 0. 2 1 1. then: For t 5. INVGWTH 6 4 0. Principles of Econometrics.5 0.8FFRATE8 0.5 0. 0 1 2 .6 1.5 b0 b1 b2 0.5. .1 For t 8. and a value of INVGWTH 1 becomes the new equilibrium value.5 1 For t 9.4 FFRATE5 0.6 FFRATE7 0.9. respectively.8 1. we use the impact and interim multipliers to examine the effect of the increase.8 For t 6.6.5 0.8 1.6 FFRATE6 0. INVGWTH 8 4 0.8 FFRATE6 0. relative to the initial value of INVGWTH of 2% (when t 4) . INVGWTH has declined by 0. 7 and 8 are given by 0.2 FFRATE4 4 0.9 . The impact and interim multipliers are 0 . 4e 310 Exercise 9.8 1.5 0.4 FFRATE9 0.4 1.4 1.5 0.2 .4 1.4 1. and 1 in periods 5.5 0. 0.6. 0. 6.6 1 0.6 FFRATE5 0.8 FFRATE7 0.2 1. the estimated effects in periods 0. INVGWTH 9 4 0.5 0.2 FFRATE3 4 0.8.5 0.5 0.4 For t 7.6 1 0. 6. respectively. 0.2 FFRATE5 4 0.8 1. 7 and 8. With an increase of 0.5 0.8 FFRATE4 0. The interim multipliers for t 8 and t 9 are the same as the total multiplier.2. 1 .5 1 Since FFRATE increased from 1% to 1.4 FFRATE8 0. we find that.6 1.5b0 0.6 FFRATE3 0.1 (continued) (c) If FFRATEt 1. INVGWTH 7 4 0.5 0.5 b0 b1 b2 b3 1 . 6. 0. namely.5 0. and 0 1 2 3 periods 5.9 and 0.5 b0 b1 5.2 FFRATE2 4 0.5 for t 5.6 FFRATE4 0. Examining the forecasts given above.4 FFRATE6 0.4 FFRATE7 0. INVGWTH 5 4 0.6 1.8 FFRATE5 0. and was then kept at its new level. Thus.8 1 0. 7 and 8. Exercise Solutions.Chapter 9.2 FFRATE6 4 0.4 1.7.2 1 1. for 0 1 .2 1 1.2 1. our forecasts agree with the estimates we get from using the impact and interim multipliers.6 .5% in period 5. 660 .8.99) 1.660 and t(0. The 5% and 10% critical values for a one-tail test are t(0.95. We use * to denote significance at a 10% level and ** to denote significance at the 5% level. 2 . b2 is significant at the 10% level for a two-tail test.560 0.030** t(0. We find that b1 is significant for both types of test and for both significance levels.900 0.984 .290 .802 5. advertising has a positive impact on sales revenue.3946 2. The greatest impact is generated after one week.501.90.265 7. 4. with standard error given by se(b0 b1 ) var(b0 ) var(b1 ) 2cov(b0 .2141 The 95% confidence interval for the one-period interim multiplier is b0 b1 tc se(b0 b1 ) 5. No * implies a lack of significance.984 1.984 and t(0.011** 0.0406) 1.1.99) 1. Relevant information for the significance tests is given in the following table.842 3. Principles of Econometrics.1606 2 ( 1.909 (b) The null and alternative hypotheses are H 0 : i 0 against H1 : i 0 . respectively.235.802 2. There is a positive effect in the current week and in the following two weeks.Chapter 9. The total effect of a sustained $1 million increase in advertising expenditure is given by total multiplier b0 b1 b2 1. respectively. The 5% and 10% critical values for a two-tail test are t(0.053) .842 1.4699 2. and the t-value is calculated from t bi se(bi ) for i 0. 4e 311 EXERCISE 9.006** 1.99) 1.2 (a) Overall.181 ( 0. but no effect after 3 weeks.185) The one-period interim multiplier is b0 b1 1. Exercise Solutions.975.587 0.975. the 95% confidence interval for the impact multiplier is given by b0 tc se(b0 ) 1. b0 is only significant at the 10% level using a one-tail test.122 0.644 1. b1 ) 1.644 .99) 1.984 1.99) 1. Standard Error t-Value b0 1.95.060* 0.1922 1.214 (3.1809 b1 b2 Coefficient (c) Using tc Two-tail p-value One-tail p-value 1.474 1.842 3.061* 1. and significant at the 5% level using a onetail test. 909 1.842 3. b2 ) 2cov(b1 .0984 2 ( 1. b1 ) 2cov(b0 .2(c) (continued) The total multiplier is b0 b1 b2 1. Principles of Econometrics.984 1.9.1606 1.909 . with standard error given by se(b0 b1 ) var(b0 ) var(b1 ) var(b2 ) 2cov(b0 .009 (5. 4e 312 Exercise 9.009 The 95% confidence interval for the total multiplier is given by b0 b1 b2 tc se(b0 b1 b2 ) 7.907.0188 1. Exercise Solutions.0406) 2 0. b2 ) 1.802 2.4214 2 ( 1.265 7.911) .3946 2.0367) 1.Chapter 9. SALES 106 ˆ b0 ADV106 b1 ADV105 b2 ADV104 25.93 For the second allocation.842 4 3.802 6 48.358 2.34 3. SALES 106 ˆ b0 ADV106 b1 ADV105 b2 ADV104 25.313 44.47 SALES 108 ˆ b0 ADV108 b1 ADV107 b2 ADV106 25.265 1.265 1.265 2 45.53 SALES 107 ˆ b0 ADV107 b1 ADV106 b2 ADV105 25.16 SALES 107 ˆ b0 ADV107 b1 ADV106 b2 ADV105 25.265 1.48 SALES 107 ˆ b0 ADV107 b1 ADV106 b2 ADV105 25.358 39.34 3.802 6 2.358 2.34 1.08 313 .358 2.842 6 2.842 6 3.802 2 2.265 1.802 1.842 2 3.265 1. Exercise Solutions.3 (a) For the first allocation.802 1.34 1.23 SALES 108 ˆ b0 ADV108 b1 ADV107 b2 ADV106 25.Chapter 9.34 1.34 2.34 3.358 51.39 SALES 108 ˆ b0 ADV108 b1 ADV107 b2 ADV106 25.313 33. Principles of Econometrics.34 3.802 1.802 4 2.34 1. SALES 106 ˆ b0 ADV106 b1 ADV105 b2 ADV104 25.265 1.313 37. 4e EXERCISE 9.15 For the third allocation.358 43.265 6 38. 950 .10 and 125. b2 ) 2 2 4 cov(b2 .4989 4.7661) 16 ( 1.93 1. suggesting that the effect of advertising on sales revenue is greatest one week after the advertising expenditure is made.1501 se( f ) 81.008 (30. provides the highest sales revenue in t 108 .1317) 81.1606 4 1.3891 2.3891 2. b1 ) 2. b2 ) 2.69.52.52) The estimated variance of the forecast error for the second allocation is var( f ) ˆ2 var( ˆ ) 62 var(b1 ) 2 6 cov( ˆ .850 The 95% confidence interval for the first allocation is SALES 108 tc se( f ) 38.008 The 95% confidence interval for the second allocation is SALES 108 tc se( f ) 48.66.984 6.15 1.4214 8 ( 0.9261 6.3891 2.1501 9.34.984 9. Thus.1606 12 ( 0. in which the marketing executive spends all $6 million in t 107 .7661) 42.9261 se( f ) 42. b1 ) 2 2 cov( ˆ . The second allocation.5598 16 2. 121.3(a) (continued) The total sales from each of the 3 allocations are 134. The coefficient for the first lag is higher than the coefficients of the other lags.4214 12 ( 0.Chapter 9.5598 36 1. the first allocation leads to the largest sales forecast over the 3 weeks. Principles of Econometrics. This outcome occurs because the first allocation allows time for the full effect of the $6 million expenditure to be realized.850 (25. (b) The estimated variance of the forecast error f allocation is var( f ) ˆ2 SALES108 SALES 108 for the first var( ˆ ) 62 var(b2 ) 2 6 cov( ˆ . respectively. Exercise Solutions.0367) 24.02) The estimated variance of the forecast error for the third allocation is var( f ) ˆ2 var( ˆ ) 42 var(b1 ) 22 var(b2 ) 2 4 cov( ˆ . b1 ) 2.4989 se( f ) 24. 4e 314 Exercise 9.63.28.1317) 4 ( 0.5598 36 2. 4e 315 Exercise 9. If maximizing expected profits at t 108 is the objective.950 (35. a risk averse marketing executive may prefer the third allocation because its expected profit is only slightly less than that for the second allocation.28.54. and it has a much lower standard error of forecast error.26. This is reflected in the forecast intervals.08 1. where sales for the second allocation could be as low as 30. Exercise Solutions.Chapter 9. whereas for the third allocation the lower limit of the forecast interval is 35. Principles of Econometrics.26. then the second allocation is best.3(b) (continued) The 95% confidence interval for the third allocation is SALES 108 tc se( f ) 45.984 4. However. .90) The most favorable allocation is the second or the third. 5436 0.96 and Z (0. Z (0.025) 1. Comparing this value to the critical Z values for a two tail test with a 5% level of significance.6 1 2 0.96 . Z (0.2 . Comparing this value to the critical Z values for a two tail test with a 5% level of significance.975) 1.Chapter 9. (ii) The test statistic for testing H 0 : 2 0 against the alternative H1 : 2 0 is Z T r2 10 0.1008 1. we do not reject the null hypothesis and conclude that r1 is not significantly different from zero.96 .4 -.201 .4 (a) Using hand calculations T r1 eˆt eˆt t 2 T t 1 (b) (i) eˆt2 T 1 0.0979 1.6 . r2 eˆt eˆt t 3 T t 1 The test statistic for testing H 0 : 1 eˆt2 2 0.0 -.4 . With this small sample.96 10 the autocorrelations are a long way from being significantly different from zero.96 and Z (0.0653 0 against the alternative H1 : 1 0 is Z T r1 10 0. .2 -. Principles of Econometrics.0634 0.0634 . 4e 316 EXERCISE 9. Exercise Solutions.0653 0.62 .025) 1.207 .975) 1.5436 0. we do not reject the null hypothesis and conclude that r2 is not significantly different from zero. . The significance bounds are drawn at 1. 1.8558 0.4882 7.0916 1.8558 0. and the test statistic is given by zk T rk 1.1 -.3369 30. The test results are provided in the table below.96 Reject H 0 r3 0.96 .719 1.2 1 2 3 250 0.0916 To test whether the autocorrelations are significantly different from zero.Chapter 9.124 . It leads us to the . thus.3 . Principles of Econometrics.5 . the null and alternative hypotheses are H 0 : k 0 and H1 : k 0 .8558 0.96 .5802 333.327 1.3369 5.2 . Exercise Solutions.4882 333. the critical values are we reject the null hypothesis if zk Autocorrelations z-statistic Critical value Decision r1 0. 15.4 . 1.448 1.96 Do not reject H 0 The significance bounds for the correlogram are same conclusion as the hypothesis tests. At a 5% level of significance.8114 rk .96 .4882 112.0 -. 4e 317 EXERCISE 9.96 Reject H 0 r2 0.5 (a) The first three autocorrelations are 250 r1 t 2 Gt G Gt 250 t 1 250 r2 t 3 Gt G Gt 250 t 1 250 r3 t 4 Gt G 250 t 1 2 2 G 2 Gt G Gt G Gt G 1 3 G 2 Gt G 162.9753 333.1 . 4e 318 Exercise 9. .5 (continued) (b) The least squares estimates for 250 ˆ 1 t 2 Gt G1 Gt 250 t 2 ˆ G1 1 G Gt 1 and 1 G1 G1 2 are 162. Exercise Solutions.662249 0.8480 The estimated value ˆ is slightly larger than r1 because the summation in the denominator for r1 has one more squared term than the summation in the denominator for ˆ .1119 0. Principles of Econometrics.664257 0. The means are also slightly different.Chapter 9.974 333.4892 1 1.48925 1. we estimate that a one percentage point increase in the mortgage rate in period t relative to what it was in period t 1 decreases the number of new houses sold by a number between 20. When autocorrelation is ignored.10 ( 86.98 ( 86. The null and alternative hypotheses are H0 : 0 and H1 : 0 . we reject the null hypothesis and conclude that there is evidence of autocorrelation.971 16.82) Ignoring autocorrelation gave a lower value for the coefficient of interest and a slightly larger standard error. 215) 1. (b) The two tests that can be used are a t-test on the significance of the coefficient of eˆt 1 and 2 the Lagrange multiplier test given by T R .510 units.97 . A 95% confidence interval for the coefficient of DIRATEt 1 is b2 tc se(b2 ) 53.1077 23. Testing the significance of the coefficient of eˆt 1 .980.975.61 1. 4e 319 EXERCISE 9.98. Principles of Econometrics. The LM test value is given by LM T R2 218 0. we reject the null hypothesis and conclude that there is evidence of autocorrelation.841.97 .40.Chapter 9. we find 0. since the t-statistic is less than 1. our inferences about the coefficient could be misleading because the wrong standard error is used.distribution is 3.971 14.3306 0 0. 20.6 (a) A one percentage point increase in the mortgage rate in period t relative to what it was in period t 1 decreases the number of new houses sold between periods t and t 1 by 53.51 1.040 and 86. resulting in a confidence interval with a similar lower bound but a larger upper bound. Since the test statistic is greater than the critical value.09 The 5% critical values are t(0. Exercise Solutions.0649 t 5. . 30. (c) The 95% confidence interval for the coefficient of DIRATEt 1 is given as: ˆ 2 tc se ˆ 2 58.48 The 5% critical value from a 2 (1) .04) With 95% confidence. .0256 1.7 (a) Under the assumptions of the AR(1) model.4 2 0. Exercise Solutions.190 When the correlation between the current and previous period error is weaker. the correlations between the current error and the errors at more distant lags die out relatively quickly.94 1 1 0. Thus.9 4 0.9 2 0.190 in part (b)(iii).Chapter 9.4 4 0. et k ) (i) corr(et . as is 1.44 1 1 0. et 4 ) (iii) 2 e 2 v 1 2 k .6561 in part (a)(ii) with 4 0. as is illustrated by a comparison of 4 0.0256 in part (b)(ii). 4e 320 EXERCISE 9. et 1 ) (ii) corr(et . 0.6561 5. et 1 ) (ii) corr(et . Principles of Econometrics. the larger the correlation illustrated by a comparison of 2 e . corr(et . Also.263 0. the greater the variance 5. et 4 ) (iii) (b) 2 e 2 v 2 1 (i) corr(et .263 in part (a)(iii) with 2 e 2 e . 225103 (0.2819 0.244606 ˆ3 (c) 0.1 0.7902 ( 0.6060 1.9106) t(0.1213) 2 0.6060 0.23542 0.81) These forecast intervals are relatively wide.5 0. 1. containing both negative and positive values.5758) INF 2010Q2 ˆ3 0.2354 1.Chapter 9.3) 0.2) 0.225103 (1 0.3638.1677 0.0576.1 0.4946 ( 0. Exercise Solutions.1.84 ˆ1 0.1213 0. .9897 0.1213 0. 4e 321 EXERCISE 9.47445 ˆ 22 ˆ v2 1 ˆ 12 ˆ2 0. 1.975.8 (a) The forecasts for inflation are INF 2009Q4 0.2819 0.1001 0.2) 0.4874 ˆ 32 ˆ v2 0.2354 0.4577.1677 1.9897 0.237577 For 2010Q2 ˆ2 ˆ 1 2 2 ˆ2 1 1 0.1213 1.4305) INF 2010Q1 t(0.2819 ( 0.0 0.4) 0.9265 1.4864 1.0 0. Thus.225103 For 2010Q1 0.7902 ( 0.9897 0.4864 INF 2010Q1 0.23542 1 0. Principles of Econometrics.4874 ( 0.4864 0.4745 ( 0.4946 The forecast intervals are INF 2009Q4 t 0.975.1677 0.5 0.7902 ( 0.23542 ) 0.81) ˆ2 0.1001 0.975.5 0.6060 INF 2010Q2 0.1001 0. the forecasts we calculated in part (a) do not provide a reliable guide to what inflation will be in those quarters.2354 0.0 0.4864 0.9265 (b) The standard errors of the forecast errors are For 2009Q4 ˆ 12 ˆ v2 ˆ1 0. Exercise Solutions. Principles of Econometrics. 4e 322 EXERCISE 9.Chapter 9.9 (a) The ARDL model can be written as 1 1 yt L 2 1 L2 L 2 0 1 L 2 1 yt L3 3 L2 3 L L4 yt 4 2 L3 4 L2 3 L3 4 x 0 t L4 1 L3 1 4 1 L 2 L2 3 L3 4 L4 1 x 0 t L4 xt from which we obtain 1 1 1 1 L 2 L2 3 L2 3 L3 2 3 4 L2 3 4 L4 1 L4 1 0 0 1 L 2 L2 L3 3 4 L4 Thus. 1 1 and 1 0 0 L0 1 L 2 0 L 0 L2 L3 0 L3 0 L4 4 L4 0 0 1 1 L 1 0 L 2 L 2 L2 L2 1 1 3 3 L3 L2 2 0 L3 4 1 2 L2 4 L4 L3 2 1 L4 1 3 L4 L3 2 2 L4 L3 3 1 L4 3 0 4 0 L4 Equating coefficients of like powers in the lag operator yields 0 0 0 1 1 0 2 1 1 2 0 3 1 2 2 1 3 0 4 1 3 2 2 3 1 4 0 5 1 4 2 3 3 2 4 1 s 1 s 1 0 0 1 1 0 2 1 1 2 0 3 1 2 2 1 3 0 0 4 1 3 2 2 3 1 4 0 0 5 1 4 2 3 3 2 4 1 0 0 2 s 2 0 3 s 3 4 s 4 0 s 1 s 1 2 s 2 3 s 3 4 s 4 . 7 -. The increases at lags 4 and 8 suggest a quarterly effect.2 MULTIPLIERS -. (c) If the unemployment rate is constant in all periods. and then drops away quickly at lag 1.150 –0. Principles of Econometrics. In absolute value terms. Weight Estimate 0 1 2 3 4 5 6 7 8 9 10 11 12 –0.140 –0. an unemployment change has its greatest effect immediately.6 -.120 –0.135 –0.1 -.8 0 1 2 3 4 5 6 7 8 9 10 11 12 13 LAG The multipliers are negative at all lags. 4e 323 Exercise 9.186 –0.188 –0.173 –0. and then drops away again.115 -.4 -.517 0 in all periods and the .122 –0. then DU estimated inflation rate is ˆ ˆ 1 ˆ1 ˆ 2 ˆ 3 ˆ 4 0.5 -. It increases again at lags 3 and 4. Exercise Solutions. After that the effect is small.1677 0.Chapter 9.174 –0.2354 0.315 –0.162 –0. although there is a slight increase at lag 8.2819 0.1001 1 0.790 –0.9 (continued) (b) The estimated weights up to 12 lags and their graph are given below.3 -.1213 0. 9 4 4 5 10 6 10 7 10 7 10 8 10 9 10 10 The one and two-quarter delay multipliers are ˆ ˆ 1 DURGWTH t INGRWTH t 1 2 DURGWTH t INGRWTH t 2 0. then growth in the consumption of durables will increase by 0.0370 These values suggest that if income growth increases by 1% and then returns to its original level in the next quarter.Chapter 9.9 3. Principles of Econometrics.9) 0.10 (a) The forecasts for DURGWTH are DURGWTH 2010Q1 0.7524) 0.1631 (0.7524 DURGWTH 2010Q 2 0.1631 (0.7422 (0.6) 0.0370 0. 4e 324 EXERCISE 9.7422 0.8) 0.1) 0. Exercise Solutions.6) 0.3479 (0.0103 0.9 1.6 10 2.3479 (0.0 4. the formulas given in that section for the lag weights are relevant.6 10 4. .2268 0.037% two quarters later.3 6.7422 (0.0060 9. They are 0 0 1 1 1 0 s 1 s 1 s 2 The lag weights for up to 12 quarters are as follows.1 1.6901 (b) Since this model has the same lags as the example in Section 9.227% in the next quarter and decrease by 0.2268 0. (c) Lag Estimate 0 1 2 3 4 5 6 7 8 9 10 11 12 0.8 of POE4.8 10 1.0103 0. 9373 This value suggests that if income growth increases by 1% and is maintained at its new level.969% in the next quarter and increase by 0. at the new equilibrium. growth in the consumption of durables will increase by 0. then growth in the consumption of durables will increase by 0. the total multiplier can be obtained by summing all the coefficients in that table. .0370 0. Exercise Solutions.Chapter 9. Principles of Econometrics.937%.7422 0.10(c) (continued) The one and two-quarter interim multipliers are ˆ ˆ ˆ 0 ˆ 0 1 1 0. then. Doing so yields ˆ j 0 j 0. Since the coefficients in the table in part (b) become negligible by the time lag 12 is reached.932 These values suggest that if income growth increases by 1% and is maintained at its new level.2268 0.932% two quarters later.969 ˆ 2 0. 4e 325 Exercise 9.969 0. we have (1 L)et vt et (1 L ) 1 vt (1 L vt vt 2 2 3 3 L 2 1 vt L )vt 3 2 vt 3 3 3 L ) ( L 2 2 L 3 3 L ) .11 (a) (c) To write the AR(1) in lag operator notation. Principles of Econometrics. 4e 326 EXERCISE 9. we have et et 1 vt et et 1 vt (1 L)et (1 L )(1 vt Since we can show that (1 (1 L)(1 L L) L) 1 2 2 1 1 1 3 3 L 2 L L ) L2 3 3 (1 L by showing L 2 2 L 1 Thus. Exercise Solutions.Chapter 9. 1954 0.034526 1.0664 0.3971 0.4620 3. (i) A 95% confidence interval for ˆ t 0.2543 0.1974 1.5) se b0 b1 b2 The critical value is t 0.1726 0.1830 0.0707 0.0365.0015 0.815 1.58427 0.0062 0. Since t 1.662 .975.4165 3.975.336 1.1.4416 0.3492 0.662 .4229 0.5592 3.5111 3.1974 0.5113 3.0328 ( 0.1728 0.5472 0.0169 2 3 4 5 0.336 .0021 0.5 The test statistic is t b0 b1 b2 ( 0.1940 0.3844 Note: AIC* = AIC 1 ln(2 ) and SC* = SC 1 ln(2 ) The AIC is minimized at q (b) 2 while the SC is minimized at q 1 .987 0.3119 1 1 q 2 q 3 q 4 q 5 6 0.987 0.4587 0.2999 0. 0.1322) (ii) The null and alternative hypotheses are H0 : 0 1 2 0.0944 6 AIC AIC* SC SC* q 3.6208 3. we reject the null hypothesis and conclude that the total multiplier is greater than 0. (iii) The estimated normal growth rate is Gˆ N 0.0584 0. Exercise Solutions.437344 1.1972 0.Chapter 9.2753 3.2052 0.5990 0.6037 3.253.88 se ˆ 0 0 0 is given by 0.4370 0. Principles of Econometrics.5239 0.5843 0. The 95% confidence interval for the normal growth rate is Gˆ N t 0.1699 0.6002 0.0475 0.1693 0.4188 0. 4e 327 EXERCISE 9.2135 0.3490 0.419) .1940 0.0417 (1.5828 0.5.0222 0.0713 0.1132 0.0065 0.5935 3.2223 0.3673 3.0828 0.2205 3.95.062656 1.0192 0.5809 3.1768 0.12 (a) Coefficient Estimates and AIC and SC Values for Finite Distributed Lag Model q ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 0 0 q 0.5991 3.88 se Gˆ N 1.2626.815 0.4314 0.88 0.0225 0.5 H1 : 0 1 2 0.0662 0. The p-value is 0. 634 8. Principles of Econometrics.4015 0.3756 0.9 0.267 Total Multiplier 6.6 0.3 1.067 8.4288 SC 1 ln(2 ) 3.211 3.13 (a) The graphs for SALES and ADV follow. SALES 34 32 30 28 26 24 22 25 50 75 100 125 150 100 125 150 ADV 1.0 0.5949 0.5 25 50 75 (b) Lag SC 0 1 2 3 4 5 0. Both appear not to be trending and both fluctuate around a constant mean. .863 8.433 3.Chapter 9.239 3.020 7.265 3. Exercise Solutions.214 3.7 0.4269 0.595 The total multiplier is sensitive to lag length up to lag 3.8 0. 4e 328 EXERCISE 9.1 1. for lag 3 and longer lags there is little variation.4 1.2 1.3736 0.275 8. 3. The estimates for this lag length appear below. The declining lag weights are sensible. with the exception of the weight at lag 3 which is greater than that at lag 2.976 0.5267 The 95% confidence interval for the two-week delay multiplier is b2 t(0.4734 1.975.6.975.488. 4e 329 Exercise 9.8249 (3.860) (iii) Two-week delay multiplier: b2 SALES t ADVt 2 1. The lag weight at lag 2 is not significantly different from zero at a 5% level. however.4. The lag structure is such that the greatest impact from advertising on sales is felt immediately and the lag weights decline thereafter. the SC reaches a minimum when the lag length equals three.Chapter 9.445) (ii) One-week interim multiplier: b0 b1 2. the increase at lag 3 is not expected.147)se(b0 b1 ) 5.13 (continued) (c) Of the six possible lag lengths.2298 1.4734 The 95% confidence interval for the one-week delay multiplier is b1 t(0.7564 2. Exercise Solutions.147)se(b2 ) 1.600.2298 The 95% confidence interval for one-week delay multiplier is b0 b1 t(0.541) . Principles of Econometrics.5267 1.502.976 1. the remaining lags weights are significant.976 0. We expect the effect of advertising to diminish over time.9976 (0.147)se(b1 ) 2.4734 5.975. (d) (i) The one-week delay multiplier is: b1 SALES t ADVt 1 2.0194 ( 0. Its estimate is b0 (b0 b1 ) (b0 b1 b2 ) 2. the total increase over 3 weeks is 3 0 2 1 2 .976 0.8387 (5. We wish to test H0 : 3 0 2 1 2 6 versus H1 : 3 0 2 1 2 6 The value of the t-statistic is t 14.655 . with the forecast in the last period equal to the intercept.5267 6. ADV157 0. The required sample values of ADV are ADV155 0.7564 2. by 0 1 in the second week.889. the large initial expenditure on advertising leads to a large initial increase in SALES which then declines over the forecast horizon.743 with se(3b0 2b1 b2 ) 1.95. The forecast values for each part are presented in the table below: Forecast Values ($millions) t 158 (i) (ii) (iii) 24.150 t 159 22.7426 6 1. Principles of Econometrics.850 In the first set of forecasts.147)se(b0 b1 b2 ) 6.160 and then 161. .018 31.7565 14.8777 ADVt 3 For forecasting 1.7035 5. we reject H 0 and conclude that the CEO’s strategy will increase sales by more than $6 million over the 3 weeks.7564 ADVt 2. ADV156 0.912 27.4734 1. and by 0 1 2 in the third week.634).7035 .4734 ADVt 1 1.998 .975.5267 ADVt 2 1.2162 2. Having a uniform expenditure of $1 million in each year leads to SALES that are more uniform and which achieve a value equal to the intercept plus the total multiplier in the final period (27.216 26. Exercise Solutions.7564 5.090 27.159.099.419 27.Chapter 9. 4e 330 Exercise 9. SALES gradually declines as the effect of the advertising expenditure during the sample period wears off.147) 1.727 27.197 27.414) A $1 million increase in advertising expenditure in each week will increase sales by 0 in the first week.681.216 + 8.13 t(0.850 = 19.2298 6.847 19. In the second set of forecasts.13 Since 5.13(d) (continued) (iv) Two-week interim multiplier: b0 b1 b2 2. (f) The estimated equation is SALES t 19. 2.394 35. 3 and 4 weeks into the future we set t 158.7565 The 95% confidence interval for the two-week interim multiplier is b0 b1 b2 (e) t(0. Thus.8.7565 1.248 t 160 t 161 21. the coefficient of ln( Pt ) . The different signs attached to the delay multipliers. at period t 3 it is 1. 4e 331 EXERCISE 9.1006) (0.5546 1. the relatively large weight at t 3 .5572 0. This result is symptomatic of collinearity in the data. b4 .14% and.22%. In period t 1 . are not significant at a 5% level.2985) The interim and delay elasticities are reported in the table below.77%. Exercise Solutions.14 (a) The estimated model is ln( AREAt ) 3.7746 0. namely. b2 . In period t 2 there is practically no change in the area planted.59% and in period t 4 there is a decrease of 0. The pattern is likely attributable to imprecise estimation.1271 Only b0 . after t 4 periods there is a 1.3221) 0.3153) (0.5868ln( PRICEt 3 ) 0.0143ln( PRICEt 4 ) (0.0143 0.7746ln( PRICEt ) 0. Lag Delay Elasticities Interim Elasticities 0 1 2 3 4 0.Chapter 9.3185) (0. The interim multipliers represent the full effect in period t s of a sustained 1% increase in price in period t. Thus.3129) (0.2175 0. In period t 3 there is an increase in area planted by 0.55%.77% in area planted.13% increase. b3 . at period t 2 it is 0. Interpreting the delay multipliers. . if the price increases by 1% in period t. there is an immediate increase of 0. When collinearity exists. b1 . The total increase when period t 1 is reached is 0.8241 0. is significantly different from zero at a 5% level of significance.1414 1.2175ln( PRICEt 1 ) 0. there is a decrease in area planted of 0. Principles of Econometrics. least squares cannot distinguish between the individual effects of each independent variable. All coefficients for lagged values of ln( Pt ) . and the interim multipliers that decrease and then increase are not realistic for this example.56%. resulting in large standard errors and coefficients which are not significantly different from zero.0026ln( PRICEt 2 ) (se) (0.7746 0. that is one period after the price shock. there is an immediate increase in the area planted of 0.5868 0.01%. if the price is increased and then decreased by 1% in period t.0026 0. 0261 0.42467 4 0. 0. Principles of Econometrics.3250 These lag weights satisfy expectations as they are positive and diminish in magnitude as the lag length increases.4247 and a1 0 a0 0.09963 0.42467 2 0.Chapter 9.42467 1 a0 a1 2 a0 2a1 0.1258 4 a0 4a1 0.42467 0. 4e 332 Exercise 9.09963 0.14 (continued) (b) Using the straight line formula the lag weights are i 0 0 0 1 0 2 0 2 1 i 3 0 3 1 i 3 4 0 4 1 i i 1 1 2 4 Substituting these weights into the original model gives ln( AREAt ) 0 ln( PRICEt ) 3 0 1 0 1 ln( PRICEt 1 ) ln( PRICEt 3 ) 0 4 1 0 2 1 ln( PRICEt 2 ) ln( PRICEt 4 ) et ln( PRICEt ) ln( PRICEt 1 ) ln( PRICEt 2 ) ln( PPRICEt 3 ) ln( PRICEt 4 ) 0 ln( PRICEt 1 ) 2ln( PRICEt 2 ) 3ln( PRICEt 3 ) 4ln( PRICEt 4 ) 1 z z 0 t0 1 t1 et et where zt 0 ln( PRICEt ) ln( PRICEt 1 ) ln( PRICEt 2 ) ln( PPRICEt 3 ) ln( PRICEt 4 ) zt1 ln( PRICEt 1 ) 2ln( PRICEt 2 ) 3ln( PRICEt 3 ) 4ln( PRICEt 4 ) (c) The least square estimates for (d) The estimated lag weights are ˆ ˆ ˆ ˆ ˆ 0 and 1 are a0 0. The linear constraint has fixed the original problem where the signs and magnitudes of the lag weights varied unexpectedly.2254 3 a0 3a1 0.09963 0. Exercise Solutions. with the biggest impact being felt immediately and with a declining impact being felt in subsequent periods. .42467 3 0.0996 . They imply that the adjustment to a sustained price change takes place gradually.09963 0. 14 (continued) (e) The table below reports the delay and interim elasticities under the new equation. .4247 0.7497 0. and the 3-year interim multipliers are very similar.1009 1. It is interesting that the total effect.0261 0.1258 0. Lag Delay Elasticities Interim Elasticities 0 1 2 3 4 0.3250 0. is a more reasonable one. with the restricted weights leading to a smaller initial impact. Principles of Econometrics. Exercise Solutions. is almost identical in both cases. This result.Chapter 9.4247 0. 4e 333 Exercise 9. The earlier interim multipliers are quite different however.9751 1. given by the 4-year interim multiplier.1270 These delay multipliers are all positive and steadily decrease as the lag becomes more distant.2254 0. compared to the positive and negative multipliers obtained earlier. 975.15 The least-squares estimated equation is ln( AREA) 3.1660) 1 vt .0613) (0.4 . yielding a p-value of 0.8884ln( PRICEt ) (se) (0.2 -. 4e 334 EXERCISE 9.96 significantly different from zero.0193.8933 0.3413) (ii) Using HAC standard errors b2 t 0. Principles of Econometrics.1.0057.32 se(b2 ) 0.Chapter 9. we reject the null hypothesis and conclude that there is evidence of autocorrelation at the 5 percent significance level. (d) The estimated equation under the assumption of AR(1) errors is ln( AREAt ) 3. Exercise Solutions.0369 0.8988 0.7761ln( PRICE ) (0.2109.7761 2. Since the p-value is less than 0.3782) HAC se's (a) The correlogram for the residuals is .3 -.5465) The wider interval under HAC standard errors shows that ignoring serially correlated errors gives an exaggerated impression about the precision of the least-squares estimated elasticity of supply.2771) least squares se's (0.336 . (c) The 95% confidence intervals are: (i) Using least square standard errors b2 t 0.3 .2775 (0.05.1 -. Autocorrelations 1 and 5 are (b) 0 and H 0 : 0 .1 .5 .7761 2.32 se(b2 ) 0.0624) (0. 10 34 12 14 16 18 20 22 24 0.0369 0.4 2 4 6 8 The significant bounds used are 1.1.0 -.3782 (0.4743 .4221et (0.0922) (0.975. and the test statistic is The null and alternative hypotheses are H 0 : LM 5.2593) et 0.2 . we test H 0 : 1 1 0 against H1 : 1 1 0.4179) This confidence interval is narrower than the one from HAC standard errors in part (c).1) model to be equal to the AR(1) model in part (d).1 .96 33 0. The test value is t ( ˆ1 ˆ 0 ) 1 se 1 + 0.3 -. 4e 335 Exercise 9.6109 ln( PRICEt 1 ) (0. We conclude that ˆ is significantly different from zero at a 5% level. although we cannot infer much from this difference because the least squares standard errors are incorrect.3588.3662 0.542 .2812 1 0 1.0559 with p-value of 0.0423 0.1660 2.6109 ( 0.2 -. (e) We write the ARDL(1.300.1 -.Chapter 9.2798) (0.975.2 .7766 ln( PRICEt ) 0.8884 2.0164. we fail to reject the null hypothesis and conclude that the two models are equivalent.1666) (0.2966) For this ARDL(1. we need to impose the restriction 1 1 0 . A 95% confidence interval for the elasticity of supply is b2 t 0. The significance bounds used are 1.423 confirms this decision. with a p-value of 0. Thus. reflecting the increased precision from recognizing the AR(1) error.30 se(b2 ) 0.15(d) (continued) The t-value for testing whether the estimate for is significantly different from zero is t 0. It is also slightly narrower than the one from least squares.6557) (0. Principles of Econometrics.4 .1) model as ln( AREAt ) 1 ln( AREAt 1 ) 0 ln( PRICEt ) 1 ln( PRICEt 1 ) et The estimated model is ln( AREAt ) 2.4043ln( AREAt 1 ) 0. .3412 .4221 0.2593 (0. Thus.1. The LM test with a p-value of 0.0 -. The correlogram presented below suggests the errors are not serially correlated. Exercise Solutions.4043 0.4 2 4 6 8 10 12 14 16 18 20 22 24 .3 .7766) 0. 02.69) For T (e3.63167) ln( AREA) T 2 t 0.20132 .0539 exp 0. The interim elasticities show the percentage change in area sown in the current and future periods when price increases by 1% and is maintained at the new level.3112 0.45830) The corresponding intervals for AREA obtained by taking the exponential of these results are: (c) For T 1 : (e3.0452 0.0196 The lag elasticities show the percentage change in area sown in the current and future periods when price increases by 1% and then returns to its original level.96 The standard errors of the forecast errors for ln( AREA) are se(u1 ) ˆ se(u2 ) ˆ 1 ˆ 12 0.56. The corresponding forecasts for AREA using the natural predictor are n AREAT 1 exp(4.46630 .63167 ) (32.4. Principles of Econometrics.3073 The 95% interval forecasts for ln( AREA) are: ln( AREA) T 1 t 0. they are c AREAT 1 n AREAT 1 exp ˆ 2 2 2 n AREAT 2 exp ˆ 2 2 c AREAT (b) 57.102.16 (a) The forecast values for ln( AREAt ) in years T 1 and T 2 are 4.0485 –0.28490 1 0.45830 ) (24.04899) 57. e4.3073 (3.7766 0.1200 –0.34 2 exp(3. 4e 336 EXERCISE 9.4797 0.71 47.404282 0.82981) 46.2916 –0.04899 and 3.29 se(u1 ) 4.2848992 2 46.3395 exp 0.2969 –0.20132. e4.82981.28490 (3.28490 0.86.29 se(u2 ) 3.82981 2.34) 2: The lag and interim elasticities are reported in the table below: Lag 0 1 2 3 4 s 0 = 0 1 1 2 1 1 3 1 2 4 1 3 1 0 Lag Elasticities Interim Elasticities 0.4.975.7766 0.2848992 2 59.04899 2.Chapter 9. .4663.0452 0. respectively.975.3597 0. Exercise Solutions.05 n AREAT Using the corrected predictor. 40428 0.16 (continued) (d) The total elasticity is given by ˆ j 0 j ˆ 1 1 ˆ1 0 0.2783 If price is increased by 1% and then maintained at its new level.61086 1 0. 4e 337 Exercise 9. Exercise Solutions. then area sown will be 0.77663 0.28% higher when the new equilibrium is reached. . Principles of Econometrics.Chapter 9. 10 .1245 .0247. The significance bounds are drawn at 1.05.1332Gt 2 0.Chapter 9. .1533Gt 3 (0.05 .00 -.0627 1 0.10 -.17 (a) The estimated model is Gt 0.0635) The correlogram of the residuals is shown below. Since the p-value is less than 0. There are two significant correlations at the long lags of 10 and 16.405 and the corresponding p-value is 0. they are relatively small. There are a few significant correlations at long lags (specifically at lag orders 5.0676 2 0. we reject the null hypothesis that autocorrelation does not exist and conclude that there is evidence of autocorrelation at the 5% significance level. 11 and 19).15 -. but apart from lag 5. (b) The estimated model is Gt (se) 0.0633 0. The significance bounds are drawn at 1.96 247 0.4432Gt 0. 10.1995Gt 0.96 248 0. . but they are relatively small.0636 The correlogram of the residuals is shown below.7316 0. Exercise Solutions.20 2 4 6 8 10 12 14 16 18 20 22 24 The test value for the LM test with two lags is LM 7. Principles of Econometrics.1247 .8386 0.4249Gt (se) 1 0. 9.05 -. 4e 338 EXERCISE 9.15 . 287) (–0.247) 1.9696 .1515 (–0. 3.15 . The t-value used to compute the forecast intervals was t(0. 3.Chapter 9.975.00 -.3371 1.969) 1.9899 1.05 -.511.916 and the corresponding pvalue is 0.17(b) (continued) . Principles of Econometrics.754) (–0.613.15 1. Exercise Solutions.18 0. Period Forecasts Standard Errors Forecast Intervals Actual Figures 2009Q4 2010Q1 2010Q2 1.0827 1.6214 1. 3.05. we conclude there is no evidence of autocorrelation at the 5% significance level.10 -.15 2 4 6 8 10 12 14 16 18 20 22 24 The test value for the LM test with two lags is LM 0.567.7014 0.914 The actual figures fall within the intervals. . Since the p-value is greater than 0. we do not reject the null hypothesis of no autocorrelation.10 .632.05 . 4e 339 Exercise 9. (c) The results are presented in the table below. 18 (a) The estimated AR(2) model is SALES t 11. . 4e 340 EXERCISE 9.328 MSPE Exp. LM tests at various lags similarly show no evidence of serial correlation. Exercise Solutions.900 28.020 28.479 Using this criterion. In two periods the absolute value of the forecast error is lower for exponential smoothing and.3658 The mean-square prediction errors for each set of forecasts is MSPE AR(2) 2.6542 –0.3946 SALESt 1 0.6187 –1.963 26.3925 28.2011 28. the AR(2) model has led to the more accurate forecasts.614 0.2596 1. Sm.5187 26.1926 SALESt 2 The correlogram below shows no evidence of serially correlated errors.6896 27.7452 –1.5364 27. Principles of Econometrics. Both methods tend to over or under forecast at the same time. Forecast Error 154 155 156 157 28. Forecast AR(2) Forecast Error Exp.5705 2.7619 2.Chapter 9. in the other two periods.430 25.9021 28.1179 –0. (b) to (e) The following table contains the one-period ahead forecasts and forecast errors for both the AR(2) and exponential smoothing models after re-estimating both models for each period.1064 1. (f) Forecast Period Observed Value AR(2) Forecast Exp.6452 26. Sm. 2. the forecast errors for the AR(2) model are smaller. Sm. 0 -80 -.3327) (0.8550 DIRATEt are significantly different from zero against the alternative H 0 : 11.05.2 -120 -. Since the p-value is greater than 0.2 -40 .8408 19.7878 DIRATEt (16. and conclude that the data are compatible with the hypothesis H 0 : 1 1 2. the series for DHOMES and DIRATE do not appear to be trending but fluctuate around constant means.4 0 .3350 DHOMES t (se) (3.400 10 1.212) 2 2 1 28. trends downwards.9283) All estimates except for the intercept and DIRATEt at the 5% level.19 (a) The four graphs are as follows HOMES IRATE 1. overall.5394. IRATE wanders up and down.8 80 .6 -.975.615 1.Chapter 9.8 -200 1992 1994 1996 1998 2000 2002 2004 2006 1992 2008 1994 1996 1998 2000 2002 The series for HOMES and IRATE exhibit trends.971 . we do not reject the null hypothesis. but. HOMES trends upwards until 2005 and then trends downwards. 4e 341 EXERCISE 9.0649) 1 50. (c) The test statistic for testing H 0 : t ˆ ˆ 1 1 se ˆ 1 ˆ 1 ˆ 1 1 2 ˆ 2 The 5% critical value is t(0.2621 2 (17. (b) The estimated model is DHOMES t 2.1278) 1 1 2 is 0. and the corresponding p-value is 0. Principles of Econometrics. .200 9 1.4 -160 -.4912 0. On the other hand. Exercise Solutions.000 8 800 7 600 6 400 5 4 200 1992 1994 1996 1998 2000 2002 2004 2006 1992 2008 1994 1996 1998 2000 2002 2004 2006 2008 2004 2006 2008 DIRATE DHOMES 120 .6 40 . (f) The estimated ARDL model is DHOMES t (se) 2.1343 .974) 5 1 46.05 -.10 .8536 with a corresponding p-value of 0.324 DIRATEt (15. . Also.0635) 0.96 213 0.3073DHOMESt (3.96 216 0. And there are no coefficients (except the constant) that are not significantly different from zero.094) Using the significance bounds 1.Chapter 9. Principles of Econometrics.2069 DHOMESt 1 (0. These four things – the lack of serial correlation.15 2 4 6 8 10 12 14 16 18 20 22 24 (e) The LM 2 test value with two lagged errors is 4.92) the coefficient of DIRATEt 2 was not significant. the exclusion of a lag with an insignificant coefficient.133 . . we would conclude there is evidence of serial correlation. If we used a 10% significance level.9215 0. the model can be written as DHOMESt 1 DHOMESt 1 1 DIRATEt 1 1 1 DIRATEt vt 2 which is equivalent to the AR(1) error model DHOMESt (d) 1 DIRATEt et et 1 e 1 t 1 vt The correlogram of residuals is displayed below. Exercise Solutions. In (9.0633) 64. the improved AIC and SC. The significance bounds are 1.92). 4e 342 Exercise 9. the AIC and SC values for this model are slightly lower than those for the model in (9. and the inclusion of significant lags.0883. It suggests that there are two significant correlations at lags at 5 and 21.00 -. At a 5% significance level.10 -. the correlogram of residuals for this model does not suggest any autocorrelation except at lag 21 which is sufficiently distant to ignore.19(c) (continued) If H 0 is true.20 . we fail to reject the null hypothesis that the errors are serially uncorrelated.2841 (0.05 . lead us to conclude the new model is an improvement.631DIRATEt 3 (16.15 . 9215 0.9215 0.0633) 64.20 (a) Recognizing that DHOMESt HOMESt HOMESt HOMESt HOMESt 1 . Exercise Solutions.0) 46.631DIRATEt 3 (16.9215 0.02) 372 5 .631 (0.3073 324 0.2069 DHOMESt 1 5 (0.6927 380 0.324 (0.3073 370 0.631DIRATEt 3 The forecasts for April.9215 0.2069 ( 38) 64.631 ( 0.094) (15.3073DHOMESt (3.324 DIRATEt 1 2 0. May and June 2010 are HOMES APRIL 2.Chapter 9. Principles of Econometrics.04) 380 HOMES JUNE 2.6927 370 0. we can write the equation as 1 1 ( HOMESt 0 HOMESt 2 ) 1 DIRATEt 1 3 DIRATEt DHOMESt 5 5 vt 3 Rearranging yields HOMESt 1 HOMESt ( 1 HOMESt 2 1 3 DIRATEt 3 vt 1) HOMESt 1 1 HOMESt 2 5 DIRATEt 1 3 DIRATEt 3 vt 0 (b) HOMESt 1 DIRATEt 0 = 1 1 5 DHOMESt DHOMESt 5 5 The estimated equation is DHOMES t (se) 2.974) The equation to be used for forecasting is HOMES t 2.6927 HOMESt 1 0.6927 411 0.2069 ( 9) 64.2841 (0. 4e 343 EXERCISE 9.0635) 0.631 ( 0.324 ( 0.324 (0.0) 46.1) 370 HOMES MAY 2.324 DIRATEt 1 46.2069 DHOMESt 46.3073 411 0.3073HOMESt 64.2069 ( 15) 64.02) 46.9215 0. 975.6927 2 2 1 47.975. 494) HOMES JUNE t 0.827 (236.6927 2 1 12 68.6927 2 ˆ 1 0. 4e Exercise 9.20 (continued) (c) The standard errors of the forecast errors are se(u1 ) ˆv se(u2 ) ˆv 1 se(u3 ) ˆv 47.502 0.971 47.502 ˆ ˆ 1 1 1 1 2 12 2 ˆ 47.508) 344 .785 12 1 0. Exercise Solutions.971 57.208 se(u3 ) 372 1.502 1 0.Chapter 9.208 se(u3 ) 370 1.785 (266.827 The three forecast intervals are HOMES APRIL t 0.208 se(u2 ) 380 1.3073 1 2 2 12 57.975. 464) HOMES MAY t 0. Principles of Econometrics.502 (276.971 68. 3742 0. The significance bounds used are 1.0082Gt (0.0846) 0.141 (i) The estimated model with DU t DU t -value 2 0.1832Gt (0.3230 DU t 1 (0. Exercise Solutions.0587) (0. No serious problems of error autocorrelation are apparent. we conclude there is no evidence of serial correlation in the errors. 4e 345 EXERCISE 9. 2 Lags 1 2 3 4 (d) 10 (se) DU t (se) 2 0.1841Gt (0.1823Gt 0.0720) (0. Using any significance level up to 10%.0374) added is 0.3870 0.0311) 0.3876 0.0586) (0.0991Gt (0.0370) 1 0.873 0.0971Gt (0.1060) (ii) The estimated model with Gt p-value 0.0 -.3 .0458 DU t (0.271 3.0307) 0.2 -.96 96 .896 6.3501DU t 1 (0.Chapter 9.2 . The only slightly significant autocorrelation is at lag 13.1 -.1 . Principles of Econometrics.2 .189 added is 0.0992Gt 1 (0. 0.1. In all cases the p-values are greater than 0.0360) 2 1 .3 2 (c) 4 6 8 12 14 16 18 20 22 24 The following table gives the LM test results for lags up to 4.680 0.170 0.273 0.3391DU t (0.0979) 1 0.0990) 2 0.0314) (0.0368) The residual correlogram for lags up to 24 is presented below.21 (a) The estimated equation is DU t (se) (b) 0. 0758) (0. Whenever DU t 2 or Gt 2 or both were added to the original equation. Also. the coefficient estimates found to be significant at the 5% percent level were those for DU t 1 . Exercise Solutions. we concluded that error autocorrelation is not significant.1821Gt (0. their estimated coefficients were insignificant.0030Gt 2 (0. in part (d).3778 0. adding DU t 2 and/or Gt 2 did not improve the model.3208 DU t (0.0376) 1 0. For these reasons. 4e 346 Exercise 9. Principles of Econometrics. we conclude that the Okun’s law specification given in (9.0970Gt (0.1065) added is 0.0429 DU t 2 2 (0.0316) 0.59) is satisfactory. (e) In parts (b) and (c). Their coefficients were not significantly different from zero.21(d) (continued) (iii) The estimated model with both DU t DU t (se) 0. Both the correlogram and the LM tests supported such a conclusion.0389) For all three estimated equations.Chapter 9.1103) 1 and Gt 2 0. . Gt and Gt 1 . Exercise Solutions.46%.4496 INCGWTH t (0.0996) (0. with a p-value less than 0. There is also some slight evidence of serially correlated errors at some longer lags (6. 4e 347 EXERCISE 9.4496 suggests that a 1% increase in the income growth rate increases the consumption growth rate by 0. 10 and 11).9738 0. they do appear to fluctuate around their respective constant means. . While both exhibit considerable serial correlation. Principles of Econometrics. INCGWTH CONGWTH 5 7 4 6 5 3 4 2 3 1 2 0 1 -1 0 -2 -1 -3 -2 25 (b) 50 75 100 125 150 175 200 25 50 75 100 125 150 175 200 The estimated model is CONGWTH t (se) 0.00005 – a strong indication of serially correlated errors.Chapter 9.22 (a) The times series graphs for CONGWTH and INCGWTH follow.93 . we find (2) 21. 2 For the LM test.0497) The estimate 0 0. The correlogram below shows significant serial correlation in the errors at lag 2. ) .1328 for the model in part (b).1250 and 0.0000 .) However. p -value 0.1508 . compared to 0.0119 for the model discussed in part (b).9273 and 1.0635) 1 0.2714 is significantly different from zero at the 5% significance level (t 4. the correlogram of the residuals displayed below suggests there is still significant serial correlation in the errors at lags 1 and 2.0197 and 2.0697 for the above model.4249 0.9940 for the above model. (2) We conclude that the model is an improvement over that in part (b).3216 INCGWTH t (0.1250 and 0. the lower values suggest this model is an improvement. See footnote 12 on page 366 of POE4. The AIC and SC values for this model are 0.2174 and 0.0452 and 0. the lower values suggest this model is an improvement. The LM test also rejects the null 2 hypothesis that the errors are not serially correlated 34.0697 for the model in part (c). respectively.6716 0.1188) (0.2806CONGWTH t (0.3501 INCGWTH t (0. Principles of Econometrics.45.0995 and 2.0750 for the model discussed in part (c). Exercise Solutions.2806 is significantly different from zero at the 5% significance level (t 4.0509) The estimate 2 0. (d) The estimated model after adding CONGWTH t CONGWTH t (se) 0. and 2.1594CONGWTH t (0.0615) 2 0. (The corresponding EViews AIC and SC values are 2.0197 and 2. (The corresponding EViews AIC and SC values are 1.Chapter 9. 4e 348 Exercise 9.0530) The estimate 1 0.0653) 1 2 is 0.22 (continued) (c) The estimated model after adding CONGWTH t CONGWTH t (se) 1 is 0. The AIC and SC values for this model are 0.0750 .2714 CONGWTH t (0.57) . and 2.1254) (0. but it is still not satisfactory. compared to 0. respectively.27) . 4e 349 Exercise 9. suggesting that serially correlated errors are still a problem. (e) The estimated model after adding INCGWTH t CONGWTH t (se) 1 is 0.3004 0. r1 2 (2) value of 15.9277.22(d) (continued) In the correlogram of the residuals given below. (The EViews values are 1.Chapter 9.143 . The AIC and SC values for this model are 0.0699) 0.1219) (0. respectively.) .3493INCGWTH t (0. Exercise Solutions. We conclude that adding CONGWTH t 2 has improved the model.2334 is significantly different from zero at the 5% significance level (t 4.0233CONGWTH t 1 (0.2101CONGWTH t 2 (0.2334 INCGWTH t 1 (0. lower than that for the model discussed in part (d). but the existence of serially correlated errors means that it is still not satisfactory. with corresponding p-value = 0.8444 and 1.2170 .0539) The estimate ˆ 1 0.3320 0. Principles of Econometrics.0610) 0. although its magnitude.0491) 0. the first autocorrelation is significantly 0.0004.33) . is not large. The LM test gives a different from zero.46. the extra coefficient was not significantly different from zero. (2) suggest serial correlation is a problem.1591) do not test values.0596) 0.593 ( p -value 0. (g) Dropping CONGWTH t 1 from the model in part (e) and re-estimating gives CONGWTH t (se) 0. (EViews values are 1.3407 0.3099 and 0.2433 . 4e 350 Exercise 9.0480) The AIC and SC values are 0.1188) (0.) The correlogram below shows some evidence of serially correlated errors at lag 4.2414 INCGWTH t 1 (0. We conclude that this model is an improvement over that in part (d).Chapter 9.9301) .145 ( p-value 0.0454) 0.8348 and 1. In both cases. (f) Adding CONGWTH t 3 or INCGWTH t 2 did not improve the model in part (e).8957) However. and (4) 6. the correlgrams and LM statistics led to the same conclusion about serially correlated errors as was reached in part (e).220 ( p-value 0. but the LM 2 2 0. Exercise Solutions.3555 INCGWTH t 2 (0.9015. and the AIC and SC values increased.204 ( p -value 0.1255) suggesting serial correlation is no longer a problem.22(e) (continued) The correlogram above shows a significant but not large autocorrelation at lag 4. Furthermore. performing the LM test with 2 and 4 lags gives (2) and 2 (4) 7. . Principles of Econometrics. 2 0. respectively – values that are lower than those for the model estimated in part (e).2143 CONGWTH t (0. 6 0. 2.0077 0.34074 0.041) (–0.9 1.Chapter 9. the standard errors of the forecasts and the forecast intervals are given in the following table.35545 I 2010Q 2 0.24144 0.24144 I 2010Q1 0.0454) 0.2414 INCGWTH t 1 (0.7 0.23 The estimated equation is CONGWTH t 0.59954 ˆ3 ˆv 1 2 2 0. Exercise Solutions.24144 0.6132 (0.21428 C 2010Q1 0.0596) 0.1188) (0.00767 The standard errors of the forecast errors are ˆ1 ˆv 0.21428 1.3 0. 2.98424 C 2010Q 3 0. Principles of Econometrics.6 0. .35545 I 2010Q1 0.95.21428 1.9842 1. the forecasts are obtained as follows C 2010Q1 0.34074 0.35545 0.34074 0.34074 0. Period Forecasts Standard Errors Forecast Intervals 2010Q1 2010Q2 2010Q3 1. showing that there is a great deal of uncertainty about future consumption growth.24144 I 2009Q 4 0.6528 .059.61315 t(0.021) Using C as an abbreviation for CONGWTH and I as an abbreviation for INCGWTH.34074 0.193) ˆ j where t(0.214277 2 0.04987 C 2010Q 2 0.21428 C2009Q 4 0.0499 0.04987 0. The intervals are relatively wide.3555 INCGWTH t 2 (0.2143 CONGWTH t (se) (0.24144 0.21428 1.24144 I 2010Q 2 0.95.0480) The forecasts. 1.193) 1.0 0.34074 0.5995 0.59954 The forecast intervals are given by C j 1 0.21428 C2009Q 3 0.35545 I 2010Q 3 0. 4e 351 EXERCISE 9.8 0.59954 ˆ2 ˆv 0.3407 0.8 1.5995 0.35545 0.007.006.975) (–0.35545 0. 6731 The total multiplier estimate is ˆ j 0 ˆ j 0 ˆ 1 1 ˆ2 0. Lag Delay Multiplier Interim Multiplier 1 2 3 0. by 0.36% in the current quarter.35545 0. Exercise Solutions.24% in the next quarter and by 0. namely 0.2414 0. without the error term.Chapter 9.60% in the next quarter and by 0. The interim multipliers show that if the growth rate of income is increased by 1% and then maintained at this new level.3555 0.24 (a) The model in (9.7597 The delay multipliers show that if the growth rate of income is increased by 1% and then returned to its original level. When a new equilibrium is reached consumption growth will have increased by the total multiplier.08% in the quarter after that.5969 0.36% in the current quarter.94). is given by CONGWTH t 2 CONGWTH t 2 0 INCGWTH t 1 INCGWTH t 1 It can be written in lag operator notation as (1 2 L2 )CONGWTH t ( L ) INCGWTH t 0 1 (1 2 or CONGWTH t (1 2 L2 ) 1 L2 ) 1 ( 0 1 L) INCGWTH t Equating this equation with the infinite lag representation CONGWTH t ( 0 1 L 2 L2 3 L3 4 L4 s Ls ) INCGWTH t implies (1 2 L2 ) 1 ( 0 1 L) 0 1 L 2 L2 3 L3 4 L4 Thus.21428 0.86%. by 0.67% in the quarter after that. 0 1 L (1 2 0 L2 )( 0 L 2 1 2 0 1 L2 L2 L 2 3 L3 L2 3 4 4 L4 L4 L3 2 2 2 0 3 2 1 L3 L4 giving 0 (b) 0 1 1 2 2 1 s 2 s 2 s 2 The estimated multipliers are presented in the table below.24144 1 0. then the growth rate of consumption will increase by 0.3555 0. . 4e 352 EXERCISE 9.0762 0. Principles of Econometrics. then the growth rate of consumption will increase by 0. 158 . The significant bounds are 2 160 0.0 2.0215 1. and a lower mean after that.56 . However. Thus. .0 25 50 75 100 125 150 Neither of the series appears to be trending over the given time period.0 1. The LM test for AR(2) errors yields a test value of LM 33.0942) The coefficient of WGWTH suggests that an increase in wage growth of 1% results in a 1. with corresponding pvalue of 0. 2.0254WGWTH t (0. Principles of Econometrics. 3 and 4.25 (a) INF 5 4 3 2 1 0 -1 -2 -3 25 50 75 100 125 150 WGWTH 3.025 percent increase in the inflation rate. up to about observation 50 (1982Q3).5 3. (b) The estimated equation is INF t (se) 0. Exercise Solutions.0 0.5 2.Chapter 9. 4e 353 EXERCISE 9. The residual correlogram that follows shows significant autocorrelations at lags 1.5 1. Both appear to have a higher mean for the earlier period.5 0. we conclude that the errors are autocorrelated.0000. an assumption of a constant mean over the whole period could be questioned for both series. Chapter 9. we have INFt (1 1 ( ) 1 (1 0 1 1 L 2 L) 1 0 L2 3 WGWTH t L3 et )WGWTH t et and (1 1 L) 1 ( 0 0 1 L 2 L2 3 L3 ) or. the impact multiplier is given by ˆ 0 ˆ 0 0. 4e 354 Exercise 9. Exercise Solutions.4914 1.5405 INFt (se) 0. but a total multiplier that is approximately the same. we need to rewrite the model in terms of the infinite distributed lag representation INFt WGWTH t s s 0 et s Working in this direction.0652) (0.4914 .25(b) (continued) (c) The estimated equation is INFt 0.069 1 0.1021) To find the impact and total multipliers. (1 0 1 0 L)( 1 0 0 1 L 1 2 L L2 0 1 L 2 3 L2 3 L3 ) L3 0 1 2 1 1 L2 L 1 1 3 2 1 L2 2 1 L3 L3 Equating coefficients of equal powers in the lag operator gives 0 0 j 1 j 1 0 for j 1 Thus.0352 0. .5405 In part (b) the total multiplier and the impact multiplier were both equal to 1.0254. Introducing a lagged value of INF has led to an impact multiplier that is much less.4914WGWTH t 1 (0. And the total multiplier is given by j 0 j 0 0 1 2 0 1 3 0 1 0 1 1 0. Principles of Econometrics. a significant autocorrelation remains at lag 3. the coefficient 1 INFt 1 2 INFt 2 3 INFt 3 0WGWTH t ˆ was not significantly different from zero (p-value = 0. and the results of the various LM tests. but those at lags 1. Adding INFt 2 does nothing to improve the situation.2174 INFt (0. and did not introduce any serial correlation in the errors. Adding INFt 3 eliminates the serial correlation at all lags. The LM tests confirm that serial correlation remains.017 0. with 2 values that are significant at the 5% level for error processes involving 2 and 3 lags. and then INFt 2 . .028 and a fall in the AIC of 0.0691) 1 0.064 2. INFt 2 . respectively.Chapter 9. Exercise Solutions. ( INFt 1 ) ( INFt 1 .4497). Adding WGWTH t 1 did not improve the equation. INFt 2 ) ( INFt 1 . 4e 355 Exercise 9. INFt 2 ) ( INFt 1 .007 0.25 (continued) (d)&(e) The residual correlograms for models with INFt 1 added. INFt 2 . INFt 3 ) LM Test and p Values Lags included in test Lags included in equation LM value 2 p value LM value 3 p value ( INFt 1 ) ( INFt 1 .565 and 0.505 After adding INFt 1 . and then INFt 3 .143 0.4537 INFt (0.137 1. There are no significant autocorrelations at the 5% level and the p-values for the LM test for processes involving 2 and 3 lags are 0.246 12.3728WGWTH t (0.1068) In the model INFt vt .040 0. 2 and 4 are no longer significant.0504 0. Its coefficient was not significantly different from zero and the AIC and SC both increased. Principles of Econometrics.009.505.439 8. and so it was worth 2 considering dropping it. The significant autocorrelation at lag 3 remains and the LM test values do not improve.007 0. Omitting it led to a fall in the SC of 0.342 0. (f) The estimated equation is INF t (se) 0.0676) 3 0.565 12. are given below. INFt 3 ) 6. 91 0.2174 0.3728WGWTH t The forecast values are INF 2010Q 2 0.5711 ( 0.1.3728WGWTH 2011Q1 0.5435 INF 2010Q3 0.3728 0.5111 ( 0.5309 INF2011Q1 0.598.4537 INF2010Q 4 0.5928 ( 0.0504 0.5111 12 0.975. but it still leaves a great deal of uncertainty.976 0.633) INF 2010Q 4 t(0.4537 INFt 0.3728 0. Principles of Econometrics.585.1.2174 0.2174 INF2010Q1 0.0504 0.3728WGWTH 2010Q3 0.0504 0.0504 0.4537 0.51115 1 0. containing both positive and negative values.65 0.26 The estimated equation used for forecasting is given by: INF t 0.5435 1.0504 0.5435 0.4537 0.2174INF2009Q3 0.4537 INF2010Q3 0.0504 0.4 0.629) These forecast intervals are very wide.4537 INF2010Q1 0.714.976 0.4578 1.453692 1 1 12 ˆ3 1 ˆ 2 12 3 0. Knowing wage growth might help predict inflation.2174 INFt 1 3 0.0504 0.153) se(u3 ) 0.5239 0.38 0.659) INF 2011Q1 t(0.51115 1 0.453692 12 0. . 4e 356 EXERCISE 9.975.5435 0.1.5309 0.7 0.4578 The standard errors of the forecast errors are se(u1 ) ˆv se(u2 ) ˆ v 1 ˆ 12 se(u3 ) ˆ v 1 ˆ 12 ˆ4 se(u4 ) ˆ v 1 ˆ 12 ˆ4 0.5309 1.5 0.2174 0.976 0.3728 0. Exercise Solutions.2174 0.4537 0.2174 INF2009Q 4 0.5928 The 95% forecast intervals are INF 2010Q 2 t(0.3728 0.5711 0.6 0.5613 ( 0.466.Chapter 9. and hence do not contain much information about likely values of future inflation.4537 0.3728WGWTH 2010Q 2 0.0504 0.975.0504 0.153) se(u1 ) 0.2174 INF2010Q 2 0.976 0.5239 INF 2010Q 4 0.3728WGWTH 2010Q 4 0.38 0.975.553) INF 2010Q 3 t(0.4537 INF2010Q 2 0.5239 1.5613 0.153) se(u2 ) 0.153) se(u4 ) 0.1.453694 12 0. estimated inflation is ˆ 0.153) 0 when 0.Chapter 9. .153247 0. we have. Exercise Solutions.0504 0.09345 0.1532 1 ˆ ˆ 1 0.27 (a) The equation is INFt 1 INFt 1 3 INFt WGWTH t 3 vt 0 Applying the lag operator to this equation. (1 1 L 3 L3 ) 1 1 1 3 and (1 1 L 3 L3 ) 1 ( 0 0 1 L 2 L2 3 L3 ) or. 0 (1 1 L 0 1 3 L L3 )( 2 1 1 L2 0 3 0 0 3 L L3 4 L3 1 3 L 2 0 1 2 L2 3 L3 4 L4 L4 0 1 ) L 1 1 L2 2 1 L3 3 1 L4 L4 1 1 L2 3 2 1 L3 0 3 4 3 1 1 3 L3 Equating coefficients of equal powers in the lag operator gives 0 0 j 1 1 j 1 0 1 0 1 1 Thus. There is no evidence to suggest that inflation will be nonzero when wage growth is zero.05 . or.28758 t ˆ se( ˆ ) 0. alternatively.45369 0. Principles of Econometrics. Thus.21743 1 0 . since ˆ se( ˆ ) . we can use t ˆ se( ˆ ) .976 . 0.0504 ˆ 0. we do not reject H 0 .533 0. the critical values are t(0.975. expressions that can be used to calculate (b) 0 0 j 1 1 j 1 1 0 3 j 3 0 0 for j 3 j 3 3 2 2 and the s are 1 1 for j 3 When WGWTH remains constant at zero. 4e 357 EXERCISE 9.539 1. The test values from these two alternatives are To test H 0 : we can use t At 3 t ˆ se( ˆ ) 0. (1 L 1 3 L3 ) 1 INFt WGWTH t 0 and INFt (1 1 L ( 3 0 1 L3 ) 1 (1 1 2 L 2 L 3 3 L 3 L L3 ) 1 WGWTH t 0 )WGWTH t et Thus. 9322 7 1 6 3 4 0. we find i i 0 ˆ i estimate of the inflation rate is INF 0.0426 0.1691 0.0130 1.0694 12 1 11 3 9 0.0767 0.0572 0. 4e 358 Exercise 9.1301 An EViews program that can be used to compute the total multiplier is vector(200) b b(1)=c(4) b(2)=c(2)*b(1) b(3)=c(2)*b(2) for !i=4 to 200 b(!i)=c(2)*b(!i-1)+c(3)*b(!i-3) next scalar tot_mul=@sum(b) (d) The delay and interim multipliers for up to 12 quarters are Delay multiplier Estimate Interim multiplier 0 0 0.25 1.9749 8 1 7 3 5 0.1335 .0824 1.0255 1.25 is INF ˆ ˆ 0. Thus an . Exercise Solutions.0067 9 1 8 3 6 0.0164 1.25 i 0 Computing the total multiplier i i 0 ˆ numerically.8239 5 1 4 3 2 0.0893 0.1159 0.1335 0.8811 6 1 5 3 3 0.3728 0.0209 1. Principles of Econometrics.0511 0.3728 1 1 0 0.7345 4 1 3 3 1 0.6187 3 1 2 3 0 0.0531 11 1 10 3 8 0.27 (continued) (c) The rate of inflation when wage growth is constant at 0.0322 10 1 9 3 7 0.1532 0.0318 1.5419 2 1 1 0.Chapter 9. 27(d) (continued) The graph for the delay multipliers for up to 12 quarters follows .Chapter 9.05 .3 0.4 0.15 .2 3 0.0661 T 5 T 4 1 0.20 . Principles of Econometrics. inflation increases at a diminishing rate. However.3 0. with the exception of a spike at lag 3.10 .3 0. After 12 quarters.1.0462 2 0.2 Estimate 0 0.8 0. approaching the total multiplier which is approximately 1. (f) The estimated changes in inflation are given in the following table.6 0.30 BETA . (e) The graph for interim multipliers for up to 12 quarters is 1. Quarter T 1 Change in Inflation 0.0526 3 .2 2 0.3 0.0 0.3 0 1 2 3 4 5 6 7 8 9 10 11 12 LAG If wage growth increases to a new level and then is held constant at that new level.00 0 1 2 3 4 5 6 7 8 9 10 11 12 LAG An increase in wage growth increases the inflation rate.0746 T 3 T 2 0. 4e 359 Exercise 9. Exercise Solutions.2 4 0.9 INTERIM 0.25 .1457 0 0.2 1 0.1 1.35 .5 0.40 . the effect decreases as the lag increases.7 0. the effect is nearly zero. CHAPTER 10 Exercise Solutions 360 . MDHOUSE. REG1. so that under this criterion we cannot conclude that we have strong instruments.10.01 we reject the null hypothesis that the coefficient of VHAT is zero using the 0. and are willing to accept a test size of 0. If we adopt the Maximum Test Size criterion.99 < 2. and REG3. and the alternative hypothesis is that MDHOUSE is endogenous. 4e 361 EXERCISE 10. (b) The model in column (1) contains one potentially endogenous variable.3 By the Staiger-Stock rule of thumb we are satisfied because the calculated F is greater than 10. The 2-tail critical value of the t-distribution with 46 degrees of freedom is 2.1 and 10E. There are 4 potential instruments. Column (2) contains the first stage regression results including all instruments. all of which might well affect not only rent paid but also the median house value. The Hausman test is a t-test for the significance of the coefficient of VHAT. (c) The regression based Hausman test for endogeneity augments the regression of interest with the least squares residuals from the first stage regression. REG2. We test for strong instruments by computing the joint F-test of significance of these variables in the first stage regression. The omitted factors from this regression include macroeconomic forces. Principles of Econometrics. Exercise Solutions.Chapter 10. interest rates. A more informative answer is obtained by examining the critical values for the weak instrument tests of Stock and Yogo in Table 10E.2.99. Column (3) contains the first stage regression omitting FAMINC.10 for a 5% test. If there is correlation between median house value and the regression error term then median house value is endogenous. Under this metric we can conclude that the instruments are strong. The calculated value of the t-statistic is 3.6 4 85. We conclude that MDHOUSE is endogenous. and a relative bias of 0.6 4 3767.01. Using the sum of squared residuals SSE in columns (2) and (3) we can compute the F-statistic as F SSE R SSEU J ˆ U2 8322. for a test of the coefficient on the endogenous variable. then the critical value for the F-statistic is 24. Since 3.2 3767. etc.. then the relevant F-critical value is 10.6 50 6 4554. comparing the bias of the IV estimator to the bias of the least squares estimator.20 for a nominal 5% test. The null hypothesis is that the instruments are weak. such as unemployment rates.1 (a) The price of housing and rent paid are determined by supply and demand forces in the market place. If we adopt the Maximum Relative Bias criterion. In order to carry out instrumental variables estimation we require at least one strong instrument.27. population growth. In order to make such a conclusion we would have to be willing to accept a test size of 0.26. The null hypothesis is that the variable MDHOUSE is exogenous. for in that case the relevant F-critical value is 10.05 level of significance. L=4].58 [B=1. .6 13. the 0. The coefficient of PCTURBAN is now insignificant.815. because it is an explanatory variable in the first stage regression. Secondly. and it is a property of the least squares residuals that they are uncorrelated with model explanatory variables. The value of the test statistic is NR 2 50 0.95 percentile of the 23 distribution is 7.11 is significant at the 0. 4e 362 Exercise 10.61 is very significant.05 level. We conclude that at least one of the extra instruments is not valid. and its standard error is larger. Principles of Econometrics. indicating a larger effect than we first estimated. The test does not identify which instrumental variable might be the problem. but the t = 6. as explained below equation (10D. That the estimates for the structural parameters are the same in columns (4) and (5) is not an accident.Chapter 10. . The first stage least squares residuals VHAT are uncorrelated with PCTURBAN. Also. whereas the least squares estimate’s t-value of 2. The resulting statistic.3 . VHAT is uncorrelated with the fitted value of MDHOUSE that is used to compute the 2SLS/IV estimates. under the null hypothesis that the surplus instruments are valid (uncorrelated with the regression error) is distributed as 2L B 4 1 3 . the IV estimate of the coefficient of PCTURBAN is much smaller than the corresponding least squares estimate. The standard error of the IV coefficient is larger (0. From Table 3 at the end of the book.8) (e) The test for the validity of the 3 surplus instruments (the overidentifying restrictions) is computed as NR 2 from the artificial regression of the 2SLS/IV residuals on all available instruments. the IV estimate of the effect of MDHOUSE on RENT is larger in magnitude.339) than the corresponding least squares estimate. and therefore that the IV estimates in column (5) are questionable.226 11. Exercise Solutions. First.1 (continued) (d) We note two important changes when we compare the least squares estimates in column (1) and the instrumental variables estimates in column (5). Exercise Solutions. In the simple regression model 2 ˆ ~N 2 The error variance 2 2. The problem with doing 2SLS with two least squares regressions is that in the second estimation the estimated variance is ˆ 2wrong ˆ yi ˆ xˆ 2 i 1 N 2 2 The numerator is the SSE from the regression of yi on xˆi . or SSE2 SLS . 4e 363 Insert: Correction of IV standard errors (Bonus material) In the simple linear regression model yi ei the 2SLS estimator is the least 1 2 xi ˆ ei where xˆi is the predicted value from a squares estimator applied to yi 1 2 xi reduced form equation. the 2SLS estimators are ˆ ˆ xˆi x 2 1 yi xˆi y y 2 x ˆ x 2 In large samples the 2SLS estimators have approximate normal distributions. is ˆ 2wrong ˆ 2wrong se wrong ˆ 2 xˆi x 2 xˆi x ˆ wrong 2 xˆi x 2 Given that we have the “wrong” standard error in the 2nd regression.Chapter 10. xˆi 2 x should be estimated using the estimator ˆ 22 SLS yi ˆ ˆ x 2 i 1 2 2 N with the quantity in the numerator being the sum of squared 2SLS residuals. which is SSEwrong. we can adjust it using a correction factor se ˆ 2 ˆ 22 SLS ˆ 2wrong se wrong ˆ 2 ˆ 2 SLS ˆ wrong se wrong ˆ 2 . calculated in the 2nd least squares estimation. the correct 2SLS standard error is se ˆ 2 ˆ 22 SLS ˆ 22 SLS xˆi x 2 xˆi ˆ 2 SLS x 2 xˆi x 2 and the “wrong” standard error. So. Thus. Principles of Econometrics. they are probably exogenous relative to the supply equation. and then decreasing work effort thereafter. and this endogeneity will result in the failure of the least squares regression. as increased household income reduces the need for the wife’s income. Because they are “demand” factors rather than “supply” factors. we expect the coefficient of WAGE to be positive. The standard errors based on this two step process are incorrect. WAGE 1 6 2 EDUC NWIFEINC 3 AGE 1 4 EXPER KIDSL6 2 5 EXPER 2 KIDS 618 ut Obtain the fitted values of the reduced form equation WAGE . and uncorrelated with the supply equation error term.Chapter 10. Exercise Solutions. these variables may be correlated with the error term. 4e 364 EXERCISE 10. as increased wage offers induce a greater quantity of labor supplied. The coefficient of EDUC in this supply equation reflects the competing forces of (i) more persistent and intelligent workers may have an inclination to work more.1 for more and a correction factor. In this case WAGE is endogenous. Since one’s ability is usually correlated with their education and wage. The coefficient of NWIFEINC should be negative. Principles of Econometrics. and WAGE and EXPER2. (c) To satisfy the logic of instrumental variables they must be correlated with the endogenous variable and uncorrelated with the error term. An endogenous variable on the right hand side of an equation makes the least squares estimator inconsistent. An argument could also be made on the basis that a measure of ability is not included in the equation. .2 (a) In this labor supply function of married women. Ability bias is a form of omitted variable bias where the effect of an individual’s ability is not measured but captured in the error term. The presence of children should have a negative effect on the labor supply of married women. we satisfy the requirement L B. (d) The supply equation is identified because we have only specified one endogenous variable and there is at least one instrumental variable. (e) Estimate the reduced form equation by least squares. The estimated parameters for this last regression will be the 2SLS/IV estimators. Recall that supply and demand jointly determine the hours and wages. just like HOURS is endogenous. as we anticipate a life-cycle work pattern of increasing labor effort up to some point in middle-age. With EXPER and EXPER2 as instrumental variables. The coefficient of AGE might be positive or negative. since workers with more experience can demand higher wages. (b) This supply equation cannot be consistently estimated by least squares. Replace its endogenous counterpart in the original supply model and apply least squares. . or (ii) more educated workers may be more efficient and choose to work less. We expect there to be a correlation between WAGE and EXPER. See the solution to Exercise 10. 52 > 2.3.64 and a p value of 0.3.9799 -10 -15 -20 -25 -30 0 40 80 120 160 200 240 280 320 360 x Figure xr10.95. obtain the least squares residuals. Or.0090 0.05 critical value is F(.73. eˆt and regress their squared values on MONEY GROWTH.3(b).0331MONEY 1. Under the null hypothesis of homoskedasticity the 2 test statistic has the (1) distribution. Since F = 10. 3 1 . The LM statistic is NR2 from this regression. 4e 365 EXERCISE 10.0000. In this case the value of the test statistic is 17. x.000024. we reject the null hypothesis and conclude that this data does not follow the quantity theory of money. page 214. Based on the figure it appears that the problem arises because of one severely unusual observation.2342 1. A scatter diagram of the least squares residuals against the variable MONEY GROWTH.3 (a) The estimated least squares regression with standard errors in parentheses is INFLATION 0. The F(.73 we reject the strong null hypothesis. which suggests that heteroskedasticity exists. we obtain an F statistic of 12. It shows a tendency for the residuals to get larger in magnitude as x increases.0000.73) critical value is 2. Principles of Econometrics. 3 1 against the alternative that at least one of these equalities is not true gives an F statistic of 10.95. The .3(b) Scatter plot for least squares residuals To use the LM test for heteroskedasticity described in Chapter 8 of POE. 10 5 0 -5 residual (b) 0.73) = 3.Chapter 10. Thus we can reject the null hypothesis of homoskedasticity. Exercise Solutions. is shown in Figure xr10.12.838 and a p-value of 0. . 0. 2 1.52 and a p value of 0.6620OUTPUT (se) 0.05. (ii) Testing the weaker hypothesis H 0 : 2 1.2506 (i) Testing H 0 : 1 0. We reject the weak joint null hypothesis. since the p value is less than the level of significance. 2 1.Chapter 10. 0.0351MONEY 1.175914 For b1 and b3. This suggests that the least squares method will understate the precision of the estimate under heteroskedasticity. 4e 366 Exercise 10.0000. we reject the weak joint null hypothesis. For b2 the robust standard errors are much smaller than the least squares standard errors so least squares overstates the precision of the estimate under heteroskedasticity.3942OUTPUT 1.250566 0. 3 1 using 2SLS estimates which have not been corrected for heteroskedasticity. we reject the null hypothesis again. Testing the weaker hypothesis using robust standard errors returns an F statistic of 2. Since the p value is less than the level of significance. we obtain a F statistic of 9. we reject the null hypothesis and conclude that this data does not follow the quantity theory of money.0003.023694 b3 = 1. Testing the same hypothesis using robust standard errors gives an F statistic of 9.05. (ii) Testing the weaker hypothesis H 0 : 2 1.7457 and a p value of 0. Since the p value is less than the level of significance.0331 0. Exercise Solutions. In this case we do not reject the null hypothesis and conclude that the weaker joint hypothesis is not rejected.619615 b2 = 1. Since the p value is less than the level of significance.0001.26 and a p value of 0. 0.05.6620 0.2342 0.1072.5515 (i) Testing the strong hypothesis H 0 : 1 0. Principles of Econometrics. .3 (continued) (c) The robust standard errors are compared to the least squares standard errors in the following table Coefficient Estimate Least Squares Standard Errors Robust Standard Errors (White’s) b1= 0.0098 0.009042 0. the robust standard errors are much smaller than the least squares standard errors.2331 and a p value of 0.05. (d) The IV/2SLS estimated model of the inflation equation is INFLATION (se) (e) 1. 3 1 using 2SLS estimates which have not been corrected for heteroskedasticity gives an F statistic of 8.979925 0.3028 and a p value of 0. 0.8582 0.0940 1. The rule of thumb states that the F-test value must be greater than 10 to be “strong”. we follow the steps outlined in Section 10. The test pvalue is 0.64 (p = 0.3. and test whether the coefficient of the residuals vˆ is significantly different from zero.4. (g) To test the null hypothesis H 0 : all the surplus moment conditions are valid. the first step is to obtain the residuals from the reduced form equation of the endogenous variable in question. which is the original model specification augmented with vˆ . we do not reject the null hypothesis and cannot conclude that OUTPUT is endogenous.3447 respectively. Since the p value is larger than the level of significance.4834. where is the coefficient of the residuals vˆ. (h) Applying the joint F-test described in Section 10. and the robust-t using White’s robust standard errors INFLATION 1.Chapter 10. respectively.3942OUTPUT (t ) 0. 0.3388vˆ ( 0. e) 0.4. The estimated auxiliary regression is reported with t-value for the key coefficient using the usual least squares standard errors.2 requires us to test the null hypothesis H0 : 1 0 in the reduced form equation from part (f). However. The F-test values are 2 3 4 4.95) (t ) robust The final step of the Hausman test requires us to test the null hypothesis H 0 : 0 against the alternative hypothesis H 1 : 0 .0351MONEY 1.4552 2 (L B) 2 (4 1) 7. We reject the null hypothesis that all the coefficients are zero at the 5% level. In this case we estimate the reduced form equation OUTPUT 1 2 MONEY 1 INITIAL 2 SCHOOL 3 INV 4 POPRATE v Obtain the least squares residuals vˆ OUTPUT OUTPUT estimate an auxiliary regression.9512 and 0.55) ( 0. The test statistic obtained is NR 2 The critical value 76 0. . This is equivalent to testing the null hypothesis H 0 : cov(OUTPUT .0178) for the least squares and robust tests. the results of both of our joint F-tests suggest that we should be concerned that we are using “weak” instruments. simply rejecting the null hypothesis is not adequate evidence of “strong” instruments. Exercise Solutions.032305 2.0022) and 3.0940 1.21 (p = 0. The robust-t statistic and the p value for this sample are 0.3 (continued) (f) To perform the Hausman test.8147 is much larger than the test statistic therefore we do not reject the null hypothesis that all surplus moment conditions are valid. For this part we will keep to the assumption that MONEY is exogenous and OUTPUT is endogenous. 4e 367 Exercise 10. Principles of Econometrics.05. . Exercise Solutions.25.20 6. Note however that these tests are not valid under heteroskedasticity.96 0. 4e 368 Exercise 10.10 10.Chapter 10. The critical values for the weak instrument test using the “maximum IV test size criterion” are in Table 10E.34 Once again we see that we cannot reject the null hypothesis that the instruments are weak even if we can tolerate up to 0.30 5.2.27 0.31 We cannot reject the null hypothesis that the instruments are weak even if we can tolerate a 5% test on the coefficient of the endogenous variable having actual size up to 0. For this case they are L 4 0.20 10.05 16.10 24.25 8.26 0.30 of the least squares estimator’s bias.3 (continued) Bonus material: Using the analysis of weak instruments in Appendix 10E we can be more precise.15 13. The critical values for the weak instrument test using the “maximum IV relative bias criterion” are in Table 10E.1. Principles of Econometrics.71 0.85 0. For this case they are L 4 0.58 0. 110162 Min Max -2. which is 1.1770892 -. However.50074 5. . 1.9490 xt (se) (0. Dev.4822 As shown in the figure below.4822 2.42647 -2.0009.110162 1.4 (a) Variable Obs Mean x e y ey 25 25 25 25 -. the estimate for 2 .42647 2.0903) The estimate for 1 . but do reject the null hypothesis of 2 1 .98294 3.4(b) Data values and regression function E(y) (c) The least-squares estimated equation is given by yˆt 1. Principles of Econometrics. We do not reject the null hypothesis of 1 1 .65135 -1. the data tend to fall below the regression line for x 0 and above the regression line for x 0.0996) (0. -4 -2 0 2 4 6 (b) As a check of your work. The t-statistics for testing whether 1 and 2 are 1 are 0.1671932 . Exercise Solutions.216547 1.50. 4e 369 EXERCISE 10.00872 and 10. the summary statistics are -2 -1 0 1 x values-generated data y 2 3 ey Figure xr10.6557176 . respectively. is very close to the true value of 1.0009 1.8229108 Std.158174 2. which is 1.949. is quite different from the true value of 1.57634 -3.Chapter 10. -4 -2 0 2 4 6 (d) -2 -1 0 1 x values-generated data y Figure xr10. In real problems the variable e is not observable and therefore we cannot calculate the correlations between x and e.90968 0. Principles of Econometrics.00000 0.41531 1.4(d) shows a fitted regression line that runs through the “center” of the observations.Chapter 10. . it is not a good estimate of the true regression function. and eˆ is as follows.00000 0. 4e 370 Exercise 10.4 (continued) In contrast to the plot in part (b).4(d) (e) 2 3 Fitted values Fitted regression and observations The sample correlation matrix of the variables x. x e eˆ x e eˆ 1. Thus.00000 1. and e and eˆ .00000 There is a high correlation between x and e. e. Exercise Solutions. Figure xr10. The zero correlation between x and eˆ is a characteristic of the least squares estimation procedure. 0392 INCOME (se) (1.1720)(0. using the incorrect standard errors from part (d) makes the estimated slope appear statistically significant at the 0. At the 0.0392 INCOME (se) (1.8435)(2.6417 AVERAGE _ INCOME ( 1.9883 0.38) The standard errors are lower than those in part (b). The result of the estimation is SAVINGS 0.0220 1.5240) (0.5 (a) The least-squares estimated equation is SAVINGS (b) (se) (0. 4e 371 EXERCISE 10.3428 0. Principles of Econometrics.Chapter 10.05 level of significance. In particular.9550) To perform the Hausman test we estimate the artificial regression as SAVINGS 0.0201) (t ) (0.46) The estimated equation using the instrumental variables estimator.0112) (t ) (5.757) To perform the Hausman test we test the null hypothesis that the coefficient for vˆ is zero.8561) (0.80) The second stage regression replaces INCOME with the fitted value from the reduced form.0052 INCOME 0.07) ( 0.0154) (0.5430) ( 3. Exercise Solutions.01 level of significance we reject the null hypothesis and conclude that x and e are correlated. which causes the t-statistics to be higher.2530) (0. The t-statistic is –3. (d) The reduced form estimation yields INCOME (t ) 35.3918 INCOME 0.0755vˆt (se) (1.0165) (t ) (0.9883 0.757.9883 0.6484) (1.0200) (t ) (0.79) (2. .83) (5. with instrument z = AVERAGE_INCOME is SAVINGS (c) 4. 3876 1.6 (a) The correlation between x and e is rxe cov( x.4016 1.4535 0.9 (2)(1) 0.7775 3.0787 0.2036 0. Principles of Econometrics. 4e 372 EXERCISE 10. (c) Figure xr10.1727 0.6(c) shows us that the data tends to fall below the regression line for x 0 and above the regression line for x 0.0237 . Exercise Solutions. only slightly higher than the true value in (a).100 1 .3608 0.3722 1.0183 Standard error 0.0341 1.1211 0.500 Estimate 2. e) var( x) var(e) 0.0169 3.65136.6364 The sample correlation between x and e is 0.0533 0.500 1 .20 1 .0078 3.10 1 .10 1 . -5 0 5 10 (b) -4 -2 0 x values-generated data y values-generated data Figure xr10.Chapter 10.20 1 .100 1 .6(c) (d) 2 4 ey Fitted regression and observations The results from least squares are presented below. 1 2 Sample Range 1 . 0000 The nonzero correlations between x and z1 (0.0034 1.10 1 .0000 0.9771 3.6(d) (continued) The estimate for 1 moves closer to the true value as the sample size increases from 10 to 20 and to 100. the estimates move away from the true value as the sample size increases.0640 1. z2 . However. . (e) The sample correlations between z1 .1132 0.9961 0. for both cases the standard errors decrease as the sample size increases.500 1.0263 0.2526 0. (f) Using z1 as an instrumental variable the estimates are Sample Range 1 2 Estimate Standard error 1 . However. it does not get closer when the sample increases from 100 to 500.20 1 . Principles of Econometrics. The estimates do not get closer to the true values as sample size increases because of the inconsistency caused by the correlation between x and e. because the correlation between x and z1 is greater than the correlation between x and z2 .0277). x and e are z1 z2 x e z1 z2 x 1.Chapter 10. The inconsistency does not disappear as the sample size increases.0504 The IV estimates for both 1 and 2 are getting closer to the true values as the sample size increases.2894).2500 0. z1 is the better instrumental variable.6208).0451 1 . mean that both z1 and z2 will be satisfactory instrumental variables.100 1 .2894 0.7144 3. and between x and z2 (0. 4e 373 Exercise 10.100 1 .20 1 . coupled with the essentially zero correlations between e and z1 ( 0.10 1 .6208 0.500 2.9363 0.4277 0.0000 0. As expected.0000 0.0315 0. reflecting the consistency of the IV estimator.0034) and e and z2 (0.1966 0.6514 e 1.0810 2.0277 1.0153 0. For 2 . Exercise Solutions.1051 0. 9921 1.06 0. Exercise Solutions.20 1 .0449 Using both z1 and z2 as instrumental variables the estimates are getting closer to the true values as the sample size increases.20 1 . Principles of Econometrics.10 1 .6975 0.1470 0.0424 1 .0295 11.9808 3.500 2. particularly those for 2 .6 (continued) (g) Using z2 as an instrumental variable the estimates are Sample Range 1 2 Estimate Standard error 1 .0311 0.20 1 .10 1 .20 1 .2555 0. The results are very similar to those obtained using only z1 as an instrumental variable.500 1.0026 0.0090 0.500 1. When the sample size is larger.10 1 . This result occurs because the correlation between z1 and x is much higher than the correlation between z2 and x . .0852 2.1349 1.0446 1 .100 1 .9503 0.0997 0.2549 0.100 1 .471 0. (h) Using both z1 and z2 as instrumental variables the estimates are 1 2 Sample Range Estimate Standard error 1 .0491 1.4337 0. the estimates move closer to the true values.500 –2.2433 2.Chapter 10.70 2.1987 0. Comparing the results using z1 alone to those using z 2 alone.100 1 . although there has been a slight improvement in precision for sample sizes T = 100 and 500.9902 3.8923 3. 4e 374 Exercise 10.1110 1.1014 Using z2 as an instrumental variable gives estimates that are very far away from the true values when the sample sizes are small (less than 20 for 2 and less than 10 for 1 ).0932 0. those using z1 alone lead to more precise estimation even when the sample size is small.7114 3.100 1 .0666 51.0887 0.10 1 . 975.4380CAP 0.0087 0. The error variance is ˆ 2 = 7.0000 0.0322 0.0647 The variance of the prediction error for this case is var( f ) var(e0 ) var(b1 ) XPER02 var(b2 ) CAP02 var(b3 ) LAB02 var(b4 ) 2 XPER0 cov(b1 . the predicted wine output is Qˆ 0 1.8347. respectively.Chapter 10. all estimates of the slope coefficients are significant.0468 0.756 and therefore se( f ) var( f ) 2.0040 0.0100 Using XPER = 10 and LAB and CAP equal to their sample averages. Alternatively. 14. the predicted value and the standard error of the prediction error can be obtained using automatic software commands.1468 XPER 0.0998 The signs of the estimates are positive as expected.2392 10.8347 0.7 (a) The least squares estimated equation is Qˆ 1.785 (3.5965. The sample averages for labor and capital are 10. Qˆ 0 tc se( f ) 9.785 .9939 2. b4 ) 2 XPER0 CAP0 cov(b2 .2392 LAB (se) 1. b4 ) Substituting the values of the variances and covariances we obtain var( f ) = 7.1468 10 0. thus.0049 0.62) .1138 0.7623 0.0012 0.0634 0. Exercise Solutions. All the standard errors are relatively low. except that for the constant term.0467 9.1176 0.0647 1.7623 0.4380 7. b3 ) 2 XPER0 LAB0 cov(b2 .0138 0. b2 ) 2CAP0 cov(b1 . b4 ) 2CAP0 LAB0 cov(b3 . b3 ) 2 LAB0 cov(b1 .0550 0. 4e 375 EXERCISE 10.51. The variance-covariance matrix for the estimates is b1 b2 b3 b4 (b) (i) b1 b2 b3 b4 1.0467 and 7. The 95% interval prediction uses tc = t(.71) = 1.9939. Principles of Econometrics. 15 8. The interval prediction is 12.001 and se( f ) 2. 16.5121 XPER 0.53 We can reject the null hypothesis that the instrument is weak if we are willing to accept a test size on the coefficient of the endogenous variable of up to 0.66 0. Qˆ 0 12. (c) The estimated artificial regression is Qˆ 2.9939 2.1978 The Hausman test to test whether the variable XPER and the error term are correlated is the same as testing whether the coefficient for vˆ is zero. the values of the estimated coefficients for XPER and CAP have changed considerably.7(b) (continued) (b) (ii) Using your computer software.5121 XPER 0. (iii) For 30 years experience.95.2400 LAB (se) 2. but that for LAB is approximately the same.91 0.9939 2.3321CAP 0.533 and se( f ) 2.4158 vˆ (t ) 2.1 the test critical values are L 1 0. we can calculate the predicted wine output and the standard errors given 20 years experience as Qˆ 0 10.12) .7230 (t ) 0. or at a 5% significance level when using one-tailed tests. the estimates have the expected positive signs.1209 2. Exercise Solutions.957 (6. using the Stock-Yogo “maximum test size criterion” from Table 10E. Relative to the least squares results.4867 0.99 As in part (a).25 5.957 .90) . Principles of Econometrics. The results suggest that the coefficient for vˆ is significant.533 1.802 . 17. For the lower maximum test size of 0. All coefficients are significant at a 10% level of significance.001 1. A 95% interval prediction is 10. (d) The IV estimated equation is Qˆ 2. .96 0.2400 LAB 0.4867 0.15.2205 2.32 0.81361.15 1. The p-value of the test is 0.Chapter 10. which is close to the rule of thumb value of 10.11.1545 0.3321CAP 0. Bonus material: The first stage F-value is 9.10 16.802 (4. However.38 0. 4e 376 Exercise 10.031 so at a 5% level of significance we can conclude that there is correlation between XPER and the error term.20 6.10 we are unable to reject the null hypothesis that the instrument is weak. 14. (iii) For 30 years experience.7 (continued) (e) The following results are obtained using automatic software commands for forecasting. 27.64) A comparison of these prediction intervals with those from least squares estimation suggests that ignoring the correlation between XPER and the error term will: yield intervals that are too narrow.9939 3.621 (5.14.Chapter 10. Exercise Solutions. The interval prediction is 17.891 (8. Principles of Econometrics. over-predict wine output for a manager with 10 years experience.89 1.890 and se( f ) 4.891 .56) (ii) For 20 years experience.6475 1. (i) For 10 years experience. Qˆ 0 17.9939 3.768 1. Qˆ 0 7.73.621 . Qˆ 0 12.9939 4.6475 and se( f ) 3. under-predict wine output for managers with 20 and 30 years experience.768 and se( f ) 3. which in turn leads to false reliability about wine output. The interval prediction is 12.468 (0.19.99) . 4e 377 Exercise 10. The interval prediction is 7. .55.468 . 0870 0.650176 -0.5571KIDS 618 0. which implies that these instruments are not “strong” instruments.000747 0.0000. The F-statistic of the joint hypothesis H 0 : 7 0 is 8. using the rule of thumb. 4e 378 EXERCISE 10.Chapter 10.357997 0.005415 0.044344 -1.9394 The Hausman test is used to test for endogeneity by considering the null hypothesis that the coefficient of vˆ is significantly different from zero.190173 Denote the residuals from the reduced form as vˆ log(WAGE ) log(WAGE ) . augmented with the residuals from the reduced form with tstatistics in the parentheses.013372 0. Exercise Solutions.164098 Mean dependent var 1. The estimated supply equation. so we can conclude that there is correlation between ln(WAGE) and the error term.32E-06 0.25.632765 1.630591 -0. is HOURS (t ) 2432.2613 0.0003.69E-06 0. the F-statistic is less than 10.000402 -1.040710 -0.017648 5. At a 5% level of 8 significance. Principles of Econometrics.5585 5.615970 -0.124729 6.7611 5.099884 -0.4980 2.860055 0.0000 0.7841AGE 7. yielding a p-value of 0.055873 -0.5272 0. These estimation results are: Dependent Variable: LOG(WAGE) Method: Least Squares Sample: 1 753 IF LFP=1 Included observations: 428 Coefficient Std.088603 0. .0187 210.318296 0. Error t-Statistic Prob.0025 0. -0. (b) The estimated reduced form equation is shown in part (a).0092 NWIFEINC 1623.4716 2.20 1544. we reject the null hypothesis and conclude that these instruments have a significant correlation with ln(WAGE).8 (a) The Hausman test is carried out by first estimating the reduced form for ln(WAGE).0636 C EDUC AGE KIDSL6 KIDS618 NWIFEINC EXPER EXPER^2 R-squared 0. The estimates suggest that the coefficient for vˆ is significant since the p-value of the test is 0.449 EDUC 10.003520 -0.60vˆ 2.5287 0.5159 0.1363 1.715373 3.834 KIDSL6 47.82ln(WAGE ) 177. However.027891 3.015097 0.3388 5. and should be strongly correlated with EDUC. HEDUC and SIBLINGS we firstly test for significant correlation between the endogenous and instrumental variables.05 critical value for the (1) is 3.0020 0.15 11.1 the test critical values are 0.8581. For the lower maximum test sizes we are unable to reject the null hypothesis that the instruments are weak.4. we can be more precise. The . A valid instrument must be uncorrelated with the regression error. The test statistic calculated is NR 2 428 0. FATHEREDUC.10 19. the test statistic has 2 a chi-square distribution with 1 degree of freedom. though perhaps in larger families each child attends school for fewer years. It is likely that MOTHEREDUC.93 L 2 0. the F7 8 9 statistic is 60. Principles of Econometrics. 4e 379 Exercise 10.8(b) (continued) Bonus material: Using the Stock-Yogo critical values for testing if instruments are weak. FATHEREDUC and HEDUC are correlated with a woman’s years of education. This F-statistic is much greater than 10 implying that at least some of our instruments are strong instruments and we should not be too concerned that we are using weak instruments. we do not reject the null hypothesis indicating that our surplus moment conditions are valid. To test the suitability of the instruments MOTHEREDUC.3543.25 7. The reduced form equation which we use to conduct the joint test is EDUC 1 6 2 AGE 3 KIDSL6 MOTHEREDUC 7 4 KIDS 618 FATHEREDUC 5 NWIFEINC 8 HEDUC 9 SIBLINGS v For a joint hypothesis test of all possible instruments.Chapter 10. (d) The potential endogeneity problem with EDUC is that a measure of ability is omitted from the equation. The argument for SIBLINGS is less obvious. Exercise Solutions. Since 0.84. It is likely that more able people also attend school longer.25 We can reject the null hypothesis that the instrument is weak if we are willing to accept a test size on the coefficient of the endogenous variable of up to 0.84. The p-value of the test is 0. thus inducing a correlation between the regression error and EDUC. H 0 : 6 0. we test the null hypothesis that all surplus moment conditions are valid.59 0. Using the “maximum test size criterion” from Table 10E. (c) Following the steps outlined for testing surplus instrument validity in Section 10. Under the null hypothesis. .8581 < 3.67 with a p-value of 0.0000.75 0.20 8. and ability in particular.25. and is thus in the error term of the supply equation.2. The only problem with MOTHEREDUC and FATHEREDUC as instruments is that more intelligent parents (and more educated) have more intelligent (and educated) children. 0014 0.6757 1.1031 1.0235 2.0001 0.3656 All instruments are significantly different from zero except for SIBLINGS.0079 0.0000 0.6375 . we use the regression estimation results.0000 0.0361 0.0200 0. This test is conducted in part (h) of this exercise.3637 0.4715 0.Chapter 10.8647 3.5206 2.0945 0.0256 0.0450 0.9253 0.0440 0.9702 0.0320 6.9093 0.1291 0.2437 0.0067 1.0000 0.9449 0.0109 0.0058 0.5795 2.0007 0.2161 1.0069 0.8(d) (continued) Testing the strength of each individual instrumental variable.6713 3.0403 0.0342 0.0013 0.3.0373 1. Furthermore.0008 0.9812 0. This suggests that HEDUC and MOTHEREDUC are strong instrumental variables and FATHEREDUC is a weaker instrumental variable.2428 0.5551 0.1179 0.3416 0.0003 0.5378 0.0025 0.3547 10.0000 0.9057 0.0009 0.0939 0.7857 3.2427 11. Principles of Econometrics. The other requirement of an instrumental variable is instrument validity.0012 0.004 0. Exercise Solutions.0525 0.4794 0.00002 0.7882 0.1096 0.953 0.5560 1. (e) The estimates of the reduced form equations EDUC or ln(WAGE ) 6 1 2 EXPER 10 AGE 7 HEDUC 3 KIDSL6 EXPER 2 11 8 SIBLINGS KIDS 618 5 MOTHEREDUC 9 4 NWIFEINC FATHEREDUC v are presented in the following table 1 2 3 4 5 6 7 8 9 10 11 Dependent variable EDUC Dependent variable ln(WAGE) Parameter estimate t-statistic p-value Parameter estimate t-statistic p-value 5.0988 0.1697 0. This can only be tested on surplus instruments and when all other endogenous variables have been fully specified.1675 2.0035 0.6371 3. These are presented in the following table 6 7 8 9 Instrumental variable t-statistic p-value MOTHEREDUC FATHEREDUC HEDUC SIBLINGS 3.5625 0. 4e 380 Exercise 10.5244 0. the t-statistics of HEDUC and MOTHEREDUC are greater than 3.0002 0.3167 0. We should use the Cragg-Donald test statistic in equation (10E. are Minimum eigenvalue statistic = 3. .13 for the EDUC and ln(WAGE) reduced form equations respectively. Both F-tests are significant at a 1% level of significance. Exercise Solutions. We see that we cannot reject the null hypothesis that the instruments are weak. 11 0.78 -----------------------------------+--------------------------------| 10% 15% 20% 25% 2SLS Size of nominal 5% Wald test | 21. MOTHEREDUC. FATHEREDUC. and the critical values reported by Stata. The two F-values in part (e) are not adequate.02 and 4. H 0 : 6 .08 4. Bonus Material: This example illustrates the problems of evaluating instrument strength when there is more than one endogenous variable..10 7.13616 Critical Values # of endogenous regressors: 2 Ho: Instruments are weak # of excluded instruments: 6 --------------------------------------------------------------------| 5% 10% 20% 30% 2SLS relative bias | 15. which we compare to the critical values using the IV relative bias or IV maximum test size criteria.33 9. we should be concerned about using weak instruments for ln(WAGE) since 4.72 9. The Stata 11.” Its value is 3. SIBLINGS. results in F-statistics of 41.1 calculation of this value. HEDUC.68 12.14.48 6..3).42 --------------------------------------------------------------------- Stata calls the Cragg-Donald statistic the “Minimum eigenvalue statistic.8(e) (continued) The F-test of joint significance of EXPER.Chapter 10. Principles of Econometrics. EXPER2.13 < 10. However. 4e 381 Exercise 10. 0000 0.709783 0. is Dependent Variable: HOURS Method: Two-Stage Least Squares Included observations: 428 C LOG(WAGE) EDUC AGE KIDSL6 KIDS618 NWIFEINC Coefficient Std.0001 0.3164 -9.035707 0.0880 .242835 -248.51466 9.066 -123.25 with a p-value of 0. The estimated artificial regression is Dependent Variable: HOURS Method: Least Squares Included observations: 428 C LOG(WAGE) EDUC AGE KIDSL6 KIDS618 NWIFEINC VHAT_EDUC VHAT_LNWAGE Coefficient Std.0117 0.263858 169.247554 5.0854 0.69185 0.0269 0.3190 0.724484 -2.699531 -1.9432 56. (g) The 2SLS estimated model.457498 3.464577 -0. 1836.006934 2.11899 5. These variables are added to the HOURS equation.8949 -39.2273 0.362 432.6870 -1538.672 1452.0020 0.997731 -1.8949 -39. Error t-Statistic Prob.955198 3.1999 32.Chapter 10.0000 To test the null hypothesis that both EDUC and ln(WAGE) are endogenous. 4e 382 Exercise 10. we conduct a joint test on the coefficients of the two residual terms.65773 -0.242835 -248.371264 -2.112617 -6.0000 0.0144 0.4846 0.004012 38.209074 -2.839361 -1.066 -123.32368 32. Error t-Statistic Prob.4070 249. and therefore we reject the null hypothesis and conclude endogeneity exists in at least one of EDUC and ln(WAGE).0000.0033 0.65773 -0.8769 4. using all instrumental variables. We arrive at a F-statistic of 18. Principles of Econometrics.77349 254.221331 -0.011856 120.80008 0. 1836.0008 0. Exercise Solutions.3164 -9.7186 55.3748 430.1438 0.011856 747.8 (continued) (f) The Hausman test uses the residuals from the reduced form equation for EDUC (called VHAT_EDUC) and the residuals from the reduced form for ln(WAGE) (called VHAT_LNWAGE).531383 -1.826912 -3.672 1452.359770 98. From the model estimates. we find that each additional year of education decreases labor supply by 123 hours and a 1% increase in wages increases labor supply by about 15 hours.8(g) (continued) Most estimates have the expected sign. The p-value of this test is 0. experience squared.05 critical value of 2 (L B) 2 (6 2) 9. 4e 383 Exercise 10. and husband. we do not reject the null hypothesis that the surplus instruments are valid since 2. father.006232 2.6673 With a . the equation omits a measure of ability. and all other explanatory variables are insignificant. The instruments are jointly strong for education. We cannot reject the validity of the 4 surplus instruments using the Sargan NR2 test for the validity. and ability is likely positively correlated with both wages and years of education. as well as the number of siblings. We might expect that a 100 dollar increase in NWIFEINC is associated with a decrease in the labor supply of 1. Another surprising result is that AGE.49 . kids and other sources of income on the labor supply of married women. (i) We have used a sample of 428 working women from 1975 to determine the influence of wage. with a t value of 3. . Education and (log) wage statistically significant at a 5% level.05 level of significance leaving only ln(WAGE) and EDUC as statistically significant. the number of children in a household and the age of the woman have no influence over labor supply. The household income from other sources than the woman’s employment (NWIFEINC) is statistically significant at a 10% level. KIDS618. and correlated with the regression error. Also. A Hausman test verifies our prior reasoning: we reject the null hypothesis that these two variables are not correlated with the regression error term. To carry out two-stage least squares we required instruments that are correlated with the endogenous variables. KIDSL6. we know that hours and wages are jointly determined by supply and demand. We employed experience. age.6149.22. The instruments are not strong jointly for log(wage).49. The two stage least squares estimation found that several of the explanatory variables were not statistically significant implying that the drivers behind the labor supply of married women are wages and the education levels. and thus wages are an endogenous variable. (h) To test the validity of the overidentifying instruments we regress the two-stage least squares residuals upon all exogenous and instrumental variables and calculate the test statistic NR 2 428 0. Because this is a supply equation. Because there are two endogenous variables.18 hours. yet uncorrelated with the regression errors. we need at least two instrumental variables. the years of education of mother.67 < 9. A negative sign of the coefficient estimate for EDUC suggests that women with more education work fewer hours than those who do not. Exercise Solutions. Principles of Econometrics. And lastly. education. The presence of a regression error that is correlated with one or more right-hand side explanatory variables means that the usual least squares estimator is both biased and inconsistent. and NWIFEINC are not significant at a 0. with the only exception being EDUC. but experience is a strong single IV.Chapter 10. 0416 0. 2.0001 The 2SLS estimate of the coefficient of ln(P) is larger and is significant at the .9 (a) The least squares estimates of the supply equation are: Dependent Variable: LOG(QPROD) Method: Least Squares Sample: 1960 1999 Included observations: 40 C LOG(P) LOG(PF) TIME LOG(QPROD(-1)) Coefficient Std.025654 0.Chapter 10.304855 0.009110 -0.133000 0. ln(PB).114962 2. as the price of feed (inputs) increases we expect production to decrease.0064 0. Error t-Statistic Prob.900298 2.020679 0.067941 0.0369 0. Principles of Econometrics.0000 All coefficients are significant at a 5% level of significance except for the coefficient of ln Pt .042646 0. 4e 384 EXERCISE 10. .163530 0.0086 0.871371 4. All signs are as expected: as the price of broilers increases we expect production to increase.0069 0. 2. and we expect that an increase in production in the previous period will be associated with an increase in production in the current period.139139 2. Error t-Statistic Prob.090195 0. Other coefficients maintain their signs and significance.786393 2.8941 0.732689 0.974702 0.007202 0.0366 0. (b) Using 2SLS and instrumental variables ln(Y).639905 0.289120 -0. POPGRO.05 level.173826 -2.598974 1.058689 0.109688 0.134083 -2. ln( Pt 1 ) and ln(EXPTS): Dependent Variable: LOG(QPROD) Method: Two-Stage Least Squares Sample: 1960 1999 Included observations: 40 C LOG(P) LOG(PF) TIME LOG(QPROD(-1)) Coefficient Std.005149 0.106635 2.799153 0.871012 0.0123 0.169632 6. Exercise Solutions. over time we expect the quantity produced to increase to feed an increasing demand due to population growth. which is disappointing in this supply relationship.011171 0. Error t-Statistic Prob.54092 1.153288 0. The Hausman test checks the significance of the variable VHAT.457227 2.938098 1. ln(PRICE) is an endogenous variable in this supply equation.143641 6.022306 0.096418 -3.4812 0.0003 0. Error t-Statistic Prob.160882 5.792487 0. 4e 385 Exercise 10.085777 0.0002 0.712994 2.284871 -1.235581 0. -11.0035 0.137022 -4. Dependent Variable: LOG(P) Method: Least Squares Sample: 1960 1999 Included observations: 40 C LOG(Y) LOG(PB) POPGRO LOG(P(-1)) LEXPTS LOG(PF) TIME LOG(QPROD(-1)) Coefficient Std.289120 -0.9246 0.0329 0.040669 0. Dependent Variable: LOG(QPROD) Method: Least Squares Sample: 1960 1999 Included observations: 40 VHAT C LOG(P) LOG(PF) TIME LOG(QPROD(-1)) Coefficient Std.185386 3. as suspected.0618 0.472287 -2.062288 0.117771 0.0303 0.232477 2.342212 1.148438 -0.100821 0.624824 0.095370 0.710735 0.9 (continued) (c) The first step in the Hausman test is to estimate the reduced form equation.954765 0.212285 0.020679 0.210586 0.88 and is significant at the .564755 0.0002 0.740066 0.974702 0. Exercise Solutions.020084 0.882331 4.679849 0.163530 0.0000 The estimated coefficient of VHAT has a t-value of 3. Thus we conclude that.092164 0. -0.977487 0.0569 0.Chapter 10.5763 Save the residuals (VHAT) from the reduced form and add to the original regression equation.269863 1.1510 0.001 level of significance.061159 0. .0089 0.0005 0. Principles of Econometrics.004990 0.598974 0.021011 4. The p-value is 0.37 10. The critical values provided by Stata 11. As domestic price rises the exports of chicken should fall.Chapter 10. as domestic price falls.92 with and the p-value is 0. shown in part (c). To check instrument validity we test the null hypothesis that the excess moment conditions are valid. . (e) One might expect the log of exports of chicken could also be endogenous. Obtain the two-stage least squares residuals.84 --------------------------------------------------------------------- We cannot reject the null hypothesis that the instruments are weak using either the relative bias or maximum test size criteria.25 -----------------------------------+--------------------------------| 10% 15% 20% 25% 2SLS Size of nominal 5% Wald test | 26.0072.05 level and ln(Y) is significant at the .9 (continued) (d) To test that the instruments are adequate we must identify at least one strong instrumental variable. and the critical chi squared value is 9. 4e 386 Exercise 10. Regress these on all exogenous variables and instruments. These are not extremely strong.10 level. Consult the Stock-Yogo paper.87 15. Thus based on this test we fail to reject the validity of the overidentifying restrictions. Bonus material: The Stock-Yogo critical values for this example are not included in Tables 10E. Exercise Solutions.1 are Critical Values # of endogenous regressors: 1 Ho: Instruments are weak # of excluded instruments: 5 --------------------------------------------------------------------| 5% 10% 20% 30% 2SLS relative bias | 18.1 and 10E. Principles of Econometrics. We conclude that the instruments we have are significantly correlated with the endogenous variable ln(P) but may not be strong enough so that two-stage least squares will reliable.2.83 6. exports should rise. If exports and domestic price are jointly determined then ln(EXPTS) is endogenous and not a valid instrument.01 level. The F-test on all of the instrumental variables in the reduced form equation.09 10. yields an F-statistic of 3.671 has a 24 distribution if all surplus instruments are valid.98 8.4523. Thus the instruments are jointly significant at the . In the reduced form we find that ln( Pt 1 ) and ln(EXPTS) are significant at the .77 5.49. but do not attain the rule of thumb value of 10. The test statistic NR 2 3. CHAPTER 11 Exercise Solutions 387 . and to then estimate 1 from ˆ 1 ˆ 2 ˆ 1 . . one way to estimate 1 If Pˆ ˆ1X 18 X Qˆ ˆ2X 5 X ˆ ˆ 2 ˆ1 and then 1 5 18 0.Chapter 11. Principles of Econometrics.2778. 4e 388 EXERCISE 11. Thus. Exercise Solutions.1 The ratio of the expressions for 2 1 2 1 1 and 2 is 1 1 1 2 1 1 is to first obtain estimates ˆ 1 and ˆ 2 by applying least squares to the reduced form equations. 9614 23 8.4611 PS * 13.4284 2.0009 PF * 0.8411+2.9587 Qˆ 2.0140 DI * 0.6705 Qˆ 3.4284 2.3380 ˆ ˆ 3 ˆ 1 ˆ P 2 ˆ PF for P yields 3 PF 2 1 Qˆ 0.2960 PS * 0.5801 2.2719+2.6705 Qˆ 3.4611 22 13. 120 100 80 P (a) Demand Supply 60 40 20 0 0 4 8 12 16 20 24 28 32 Q Figure xr11.9614 PF * 59.3380 1.0328 0.3745 1.2(a) is a sketch of the demand and supply equations for the given set of exogenous variable values. 4e 389 EXERCISE 11.3380 59.3745 11.9587 Qˆ Figure xr11.3745 5.6706 Qˆ Similarly. Exercise Solutions.Chapter 11.3745 ˆ4 DI ˆ2 1 Qˆ 0.3899 3.2 Let the estimated demand curve be Qˆ ˆ1 ˆ 2P ˆ 3 PS ˆ 4 DI Solving for P and inserting values PS and DI * . we have ˆ1 ˆ2 P ˆ3 PS ˆ2 1 ˆ Q ˆ2 4.2(a) Demand and supply graph . Principles of Econometrics.2795 0. solving the supply curve Qˆ ˆ P ˆ 1 ˆ Q ˆ 1 2 2 20.5 111.3899 DI * 11.2719+2.9587 Qˆ 2. Figure xr11.1672 3.5 0.2960 22 5.9881 0.2b.2604 PEQM _ RF 32.0009 23 2.7822 0.2795 0.5070 23 18.0140 3.8407 . Exercise Solutions.5124 1.2(d) Demand and supply graph following a change in income .3539 23 62.7081 22 7.5 41. 4e 390 Exercise 11.2(d) is a plot of the two demand curves and the supply curve.2 (continued) (b) The equilibrium values can be found by equating the demand and supply equations at the given exogenous variable values.3745 P 1. the predicted equilibrium values are QEQM _ RF 7.Chapter 11.0328 0. Using the reduced form estimates in Tables 11.8951 0. One can equate either the equations derived in (a) or those with quantity as the left-hand side variable.5 1.3380 P Solving these two equations.3380 P 1.8154 These values are very close to those calculated in part (b).2509 and PEQM 62.3745P Supply: Qˆ 20. Principles of Econometrics.6564 22 2. we have QEQM (c) 18. The latter are Demand: Qˆ 4. 140 120 100 80 Demand* Supply Demand** P (d) 60 40 20 0 0 4 8 12 16 20 24 28 32 Q Figure xr11.2a and 11.6025 3. 3745P 1.5 46.5 3.4561 Using the reduced form estimates we first calculate the quantity demanded after income is increased from DIi = 3.2604 18. We calculate this as D % QEQM QEQM QEQM % DI DI DI 2.5 0.5 3.4154 The elasticity calculated using the graphical solution is similar to the elasticity calculated using reduced form estimates. or PF appears in the demand equation or the supply equation or both. They are not exactly the same because the result from the reduced form estimates does not take into account whether each of the variables PS.4276 18.Chapter 11.5) 3. Exercise Solutions.3786 18.2509 2. The graphical solution uses information on the location of these exogenous variables in the demand and supply equations.2960 22 5. The new demand equation is Qˆ 4.4276.2604 % DI DI DI (4.2 (continued) (e) The new equilibrium price and quantity are given by equating the new demand equation with the old supply equation.5 to DIi = 4. Combining this value with the equilibrium quantity demanded from part (c).7962 0.2509 (4.3745P ** Therefore the new equilibrium is QEQM ** 20.8407 7. we calculate the income elasticity of demand at the given exogenous variable values as D % QEQM QEQM QEQM 20.0377 QEQM 20.8784 62.5 0.3786 The income elasticity of demand is the percentage change in quantity demanded due to a percentage change in income and can be derived from the equation D % Q % DI Q Q DI DI The income elasticity of demand implied by the shift in part (d) is the percentage change in equilibrium quantity demanded given a percentage change in income at the given exogenous variable values.6295 and PEQM 69.0140 4. DI. and the changes in equilibrium price and quantity are (f) PEQM 69.2795 0. Principles of Econometrics.6295 18. This new equilibrium quantity demanded is ** QEQM 20.8784. 4e 391 Exercise 11. .5) 3. 3 (a) The wage equation cannot be estimated satisfactorily using the least squares estimator because it is part of a simultaneous equation system. See the insert on the following page for how to correct the standard errors . Exercise Solutions. For this simultaneous equation system. the steps are: Least squares estimation of the reduced form equation for HOURS. to be identified the equation must have M 1 = 2 1 = 1 variable absent (M being the number of equations in the simultaneous model system). Note that the standard errors calculated using this method will not be correct. Replace HOURS with HOURS in the wage equation. is omitted. The steps for conducting a two-stage least squares regression are outlined in Section 11. Principles of Econometrics. Having identified an auxiliary relationship. The wage equation is subject to endogeneity and the least squares estimator is biased and inconsistent.1. In this context. where the exogenous variables are EDUC. but the estimator is consistent. EXPER and KIDS Calculate the predicted values for the variable HOURS .5. (b) The wage equation is identified because 1 variable. 4e 392 EXERCISE 11. which has ln(WAGE) as an explanatory variable and HOURS as the dependent variable.Chapter 11. KIDS. and then estimate this new wage equation by least squares. tells us that ln(WAGE) and HOURS are endogenous variables. there are two simultaneous equations. (c) The alternative to least squares estimation is two-stage least squares estimation. Therefore. So. the correct 2SLS standard error is ˆ 22 SLS ˆ 22 SLS se ˆ 2 xˆi x 2 xˆi ˆ 2 SLS x 2 xˆi x 2 and the “wrong” standard error.Chapter 11. calculated in the 2nd least squares estimation. 4e 393 Insert: Correction of IV standard errors (Bonus material) ei the 2SLS estimator is the least In the simple linear regression model yi 1 2 xi ˆ ei where xˆi is the predicted value from a squares estimator applied to yi 1 2 xi reduced form equation. or SSE2SLS . Principles of Econometrics. Exercise Solutions. The problem with doing 2SLS with two least squares regressions is that in the second estimation the estimated variance is ˆ yi ˆ 2wrong ˆ xˆ 2 i 1 2 2 N The numerator is the SSE from the regression of yi on xˆi . In the simple regression model 2 ˆ ~N 2 The error variance 2 2 . which is SSEwrong. Thus. xˆi 2 x should be estimated using the estimator yi ˆ 22 SLS ˆ ˆ x 2 i 1 N 2 2 with the quantity in the numerator being the sum of squared 2SLS residuals. is se wrong ˆ ˆ 2wrong ˆ 2wrong 2 xˆi x 2 xˆi x ˆ wrong 2 xˆi x 2 Given that we have the “wrong” standard error in the 2nd regression. the 2SLS estimators are ˆ ˆ xˆi x 2 1 yi xˆi y 2 x ˆ x 2 y In large samples the 2SLS estimators have approximate normal distributions. we can adjust it using a correction factor se ˆ 2 ˆ 22 SLS ˆ 2wrong se wrong ˆ 2 ˆ 2 SLS ˆ wrong se wrong ˆ 2 . Two-stage least squares should be used to estimate the parameter because there is an endogenous variable. on the right-hand side of the second equation. The parameter is identified because x is absent from the second equation.4 (a) Least squares should be used to estimate the parameter explanatory variables in the first equation. The parameter (b) because there are no endogenous is identified because it can be consistently estimated.Chapter 11. Principles of Econometrics. There are M = 2 equations in this model. which implies that M – 1 = 1 variables should be absent for the model to be identified. Exercise Solutions. 4e 394 EXERCISE 11. and it is present in the first equation. y1. . 08252 t 13.3552 2.Chapter 11.05.014DI se 0.16) PS 1.3552 2.296 PS 5.5 (a) The demand and supply curve estimates are XR 11-5: 2SLS estimations -------------------------------------------(1) (2) DEMAND_2SLS SUPPLY_2SLS -------------------------------------------C -4. 4e EXERCISE 11.273 0.2795 20.0328*** (5. we have.02492 0.02) PF -1.3745 P 1. Demand Qˆ 4.1648 0.3380 P 1.56 12.08) -------------------------------------------N 30 30 -------------------------------------------Standard errors in parentheses * p<0.03 0.3380*** (0.280 0.001PF se 0.3745* (0.196 Supply Qˆ 20.0009*** (0. ** p<0.2960** (0.28) 0.22) P -0.54) (1.13 395 .284 t 2.36) DI 5. Exercise Solutions. *** p<0. Principles of Econometrics.001 Reporting these equations in the usual format.0140* (2.01. Both elasticities have a magnitude greater than 1 which indicates that both supply and demand considered elastic and therefore responsive to prices. are calculated as S % Q % P Q Q P P 2 P Q D % Q % P Q Q P P 2 P Q Using our estimates ˆS ˆ P 1 Q ˆD ˆ1 P Q 0.458 1. a percentage increase in price leads to a larger than 1% change in supply and demand.724 18.1485 18. 4e 396 Exercise 11. at the mean.5 (continued) (b) The price elasticities of supply and of demand.Chapter 11. we expect S to be positive because quantity supplied increases as price increases and we expect D to be negative because quantity demanded decreases as price increases.458 62.724 1.3380 0. .3745 62. Exercise Solutions. Principles of Econometrics.2725 The signs of the elasticities are as expected. 001PF se t 0. The standard errors of the least squares estimates are all smaller than those in Table 11.0328*** (1. Economic reasoning suggests that it should be negative since the quantity demanded decreases when price increases.001 Reporting these equations in the usual format.191 t 0.0910 (3.3380*** (0. the coefficients are almost equal to the estimates in 11.54 13.3380 P 1.08) -------------------------------------------N 30 30 -------------------------------------------Standard errors in parentheses * p<0. ** p<0.01.313 0.22) P 0. On the other hand. . 4e 397 EXERCISE 11.21) DI 0.19) PF -1.0233 (0.0009*** (0.2143 1. The intercept and coefficient of P have the opposite sign to their two-stage least squares counterparts and the coefficient estimates of PS and DI are much smaller than those in Table 11.0764 (1.Chapter 11. the least squares demand coefficient estimates are very different to the estimates in Table 11. we have.3b.7100** (0.06419 Supply Qˆ 20.3a.3a.05.7100 PS 0.02175 0.07639 15.02) ps 0.10 Considering the supply equation first.02330 P 0. All coefficients have signs which agree with economic reasoning except for the positive coefficient of P in the least squares demand equation.6 The least squares estimates of the demand and supply equations are XR 11-6: LS estimations -------------------------------------------(1) (2) Demand_ls Supply_ls -------------------------------------------- C 1. *** p<0.08) 0.07684 0.71) 20. Once again. but even though they are smaller the coefficients of P and DI are not significantly different from zero. Demand Qˆ 1.091 0. Exercise Solutions. the least squares standard errors are smaller than the two-stage least squares standard errors.07644 DI se 0. Principles of Econometrics.3b.03 0.3032 3. 4081 0. increasing the price of truffles. so we expect 2 0 .1750 1. Exercise Solutions.0000 or Pˆ se 11. Rearranging the supply equation.746707 -0.272869 3.461081 13. and if truffles are a normal good. (b) The estimated demand equation is Dependent Variable: P Method: Two-Stage Least Squares Included observations: 30 Instrument list: C PS DI PF C Q PS DI Coefficient Std.4284 2.0046 0. Q P 1 Q 1 3 1 PS 2 4 P DI 3 PS 4 DI e d . yields ed 2 1 2 Q 3 PS 4 DI u d According to economic theory.59161 1.1156 2. Q P 1 Q 1 3 1 2 P 3 PF e s . Thus we expect 2 0 . it is expected that there is an inverse relationship between price and quantity demanded.3899 DI 13.5916 1.6705Q 3. so we expect 3 0. If disposable income increases.38992 13.840843 -2.874899 0.7 (a) Rearranging the demand equation. so we expect 3 0 . An increase in the price of a factor of production reduces supply and increases equilibrium price. We expect 4 0.7467 .670519 3. there is a positive relationship between quantity supplied and price.102517 4. -11. Principles of Econometrics. Error t-Statistic Prob.115572 2.42841 -2.Chapter 11. If the price of a substitute increases the demand for truffles increases. yields PF e s 2 1 2 Q 3 PF u s According to economic theory.174955 1.4611PS 13. 4e 398 EXERCISE 11. then demand increases and equilibrium price increases.0315 0. 1560 The signs are as we expected in part (a) and all coefficients are significantly different from zero since all p-values are less than the level of significance of 0.936711 2. (c) The price elasticity of demand at the mean is calculated as % Q % P D Q Q P P 1 2 P Q Using our estimates ˆD 1 ˆ 2 (d) P Q 1 62.9367Q . Error t-Statistic Prob.0000 or Pˆ 58.2158 0.3899 3.155964 -10.859161 0.8592 0. 4e 399 Exercise 11.2470 2.958486 5.79822 2.9585 23 9. The lines are given by linear equations: Demand: Pˆ 11.7(b) (continued) The estimated supply equation is Dependent Variable: P Method: Two-Stage Least Squares Included observations: 30 Instrument list: C PF DI PS C Q PF Coefficient Std.6705 18.7982 2.724 2.0000 0.6705Q 3.9367Q 2.7(d) is a sketch of the supply and demand equations using the estimates from part (b) and the given exogenous variable values.4284 2.5 111.458 1. Principles of Econometrics.9585 PF se 5.05.4611 22 13. -58.215772 0.96905 0.61027 18.2725 Figure xr11.03526 13.5801 2.7982 2.Chapter 11. Exercise Solutions.6705Q Supply: Pˆ 58.9367Q 2.0000 0. 9367QEQM QEQM =18. thus PEQM 111.2b.8427 Using the reduced form estimates in Tables 11.5 0.2503 When QEQM is substituted into the demand equation (substituting into the supply equation will yield the same result) we find the equilibrium value of P.2604 PEQM _ RF 32.2503 62. . 4e 400 Exercise 11.5070 23 18.5 1.5124 1.2470 2. Exercise Solutions.5801 2.5801 2.6705QEQM 9.6025 3. we find them to be almost equal. Comparing the equilibrium values calculated using the results from part (d) to those calculated using the reduced form estimates.7(d) Demand and supply graph (e) The estimated equilibrium values from part (d) are given by equating the supply and demand equations after substituting in the given exogenous variable values.2a and 11.7(d) (continued) 120 100 80 Demand Supply 60 40 20 0 0 5 10 15 20 25 30 Q Figure xr11.1672 3.6564 22 2. the predicted equilibrium values are QEQM _ RF 7. Therefore equating these equations yields 111.8951 0.8154.7081 22 7.6705 18.Chapter 11.3539 23 62. Principles of Econometrics. the coefficient of Q has the opposite sign and the estimated intercept and the coefficient of PS are much smaller. All coefficient signs are correct. 4e 401 Exercise 11.7 (continued) (f) The estimated least-squares estimated demand equation is Pˆ se 13. Exercise Solutions.Chapter 11.3607 PS 12. The sign for the coefficient of Q is incorrect because it suggests that there is a positive relationship between price and quantity demanded.3582 DI 9. . The estimated supply equation is Pˆ se 52. Principles of Econometrics.1712 0. and the coefficient values do not differ much from the estimates in part (b).1512Q 1.1482 All estimates in this supply equation are significantly different from zero.6613Q 2.0872 0.4988 0. Compared to the results from part (b).6195 0.8763 2.0238 0.9217 PF 5.8254 All estimated coefficients are significantly different from zero except for the intercept term and the coefficient of Q.5940 1. 4219 0.1402 24130 43. Error t-Statistic Prob.9720 0.40781 -14. (b) 2 3 0 : A higher wage leads to an increased quantity of labor supplied. Exercise Solutions.729976 -342. : This sample has been taken for working women between the ages of 30 and 60.3919 11671 8.0205 -0. Also. 4 0. women who work are younger.44486 -7.3662 21698 7.397965 -3. : The effect of an increase in education is unclear.21544 17.4678 0. (c) The least squares estimated equation is Dependent Variable: HOURS Method: Least Squares Sample: 1 753 IF LFP=1 Included observations: 428 C LNWAGE EDUC AGE KIDSL6 KIDS618 NWIFEINC Coefficient Std. the standard deviation across all variables is smaller for working women. 2114.2831 0. 7 NWIFEINC measures the sum of all family income excluding the wife’s income.1307 54.730889 -1.161385 0.0002 0. 6 0 : The presence of children in the household reduces the number of hours worked because they demand time from their mother.82925 0.7483 0.424845 -3.8 (a) The summary statistics are presented in the following table Mean Standard Deviation Variable LFP = 1 LFP = 0 LFP = 1 LFP = 0 AGE KIDSL6 FAMINC 41.003656 6.004246 340.697 -17.321086 -0. 4e 402 EXERCISE 11. 5 0 : As income from other sources increases. have fewer children under the age of 6 and have a higher family income.529450 100.803925 -1. it becomes less necessary for the woman to work.7211 0.0000 0.5048 -115.Chapter 11.217309 -0.6369 12728 On average.96793 5.2461 .1629 0. Principles of Econometrics. It is not certain whether hours worked increases or decreases over this age group.0059 30.0007 0. 088603 0.015097 0. HOURS se 2115 17. 4e 403 Exercise 11.650176 -0.8(c) (continued) or.0999 100% = 9. .040710 -0.22 17. (d) The estimated reduced form equation is Dependent Variable: LNWAGE Method: Least Squares Sample: 1 753 IF LFP=1 Included observations: 428 C EDUC AGE KIDSL6 KIDS618 NWIFEINC EXPER EXPER^2 Coefficient Std.0636 An additional year of education increases wage by 0.715373 3.055873 -0.5 KIDSL6 115.124729 6. Assuming that this supply equation is part of a demand and supply simultaneous equation system.0000 0.632765 1.41ln(WAGE ) 14. -0.5272 0.027891 3.860055 0.00366 The negative coefficient for ln(WAGE) is unexpected.0 30.017648 5. Therefore only one exogenous variable needs to be absent from the supply equation for it to be identified.44 EDUC 7. Exercise Solutions.013372 0. using a one tail test.25.32E-06 0.0 KIDS 618 0.003520 -0.1 54.69E-06 0.730 AGE 340. the F value is less than the rule of thumb value for strong instrumental variables of 10. we expected this coefficient to be positive.5287 0.01 level.000402 -1.005415 0.99%.Chapter 11.5159 0. We see that EXPER is significant at the .530 342.357997 0. The F-test of their joint significance yields an F value of 8.318296 0.0025 0. which gives a p-value of 0.00425 NWIFEINC 100.2613 0.83 0. While the joint test leads us to reject the null hypothesis that the coefficients of both EXPER and EXPER2 are zero. (e) The presence of EXPER and EXPER2 in the reduced form equation and their absence in the supply equation serves to identify the supply equation.000747 0.97 5.615970 -0. Principles of Econometrics.0003.0870 0. and having 2 exogenous variables absent is sufficient for this requirement if these variables are strongly significant.630591 -0. and EXPER2 is significant at the 5% level. Error t-Statistic Prob.099884 -0. M – 1 = 2 – 1 = 1.044344 -1. written out in full. The other coefficients have signs that are not contrary to our expectations.14259 9.818 -177. 2432.1543 or HOURS se 2432 1545ln(WAGE ) 177 EDUC 10.191596 -0.213426 -3.92 0.427088 0.2 480.051961 -1.577347 176.4039 0.835539 -1.1 9.0001 0.009249 594.00648 The statistically significant coefficients are the coefficients of ln(WAGE) and EDUC. Error t-Statistic Prob. .1718 480.7 58.00925 NWIFEINC 177 56.55708 -0.78 AGE 594.91786 0.0024 0.0014 0.8 (continued) (f) The two-stage least squares estimated equation is Dependent Variable: HOURS Method: Two-Stage Least Squares Sample: 1 753 IF LFP=1 Included observations: 428 Instrument list: C EDUC AGE KIDSL6 KIDS618 NWIFEINC EXPER EXPER^2 C LNWAGE EDUC AGE KIDSL6 KIDS618 NWIFEINC Coefficient Std.Chapter 11. 4e 404 Exercise 11.78409 -210. The sign of ln(WAGE) has changed to positive and so is now in line with our expectations.2608 0.4490 -10. Exercise Solutions.2341 0.125999 -1.093425 3.9340 56.8339 -47.198 1544. Principles of Econometrics.006481 4.7387 58.56 KIDS 618 0.577 211KIDSL 6 47. 59. Critical Values # of endogenous regressors: 1 Ho: Instruments are weak # of excluded instruments: 2 --------------------------------------------------------------------| 10% 15% 20% 25% 2SLS Size of nominal 5% Wald test | 19. If we use the Stock-Yogo critical value we can be more precise.Chapter 11. the F value is less than the rule of thumb value for strong instrumental variables of 10.25.25 --------------------------------------------------------------------- . choosing the criteria based on the size of nominal 5% test having maximum size of 15%. Indeed we cannot reject the hypothesis that the instruments are weak unless we are willing to accept a 25% rejection rate for a nominal 5% test. 4e 405 Exercise 11. we noted that the F-test of their joint significance yields an F value of 8.8 (continued) Bonus material: Additional analysis of identification (e) In the solution above. While the joint test leads us to reject the null hypothesis that the coefficients of both EXPER and EXPER2 are zero. against the alternative that they are not.0003. which gives a p-value of 0. We cannot reject the null hypothesis that the instruments are weak based on this criterion.59 8.93 11.75 7. the critical value for the F-statistic is 11. Testing the null hypothesis that the instruments are weak. Principles of Econometrics. Exercise Solutions. If = 1.Chapter 11. then 5 0 . (iv) The approximation 100 ln y % y is accurate if the changes in the variable are not too large. Thus both equations are “identified” according to the order condition. The law of demand implies that 3 0 . since its interpretation is the same as in the log-log demand model. Because the variables in the equation are time series. Principles of Econometrics. thus we anticipate 3 0 . so we expect 4 0 . (iii) 2 is the price elasticity of supply. The exogenous variables are ln(Y) and ln(PB) as income and the price of beef are determined outside the model. because these variables are jointly determined by supply and demand. the variables are growth rates. TIME and lagged production QPRODt 1 . (b) (i) The intercept falls out of the model. (iv) The law of supply suggests 2 0 . The serial correlation problem is solved. 4e 406 EXERCISE 11. just attached to transformed variables. which is a necessary but not sufficient condition. (ii) The exogenous variables are the price of broiler feed (PF). (iii) The generalized least squares transformation is discussed in Appendix 9A. (ii) The parameters of interest are not affected. there must be at least M = 2 1 = 1 variable omitted from an equation for identification. TIME and lagged production QPRODt 1 . thus we expect 4 0 . . An increase in the price of the substitute good (beef) will increase the equilibrium price and quantity of poultry. If a year of high production follows a previous year of high production. If there is technical progress there should be more output from unchanged inputs. Exercise Solutions. as price and quantity are jointly determined by supply and demand. (c) (i) The endogenous variables in this supply equation are ln(QPROD) and ln(PRICE). then the transformed error vtd is not serially correlated.9 (a) The endogenous variables in this demand equation are ln(Q) and ln(P). (d) In this system of M = 2 equations. The supply equation omits two variables: the changes in income Y and the price of beef. (vi) Since poultry is a normal good we anticipate 2 0 . An increase in the price of an input reduces equilibrium quantity. (v) The parameter 2 is the income elasticity of demand. or exogenously. and the variables are in “differenced” form. In the demand equation there are 3 variables omitted: price of broiler feed (PF). Exercise Solutions.173381 .069893 -0. Error t-Statistic Prob. -2.009202 0.632990 0.031340 0.4034 0.0266 0.134%. Principles of Econometrics.05 level) effects on the equilibrium growth rate of price. 17. YEAR DLP DLPF_SEF DLP_LB DLPF DLP_UB 2000.142191 -0. The effect of growth in the price of feed [log(PF)] has a positive and significant (at the .0038 0.9 (continued) (e) The estimated reduced form for the change in ln(P) = DLP is given below Dependent Variable: DLP Method: Least Squares Sample: 1950 2001 IF (YEAR>1959) AND (YEAR<2000) Included observations: 40 C DLY DLPB LOG(PF) TIME LOG(QPROD(-1)) Coefficient Std.259794 1.6%.167566 1.1673 0.10 level) effect on equilibrium growth rate of price.0322 [34 degrees of freedom].283072 0.000 -0. The 95% interval estimate is [ 11.2081 (i) The reduced form shows that increases in the growth of income (DLY) and in the growth of beef price (DLPB) have positive and significant (at the .411132 3.34%].Chapter 11.536048 0.317904 1.077109 0.110700 0.844021 -0.007787 0.963925 0.0739 0. 4e 407 Exercise 11. The other variable are not significant.07%.102618 2. based on the reduced form in (i) is 3.846152 1. The actual value is inside this rather wide interval. The predicted value. (ii) The actual growth in price in 2000 was 2.453689 0.026290 0.195732 0. using the t-critical value 2.202478 -1. Chapter 11, Exercise Solutions, Principles of Econometrics, 4e 408 Exercise 11.9 (continued) (f) The estimated reduced form for ln(P) is Dependent Variable: LOG(P) Method: Least Squares Sample: 1950 2001 IF (YEAR>1959) AND (YEAR<2000) Included observations: 40 C LOG(PF) TIME LOG(QPROD(-1)) DLY DLPB Coefficient Std. Error t-Statistic Prob. -2.811041 0.272105 -0.031646 0.437906 0.246556 0.400223 1.852558 0.092998 0.011098 0.244199 0.763420 0.236064 -1.517384 2.925939 -2.851342 1.793231 0.322963 1.695400 0.1384 0.0061 0.0074 0.0818 0.7487 0.0991 (i) The estimates show that increasing the price of feed [log(PF)] has a positive and significant [at the .01 level] effect on equilibrium ln(PRICE). The effect of TIME is negative and significant at the .01 level, implying significant technological progress. Lagged production and growth in the price of beef have positive and significant (at the 0.10 level) effects on ln(PRICE). (ii) The real price of chicken is $0.946. The 95% interval estimate is [$0.701, $0.987]. The point prediction [using the natural predictor] is $0.831. The observed value is within the interval. YEAR P PHAT_LB PHAT PHAT_UB 2000.000 0.945990 0.700588 0.831498 0.986869 To obtain this prediction interval we follow the procedure outlined in Chapter 4.5.5 of POE, page 155. Chapter 11, Exercise Solutions, Principles of Econometrics, 4e 409 Exercise 11.9 (continued) (g) The two stage least squares estimates are: Demand Dependent Variable: DLQ Method: Two-Stage Least Squares Sample: 1950 2001 IF (YEAR>1959) AND (YEAR<2000) Included observations: 40 Instrument list: C LOG(PF) TIME LOG(QPROD(-1)) DLY DLPB DLY DLP DLPB Coefficient Std. Error t-Statistic Prob. 0.856237 -0.453350 0.311649 0.150318 0.110838 0.106347 5.696173 -4.090210 2.930493 0.0000 0.0002 0.0058 Supply Dependent Variable: LOG(QPROD) Method: Two-Stage Least Squares Sample: 1950 2001 IF (YEAR>1959) AND (YEAR<2000) Included observations: 40 Instrument list: C LOG(PF) TIME LOG(QPROD(-1)) DLY DLPB C LOG(P) LOG(PF) TIME LOG(QPROD(-1)) Coefficient Std. Error t-Statistic Prob. 2.784102 0.227421 -0.147371 0.018584 0.628437 1.158856 0.245024 0.077867 0.009831 0.164478 2.402458 0.928162 -1.892606 1.890218 3.820798 0.0217 0.3597 0.0667 0.0670 0.0005 The demand equation estimates are the correct signs and significant at the .01 level. The income elasticity of demand is estimated to be 0.856. The price elasticity of demand is estimated to be 0.453, and the cross-price elasticity of demand is 0.312. The supply estimates reveal that the price elasticity of supply is not estimated very precisely, and it is statistically insignificant. The estimated coefficient of the price of feed implies that a 1% increase in the price of feed decreases supply by 0.147 percent. The estimate is significant at the .10 level. The estimated coefficient of TIME is positive and significant at the .10 level, showing that technology has increased the quantity produced by about 1.8% per year. Finally, lagged production is very significant and positive. Chapter 11, Exercise Solutions, Principles of Econometrics, 4e 410 Exercise 11.9 (continued) (h) Adding the log of exports as an instrument yields the following estimates of the supply equation. Dependent Variable: LOG(QPROD) Method: Two-Stage Least Squares Sample: 1950 2001 IF (YEAR>1959) AND (YEAR<2000) Included observations: 40 Instrument list: C LOG(PF) TIME LOG(QPROD(-1)) DLY DLPB LEXPTS C LOG(P) LOG(PF) TIME LOG(QPROD(-1)) Coefficient Std. Error t-Statistic Prob. 3.342003 0.408017 -0.194669 0.024716 0.542197 1.240020 0.193525 0.074501 0.009232 0.170358 2.695120 2.108346 -2.612968 2.677104 3.182687 0.0107 0.0422 0.0131 0.0112 0.0031 The effect of using this instrument is to increase the magnitudes of the coefficients, and reduce their p-values, except for lagged production. Exports are a good instrument in the sense that they should be strongly correlated with the endogenous variable PRICE. However, if exports are jointly determined with price and domestic consumption, then exports are endogenous and correlated with the supply equation making it an invalid instrument. Using exports as an instrument means that we have two surplus instruments. Testing their validity using Sargan’s NR2 test yields a p-value of 0.657, indicating that we cannot reject the validity of the overidentifying (surplus) instruments. Bonus Material on instrument strength (g) In part (g) we noted that the demand equation estimates are of correct sign and significant. However, for the demand equation the first stage F-statistic is only 3.73, which is far less than the desired rule of thumb. The Stock-Yogo critical values are 5.34 for maximum relative bias of 30% [Table 10E.2], and 8.31 for 25% test size for a nominal 5% test [Table 10E.1]. Similarly for the supply equation the first stage F-is 1.88. The Stock-Yogo critical value is 7.25 for 25% test size for a nominal 5% test [Table 10E.1]. Chapter 11, Exercise Solutions, Principles of Econometrics, 4e 411 Exercise 11.9 (continued) Bonus Material on instrument strength (h) The log of exports is statistically significant in the following first stage regression (Stata output) First-stage regression of lp: OLS estimation -------------Estimates efficient for homoskedasticity only Statistics consistent for homoskedasticity only Total (centered) SS Total (uncentered) SS Residual SS = = = 1.763650454 3.267532368 .1579906212 Number of obs F( 6, 33) Prob > F Centered R2 Uncentered R2 Root MSE = = = = = = 40 55.90 0.0000 0.9104 0.9516 .06919 -----------------------------------------------------------------------------lp | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------lpf | .2777852 .0864871 3.21 0.003 .1018258 .4537447 time | -.0248097 .0106693 -2.33 0.026 -.0465165 -.0031028 lqprod_1 | .0330933 .278201 0.12 0.906 -.5329108 .5990974 dly | .1278641 .7112976 0.18 0.858 -1.319282 1.57501 dlpb | .4939115 .2225954 2.22 0.033 .0410377 .9467853 lexpts | 1.042761 .4141909 2.52 0.017 .2000836 1.885439 _cons | .5263764 2.173379 0.24 0.810 -3.895396 4.948148 -----------------------------------------------------------------------------Included instruments: lpf time lqprod_1 dly dlpb lexpts ------------------------------------------------------------------------------ The first stage F-for the significance of the instruments is 3.56 which is far less than the desired rule of thumb. The Stock-Yogo critical values are 5.39 for maximum relative bias of 30% [Table 10E.2], and 7.80 for 25% test size for a nominal 5% test [Table 10E.1]. We cannot reject the null hypothesis that the instruments are weak. Chapter 11, Exercise Solutions, Principles of Econometrics, 4e 412 EXERCISE 11.10 (a) The two-stage least squares estimation of the supply equation (11.14) is Dependent Variable: LQUAN Method: Two-Stage Least Squares Included observations: 111 Instrument list: MON TUE WED THU STORMY Coefficient Std. Error t-Statistic Prob. 8.628354 0.001059 -0.363246 0.388970 1.309547 0.464912 22.18256 0.000809 -0.781321 0.0000 0.9994 0.4363 C LPRICE STORMY or ln(QUAN ) 8.628 0.00106ln( PRICE ) 0.363STORMY se 0.3890 1.31 0.465 The signs of these estimated coefficients are as expected. The coefficient ˆ 2 is positive suggesting that there is a positive relationship between price and quantity supplied. However, this coefficient is virtually zero and is not significant at any level. The coefficient ˆ 3 is negative agreeing that less fish are supplied in stormy weather, but it is also not significant. The elasticity of supply is estimated as the coefficient of ln(PRICE) since this is a log-log equation. Thus, S = 0.0011 implying that supply is inelastic. (b) The new demand equation is ln(QUAN ) 1 2 ln( PRICE ) 6 RAINY 3 7 MON TUE 3 WED 4 THU 5 COLD e d The algebraic reduced form for ln(PRICEt) is ln( PRICE ) 12 22 72 MON RAINY TUE 32 82 WED 42 COLD v2 THU 52 62 STORMY Chapter 11, Exercise Solutions, Principles of Econometrics, 4e 413 Exercise 11.10 (continued) (c) The estimated reduced form equation is Dependent Variable: LPRICE Method: Least Squares Sample: 1 111 Included observations: 111 C MON TUE WED THU STORMY RAINY COLD Coefficient Std. Error t-Statistic Prob. -0.290228 -0.121576 -0.056677 -0.028360 0.040420 0.312658 -0.016733 0.080989 0.082069 0.108589 0.106981 0.108520 0.105824 0.081793 0.093620 0.074359 -3.536396 -1.119604 -0.529786 -0.261330 0.381961 3.822553 -0.178737 1.089155 0.0006 0.2655 0.5974 0.7944 0.7033 0.0002 0.8585 0.2786 or ln( PRICE ) se 0.2902 0.1216 MON 0.0567TUE 0.0284WED 0.0404THU 0.0821 0.1086 0.1070 0.1085 0.3127 STORMY 0.0167 RAINY 0.0810COLD 0.0818 0.0936 0.0744 0.1058 The degrees of freedom for the F-test of the joint significance of all variables except for STORMY are (6, 103). The test returns a p-value of 0.7229 which is much larger than the level of significance, 0.05. This implies that we cannot reject the null hypothesis that all coefficients are equal to zero. Thus, the instrumental variables are not adequate for estimation of the supply equation. The value of the F-statistic is only 0.61, far below the rule of thumb value of 10. Chapter 11, Exercise Solutions, Principles of Econometrics, 4e 414 Exercise 11.10 (continued) (d) The least squares and two-stage least squares estimates of the demand and supply equations are: XR 11-10(d): 2SLS and LS estimations ---------------------------------------------------------------------------(1) (2) (3) (4) DEMAND_LS DEMAND_2SLS SUPPLY_LS SUPPLY_2SLS ---------------------------------------------------------------------------C 8.6169*** 8.4417*** 8.5009*** 8.5848*** (0.16) (0.22) (0.10) (0.32) LPRICE -0.5446** (0.18) -1.2228* (0.53) MON 0.0316 (0.21) -0.0333 (0.23) TUE -0.4935* (0.20) -0.5328* (0.22) WED -0.5392* (0.21) -0.5756* (0.22) THU 0.0948 (0.20) 0.1179 (0.22) RAINY 0.0666 (0.18) 0.0720 (0.19) -0.0616 (0.13) 0.0681 (0.17) COLD -0.4381* (0.19) -0.1489 (1.06) STORMY -0.2160 -0.3130 (0.16) (0.39) ---------------------------------------------------------------------------Standard errors in parentheses * p<0.05, ** p<0.01, *** p<0.001 Discussion of demand equation: The estimated sign for ln(PRICE) is as expected and the coefficient is statistically significant in both estimations. We see that the two-stage least squares estimate is more negative suggesting that using least squares has an upwards bias on the coefficient of ln(PRICE). The coefficient estimates of the dummy variables TUE, WED and THU have the same signs in both estimations and the dummy variable MON has a different sign. MON and THU are not significantly different from zero in both model estimations. The signs of the 2SLS estimates of the coefficients of RAINY and COLD are not as expected, since rainy and cold days are meant to deter people from eating out. However, we note that these coefficient estimates are not significantly different from zero at a 5% level of significance in both model estimations. Discussion of supply equation: The coefficient for ln(PRICE) does not have the expected coefficient in either estimation. The negative coefficient estimate is not consistent with economic theory which says that quantity supplied and price are positively related. In addition to having the wrong sign, the two-stage least squares estimate has a large standard error which could be a result of inadequate instrumental variables. The coefficient for STORMY has the expected negative coefficient but is not significantly different from zero in both estimations. Chapter 11, Exercise Solutions, Principles of Econometrics, 4e 415 Exercise 11.10 (continued) (e) The augmented supply equation is ln(QUAN ) 1 2 ln( PRICE ) 3 STORMY 4 MIXED e s The demand equation is as specified in part (b). The least squares estimated reduced form equation for ln(PRICE) is Dependent Variable: LPRICE Method: Least Squares Included observations: 111 C MON TUE WED THU STORMY RAINY COLD MIXED Coefficient Std. Error t-Statistic Prob. -0.373905 -0.114093 -0.076200 -0.060763 0.033436 0.416731 -0.004910 0.063500 0.231099 0.084282 0.104875 0.103509 0.105366 0.102202 0.086674 0.090483 0.072045 0.079313 -4.436355 -1.087897 -0.736169 -0.576681 0.327152 4.808031 -0.054265 0.881394 2.913736 0.0000 0.2792 0.4633 0.5654 0.7442 0.0000 0.9568 0.3802 0.0044 or ln( PRICE ) se 0.3739 0.1141MON 0.0762TUE 0.0608WED 0.0334THU 0.0843 0.1049 0.1035 0.1054 0.1022 0.4167STORMY 0.0049RAINY 0.0635COLD 0.2311MIXED 0.0867 0.0905 0.0720 0.0793 The F-test for the joint significance of the coefficients of MON, TUE, WED, THU, RAINY and COLD has an F-statistic value of 0.5432 and a p-value (F(6, 102)) of 0.7742. Since this p-value is larger than the level of significance, 0.05 and ( 0.542 F(0.95,6,102) 2.189 ), we cannot reject the null hypothesis that these coefficients are equal to zero. Therefore, since the instrumental variables that are required to identify the supply equation are not statistically significant, the addition of MIXED does not increase the chances of estimating the supply equation by two-stage least squares. Chapter 11, Exercise Solutions, Principles of Econometrics, 4e 416 Exercise 11.10 (continued) (f) The least squares and two-stage least squares estimates of the demand and supply equations are: XR 11-10(f): 2SLS and LS estimations ---------------------------------------------------------------------------(1) (2) (3) (4) DEMAND_LS DEMAND_2SLS SUPPLY_LS SUPPLY_2SLS -0.4021 (0.20) 1.0723 (1.41) -0.2738 (0.19) -0.9178 (0.65) ---------------------------------------------------------------------------C 8.6169*** 8.5130*** 8.5570*** 9.1348*** (0.16) (0.19) (0.13) (0.57) LPRICE -0.5446** (0.18) -0.9470* (0.41) MON 0.0316 (0.21) -0.0069 (0.21) TUE -0.4935* (0.20) -0.5168* (0.21) WED -0.5392* (0.21) -0.5608** (0.21) THU 0.0948 (0.20) 0.1085 (0.21) RAINY 0.0666 (0.18) 0.0698 (0.18) -0.0616 (0.13) 0.0153 (0.15) COLD STORMY MIXED -0.1062 -0.4541 (0.17) (0.39) ---------------------------------------------------------------------------Standard errors in parentheses * p<0.05, ** p<0.01, *** p<0.001 Discussion of demand equation: The estimated sign for ln(PRICE) is as expected and the coefficient is statistically significant in both estimations. We see that the two-stage least squares estimate is more negative suggesting that using least squares has an upwards bias on the coefficient of ln(PRICE). The coefficient estimates of the dummy variables TUE, WED and THU have the same signs in both estimations and the dummy variable MON has a different sign. MON and THU are not significantly different from zero in both estimations. The signs of the coefficients of RAINY and COLD are not as expected with the exception of the least squares coefficient estimate of COLD. We can note that these coefficient estimates are not significantly different from zero at a 5% level of significance in both estimations. Discussion of supply equation: The coefficient for ln(PRICE) has the expected sign in the twostage least squares estimation and an unexpected sign in the least squares estimation. This is a slight improvement on the results obtained in part (d). However, this coefficient remains statistically insignificant in the two-stage least squares estimation. The coefficients for STORMY and MIXED have the expected negative coefficient but are not significantly different from zero in both estimations. Chapter 11, Exercise Solutions, Principles of Econometrics, 4e 417 EXERCISE 11.11 (a) The estimated reduced form equation is Dependent Variable: LPRICE Method: Least Squares Sample: 1 111 IF CHANGE=1 Included observations: 77 C MON TUE WED THU STORMY Coefficient Std. Error t-Statistic Prob. -0.481515 0.037860 0.067762 0.142633 0.257394 0.443856 0.097537 0.122927 0.123268 0.129110 0.126655 0.082429 -4.936764 0.307990 0.549712 1.104736 2.032245 5.384721 0.0000 0.7590 0.5842 0.2730 0.0459 0.0000 or ln( PRICE ) se 0.4815 0.0379MON 0.0975 0.1229 0.0678TUE 0.1426WED 0.2574THU 0.1233 0.1291 0.1267 0.4439 STORMY 0.0824 The p-value for testing the null hypothesis H 0 : 62 0 is 0.0000. Since this value is less than the level of significance, 0.05, we reject the null hypothesis and conclude that this coefficient is significantly different from zero. The F-test value is 29.00, well above the rule of thumb threshold of 10. It is important to test for the statistical significance of STORMY because it is the supply equation’s shift variable. It is required to be statistically significant for the demand equation to be identified. If STORMY is not statistically significant, then the two-stage least squares regression and the estimation procedure will be unreliable. Bonus material: The Stock-Yogo test for weak instrument critical value, using the criteria of test size, is 16.38 [Table 10E.1] if we can tolerate a test with Type I error of 10% for a 5% nominal test. Chapter 11, Exercise Solutions, Principles of Econometrics, 4e 418 Exercise 11.11 (continued) (b) The null hypothesis of this Hausman test is H 0 : cov(ln( PRICE ), e) 0 , which is tested by testing for the significance of the coefficient of vˆ2 in ln(QUAN ) 1 2 ln( PRICE ) 3 MON TUE 4 WED 5 THU 6 vˆ2 error In the results below vˆ2 is denoted VHAT2. Dependent Variable: LQUAN Method: Least Squares Included observations: 77 after adjustments C LPRICE MON TUE WED THU VHAT2 Coefficient Std. Error t-Statistic Prob. 8.362892 -1.018517 0.295299 -0.345265 -0.361873 0.394476 0.821220 0.180431 0.316737 0.209392 0.209403 0.220716 0.226110 0.375889 46.34949 -3.215656 1.410268 -1.648803 -1.639538 1.744621 2.184743 0.0000 0.0020 0.1629 0.1037 0.1056 0.0854 0.0323 0 are 2.1847 and 0.0323 The t-statistic and p-value for the null hypothesis H 0 : respectively. Since this p-value is less than the level of significance, 0.05, we reject the null hypothesis and conclude that ln(PRICE) is endogenous. The robust version of this test yields t-statistic of 2.27, and thus our conclusion is unchanged. Chapter 11, Exercise Solutions, Principles of Econometrics, 4e 419 Exercise 11.11 (continued) (c) The least squares and two-stage least squares estimates of the demand equation are: XR 11-11(c): 2SLS and LS estimations -------------------------------------------(1) (2) DEMAND_LS DEMAND_2SLS -------------------------------------------C 8.5327*** 8.3629*** (0.17) (0.19) LPRICE -0.4354* (0.18) -1.0185** (0.34) MON 0.2928 (0.21) 0.2953 (0.22) TUE -0.3500 (0.21) -0.3453 (0.22) WED -0.4081 (0.23) -0.3619 (0.23) THU 0.2690 0.3945 (0.22) (0.24) -------------------------------------------N 77 77 -------------------------------------------Standard errors in parentheses * p<0.05, ** p<0.01, *** p<0.001 These estimates have the expected signs. The two-stage least squares and least squares estimates are very similar in values with the exception of the coefficient of ln(PRICE). Both estimation procedures conclude that the day indicator variables are not significant at a 5% level of significance. Compared to Table 11.5 all estimated coefficients have the same sign except for the coefficient of MON. Also the intercept estimate and the coefficient estimate of ln(PRICE) are similar but the coefficient estimates for TUE, WED and THU are quite different. Furthermore, all of the part (c) two-stage least squares estimates of the weekday indicator variables are insignificant whereas Table 11.5 shows that TUE and WED are statistically significant. 2760 0.782984 -0.0322TUE 0.171103 -0.3780 or ln( PRICE ) se 0.166718 -0.8636 0. although the only estimated coefficient with the opposite sign is the coefficient of THU.091936 -0.895749 0.185680 0. Exercise Solutions. Error t-Statistic Prob.1647 0. TUE and WED.1711MON 0. in part (a) the only variables which were not statistically significant were MON. Comparing these results to Table 11.244165 0.149337 0.218527 0.1493 0.164665 0.171072 0.4b and not statistically significant in the above regression.032193 -0.482796 0.112051 0.1711 0. In addition. However.1667 These results are very different to those obtained in part (a). STORMY is statistically significant in Table 11.110988 -1. all of the estimated coefficients have very different values.11 (continued) (d) The estimated reduced form equation is Dependent Variable: LPRICE Method: Least Squares Sample: 1 111 IF CHANGE = 0 Included observations: 34 C MON TUE WED THU STORMY Coefficient Std. -0. In part (d) all exogenous variables are statistically insignificant. All weekday indicator variables are statistically insignificant in both estimated regressions.1493 STORMY 0.2442THU 0.2185 0.0103 0.190059 -0. .010302 -0.9274 0.Chapter 11.1901WED 0.1121 0.1857 0. 4e 420 Exercise 11.173381 -1.4(b). Principles of Econometrics. All the coefficients of the weekday indicator variables have opposite signs and the coefficient for STORMY is smaller.4402 0. 776691 -0.109989 0. 4e 421 Exercise 11. Since this p-value is greater than the level of significance.0000 0.9680.05.872323 -0. we do not reject the null hypothesis and conclude ln(PRICE) does not show signs of endogeneity.436043 -0.1121 0.868443 -0.324423 -1.6262 0. Error t-Statistic Prob.642430 -1.427418 0.49548 -0. The t-statistic and p-value for the null hypothesis H 0 : respectively. 0. the Hausman test is a test for the endogeneity of ln(PRICE).11 (continued) (e) As described in part (b). which is tested by testing for the significance of the coefficient of vˆ2 in ln(QUAN ) 1 2 ln( PRICE ) 3 MON TUE 4 WED 5 THU 6 vˆ2 ed The variable V2HAT = vˆ2 Dependent Variable: LQUAN Method: Least Squares Sample: 1 111 IF CHANGE = 0 Included observations: 34 C LPRICE MON TUE WED THU V2HAT Coefficient Std. 8. Principles of Econometrics.714966 32.607449 0.7481 0.676883 0.040512 0. Exercise Solutions.901467 -0.354528 -0.719563 2.971814 -1.842788 -0.0590 0.270090 2.1625 0.0405 and 0. simultaneity between demand and supply does not exist. .492699 -0.9680 0 are 0. This is consistent with Graddy and Kennedy’s expectation that when inventory changes are small.Chapter 11.548862 0. 4e 422 Exercise 11.05.71) -------------------------------------------N 34 34 -------------------------------------------Standard errors in parentheses * p<0. The major difference between the two sets of estimates is that. *** p<0.01. which is likely to exhibit simultaneous equation bias.41) -0. the coefficient of ln(PRICE) is always significantly different from zero in part (c) and only significant in the least squares part (f) estimation.63) MON -0. .3786 -0.7767*** (0. we find that the coefficient estimates for ln(PRICE) appear to be quite similar with the exception of the least squares coefficient estimate of ln(PRICE) in part (c).40) (0.Chapter 11. Comparing these values to those in part (c).27) LPRICE -0.9754* (0.26) (0. Exercise Solutions. TUE and THU whereas none of the two-stage least squares coefficient estimates are significantly different from zero.7756*** 8.44) -0.42) -0.9118 (0.54) TUE -0.8428 (0.60) THU -0.001 All the estimates have the expected signs and are almost identical. The estimated values of the coefficients of the weekday indicator variables are very different. as a consequence of the smaller least squares standard errors. Principles of Econometrics.42) WED -0.3545 (0.8904* (0. ** p<0.8409 (0. Also.11 (continued) (f) The least squares and two-stage least squares estimates of the demand equation are: XR 11-11(f): 2SLS and LS estimations -------------------------------------------(1) (2) DEMAND_LS DEMAND_2SLS -------------------------------------------C 8.8684 (2.9015 (0.48) -0.8723 (0. all of the least squares coefficient estimates are significantly different from zero except those for MON. Chapter 11. 4e 423 Exercise 11. This is supported by our estimates which show that the two stage least squares and least squares coefficient estimates of ln(PRICE) are similar when CHANGE = 0 but very different when CHANGE = 1. causing large changes in inventory. . Principles of Econometrics. and part (f) models the demand for fish for small changes in inventory. It has been postulated that when more fish are sold and bought. This discrepancy suggests that a coefficient bias exists when CHANGE = 1 due to endogeneity. sellers are more responsive to prices and therefore endogeneity is present and on the days where there is little change in inventory endogeneity should not be present.11(f) (continued) Part (c) models the demand for fish when there are large changes in inventory. Also note that the least squares estimate of the price elasticity of demand when CHANGE = 0 is similar in magnitude to the two-stage least squares estimate of the price elasticity of demand when CHANGE = 1. Exercise Solutions. CHAPTER 12 Exercise Solutions 424 . Chapter 12. Exercise Solutions. yt 2 ] E[vt vt 2 1 = E[ 2 vt2 2 = 2 = 2 2 [ [1 2 /(1 vt 4 2 t 3 v 2 4 2 )] 3 2 vt 6 2 t 4 v ] ][vt 3 ] 2 vt 2 3 vt 4 ] . j k since E[vt2 ] E[vt2 1 ] ] 2 v )] ] [1/(1 2 )] is the sum of a geometric progression. The covariance between yt and yt is: 2 cov[ yt . The variance of y is: where [1 2 var[ yt ] E[vt vt = E[vt vt = E[vt2 2 2 t 1 2 = 2 [1 = 2 [1/(1 2 1 vt 2 1 vt v 2 2 ][vt 2 4 2 t 2 v 2 ]2 2 2 vt 1 ] vt ] 2 since E[vt j vt k ] 0.1 (a) The AR(1) model yt yt y1 y0 y2 y1 v2 yt vt vt can be rewritten as a function of lagged errors: 1 v1 ( y0 2 vt 1 t vt 2 v1 ) v2 2 y0 v1 v2 y0 The mean of yt is: E[ yt ] E[vt vt 2 1 vt ] 0. Principles of Econometrics. 4e 425 EXERCISE 12. 2 since the error vt has zero mean and the value of t y0 is negligible for a large t. The mean of yt is: E ( yt ) E (v1 v2 y0 vt ) since E (vt ) 0 y0 The variance of yt is: var( yt ) E ( yt )]2 E[ yt E[ y0 y0 ]2 vt E[v1 v2 vt ]2 E[v12 vt2 ]2 t since E (vt2 ) v1 v2 v22 2 v E (vt2 1 ) The covariance of yt and yt 2 v 2 is: cov[ yt .Chapter 12. Principles of Econometrics.1 (continued) (b) The random walk model yt y1 y0 y2 y1 v2 yt yt yt vt can be written as a function of lagged errors: 1 v1 ( y0 vt 1 v1 ) v2 t y0 s 1 2 y0 s 1 vs vs where y0 is the initial value. 4e 426 Exercise 12. yt 2 ] E[vt vt = E[vt2 2 = 2 v 1 vt vt2 3 [t 2] 2 vt vt2 4 3 ][vt ] 2 vt 3 vt 4 ] . Exercise Solutions. 41. the null of nonstationarity is rejected.Chapter 12.41. For X: since the tau ( 3 099) is greater than the 5% critical value of 3. and we infer that Z is not stationary. the null of nonstationarity is not rejected. and we infer that Y is not stationary. and we infer that X is not stationary.178) is less than the 5% critical value of 2. the null of nonstationarity is not rejected. the null of nonstationarity is not rejected.86. Exercise Solutions. For Z: since the tau ( 913) is greater than the 5% critical value of 3.2 For W: since the tau ( 3. and we infer that W is stationary. . For Y: since the tau ( 975) is greater than the 5% critical value of 2. Principles of Econometrics.86. 4e 427 EXERCISE 12. is stationary. where 1 yt is integrated of order 1. since it had to be differenced once stationarity. yt yt 1 2 ) vt . Principles of Econometrics. 4e 428 EXERCISE 12. Hence yt is integrated of order 1.Chapter 12. from both sides of the equation: yt vt ~ N (0. yt is integrated of order 2. . In other words. Exercise Solutions.3 Consider the time series of form: yt Subtract yt 1 yt 1 vt . Now consider the time series of form: yt Subtract yt Thus 1 2 yt from both sides of the equation: yt 1 yt yt yt 1 to achieve vt 2 yt vt . since its first difference yt yt 1 yt 1 yt 1 yt 2 . because it had to be differenced twice to achieve stationarity. The data appears to be fluctuating and it may be stationary.86.Chapter 12.4(a) Plot of time series for oil (b) Since the data appears to fluctuating around a constant term.625) Since the tau ( 3. and we infer that OIL is stationary.942 ( 3.269OILt 1 0. Principles of Econometrics. the null of . it is integrated of order 0.4 (a) A plot of the data is shown below. 4e 429 EXERCISE 12. OILt (tau ) 0. (c) Since OIL is stationary.625) is less than the 5% critical value of nonstationarity is rejected. Exercise Solutions. we use the Dickey Fuller test which includes a constant term. 2. Figure xr12. the null of nonstationarity is not rejected.Chapter 12.035BONDt 1 27.459 BONDt 1 ( 1.5(a) Plot of time series for bond yields (b) Since the data appears to fluctuating around a trend. BONDt (tau ) 0.5(a) Plot of time series for bond yields . (c) The first difference of the series DBOND BOND BOND( 1) is shown below. we use the Dickey Fuller test which includes a constant term and a trend. 4e 430 EXERCISE 12.835) An augmented Dickey-Fuller test with one lagged term BONDt 1 was needed to ensure that the residuals were not autocorrelated.015t 0.5 (a) A plot of the data is shown below. Figure xr12.866 0. Exercise Solutions. Figure xr12. Since the tau ( 835 is greater than the 5% critical value of 3 . The data appears to be trending and hence may be nonstationary. Principles of Econometrics. and we infer that BOND is not stationary. we use the Dickey Fuller test which includes a constant term. we conclude that BOND is integrated of order 1.955) Since the tau ( 5.532 DBONDt 1 0.Chapter 12. DBONDt (tau ) 0. the null of Since BOND has to be differenced once to achieve stationarity. (d) 2 86. Exercise Solutions. Principles of Econometrics.955 is less than the 5% critical value of nonstationarity is rejected.871 ( 5.5(c) (continued) Since the data appears to fluctuating around a constant. 4e 431 Exercise 12. and we infer that DBOND is stationary. . CONSUMPTION t (tau ) 0.6 (a) Plots of the data CONSUMPTION and INCOME are shown below. Figure xr12.41.894) is greater than the 5% critical value of 3.6(b) Plots of differenced series . Since INCOME appears to trending. Principles of Econometrics. Exercise Solutions. 4e 432 EXERCISE 12. Figure xr12.188t (1.024CONSUMPTION t 1 67.550) is greater than the 5% critical value of 3. and we infer that INCOME is not stationary.248t ( 0. we the Dickey Fuller test which includes a constant term and a trend. the null of nonstationarity is not rejected.Chapter 12.176 16.6(a) Plots of time series for CONSUMPTION and INCOME Since CONSUMPTION appears to be trending.300 54.040 INCOMEt 1 2378. (b) To determine the “order of integration” we need to test the first differences.894) Since the tau ( 0. A plot of the differences in CONSUMPTION (DC) and in INCOME (DI) is shown below. the null of nonstationarity is not rejected. and we infer that CONSUMPTION is not stationary. INCOME t (tau ) 0.550) Since the tau (1. we use the Dickey Fuller test which includes a constant term and a trend.41. 228) In this case. Figure xr12. since the tau value ( 6.617 DCt 1 0. and we infer that DC is not stationary. Its graph appears below. is also known as the second difference of CONSUMPTION. Principles of Econometrics. and. the null of nonstationarity is rejected. The order of integration is determined from the number of times a series has to be differenced to render it stationary. In this case the test results are sensitive to the number of augmentation terms that are included. DDC DC DC ( 1) .6(b) Plot of second difference of CONSUMPTION . DCt that the residuals are not autocorrelated leads to a different result.579) In this case.6(b) (continued) Since DC appears to fluctuating around a constant. we have to difference the series again to check for stationarity.Chapter 12. Using an augmented Dickey-Fuller test with two lagged terms DCt 1 . DC t (tau ) 0. it follows that CONSUMPTION is integrated of order 1.714 DCt 1 860. With no augmentation terms the estimated equation is DC t (tau ) 0.295DCt 1 359. since the tau ( 2.74 ( 6.944 0. The difference of a difference. we use the Dickey-Fuller test which includes a constant term. To find the order of integration. We present results for the case with no augmentation terms and the case with 2 augmentation terms. 4e 433 Exercise 12. the null of nonstationarity is not rejected.579 is less than the 5% critical value of 2 86. in turn. the number of augmentation terms included will depend on the selection criterion used by your software. Since we concluded that the first difference of CONSUMPTION is stationary. Exercise Solutions.428 DCt 2 to ensure 2 ( 2.228 is greater than the 5% critical value of 2 86. and we infer that DC is stationary. 94.316eˆt 1 (tau ) ( 3. if we have concluded that CONSUMPTION and INCOME are both I(1).339 DDCt 1 0. Turning now to INCOME. 2 86. ˆ DI t 1.252 (tau ) ( 10.661 is less than the 5% critical value of 1. then any estimated relationship between them will be spurious because they are not of the same order of integration.524 DDCt 1 ( 13. In this case. 4e 434 Exercise 12. that CONSUMPTION is integrated of order 2. in this case we conclude that the variables CONSUMPTION and INCOME are cointegrated. Principles of Econometrics.9884 INCOME The estimated Dickey-Fuller test equation for the residuals from this regression is eˆt 0. since DI appears to fluctuating around a constant.37 . and so we reject the null hypothesis that the residuals are not stationary. DDC t (tau ) 2. The estimated equation is CONSUMPTION 9084 0. we use the Dickey Fuller test which includes a constant term.187 DI t 1 1464.909 is less than the 5% critical value of 3. the null of Since INCOME had to be differenced once to render it stationary. the null of nonstationarity is rejected.909) The tau value of 3.6(b) (continued) Since the second difference of CONSUMPTION (DDC) appears to fluctuating around zero. it follows that INCOME is integrated of order 1.676) Since the tau ( 10. it follows that since CONSUMPTION had to be differenced twice to be stationary. However. and we infer that DDC is stationary. Exercise Solutions.Chapter 12. Given the residuals are stationary. then we need to test the stationarity of the residuals from a regression of CONSUMPTION on INCOME to determine whether the variables are spuriously related or cointegrated.661) Since the tau ( 13.676 is less than the 5% critical value of nonstationarity is rejected. and we infer that DI is stationary. (c) If we conclude that CONSUMPTION is I(2) and INCOME is I(1). . we the Dickey Fuller test without a constant term. 41. DTX are shown below.471t 0. Principles of Econometrics. Figure xr12. For USNAG: since the tau ( 2.7(a) Plots of time series for TXNAG and USNAG Both series – TXNAG and USNAG – are trending upwards. the null of nonstationarity is not rejected. Exercise Solutions. TXNAG TXNAG ( 1) and DUS USNAG USNAG ( 1) Figure xr12. The Dickey-Fuller tests with a constant are: .798 USNAGt 1 ( 1.7(b) Plots of first differences DTS and DUS Both series – DTX and DUS – are fluctuating around a constant. the null of nonstationarity is not rejected.41. (b) Changes in the variables. and we infer that USNAG is not stationary.069USNAGt 1 ( 2.792) is less than the 5% critical value of 3. TXNAG t (tau ) USNAG t (tau ) 0.213) 0.792) For TXNAG: since the tau ( .763 TXNAGt 1 5313.024TXNAGt 1 123.213) is less than the 5% critical value of 3.Chapter 12.960 0.804t 0.7 (a) The series TXNAG and USNAG are graphed below. The Dickey-Fuller tests with a constant and a trend are shown below. and we infer that TXNAG is not stationary.248 33. 4e 435 EXERCISE 12. 015eˆt 1 0. .258 (3.191) 0. Principles of Econometrics.1097 0.549) DUS t (tau ) 0.Chapter 12.10. at the 5% and 10% levels of significance. we conclude that the change variables DTX and DUS are stationary. using a level of significance of 11%.587) 1 120.37.226 DTX t 1 8. the null of no cointegration is not rejected. Exercise Solutions. for illustrative purposes. 4e 436 Exercise 12.1017 Because the p-values are greater than 0. This is an example where it would be prudent to gather more information so that a more decisive inference about the property of the data can be made.780 eˆt 1 (tau ) ( 0.7(b) (continued) DTX t (tau ) ( 2.547 p -value 0.096USNAGt 2859. eˆt TXNAGt (t ) eˆt 0.230 DUSt ( 2. (d) The regression of DTX on DUS is as follows DTX (t ) 0. However.739 (19. that TXNAG and USNAG are I(1) variables. (e) In (c) we are testing the relationship between nonstationary variables with a view to establishing their long run relationship. we do not reject the null hypothesis of nonstationarity. The variables TXNAG and USNAG are spuriously related.117 p -value 0. In (d) we are testing the relationship between stationary variables with a view to establishing their short run relationship.412) This result shows that the change in TXNAG is significantly related to the change in USNAG.811) is greater than the 5% critical value of 3.811) Since the tau ( 0. we can check whether they are cointegrated or spuriously related by testing the property of the regression residuals.036 DUS 23. (c) Assuming. 104 0. was needed to ensure that the residuals were not autocorrelated.722 INFt 4 0. Exercise Solutions. Figure xr12.797 2. The variable INF is not stationary.8(a) Plots of time series for GDP and INF Since GDP is trending.Chapter 12. we apply the Dickey-Fuller test with a constant and a trend: GDP t (tau ) 0.277 INFt 7 0. The variable GDP is not stationary.961) is greater than the 5% critical value of nonstationarity is not rejected.961) Since the tau ( . 3.194 INFt 2 0.350) 0. .553 INFt 3 ( 1. Principles of Econometrics.026 INFt 1 0. the null of Since INF is wandering around a constant. INFt 1 to INFt 8 .real gross domestic product (GDP) and the inflation rate (INF) are shown below.024GDPt 1 105.406 INFt 5 0. Since the tau ( 1.41.100 INFt 6 0.400 INFt 8 An augmented Dickey Fuller test with 8 lagged terms.608 INFt 1 0.8 (a) The data series . we apply the Dickey-Fuller test with a constant.552 GDPt 1 ( 1.866t 0. INF t (tau ) 0. 4e 437 EXERCISE 12.350) is greater than the 5% critical value of 2.86. the null of nonstationarity is not rejected. 8(b) Plots of first differences for GDP and INF Since DG is fluctuating around a constant.126 DPt 6 0.869 ( 5. It follows that since INF has to be differenced once to be stationary. 4e 438 Exercise 12. .429 DGt 1 43. we apply the Dickey-Fuller test with a constant.631DPt 1 0. Exercise Solutions. Graphs of the first differences of GDP (DG) and INF (DP) are shown below.627) 0.41.186 DPt 4 0.41.228) Since the tau ( 5.Chapter 12.016 DPt 1 0. we need to examine the time-series property of the differenced series. It follows that since GDP has to be differenced once to be stationary. that INF is integrated of order 1.210 DPt 5 0.412 DPt 7 Since the tau ( . Figure xr12.524 DPt 2 3 ( 4. that GDP is integrated of order 1. Since the first difference of INF (DP) is fluctuating around the zero line. The variable DG is stationary.228) is less than the 5% critical value of 3.627) is less than the 5% critical value of 3.196 DPt 0. DG t (tau ) 0. The variable DP is stationary. Principles of Econometrics. we apply the Dickey-Fuller test without an intercept: DP t (tau ) 0. the null of nonstationarity is rejected.8 (continued) (b) To determine the order of integration of these series. the null of nonstationarity is rejected. 44 0.27 0.14 0.181 INFt 5 0.563 INF104 0.414 INFt 8 The forecast for INF for 2010:1 is: F INF105 INF104 0.023 0. re-estimated assuming the coefficient of GDPt 1 is zero.395 INFt 1 0.712 INFt 3 7 4 0.04 0. we can use the fact that INF is a random walk process.414 INF97 2.601 23.414 0.712 INF101 0.3 30. Re-estimating the model from part (a). Principles of Econometrics.023 0.59 0.44 Similarly.086 INFt 0. we obtain the following estimated forecasting model: GDP t 30. Exercise Solutions. as a forecasting model.181 INF103 0.395 INF100 0.023 0.225 105 0.601 0.505 INF102 0.086 0.225 t 0.8 (continued) (c) We can use the fact that GDP is a random walk process.6 14421.563 INFt 0.601 0.505 INFt 2 6 0.712 0. we obtain the following forecasting model: INF t 0.553 GDPt 1 Given t = 104 is 2009:4.11 0.Chapter 12.40 .284 INF98 0. as a forecasting model. Using the unit root test equation from part (a).086 INF99 F INF105 0.284 0.23 0.625 0.505 0.553 GDP104 14277. the forecast for GDP for 2010:1 is F GDP105 GDP104 30.563 0.553 162.01 0.395 3.181 0.284 INFt 0. 4e 439 Exercise 12.01 0. 045CANADAt 1 0. Figure xr12. Exercise Solutions. the series appears to be fluctuating around a constant.392) Since the tau ( 2. the null of nonstationarity is not rejected.392) is greater than the 5% critical value of 3.Chapter 12.9(a) Plot of the time series CANADA Over the sample period 1971:01-1987:12. despite the appearance of different “trendlike” behavior.41. . Over the sample period 1988:01-2006:12.244 CANADAt 1 ( 0. The variable CANADA is not stationary.897) is greater than the 5% critical value of 2. the null of nonstationarity is not rejected. The Dickey-Fuller test with a constant and a trend is shown below: CANADAt (tau ) 0.186 CANADAt 1 ( 2.007CANADAt 1 0. the series appears to be fluctuating around a trend.9 (a) A plot of the data CANADA is shown below. (b) The results for the two sample periods are consistent. The Dickey-Fuller test with a constant is shown below: CANADAt (tau ) 0.009 0.042 0. 4e 440 EXERCISE 12.897) Since the tau ( 0.86. Principles of Econometrics.000103t 0. The variable CANADA is not stationary. 94.Chapter 12.86. 4e 441 Exercise 12. is shown below. the null of It follows that since CANADA has to be differenced once to be stationary. .9 (continued) (c) For the whole sample period. Figure xr12. Principles of Econometrics.559) Since the tau ( 1. A plot of the first difference.226 CANADAt 1 ( 1.559) is greater than the 5% critical value of 2.007CANADAt 1 0. The variable CANADA is not stationary. we use a Dickey-Fuller test with a constant but without the trend term (since its effect is insignificant). DC CANADA CANADA( 1) .9(c) Plot of first difference for CANADA Since DC is fluctuating around the zero line.776 DCt 1 ( 16. The variable DC is stationary. CANADAt (tau ) 0. the null of nonstationarity is not rejected.461) Since the tau ( 6. that CANADA is integrated of order 1.461) is less than the 5% critical value of nonstationarity is rejected. we apply the Dickey-Fuller test without an intercept: DC t (tau ) 0. Exercise Solutions. .008 0. The variable CSI is stationary. .483) Since the tau ( 3.001) Since the tau ( 3.10(b) Plot of time series CSI (c) Since the CSI is stationary. it suggests that the effect of news is temporary. 4e 442 EXERCISE 12. This assumption is not correct (see graph in part (b)). hence consumers “remember” and “retain” news information for only a short time. Principles of Econometrics.483) is less than the 5% critical value of nonstationarity is rejected. the null of Dickey Fuller Test 3 (constant term and trend term) CSI t (tau ) 0.10 (a) Results from the three Dickey-Fuller tests are: (1) Dickey Fuller Test 1 (no constant term and no trend term) CSI t (tau ) 0.500 ( 3.309 0.068CSI t 1 5. Figure xr12. This is because Test 1 assumes that when the alternative hypothesis of stationarity is true. (2) Dickey Fuller Test 2 (constant term but no trend term) CSI t (tau ) 0. The variable CSI is stationary. Exercise Solutions.004t ( 3.001) is less than the 5% critical value of nonstationarity is rejected. (b) The graph suggests that we should use the Dickey-Fuller test with a constant term.86.299) is greater than the 5% critical value of 1. the series has a zero mean. the null of The result of the Dickey-Fuller test without an intercept term is not consistent with the other two results. The variable CSI is not stationary.41. the null of nonstationarity is not rejected. (3) 2.Chapter 12. 3.051CSI t 1 4.299) Since the tau ( 0.94.001CSI t 1 ( 0. 37. the null of no cointegration is not rejected. Exercise Solutions. .078) Since the tau ( 2. Variables MEXICO and USA are spuriously related.229) eˆ 0.078) is greater than the 5% critical value of 3. (3) Cointegration Test 3 (regression model has an intercept term and a trend) eˆ MEXICO 1.396) Since the tau ( 2. the null of no cointegration is not rejected. but no trend) eˆ MEXICO 0.11(a) Plots of MEXICO and USA The 3 tests for cointegration are: (1) Cointegration Test 1 (regression model has no intercept and no trend) eˆ MEXICO 0.Chapter 12.143) eˆt 0. Variables MEXICO and USA are spuriously related.11 (a) The data for MEXICO and USA are plotted below.948) is greater than the 5% critical value of 2. (2) Cointegration Test 2 (regression model has an intercept term.852USA 12.088eˆt 1 (tau ) ( 2.76.396) is greater than the 5% critical value of 3.062eˆt 1 (tau ) ( 1.948) Since the tau ( 1.283USA 8. Figure xr12.166 0.751) eˆt 0.268t (t ) (9.42. Variables MEXICO and USA are spuriously related.135 (t ) (50.107eˆt 1 (tau ) ( 2. 4e 443 EXERCISE 12. Principles of Econometrics. the null of no cointegration is not rejected.995USA (t ) ( 195. however. Alternatively. including the one without a constant and a trend.11 (continued) (b) Since none of the tests supported the existence of cointegration. Principles of Econometrics. If USA is exogenously determined. A constant term is unnecessary in this example because the two series have been standardised to the same base value.Chapter 12. 4e 444 Exercise 12. then one can estimate a dynamic model for MEXICO. not their growth rates. the relationship between MEXICO and USA can be examined by working with the stationary form of the variables which in this case is their first differences. using the econometric techniques discussed in Chapter 9. if USA is endogenously affected by MEXICO. Note. that the cointegration tests examine the relationship between the levels of the series. using the econometric techniques discussed in Chapter 13. The introduction of a trend in Cointegration Test 3 allows MEXICO to ‘diverge’ from USA. Exercise Solutions. the results do not support the theory of convergence in economic growth. . Cointegration Test 1 is the most straightforward test of the co-movement of MEXICO and USA. one can estimate a VAR model. (c) If the variables are not cointegrated. Yt has to be differenced twice to achieve stationarity.001 0.dat is shown below. Since the tau ( .088) is greater than the 5% critical value of .371) ( Yt ) (tau ) ( 2Yt ) (tau ) 0. thus Yt is integrated of order 2. Since the tau ( .088) 0. Since D 2Yt is fluctuating around zero.Chapter 12. Since the tau ( . The variable Yt is not stationary. the ADF test includes a constant and a trend. Exercise Solutions. Since DYt is fluctuating around a constant.94.011 Yt 0. the null of nonstationarity is rejected. and the second difference D2Y 2 Yt Yt Yt 1 of the data.940) Since Yt clearly has a trend. Figure xr12. .41. the ADF test includes a constant. the null of nonstationarity is not rejected.00006t 0.86.001Yt 1 0. The variable DYt is not stationary.940) is less than the 5% critical value of .991 Yt 1 (tau ) ( 3. the ADF test does not include a constant.001 1 ( 1. The variable D 2Yt is stationary. 4e 445 EXERCISE 12. the first difference DY Yt Yt Yt 1 .12 A plot of the data in inter2.12 Plots of Y and its first and second differences The ADF unit root tests are shown below. In other words.371) is greater than the 5% critical value of 3. Principles of Econometrics. Yt 0. The graph shows the level Yt .987 2Yt 1 ( 16. the null of nonstationarity is not rejected. 13 Plots of UK and Euro price indices The data are clearly not stationary and so we use the ADF test which includes a constant and a trend. The variable EURO is not stationary. the null of nonstationarity is not rejected.86. Exercise Solutions. the null of nonstationarity is not rejected.996 which is greater than the 5% critical value of 3. the null of nonstationarity is rejected.41. we perform an ADF test on the difference DUK = UK – UK( 1). the null of nonstationarity is rejected. We conclude that the differenced variable DUK is stationary. . Thus.86. Thus. EURO is I(1). For UK: An ADF test with a constant and a trend and 5 augmentation terms gives a tau value of 0. The variable UK is not stationary.13 (a) A plot of the price indices in the United Kingdom and in the Euro Area is shown below.41. For EURO: An ADF test with a constant and a trend and 6 augmentation terms gives a tau value of 2. Thus. Thus. An ADF test on DUK with a constant term and no augmentation terms gives a tau value of 13.49 which is less than the 5% critical value of 2. UK is I(1).56 which is less than the 5% critical value of 2. Figure xr12. Principles of Econometrics. Thus.916 which is greater than the 5% critical value of 3. 4e 446 EXERCISE 12.Chapter 12. we perform an ADF test on the difference DEURO = EURO – EURO( 1). An ADF test on DEURO with a constant term and no augmentation terms gives a tau value of 11. To assess whether EURO is I(1). Thus. We conclude that the differenced variable DEURO is stationary. To assess whether UK is I(1). 37.052) Estimating the error correction model using the residuals from the long-run equation.179.00643 UK t (0. . the residuals from the “long-run” equation have low t-values. the null hypothesis of no cointegration is not rejected.13 (continued) (b) The least squares equation relating UK and EURO is UK t 0.799 EUROt 20. and no augmentation terms.051 Testing the residuals from this equation for stationarity using an ADF test equation with no constant or trend.16) In both cases. This conclusion is supported by the results from an error correction model. Estimating the error correction model directly using nonlinear least squares.8367 EUROt (11. The variables UK and EURO are spuriously related.349) 1 102. we obtain UK t (t ) 0.8706 EUROt (13.Chapter 12. we obtain UK t (t ) 0.428) ( 0.017) 1 0.348) 1 0. Since 0. Exercise Solutions. Principles of Econometrics.7 0. implying they are not significantly different from zero. 4e 447 Exercise 12.00644eˆt (0. we obtain a tau value of 0.179 is greater than the 5% critical value of 3.0385EUROt ( 0. CHAPTER 13 Exercise Solutions 448 . 4e 449 EXERCISE 13. Exercise Solutions. t (b) & (d) 4. t 3. Principles of Econometrics. t 3. y 0 y2 11 1 y 12 1 x 11 y x2 21 1 y 22 1 x 21 y y3 11 2 y 12 2 x3 21 2 y 22 2 y4 11 3 y 12 3 x4 21 3 y 22 3 t 2.1 For the first-order VAR model below: (a) & (c) yt 11 t 1 y 12 t 1 x 1t xt 21 t 1 y 22 t 1 x 2t Effects of a shock to y of size t 1. y1 x1 t 2.Chapter 13. on y and x: y 11 0 y 21 12 21 ) 22 21 y y ) y ( 11 11 12 21 ) y 12 ( 11 11 12 21 ( 21 11 22 21 ) y ) y 22 ( 21 11 22 21 ) y ( 21 12 22 22 ) x ( 21 12 22 22 ) x on y and x: x 0 x1 x y2 11 1 x2 21 1 y3 11 2 x3 21 2 y4 11 3 x4 21 3 y 12 1 x 11 0 12 x y 22 1 x 21 0 22 x y 12 2 y 22 2 y 12 3 y 22 3 x ( 11 12 x ( 21 12 x 11 x 21 12 x 22 12 22 ) 22 22 x x ) x ( 11 12 12 22 ) x 12 ( 11 12 12 22 ) x 22 . t 4. y1 12 ( 11 11 x ( 21 11 x 11 x 21 0 22 x Effects of a shock to x of size t 1. 4e 450 EXERCISE 13. Principles of Econometrics. since Et [ 11 t ] 12 t y x .2 1–step ahead forecasts: ytF 1 Et [ 11 t y 12 t x 1t 1 xtF 1 Et [ 21 t y 22 t x 2t 1 ] y x . since Et [ 21 t 22 t 1t 1 ] 0 2t 1 ] 0 2–step ahead forecasts: ytF 2 Et [ 11 t 1 Et [ 11 = y y 12 t 12 t Et [ 21 t 1 y Et [ 21 ( 21 x 1t 1 12 t x) 12 t 22 ( ( y 21 t 22 t x 2t 1 ) 22 t x 2t 1 ) 2t 2 x 2t 1 ) 1t 2 1t 2 ] x 22 t ] 2t 2 y 12 21 t 22 t 1 11 t y ) y 12 x 11 t ] 1t 1 x 11 t ( 1t 2 x 11 t y 11 xtF 2 ( x 12 t 1 ) 22 y ( y 21 t ] x) 21 t 22 t 3–step ahead forecasts: ytF 3 Et [ 11 t 2 y Et [ 11 x 12 t 2 11 ( 11 t 21 ( 12 11 + xtF 3 12 21 21 t 2 Et [ 21 ( y ( 22 ( + x y 11 22 ( 21 ( 1t 1 x 11 t 12 t x 12 t 12 t x x t 3 y ( y ( y 11 t 22 1t 1 x 12 t 22 ( ) ( ) 1t 1 12 x) 12 t 21 t y 12 t x) 11 t 21 x y ( y 22 t y x 21 t 22 t 2t 1 ) 2t 2 1t 3 ] 2t 3 ] x 22 t x) ] y 21 t 22 t ] 12 t 11 t 12 21 t x) y 11 t 11 t ) 1t 1 12 22 t 2 11 ] 12 t 11 t Et [ 21 y y 11 1t 3 ( 12 ) 22 y ( y x 21 t ( 22 t y 21 t x )) 21 t 22 ( 22 t y 21 t x) ] 22 t x 22 t 2t 2 ) 2t 1 1t 2 ) ) 2t 2 .Chapter 13. Exercise Solutions. 2 (continued) 1-step ahead forecast errors and variances: FE1y yt 1 Et [ yt 1 ] 1t 1 . var( FE1x ) 2 x 2-step ahead forecast errors and variances: FE2y yt 2 Et [ yt 2 ] [ 11 1t 1 12 2 t 1 1t 2 ] var( FE2y ) 2 11 2 y 2 12 2 x 2 y FE2x xt 2 Et [ xt 2 ] [ 21 1t 1 22 2 t 1 2t 2 ] var( FE2x ) 2 21 2 y 2 22 2 x 2 x 12 2 t 1 1t 2 3-step ahead forecast errors and variances: FE3y yt Et [ yt 3 ] [ 3 11 var( FE3y ) FE3x xt 3 4 11 Et [ xt 3 ] [ 21 var( FE3x ) (a) 11 1t 1 2 2 21 11 2 x 2 11 12 2 t 1 2 y 2 2 21 12 2 x 2 y 2 2 12 21 2 y 2 2 12 22 1t 2 ) 22 2 21 2 y 2 2 22 21 21 1t 1 2 y 2 x 2 12 22 2 t 1 4 22 2 x 2 22 2t 2 2 x 1t 3 2 y 2t 2 2 x 2t 3 2 x + 4 11 ( 2 11 2 2 12 21 1) ( 4 11 2 y 2 2 11 12 2 x 2 11 2 y 2 2 12 21 2 y 2 2 12 22 2 x 2 12 2 x 2 y ) ( 2 2 11 12 2 2 12 22 2 12 ) ( 4 11 2 y 2 2 11 12 2 x 2 11 2 y 2 2 12 21 2 y 2 2 12 22 2 x 2 12 2 x 2 y ) 2 x 2 22 2 x + 2 x ) 2 x 2 22 2 x + 2 x ) The contribution of a shock to y on the 3-step forecast error variance of x is: 2 y (d) 2 2 11 12 22 2 t 1 The contribution of a shock to x on the 3-step forecast error variance of y is: 2 x (c) 2 y 21 1t 1 12 The contribution of a shock to y on the 3-step forecast error variance of y is: 2 y (b) ( 11 1t 1 ( 2 2 21 11 2 21 2 2 22 21 ) ( 2 2 21 11 2 y 2 2 21 12 2 x 2 21 2 y 2 2 22 21 2 y 4 22 The contribution of a shock to x on the 3-step forecast error variance of x is: 2 x ( 2 2 21 12 4 22 2 22 +1) ( 2 2 21 11 2 y 2 2 21 12 2 x 2 21 2 y 2 2 22 21 2 y 4 22 ] ] . Exercise Solutions. Principles of Econometrics. 4e 451 Exercise 13. var( FE1y ) 2 y FE1x xt 1 Et [ xt 1 ] 2t 1 .Chapter 13. 6( yt 1 1 1 0.1xt 1 1 To rewrite the VAR model in the VEC form.5) (1 0. first expand the terms: yˆt yt 1 2 0.7 1.3) (1 0.8 xt 1 ) 1 0. Now factorize out this coefficient to obtain the cointegrating equation: yˆt xˆt 0.3 2.5 xt 1 1 0. factorize out the cointegrating equation to obtain the errorcorrection coefficient.3 0.3( yt xˆt 0.5 xt xˆt (3 0.6.5 3. first rearrange terms so that the left hand side is in first-differenced form: yˆt yt 1 yt xˆt xt 1 xt 1 1 0.5) yt 1 3. 0.6 0.1xt 1 1 Simplifying gives the VAR model: (b) yˆt 2. The VEC model is: yˆt 0.1xt 1 1 Then rearrange in VAR form: yˆt (2 0.6 yt 1 1 0.5 0.8 xt 1 ) 0.6 yt 1 1 0. Exercise Solutions. Principles of Econometrics. that is 0.3) yt 1 2. 4e 452 EXERCISE 13.52 xt 1 1 Next recognize that the error correction term for the first equation is the coefficient in front of the lagged variable yt 1 .5 yt xˆt xt 1 3 0.5 yt 1 3.6 ( 1 0.3.3 yt 1 2.3( yt 0.3 (a) To rewrite the VEC in VAR form.5 xt xˆt 2.8 xt 1 ) .52) xt 1 1 For the second equation.3 yt 0.7 yt 0.Chapter 13.24 xt 0. 4 (a) Consider the following estimated VAR model. Principles of Econometrics. yt ˆ y 11 t 1 ˆ x 12 t 1 vˆ1t xt ˆ y 21 t 1 ˆ x 22 t 1 vˆ2t The forecasts for yt and xt are: 1 ytF 1 ˆ y 11 t ˆ x 12 t xtF 1 ˆ y 21 t ˆ x 22 t The forecasts for yt (b) 1 2 and xt are: 2 ytF 2 ˆ yF 11 t 1 ˆ xF 12 t 1 xtF 2 ˆ yF 21 t 1 ˆ xF 22 t 1 Consider the following estimated VEC model.Chapter 13. yt ˆ 11 ( yt 1 ˆ x ) vˆ 1 t 1 1t xt ˆ 21 ( yt 1 ˆ x ) vˆ 1 t 1 2t Rearrange terms as: ˆ 11 ˆ 1 xt 1 vˆ1t ( ˆ 21 ˆ 1 1) xt 1 vˆ2t yt ( ˆ 11 1) yt xt ˆ 21 yt The forecasts for yt 1 1 and xt 1 1 are: ytF 1 ( ˆ 11 1) yt xtF 1 ˆ 21 yt The forecasts for yt 2 and xt ˆ 11 ˆ 1 xt ( ˆ 21 ˆ 1 1) xt 2 are: ˆ 11 ˆ 1 xtF 1 ytF 2 ( ˆ 11 1) ytF 1 xtF 2 ˆ 21 ytF 1 ( ˆ 21 ˆ 1 1) xtF 1 453 . 4e EXERCISE 13. Exercise Solutions. 9908 . (b) The estimated relationship with a constant included is AUS (t ) 1.37.05) The 5% critical value is 3. Given conclude that cointegration exists. For AUS with no augmentation terms.139et 1 (tau ) ( 3.7) of the text.072 1. we obtain tau 0.Chapter 13. Principles of Econometrics.9) of POE4. real GDP of Australia and real GDP of the US are shown in Figure 13. we obtain tau corresponding p -value 0. The cointegration equation excluding the constant term is in equation (13.66) (164) The test for cointegration using the residuals from this equation is eˆt 0.5 (a) The data.400 with intercept and trend. there is insufficient evidence to One could argue that a negative intercept is not sensible because the real GDP for Australia will be positive even when the GDP for the US is zero. Both series are clearly nonstationary which is confirmed by the Dickey-Fuller test with an 0. 3. The test of stationarity in the residuals is in equation (13.1. (c) The estimated VEC model is reported in equation (13. Exercise Solutions.9866 .05 3. and vice versa. . For USA with one augmentation term.265 with corresponding p -value 0. 4e 454 EXERCISE 13. It leads to a reversal of the above test decision.37 .001USA ( 2.8). And.206760 0. This change was made because the output comes from EViews in which C is a reserved name. We test for this possibility in part (b).956790 2. indicating that C and Y would not be adequately described as trend stationary variables. Error t-Statistic Prob.41 1.041294 0.41.843330 0.000121 -1. we conclude that both C and Y are nonstationary series.0028 0.184933 0.Chapter 13.0033 0.185046 0. Augmented Dickey-Fuller Test Equation Dependent Variable: D(CN) Method: Least Squares Sample (adjusted): 1961Q1 2009Q4 Included observations: 196 after adjustments CN(-1) D(CN(-1)) D(CN(-2)) D(CN(-3)) C @TREND(1960Q1) Coefficient Std. are shown below.777962 1.0945 0.643680 2.113205 0.069927 0. Error t-Statistic Prob. in particular.014761 0. with a constant and a trend.680440 1. Principles of Econometrics.024805 0. These residuals are displayed in Figure xr13.104296 0.1520 Augmented Dickey-Fuller Test Equation Dependent Variable: D(Y) Method: Least Squares Sample (adjusted): 1960Q2 2009Q4 Included observations: 199 after adjustments Y(-1) C @TREND(1960Q1) More light is shed on this issue by examining the residuals from separate regressions of C and Y on a constant and a trend. they are not trend stationary.98 and 3.626078 3.68 .0089 0.0770 0. Note that C is labeled as CN in the output. The residual series appear to move in similar directions suggesting that cointegration may be a possibility.0035 0. They appear to be nonstationary. -0.977019 2. 4e 455 EXERCISE 13.070465 0.316165 0. The critical value for a 5% significance level is 3.000174 0. Because 3.0093 0.6 (a) Output for the Dickey-Fuller test equations for C and Y. the next step is to check whether they are cointegrated.41 2.000335 0.013871 0.201274 0.6(a). Given that C and Y are nonstationary. .438159 0.069953 0.000118 -2. -0.0050 Coefficient Std. Exercise Solutions.031421 2. 6(a) Trend lines and residuals from those trend lines fitted to C and Y (b) Results from the potentially cointegrating equation with a constant and no trend are: Dependent Variable: CN Method: Least Squares Sample: 1960Q1 2009Q4 Included observations: 200 Variable C Y Coefficient Std.132 351.40416 1. -0.06716 -2.458 0.305 0.03529 0.08765 -0.03051 0. The relationship between C and Y could be a spurious one. . Exercise Solutions.0000 The tau value (unit root t-value) of 2. indicating that the errors are not stationary and hence that we have no cointegration. Principles of Econometrics.37. we obtain the following output.0045 0.873 is greater than 3.00295 -16. 4e 456 Exercise 13.873 -4.6(a) (continued) Figure xr13.Chapter 13.29941 0.02505 0.0000 0.0000 Testing for a unit root in the residuals from this equation. Augmented Dickey-Fuller Test Equation Dependent Variable: D(EHAT) Method: Least Squares Sample (adjusted): 1960Q3 2009Q4 Included observations: 198 after adjustments Variable EHAT(-1) D(EHAT(-1)) Coefficient Std. Error t-Statistic Prob. -0. Error t-Statistic Prob. we include lags when the lag coefficients of one or more of the variables is significant. Also.0020 3.230 12. However.91299 0.0000 0.00020 0.03471 0. -0.22216 0.241 is greater than the 5% critical value of 3.0493 0. increasing the lags to 3 eliminates serial correlation in the errors of the equation for Ct .Chapter 13.14301 0.42 .106 0.11) on page 504 of POE4.07094 -3. 1. If we only include lags where the coefficients of both lagged variables are individually significant. In this case the estimated potential cointegrating relationship is Dependent Variable: CN Method: Least Squares Sample: 1960Q1 2009Q4 Included observations: 200 Variable C @TREND Y Coefficient Std. . then a lag order of 1 is suitable.11248 -0.07229 0. 4e 457 Exercise 13.454 30. however.132 0.73322 0. (c) The results from estimating a VAR model with lags of order 1 for the pair of I(0) variables { Ct .978 3. Error t-Statistic Prob. at a 10% level of significance there is evidence of cointegration. or a joint test of both coefficients at a given lag yields a significant result.0014 0. Principles of Econometrics. and the coefficient of the trend in the unit-root test equation for C was significant at a 5% level of significance.0000 0. If. Error t-Statistic Prob.18700 0. The results from estimating a VAR(3) are as follows.0000 The results from the unit-root test on the residuals from this equation are: Augmented Dickey-Fuller Test Equation Dependent Variable: D(EHAT_T) Method: Least Squares Sample (adjusted): 1960Q4 2009Q4 Included observations: 197 after adjustments Variable EHAT_T(-1) D(EHAT_T(-1)) D(EHAT_T(-2)) Coefficient Std.241 -1. it is worth checking for a cointegrating relationship that includes a trend term. Exercise Solutions. suggesting the The value tau residuals are nonstationary and that C and Y are not cointegrated.02435 10.00248 0. Yt } are provided in equation (13. then a VAR with lags of order 3 is suitable.6(b) (continued) Since both C and Y are trending. We now ask whether the model can be improved upon by adding more lags. 541] D(Y(-2)) 0.08161) [2.07988) [1.037] 0.6(c) (continued) Vector Autoregression Estimates Sample (adjusted): 1961Q1 2009Q4 Included observations: 196 after adjustments Standard errors in ( ) & t-statistics in [ ] D(CN) D(Y) D(CN(-1)) 0.410] 0.12820 (0. The following table contains the results from joint Wald tests of both coefficients at each lag in a VAR of lag order 3.910] D(CN(-2)) 0. 4e 458 Exercise 13.183] 0.20484 (0.01935 (0. and both equations jointly.10990) [ 0.16620 (0.22013 (0. Lags of Y beyond 1 are not significant in either equation. Results are provided for each equation separately. the joint test is for whether the two coefficients at a given lag are zero.42060 (0.07908) [2. When testing within each equation separately.197] C Lags for C of orders 2 and 3 are significant at a 5% level in the consumption equation.141] 0.170] D(CN(-3)) 0.01868 (0. Principles of Econometrics.247] D(Y(-3)) 0.06300) [ 0.00523 0.01834 (0.067] D(Y(-1)) 0.08145) [ 0.08062) [ 2.10650) [2. When testing the two equations jointly.13063 (0.17263 (0.635] 0.06048) [0. we are testing whether the four coefficients at a given lag are all zero.00342 0.08484) [ 0.03343 (0.307] 0.00124 [4.05987) [2.00092 [3.10757) [3.02100 (0.Chapter 13.303] 0. and lags of C of orders 1 and 3 are significant at a 5% level in the income equation. Exercise Solutions.699] 0. . 651329 [0. although coefficients at both lags 1 and 3 are significant at a 10% level. 4e 459 Exercise 13. Principles of Econometrics.1608] 8.16349 [0.0003] 0. Exercise Solutions.03622 [0.0813] 4 .579234 [0.0343] 2 16. Adding lags of order 4 did not lead to any significant coefficients.Chapter 13.0977] 6. The joint test involving two equations uses estimation within a seemingly unrelated regression (SUR) framework that is discussed in Chapter 15.740683 [0. In the joint test for both equations. only lag 1 coefficients are significant at the 5 % level.9215] 4. The SUR framework is needed to get the covariances between coefficient estimates from different equations.1013] 2 31.0015] 4. VAR Lag Exclusion Wald Tests Sample: 1960Q1 2009Q4 Included observations: 196 Chi-squared test statistics for lag exclusion: Numbers in [ ] are p-values Lag 1 Lag 2 Lag 3 df D(CN) D(Y) Joint 13. but testing hypotheses involving coefficients in different equations requires the SUR framework.563613 [0. The separate equation joint tests suggest that the estimates for coefficients at lags 1 and 3 are significant at a 5% level for the C equation.05791 [0.0000] 6. Least squares and SUR estimates of VAR equations are the same because the same explanatory variables appear in each equation. but only those at lag 1 are significant in the Y equation.97885 [0.6(c) (continued) The 2 Wald test for a single equation is that described in Appendix 6A of POE4.29717 [0. 550) (a) The correlogram (up to order 4) for the residuals is shown in the first column of the diagram below.7 The cointegrating equation between x and y (normalised on y) is : yˆ t 0. None of the autocorrelations exceed the significance bounds. Also. the column labeled ‘Prob’ shows that the probability values are all greater than 5% and hence that there is no evidence of autocorrelation up to order 4.495 xt (t ) (37.Chapter 13.576) indicates that y falls. Principles of Econometrics.450) indicates that x rises. This behavior (negative change in y and positive change in x ) is necessary to “correct” the cointegrating error. . when there is a positive cointegrating error: ( rest 1 0 when yt 1 0. 4e 460 EXERCISE 13. (b) The negative error correction coefficient in the first equation ( 0. while the positive error correction coefficient in the second equation (0.495 xt 1 ). Exercise Solutions. Exercise Solutions.Chapter 13.641 z z . (b) Expressions for the impulse responses were derived in Exercise 13. Since there are no autocorrelations that exceed the significance bounds and the p-values (under ‘Prob’) are all greater than 5%. 4e 461 EXERCISE 13. Principles of Econometrics. we can infer that there is no evidence of significant autocorrelation up to order 4. Since there are no autocorrelations that exceed the significance bounds and the p-values (under ‘Prob’) are all greater than 5%. The correlogram of the residuals from the z equation are shown below. w1 z1 t 2.743 w 0.1. w1 z1 t 2.8 (a) The correlogram of the residuals from the w equation is shown below. w of size w on w and z : on w and z w 0 0.155 w z of size w 0 z w2 0. Effects of a shock to t 1. we can infer that there is no evidence of significant autocorrelation up to order 4.214 z2 0. w2 z2 Effects of a shock to t 1. 2 z 2-step ahead forecast errors and variances: FE2 w wt Et [ wt 2 ] [ 2 var( FE2 w ) 0. 1-step ahead forecast errors and variances: FE1 w wt FE1 z zt w t 1 Et [ wt 1 ] 1 z t 1 Et [ zt 1 ] 1 var( FE1 w ) . Exercise Solutions.7432 The contribution of a shock to 2 w on the 2-step forecast error variance of w is: 2 0.6412 w z 12 t 1 z 22 t 1 2 z t 2 ] 2 z z w on the 1-step forecast error variance of w is: 2 w w The contribution of a shock to z on the 1-step forecast error variance of w is: 2 0 w The contribution of a shock to w on the 1-step forecast error variance of z is: 2 0 z The contribution of a shock to 2 z on the 1-step forecast error variance of z is: 2 z z The contribution of a shock to 2 w (0.Chapter 13. Principles of Econometrics.1552 ) ( 0.2142 w 2 0.2142 w 2 2 z w ) 2 2 z w ) w on the 2-step forecast error variance of z is: ( 0.1552 The contribution of a shock to 2 z on the 2-step forecast error variance of w is: (0.7432 1) (0.7432 FE2 z zt var( FE2 z ) 0.6412 w 2 2 z z ) z on the 2-step forecast error variance of z is: 1) ( 0. 4e Exercise 13. 2 w var( FE1 z ) .2.6412 2 2 z z ) 462 .8 (continued) (c) Expressions for the variance decompositions were derived in Exercise 13.1552 The contribution of a shock to 2 2 w 11 t 1 2 2 w t 2 ] 2 z w w 21 t 1 0.641 0.2142 ) (0.2142 w Et [ zt 2 ] [ 2 0.1552 2 w 0.7432 The contribution of a shock to 2 z 2 w 2 z (0. 017) 1 (1.514 Pt (2. 4e 463 EXERCISE 13. (b) The error correction coefficients are 0.005 M t 1 (0.004 M t of 1.663) is less than the 5% critical value of 3.418 rest 0.004M t The unit root test confirms that the residuals are stationary: rest 0.127) (7.796) 0.215) 1 0.039 .37. This means that both variables will “error-correct” to achieve equilibrium.802) 1 .340 M t (4.067(rest 1 ) 0.Chapter 13. 1 (tau ) ( 3. The coefficient 0. Exercise Solutions.9 (a) The cointegrating relationship between P and M is Pt 1. (c) The cointegrating residuals are obtained as: rest Pt 1. The residual series is an I(0) variable.086rest 1 0.336 Pt (3.016 and 0. They are both significant and of the right signs.663) Since tau ( 3. the null hypothesis of no cointegration is rejected.039 .067.004 is consistent with the quantity theory of money. The system is stable. (d) The VEC model estimated using the cointegrating residuals is: Pˆt (t ) Mˆ t (t ) 0.999) 0.016(rest 1 ) 0. Principles of Econometrics. (d) The response of DU at time t 2 is DU t (e) 2 0. Principles of Econometrics.373 0.373DPt 1 0. (b) The response of DU at time t 1 following a unit shock to DU at time t is 0.098 .098 The response of DP at time t 2 is DPt 2 0.098 0.098) suggest an inverse relationship between a change in the unemployment rate (DU) and a change in the inflation rate (DP).098 0.046 and 0.180.180 0. (c) The response of DP at time t 1 following a unit shock to DU at time t is 0. that DU increases but DP falls.054 These results suggest. following a shock to unemployment.180 0. .180 0.Chapter 13.10 (a) The coefficients ( 0. Exercise Solutions. 4e 464 EXERCISE 13.037 0.046 DPt 1 0.046 0.098DU t 1 0.180 DU t 1 0. 37.438) ( 2. The estimated VEC model is below. the null hypothesis of no cointegration is rejected and we infer that STERLING and EURO are cointegrated.209 0.209 EUROt 1 (2. (t ) R2 0.637) ( 2.290) Note that the error correction term for the second equation is not significant. in the event of a disequilibrium between EURO and STERLING.973) The unit root test of the regression residuals (res) is: res t 0. Principles of Econometrics.518) Since the tau ( 3. not EURO. for estimating a VEC model.977) 1 0.250(rest 1 ) 0.347 EUROt 1 (2. Hence. we should use the data in the levels (EURO and STERLING) and their changes. This suggests that.429 EURO. A VAR model is concerned with the relationship between stationary variables.939 (37. that STERLING adjusts to restore equilibrium. 4e 465 EXERCISE 13. once we establish that the variables are I(1) and not cointegrated.Chapter 13. or changes if the variables are I(1) and not cointegrated. In Figure 13.817) 0. Those stationary variables could be levels if the variables are I(0). once we establish that they are indeed nonstationary and cointegrated.090(rest 1 ) 0.201) 1 0. . Exercise Solutions. (b) The least squares regression between EURO and STERLING is: STERLING 0.518) is less than the critical value of .633 STERLINGt ( 0.375 STERLINGt ( 2.236rest 1 (tau ) ( 3.11 (a) A VEC model is concerned with the short-run relationship between changes in nonstationary variables and departures from the long-run cointegrating relationship between the levels of those variables. STERLING t (t ) EURO t (t ) 0.7 the variables appear to be I(1) and so we would use the changes in the data. The intercept terms were not significant and hence not included. Principles of Econometrics.278) 0.484 EUROt 1 ( 3.707) 1 0. 4e 466 Exercise 13. Exercise Solutions.700) 1 0.672 EUROt 1 ( 2. STERLING t (t ) EURO t (t ) 0. This is confirmed by the correlograms of residuals.Chapter 13.467) The order of the lag is 1 as all the second order terms were not significant.373 STERLINGt (2. Residuals from STERLING equation: Residuals from EURO equation: .11 (continued) (c) The least squares regression of a VAR model between the change in EURO and the change in STERLING is shown below.283 STERLINGt (4. 627032 0.93709] 2.463655 -2.434256 (1.07637] DV(-1) -0.08770) [ 4.0000 Dependent Variable: DV Method: Least Squares Sample (adjusted): 1891 1979 Included observations: 89 after adjustments C DV(-1) SP(-1) SP .Chapter 13.363245 1.77289) [ 1.15562) [-1.129639 1.190763 0.092796 0.605104 (1.300147 (0.11261) [-0.288887 0.28289) [ 2.357491 (0.030508 0.067881 1.693724 1.133686 5.0023 0.301399 (0. Vector Autoregression Estimates Sample (adjusted): 1891 1979 Included observations: 89 after adjustments Standard errors in ( ) & t-statistics in [ ] SP DV SP(-1) 0. Error t-Statistic Prob. 1.3057 0.140131 0.351182 0.2371 0.135393 0.03066] Dependent Variable: SP Method: Least Squares Sample (adjusted): 1891 1979 Included observations: 89 after adjustments C SP(-1) DV(-1) DV Coefficient Std. 1.12119) [ 2.48689] 0.0358 0. Exercise Solutions.078989 0.6441 0.14414] C 3.248009 0.053399 -0.578864 0.351182 0.3563 0.139810 5.016231 (0. 4e 467 EXERCISE 13.0000 Coefficient Std.927432 3. Principles of Econometrics.357627 0.115169 0.12 The results for a first-order VAR and the ARDL equations are as follows. Error t-Statistic Prob.92877] -0.100057 0. Note. There are multiple values of the ij that will lead to the same reduced form estimates. The pair of ARDL equations represents two simultaneous equations with endogenous variables SPt and DVt .605 21 ˆ 21 ˆ 23 ˆ 11 1 ˆ 13 ˆ 23 0. we have ˆ ˆ ˆ 10 ˆ 10 ˆ 13 ˆ 20 1 ˆ 13 ˆ 23 3. The differences should not be surprising since the coefficients in the VAR and ARDL models have are quite different interpretations. In this model. These concepts were discussed in Chapter 11. Thus. In the language of Chapter 11. we find the coefficients of corresponding variables in the VAR and ARDL models are quite different. deriving estimates of the reduced form coefficients from the structural coefficients estimates. The VAR equations are the reduced form equations from the simultaneous system.3575 22 ˆ 22 ˆ 23 ˆ 12 1 ˆ 13 ˆ 23 0. the lack of identification means that the ARDL results should not be used to infer the contemporaneous role of dividends on share prices. (a) As long as vts and vtd are serially uncorrelated. with the exception of the coefficient of DVt 1 in the equations for SP. lagged values of SP and DV will be uncorrelated with vts and vtd . Exercise Solutions. the structural equations are unidentified. and least squares estimation of the VAR yields consistent estimates. Principles of Econometrics. however. that we are unable to derive structural estimates from the reduced form estimates. . There are only 6 reduced form coefficients and 8 structural coefficients.01623 These estimates are identical to those obtained by directly estimating the reduced form equations.Chapter 13. deriving the reduced form estimates from the structural leastsquares estimates yields the same results as least squares estimation of the reduced form.3014 ˆ 12 ˆ 12 ˆ 13 ˆ 22 1 ˆ 13 ˆ 23 0. To derive the reduced form coefficients from those in the structural ARDL system. The solution is SPt DVt 10 1 13 13 20 1 20 13 1 23 23 11 10 23 13 13 21 1 21 12 1 11 SPt 13 1 23 23 13 SPt 13 22 1 1 23 22 DVt ets 1 1 23 23 13 12 DVt 23 d 13 t e 13 23 s 23 t e 1 1 13 etd 23 Thus. we solve the two ARDL equations simultaneously for SPt and DVt .3001 ˆ 20 ˆ 20 ˆ 23 ˆ 10 1 ˆ 13 ˆ 23 2.12 (continued) Comparing the two sets of estimates. 4e 468 Exercise 13. although the contemporaneous variables (SP and DV) appear to be significant in the ARDL equations. It is important to include sufficient lags to eliminate serial correlation in the errors.434 ˆ 11 ˆ 11 ˆ 13 ˆ 21 1 ˆ 13 ˆ 23 0. This correlation leads least squares estimates of the ARDL equations to be inconsistent. . the VAR results show that the lagged rate of change in dividends has no significant influence on the rate of change in share prices. Principles of Econometrics. 4e 469 Exercise 13. (c) Using a 5% significance level.12 (continued) (b) In the derivation above we showed that vts ets 1 d 13 t e 13 23 and vtd s 23 t e 1 13 etd 23 Solving these two equations for ets and etd shows that ets and etd both depend on vts and vtd . Exercise Solutions. We also have the bigger problem of structural coefficients that are unidentified. but the lagged rate of change in share prices has a significant effect on the rate of change in dividends. Since SPt and DVt depend directly on vts and vtd through their reduced form equations. ets and etd will both be correlated with SPt and DVt .Chapter 13. 49389 0. 4e 470 EXERCISE 13. Principles of Econometrics. Exercise Solutions. (b) The long run model with LEURO as the left-hand-side variable is LEUROt 1 2 LUSAt et We wish to investigate whether et is an I(0) variable. The wandering nature of the residuals in the graph suggests they are nonstationary. t-Statistic Prob. and we infer that LUSA and LEURO are not cointegrated. -0.794 0.06534 -1.0000 0. Error t-Statistic Prob. Err. The results from the least squares residuals and an ADF test on the residuals follow.047538 68. Dependent Variable: LEURO Method: Least Squares Sample: 1995Q1 2009Q4 Included observations: 60 LUSA C Coefficient Std.010277 0.37.0781 Since the tau ( . 0.794) is greater than the critical value of .706170 1. their relationship could be spurious.13 (a) Growth in GDP of the two economies appears to move together.354549 0. the null hypothesis of no cointegration is not rejected.Chapter 13.11720 0.0000 Augmented Dickey-Fuller Test Equation Dependent Variable: D(EHAT) Method: Least Squares Sample (adjusted): 1995Q2 2009Q4 Included observations: 59 after adjustments Variable EHAT(-1) Coefficient Std.71315 28. . Vector Autoregression Estimates Sample (adjusted): 1995Q3 2009Q4 Included observations: 58 after adjustments Standard errors in ( ) & t-statistics in [ ] DLEURO DLUSA DLEURO(-1) 0.361594 (0.92280] 0.Chapter 13. Exercise Solutions. the model is LEUROt e1t 10 11 LUSAt 1 12 LEUROt 1 LUSAt 20 21 LUSAt 1 22 LEUROt 1 e2t The estimated first-order VAR and the residuals are shown below. The LEURO residuals are generally small relative to those for LUSA. we specify a VAR model in first differences.375194 (0.11914) [ 3.356554 (0.13 (continued) (c) Because we concluded that LEURO and LUSA are not cointegrated.000222 (0.15238) [ 1. Principles of Econometrics. but that does not appear to be the case for LEURO.12837) [ 2.03512] 0. Using lags of order 1. . with the exception of those at the end of the period where they are larger.18565] These results show that LEURO and LUSA affect each other via the lagged terms.003352 (0.00082) [ 0.17164] DLUSA(-1) 0.00105) [ 3. 4e 471 Exercise 13.71312] C 0.27045] 0. It seems reasonable to assume the LUSA residuals have constant variance. for a short-run relationship.16419) [ 2.261045 (0. CHAPTER 14 Exercise Solutions 472 . ht ) because zt ~ N (0.1) and hence zt ht ~ N (0. 4e 473 EXERCISE 14. ht ) since time t-1. 0 2 1 t 1 e ht ht is known at . Principles of Econometrics.Chapter 14.1 (a) The conditional mean E (et | I t 1 ) 0 because: Et 1 et Et 1 zt ht Et 1 zt Et where Et 1 ( ) is an alternative way of writing E ( | I t 1 ) 1 ht since zt is independent of ht 0 (b) since Et 1[ zt ] 0 The conditional variance E (et2 | I t 1 ) ht because: Et 1 et2 Et 1 Et 1 ht (c) 2 zt ht zt2 Et 1 ht since Et 1 zt2 1 and Et 1 ht et | I t 1 ~ N (0. Exercise Solutions. 474 . Principles of Econometrics.2 (a) If 0 .Chapter 14. Exercise Solutions. the conditional mean of yt Et yt Et 1 = 0 is: 1 since Et et 0 (b) et 1 1 1 0 is: 2 1 t e 0 2 1 t e et 1 since Et et 1 0 The extra information used to forecast returns is the “news” captured in et . 4e EXERCISE 14. the conditional mean of yt Et yt Et 1 0 If 0 . ht 1 (1)2 2 1 t 1 0 . 4e 475 EXERCISE 14.3 (a) (b) If If 2 1 t 1 0 . 0 and ( 1)2 1 1 0 cases lies with the contribution of the . Principles of Econometrics. ht e and 1. dt 1 dt 1et2 1 and e when et 1 1 .Chapter 14. ht 1 ( 1)2 when et 1 when et 1 0. ht 1 when et 1 0 . dt 1 1 ht 1 ( 1)2 0 ht 1 (0)2 0 ht 1 (1)2 1 1 The key difference between the asymmetric factor. ht 1 (0)2 when et 1 1. Exercise Solutions. dt when et 1 1 . ..Chapter 14. t 1 ) /(1 1 ) Thus the GARCH(1.1) may be re-written as an ARCH(q) where q is a large number (infinity).. ht / (1 1 ) 1 et2 1 2 1 t 2 e 2 2 1 t 3 e .. .1) model: ht Lag the expression: ht e h 1 t 1 1 2 1 t 2 1 t 2 ht 2 1 t 1 1 e h And substitute: e (1 e 2 1 t 1 ) 1 2 1 t 2 ( e h ) 1 t 2 2 1 1 t 2 2 1 t 2 2 1 1 t 2 2 1 e h Continue with the recursive substation: ht The last term drops out as progression: (1 (1 1 (1 1 2 1 t 1 ) e 2 1 e ) 1 et2 1 2 1 t 2 e 2 1 t 3 ( e 2 2 1 t 3 h ) 1 t 3 3 1 t 3 e h becomes negligible while the first term is the sum of a geometric 1 1 2 1 . Principles of Econometrics.. Exercise Solutions.4 2 1 t 1 GARCH(1.. 4e 476 EXERCISE 14. Exercise Solutions. . 5. (b) The correlogram of squared returns (up to order 12) is given below. 4e 477 EXERCISE 14. and the p-values are all less than 0.05.05. there is indication of significant lagged variance effects. Principles of Econometrics. there is no indication of significant lagged mean effects.5 (a) The correlogram of returns (up to order 12) is presented below. 6 and 8. There is evidence of significant autocorrelation since the autocorrelations exceed their significance bounds at lags 1. In other words. In other words. There is no evidence of autocorrelation since none of the autocorrelations exceed their significance bounds and the p-values are all greater than 0. 4.Chapter 14. 1998.431) is greater than the 5% critical value of 3. These periods coincide with the periods of big and small changes in returns noted in (a). (b) The distribution of returns is not normal since it is negatively skewed (skewness = 0. Exercise Solutions. 2000 and 2002. The Jarque-Bera statistic is a test of normality which jointly tests whether skewness is significantly different from zero and whether kurtosis is significantly different from 3. The statistic is distributed as a 2 distribution with 2 degrees of freedom.841 and hence it also suggests the presence of first order ARCH effects.287 is greater than the 5% critical value (5. 2000 and 2002) and periods of small changes (notably around 1989 and 1995). . (e) The plot of the conditional variance shows that volatility is high around 1990. The t-statistic on the ARCH effects of 2.51) and the kurtosis is greater than 3 (kurtosis = 4. Since the calculated value of 20.198 is significant.6 (a) The time series of returns shows that there were periods of big changes (around 1990. we reject the null hypothesis that the distribution is normal. (c) The t-statistic on the squared residuals term indicates the presence of first order ARCH. 1998. Principles of Econometrics.99).159). and it is especially low around 1989 and 1995.Chapter 14. 4e 478 EXERCISE 14. The Lagrange Multiplier test (11.879. This is the unconditional distribution. (d) The results show that the mean value of returns is 0. 4e 479 EXERCISE 14.227 0. (b) The results show that the average value of the change in the exchange rate s is 0. The significance of the coefficient of ht 1 (0.042. From the variance equation.Chapter 14. the significance of the coefficient of the lagged squared residual term (0.615 0.29 0. (c) The forecast for the exchange rate is 0.221 is significantly different from the critical value of 5.615 0. Furthermore.800) indicates the significance of lagged volatility effects. Exercise Solutions.149) indicates that lagged news/shocks affect volatility. Principles of Econometrics.7 (a) The unconditional distribution of the series is not normal.800(20.149e2010:06 = 0.248)2 = 23.61 5.898(20.61). The value of 192. since h2010:06 since e2010:06 20. The forecast for the conditional variance is F hˆ2010:07 2 0.61).042.042 . It has a kurtosis of 6.484 which is very different from the kurtosis of 3 for normality.149(5.99. the Jarque-Bera statistic which tests whether skewness is significantly different from zero and whether kurtosis is significantly different from 3 is very large. 0. 253( 12 ) 3.123) 3.268) 3.253( 12 ) 3.442 0.Chapter 14. 4e 480 EXERCISE 14.695 (b) Results for the T-ARCH model are given in the text.437 (0.442 0.8 (a) The value of the conditional variance when et hˆt 1 is: 3. (c) The value of the conditional variance when et hˆt (d) 1 is: 1 1 is: 3. it suggests that the asymmetric T-ARCH model is better than the symmetric ARCH model.828 Since the coefficient on the asymmetric term (0.560 The value of the conditional variance when et hˆt 1 3.437 (0. it suggests that volatility is greater when the shock is negative which is consistent with financial economic theory.695 The value of the conditional variance when et hˆt 1 1 1 is: 3.123 0. Principles of Econometrics. . Exercise Solutions.268) is significant. Since the coefficient on the asymmetric effect is positive. 4e 481 EXERCISE 14.Chapter 14. Exercise Solutions. . Principles of Econometrics.211 ht suggests that the GARCH-in- mean model is better than the GARCH model in a financial econometric sense. The contribution of volatility to the term premium is captured in the term 0. (b) The estimated GARCH-in-mean model is given in the text. (c) The significance of the GARCH-in-mean term 0. The positive sign suggest that returns increase when volatility rises which is consistent with financial economic theory.9 (a) The estimated GARCH model is given in the text.211 ht . It shows that volatility of returns changes over time.929) is also not significant at the 5% level.010 (1.394 0.10(b) Histogram for returns to gold shares (c) The regression of squared residuals on a constant and the lagged squared residuals is: eˆt2 (t ) 0.776 is greater than 3. Exercise Solutions.Chapter 14. Figure xr14.10(a) Plot of returns to gold shares (b) The histogram of returns is given below.00) and the kurtosis of 4. . 4e 482 EXERCISE 14. It is not significant when compared with the 5% critical value of 3.99.101eˆt2 1 . and hence we reject the null hypothesis that the distribution is normal. Note that the t-statistic (1. The Jarque Bera statistic (59.929) The Lagrange Multiplier test statistic for the presence of first-order ARCH is 2.10 (a) A plot of the returns is shown below.841. Since the distribution is negatively skewed (skewness is 1.926) is significantly different from the 5% critical value of 5. Principles of Econometrics. R2 0. There are periods of big changes (for example around June 2006) and periods of small changes (for example around December 2005). It is the unconditional distribution.048. Figure xr14. the distribution of returns is not normal. 259) (1. (t ) (0.775) 1 (1.138 0.903) However.655) hˆt 0.10 (continued) (d) (e) The estimated GARCH(1.200eˆt2 1 (1. However. these results do not support such a model since the t-statistic on the not significant. GOLD t 0. The coefficients are of the correct sign and magnitude.167 ht . Exercise Solutions.534hˆt (1.201eˆt2 1 0. 4e 483 Exercise 14. (0.Chapter 14. they are not significant.118 0.822) hˆt 0.1) model is presented below.120 0.539hˆt 1 (1. ht term is .037.837) 0.887) An estimated GARCH in mean model could improve the forecast of returns: GOLD t (t ) 0.232) (1. Principles of Econometrics. Chapter 14, Exercise Solutions, Principles of Econometrics, 4e 484 EXERCISE 14.11 (a) The monthly rate of inflation is shown below. Figure xr14.11(a) Plot of monthly rate of inflation. (b) The estimated T-GARCH-in-mean model is given in the text. (c) The negative asymmetric effect ( 0.221) suggests that negative shocks (such as falls in prices) reduce volatility in inflation. This result is consistent with an economic hypothesis that volatility tends to be low when inflation rates are low. (d) The positive in-mean effect (1.983) means that inflation in the UK increases when volatility in prices increases. Chapter 14, Exercise Solutions, Principles of Econometrics, 4e 485 EXERCISE 14.12 (a) The estimated GARCH(1,1) and ARCH(5) models are shown below. Dependent Variable: RETURN Method: ML - ARCH (Marquardt) - Normal distribution Sample (adjusted): 1/03/2008 12/31/2008 Included observations: 260 after adjustments Convergence achieved after 45 iterations Bollerslev-Wooldrige robust standard errors & covariance Presample variance: unconditional GARCH = C(2) + C(3)*RESID(-1)^2 + C(4)*GARCH(-1) C Coefficient Std. Error z-Statistic Prob. -0.000633 0.001507 -0.420130 0.6744 1.317690 2.777856 22.98351 0.1876 0.0055 0.0000 Variance Equation C RESID(-1)^2 GARCH(-1) 1.88E-05 0.107483 0.875546 1.42E-05 0.038693 0.038095 Dependent Variable: RETURN Method: ML - ARCH (Marquardt) - Normal distribution Sample (adjusted): 1/03/2008 12/31/2008 Included observations: 260 after adjustments Convergence achieved after 21 iterations Bollerslev-Wooldrige robust standard errors & covariance Presample variance: unconditional GARCH = C(2) + C(3)*RESID(-1)^2 + C(4)*RESID(-2)^2 + C(5)*RESID( -3)^2 + C(6)*RESID(-4)^2 + C(7)*RESID(-5)^2 C Coefficient Std. Error z-Statistic Prob. -0.001689 0.001300 -1.299682 0.1937 3.161446 1.021026 0.470738 1.543519 2.173448 2.352776 0.0016 0.3072 0.6378 0.1227 0.0297 0.0186 Variance Equation C RESID(-1)^2 RESID(-2)^2 RESID(-3)^2 RESID(-4)^2 RESID(-5)^2 0.000208 0.095248 0.016531 0.118779 0.243126 0.387344 6.57E-05 0.093286 0.035116 0.076953 0.111862 0.164633 Chapter 14, Exercise Solutions, Principles of Econometrics, 4e 486 Exercise 14.12(a) (continued) The GARCH(1,1) model is preferred because it is a parsimonious way of capturing a large order ARCH model (especially when the intervening terms in the ARCH model are not significant – see RESID(-1)^2, RESID(-2)^2 and RESID(-3)^2). (b) (c) Based on the GARCH(1,1) model, the expected return and volatility next period are: E st 1 E ht 1 0.001 0.000 0.107eˆt2 0.875hˆt The forecasted return and volatility next period are: stF 1 htF1 0.001 0.000 = 0.000 0.107( st E ( st )) 2 0.875(0.001) 0.107( 0.001 0.001) 2 0.875(0.001) 0.001 (d) The estimated TARCH in mean model is shown below. Dependent Variable: RETURN Method: ML - ARCH (Marquardt) - Normal distribution Date: 10/04/10 Time: 11:39 Sample (adjusted): 1/03/2008 12/31/2008 Included observations: 260 after adjustments Convergence achieved after 20 iterations Bollerslev-Wooldrige robust standard errors & covariance Presample variance: unconditional GARCH = C(3) + C(4)*RESID(-1)^2 + C(5)*RESID(-1)^2*(RESID(-1)<0) + C(6)*GARCH(-1) @SQRT(GARCH) C Coefficient Std. Error z-Statistic Prob. 0.037767 -0.002399 0.198138 0.004756 0.190609 -0.504410 0.8488 0.6140 1.750107 -1.171177 3.713889 24.57561 0.0801 0.2415 0.0002 0.0000 Variance Equation C RESID(-1)^2 RESID(-1)^2*(RESID(-1)<0) GARCH(-1) 2.90E-05 -0.023328 0.228222 0.877078 1.66E-05 0.019919 0.061451 0.035689 Chapter 14, Exercise Solutions, Principles of Econometrics, 4e 487 Exercise 14.12(d) (continued) Since the ARCH term is insignificant, we re-estimate the model: Dependent Variable: RETURN Method: ML - ARCH (Marquardt) - Normal distribution Sample (adjusted): 1/03/2008 12/31/2008 Included observations: 260 after adjustments Convergence achieved after 41 iterations Bollerslev-Wooldrige robust standard errors & covariance Presample variance: unconditional GARCH = C(3) + C(4)*RESID(-1)^2*(RESID(-1)<0) + C(5)*GARCH(-1) @SQRT(GARCH) C Coefficient Std. Error z-Statistic Prob. 0.043067 -0.002513 0.200922 0.004839 0.214346 -0.519271 0.8303 0.6036 1.671506 3.464072 25.46199 0.0946 0.0005 0.0000 Variance Equation C RESID(-1)^2*(RESID(-1)<0) GARCH(-1) 2.60E-05 0.195996 0.873353 1.56E-05 0.056580 0.034300 The in-mean effect is not significant. When news is good, the contribution of RESID(-1)^2 is insignificant, while when news is negative, the contribution is 0.196. Chapter 14, Exercise Solutions, Principles of Econometrics, 4e EXERCISE 14.13 (a) Model where only own lagged effects matter (here specified as a lag-order 1 model): EUROt (b) 1 ht 1 1t 1 ; 1t ~ N (0, 2 1 ) EUROt 1 1t ; 1t | I t 1 ~ N (0, ht ) h 2 t 1 Model where own lagged growth and lagged USA growth matter: 1 1 EUROt USAt 1 1 1t 1 ; 1t ~ N (0, 2 1 ) Model where shocks affect expected returns: EUROt ht (e) 1 2 1 1t 1 0 EUROt (d) EUROt Model where only own lagged effects matter but with time-varying variance: EUROt (c) 1 1 1 2 1 1t 1 0 EUROt 1 ht 1 1t ; 1t | I t 1 ~ N (0, ht ) h 2 t 1 Model where shocks from the EURO and USA affect the expected EURO return: USAt EUROt ht USAt 2 0 2 1 2 1 1t 1 1 1 2t EUROt 1 h 2 t 1 ; 2t ~ N (0, USAt 1 2 3 2t 1 1 1 2 2 ) ht 1t ; 1t | I t 1 ~ N (0, ht ) 488 CHAPTER 15 Exercise Solutions 489 Chapter 15, Exercise Solutions, Principles of Econometrics, 4e 490 EXERCISE 15.1 (a) The negative coefficient of POP suggests that countries with higher population growth tended to have lower growth in per capita GDP. The increasing population has not led to a more than compensating gain in GDP, leading to a fall in the ratio of GDP to population. A positive coefficient for INV implies more investment leads to a higher growth rate, as one would expect. The negative coefficient for IGDP suggests that a lower initial level of GDP provides greater scope for growth in per capita GDP – a reasonable outcome. Finally, the positive sign on the human capital variable suggests that a greater level of education leads to a higher growth rate. This outcome also conforms with our expectations. (b) The coefficient for human capital for the period 1960 is significantly different from zero (the t-ratio is greater than 2 and the p-value is less than 0.05) while those for the periods 1970 and 1980 are not. Thus, human capital appears to influence growth rate only for 1960 but not for 1970 and 1980. (c) The null hypothesis is H 0 : 12 13 23 0 where ij refers to the covariance between the errors in equations i and j. The test statistic is LM T (r122 r132 r232 ) 86 (0.10842 0.1287 2 0.3987 2 ) 86 (0.0118 0.0166 0.1590) 16.11 The 5% critical value for a 2 -distribution with 3 degrees of freedom is 2 (0.95,3) 7.81 . We reject the null hypothesis. Thus, SUR is preferred over separate least squares estimation. (d) The null hypothesis being tested is that the impact of each explanatory variable on the growth rate is the same in each of the three periods. The intercepts are left unrestricted. (e) The 2 test statistic value is 12.309. At a 5% significance level with 8 degrees of freedom the critical value is 2 (0.95,8) 15.51 . Since the test statistic value is not greater than the critical value, we do not reject the null hypothesis. The p-value for this test is 0.1379. (f) 2 J 12.309 8 1.539 where J is the number of The F test statistic value is F equalities in the null hypothesis. The corresponding 5% critical value for (8, 243) degrees of freedom is F(0.95,8, 243) 1.977 . Since the test statistic value is less than the critical value, we do not reject the null hypothesis. The p-value for this test is 0.1443. Chapter 15, Exercise Solutions, Principles of Econometrics, 4e 491 EXERCISE 15.2 (a) The restrictions are that, for each explanatory variable, the coefficients are the same across equations. Only the intercept coefficient varies across equations. (b) The main difference between these results and those in Exercise 15.1 is the magnitude of the standard errors. After imposing the restrictions the standard errors decrease for all coefficients. In particular, the standard errors for the coefficients of POP and SEC decrease substantially. The magnitude of each restricted coefficient estimate lies between the highest and lowest values for the corresponding unrestricted estimates. (c) The 2 test statistic value is 93.098. At a 1% level of significance and 2 degrees of freedom the critical value is 2 (0.99, 2) 9.21 . Since the test statistic value is greater than the critical value, we reject the null hypothesis. The p-value for this test is 0.00000. Chapter 15, Exercise Solutions, Principles of Econometrics, 4e 492 EXERCISE 15.3 (a) In Exercise 15.2 the error variances for the different years were assumed different, and correlation between errors for the same country, in different years, was permitted. If the observations are all pooled with dummy variables inserted for each of the years 1960, 1970 and 1980, and the model is estimated using least squares, implicit assumptions are that the error variance is the same for all observations, and all error correlations are zero. (b) The estimated equation is Gˆ 0.0315 D60 (se) (0.0147) 0.0205 D70 0.0029 D80 0.4365 POP 0.1628 INV (0.0153) (0.0158) (0.1823) (0.0208) 1.43 10 6 IGDP 0.0149 SEC (9.42 10 7 ) (0.0098) The estimates obtained in this exercise are very similar to those in Exercise 15.2. They will not be exactly the same because the estimation procedure in Exercise 15.2 is a generalized least squares one that uses information on different error variances and correlated errors. (c) The test statistic value for RESET is 1.2078 with a p-value of 0.3006. The p-value is greater than a significance level of 0.05. Thus, RESET does not suggest the equation is misspecified. Chapter 15, Exercise Solutions, Principles of Econometrics, 4e 493 EXERCISE 15.4 (a) From equation (15.14) in POE4, we have yit x eit 2 it where yit yit yi , xit xit xi , and eit eit ei . The fixed effects estimator for least squares estimator applied to this equation. It is given by N ˆ T i 1 t 1 N T 2, FE i 1 t 1 (b) N xit yit where yit 1 xit2 N ˆ xit i 1 t 1 yit xit xi yi 2 vit ˆ xi . This estimator is given by xit T xi is the is the least squares estimator applied to the equation x i 1 t 1 N 2, RE 2 2 it T T i 1 t 1 1 ˆ ˆ yi , and xit yit xit i 1 t 1 N The random effects estimator for yit T 2 x yit xit x y 2 Now, 1 NT x N T i 1 t 1 xit ˆ xi ˆ xi x ˆx x and xit x xit A similar result holds for yit N ˆ 2, RE T i 1 t 1 ˆ xi xit x x y . Thus, xit ˆ xi N T i 1 t 1 (c) ˆx x xit x ˆ xi yit x ˆ yi x y y 2 The pooled least squares estimator is given by N ˆ 2, PLS T i 1 t 1 N xit T i 1 t 1 x yit xit x y 2 The pooled least squares estimator uses variation in xit and yit around their overall means; it does not distinguish between variation within and between individuals. The fixed effects estimator uses only variation from individual means, known as within variation. The random effects estimator uses both overall and between variation, weighted according to the value of ˆ ; between variation uses xi x and yi y . Chapter 15, Exercise Solutions, Principles of Econometrics, 4e 494 EXERCISE 15.5 (a) The three estimates for (i) 2 are: Dummy variable / fixed effects estimator b2 0.0207 se b2 0.0209 (ii) Estimator from averaged data ˆA 2 0.0273 se ˆ 2A 0.0075 (iii)..Random effects estimator ˆ 2 0.0266 se ˆ 2 0.0070 The estimates from the averaged data and from the random effects model are very similar, with the standard error from the random effects model suggesting the estimate from this model is more precise. The dummy variable model estimate is noticeably different and its standard error is much bigger than that of the other two estimates. (b) To test H 0 : 1,1 1,2 1,40 against the alternative that not all of the intercepts are equal, we use the usual F-test for testing a set of linear restrictions. The calculated value is F 3.175 , while the 5% critical value is F(0.95,39,79) 1.551 . Thus, we reject H0 and conclude that the household intercepts are not all equal. The F value can be obtained using the equation F ( SSE R SSEU ) J SSEU ( NT K ) (195.5481 76.15873) 39 76.15873 (120 41) 3.175 Chapter 15, Exercise Solutions, Principles of Econometrics, 4e 495 EXERCISE 15.6 (a) Fixed effects estimates of the model are given below Dependent Variable: LNPRICE Method: Panel Least Squares: Fixed Effects Periods included: 4 Cross-sections included: 754 Total panel (balanced) observations: 3016 Variable C REGULAR RICH ALCOHOL NOCONDOM BAR STREET Coefficient Std. Error t-Value Prob. 5.46126 0.03722 0.08264 -0.05686 0.17028 0.29846 0.45516 0.13028 0.01685 0.02053 0.02614 0.02582 0.13445 0.13047 41.920 2.209 4.025 -2.175 6.596 2.220 3.489 0.0000 0.0273 0.0001 0.0297 0.0000 0.0265 0.0005 (i) Sex worker characteristics are omitted because they are time-invariant over the time in which the 4 transactions took place. Their effect cannot be separated from the individual effects given by the coefficients of the fixed-effects dummy variables. (ii) All coefficient estimates are significantly different from zero at a 5% level. (iii) The estimated risk premium for not using a condom is approximately 17%. The exact estimate is 100 exp(0.170282) 1 % 18.6% . The price is approximately 3.7% higher for regular customers and approximately 8.3% higher for rich customers. It is 5.7% lower for customers whio have consumed alcohol. The origin of the transaction has a relatively large effect on the price. For transactions that originated in a bar, there is a 29.8% premium (approximately) and for transactions originating in the street, the premium is approximately 45.5%. (b) Random effects estimates are presented on the next page. Treating the effects as random instead of fixed and adding the sex worker characteristics has had a dramatic effect on some of the common coefficients. Rich clients are now estimated to pay 11.6% extra instead of 8.3%. Those who have consumed alcohol are now estimated to pay a higher price instead of a lower price, although this coefficient is not significantly different from zero. The premium for not using a condom has declined slightly to 13.9%. There have been large changes in the coefficients of BAR and STREET. The random effects specification suggests that transactions originating in a bar are much more expensive than those originating on the street, whereas the reverse was true with the fixed effects specification. The price of commercial sex is lower for older sex workers, higher for attractive workers, and higher for secondary educated sex workers. 04447 4.63196) 1 88.0000 .0000 0.02455 5. compared with protected sex with an unattractive uneduated sex worker.11601 0.02363 0.02577 0.2 %. Dependent Variable: LNPRICE Method: Panel EGLS (Cross-section random effects) Periods included: 4 Cross-sections included: 754 Total panel (balanced) observations: 3016 Swamy and Arora estimator of component variances Variable Coefficient Std.540 ATTRACTIVE 0.21839 p-value 0.02577 0.00275 0.8602 0.904 ALCOHOL 0.D.46425 0.738 STREET 0.01489 0. ˆu Cross-section random 0.27683 0.01489 0.2976 0. Stata11 gives slightly higher standard errors which lead to smaller t-values and larger p-values. Error 0.460 5.685 SCHOOL 0.0000 0.597 5.3069 0.01618 0.21615 0.09914 1. Variable C REGULAR RICH ALCOHOL NOCONDOM BAR STREET AGE ATTRACTIVE SCHOOL Coefficient 5.0000 0.0000 0.767 p-value 0.042 AGE -0. Error t-value C 5.13898 0.861 Effects Specification S.608 NOCONDOM 0.91037 0.02003 0.0000 Rho 0.46425 0.02448 0. the extra pecentage premium for having unprotected sex with an attractive secondary-educated sex worker.357 4.12782 46. The exact calculation is 100 exp(0. Exercise Solutions.Chapter 15.353 1.0000 0. 4e 496 Exercise 15.488 RICH 0.022 -9. as presented in the following table.1367 0.662 BAR 0.553 4.6(b) (continued) Other things held constant.0000 0.91037 0.21615)% 63.09989 0.06024 0.02496 0.0000 0.0000 0.02363 0.1398 Note: The above results are those computed by EViews7.02503 0.240 REGULAR 0.13032 0.1% .00270 -9.01965 5.790 0.05908 4.09798 4.04534 t-value 45.1444 0.11601 0.0000 0.21615 Std.0000 0.5430 0.13898 0. is approximately 100 (0.596 4.01587 1.10329 -0.5508 0. Principles of Econometrics.27683 0.13898 0.10329 0.10108 0.54163 ˆe Idiosyncratic random 0.648 1.0000 0.27683 0.0000 0. 00315 0.46510 0.5% . we reject a null hypothesis that the individual random effects are uncorrelated with the variables in the model.0000 0.10343 0.63373 0.k bRE .0000 0.21810 Std.0959 0.06770 0.0655 0.k t-value p-value 0.0000 0. or.0 %.089992 0.0000 0. Thus.00309 0. Error 0.02537 0.k se bFE .0000 0.0162 0.129 6. then NOCONDOM is likley to be correlated with the individual effect.007999 0.801 4.937 -1.901 -7.149 0.0000 0.188 4.092074 0.071746 0.0000 0.22563 0.01954 0.432 p-value 0. There have been no dramatic changes in the coefficient estimates and REGULAR. is approximately 100 (0.084809 2. At a 5% level of significance. Hausman-Taylor estimates with NOCONDOM endogenous Variable C REGULAR RICH ALCOHOL NOCONDOM BAR STREET AGE ATTRACTIVE SCHOOL ˆu ˆe Coefficient 5.351873 0. In this case. EViews REGULAR RICH ALCOHOL NOCONDOM BAR STREET (d) Stata bFE . The fixed effects estimates are more reliable in this instance because they are consistent.346 4.8975 0.15619 -0.456 -9.031298 -0.821 3.033371 -0.22563)% 67.10909 0.005647 0.01585 0.6 (continued) (c) Results for the Hausman test on each difference between the fixed effects and random effects estimates are given in the following table for both EViews and Stata standard errors.k t-value p-value se bFE . The results are very similar to those obtained in part (b).691 1. there is a significant difference between all coefficients except those for BAR.k bRE .165790 0.k bRE . Principles of Econometrics.006340 0.619 -7.004684 0.582 0.1311 0.0000 .619 4.0000 If a sex worker has individual characteristics that make her a risk taker.0000 0. risk averse. ALCOHOL and STREET continue to be insignificant at a 5% level of significance.02660 0.225 4. The estimates obtained using the Hausman-Taylor estimator assuming NOCONDOM is endogenous are given in the table below. The exact calculation is 100 exp(0.004475 0. compared with protected sex with an unattractive uneduated sex worker.406 -5.0000 0.532 1. Exercise Solutions.13894 0.16099 0.009173 0.28352 0. the extra pecentage premium for having unprotected sex with an attractive secondary-educated sex worker. 4e 497 Exercise 15.082490 2.0000 0.266 0.510 -8.0037 0.0719 0.007777 0.02640 0. conversely.67014) 1 95.842 4.93145 0.666 5.005939 0.02442 0.Chapter 15.28352 0.0000 0.05091 t-value 42.013590 -0.913 -1.0001 0.0000 0.16099 0.10263 0. 69 0.39 13. .70 SMALL 8.8 points worse than girls.774 0.6509 2. (c) The F-value for testing for significant school effects is F 18. we reject the null hypothesis that there are no school effects.000 The general conclusions made in part (a) when school fixed effects were not included remain the same.39 3.1134 2.03 6. Having significant school effects that have not changed our general conclusions about the coefficients suggests that the school effects are not highly correlated with the explanatory variables.4692 0.30 –0.696 0.8404 WHITE_ASIAN 17.000 0. Err.1 points. Err. The estimated effect of small classes is slightly larger at 9.42015 –6.000 0. Students of experienced teachers score slightly better than those of inexperienced teachers. The presence of a teacher’s aide continues to be insignificant.7 (a) The results from estimating the equation with MATHSCORE as the dependent variable and no fixed or random effects are as follows: Pooled Least Squares Estimates Coef.00 0. the estimate is significant but not very large (0. Boys score 6. t-value p-value 466.000 0. Gender and race have a big impact.42210 TCHEXPER 0. suggesting that having aide does not improve the score.14 –6.29 6.88 –5. The most dramatic effect is the increase in the coefficient of WHITE_ASIAN from 17.3491 0.1072 1.000 We find that being in a small class increases the math score by 8.3177 268. Exercise Solutions.6312 23.6 points worse than girls. Boys score 7.000 0.000 0.Chapter 15.2275 1. other things equal.3496 0. Std.000 0. Having an experienced teacher has a significant but very small effect.66 points). it has an F-distribution with (78.7476 1.066 . 4e 498 EXERCISE 15.1084 1.23 0.52689 0. t-value p-value C 469. Thus.17 9.3 points.65787 BOY –7.000 0.5254 1.000 0.1 to 23.7 points.0000. (b) Including fixed effects leads to the following set of estimates Fixed Effects Estimates C SMALL AIDE TCHEXPER BOY WHITE_ASIAN Coef. Correct to 4 decimal places the corresponding p-value is 0.0833 AIDE –0.77 5.3970 1.1579 1. The coefficient for teacher’s aide is not significant. Assuming there are no school fixed effects. Std.96 10.1 points better.3109 216.1241 1. Principles of Econometrics. and white Asians score 17. 5682) degrees of freedom. Exercise Solutions.719 0.445 0.0418 TCHEXPER –0.84122 t-value p-value 1. Similarly.000 0. t-value p-value 466. and no conclusions change.3965 1.7145 22.9466 3. If the Asian students tend to be concentrated in particular schools.1076 1. 4e 499 Exercise 15.01302 0.000 0.4353 19. Principles of Econometrics. We conclude that the random effects estimates are consistent and more efficient. and coefficient estimates and standard errors with slight differences.48505 0.68 6.215 0. Err. The insignificant differences between the fixed and random effects estimates suggest that the explanatory variables are not correlated with the school effects.k 0. Because of random assignment of SMALL and AIDE.k bRE .8714 41.1135 2. That for BOY is not included because in this case se bFE .7 (continued) (d) Random effects estimates are presented in the following table.0173 WHITE_ASIAN 1.3009 0. Other software such as EViews may produce a slightly different estimate for ˆ u .148 .04447 0. some schools could have a predominance of experienced teachers in which case TCHEXPER would be correlated with the school effects. Hausman Test Results bFE .57 9.36 4.347 0.000 The random effects estimates are very similar to those obtained using fixed effects.k se bRE .239 0.941 –1.2156 bRE .43742 –6.Chapter 15.03 10.327 1.3484 0.000 0. we would not expect the other variables to be correlated with the school effects.0759 1.k se bFE .k .000 0.1523 151. There are only minor differences.k SMALL 0. (e) Results from the Hausman test for the differences between the fixed and random effects estimates are given in the following table.07 –6. These estimates are those obtained from Stata version 11.42 0. and because gender is likely to be random. Random Effects Estimates C SMALL AIDE TCHEXPER BOY WHITE_ASIAN ˆu ˆe Coef. Std.185 0. then WHITE_ASIAN could be correlated with the school effects.0487 AIDE 0.66 0.03930 0. 70 8.7 (continued) (f) Random effects estimates of a model with AIDE omitted and TCHMASTERS and SCHURBAN included follow.Chapter 15. Err.000 0.68 –0.3960 –1.093 0. Principles of Econometrics. 4e 500 Exercise 15.1104 1.38 –6.02 10. and the effect of a teachers master’s degree seems to be negative! Fixed effects estimation of this model will break down because of perfect collinearity between SCHURBAN and the school effects.4264 5.1134 2.000 0. t-value p-value 467. Exercise Solutions.2218 0.32 4. there are no dramatic changes in the coefficient estimates for the variables that were in the earlier model. Std.2167 1.48351 –6. Again.000 0.2880 –2.6982 22.7813 41.2199 130.21 0.9379 .9455 0.05 –1. Neither TCHMASTERS nor SCHURBAN is significant at a 5% level of significance.000 0. Random Effects Estimates of Extended Model C SMALL TCHEXPER BOY WHITE_ASIAN TCHMASTERS SCHURBAN ˆu ˆe Coef.60 7.1012 3.5810 1.833 19.000 0. 0774 (0.1234 (0.7346 (0.0312) 0.0728) 0.2407) 0.0530) EXPER2 0.1265 (0.3288 (0. we are assuming that all women have identical coefficients (there is no individual heterogeneity) and the coefficients are the same in each year.1445 (0.2255 (0.2522) (0. however.1264) UNION 0.0814 (0. For this estimation.3261 (0.0822 (0.0328)* 0. There are no substantial year-to-year changes in the magnitudes of the coefficients. The standard errors are less.2688)* EXPER 0.1365 (0.1105) SOUTH 0.0575 (0.0241) 0.1506) 1. Intercept 0.0344) 0.2128 (0.1264) 0.0323) 0.0312) (0. Having a separate equation for each year does allow the coefficients to be different in different years.0524) Variable ( 102 ) D88 Part (f) Diff.2384 (0.1365 (0.3066 (0.1102) 0.0814 (0.0312) 0. reflecting the greater precision from a larger number of observations.1102 (0.1234 (0.9482 (0.5468 (0.1258) (0.2495)* 0.1258) 0.Chapter 15. The coefficient estimates for the two years and their standard errors are very similar.1187 (0.1102) (0. For these individual year estimations.0272) 0.1105) 0.2010) 0.1187 (0.9348 (0.6050) * Cluster-robust standard errors (a) The estimates for this model for the two years 1987 and 1988 are presented in the second and third columns of the table.0367)* 0.0530) 0.1229 (0.1069) 0.0822 (0. with the coefficients and standard errors for EXPER 2 reported as 100 times greater than their actual values. Eff.0338) 0.1274 (0. however. (b) The estimates for this model are presented in the fourth column of the table.0774 (0. Principles of Econometrics. the magnitudes of the coefficients are similar to those obtained for the 1987 and 1988 equations.0211) 0.0295) 0. Again.0330) 0. . the model does not account for differences that might be attributable to individual heterogeneity.0575 (0. with standard errors are in parentheses below the estimated coefficients.8993 (0. We are also assuming the variance of the error term is the same for all individuals and in both years.1067) 0.3453 (0.8 The coefficient estimates for the different parts of this question are given in the following table.0387) 0.3261 (0.3453 (0.1096)* 0. Part (g) Dum. Part (a) 1987 LS Part (a) 1988 LS Part (b) Parts (d)(e) Pool LS Fix.0312) 0.0524) 0. we are assuming that all individuals have the same regression parameter values. 4e 501 EXERCISE 15.0330) (0. Exercise Solutions.3089 (0.1270 (0. 88 Part (h) Dif-Dum 0.0382) 0. Using cluster-robust standard errors allows for the variances to be different for different individuals in both 1987 and 1988.t 2 3 EXPERi .5285 22. while the 5% critical value is F(0.Chapter 15. the coefficients of EXPER and EXPER 2 have more than halved in the fixed effects model.t 2 it DEXPER 4 1 SOUTH i .95.t 4 UNIONi . The estimates for this model are presented in the sixth column of the table. The estimates do suggest some bias could have been present. and that the coefficients are identical in both years. (d) The estimates of the fixed effects model are presented in the fifth column of the table. 4e 502 Exercise 15.t DSOUTH it 5 1 ei . Without cluster-robust standard errors we are assuming that the error variance is the same for all individuals and in both years. (f) Writing down the lagged model and subtracting it from the original model yields ln(WAGEi . we use the usual F-test for testing a set of linear restrictions.43925 (1432 716 4) The existence of individual heterogeneity means the estimates of the remaining coefficients will be biased if such heterogeneity is correlated with explanatory variables such as experience and SOUTH. In parts (a) and (b).t 3 4 EXPERi2. The F value can be obtained using the equation ( SSE R SSEU ) J SSEU ( NT K ) F (285. and it permits correlation between errors in 1987 and 1988 for the same individual. They are identical to the fixed effects estimates obtained in part (d).675.t UNIONi . differences in the behaviour of individuals have not been accounted for since a single intercept value is estimated for all i.31 . .t ) 1i ln(WAGEi . assumptions that were not made in part (a).t 1 ) DLWAGEit 2 1i 2 EXPERi . and that there is no correlation between errors in the different years for the same individual. we reject H0 and conclude that the intercepts for all women in the sample are not all equal. 712) 1.43925) 715 11.675 22. Thus. The calculated value is F = 11.8 (continued) (c) The fixed effects model accounts for differences in behaviour (individual heterogeneity) by allowing the intercept to change for each individual. However. Principles of Econometrics. For example. 715.716 against the alternative that not all of the intercepts are equal.t DEXPERit 3 1 EXPERi2.2 1. The cluster-robust standard error for the coefficient of SOUTH is approximately double that of its more restrictive counterpart. To test H 0 : 1. Exercise Solutions.t 5 DUNION it 1 ei . (e) Cluster-robust standard errors for the fixed-effects estimated model are given below the conventional ones in column 5 of the table.t 5 SOUTH i .1 1. this fixed effects model assumes that the variance of the error term is the same for both years. The cluster-robust standard errors are similar to the conventional ones except for the case of SOUTH.t 1 Deit By taking the first differences we remove the heterogeneity term. this model estimates that the average growth rate was 7. 0. Principles of Econometrics. (h) The estimates for this model are presented in the last column of the table. 4e 503 Exercise 15. Thus. Subtracting D88 = 0 from D88 = 1.8 (continued) (g) The estimates for this model are presented in the next-to-last column of the table. Thus. The coefficient for the dummy variable. Exercise Solutions. D88. . and is identical to the estimate found in part (g).Chapter 15.1402 is greater than 0. is not significant at a 5% level of significance since its p-value. yields the constant term 1.74% from 1987 to 1988. This dummy variable describes the growth rate of real wages averaged over all individuals. in this model the intercept term represents the average growth rate of real wages.05. 1648 (0.1610) EDUC 0.0066) EXPER 0.0575 (0. with standard errors are in parentheses below the estimated coefficients.1324 (0.1199 (0.1509 (0.1881) 0.1432 (0.0330) 0.0064) 0. however.1406) 0.1029 (0.1258) 0.0265) 0.0830 (0.0822 (0.2522) 0.0303) UNION 0. We are also assuming the variance of the error term is the same for all individuals and in both years. For these individual year estimations. The coefficient estimates for the two years and their standard errors are similar.1528) 1.1432 (0.0319) 0.0366) 0. 4e 504 EXERCISE 15.0334) 0.2381 (0.0875 (0.0834 (0. the model does not account for differences that might be attributable to individual heterogeneity.0778 (0.3086 (0.0235) Variable ( 102 ) 0. .0771 (0.3261 (0.1319 (0. Again.0964) 0.1368 (0.1170 (0.0248) 0.0345) (a) The estimates for this model for the two years 1987 and 1988 are presented in the second and third columns of the table.0958) 0. The standard errors are less.0260) 0.1852 (0. Exercise Solutions. Part (f) Ran.0354) 0.1852 (0.1790 (0.0190) 0. There are some changes but no substantial yearto-year changes in the magnitudes of the coefficients. we are assuming that all individuals have the same regression parameter values.0771 (0. Eff.5468 (0.2033 (0.0702) BLACK 0.2216 (0.0350) 0.0654) 0.Chapter 15.0776 (0.0292) 0.0063) 0.1701 (0.1509 (0.9 The coefficient estimates for the different parts of this question are given in the following table. Part (a) 1987 LS Part (a) 1988 LS Part (b) Pool LS Part (c) PLS (cl se) Part (e) Fix.0834 (0.2268 (0. the magnitudes of the coefficients are similar to those obtained for the 1987 and 1988 equations. Eff.0312) 0.0060) 0.0306) 0.0327) 0.0314) SOUTH 0. with the coefficients and standard errors for EXPER 2 reported as 100 times greater than their actual values.0762 (0. Having a separate equation for each year does allow the coefficients to be different in different years.1562 (0. we are assuming that all women have identical coefficients (there is no individual heterogeneity) and the coefficients are the same in each year.1199 (0.0722) 0. (b) The estimates for this model are presented in the fourth column of the table. reflecting the greater precision from a larger number of observations.0372) 0.0045) 0.1234 (0.2227) 0. however. Principles of Econometrics.2381 (0. Intercept 0.0205) EXPER2 0. and that the errors are uncorrelated over individuals and between the two years for each individual.0206) 0.1102) 0.0233) 0. For this estimation.1350 (0.0758 (0.1309 (0. This was the test performed in Exercise 15. Other sources of heterogeneity are possible in the fixed effects model but the effects of each source cannot be estimated separately.9 (continued) (c) Pooled least squares estimates of the coefficients with cluster-robust standard errors are presented in column 5 of the table. suggesting that ignoring heteroskedasticity and within individual correlation can lead to an overstatement of the precision of our estimates. Other differences are that the fixed effects model assumes that the variance of the error term is the same for both years.Chapter 15. we rewrite the equation in part (d) as ln(WAGE ) 1i 3 EXPERit 4 EXPERit2 6 SOUTH it UNION it 7 eit where 1i 1i 2 EDUCi 5 BLACKi It is the 1i that are estimated by the fixed effects model. This characteristic means that the coefficients of EDUC and BLACK cannot be estimated separately from the fixed effects 1i . The variables EDUC and BLACK have an i subscript and no t subscript because they do not change over time.8(d).2 1. Without cluster-robust standard errors we are assuming that the error variance is the same for all individuals and in both years. In parts (a) and (b) where fixed effects are not specified. An individual’s level of education and color do not change. Principles of Econometrics. Exercise Solutions.) . Omission of EDUC and BLACK is necessary to avoid perfect collinearity. Alternatively. 4e 505 Exercise 15. Omitting EDUC and BLACK raises a question about the definition of the intercept.1 1. In 0 . Using clusterrobust standard errors allows for the variances to be different for different individuals in both 1987 and 1988. We can test whether the 1i are identical for all women in the sample. it is implicitly assumed that EDUC and BLACK are the only sources of individual heterogeneity. (We have dropped the subscript i from 1i . and that there is no correlation between errors in the different years for the same individual.716 against the alternative that not all of the intercepts are equal. and it permits correlation between errors in 1987 and 1988 for the same individual. In this latter case we are testing whether EDUC and BLACK are the only sources of individual heterogeneity. we must be clear about which intercepts we want to test. we can test whether the context of the current model it implies 2 5 the 1i are identical for all women in the sample. namely. To appreciate the issue. To test whether the intercepts are identical for all women in the sample. assumptions that were not made in part (a). and that the other coefficients are identical in both years. H 0 : 1. We proceed with this test. The cluster-robust standard errors are slightly larger than the regular ones from least squares estimation. Note that another way of writing the restriction in the null hypothesis is to say that 1i 1 2 EDUCi 5 BLACK i for all i. (d) The fixed effects model accounts for differences in behaviour (individual heterogeneity) by allowing the intercept to be different for each individual. (e) The estimates of the fixed effects model with EDUC and BLACK omitted are presented in the next-to-last column of the table. 95) 1.0893) It is not possible to estimate a return to education from the fixed effects model in part (e) because EDUC does not change over time and is therefore perfectly collinear with the dummy variables. When there is no within-individual variation.43925 (1432 716 4) 9.95.8772 22.30) on page 554 of POE4. In this case we can find an estimate of the return to education by using the variation in education across individuals.131 .645 . A 95% interval estimate can be calculated as ˆ (h) 2 t(0.975.962 0. we reject H 0 and The 5% critical value is F(0. Because 9.4 This value clearly exceeds the critical value z(0.43925) 713 22. Its value is N LM NT 2(T 1) T i 1 N t 1 T 2 eˆit 1 i 1 t 1 eˆit2 1432 408. 4e 506 Exercise 15. it fails. Exercise Solutions. Thus.0000 indicating that it is significant at a 1% level of significance. To test the null hypothesis that there are no random effects we test : H 0 : u2 0 against the alternative H1 : u2 0 where u2 is the variance of the random effect u.1425) se ˆ 2 0. the number of restrictions is 713. On the other hand the random effects estimator in part (f) uses both variation within individuals and variation between individuals to obtain estimates of slope coefficients.9(e) (continued) The restricted model is that estimated in part (b).098 1.8772 21. Because we are replacing 716 intercepts 1. (f) The estimates of the random effects model are presented in the last column of the table. Its pvalue is 0. .76%. The value of 1i with three parameters the F-statistic is F ( SSER SSEU ) J SSEU ( NT K ) (226.3288 1 2 1 226. The test statistic is that given in equation (15. (g) The return on an additional year of education in the random effects model is 7. 712) conclude that EDUC and BLACK are not the only sources of individual heterogeneity. Principles of Econometrics.Chapter 15. 713.0658. The fixed effects estimator uses only the variation within each individual to estimate the slope coefficients.077557 1. as is the case with education.131 .0.005969 (0. 2 and 5 .098 1. we reject the null hypothesis and conclude that random effects are present. Thus.Chapter 15.477) (p-value = 0. are EXPER: t 0.05.018355 0.k ) 2 se(bRE .565 (p-value = 0.000850 SOUTH: t 0.091) 0. If there was a significant difference between the sets of estimates we would choose the fixed effects estimator because the random effects estimator would be biased.025826 EXPER2: t 0. there is not evidence that the random effects model is an incorrect specification.692 (p-value = 0.711 0. Exercise Solutions.626) All p-values are greater than 0.487 (p-value = 0.118) UNION: t 0. k se(bFE . with p-values in parentheses. We do not reject the null hypothesis that the difference between the estimates is zero.034791 0. the random effects model is better than the fixed effects model because it allows us to estimate the coefficients for the time invariant variables and it is more precise in large samples. EXPER2.020563 1.k bRE .122081 1. leading us to conclude that the difference between the two sets of estimates is not significant.k )2 1/ 2 The results.9 (continued) (i) The t-test values for the Hausman tests on the coefficient differences for EXPER. . SOUTH and UNION are calculated using the general formula t bFE . When its assumptions hold. Principles of Econometrics.000414 0. 4e 507 Exercise 15.191053 0. Thus. Moreover.0277) 0. (iii) Higher population density is linked with a higher residential crime rate. (c) The estimated equation is LCRMRTE (se) 3. once we allow for county heterogeneity.23% decrease in the crime rate.2082LPRBPRIS (0. (ii) The coefficient on LPRBARR suggests that a 1% increase in the probability of being arrested results in a 0.0586 LAVGSEN (0.0393) 0. the deterrent effect of being arrested is much less. The coefficient of LPRBARR suggests that a 1% increase in the probability of being arrested results in a 0. (b) The estimated equation is LCRMRTE (se) 6.0606) (0. This estimated elasticity is less than half of the estimated elasticity from part (b).0619) (i) LPRBARR.1666LWMFG (0. (iv) Young males are the most likely demographic group to be involved in illegal activities.2921LWMFG (0.0183LAVGSEN 0.0310) (0.0376) (0. We expect the coefficients of these variables to be negative. an increase in the percentage of young males should increase the crime rate.3654) (0. . (ii) If wages in the private sector increase the return to legal activities increases relative to the return to illegal activities. all coefficients are elasticities. Therefore crime rates should drop. The variable LWMFG. which represents wages in the private sector.4466LPRBCONV (0. (ii) Since the model is in log-log form. Exercise Solutions.0861 0. Principles of Econometrics. All coefficients are significantly different from zero at a 5% level of significance except for the coefficient of LAVGSEN.10 (a) (i) If deterrence increases crime rates should drop. (i) All estimated coefficients have the expected sign except for LAVGSEN. has a positive coefficient that is not consistent with our expectations. 4e 508 EXERCISE 15.2288 0.Chapter 15.6566LPRBARR 0. Thus. LPRBPRIS and LAVGSEN are explanatory variables that describe the deterrence effect of the legal system.0222) (0.1378LPRBCONV 0. all estimated coefficients are significantly different from zero at a 5% level of significance except for the coefficient for LAVGSEN.0403) 0.0727) 0.0553) The reported intercept term is the average of the fixed effects. We find that all of these coefficients are negative except for the coefficient of LPRBPRIS.66% decrease in the crime rate.3236) (0.2313LPRBARR 0.1431LPRBPRIS (0. LPRBCONV. 0240 0.1113 0.5762 0.0612 0.1591 0. Exercise Solutions.0101 2.0694 0. while the 5% critical value is F(0.5906 and a p-value of 0.0600 0.287 . a null hypothesis that the coefficient of LAVGSEN is zero is not rejected.1056 0. we reject H0 and 1.3550 0. .0176 0. 4e 509 Exercise 15.0160 0.0589 0.0352 0. Ignoring county effects can lead to misleading conclusions.0274 0.0669 0.4662 0.1330 0.Chapter 15.0183% increase in the crime rate.3052 0.4245 0.0657 0.1952 0. Principles of Econometrics.0288 0.0217 0.3377 0.0273 0.6769 0.555.535) 1. a two tail t-test on the significance this estimate yields a t-statistic of 0.0877 0.0384 0.0315 0.0574 0. Estimates Variable Intercept LPRBARR LPRBCONV LPRBPRIS LAVGSEN LWMFG LDENSITY LPCTYMLE D82 D83 D84 D85 D86 D87 Standard Errors LS FE LS FE 3.4346 0.749.0367 0.1194 0. There is no support for the idea that longer prison sentences are a deterrent to crime.0585 0.14881 (630 90 5) 33.0419 0. The calculated value is F = 33.0577 0.7694 1.8144 16.0528 0.2435 0.0617 1.0579 0.95.0772 (i) It is apparent that the coefficient estimates obtained by using least squares are very different to those obtained using the fixed effects method.0426 0. The magnitudes change considerably and there are some sign reversals. The F value can be obtained using the equation F (e) ( SSE R SSEU ) J SSEU ( NT K ) (106.90 conclude that the county level effects are not all zero.0253 0.0121 0.89.0977 0.2827 0.0705 0.1 1.0216 0.2460 0. Thus. Thus.14881) 89 16. we use the usual F-test for testing a set of linear restrictions.10(c) (continued) (c) (iii) The coefficient on LAVGSEN suggests that a 1% increase in the average prison sentence results in a 0.2782 0.2 1. (d) To test H 0 : against the alternative that not all of the intercepts are equal.0840 0.1083 0.7494 The coefficient estimates and standard errors from least squares (LS) and fixed effects (FE) estimation are presented in the following table. However.0652 0.1586 0. (iii) The coefficient of LWMFG represents the elasticity of the crime rate with respect to the average weekly wage in the manufacturing sector. LPRBPRIS. However. LPRBCONV. respectively. (f) According to the fixed effects estimates. respectively. In the least squares estimated model with no fixed effects the F and p values for the test are 1. The least squares estimation suggests that a 1% increase the average weekly wage in the manufacturing sector results in a 0. leading us to conclude that there is no evidence of time effects.0160% increase in the crime rate. Exercise Solutions. will be the most effective in dealing with crime.2442. The fixed effects estimation suggests that a 1% increase in average weekly wage will result in a 0. Principles of Econometrics. the explanatory variables which have the expected signs and a significant effect on the crime rate are LPRBARR. LWMFG.Chapter 15. leading us to conclude that there are time effects. those that have the largest effect on crime rate. but lengthening the term of imprisonment is not. Improving policing and court policies that increase the probability of arrest. . On the other hand. An examination of the coefficients for the year dummies in the fixed effects model does show some evidence of a trend effect. Out of these variables. there are dramatic increases in the coefficients for 1986 and 1987. 4e 510 Exercise 15. The coefficients for 1982.0000.5762% decrease in the crime rate. although this estimate is not significantly different from zero. 1983 and 1984 are all small and not significantly different from zero. from a small increase in 1985.10(e) (continued) (e) (ii) The outcome of the test for the joint significance of the dummy variables is different for each of the two models. and this importance is confirmed if we carry out a further test for their inclusion in the model with time effects. Opportunities for higher wages and the avoidance of high-density population areas are also likely to be productive directions for public policy. Since we have established the importance of the county effects. conviction and imprisonment are likely to be effective. and are reasonable to implement as public policy. and so there does not appear to be a trend effect in these early years.118 and 0. the F and p values for the test are 9. our final conclusion is that the least squares test result is misleading and the year effects are important.324 and 0. suggesting an upward trend in the crime rate. LEDNSITY and LPCTYMLE. in the fixed effects model. 0121t (se) (0.00095) The coefficient of t represents the growth rate of GDP. (b) The estimated equation is LY 0.0000. The coefficient of LK suggests that a 1% increase in capital is associated with a 0. expressed in decimal form. However.and p-values of 0.8624% increase in GDP.0983) (0. we do not reject the null hypothesis and conclude that there is no evidence against the hypothesis of constant returns to scale.11 (a) The estimated equation is LY 0. Since this p-value is much larger than the level of significance 0. 1 (constant returns to scale) against the Testing the null hypothesis H 0 : 2 3 1 yields F. (c) Substituting the restriction LY 1 2 2 1 into the model from part (b) yields 3 LK (1 2 ) LL t e LL) t e Rearranging this equation LY LL 1 2 ( LK Converting it into a more familiar form ln(Y ) ln( L) 1 2 ln( K ) yields ln Y L LYL K L 1 2 ln 1 2 LKL t e t e 2 ln( L) t e . The addition of t to the model has very little effect on the estimates of 2 and 3 .21% per year.1351LL 0.1373LL (se) (0. The p-value for testing the significance of this estimate is 0.00488) (0. and so we can conclude that the coefficient is significantly different from zero at a 1% level of significance. alternative hypothesis H1 : 2 3 respectively. they are almost identical to those obtained in part (a).Chapter 15.3787 0. the coefficients represent elasticities. The coefficient of LL suggests that a 1% increase in labor is associated with a 0.00661) (0.00684) Since this is a log-log equation.8624 LK 0.0048) (0. Because it represents the growth rate not attributable to changes in capital and labor.1373% increase in GDP. Exercise Solutions. Principles of Econometrics. 4e 511 EXERCISE 15. These estimates suggest that the average growth rate of GDP over the period 1960-1987 is 1. we may question whether a negative growth rate is a realistic outcome.2995 0.9483.0952) (0. it is often viewed as growth from technological change.0042 and 0.05.8743LK 0. The estimated equation is.81.0415) (0. The calculated value is F = 211.0119 t (se) (0. 4e 512 Exercise 15.5164) (0. The F value can be obtained using the equation F ( SSE R SSEU ) J SSEU ( NT K ) (292.0124) To test H 0 : 1.1333LL 0. The fixed effects model without the restriction for constant returns to scale is the preferred specification.000551) These results are very different from those in part (c).95.46 and 0.1322 The fixed effects estimates are markedly different from those estimated in part (b).000327t (se) (0.8731LKL 0. (f) The estimated equation is LYL 3.13. (e) 1 (constant returns to scale) against the Testing the null hypothesis H 0 : 2 3 1 yields F. It is preferred because.3751 0. we use the usual F-test for testing a set of linear restrictions. However. with the average of the fixed effects reported as the intercept.5435LKL 0. All estimates have the same sign. the coefficient of t has changed sign to positive.82 0.51557 (2296 82 3) 211. the standard errors of this model are much larger with the exception of se ˆ . we should allow for country fixed effects and we do not have any evidence to support the presence of constant returns to scale. The outcome of this hypothesis test is clearly very sensitive to whether or not we include fixed effects.00747 t (0. while the 5% critical value is F(0.00093) against the alternative that not all of the intercepts are equal.0127) (0. the trend coefficient is positive in line with our expectation that technological change should have a positive effect on output. Thus.7529 33.1245 0.1 1.5316LK (se) (0. Exercise Solutions.1030) (0. relative to the estimates in part (c).00476) The estimate of (d) 2 (0.0000.00094) is identical to the estimate obtained in part (b) to two decimal places.and p-values of 107.Chapter 15. we reject H0 and conclude that the country level effects are not all equal. Since this p-value is smaller than the level of significance 0. according to our hypothesis tests.279 .11(c) (continued) The estimated equation is LYL 0.05. the intercept is much larger and the coefficient estimates are much smaller. Furthermore.2 1. alternative hypothesis H1 : 2 3 respectively. more in line with our expectations. . LY 8.4530 0. In particular. we reject the null hypothesis and conclude there are not constant returns to scale. Principles of Econometrics. Also. The elasticity of output with respect to capital is much smaller and the standard errors of both elasticities are much larger.0336) (0. 2211) 1.51557) 81 33. 0569 0.0561 0.2662 0.0562 0.0290 0.0560 0.2958 0.0696 0.0561 0.1784 0.0560 0.0562 0.0829 0.0741 0.2353 0.0584 0.1352 0.0561 D14 D15 D16 D17 D18 D19 D20 D21 D22 D23 D24 D25 D26 D27 D28 0.0255 0.0239 0.2564 0. 4e 513 Exercise 15.0561 0. Exercise Solutions.0560 0.0212 0.0560 0.1375 0.0562 0.2985 0.8741 0. .0617 0.1735 0.2907 0. Using time dummy variables allows the rate of growth between years to be different for each year.0561 0. Principles of Econometrics.Chapter 15.0066 0.0600 0.0396 0.0048 0. Since we have an intercept and then time dummies for all years except the first. Variable Estimate se Variable Estimate se Intercept LK LL D2 D3 D4 D5 D6 D7 D8 D9 D10 D11 D12 D13 0.0561 0.0562 0.2054 0.0560 0.0192 0.0563 0.2980 0.0562 0.0560 0.1024 0.0561 0.0560 0.0563 0.0560 0.0560 0.1566 0.0562 0.0563 The single time trend variable restricts the year-to-year growth rate to be the same between all years.0562 0.11 (continued) (g) The estimates are presented in the following table.1430 0.1011 0. each coefficient of a time dummy gives the growth rate between the year of the time dummy and the first year.0560 0.2850 0. 000843 –0.009953 0.3574)se 7. 6.063114 –0.28) . Pooled Least Squares Estimates Variable Coefficient Std.3574) se 5. 8.007989 0.450940 0.000840 0.03 p-value 0.000 0.38 13.04) Relaxing the assumption that the errors are homoskedastic and that they are uncorrelated. 8.000843 –0.02) ˆ t ˆ (0. we use pooled least squares without cluster-robust standard errors.94. The results are as follows.103035 0.65 p-value 0.001229 –0.0823 1. 4e 514 EXERCISE 15.00 –9.2765 (6.000 0.9606 0.000 are ˆ t ˆ (0.000 0. 6.074821 0. this quantity becomes 100 (b) 3 10 4 3 2 4 EXPER .9606 0.000 0. When 1i 1 and the errors are homoskedastic and uncorrelated. Error t-value C EDUC EXPER EXPER2 HOURS BLACK 0.40.12 (a) The percentage return to experience is WAGE WAGE 100 EXPER LWAGE 100 100 EXPER When EXPER 5 .000412 0.81 –1.12.3574)se 5.001925 0.9606 0.450940 0.028968 The 95% confidence intervals for and 4.Chapter 15.4821 1.44 –4.000 0.99 –0.31 27. we use pooled least squares with cluster-robust standard errors.90 –3.34 –2.134715 0.4821 1.975.003 0.000 0.014922 The 95% confidence intervals for (c) and 7.4887 (4.134715 0.975.5526 (6.074821 0.975.000 are ˆ t ˆ (0.000 0.57) ˆ t ˆ (0. Pooled LS Estimates with Cluster-Robust Standard Errors Variable Coefficient Std.005526 0. Error t-value C EDUC EXPER EXPER2 HOURS BLACK 0. Exercise Solutions.9606 0.002765 0.001229 –0.316 0.54 6. The results are as follows.6100 (3.662 0.061691 0. Principles of Econometrics.0823 1.06 7.975.88.063114 –0.3574)se 7.000323 0. Those with higher ability and greater motivation are likely to have more years of education and to work longer hours. we use random effects estimation. Exercise Solutions.09. the estimated return to education is slightly higher and has similar precision to that from estimation with cluster-robust standard errors.56 13.9606 0.76) ˆ t ˆ (0.005561 0.076867 0.) Random Effects Estimates Variable Coefficient Std.7680 1.Chapter 15.000 are ˆ t ˆ (0.34083 0. and they are omitted.19394 0.000 0. .000695 0. 4e 515 Exercise 15. 5.005390 –0.6867 1. Error t-value C EDUC EXPER EXPER2 HOURS BLACK ˆu ˆe 0.3574)se 7.76 –4. Principles of Econometrics.9606 0.000219 0. and the eit are homoskedastic and uncorrelated.3476 (4. EViews’ standard errors are slightly smaller.45) Compared to the earlier results.61.127097 0. (The standard errors are those computed by Stata.000 0. it is not possible to get fixed effects estimates of their coefficients.98 10.3574) se 4.12(c) (continued) Using cluster-robust standard errors has led to wider confidence intervals for both quantities of interest. Results for Hausman tests on the coefficients separately appear on the next page. (d) When 1i is a random variable with mean 1 and variance u2 .029818 The 95% confidence intervals for and 7.000 0.975.26 p-value 0. 8.000 0.083254 0.629405 0. but not for EXPER and EXPER2. The results are given in the table below.000 0.005496 0. then it is likely that EDUC and HOURS will be correlated with the individual effects.20 –7.001140 –0.059082 –0. The estimated return to experience is much smaller than previously. Ignoring the heteroskedasticity and within-individual correlation leads to an overstatement of the precision with which we are estimating the returns to education and to experience. There is a significant difference between the fixed and random effects estimates of the coefficient of HOURS. (e) If the individual effects capture characteristics such as motivation and ability.5496 (6.975. and is estimated with more precision.62 –5. A noticeable difference between these results and the earlier ones is the larger and now significant estimate for the coefficient of HOURS. Because EDUC and BLACK are time invariant. 000940 0.58 –1.000249 t-value p-value –0. Error t-value p-value C EDUC EXPER EXPER2 HOURS BLACK ˆu ˆe 0.000737 0. Those from EViews yield results with slight differences.058326 –0.000225 0. The results are presented in the table below.825 0.3574)se 11.35) ˆ t ˆ (0. The interval estimate for the return to experience is very similar to that from the random effects estimator. Hausman-Taylor Estimates Variable Coefficient Std.000 0. we use the Hausman-Taylor estimator.975. Exercise Solutions.9606 0.783 0.471 0.009 0.000012 –0.3595 (4.000677 0.215297 0.000 To accommodate the fact that EDUC and HOURS are correlated with the random effects.k bRE .005732 0.0012. Compared to the random effects estimates that did not allow for endogeneity.085 0. The coefficient of BLACK has gone down (in absolute value).2161 (2.12(e) (continued) The overall 2 test yields a value of 2 (3) 15.02.553607 0.110916 0.975.090999 0.042161 0. 4e 516 Exercise 15.9606 4.3574)se 4.94 –8.k bRE .638 0.000 0.35747 0.k se bFE . 19.43) Although the point estimate for the return to education is higher than previous estimates.000 0.8 .63 10.0916 1.221 –3.72 0.19384 The 95% confidence intervals for and are ˆ t ˆ (0.7231 1. we find that the estimated return to education has increased dramatically.k EXPER EXPER2 HOURS (f) –0. We conclude therefore that the random effects estimates and the fixed effects estimates are significantly different.052886 0. the interval estimate is so wide we cannot make any firm conclusion about its value.001440 0.83. with p-value of 0. Other coefficient estimates and their standard errors are similar in magnitude.18 –4.001110 –0.39 2. and hence there is correlation between the 1i and the variables in the model. but its standard error has also increased. but so has its standard error.000053 0.006318 –0. 5.697 0.Chapter 15.) Hausman Test Results bFE . (These results use the random effects standard errors from Stata. . Principles of Econometrics. 31003 .49 6328. Exercise Solutions.855 85.650 419.02969 K i 50627.085 102. i (c) 2 e 7.Chapter 15.325 4333.245 5.790 1971.921 1941.290 608.825 670. 2530.210 70.0099) (0.545 231. They are given by: INV (se) 53.889 55.3825 0.765 121.470 693.13 (a) Fixed effects estimates for the slope coefficients are bFE .437 0.1093V 0.944 0.803 86.0164) .13460Vi 1012549.87 20 50627.160 648.865 149.900 104. Principles of Econometrics.3080 K (25.285 314.124 3.475 42.845 333.945 294.042 20 6328.411 47.3 0.910 Ki 68.020 41.848 61. Vi INV i 6.11013 and bFE .2 The error variance estimate is ˆ (b) 0.698) (0.022 486.29952 2 The sample means are given in the following table.042 .892 1 2 3 4 5 6 7 8 9 10 11 57.85862 (e)&(f) Pooled least squares applied to the transformed variables and random effects estimates of the original equation yield identical results. 4e 517 EXERCISE 15.640 Results from regressing INV i on Vi and Ki are INV i SSE ˆ2 (d) 50.435 297.595 410.942 400.437 Substituting in the estimated values yields 1 2530.49 8 0. 1867) 2 0.434 2.911 1.81 . Thus. agreeing with our a priori expectations.034 0. reflecting the increased precision obtained by allowing for the contemporaneous correlation. For testing the null hypothesis that the errors are uncorrelated against the alternative that 2 they are correlated.870 0.092 0.870 0.t 30 3 t 1 0.0213 r122 ( 0.4506)2 0.021 0.065 0.t eˆ2. suggesting that the prices and income are relevant variables.453 0.232 0.14 (a).108 0.2405 where The 5% critical value for a 2 test with 3 degrees of freedom is 2 (0.017 0.567 1. or in the price elasticity for commodity 3.t eˆ3.0448 r132 ( 0.188 0.217 Constant Price-2 Income 2. 4e 518 EXERCISE 15.3708 ˆ 23 1 30 eˆ2.501 0.468 0.103 All price elasticities are negative and all income elasticities are positive. we obtain a value for the (3) test statistic LM T (r122 r132 r232 ) 30 (0.t eˆ3.964 0.229 1.871 5. Principles of Econometrics. we reject the null hypothesis and conclude that the errors are contemporaneously correlated.999 0.Chapter 15.648 1.1867)2 0.3708 0.t 30 3 t 1 0.0413 r232 ( 0.248 Constant Price-3 Income 4.t 30 3 t 1 0.546 0.3943)2 (0.0448) 2 (0.215 0.3943)2 (0.125 0.530 0.4506) 2 (0. Relative to the least squares estimates.261 1.(b) Least squares and SUR estimates and standard errors for the demand system appear in the following table Estimates Standard Errors Coefficient LS SUR LS SUR Constant Price-1 Income 1.453 1.95. Also.130 0.463 0.867 1.77 ˆ 12 1 30 eˆ1.144 3.136 1. . the SUR estimates for the price elasticities for commodities 1 and 2 have increased (in absolute value) noticeably.3) 7. all elasticity estimates are significantly different from zero.2405) 18. The SUR standard errors are all less than their least squares counterparts.0144 0. There have been no dramatic changes in the income elasticities.0413)2 (0.354 0.0144 ˆ 13 1 30 eˆ1.0213)2 (0. Exercise Solutions. Principles of Econometrics.686 with a p-value of 0.13. The test values are F 1. Thus. 4e 519 Exercise 15.14 or 2 5. Exercise Solutions.895 with a p-value of 0.14 (continued) (c) We wish to test H 0 : 13 1. 23 1. . This test can be performed using an F-test or a 2-test. we do not reject the hypothesis that all income elasticities are equal to 1.Chapter 15. 33 1 against the alternative that at least one income elasticity is not unity. Both are large-sample approximate tests. 8447 (0.1250) 0.4338 (0.7760 (0.7266 (0.1067) 0.3475 (0.1739) 0.2469) 1.2900) 0. Exercise Solutions.1878 (0.0419 (0. Constant Austria LS SUR Belgium LS SUR Canada LS SUR Denmark LS SUR France LS SUR Germany LS SUR ln Y POP ln PMG PGDP ln CAR POP 3.4525) 3.0912) 0.1145 (0. following the usual laws of demand. 4e 520 EXERCISE 15.7939 (0.4019 (0. and the income and price coefficients for Denmark. .1501) 0.1218) 0.0708) 3. Principles of Econometrics.2847) 3.1090) 4.0417 (0.0635) 0.1282) 0.5171 (0.1376) 0.1279) 0.1154) 0.1529) 0.0007 (0.3119) 0.7607 (0.2194) 0.9170 (0.0784) 0.0931) 3.2115) 0.3730) 3.0893) 0. with standard errors in parentheses.1371 (0.0712) 0.1624 (0.1320 (0.3738 (0.0646) 0.7008 (0.0615) 0.2045) 0.3322) 0.2224 (0.1260 (0.6735 (0.2810) 2.2721) 4. Exceptions are the price coefficient in the equation for Belgium.0465) The signs of the coefficients are consistent across countries and estimation techniques. The use of SUR has led to a reduction in the standard errors relative to those for least squares.1092 (0.8451 (0.0773) 0.15 (a) The least squares (LS) and SUR estimates are given in the following table.0928 (0.1702) 1.3924 (0.Chapter 15. An increase in per capita income leads to an increase in gasoline consumption per car.5199 (0.3235) 0.1264) 0.8582 (0.3618 (0.5212 (0.1827) 0.5264 (0.0666) 0.1591) 1.1193 (0.0390 (0.4385 (0. The negative sign for number of cars per capita is likely to occur because each car gets driven less as the number of cars per person increases.0933) 0. An increase in price leads to a decline in gas consumption. although their magnitudes vary considerably.2398) 0.0848) 0.9890 (0.1342 (0.1059) 3.2635 (0.1579) 0. presumably because a higher income leads to more travel and/or the purchase of a bigger car.2368 (0.3036 (0.0751) 0.0433) 0.7932 (0.2043 (0.3629 (0. Most estimated coefficients are significantly different from zero.1226) (0.4826 (0.1943 (0.1131) 0.1920 (0.1671 (0. 12216 eˆ5 eˆ6 1 The degrees of freedom are 6 (6 1) / 2 15 . .11169 0.15686 0. The 5% critical value with 15 degrees of 2 25. Exercise Solutions. Principles of Econometrics.10779 0.Chapter 15.19566 0. We therefore reject the null hypothesis.045 The rij are readily available from the least squares residual correlation matrix presented in the following table.0000. (c) (i) The test statistic value for testing the null hypothesis that corresponding slope coefficients in different equations are equal is 2 686.0000. 4e 521 Exercise 15.52806 1 0. eˆ1 eˆ1 eˆ2 eˆ3 eˆ4 eˆ2 1 eˆ3 eˆ4 eˆ5 eˆ6 0.22654 0.00 .26571 0. Different countries have different slope coefficients. We do not reject the null hypothesis and conclude that there is no freedom is 15 evidence of contemporaneous correlation.92 with a very small p-value of 0.14583 0.22322 0. Thus. We cannot conclude that ln(CAR POP ) is an irrelevant variable in all countries.17883 1 0.17015 0.15 (continued) (b) The test statistic for testing for contemporaneous correlation is LM T M i 1 rij2 i 2 j 1 19 (r122 r132 r142 r152 r162 r232 r242 r252 r262 r342 r352 r362 r452 r462 r562 ) 19. we reject the null hypothesis.56349 1 0. (ii) The test statistic value for testing the null hypothesis that 2 4 0 for all equations is 252.21192 0.03 with a very small p-value of 0.07246 1 0. 424 0. cattle numbers should increase.402 0. the feed situation gets better.509 1.077 8. As rainfall increases.361 0.457 37.194 0.145 1. (b) The three equations should be estimated jointly as a set rather than individually if the errors eit .979 1.946 1.136 22. Alternatively.95. 3) 52.600 45.144 0. The test statistic value is LM T (r122 The 5% critical value for r132 2 12 13 r232 ) 26 23 0 and H1 : at least one covariance is 0.081 0.21 7. Exercise Solutions.104 0.75112 2 (0.465 0. 2. as it would be profitable for the farmers to hold more cattle.271 0. All estimates have the expected signs and are significant at a 5% level. Hence. i 1. and the estimates are generally significant.548 45.309 0.535 0. are contemporaneously correlated. Principles of Econometrics.308 2. As average price increases. . (e) Like the least squares results. if the firm cannot adjust cattle numbers immediately to a desired level.907 0.051 0.351 0. the signs of the SUR coefficients agree with our a priori expectations.16 (a) One would expect all the coefficients to have positive signs.906 0.081 Constant Price-2 Rainfall-2 Lagged cattle-2 17.626 0.135 0. particularly for equation 1.188 0.576 1. The more cattle that are carried on the property in the previous year.265 1.013 1.363 0. a comparison of the magnitudes of the LS and SUR estimates does show some differences. The standard errors for the SUR estimates are uniformly less than those for LS.613 0.368 0.3 .541 0.818 0.706 0. (c) Least squares and SUR estimates and standard errors are given in the table. Estimates Coefficient (d) LS Standard Errors SUR LS SUR Constant Price-1 Rainfall-1 Lagged cattle-1 16. 4e 522 EXERCISE 15.536 0.619 0.084 The relevant hypotheses are H 0 : nonzero. except for the intercepts. we reject H0 and conclude that contemporaneous correlation exists. the magnitudes are feasible.047 Constant Price-3 Rainfall-3 Lagged cattle-3 16.527 42.406 13.87622 0.987 0.126 30. However. and the farmer can run more cattle per acre.662 77.81 . a partial adjustment model might be appropriate.82232 test with 3 degrees of freedom is 0.Chapter 15. in which case the coefficients of lagged cattle numbers would lie between zero and one.319 0. the greater the number likely to be carried in the current year. supporting the theoretical result that the coefficients produced by SUR are more reliable than those from LS. 912.84. (d) In this case the elasticity of substitution estimate is 0. the standard errors suggest more precise estimation from using the generalized least squares method. The 5% critical value for 2(1) is 3.0528 (0.0862 (0.975.0862 and (2) 0. we reject H0 and conclude that a Cobb-Douglas function would not be adequate. and from imposition of the restriction.06063 (0.0405 (0.606 .9124 (0.0415 (0. Principles of Econometrics. Since 2.9047 (0.0384) 0.0687) 0.9483 . (b) For testing H 0 : 12 0 against the alternative H 0 : 2 12 12 0 . This value lies between the two unrestricted generalized least squares estimates obtained in part (c).026 and t(0. standard errors are in parentheses below the estimated coefficients.0465 (0. The estimates of the elasticity of substitution are slightly less than those obtained by least squares. respectively. Hence.9124 (0.9124 1 0.026 . and is closer to the second one that has the smaller standard error. statistic is LM (c) The SUR estimates appear in the second column of the table.0732) 0.Chapter 15. the value of the chi-square 2 T r 20 0. Exercise Solutions.0336) (2) 0.0838) 0.03362 The 5% critical values are t(0.37) 2.0759) 0. (c) and (d) are given in the table below.37) 1 is 2.606 2.025. the standard errors obtained in part (d) are less than those in part (c) with the exception of that for 1. Also. (f) The t value for testing H 0 : t ˆ 1 se( ˆ ) 1 against the alternative H1 : 0.9422 (0.17 Results for parts (a).0731) 0. (e) The standard errors obtained in part (c) are less than their counterparts in part (a). Thus.026 .0480) 0. (a) LS (c) SUR (d) Restricted SUR 0.0541 (0. These values are reasonably close and the magnitudes of their standard errors suggest that they could be two different estimates of the same parameter.0719) 0.417 .0336) 1 2 (a) The separate least squares estimates of the elasticity of substitution are (1) 1.906 16.0696) (1) 1. 4e 523 EXERCISE 15.9483 (0. 2. we reject H0 and conclude that there is contemporaneous correlation between the errors.1047) 0. 0000 K are . (i) Variable Intercept ln(AREAit) ln(LABORit) ln(FERTit) SSE 1it 1 1.447 (43.Chapter 15.4221 (0.298) (7.0000 0.2484 (0.18 (a) The estimates and their standard errors are presented in the following table. and hypothesis tests support the inclusion of both farm effects and year effects.5841 (0.0432) 26.0383) 40.f.f.08242 (b) From the table.0380) 36.0669) 0. The critical values are for a 5% level of significance.0663) 0.2075 (0.514 3.0682) 0.5468 (0. Test (i) versus (ii) (i) versus (iii) (i) versus (iv) (ii) versus (iv) (iii) versus (iv) F-value d. but those in (i) and (iii) are different to those in (ii) and (iv).) column as ( J .0618) 0. Critical F p-value 3. The F-values are calculated using the formula F ( SSE R SSEU ) J SSEU ( NT K ) The values for SSE are provided in the table above.309 5. Principles of Econometrics. we see that the estimates in parts (i) and (iii) are similar.3617 (0.2586 (0.394 1.0703) 0.870 4.0000 0. The results of the tests are given in the table below.420 2. but that they are very sensitive to the assumption made about the intercept changing with farms.2557) 0.4964 (0.036 1.0952 (0. NT K ) in the table below.0000 0.56536 (ii) 1it 1i 0. Exercise Solutions.305) (7.0890 (0.939 8.4328 (0.6243 (0. 4e 524 EXERCISE 15.3263) 0. In the models with fixed effects the reported intercept is the average of the fixed effects. and those in (ii) and (iv) are similar. This suggests that the estimates are not sensitive to assumptions about the intercept changing with time.3759 (0.0415) 23. The values for J and NT given in the degrees of freedom (d.0802) 0.3067) 0.298) (43. (c) The fixed effects model from part (iv) is preferred because it accounts for behavioural differences between the farms and differences over time.0000 0.040 0.2095 (0.2473) 0.298) 1.0755) 0.2412 (0.421 2.66229 (iii) 1it 1t (iv) 1.3352 (0.0640) 0.20311 1it 1it 0.341) (50. In all cases the cluster-robust standard errors lead to wider intervals suggesting that ignoring the within-farmer error correlation and heteroskedasticity can lead to unjustified confidence in the reliability of the estimates. Principles of Econometrics.7729 0.4219 0. Exercise Solutions.1706 –0. 4e 525 Exercise 15. ln(AREA) ln(LABOR) ln(FERT) 95% Interval Estimates Coventional se Cluster-robust se lower upper lower upper 0.1071 0.4429 0.0947 0.3753 0.Chapter 15.2727 .0074 0.18 (continued) (d) The two sets of interval estimates for the elasticities are in the following table.0395 0.4757 0.8267 0. 35802 2 r1995. Variable 1995 1996 1997 Intercept 2.4363 (0.19 (a) The estimates and their standard errors are presented in the following table.1553) 0.1997 To test whether the contemporaneous correlation is significant.5693) 0.4745 (0.1995 . We reject the null hypothesis and conclude that this contemporaneous correlation is significant.1997 ) 0. The calculated value for the F-test is F = 4.1997 ) 1996.5181 (0.Chapter 15.3165 (0.95.0566) 0.175 .2162) 0.0591 (0. It says that the error terms in the three equations.43782 2 r1996.8405) 0. ei . (c) The equality of the elasticities in the different years can be tested using an F-test or a 2test.1996 44 (0. The test statistic is 1996.0353 (0.6. ei . It encompasses the idea that the individual farm behaves in a similar manner over time.2066) 0.1540) 0. Thus. for individual i.1751) 0.95. The calculated value for the 2-test is 2 = 25. Principles of Econometrics.81.9284 (0. 4e 526 EXERCISE 15.4106 (0.1152) 1.1442) 0.1483) ln(AREAit) ln(LABORit) ln(FERTit) (b) The assumption that you make when you estimate the equations in part (a) is that the error terms are correlated across time for each individual.1996 . the corresponding 5% critical value 2 for 6 degrees of freedom is (0.120) hypothesis that all elasticities are the same in all 3 years.1996 cov(ei .6596) 1.13 The 5% critical value for a chi-square distribution with 3 degrees of freedom is 7.1997 0.59 .1997 LM 2 N (r1995. 2.1997 1995.265.1997 ) 1995. we reject the null . are correlated.3037 2 ) 18.5051 (0.1996 0 .3249 (0.59. ei . The contemporaneous correlations in part (a) can be represented as cov(ei . Exercise Solutions.2612 (0.3058 (0.6) 12. we test the null hypothesis H 0 : 1995.1995 .1997 cov(ei .1996 ) 1995. the corresponding 5% critical value is F(0. CHAPTER 16 Exercise Solutions 527 . 2.1318229 | | -1. the estimated increase in the probability of a person choosing automobile transport for a 10 minute (1 unit) increase in DTIME is 0. 21.24 1 . 10. 5. 11. 15. 3. Principles of Econometrics. This problem is inherent in the linear probability model because the marginal effect of the dependent variable on pˆ is constant.7 0 . These are not plausible probabilities.4355781 | | 5.124615 | | 5. .Chapter 16.8511097 | | -8. 12. (b) The predicted probabilities (PHAT) are 1.46 0 . 16.16 0 . 14.4848 0. Therefore. +--------------------------+ | dtime auto phat | |--------------------------| | -4.16 1 . at DTIME = 2 (a 20 minute differential). 18. 13.0277642 | | 4. 4. 6.2626157 | | 9.55 1 .0129) The estimated marginal effect of DTIME on pˆ is constant and does not change with DTIME.18 0 . Exercise Solutions.0714) (0. 8.77 0 -.07 0 -.1 (a) The least squares estimation of the linear probability model is pˆ 0.3118327 | | -3.15 0 . 7.3652682 | | -5.21 1 .1 1 1.0502798 | | 3.7238488 | | 2.0703DTIME (se) (0.4 1 .1754314 | | -.99 1 . 4e 528 EXERCISE 16.122699 | | -9.8475943 | | 3.79 1 .28 1 1.6563513 | | 8.4 1 .945325 | | -4. 20. 19.8356416 | +--------------------------+ Some predicted probabilities are greater than 1 and some less than 0.143792 | | 2.066961 | | -2. 9.44 0 .7125992 | | -6.0703.6809598 | | -7.7 0 .29 0 -.1529159 | | 6.85 0 . 17. 77 0 -. The generalized least squares estimation procedure does not fix the basic deficiency of the linear probability model.16 0 . .4 1 . 20. the percentage of correct predictions is 90.44 0 .16 1 .795579 | +--------------------------+ Using the generalized least squares estimates of the linear probability model. (e) The predicted probabilities (PGLS) are 1.48%.7 0 .6998869 | | 2.6631747 | | -7.4953 0.1853157 | | -9.042934 | | 5. 21.1 1 1.18 0 .79 1 . 12.2033708 | | 2. Exercise Solutions.8058103 | | 3.4 1 . thereby producing correct standard errors for the linear probability model.24 1 .8894657 | | -4. 19.55 1 . 8.00419) Compared to part (a). 5. It only accounts for heteroskedasticity. 15.07 0 -. 2. 4e 529 Exercise 16.1233264 | | 3. (d) False. 17.0565224 | | 4.46 0 . Principles of Econometrics. 3.15 0 .4531334 | | 5.0602DTIME (se)(0.3050813 | | 9.9935836 | | -2.1 (continued) (c) Feasible generalized least squares estimation of the linear probability model yields pˆ 0.29 0 .6421104 | | 8. 11.28 1 .6902574 | | -6. 14.99 1 . 4. 9. 16.0506047 | | 6.3929496 | | -5.0325496 | | -1. these coefficients are similar in magnitude but the standard errors are much smaller. 18.7 0 . It is still possible to predict probabilities that are greater than 1 or less than 0 using generalized least squares.2304535 | | -. 7.8088195 | | -8. 6.21 1 . 13.0333) (0.34721 | | -3. 10. +--------------------------+ | dtime auto pgls | |--------------------------| | -4.Chapter 16.85 0 . 933 | | -8. 11.1 1 .4 1 .9713162 | | -4.79 1 .16 0 .1556731 | | 9.996156 | | 5. 10. Principles of Econometrics. 19.07 0 .9922287 | | -2.True -------Classified | AUTO BUS | Total -----------+--------------------------+----------AUTO | 9 1 | 10 BUS | 1 10 | 11 -----------+--------------------------+----------Total | 10 11 | 21 .0643329 | | 2. 5.55 1 . 4e 530 Exercise 16. 3. 17. 14.8179376 | | -6.Chapter 16.7477868 | | 8.16 1 .48%. Exercise Solutions.85 0 .46 0 . 2. 16. 8. 1.24 1 .18 0 .15 0 .4 1 .0121814 | | 4.21 1 . 9.0831197 | | -. 13. +-------------------------+ | dtime auto pprobit | |-------------------------| | -4.9240018 | +-------------------------+ A classification table is -------.7 0 .0035158 | | -1.282843 | | -5. 6.29 0 .77 0 .44 0 .1 (continued) (f) The percentage of correct predictions using the probit model (PPROBIT) is 90. 20.780102 | | -7.99 1 . 7.7 0 . This is identical to the percentage of correct predictions using the linear probability model.0537665 | | -9. 4. 21.8303455 | | 2. 15.0026736 | | 6.28 1 . 12.3918784 | | 5.931031 | | 3.2111583 | | -3.027532 | | 3. 18. the marginal effect of an increase in DTIME using the logit estimates is dp dDTIME ( 1 2 DTIME ) 2 ( 0. the probability of a person choosing automobile transportation given that DTIME = 3 is ( 1 2 DTIME ) ( 0.16) gives dp dx d (l ) dl dl dx ( 1 2 x) 2 . The differences are primarily a consequence of 2 the variance of the logistic distribution 3 being different to that of the standard normal (1).2376 0.2064) These estimates are quite different from the probit estimates on page 593. The logit estimate 1 is much smaller than the probit estimate.7505) (0. Both the logit and probit estimates suggest that a 10 minute increase in DTIME increases the probability of driving by about 10%.5311 2) 0.5311 0.5311 3) 0. . 4e 531 EXERCISE 16.2376 0. There is little difference in predicted probabilities from the probit and logit models.11) and replacing the standard normal density function with the logistic probability density function (16.5311DTIME 2 DTIME (se) (0.2 (a) The maximum likelihood estimates of the logit model are 1 0. whereas 2 and the standard errors are larger compared to the probit model. where l 1 2 x Given that DTIME = 2. (c) Using the logit estimates.1125 This estimate is only slightly larger than the probit estimate of the marginal effect. (b) Using (16. Principles of Econometrics.7951 The prediction obtained from the probit model is 0.Chapter 16.798.2376 0. Exercise Solutions. 4 1 .8150529 | | -6. 13.48% of the predictions are correct. 17.0486731 | | -9.7 0 . 16.16 1 .21 1 . 21.3522088 | | 5.0063392 | | 6. 19.29 0 .24 1 . 9.4 1 . 18. 12. 6.0287551 | | 3.85 0 .9261805 | | -8.15 0 .07 0 .1283255 | | 9.9623526 | | -4.7 0 .9177834 | +-------------------------+ Using the logit model.55 1 . 20.28 1 .0074261 | | -1.1 1 . 2. 90. Principles of Econometrics.44 0 .9900029 | | 5.1759433 | | -3. 5. 8.0566042 | | 2.18 0 .79 1 .Chapter 16.2422391 | | -5. 14.0708038 | | -. 15.827521 | | 2. 11. Exercise Solutions. 4e Exercise 16.77 0 .2 (continued) (d) The predicted probabilities (PHAT) are 1.99 1 . +-------------------------+ | dtime auto phat | |-------------------------| | -4. 532 .16 0 . 3. 4.7762923 | | -7. 7.9243443 | | 3.9846311 | | -2. 10.46 0 .0161543 | | 4.7423664 | | 8. 506698 -. we find that all coefficients are statistically significant with the exception of those for MATURITY and POINTS.1580211 _cons | -1.057 -. .121) (0.174) (1.0708 0.0838 NETWORTH ) (0. Interval] -------------+---------------------------------------------------------------fixrate | . 4e 533 EXERCISE 16.288) (0.2612297 maturity | -.323) 0. z P>|z| [95% Conf.0887 POINTS 0.199112 ------------------------------------------------------------------------------ or pˆ ( 1.262) (0.6225826 -0.499 FIXRATE 0.279425 1.037854 2.191) (0.24 0.90 0.0822) (0.207128 Number of obs LR chi2(6) Prob > chi2 Pseudo R2 = = = = 78 27.2575 -----------------------------------------------------------------------------adjust | Coef.0379) All the estimates have the expected signs.0591MATURITY 0.20 0.924 -1.877266 4.623) (0. Principles of Econometrics.21 0.241) 0. Exercise Solutions.19 0.46 0. (b) The estimated probit model in tabular form is Probit regression Log likelihood = -39.877 0.0711) All the signs of the estimates are consistent with expectations.083047 -2.0289 NETWORTH (0.013 -.161054 points | -.4309509 .0498) (0.1739101 -2.431MARGIN 2.953644 6.384YIELD (se) (4. The predicted values are between zero and one except those for observations 29 and 48 which are negative.2624758 1.028 -4.132 MARGIN 0.7730246 .214 -.649 -9.0341MATURITY 0. Err.2413875 -1.0591854 .793YIELD (se) (1.083) 0. Ignoring the intercept and using a 5% level of significance and one-tail tests.120677 -0.2999138 .Chapter 16.0001 0.10 0.3 (a) The least squares estimated model is pˆ 0.0157148 1.1731971 networth | .0838286 .0118) (0.300 POINTS (0.0096361 .0900934 yield | -2.48 0.4987284 . Std.383964 1.160 FIXRATE 0.027 .7718083 -.013172 margin | -. at the sample means. 0. Probit model for adjust -------.431MARGIN 2.504 The marginal effect of an increase in MARGIN at the sample means is dp dMARGIN 0.64%.294. POINTS 1. Principles of Econometrics.066.384YIELD 0.877 0. NETWORTH 3.034 using standard normal critical values.25.300 POINTS 0.498. MARGIN MATURITY 2. 4e 534 Exercise 16.Chapter 16.292.0838 NETWORTH 0. The delta-method standard error is 0.0591MATURITY 0.3 (continued) (c) The percentage of correct predictions using the probit model to estimate the probabilities of choosing an adjustable rate mortgage is 75.431 1.499 FIXRATE 0.4 percent. 1. YIELD 1.True -------Classified | Adjust Fixed | Total -----------+--------------------------+----------Adjust | 21 8 | 29 Fixed | 11 38 | 49 -----------+--------------------------+----------Total | 32 46 | 78 (d) The sample means are FIXRATE 13. a one percent increase in the difference between the variable rate and the fixed rate decreases the probability of choosing the variable-rate mortgage by 16.164 This estimate suggests that. .606. Exercise Solutions.058. and a 95% interval estimate of this marginal effect is 0. 815 -.4765344 .000 2.003 .0018099 2.0238FEMALE 0. Students from wealthier families are more likely to have college funds.6937 0. We expect the coefficient of PARCOLL to be positive. Exercise Solutions.0375) 0.or 4-year.0000 0. z P>|z| [95% Conf.2138 -----------------------------------------------------------------------------college | Coef. we expect the coefficient of GRADES to be negative.Chapter 16.72 0. so we expect the coefficient for FAMINC to be positive. however we do not have expectations on the signs of FEMALE and BLACK. students from smaller households are more likely to go to college because larger families might not have enough money to send all family members to college.1015) (0.0274882 -10.0374572 -1.6109 BLACK ) (0.0089404 famsiz | -.4 (a) 77.8% of all high school graduates attended college. .0531059 .0203089 parcoll | .34 0.037431 _cons | 2.1014679 0.0531FAMSIZ (se) (0.1425) Because students with better grades are more likely to be accepted into college.693662 .00539FAMINC 0.249231 ------------------------------------------------------------------------------ or pˆ (2. Similarly.23 0. All coefficients are consistent with our expectations.2945521 .50 0.2406761 faminc | . All coefficients are statistically significant except for FAMSIZ and FEMALE.2946GRADES 0.283459 9. we expect the coefficient of FAMSIZ to be negative. Err.1424817 3.138092 3.42 0.0018457 .184375 1.001 .81 0.2176) (0.005393 .005 .2226661 black | . Std.4765 PARCOLL 0.1750806 .2176202 2.000 -.21967 Number of obs LR chi2(6) Prob > chi2 Pseudo R2 = = = = 1000 226. Interval] -------------+---------------------------------------------------------------grades | -.0275) (0. 4e 535 EXERCISE 16. Principles of Econometrics. Therefore.42 0.2835) (0.7557933 female | .348428 -.98 0. (b) The estimated probit model in tabular form is Probit regression Log likelihood = -416.00181) (0.156 -.1972755 .1265207 .6109028 .0237927 . either 2. 956 For a white male with GRADES = 10 pˆ (2.0531 5 0.4765 1 0.2946 5 0. the probability of attending college for a black female with GRADES = 5.0238 0 0.4 (continued) (c) Using the estimates from (b).6109 1) 0. FAMINC equal to the sample mean.0531 5 0.958 For a white female with GRADES = 10 pˆ (2.39 0.Chapter 16.6937 0.00539 51.990 When this student has GRADES = 10 pˆ (2.0238 0 0.0238 1 0.6937 0.4765 1 0. FAMSIZE = 5 and PARCOLL =1 is pˆ (2.603 (ii) For a white male with GRADES = 5 pˆ (2.00539 51.00539 51.808 (d) (i) For a white female with GRADES = 5 pˆ (2.39 0.2946 5 0.2946 10 0.39 0.4765 1 0.593 .2946 10 0.2946 10 0.39 0.6937 0.00539 51.6937 0. Principles of Econometrics.6109 0) 0.6937 0.6109 0) 0.6109 1) 0.0238 1 0.0531 5 0.0531 5 0. 4e 536 Exercise 16.0238 1 0.39 0.6109 0) 0. Exercise Solutions.0238 1 0.0531 5 0.2946 5 0.6109 0) 0.4765 1 0.4765 1 0.00539 51.00539 51.0531 5 0.39 0.4765 1 0.6937 0. 2938GRADES 0.80 0.48 0.271277 3. Interval] -------------+---------------------------------------------------------------grades | -.4) = 9.1364007 .2664) (0.0106087 famsiz | -.74 0. The test p-value is 0.0000 0. Principles of Econometrics. Err. However. BLACK and FEMALE using a likelihood ratio test yields the test statistic LR = 2(lU lR) = 2( 416.61 The critical chi-squared value at a 0.22 ( 426.000 2.0073394 .527)) = 20.00734FAMINC 0.05 level of significance.95. FAMSIZ becomes statistically significant using a one tailed test and a 0.05 level of significance is 2 (0.0001 .00167) (0.793474 .315671 ------------------------------------------------------------------------------ or pˆ (2.34 0.0368791 -1. Std.2430718 faminc | . All coefficients remain significant.000 -. Since the test statistic is greater than the critical value.3446186 -. excluding PARCOLL.2664321 10.49.40 0. z P>|z| [95% Conf. BLACK and FEMALE in tabular form is Probit regression Number of obs LR chi2(3) Prob > chi2 Pseudo R2 Log likelihood = -426. BLACK and FEMALE are jointly significant and should be included in the model. (f) Testing the joint significance of PARCOLL. Exercise Solutions.7935 0.1944 -----------------------------------------------------------------------------college | Coef.4 (continued) (e) The estimated model.0081627 _cons | 2.064119 .Chapter 16.0259053 -11. 4e 537 Exercise 16.000 .001668 4.0259) (0.2938452 .0040701 . we reject the null hypothesis and conclude that PARCOLL.0641FAMSIZ ) (se) (0.0369) The signs of the remaining variables are unaffected.082 -.52689 = = = = 1000 205. 0834497 . All estimates are statistically significant except for the coefficient of FAMSIZ.000 1.53 0. Students with better grades are more likely to be accepted into a 4-year college. Similarly.0858997 _cons | 1.2135331 0.99% of 4-year college students are female and 5.63 0.88% are black.0122372 . z P>|z| [95% Conf.0168619 1. z P>|z| [95% Conf.39 0.8360657 .0674716 . (b) The estimated probit model in tabular form is Probit regression Log likelihood = -429.217 -.Chapter 16.321768 Number of obs LR chi2(3) Prob > chi2 Pseudo R2 = = = = 44 22.07 0. Students from wealthier families are more likely to have college funds.581626 .814 -. students from smaller households are more likely to go to college because larger families might not have enough money to send all the family members to college.24 0.696 -.0538604 famsiz | .1799161 faminc | .30 0.0014891 3.74% of high school graduates who attended college chose a 4-year college. Interval] -------------+---------------------------------------------------------------grades | -. 4e 538 EXERCISE 16. so we expect the coefficient for FAMINC to be positive.000 . We expected the coefficient of FAMSIZ to be negative.5 (a) 67. Exercise Solutions.0092141 .114599 3.000 -.2762361 -3.0052572 .3350676 .4263 -----------------------------------------------------------------------------fouryr | Coef.377478 -.114219 2.69285 Number of obs LR chi2(3) Prob > chi2 Pseudo R2 = = = = 778 119.002 2. Std. Interval] -------------+---------------------------------------------------------------grades | -.68213 ------------------------------------------------------------------------------ . Err.03 0. 51. Err.09 0.1217 -----------------------------------------------------------------------------fouryr | Coef.002 -1.2279592 .0208116 .0391261 0.537597 2. Principles of Econometrics.049033 ------------------------------------------------------------------------------ The estimate signs are as expected.0023386 .2384773 6.2760023 -.77 0.0000 0.0000 0.2946529 faminc | .39306 10. therefore we expect the coefficient of GRADES to be negative.0245122 -9.0081757 famsiz | .23 0.501967 _cons | 6. Std. (c) The estimated probit model for black students is Probit regression Log likelihood = -15. 002) famsiz 0.Chapter 16.085 chi2 119.000 .990102 ------------------------------------------------------------------------------ There are large differences between the coefficients of the two models.717 -----------------------------------------------------------Standard errors in parentheses * p<0.0985247 _cons | 1.005*** (0.21 0. ** p<0.231*** (0.039) 0. All coefficient estimates are significant with the exception of FAMSIZ.001 .000 -.0000 0.08529 = = = = 734 112.59 0.005*** (0.512805 .0194625 .083 (0.693 -15. 4e 539 Exercise 16.72 0. The table below summarizes the results Probit models -----------------------------------------------------------(1) (2) (3) full sample black only white only -----------------------------------------------------------grades -0.0054343 .48 0.040) _cons 1.019 (0.513*** (0.769 112.0024683 .001) 0.228*** -0.025459 -9.01.009 (0. Err.000 1.1219 -----------------------------------------------------------------------------fouryr | Coef.1807918 faminc | .035509 1.0403386 0. *** p<0. z P>|z| [95% Conf.06 0.025) faminc 0.0595997 . Exercise Solutions.836** -0.5(c) (continued) The estimated probit model for white students is Probit regression Number of obs LR chi2(3) Prob > chi2 Pseudo R2 Log likelihood = -406.322 -406.2306905 .025) (0.021 (0.074 22. Given identical values for the variables.214) 0. Interval] -------------+---------------------------------------------------------------grades | -.629 -. Std.017) 0.244) -----------------------------------------------------------N 778 44 734 ll -429.05.538** 1. Principles of Econometrics.0015133 3.2435231 6.582*** 6.2805892 -.276) (0.115) (0. the effect of unit changes in both GRADES and FAMINC on the probability of attending a 4-year college is larger for black students than it is for white students.0084004 famsiz | .238) (2. 4e 540 EXERCISE 16. z P>|z| [95% Conf.4079379 3.1949887 0.80 0.60 0.4171899 -.94 0.5124453 .549 -. Exercise Solutions.0038319 5.607127 . Interval] -------------+---------------------------------------------------------------2 | grades | -.937035 .002 .0195931 female | . if they did not attend college 2.4295461 1.1932327 -0.4744651 10.186 -.0118943 . if they attend a 2-year college 3.962637 .489 -. Principles of Econometrics. Err.0204678 .40667 _cons | 4.807583 2.0552152 -5.60 0.000 4.4991191 black | .0041956 .1688 -----------------------------------------------------------------------------psechoice | Coef.46 0.2007503 faminc | .7272638 .6 (a) The probabilities of this multinomial logit model are pˆ i1 1 1 exp( zi 2 ) exp( zi 3 ) pˆ i 2 exp( zi 2 ) 1 exp( zi 2 ) exp( zi 3 ) pˆ i 3 exp( zi 3 ) 1 exp( zi 2 ) exp( zi 3 ) where j 1.899642 -------------+---------------------------------------------------------------3 | grades | -.6161929 faminc | .000 .000 -.32 0.03 0.2652224 .000 .0129574 .003928 3.491135 3.0000 0.83 0.34 0.892571 -----------------------------------------------------------------------------(psechoice==1 is the base outcome) . if they attend a 4-year college and zi 2 12 22 GRADESi 32 FAMINCi 42 zi 3 13 23 FEMALEi 52 GRADESi 33 FAMINCi 43 FEMALEi 53 BLACKi BLACK i The estimates for this multinomial logit model are presented in the following table Multinomial logistic regression Log likelihood = -846.1337162 .1169483 .032702 5. Std.2739136 1.9744285 2.0279781 female | -.5679813 .409876 _cons | 1.8383346 -.000 -.0566698 -12.3089701 .Chapter 16.000 .94 0.2450128 black | 1.75602 Number of obs LR chi2(8) Prob > chi2 Pseudo R2 = = = = 1000 343.69 0. 6(a) (continued) The p-values in this table suggest that all estimated coefficients are significant at a 5% level of significance except for 42.5239 0. This value is calculated as pˆ i 3 pˆ i 3 0.2081 .5239 The probability that a white male with median values of GRADES (6.5239.5) attending a 4-year college rather than not attending any college is 2.0205 42.5 0. Principles of Econometrics.5680 0 0.905 for GRADES is calculated by finding zi 2 and zi 3 .1170 0 0.7320 Therefore.64) and FAMINC (42.9370 -0.5 -0.64 0.2081. the increase in the probability of attending a 4-year college of a white male with median FAMINC whose GRADES change from 6.7278 Therefore the probability ratio of a white male with median values of GRADES (6. (b) This probability is calculated by firstly finding zi 2 and zi 3 zi 2 1.9626 -0.0035) 2.1337 0 1.Chapter 16. Exercise Solutions. (d) The probability that a white male with median FAMINC and a value of 4.905 0.1921 exp(1.64) and FAMINC (42.1170 0 0. zi 2 1. (c) The probability ratio value is calculated using an expression analogous to (16.7273 6.0119 42.905 0.9370 -0.0035 Therefore.5 -0.64 0.3090 4. These three coefficients correspond to the variables FEMALE and BLACK.905 is 0.1337 0 1.21) of the text pˆ i 3 pˆ i1 exp( zi 3 ) Using the value of zi 3 from part (b) the probability ratio is pˆ i 3 pˆ i1 0.0205 42.0119 42.5 0.7320 0. pˆ i 3 exp( zi 3 ) 1 exp( zi 2 ) exp( zi 3 ) 0. 4e 541 Exercise 16.5) will attend a 4-year college is 0.3090 6.6071 0 1.5239 0.6071 0 2.7273 4.5680 0 0.64 to 4.73 to one. 52 and 43.3910 zi 3 4.9270 zi 3 4.2653 then pˆ i 3 exp( zi 3 ) 1 exp( zi 2 ) exp( zi 3 ) 0.9626 -0. 000 4. z P>|z| [95% Conf. .26 0.0182128 .5561 = = = = 749 291.208393 _cons | 5.5508474 2.0616125 -11. Principles of Econometrics.0258541 female | -.8479787 -.34 0.001 .504986 10.Chapter 16.04 0. 4e 542 Exercise 16.000 . Interval] -------------+---------------------------------------------------------------grades | -.000 -.079889 6.42 and the median family income is $42.7272205 .5304858 .0105716 .0000 0.6064622 faminc | .2677932 black | 1. Std.519 -.1313463 . Exercise Solutions.759 where the median value of grades for this sample is 6.2036463 -0. Err.64 0.80 0.500.6 (continued) (e) The estimated logit model is Logistic regression Log likelihood = Number of obs LR chi2(4) Prob > chi2 Pseudo R2 -309.67 0.422851 3.734 0.059397 ------------------------------------------------------------------------------ The probability ratio that a white male student with median will attend a 4-year college rather than not attend any college is pˆ 1 pˆ 0.3200 -----------------------------------------------------------------------------fouryr | Coef.266 2.0038987 4.37962 .069643 . if Pepsi if 7-Up if Coke and zi1 2 PRICEi1 zi 2 2 PRICEi 2 3 DISPLAYi 2 4 zi 3 2 PRICEi 3 3 DISPLAYi 3 4 3 DISPLAYi1 4 FEATUREi1 FEATUREi 2 FEATUREi 3 The estimates for this conditional logit model are presented in the following table Conditional logistic regression Log likelihood = -1822. and that 4 is not significantly significant. at a 0.097115 -1. Std. The coefficient of 3 (DISPLAY) is positive.70 0. .000 -2.167278 . 2.000 .1460705 ------------------------------------------------------------------------------ The coefficient 2 (PRICE) is negative. The coefficient 4 (FEATURE) is negative suggesting that being “featured” decreases the brand’s probability of being bought. The signs of 2 and 3 are as expected.744454 .0896 -----------------------------------------------------------------------------choice | Coef.Chapter 16.0000 0.1799323 -9.0930481 4.2800767 .13 0. z P>|z| [95% Conf. Interval] -------------+---------------------------------------------------------------price | -1.894 -. which implies that displaying the brand increases its probability of being bought. 4e 543 EXERCISE 16.05 level of significance. Principles of Econometrics. The p-values suggest that 2 and 3 are statistically significant. which suggests that an increase in own price decreases the brand’s probability of being bought.97 0.7 (a) The probabilities of this conditional logit model are pˆ i1 exp( zi1 ) exp( zi1 ) exp( zi 2 ) exp( zi 3 ) pˆ i 2 exp( zi 2 ) exp( zi1 ) exp( zi 2 ) exp( zi 3 ) pˆ i 3 exp( zi 3 ) exp( zi1 ) exp( zi 2 ) exp( zi 3 ) where j 1. 3.89 0.2267 Number of obs LR chi2(3) Prob > chi2 Pseudo R2 = = = = 5466 358. however we would expect the sign of 4 to be positive. Err.391793 display | .4624476 .0106038 .0799373 -0. Exercise Solutions.6448185 feature | -. 25 and there is no display or feature.538 exp( zi 2 ) zi1 1.33 0. This value is calculated as pˆ i 3 pˆ i1 0.7445 1. Exercise Solutions.2787 exp( zi 3 ) 1.8053 Adding the alternative specific intercept yields the following conditional logit model specifications and estimates pˆ i1 exp( zi1 ) exp( zi1 ) exp( zi 2 ) exp( zi 3 ) pˆ i 2 exp( zi 2 ) exp( zi1 ) exp( zi 2 ) exp( zi 3 ) pˆ i 3 exp( zi 3 ) exp( zi1 ) exp( zi 2 ) exp( zi 3 ) .2787 exp( zi 3 ) 1.4624 1 0.0106 0 1.4624 0 0.33 1.7 (continued) (b) The probability ratio of choosing Coke relative to Pepsi and 7-Up. if the price for each is $1.1806 zi 3 1.7445 1. These changes are calculated as exp( zij ) exp( zij ) exp( zi1 ) exp( zi 2 ) exp( zi 3 ) exp( zi1 ) exp( zi 2 ) exp( zi 3 ) pˆ ij pˆ ij zi1 zi1 . 0.33 In this scenario the alternatives are equally likely.4624 1 0. are 1. Coke is on display and there is feature.538.538 exp( zi1 ) pˆ i 3 pˆ i 2 0.2894 compared to 0. the probability of choosing either Pepsi or 7up is 0.0106 0 2.25 0. This value is calculated as pˆ i 3 pˆ i1 pˆ i 3 pˆ i 2 0.0106 0 2.4212 compared to 0. Principles of Econometrics.Chapter 16.4624 0 0.0214.4426 0.7445 1.30 0.0106 0 1.25.33 1 0. so their choice probabilities are equal. is 1.2787 in part (c).7181 where (d) Under this scenario.1806 zi 2 1. The probability of choosing Coke is 0. from (c) where zi 3 (e) 1. a decrease of 0. 4e 544 Exercise 16. from (c) zi 2 zi 2 .0107.25 0.4426 0. (c) The probability ratio of choosing Coke relative to Pepsi and 7-Up. a change of +0.7445 1. if the price for each is $1.25 0.4426 in part (c). 25 0.218952 -1.25 and a display for Coke is present.3983 Pr(choice = Pepsi) = .8492 1.42 0.2160352 ------------------------------------------------------------------------------ If the price of each is $1. exp( zi1 ) pˆ i 3 pˆ i 2 0. Exercise Solutions.849186 .0935445 5.0409 0 2.1614722 .3299 Pr(choice = SevenUp) = .3764 Pr(choice = Pepsi) = .156 -.4067008 sevenup | .21.47.0639666 1.2841 1.3983 0. z P>|z| [95% Conf.4727 0 0.05 0.4727 1 0.2036821 .2893346 .0000 0.0625595 4.4726785 .3983 0.4727 0 0.1219669 display | . 4e 545 Exercise 16.49 0. Principles of Econometrics.8492 1.0409 0 2.000 -2. Coke display) the probabilities are: Pr(choice = Coke) = . Std.0409 0 2.3115 Under the first scenario (all prices $1.54 0.0408576 .47942 feature | -. Interval] -------------+---------------------------------------------------------------price | -1.25 0.7(e) (continued) where zi1 11 2 PRICEi1 zi 2 12 2 zi 3 2 PRICEi 2 PRICEi 3 3 3 DISPLAYi1 3 4 DISPLAYi 2 DISPLAYi 3 4 FEATUREi1 4 FEATUREi 2 FEATUREi 3 The estimation results are Conditional logistic regression Number of obs LR chi2(5) Prob > chi2 Pseudo R2 Log likelihood = -1811.25 0.6560225 pepsi | .0906629 . the odds of choosing Coke relative to Pepsi is 1.2718 Under the second scenario (Coke price increase to $1.21 and the odds of choosing Coke relative to 7-Up is 1.000 .3419 Pr(choice = SevenUp) = .3543 = = = = 5466 380.623 -.1886595 -9. Err.3299 exp( zi 3 ) 1.47 exp( zi 2 ) where z i1 0.25.0907 1.63 0.Chapter 16.8492 1.80 0.0951 -----------------------------------------------------------------------------choice | Coef.2718 exp( zi 3 ) 1. These odds are calculated as pˆ i 3 pˆ i1 0.0274 zi 2 0.0347094 .2840865 .0830752 -0.2818 .000 .2208 zi 3 (f) 1.30) the probabilities are Pr(choice = Coke) = . 0052525 .03 0.9456 0.86473 = = = = 1000 357.3066 6.5215 Recomputing these probabilities when GRADES = 4.7131312 .3349328 -.0900 ( 0. y 1 .1815 The probability that a student with median GRADES chooses to attend a 2-year college is P[ y 2] 2.4236226 .3066 4. and the probability of not going to college to decrease. the probability that a student with median GRADES (6. z P>|z| [95% Conf. Interval] -------------+---------------------------------------------------------------grades | -.3066 4.2045863 -2.1971 .1755 -----------------------------------------------------------------------------psechoice | Coef.3066 6.0900 ( 0.1971365 -2.0301846 -0.1767871 4.8 (a) Using the estimates in Table 16.2952923 . for students with better grades.0000 0.64) 2.905 0. Err.694591 .2970 The probability that a student with median GRADES chooses to attend a 4-year college is P[ y 3] 1 2.0202251 -14.2046 2 1.424 -. (b) The ordered probit estimates are Ordered probit regression Number of obs LR chi2(5) Prob > chi2 Pseudo R2 Log likelihood = -839.0832822 .64) 0.080971 -1. is P[ y 1] 2.Chapter 16.5.595845 .9456 0.80 0.194864 /cut2 | -1.905 0.3666348 1.6946 se( 2 ) 0.996827 -2.3066 6.0900+0.0078435 famsiz | -.905 2] P[ y 3] 1 2.0350391 black | .0900+0.30821 ------------------------------------------------------------------------------ where 1 2.2244071 . Std.3066 4. 4e 546 EXERCISE 16.905 yields P[ y 1] 2.9456 ( 0.905 P[ y 2.000 -.7211 These results are as anticipated since we expect the probability of going to a 4-year college to increase.97 0.000 . Exercise Solutions.5958 se( 1 ) 0.000 .9456 ( 0.3066 4.3066 6.60 0.1016424 4. Principles of Econometrics.2042 0.64) will choose no college.6228381 -------------+---------------------------------------------------------------/cut1 | -2.2556518 faminc | .0026615 .0747 2.0241215 .64) 0.59 0.000 .059628 parcoll | .001322 3.17 0.64) 0. Interval] -------------+---------------------------------------------------------------grades | .3935 (mean) famsiz = 4.0709968 .0357634 .00525 FAMINC 0.42 0.53039 (mean) faminc = 51.t. Principles of Econometrics.0202) (0.2953GRADES 0.206 (mean) black = .0053038 8.r. Err.53039 (mean) faminc = 51.7131BLACK 0. predict(outcome(1)) dy/dx w.0565541 ------------------------------------------------------------------------------ .308 (mean) -----------------------------------------------------------------------------| Delta-method | dy/dx Std. predict(outcome(2)) dy/dx w.056 (mean) parcoll = .1) are Expression : Pr(psechoice==1). Err.0302) 0.3935 (mean) famsiz = 4.308 (mean) -----------------------------------------------------------------------------| Delta-method | dy/dx Std.000 . 4e 547 Exercise 16. z P>|z| [95% Conf.70 0.0606261 .0241FAMSIZ (se)(0.000 .00132) (0.r.4236 PARCOLL (0.8(b) (continued) If zˆ 0.206 (mean) black = . : grades at : grades = 6.1016) then to compute probabilities we use P[ y 1] ( P[ y ( 2] P[ y 3] 1 zˆ ) 1 2 ( zˆ ) 2 ( 1 zˆ ) zˆ ) The marginal effects (evaluated at the means and using Stata 11.0461587 .1768) (0. : grades at : grades = 6.Chapter 16. z P>|z| [95% Conf.0052913 13. Exercise Solutions.0813676 -----------------------------------------------------------------------------Expression : Pr(psechoice==2).t. Interval] -------------+---------------------------------------------------------------grades | .056 (mean) parcoll = . 4e 548 Exercise 16. (c) Testing the joint significance of FAMINC.2953 6.64 is P[ y 3] 1 1.49.000 income will attend a 4-year college when (i) GRADES = 6.82) 71. the probability of choosing a 4-year college decreases and the probability of choosing no college increases.1014807 These estimates suggest that as the student’s grades improve. FAMSIZ.86 ( 875. or the family income increases. Exercise Solutions. (d) The probability that a black student from a household of 4 members with $52.0079975 -14.3935 (mean) famsiz = 4. The p-values of these estimates indicate that all variables are statistically significant at a 0.64 0. As the family size increases.206 (mean) black = . Since the test statistic is greater than the critical value. Interval] -------------+---------------------------------------------------------------grades | -.t.Chapter 16. has a higher probability of choosing a 4-year college and a lower probability of choosing no college.05 level of significance with the exception of FAMSIZE.1171555 .6946 ( 0. Err.0241 4 0. : grades at : grades = 6. a black student.000 -. FAMSIZ.4) = 9. Also. PARCOLL and BLACK using a likelihood ratio test yields the test statistic LR 2(lU lR ) 2 839.65 0.8525 .1328304 -. predict(outcome(3)) dy/dx w. or a student whose parent/s graduated from college or has an advanced degree.53039 (mean) faminc = 51.91 The critical chi-squared value at a 0. the probability of choosing a 4-year college increases but the probability of choosing no college decreases.95. PARCOLL and BLACK are jointly significant and should be included in the model. we reject the null hypothesis and conclude that FAMINC.308 (mean) -----------------------------------------------------------------------------| Delta-method | dy/dx Std.7131 1 0.r.00525 52 0. Principles of Econometrics.4236 1) 0.05 level of significance is 2 (0. z P>|z| [95% Conf.8(b) (continued) Expression : Pr(psechoice==3).056 (mean) parcoll = . 00525 52 0.9406 The probability that a non-black student from a household of 4 members with $52.6946 ( 0.7131 1 0.2953 6.8013 Given values of FAMINC.905 0.905 is P[ y 3] 1 1.6946 ( 0.00525 52 0.0241 4 0. 4e 549 Exercise 16.2953 4.905 is P[ y 3] 1 1. Exercise Solutions.4236 1) 0. FAMSIZ and PARCOLL. Also. at a given value of GRADES we find that the probability of a black student going to a 4-year college is higher than a non-black student.905 0.0241 4 0.6309 GRADES = 4.000 income will attend a 4-year college when (i) GRADES = 6. .7131 0 0. Principles of Econometrics.64 0.6946 ( 0.64 is P[ y 3] 1 1. we find that the probability of a black or a non-black student attending a 4-year college increases as their value of GRADES decreases.00525 52 0.4236 1) (ii) 0.2953 4.8(d) (continued) (ii) GRADES = 4.4236 1) (e) 0.0241 4 0.Chapter 16.7131 0 0. 6.8875 0.000 .Chapter 16.5964 -----------------------------------------------------------------------------medaltot | Coef.000 -15.2460392 lgdp | . E [ MEDALTOTCanada ] exp 15.0000 0.89526 -14.93 0. The most noticeable difference is that these standard errors are smaller than those in Table 16.1800ln(16.19 1011 ) 15. .64 0. (c) The estimates are presented in the following table Poisson regression Log likelihood = -1278. Principles of Econometrics. Std.4853 Number of obs LR chi2(2) Prob > chi2 Pseudo R2 = = = = 357 3778.12 0.000 .3238893 -47. z P>|z| [95% Conf. 4e 550 EXERCISE 16.51.59.62564 ------------------------------------------------------------------------------ These estimates and standard errors are very similar to those in Table 16. Exercise Solutions.26045 .5 106 ) 0.9 106 ) 0.5766ln(3. (b) The Poisson regression predicts that Canada won 16 medals in the 1988 Olympics.5360921 .0154764 34. Err.9 (a) The Poisson regression predicts that Australia won 10 medals in the 1988 Olympics.0 1011 ) 10.5766ln(5.8875 0.5664252 _cons | -15.2054832 .41 The probability that Australia would win 10 medals or more in 1988 is 0. Interval] -------------+---------------------------------------------------------------lpop | .59 The probability that Canada would win 15 medals or less in 1988 is 0.1649272 .0206922 9.6.1800ln(26. This value is calculated as E [ MEDALTOTAust ] exp 15.505759 .13 0. 919205 2. All variables are statistically significant at a 5% level of significance except for HOST. Interval] -------------+---------------------------------------------------------------lpop | .8558054 host | .83 0.3581296 _cons | -15. Countries with a larger GDP have more money to spend on sports technology and training.3913984 . The model which includes SOVIET is preferred because it has a higher log-likelihood value.96 0.000 .000 . Host countries have the advantage of being acclimatized. so we expect the coefficient of ln(POP) to be positive. (e) Estimates for the Poisson regression model that adds HOST and PLANNED are: Poisson regression Log likelihood = -1265.0000 0.091907 .0360083 .Chapter 16.1382 Number of obs LR chi2(4) Prob > chi2 Pseudo R2 = = = = 357 4162.1005053 1.3901 Number of obs LR chi2(4) Prob > chi2 Pseudo R2 = = = = 357 3804.5283541 .1521344 .48198 ------------------------------------------------------------------------------ The signs of the all coefficients are as expected. Std.5625461 .6571 -----------------------------------------------------------------------------medaltot | Coef.9 (continued) (d) Estimates for the Poisson regression model that adds HOST and SOVIET are: Poisson regression Log likelihood = -1086.51544 -14. Exercise Solutions.0174452 32.5967381 planned | .5975835 soviet | 2.26 0.000 -15.000 1.1005472 1.3474621 -43. z P>|z| [95% Conf.60 0.000 .84163 .6005 -----------------------------------------------------------------------------medaltot | Coef. Countries with a larger population have a greater pool of talent. so we expect the coefficient of ln(GDP) to be positive.83 0.109 -. and having the home crowd.0774775 . Std. Therefore we expect the coefficient of HOST to be positive. z P>|z| [95% Conf.0222411 6.1957263 lgdp | . Former Soviet Union countries win more medals than the average country.248088 host | .0000 0. 4e 551 Exercise 16. .6236019 .1184734 5.5304936 .17 0.84 0.25 0.16782 ------------------------------------------------------------------------------ All estimates and standard errors are similar to those in part (d).16299 . Err.0171151 32.1195092 .0839005 24.844 -14.3437894 -43.32 0.1876441 lgdp | .1610607 .64 0. being familiar with the sporting facilities.0244232 5.1397755 . Err.234 -. Interval] -------------+---------------------------------------------------------------lpop | .1085426 .19 0.000 -15. therefore we expect that the coefficient of SOVIET will be positive.5640386 .72 0.000 . Principles of Econometrics.000 .3164959 _cons | -14.083646 . 41256 1011 ) 0.528 E[ MEDALTOTCanada ] exp 14.071 106 ) 0.6236 0 17. These predictions were calculated as E[ MEDALTOTAust ] exp 14.5625ln(6.1195 1 0. Australia would win 13 medals and Canada would win 17 medals. .1195 0 0.1398ln(30.5625ln(3.495 The prediction for Canada is reasonably close to the actual value. Principles of Econometrics.8416 0.8416 0.6236 0 12.1398ln(19.22224 1011 ) 0. Exercise Solutions. in 2000. 4e 552 Exercise 16. That for Australia is a long way from the actual value.Chapter 16.9 (continued) (f) The Poisson regression model from part (e) predicts that.689 106 ) 0. Chapter 16, Exercise Solutions, Principles of Econometrics, 4e 553 EXERCISE 16.10 Figure xr16.10(a) shows the histogram of the variable SHARE 0 100 Frequency 200 300 400 (a) 0 .05 .1 share .15 .2 Figure xr16.10(a) Histogram of SHARE There is a large number of observations at SHARE = 0; specifically, 61.96% of the observations are zero. This value can be classified as the limit value. The variable is an example of censored data. (b) The least squares estimated model is Source | SS df MS -------------+-----------------------------Model | .075988344 4 .018997086 Residual | .096739617 503 .000192325 -------------+-----------------------------Total | .172727961 507 .000340686 Number of obs F( 4, 503) Prob > F R-squared Adj R-squared Root MSE = = = = = = 508 98.78 0.0000 0.4399 0.4355 .01387 -----------------------------------------------------------------------------share | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------lpop | -.0002036 .0004342 -0.47 0.639 -.0010568 .0006495 lgdp | .0033415 .0003983 8.39 0.000 .0025589 .0041241 host | .044673 .0081264 5.50 0.000 .0287072 .0606388 soviet | .0555981 .0044663 12.45 0.000 .0468232 .0643729 _cons | -.0694966 .0061669 -11.27 0.000 -.0816127 -.0573805 ------------------------------------------------------------------------------ (i) The coefficients of ln(GDP), HOST and SOVIET have the expected signs and these variables are statistically significant at a 0.05 level of significance. The coefficient of ln(POP) does not have the expected sign and is not statistically significant. However, we must be careful when interpreting these coefficients because we are using censored data. This data yields least squares coefficients that are biased and inconsistent. Chapter 16, Exercise Solutions, Principles of Econometrics, 4e 554 Exercise 16.10(b) (continued) (b) (ii) A plot of the residuals against ln(GDP) is shown in Figure xr16.10(b). The residuals do not appear random. Where ln(GDP) is less than 21, all residuals are positive and seem to follow a decreasing linear trend. Where ln(GDP) is greater than 21 the majority of residuals are negative and appear to continue along the decreasing linear trend. Also, the variance of the residuals increases greatly as ln(GDP) increases. .12 residuals .08 .04 .00 -.04 -.08 16 18 20 22 24 26 28 30 ln(GDP) Figure xr16.10(b) A scatter plot of residuals versus ln(GDP) (iii) The skewness and kurtosis of the residuals is 3.64 and 27.15 respectively. These values are very different to the skewness and kurtosis of the normal distribution, which are 0 and 3 respectively. A Jarque-Bera test for normality on the residuals rejects the null hypothesis at a 0.01 level of significance. (c) Based on the estimates in part (b), it is predicted that Australia’s share of the Olympic medals in 2000 would be 0.060 and Canada’s share would be 0.018. The actual shares of medals won were 0.062 and 0.015 for Australia and Canada respectively. Our predictions are very close to the actual values. The predicted shares are calculated as SHARE CANADA 0.0695 0.000204ln(30.689 106 ) 0.003341ln(6.41256 1011 ) 0.04467 0 0.05560 0 0.018 SHARE AUST 0.0695 0.000204ln(19.071 106 ) 0.003341ln(3.22224 1011 ) 0.04467 1 0.05560 0 0.060 Chapter 16, Exercise Solutions, Principles of Econometrics, 4e 555 Exercise 16.10 (continued) (d) The estimated Tobit model: Tobit regression Log likelihood = Number of obs LR chi2(4) Prob > chi2 Pseudo R2 382.50206 = = = = 508 340.73 0.0000 -0.8031 -----------------------------------------------------------------------------share | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------lpop | .0012241 .0009766 1.25 0.211 -.0006947 .0031428 lgdp | .0086398 .0008358 10.34 0.000 .0069978 .0102818 host | .0366962 .0124336 2.95 0.003 .0122682 .0611242 soviet | .0625879 .0068415 9.15 0.000 .0491466 .0760292 _cons | -.2339405 .0159412 -14.68 0.000 -.2652599 -.202621 -------------+---------------------------------------------------------------/sigma | .0211255 .0010714 .0190206 .0232304 -----------------------------------------------------------------------------Obs. summary: 314 left-censored observations at share<=0 194 uncensored observations Comparing the Tobit estimates to the least squares estimates, the coefficient of ln(POP) has a different sign and is still statistically insignificant, the coefficient of ln(GDP) is larger, and the coefficients of HOST and SOVIET are similar. (e) The predicted shares of medals won in the 2000 Olympics using the Tobit model are 0.053 and 0.028 for Australia and Canada respectively. Compared to the predictions in part (c), these predicted shares are not closer to the true shares Note that the calculations for these predictions require us to use an expression like (16.40) but specific to this model. The expression used is E[ SHAREi | SHAREi 1 2 0] ln( POPi ) 3 ln(GDPi ) 4 HOSTi 5 ( 1 2 ln( POPi ) 3 ln(GDPi ) 4 ( 1 2 ln( POPi ) 3 ln(GDPi ) 4 SOVIET HOSTi 5 SOVIETi ) HOSTi 5 SOVIETi ) Chapter 16, Exercise Solutions, Principles of Econometrics, 4e 556 EXERCISE 16.11 (a) The probit results are given in the table below in column (2). Probit models: Small ---------------------------------------------------------------------------(1) (2) (3) (4) ---------------------------------------------------------------------------boy 0.004 0.004 0.003 (0.120) (0.120) (0.080) white_asian -0.513 (-1.529) -0.514 (-1.529) -0.509 (-1.501) black -0.549 (-1.633) -0.539 (-1.599) -0.483 (-1.418) -0.023 (-0.589) -0.028 (-0.719) freelunch tchwhite 0.216*** (4.067) tchmasters -0.158*** (-4.238) _cons -0.523*** -0.002 0.006 -0.140 (-30.208) (-0.006) (0.018) (-0.409) ---------------------------------------------------------------------------N 5786 5786 5786 5786 lnL -3536.323 -3534.617 -3534.444 -3519.014 ---------------------------------------------------------------------------t statistics in parentheses * p<0.05, ** p<0.01, *** p<0.001 Based on the individual t-statistics we conclude that BOY, WHITE_ASIAN and BLACK are not statistically significant. The joint test of the significance of these three variables is based on the likelihood ratio test statistic LR 2 ln LU ln LR 2 3534.617 ( 3536.323 3.411 The value of the restricted log-likelihood function ln LR comes from the model including only an intercept, in column (1). The critical values for the Chi-square distribution are given in Table 3. The test degrees of freedom is 3, because we are testing 3 joint hypotheses that the coefficients of the selected variables are 0. The 95th percentile of the Chi-square distribution for 3 degrees of freedom is 7.815. Since the value of the LR statistic is less than the critical value we fail to reject the null hypothesis that the 3 variables BOY, WHITE_ASIAN and BLACK have coefficients that are 0. Chapter 16, Exercise Solutions, Principles of Econometrics, 4e 557 Exercise 16.11(a) (continued) If the assignment of students to small classes is random, we would expect to find no significant relationship between SMALL and any variable. Our findings are consistent with the hypothesis of random student assignment. (b) The results of probit models for AIDE and REGULAR are in the following 2 tables. Probit models: Aide ---------------------------------------------------------------------------(1) (2) (3) (4) ---------------------------------------------------------------------------boy -0.003 -0.003 -0.003 (-0.085) (-0.077) (-0.091) white_asian 0.173 (0.486) 0.174 (0.489) 0.186 (0.522) black 0.224 (0.629) 0.201 (0.564) 0.267 (0.746) 0.050 (1.310) 0.047 (1.234) freelunch tchwhite 0.136** (2.662) tchmasters 0.060 (1.675) _cons -0.377*** -0.565 -0.582 -0.745* (-22.294) (-1.588) (-1.636) (-2.070) ---------------------------------------------------------------------------N 5786 5786 5786 5786 lnL -3757.086 -3755.940 -3755.082 -3749.378 ---------------------------------------------------------------------------t statistics in parentheses * p<0.05, ** p<0.01, *** p<0.001 Chapter 16, Exercise Solutions, Principles of Econometrics, 4e 558 Exercise 16.11(b) (continued) Probit models: Regular ---------------------------------------------------------------------------(1) (2) (3) (4) ---------------------------------------------------------------------------boy -0.002 -0.002 0.000 (-0.045) (-0.045) (0.007) white_asian 0.404 (1.071) 0.400 (1.064) 0.387 (1.026) black 0.386 (1.023) 0.396 (1.052) 0.277 (0.731) -0.028 (-0.741) -0.022 (-0.572) freelunch tchwhite tchmasters -0.332*** (-6.579) 0.087* (2.418) _cons -0.395*** -0.791* -0.778* -0.491 (-23.285) (-2.102) (-2.070) (-1.290) ---------------------------------------------------------------------------N 5786 5786 5786 5786 lnL -3733.530 -3732.827 -3732.552 -3709.791 ---------------------------------------------------------------------------t statistics in parentheses * p<0.05, ** p<0.01, *** p<0.001 As in part (a) none of the variables BOY, WHITE_ASIAN or BLACK is statistically significant based on the t-values. For AIDE the LR test value is 2.292, and for REGULAR the LR test value is 1.405. Thus the variables are neither individually or jointly significant, which is consistent with the notion that students were assigned randomly. (c) The variable FREELUNCH is added in column (3) of the table results. It is not statistically significant in any of the probit model estimations. Its inclusion in the model has little effect on the other coefficient estimates. (d) The variables TCHWHITE and TCHMASTERS are added in column (4) of the tables. They are statistically significant in each estimation, except TCHMASTERS is not significant in the AIDE model. The LR test for their joint significance is obtained using the likelihood ratio test statistic Chapter 16, Exercise Solutions, Principles of Econometrics, 4e 559 Exercise 16.11(d) (continued) LR 2 ln LU ln LR In each case the unrestricted log-likelihood value comes from the model in column (4) and the restricted log-likelihood value comes from the model in column (3). These values are 30.86, 11.41 and 45.52 for the models for SMALL, AIDE and REGULAR, respectively. We are testing 2 joint hypotheses, that the coefficients of TCHWHITE and TCHMASTERS are both 0. The LR test statistic has a Chi-square distribution with 2 degrees of freedom if the null hypothesis is true. The 99th percentile of this distribution is 9.210. Thus we reject the null hypothesis at the 0.01 level of significance in all three cases. In the STAR program students were randomly assigned within schools but not across schools. It is possible that schools in wealthier or predominately white school districts have more teachers who are white or who has Master’s degrees. This would explain the significance of these variables in the probit models. Chapter 16, Exercise Solutions, Principles of Econometrics, 4e 560 EXERCISE 16.12 (a) The least squares estimates are reported in the following table: Linear regression Number of obs F( 8, 991) Prob > F R-squared Root MSE = = = = = 1000 43.39 0.0000 0.3363 .32673 -----------------------------------------------------------------------------| Robust delinquent | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------lvr | .0016239 .0006752 2.40 0.016 .0002988 .0029489 ref | -.0593237 .0240256 -2.47 0.014 -.1064706 -.0121768 insur | -.4815849 .0303694 -15.86 0.000 -.5411807 -.4219891 rate | .0343761 .0098194 3.50 0.000 .0151068 .0536454 amount | .023768 .0144509 1.64 0.100 -.0045898 .0521259 credit | -.0004419 .0002073 -2.13 0.033 -.0008487 -.0000351 term | -.0126195 .003556 -3.55 0.000 -.0195976 -.0056414 arm | .1283239 .0276932 4.63 0.000 .0739798 .1826681 _cons | .6884913 .2285064 3.01 0.003 .2400792 1.136903 ------------------------------------------------------------------------------ The outcome variable is whether a borrower was delinquent on a payment. The explanatory variables are: LVR. If the loan-to-value ratio increases the estimated probability of a delinquent payment increases, holding all else fixed. If a borrower is trying to obtain a loan that that is large relative to the value of the property, this may indicate that their finances are “stretched.” The positive sign is consistent with that notion. REF. If the loan is for a refinance, to take advantage of lower rates or to cash out some of the equity, there is an indication the borrower has been reliably paying on time and more history. The estimated probability of being delinquent is smaller, and significantly so, for loans for a refinance, holding all else constant. INSUR. Mortgage insurance is required of loans with loan-to-value ratio of greater than 80%. If a mortgage carries mortgage insurance there is a large and significant reduction in the probability of a delinquent payment. Those with mortgage insurance, who pay an additional fee for it, and may go through additional scrutiny and screening, which may increase the lending standard and reduce the probability of a delinquent payment. The magnitude of the effect estimated is too large. INSUR may be picking up other effects not identified in the model. RATE. The higher the interest rate the larger the probability of a delinquent payment. Higher rate loans are more of an economic burden to the borrower, increasing the monthly payments. Also, the riskier the loan the higher the rate charged, so higher rates may indicate loans that have a higher probability of default. Chapter 16, Exercise Solutions, Principles of Econometrics, 4e 561 Exercise 16.12(a) (continued) AMOUNT. The larger the amount of borrowed money the higher the probability of a delinquent payment. Larger loans lead to larger monthly payments, increasing the chance of a delinquent payment. This effect is significant at the 10% level. CREDIT. The larger the borrower’s credit score the lower the chance of a delinquent payment. The credit score is a history of borrowing and repayments on everything from credit cards to car loans. It makes sense that those with higher scores will have less chance of making a late payment, based on their history. TERM. The longer the term of the loan the smaller the monthly payments, reducing the probability of a delinquent loan, holding all else fixed. ARM. Adjustable rate mortgages can change the monthly interest applied to the loan. If the rate is adjusted upwards, the borrower has a larger monthly payment, which significantly increases the probability of a delinquent payment, all else held constant. (b) The probit model estimates are reported below: Probit regression Log likelihood = -332.79661 Number of obs LR chi2(8) Prob > chi2 Pseudo R2 = = = = 1000 332.43 0.0000 0.3331 -----------------------------------------------------------------------------delinquent | Coef. Std. Err. z P>|z| [95% Conf. Interval] -------------+---------------------------------------------------------------lvr | .0076007 .0045911 1.66 0.098 -.0013977 .0165991 ref | -.2884561 .1259446 -2.29 0.022 -.5353029 -.0416092 insur | -1.772714 .1158088 -15.31 0.000 -1.999695 -1.545733 rate | .1711988 .0438147 3.91 0.000 .0853236 .2570741 amount | .121236 .0615491 1.97 0.049 .000602 .2418701 credit | -.0019131 .0010638 -1.80 0.072 -.0039981 .0001718 term | -.0775769 .0198396 -3.91 0.000 -.1164618 -.038692 arm | .8091109 .2077119 3.90 0.000 .402003 1.216219 _cons | .964646 1.088121 0.89 0.375 -1.168033 3.097325 The signs and significance of the coefficients is much the same as in the linear probability model. The variable AMOUNT is now significant at the 5% level. Chapter 16, Exercise Solutions, Principles of Econometrics, 4e 562 Exercise 16.12 (continued) (c) The predicted values and explanatory variable values for the 500th and 1000th observations are: +--------------------------------------------------------------------------------+ | LPM delinquent lvr ref insur rate amount credit term arm | |--------------------------------------------------------------------------------| 500. | .1827828 0 70 1 1 10.95 .854 509 30 1 | 1000. | .5785297 0 88.2 1 0 7.65 2.91 624 30 1 | +--------------------------------------------------------------------------------+ . +--------------------------------------------------------------------------------+ | PROBIT delinquent lvr ref insur rate amount credit term arm | |--------------------------------------------------------------------------------| 500. | .1404525 0 70 1 1 10.95 .854 509 30 1 | 1000. | .6167872 0 88.2 1 0 7.65 2.91 624 30 1 | +--------------------------------------------------------------------------------+ Neither of the individuals made a delinquent payment. Both models predicted a low probability of a delinquent payment (0.18 and 0.14 for linear probability and probit models, respectively) for the first borrower, who has a lower loan to value ratio, LVR, and a lower loan AMOUNT. The predicted probabilities for the second borrower were 0.58 and 0.62 for linear probability and probit models, respectively). This borrower had a high loan to value ratio (88.2) and borrowed a larger amount ($291,000). The histogram for credit score is 0 .002 Density .004 .006 .008 (d) 400 500 600 credit score 700 800 Figure xr16.12(d) Histogram for credit score Note that most borrowers have a credit score between 500 and 800. The predicted probabilities for a delinquent payment using the linear probability model are: CREDIT | Margin Std. Err. z P>|z| [95% Conf. Interval] -------------+---------------------------------------------------------------500 | .1407109 .0257668 5.46 0.000 .090209 .1912129 600 | .0965214 .0190135 5.08 0.000 .0592556 .1337872 700 | .0523319 .0303051 1.73 0.084 -.007065 .1117287 Chapter 16, Exercise Solutions, Principles of Econometrics, 4e 563 Exercise 16.12(d) (continued) For probit, the predicted probabilities of delinquency are: | Delta-method CREDIT | Margin Std. Err. z P>|z| [95% Conf. Interval] -------------+---------------------------------------------------------------500 | .0984307 .0233882 4.21 0.000 .0525907 .1442708 600 | .069189 .0136925 5.05 0.000 .0423523 .0960257 700 | .0471468 .0157545 2.99 0.003 .0162686 .078025 Note that higher credit scores reduced the predicted probabilities. For the linear probability model these changes are the same (0.0441895) for each 100 point increase in CREDIT. The effect is not equal for the probit model, being 0.0292417 for a credit score change of 500 to 600, and 0.0220422 for the credit score change from 600 to 700. (e) For the probit model these marginal effects are: | Delta-method CREDIT | dy/dx Std. Err. z P>|z| [95% Conf. Interval] -------------+---------------------------------------------------------------500 | -.0003319 .0002268 -1.46 0.143 -.0007764 .0001126 600 | -.0002546 .0001403 -1.82 0.070 -.0005295 .0000203 700 | -.0001883 .000073 -2.58 0.010 -.0003313 -.0000452 These values are small, and decreasing. Recall that these are the marginal effects of a 1 point increase in credit score, which is a relatively small amount. That the values decrease in magnitude shows that the most benefit from improved credit is for those will smaller credit scores. As an alternative to examining these marginal effects one could look at discrete changes in probabilities as in the previous question part, or scale credit to be in units of 10 points or 100 points, which would shift the decimal point in the above marginal effects accordingly. .02 .04 Density .06 .08 .1 The histogram of loan to value ratio is given below. 0 (f) 0 20 40 60 80 loan amount to value of property, percent 100 Figure xr16.12(f) Histogram for loan to value ratio Chapter 16, Exercise Solutions, Principles of Econometrics, 4e 564 Exercise 16.12(f) (continued) The most popular amount is 80%, though there is a spike at 20% as well. For the linear probability model the predicted probabilities of delinquency in the two cases are: | Delta-method LVR | Margin Std. Err. z P>|z| [95% Conf. Interval] -------------+---------------------------------------------------------------20 | -.0009097 .0434921 -0.02 0.983 -.0861527 .0843332 80 | .0965214 .0190135 5.08 0.000 .0592556 .1337872 For the probit model the predictions are: | Delta-method LVR | Margin Std. Err. z P>|z| [95% Conf. Interval] -------------+---------------------------------------------------------------20 | .0263176 .018372 1.43 0.152 -.0096909 .062326 80 | .069189 .0136925 5.05 0.000 .0423523 .0960257 In each case an increase in the loan to value ratio increases the probability of a delinquent payment. (g) The predictions from the linear probability model are summarized in the following table. The upper values are the frequencies and the lower values are the cell percentages. = 1 if | payment | late by | Linear Probability Model 90+ days | 0 1 | Total -----------+----------------------+---------0 | 727 74 | 801 | 72.70 7.40 | 80.10 -----------+----------------------+---------1 | 68 131 | 199 | 6.80 13.10 | 19.90 -----------+----------------------+---------Total | 795 205 | 1,000 | 79.50 20.50 | 100.00 The successful predictions are on the diagonal, 72.7% of those who did not have a late payment were correctly predicted; 13.1% of those with a delinquent payment were predicted correctly. 0. If we count just successful predictions of 0 and 1.90 -----------+----------------------+---------Total | 814 186 | 1. We try several thresholds: 0.80 and 0.60 | 80.2 % respectively. In the full sample only about 20% of borrowers have a late payment.50. These values for the thresholds : 0. such as 0. who forgoes a “good” loan. Principles of Econometrics.50. then the “hit rate” for each threshold is relevant.50. However. respectively. Some estimates show that the loss of foreclosure process could be more than 20%-30% of the outstanding loan amount. what is the cost of giving a loan to a person who is delinquent on payments? If we use the thresholds 0. that lenders concern more about the cost of giving the loan to a person who is likely to default in the future. what the percentages of those who were not delinquent would have been predicted delinquent using the three rules? Presumably if a person is predicted delinquent they would not receive the loan. we presume.20.50. 0. 0.2% and 15. 4e 565 Exercise 16.10 -----------+----------------------+---------1 | 79 120 | 199 | 7.6%. 87% and 80. These percentages for the thresholds: 0.80.6%. If this is the costlier error then the use of a low threshold. 0.50 6. 12. (h) This is an important and difficult question. 0.40 18.2%.2% respectively. if a loan falls into default then goes into the foreclosure process. creating an opportunity cost for the borrower.12(g) (continued) For the probit model the prediction summary is = 1 if | payment | late by | Probit Model 90+ days | 0 1 | Total -----------+----------------------+---------0 | 735 66 | 801 | 73.80 and 0. may be better. then a higher threshold.90 12.20 are 88. So. From the lender's perspective. However.60 | 100.80 and 0.00 | 19. but is slightly less successful in predicting those who make a late payment. might be best.000 | 81. If this is the most costly error. such as 0.00 The probit model is more successful in predicting those who did not have a delinquent payment. For example.80 and 0. origination fee typically is 1% of the loan amount.8% and 4. the total loss will be way higher than 1% of the loan amount since foreclosure processes are very expensive for lenders.20 the percentages of these miscalculations are 6.20 are 5.Chapter 16. it may be that a focus on the two types of errors is useful. On the other hand.2%. Exercise Solutions. .20. 60 24.00 = 1 if | payment | late by | phat > 0.20 | 86.00 566 .80 90+ days | 0 1 | Total -----------+----------------------+---------0 | 432 1 | 433 | 86.80 0.12 (h) (continued) = 1 if | payment | late by | phat > 0.60 -----------+----------------------+---------1 | 33 34 | 67 | 6.20 0.20 9.80 | 13. 4e Exercise 16.00 | 100.40 -----------+----------------------+---------Total | 378 122 | 500 | 75.Chapter 16.40 | 100. Principles of Econometrics.20 | 13.60 -----------+----------------------+---------1 | 21 46 | 67 | 4.00 12.40 -----------+----------------------+---------Total | 440 60 | 500 | 88.20 90+ days | 0 1 | Total -----------+----------------------+---------0 | 357 76 | 433 | 71.20 | 86. Exercise Solutions.20 | 86.40 5.80 | 100.40 15.60 6.40 0.00 = 1 if | payment | late by | phat > 0.40 -----------+----------------------+---------Total | 496 4 | 500 | 99.60 | 13.50 90+ days | 0 1 | Total -----------+----------------------+---------0 | 407 26 | 433 | 81.60 -----------+----------------------+---------1 | 64 3 | 67 | 12. 030) credit -0.003 (-0. but their magnitudes and significance varies.227) termi 0.816** 0.110 (1.021 ---------------------------------------------------------------------------t statistics in parentheses * p<0.149) _cons -0.083** -0.207) (-1.116 -0.009 0.114 0.003* (-1.002 0.816** (3.207) arm 0.062) crediti 0.067 (-0.003* -0.003 (1.006 0. . 4e 567 EXERCISE 16. Exercise Solutions.058 -0.001 -0.600) (1.131) (1. *** p<0.014) (1.396 42. The value labeled “ll” is the log-likelihood function value.01. exceptions being the coefficients for RATE and AMOUNT.538) (1.677 -159.329) (0.450) (1.736 2.030) (1.333) (-2.589) (1.111 (3.026 (0.13 The probit estimates for the alternative models for parts (a)-(e) are reported in the following table. such as LVRI LVR INSUR . and those with INSUR 1 (part (c)).045* -0. ending in “i” are interaction variables.434 (-0.315 -171.242) (-2. Principles of Econometrics.251) lvri -0.001 -0.335 (1.747 336.569) (-3.009 (0.434 -2.237* -0.626) armi -0.008 (0.027) (3.544** 0.111 0. we find the signs of the coefficients are consistent across all estimations.329) ref -0.148) (-0.213) (-2.310) ratei 0.279) insur -4.773) (1.765) (3.971* (-2.765) amount 0.120** 0.106 0. In most cases the coefficients in the equation for INSUR 0 are larger (in absolute value) than their counterparts in the other two equations.299) refi 0.537 2. ** p<0.583) (-1.750* 0.259*** 0.05.279) (2.451* (-2.326 -331.451* -0.750 28.641) (-2.493) rate 0.111) (3.583) ---------------------------------------------------------------------------N 1000 280 720 1000 ll -468.106 (5.831) (1.002 chi2 61.238) amounti 0.083** (-2.493) (-0. those with INSUR 0 (part (b)).001 (d) Comparing the estimates from the pooled observations (part(a)).Chapter 16. Probit models ---------------------------------------------------------------------------(1) (2) (3) (4) pooled insur=0 insur=1 full ---------------------------------------------------------------------------delinquent lvr 0.222*** 0. The variables denoted lvri.148) term -0.807) (-3. N ln L i 1 N0 ln f yi i 1 N1 ln f yi i 1 ln f yi ln L0 ln L1 Thus estimating the model separately and summing then is equivalent to estimating the full model with interactions between INSUR and the remaining variables. Now suppose we have two groups of observations among the N: those N 0 individuals for whom INSUR = 0 and N1 individuals for whom INSUR = 1.326 331.Chapter 16. (e) For a sample of N individuals the log-likelihood function is formed from the probability function in equation (16.002 ln L0 ln L1 171. . one for each variable plus a constant term.313 274. that the coefficients of INSUR and the interaction variables.002 from the “full” model in column (4) of the above table. but there is little consistency across the two equations. The value of the unrestricted log-likelihood if 331.315. the statistic has an asymptotic chi-square distribution with degrees of freedom equal to the number of hypotheses being tested. Therefore the value of the likelihood ratio test statistic is LR 2 ln LU ln LR 2 331. In this case there are J = 8 hypotheses. and conclude there is some behavioral differences between these two groups.1 The model in question has more than one explanatory variable but the principle is the same. Exercise Solutions. 4e 568 Exercise 16. In the other equations more coefficients are significant. If the null hypothesis is true.507. such as LVRI. Principles of Econometrics. are zero. The null hypothesis is rejected if the value LR is larger than the chi-square distribution critical value. which is 15.003 . we can rearrange the terms as we like.626 distribution. The restricted log-likelihood function value is 468. In this estimation example ln LU 331. N ln L i 1 ln f yi In the above notation let represent all the parameters in the probit model. The log-likelihood function is the sum of the natural logarithms of the probability function. Therefore we reject the null hypothesis that the coefficients for the insured and uninsured groups are equal.13) f ( yi ) [ ( 1 x )] yi [1 2 i ( 1 x )]1 yi . yi 2 i 0.315 2 8 2 137.002 The test critical value is the 95th percentile of the 468. The restricted model is the pooled model in column (1) of the table. The slight difference is due to rounding error. (f) The likelihood ratio test statistic is LR 2 ln LU ln LR . Because the log-likelihood is a sum.677 159.13(d) (continued) Only the coefficient for RATE is significant in the equation for INSUR 1 . 4.79 47. 2. 7. 6.5 | | 2 Skist-oil .5 | | 2 ChiSea-oil .58 47. whereas INCOME is a household variable and is the same for all 4 alternatives on any choice occasion. 3.79 47. Principles of Econometrics. The first two households’ data are 1.56 47. but that income is constant.5 | |-----------------------------------------| | 2 Skist-water .Chapter 16.14 (a) The variable NETPRICE shows variation across the alternative brands. (b) The choices among the 1500 cases are Alternatives summary for alt +-----------------------------------------------------------+ |Alternative | Cases Frequency Percent | | value label | present selected selected | |-------------------------+---------------------------------| | 1 Skist-water | 1500 548 36. 4e 569 EXERCISE 16.53 | | 2 Skist-oil | 1500 291 19. .5 | | 1 ChiSea-water .40 | | 3 ChiSea-water | 1500 475 31.5 | +-----------------------------------------+ Note that the prices of the alternatives change within each group of 4 observations.40 | +-----------------------------------------------------------+ We observe that this group of consumers has a preference for tuna packed in water.79 47.5 | | 2 ChiSea-water .58 47.56 47.67 | | 4 ChiSea-oil | 1500 186 12. 8. 5.5 | | 1 ChiSea-oil . Exercise Solutions. +-----------------------------------------+ | hhid alt netprice income | |-----------------------------------------| | 1 Skist-water .79 47.5 | | 1 Skist-oil . is facilitated by using some simplifying notation. Then the probabilities that each of the options is chosen are: pSkist -water pSkist -oil pChisea -water pChisea -oil exp xb Skist -water exp xb Skist -water exp xb Skist -oil exp xb ChiSea-water exp xb ChiSea-oil exp xb Skist -oil exp xb Skist -water exp xb Skist -oil exp xb ChiSea-water exp xb ChiSea -oil exp xb Chisea -water exp xb Skist -water exp xb Skist -oil exp xb ChiSea-water exp xb ChiSea -oil exp xb Chisea-oil exp xb Skist -water exp xb Skist -oil exp xb ChiSea-water exp xb ChiSea-oil .Chapter 16. Exercise Solutions. Principles of Econometrics. but this has been suppressed to simplify notation. which has none and serves as our base case. Each of the options has an intercept parameter except for Starkist-in-Water.14 (continued) (c) The probability that individual i chooses alternative j. for each of these 4 alternatives. Let the variables and parameters for each alternative be denoted as follows: xb Skist -water xb Skist -oil 2 12 3 NETPRICESkist -oil 3 2 xb ChiSea -water xb ChiSea -oil NETPRICESkist -water 13 14 2 2 DISPLAYSkist -water DISPLAYSkist -oil NETPRICEChiSea -water NETPRICEChiSea -oil 3 3 4 4 FEATURESkist -water FEATURESkist -oil DISPLAYChiSea -water DISPLAYChiSea -oil 4 4 FEATUREChiSea -water FEATUREChiSea -oil Each variable should have a subscript “i” to denote the individual. 4e 570 Exercise 16. 611371 -------------+---------------------------------------------------------------Skist_water | (base alternative) -------------+---------------------------------------------------------------Skist_oil | _cons | -.000 1. Err. Interval] -------------+---------------------------------------------------------------alt | netprice | -9. z P>|z| [95% Conf.160052 2.56 0. From the probabilities in the previous question part.66319 -8. The alternative-specific variables are negative and statistically significant.83 0.243529 We note that the estimated coefficients are all statistically significant.7395775 -.636453 -1.15 0.53 0.0816866 -6.3732396 -------------+---------------------------------------------------------------ChiSea_oil | _cons | -1.37 0.693445 -.280729 display | 1. we see that all else being equal. 4e 571 Exercise 16.13 0.000 -11.110919 feature | 1. Std.075652 1. the Starkist in oil.74 0. .000 -.1) are Alternative-specific conditional logit Case variable: hhid Number of obs Number of cases = = 6000 1500 Alternative variable: alt Alts per case: min = avg = max = 4 4.0732714 -8.2704 Wald chi2(3) Prob > chi2 = = 405.343511 .0 4 Log likelihood = -1537. Exercise Solutions.Chapter 16.439991 .1366656 9.1002377 -14.635486 . and Chicken of the Sea brands have a lower estimated probability of being selected than Starkist in water.14 (continued) (d) The estimates (obtained with Stata 11.8628894 -11. with the coefficient of the continuous variable NETPRICE carrying a negative sign and the indicator variables DISPLAY and FEATURE having positive signs.2425727 6.000 -1.4523589 -------------+---------------------------------------------------------------ChiSea_water | _cons | -.000 1. Principles of Econometrics.0000 -----------------------------------------------------------------------------choice | Coef.5959682 .5333423 .971961 .000 -. as shown in the third table below pi 3 PRICEi1 pi 3 pi1 2 . the first table shows that the probability of Starkist in water being selected is 0.26646827 .40557918 9.40557918 9. Principles of Econometrics. which is more than the cost of the item.40557918 2 1 .24) pij PRICEij pij 1 pij 2 For example given that DISPLAY and FEATURE are zero. shown in the final column labeled “X”.07771 . If the change is 10 cents.971961= 2. the probabilities reduce to a dependence on the alternative specific constants and NETPRICE. If we set the NETPRICE at its mean for each brand we can compute the probabilities of each choice being selected. The first table gives the marginal effect of a price change for each of the brands on the probability of choosing Starkist in water. then we anticipate a reduction in the probability of purchase of 0.Chapter 16.14 (continued) (e) The marginal effects are given in the tables below. The “own” price effect is given using equation (16. Exercise Solutions.00 change in price.24. The “cross-price” effect of a change in the price of one brand on the probability of selecting another brand is given by pij PRICEik pij pik 2 The marginal effect of an increase in the price of Starkist in water on the probability of choosing Chicken of the Sea in water is. For example.40409 The change in probability is for a $1. 4e 572 Exercise 16. The marginal effect of an increase in the net price of Starkist in water on the probability of choosing Startkist in water is pi1 PRICEi1 pi1 1 pi1 .406 with the price of each variable at its mean.971961= 1. 034964 8.98 0.046209 9.000 .I.26646827 ------------------------------------------------------------------------------variable | dp/dx Std.349111 .10559312 ------------------------------------------------------------------------------variable | dp/dx Std.941784 .165592 -10.000 .000 .41 0.02 0. z P>|z| [ 95% C.286742 .212055 .000 .471934 . 4e 573 Exercise 16.29 0.181533 .I.096762 9.68163 ChiSea_water | 1.427063 .000 -2.14(e) (continued) Pr(choice = Skist-water|1 selected) = .35 0.000 .02 0.22235943 ------------------------------------------------------------------------------variable | dp/dx Std.709778 .875935 1.709664 1. Err.47 0.212055 .000 -2.27948 .899315 . ] X -------------+----------------------------------------------------------------netprice | Skist_water | 1.47 0. Err.68163 ChiSea_water | -1.74 0.000 .66976 ChiSea_oil | .234138 .40409 .39976 .08896 .24 0.590856 .471934 .67167 ------------------------------------------------------------------------------Pr(choice = ChiSea-oil|1 selected) = .68112 Skist_oil | -1.102948 10.096762 9.68163 ChiSea_water | . ] X -------------+----------------------------------------------------------------netprice | Skist_water | .026839 8.74 0.66976 ChiSea_oil | .427063 .709665 1.060675 9. ] X -------------+----------------------------------------------------------------netprice | Skist_water | -2. z P>|z| [ 95% C.07771 .234138 .40557918 ------------------------------------------------------------------------------variable | dp/dx Std.000 .29709 -1.000 .72431 .67167 ------------------------------------------------------------------------------Pr(choice = ChiSea-water|1 selected) = .060676 9.81927 -1.24 0.98891 .08897 .336495 .000 -2.68112 Skist_oil | .94915 .034964 8.68112 Skist_oil | .000 .04886 -1.68163 ChiSea_water | . Err.286743 .046209 9.Chapter 16.747131 .51763 .13644 -.07771 .875935 1.177526 -10. z P>|z| [ 95% C.66976 ChiSea_oil | -. Err.72 0.67167 ------------------------------------------------------------------------------Pr(choice = Skist-oil|1 selected) = .181533 .102948 10. Principles of Econometrics.590856 . Exercise Solutions.211831 -11.336495 . z P>|z| [ 95% C.280583 .72 0.349111 .I.709778 .000 .67167 .I.000 .000 -1.66976 ChiSea_oil | .48 0.29 0.280583 .27948 .02684 8.6012 . ] X -------------+----------------------------------------------------------------netprice | Skist_water | .517631 .899315 .68112 Skist_oil | .099314 -9.000 . 021638 .3439 = = 419.1710712 -2. Err.0075311 -1.69 0. and conclude that INCOME has an effect on these choices. Interval] -------------+---------------------------------------------------------------alt | netprice | -9.143048 2.14 (continued) (f) Adding the individual specific variable INCOME to the model adds 3 parameters to estimate.619318 .0000 -----------------------------------------------------------------------------choice | Coef. Like alternative specific constants.815. z P>|z| [95% Conf. Thus we reject the hypothesis that the coefficients on INCOME are all zero.095587 feature | 1.7021478 The likelihood ratio test is based on the difference in the log-likelihood values for the two models.604371 -------------+---------------------------------------------------------------Skist_water | (base alternative) -------------+---------------------------------------------------------------Skist_oil | income | -.53292 -.76 0.000 -1.000 -11. coefficients of individual specific variables are different for each alternative. Exercise Solutions.0098662 _cons | -.117534 .66 0.302471 display | 1.0027101 .000 -. which is 7.0673146 .0019926 _cons | -1.068463 1.68989 -8.60 0.57 0.4607403 .7960337 -.78 0.47 0.1611304 -0.27 0. .8641534 -11.70 0.0058179 -0.99618 .012768 .000 1.0060061 -3.2484952 -------------+---------------------------------------------------------------ChiSea_water | income | -. LR 2 ln LU ln LR 2 1529. Principles of Econometrics.676 -.641 -.2119356 -5.014113 .090 -. with Starkist in water again set as the base case. Std.336417 .853 distribution.2704 15.3831243 .2429992 6.0334097 -.3439 The test critical value is the 95th percentile of the 2 3 1537.Chapter 16.0 4 Wald chi2(6) Prob > chi2 Log likelihood = -1529.1367137 9. 4e 574 Exercise 16.42 0.0275287 .1254469 -------------+---------------------------------------------------------------ChiSea_oil | income | -.000 1.0086928 _cons | -.007 -. The estimates are Alternative-specific conditional logit Case variable: hhid Number of obs Number of cases = = 6000 1500 Alternative variable: alt Alts per case: min = avg = max = 4 4. 353119 .526806 .88 0.259773 .36 0.353119 .67167 ------------------------------------------------------------------------------- .04261 .209761 .281181 .025517 8.22 0.059582 9.557554 .440775 . Err.66 0.20398375 ------------------------------------------------------------------------------variable | dp/dx Std.40 0.62312 .93977 -1. FEATURE and INCOME.723972 . Err.048578 8.440776 .I. Principles of Econometrics.66976 ChiSea_oil | .000 .161558 -10.557554 .10287145 ------------------------------------------------------------------------------variable | dp/dx Std. z P>|z| [ 95% C.281181 .11 0.526806 . z P>|z| [ 95% C.000 -1.36428 .27343699 ------------------------------------------------------------------------------variable | dp/dx Std.68112 Skist_oil | -1.22 0. and with NETPRICE at its mean for each brand.000 . z P>|z| [ 95% C.922537 . Exercise Solutions.05 0. ] X -------------+----------------------------------------------------------------netprice | Skist_water | . Err.36428 .I.674332 .000 .66976 ChiSea_oil | .41970781 ------------------------------------------------------------------------------variable | dp/dx Std.025517 8.000 .88 0.I. ] X -------------+----------------------------------------------------------------netprice | Skist_water | .91 0.036704 7.01591 .000 .110758 10.1211 -.000 .98 0. 4e 575 Exercise 16.000 -2.095308 8.000 . ] X -------------+----------------------------------------------------------------netprice | Skist_water | 1.98 0.67167 ------------------------------------------------------------------------------Pr(choice = ChiSea-water|1 selected) = .159749 . z P>|z| [ 95% C.000 .059582 9.336383 .68163 ChiSea_water | 1.66976 ChiSea_oil | . are: Pr(choice = Skist-water|1 selected) = .68112 Skist_oil | .000 .36 0.000 . Err.62932 .34254 -1. using the specified values for DISPLAY.930116 1.30648 .209243 .036704 7.000 -2.1472 .36 0.04261 .181947 -10.855809 .669008 1.000 -1.36 0.431595 .048578 8.209243 .66 0.431595 .669008 1.674332 .67167 ------------------------------------------------------------------------------Pr(choice = Skist-oil|1 selected) = .095308 8.000 .68112 Skist_oil | .68163 ChiSea_water | .000 .855809 .10131 -9.Chapter 16.67167 ------------------------------------------------------------------------------Pr(choice = ChiSea-oil|1 selected) = .259773 .I.930117 1.14 (continued) (g) The marginal effects of NETPRICE. ] X -------------+----------------------------------------------------------------netprice | Skist_water | -2.1472 .98593 .85329 -2.110758 10.4346 .159749 .213621 -11.68112 Skist_oil | .68163 ChiSea_water | -1.66976 ChiSea_oil | -.68163 ChiSea_water | .209761 .336383 . APPENDIX A Exercise Solutions 576 . (b) Recall that elasticity dQ s P dP Q s slope P Qs When P 10 . 1. which represents a 1. a 1 percent change in P is associated with a 1. Principles of Econometrics.04 percent . elasticity At the point P 50 and Q s s change in Q . Qs 3 1.5 50 72 1. elasticity 1.25 The elasticity shows the percentage change in Q s associated with a 1 percent change in P.1 (a) The slope is the change in the quantity supplied per unit change in market price.042 72 .Appendix A. a 1 percent change in P is associated with a 1. The slope here is 1.5.25 percent s change in Q . At the point P 10 and Q s 12 . 4e 577 EXERCISE A.5 10 12 Thus. Exercise Solutions.5 unit increase in the quantity supplied of a good due to a one unit increase in market price. Qs 3 1. When P 50 .5 10 12 1.5 50 72 Thus. 24 . (c) The marginal effect of the unemployment rate on inflation when UNEMP 5 is given by d ( INF ) d (UNEMP ) 6 UNEMP 2 6 52 0. 2 6 UNEMP for values of UNEMP between 1 and 10 4 3 INF 2 1 0 -1 0 1 2 3 4 5 6 7 8 9 10 UNEMP Figure xr-a.Appendix A.2(a) Curve relating inflation to unemployment (b) The impact of a change in the unemployment rate on inflation is given by the slope of the function d ( INF ) d (UNEMP ) 6 UNEMP 2 The absolute value of this function is largest as UNEMP approaches zero and it is smallest as UNEMP approaches infinity. Exercise Solutions. Thus.2 (a) A sketch of the curve INF appears below. Principles of Econometrics. This property is confirmed by examining Figure xr-a. the impact is greatest as the rate of unemployment approaches zero and it is smallest as unemployment approaches infinity. 4e 578 EXERCISE A.2(a). Appendix A.3 (a) x1 2 x1 6 (b) x2 3 (c) x4 y3 x1 2 1 6 x2 3 x7 8 x2 3 78 12 x4 ( 1 2) x16 24 y3 ( 1 2) 21 24 x x 2y 1 5 24 x 32 5 24 1 x y3 2 2 579 . Exercise Solutions. 4e EXERCISE A. Principles of Econometrics. 000 1.536.536.1536) (105 107 ) 5.1536 107 seconds (c) The distance light travels in a year is 186.000 31. 4e EXERCISE A.000 3.86 105 ) (3. Principles of Econometrics.Appendix A.4 (a) The velocity of light is 186.1536 107 ) (1.86 105 miles per second (b) The number of seconds in a year is 60 60 24 365 31.86 3.865696 1012 miles per year 580 .000 (1. Exercise Solutions. 0041 when t dWHEATt t dt WHEATt 0. when t = 49.Appendix A. WHEAT 0.0004t 2 1.0041 49 49 1. For example.0911 when t 1.7604 . when t = 49.0392 when t 0.20 ln( t ) 0.2784 0.2784 . Figure xr-a.7604 49 . WHEAT 0. For example. 4e 581 EXERCISE A.9(a) Graph of WHEAT The slope and elasticity for t Slope (b) 49 are dWHEATt dt Elasticity 0.5 0.9(b) Graph of WHEAT The slope and elasticity for t Slope 49 are dWHEATt dt Elasticity 0. Principles of Econometrics.5 0.0392 49 49 1.8 0. Figure xr-a. Exercise Solutions.1564 when t 49 The graph of the relationship between average wheat production (WHEAT) and time (t) is shown below.20 t 0.20ln(t ) 1.8 0.5 (a) The graph of the relationship between average wheat production (WHEAT) and time (t) is shown below.0004 2t dWHEATt t dt WHEATt 0.0004t 2 0. Principles of Econometrics. 1.6 3.6 The equation ln( y ) 0.00 .16 X The graph of y 1.12 .8 2.0 .2 0.02 .8 0.2 0.4 0.2 0.0 0. A graph of this function follows with y labeled as Y1.8 0.75 20 x can be rewritten as y exp( 1.14 .5 0.2 2.02 .16 . Exercise Solutions.0 Y3 (a) 1.4 0.4 2.2 1.0 .00 .8 Y1 0.16 X The equation ln( y ) 1. 1.02 .14 .08 .2ln( x) is shown below with y labeled as Y2. A graph of this function follows with y labeled as Y3.8 x0.04 . 4e 582 EXERCISE A.10 .06 .06 .8 0.06 .4 .12 .0 .12 .2 1.75 20 x) .6 1.00 .04 .6 0.4ln( x) can be rewritten as y e0.Appendix A.10 .6 0.08 X .10 .04 .4 0. 3.0 Y2 0.14 .08 . the slope is given by 0.6 3.4 For equation 2.10 For equation 2.2 x dy dx For equation 3. Principles of Econometrics.8 x0.6 (continued) (b) For equation 1.10 The slope is the change in arsenic concentration in toenails associated with a one-unit change in the arsenic concentration in the drinking water.284 2.1 1. y 1.10 .5 0.03948 and the elasticity is dy x dx y 2 0. .1924 For equation 3. (c) For equation 1. when x 0. the slope is given by 1. y 1. when x 0.4e0. y e dy dx 2 when x 1.Appendix A.8 x 0.10.6805 when x 0. Exercise Solutions.75 20 x 20e 0. when x dy x dx y 0.4 . the slope is given by dy dx 0.03948 0.0 The elasticity is the percentage change in arsenic concentration in toenails associated with a one-percent change in the arsenic concentration in the drinking water.1 0.6805 0.1. 4e 583 Exercise A. y 1. y e0. y 3.886 0.886 and the elasticity is 0.1 1.544 when x 0.75 20 x 25.2ln( x) .544 0.10.2840 and the elasticity is dy x dx y 25. 975711) (106 104 ) 0.11 584 .573239 106 ) (0.573239 5. Exercise Solutions.76530458 102 (d) 76.530458 x y (4.7328354597929 1011 x y (4.975711 104 ) (4.11 5.573239 106 ) (5.573239 106 ) (5.975711 104 ) (4.63299611 106 4632996.975711 104 (b) xy (4.573239 5.05975711 106 ) 4.573239 106 y 59757.975711) (106 104 ) 27.7 (a) x 4573239 4. Principles of Econometrics.328354597929 1010 (c) 2. 4e EXERCISE A.Appendix A. 8603 14 0.012 19. The values should be close because the tangent and the curve are virtually identical for values of x close to 2.8603 y2 f (2.01) f (1. .1403 18. (c) (d) The values located on the sketch are y1 f (1.99 2 18.99 3 1.8. Exercise Solutions.1403 The numerical derivative is m f (2.02 19.8 Figure xr-a.01 3 2.99) 0.Appendix A.02 The analytic and numerical derivatives are equal.99) 3 2 1.8 (a) The curve is displayed in Figure xr-a. 4e 585 EXERCISE A. Principles of Econometrics.8 Graph of quadratic function and tangent at x = 2 (b) The derivative is dy dx 2 6 x 14 when x 2 The tangent is sketched in Figure xr a.01) 3 2 2. APPENDIX B Exercise Solutions 586 . This result was used in the second line of the equation which would contain terms like cov( X i . X 2 .1 (a) E X E 1 X1 n 1 n (b) var X 1 E ( X1 ) E ( X 2 ) n X 2 .. X n are independent random variables. Principles of Econometrics.. X n E( X n ) n n var 1 X1 n X2 1 var X 1 n2 1 n n2 2 var X 2 Xn var X n 2 n Since X 1 .Appendix B. X j ) if these terms were not zero.... their covariances are zero.. Exercise Solutions. . 4e 587 EXERCISE B. Principles of Econometrics. 4e 588 EXERCISE B.Appendix B.Y3 ) 2cov(Y2 .Y3 ) .2 (a) E Y (b) var Y 1 3 Yi 3i 1 E 1 3 E Yi 3i 1 1 3 Yi 3i 1 var 1 var Y1 9 1 9 3 1 3 2 2 2 3 1 var Y1 Y2 Y3 9 var Y2 + var Y3 2 2 1 3 1 3 3 3 2 2 2 2 cov(Y1 . Exercise Solutions. Y2 ) 2cov(Y1 . .3 (a) The probability density function is shown below. Thus.Appendix B. P X 1 1 2 0. P X (e) 1 2 12 0 1 2 x 1 dx 1 4 x2 x 12 0 1 16 1 2 7 16 For a continuous random variable the probability of observing a single point is zero. Principles of Econometrics. f ( x) f (1) Using geometry. i.. Exercise Solutions. 4e 589 EXERCISE B. P( X (d) When x = 1 2 1 2 .5 2 1 1 (c) When x = 1. 2 1) is given by the area to the right of 1 which is . Using integration. P X 1 2 1 P X 1 2 1 1 2 3 4 7 16 Using integration. f 1) 2 1 2 1 x 1 dx 1 4 x2 1 2 1 2 x 1 4 ( 1 2) 1 1 1 4 3 4 Using geometry.e. the area is 0. P ( X P( X 1 2 1) 1 2 1 1 4 1 . (b) Total area of the triangle is half the base multiplied by the height. Appendix B. Principles of Econometrics. Exercise Solutions. 4e 590 Exercise B.3 (continued) (f) The mean is given by 2 E( X ) 0 2 x f ( x)dx 1 2 0 x2 1 6 x dx x3 1 2 x2 2 8 6 0 4 2 2 3 The second moment is given by 2 E( X 2 ) 0 x 2 f ( x)dx 2 0 1 2 x3 1 8 x 2 dx x4 1 3 x 3 2 0 The variance is given by E ( X 2 ) [ E ( X )]2 var( X ) (g) 2 3 2 3 2 2 9 The cumulative distribution function is given by F ( x) x 0 f (t )dt x 0 1 t 2 1 dt 1 2 t 4 t x 0 x x 1 4 16 8 8 3 2 3 . Principles of Econometrics. Exercise Solutions. its probability density function is 1 f ( x) (a) a b a x b The mean of X is given by E( X ) b a x f ( x)dx b2 1 b a b x a b a a2 dx x2 b a 2 1 b a b a 2 2 The second moment of X is given by b 2 E( X ) a b x2 a b a 2 x f ( x )dx b3 a 3 b a 3 b2 1 a2 3 dx x3 b a 3 1 ab The variance of X is E ( X 2 ) [ E ( X )]2 var( X ) b2 a2 3 4b 2 ab 4a 2 12 b a 2 4ab 3b2 2 3a 2 12 (b a )2 12 (b) The cumulative distribution function is F ( x) x 1 a b a dt t x b aa x a b a 6ab b a 591 . b) . 4e EXERCISE B.4 When X is a uniform random variable on ( a.Appendix B. 35. … is much less than it is in the intervals (0.25.35). (0. Moreover. Principles of Econometrics. (0.0.30.55.5 After setting up a workfile for 41 observations.30).0.15.0.0. ….10).15).05.Appendix B. A histogram of these numbers follows: These numbers are far from random. 4e 592 EXERCISE B. (0. Exercise Solutions.0.0.60). … The random number generator is clearly not a good one.40).0.10. (0. (0.25).50).45. the frequency of observations in the intervals (0. There are no observations in the intervals (0.1) interval.20.0. (0.20). the following EViews program can be used to generate the random numbers series x x(1)=79 scalar m=100 scalar a=263 scalar cee=71 for !i= 2 to 41 scalar q=a*x(!i-1)+cee x(!i)=q-m*@ceiling(q/m)+m next series u=x/m If the random number generator has worked well. . the observations in U should be independent draws of a uniform random variable on the (0.0. Appendix B. 4e 593 EXERCISE B.6 If X N 2 . . Principles of Econometrics. Exercise Solutions. a b and . and the probability density function for Y is f y b a dx dy 1 1 y b exp 2 2 a 2 1 a 2 exp 1 2a 2 2 2 1 a y ( a b) 2 This probability density function is that of a normal random variable with mean variance a 2 2 . then X aX Y g( y) 2 b a . its probability density function is given by 1 1 exp x 2 2 2 f ( x) If Y b . Y g ( X ) EY Y E X .Appendix B. g ( X ) 0 if E X .Y Y g ( X ) E X . E X and EY are expectations taken with respect to the marginal distributions of X and Y. Using iterated expectations. Exercise Solutions. Y ) .Y (Y ) E X . Principles of Econometrics.Y Y .Y Y g( X ) E X EY | X Y g( X ) E X g ( X ) EY | X Y EX g( X ) E X . and EY |X is an expectation taken with respect to the conditional distribution of Y given X. Now cov Y .7 Let E X . we can write E X . Y be an expectation taken with respect to the joint density for ( X .Y g ( X ) . 4e 594 EXERCISE B. Appendix B.546848 -3. Principles of Econometrics. as one would expect from a normal distribution.8(a) Histogram for combined observations Z1 and Z2 (b) The sample mean and variance are close to zero and one.056003 Jarque-Bera 0.8(c) Scatter diagram for Z1 and Z2 .dat (a) The histogram obtained by combining Z1 and Z2 into one series of 2000 observations. (c) The scatter diagram in Figure xr-b. Skewness Kurtosis 160 120 80 40 -0.042335 3. and the summary statistics from that series are displayed below. There is no evidence to suggest the observations are not normally distributed.8 Using uniform1. Dev. The histogram is bell-shaped.867716 0 -3 -2 -1 0 1 2 3 Figure xr-b.8(c) does not suggest any correlation between Z1 and Z2. respectively. It is a random scatter.026558 -0. 240 Series: Z Sample 1 2000 Observations 2000 200 Mean Median Maximum Minimum Std. and the p-value from the Jarque-Bera test for normality is 0.868.643559 0.008201 3. Exercise Solutions.283781 Probability 0. 4 3 2 Z2 1 0 -1 -2 -3 -4 -4 -3 -2 -1 0 1 2 3 4 Z1 Figure xr-b. 4e 595 EXERCISE B.989170 -0. Exercise Solutions.8 Using uniform2.8(c) does not suggest any correlation between Z1 and Z2.013018 3.804187 -4. (c) The scatter diagram in Figure xr-b.014171 Jarque-Bera Probability 0. as one would expect from a normal distribution. The histogram is bellshaped.000920 -0.8(c) Scatter diagram for Z1 and Z2 .000 observations.000 1. Principles of Econometrics. and the pvalue from the Jarque-Bera test for normality is 0.693411 0 -4 -3 -2 -1 0 1 2 3 4 Figure xr-b. Dev. 4 3 2 1 Z2 0 -1 -2 -3 -4 -5 -4 -3 -2 -1 0 1 2 3 4 Z1 Figure xr-b.Appendix B.200 800 400 Mean Median Maximum Minimum Std. respectively. Skewness Kurtosis 0.400 Series: Z Sample 1 20000 Observations 20000 2. 2. It is a random scatter.732266 0. There is no evidence to suggest the observations are not normally distributed. and the summary statistics from that series are displayed below.600 1.693.dat (a) The histogram obtained by combining Z1 and Z2 into one series of 20. 4e 596 EXERCISE B.8(a) Histogram for combined observations Z1 and Z2 (b) The sample mean and variance are very close to zero and one.001389 -0.000987 3.743557 1. Principles of Econometrics. Exercise Solutions.9 The cumulative distribution function for X is given by x F ( x) (a) P 0 X (b) P (1 X 1 2 2) 0 F 1 2 3t 2 dt 8 12 8 F (2) F (1) t3 8 3 x x3 8 0 1 64 23 8 13 8 7 8 597 .Appendix B. 4e EXERCISE B. 000 observations.0 0 1 2 3 4 5 6 7 8 9 10 X Figure xr-b. we use the distribution function to write U 1 exp( X ) from which we obtain X ln(1 U ) The histograms from 1000 and 10.6 0.4 0. Principles of Econometrics. Figure xr-b. particularly the one from 10.4 0. Exercise Solutions. 4e 598 EXERCISE B.2 0.8 FX 0.2 0.10(a) Exponential density function (b) 1.8 FBIGX 0.10 (a) 1.0 0.0 0.000 observations generated using X ln(1 U ) are given below.0 0 1 2 3 4 5 6 7 8 9 10 X Figure xr-b.10(c) Histogram for 1000 observations .10(b) Exponential distribution function (c) To use the inverse transformation method. They resemble the density in part (a).6 0.Appendix B. 9984 s2 0.0025 For 10.Appendix B. Exercise Solutions. Principles of Econometrics. 599 .10(c) (continued) Figure xr-b. 4e Exercise B.0272 s2 1.000 observations (d) The sample means and variances from the two samples are For 1000 observations: X 1.10(c) Histogram for 10.000 observations: X 0.9945 All four of these sample quantities are very close to 1. series x x(1)=1234567 scalar m=2^32 scalar a=1103515245 scalar cee=12345 for !i= 2 to 1001 scalar q=a*x(!i-1)+cee x(!i)=q-m*@ceiling(q/m)+m next series u1=x/m If the random number generator has worked well.0 0.U 2) These sample quantities are very close to the population values 0.11(a) Histogram and summary statistics for U1 40 Series: U2 Sample 1 1001 Observations 1001 35 30 Mean Median Maximum Minimum Std.Appendix B.7 0.999800 0.280072 -0.8 0.5 . Dev.0 Figure xr-b.2948 U2: X 0. the observations on U1 should be independent draws of a uniform random variable on the (0.5 0.6 0.2 0.2 0.294802 0.8 0. implying the random number generator is a good one.3 0. Skewness Kurtosis 25 20 15 10 0.5089 s 0.000000 0 0.23e-05 0.493013 0.499503 0.0471 0.2887 and 0.1) interval.882576 Jarque-Bera 52. Exercise Solutions.4995 s 0. Histograms of these numbers and those from U2 obtained using the seed value x(1)=95992 follow: 40 Series: U1 Sample 1 1001 Observations 1001 35 30 25 20 15 10 5 Mean Median Maximum Minimum Std.11(b) Histogram and summary statistics for U2 The histograms are approximately uniformly distributed.761368 Jarque-Bera Probability 63.4 0. the following EViews program can be used to generate the random numbers U1.1 0.508509 0. 4e 600 EXERCISE B.4 0.7 0. 0.9 1.0 0.3 0.999776 2.000000 5 0 0.5 0.2801 cor(U 1. The sample means.0 Figure xr-b.508945 0.37437 Probability 0.1 0. Dev. standard deviations and correlation are U1: X 0.042108 1. Skewness Kurtosis 0. .000379 1.11 After setting up a workfile for 41 observations. Principles of Econometrics.9 1.98931 0.6 0.000287 0. It is clear y 1 . y ) f ( y) f ( x | y) 6x2 y 2y 3x 2 and thus. Principles of Econometrics. we require f ( x.12 (a) For f ( x. f xY (e) Since f ( x | y ) 1 2 3x 2 f ( x ) . Exercise Solutions. the conditional mean and variance of X given Y 1 2 to the mean and variance of X found in part (b). 4e 601 EXERCISE B. y ) dx dy 1 . To establish the second condition. we 6 x 2 dx dy 1 0 2 x3 y 1 0 dy 2 1 0 y dy 2 y2 2 1 1 0 The marginal pdf for X is given by f ( x) 1 0 2 6 x y dy 6x 2 y2 2 1 3x 2 0 The mean of X is E( X ) 1 0 1 xf ( x )dx 0 3x 4 4 3 3 x dx 1 3 4 0 The second moment of X is 1 2 E( X ) 0 2 x f ( x )dx 1 0 3x 5 5 4 3 x dx 1 3 5 0 The variance of X is (c) 3 4 2 3 80 The marginal pdf for Y is given by f ( y) (d) 3 5 E ( X 2 ) [ E ( X )]2 var( X ) 1 0 2 6 x y dx x3 6y 3 1 2y 0 The conditional pdf f ( x | y ) is f ( x. X and Y are independent because f ( x . y ) to be a valid pdf.Appendix B. y ) consider (b) 1 1 0 0 0 for all 0 1 6 x 2 y dx dy 0 y x 1. (f) Yes. are identical . y ) 0 and 6x2 y that f ( x. 0 1 0 1 1 0 0 f ( x. y ) 6x2 y f ( x) f ( y) 3x 2 2 y . Exercise Solutions.13 (a) (b) The volume under the joint pdf is 2 y 0 0 1 2 2 dx dy x 2 0 y 2 dy 0 0 y 2 2 y2 4 dy 1 0 The marginal pdf for X is 2 f ( x) x 1 2 dy 1 2 dx 2 y 2 x 2 1 x The marginal pdf for Y is y f ( y) (c) (d) P X 12 1 2 0 0 1 dx y 2 12 x2 4 x 1 1 2 16 0 7 16 The cdf for Y is F ( y) (e) x 2 y x 20 y t 2 0 dt y t2 4 y2 4 0 The conditional pdf f ( x | y ) is given by f ( x. Principles of Econometrics. y ) f ( y) f ( x | y) 12 y 2 1 y implying f xY The required probability is P X 1 2 3 2 Y 12 0 2 3 dx 1 2 X and Y are not independent because P X (f) 12 2x 3 1 3 0 3 2 Y P X The mean of Y is E (Y ) 2 0 y f ( y ) dy y2 2 2 0 dy y3 6 2 4 3 0 The second moment of Y is 2 E (Y ) 2 0 2 y f ( y ) dy 2 0 y3 2 dy y4 8 2 2 0 1 2 . 3 2 2 3 . 4e 602 EXERCISE B.Appendix B. Exercise Solutions. y E( X | Y ) E( X ) 0 y x f ( x | y ) dx 0 2 EY E ( X | Y ) x y 0 y 2 dx f ( y )dy x2 2y 2 0 y y 2 0 y2 4 2 dy y3 12 0 2 3 We can check this result by using the marginal pdf for X to find E ( X ) : E( X ) 2 0 x f ( x) dx 2 0 x x2 2 dx x2 2 x3 6 2 2 0 4 3 2 3 603 .13(f) (continued) The variance of Y is E (Y 2 ) [ E (Y )]2 var(Y ) (g) 4 3 2 2 2 9 From part (e).Appendix B. 4e Exercise B. Principles of Econometrics. APPENDIX C Exercise Solutions 604 . E Y* (c) 1 Yi 1 2 2 1 Y2 2 2 var Y2 2 11 cov Y1 . .1 (a) aiYi where ai is a constant. Y* Y1 Y2 2 1 Y1 2 1 Y2 2 Thus. A linear estimator is one that can be written in the form Rearranging Y* yields. Y2 0 The sample mean is a better estimator because it uses more information. Exercise Solutions. . thus making it a better estimator than Y*.Appendix C. Y* is a linear estimator where ai (b) 1 2 for i 1. Y1 Y2 2 E 1 E Y1 2 1 E Y2 2 1 2 1 2 The variance of Y* is given by var Y * Y1 Y2 2 var 1 4 1 2 var Y1 2 1 4 2 1 Y1 2 var 2 1 2 (d) i The expected value of an unbiased estimator is equal to the true population mean. In general. 4e 605 EXERCISE C. 4.N . The variance of the sample mean is 2 N which is smaller than 2 2 when N 2 . increasing sample information reduces sampling variation. 2 and ai 0 for i 3. Principles of Econometrics. Y2 22 2 2 since cov Y1 . (d) Since var Y (e) If 2 var Y . 1 Y3 6 1 Y2 3 1 var Y1 4 1 4 3 1 E Y1 2 1 E Y3 6 1 2 1 3 1 6 1 Y3 6 1 var Y2 9 1 36 2 1 E Y2 3 2 1 var Y3 . The probability that the estimator Y is within one unit on either side of P 1 Y 1 P Y 1 var Y P 1 3 P 0. 4e 606 EXERCISE C. Principles of Econometrics.2 (a) Y 1 Y1 2 (b) E Y (c) var Y 1 Y2 3 1 Y3 6 1 Y1 2 E var 1 Y1 2 2 i 1 1 Y2 3 1 9 aiYi . 9 . Y3 0 7 2 18 The variance of the sample mean is var Y 2 2 N 3 6 2 18 which is smaller than the variance of Y .577 is: var Y 1 var Y 1 3 Z Z 0. 36 since cov Y1 . Exercise Solutions. then var Y 2 N 9 3 3 .5 .5 1 var Y Z 0.5345 0.577 0. Y is not as good an estimator as Y . and var Y 7 2 18 7 9 18 3. where ai are constants for i 1. 2 and 3.5345 Z var Y 1 3.Appendix C.407 is: .5 0. Y2 cov Y2 .436 The probability that the estimator Y is within one unit on either side of P 1 Y 1 P P P Y 1 var Y 1 3. X N (2000.Appendix C. 4e 607 EXERCISE C.222 pieces.332] 0.5002 ) . The probability that in a 9 hour day.000 pieces will be sold is the same as the probability that average hourly sales of fried chicken is greater than 20. more than 20.091 . Principles of Econometrics. Exercise Solutions.3 Let X be the random variable denoting the hourly sales of fried chicken which is normally distributed. P X 2222 P 2222 X P Z P Z N N 2222 2000 500 9 666 500 P[ Z 1.000/9 2. 80002 ) . Principles of Econometrics. Exercise Solutions. P X 50000 P X P Z P[ Z 50000 N N 50000 47000 8000 40 2. X N (47000.Appendix C.9911 0.0089 . Assume it is normally distributed.37] 1 0. 4e 608 EXERCISE C.4 Let the random variable X denote the starting salary for Economics majors. 649 .462 1. Exercise Solutions.649 . (b) p P t(399) 2.462 1 P t(399) 2. given H0 is true.5 (a) 170 versus H1 : 170 . is: t X 170 ˆ N t 399 The rejection region is t 1.462 0.Appendix C.462 Since t 2. we reject H0 and conclude that the new accounting system is cost effective. The alternative is H1 : 170 We set up the hypotheses H 0 : because we want to establish whether the mean monthly account balance is more than 170. Principles of Econometrics.007 . The value of the test statistic is t 178 170 65 400 2. 4e 609 EXERCISE C. The test statistic. 05 level of significance.5 1. the rejection region is t 1.765 8 N . Since t = 0. 4e 610 EXERCISE C. at the 0.895 . given H0 is true.598 < 1. we set 6 versus H1 : 6.5 X ) 2 11.895.05 level of significance. Principles of Econometrics.Appendix C.235. the students are studying more than 6 hours per week (b) A 90% confidence interval for the population mean number of hours studied per week is: X tc ˆ2 11. we do not reject H0 and therefore cannot conclude that.4286 8 1 8 ( xi 7i1 5. Exercise Solutions. is t X 6 ˆ N t7 X 1 8 xi 8i 1 1 1 3 4 4 6 6 8 12 8 ˆ2 var X t 5.895 = 3.4286 0.598 At the 0.6 (a) To decide whether the students are studying on average at least 6 hours per week.5 6 11. 7.4286 = 5. up the hypotheses H 0 : The test statistic. This would be a costly error to make because we would be dismissing a cost effective practice. p-value = 1 P (t(49) 1. a type I error occurs when we wrongly reject the hypothesis that the hiring procedures are effective. Exercise Solutions. Since 1.0344 . 4e 611 EXERCISE C.677. the rejection region is t > 1.677.861 38 50 Using a 5% significance level at 49 degrees of freedom. when H0 is true. In this example. The test statistic. is t X 450 ˆ N t 49 The value of the test statistic is t 460 450 1.861) = (1 0.9656) = 0. we reject H 0 and conclude that the current hiring procedures are effective.7 (a) To test whether current hiring procedures are effective.Appendix C. we test the hypothesis that H0 : 450 against H1 : 450. Principles of Econometrics.861 > 1. The manager is interested in workers who can process at least 450 pieces per day. (b) (c) A type I error occurs when we reject the null hypothesis but it is actually true. derive N as follows: 2 zc zc zc N 4 2 N 2 4N 1. .96 21 2 4N 423. Exercise Solutions.8 The interval estimate of a normally distributed random variable is given by Y zc N .525 N A sample size of 424 employees is needed.Appendix C. where zc is the corresponding critical value at a 95% level of confidence. The length of the interval is therefore 2 zc N . Principles of Econometrics. 4e 612 EXERCISE C. To ensure that the length of the interval is less than 4. 5 .1 Y1 Y2 3 1 E Y1 3 1 0.4 1 613 .4 pdf .2 Y1 3 E 3 3 Y2 3 1 E Y3 3 1 1 1 3 3 3 3 3 3 3 var Y var Y1 Y2 3 1 var Y1 9 Y3 1 var Y2 9 1 1 1 1 1 1 1 9 9 9 3 1 var Y3 9 2 E 0.Appendix C.3 .3 Y3 3 4 3 2 0.0 1 2 3 4 y (b) E Y 4 i 1 (c) yi P Y 4 var Y i 1 yi 1 3 (d) E Y E yi E Y 2 2 0. Principles of Econometrics.2 .3 4 0.2 3 0. .1 .4 3 P Y yi 2 2 3 Y3 E 1 E Y2 3 0. Exercise Solutions. 4e EXERCISE C.9 (a) A sketch of the pdf is shown below.1 2 0. 305 5070.20) 1.956 7083.039 2838.127 21 1.. Principles of Econometrics. 2 2 2 3 1 against H1 : 2 2 2 3 1 The test statistic.7437 0.10.351 3814. ˆ (a) (i) The null and alternative hypotheses are H0 : (ii) 10000 against H1 : 10000 The test statistic.622 3200. sample mean and standard deviation for the Fulton Fish Market data. when H0 is true.950 0. we reject the null hypothesis that the mean quantity sold is greater than or equal to 10000.739 4367.11 The sample size. is F ˆ 22 ˆ 32 F N2 1. at (22. N3 1 . Monday Tuesday Wednesday Thursday Friday 21 23 21 23 23 8070.950 = 0.762 4847.0483.Appendix C. (iv) Since 1.10.711 N Mean.069. X Std.725. 20) degrees of freedom.039) 2 (2838. we fail to reject the null hypothesis that the variances are equal. . on various days. at 20 degrees of freedom.622) 2 1.Dev.05 level of significance. when H0 is true. (v) (b) (i) p-value = P t(20) The null and alternative hypotheses are H0 : (ii) 1. (iv) Since 1. the rejection (iii) Using an region is t < 1.127 3964. the rejection (iii) Using an region is F > 2.725.950 < 2.7437 = 0. 4e 614 EXERCISE C. Exercise Solutions. (v) p-value = P F(22. are shown below. is t X 10000 ˆ N tN 1 The value of the test statistic is t 8070.05 level of significance.762 10000 5070. The value of the test statistic is F (3964.476 7283.7437 < 1. 476) 1 23 3473.739 4367.Appendix C. (v) (d) p-value = P t(42) 0.11 (continued) (c) (i) The hypotheses are H0 : (ii) 2 3 against H1 : 2 3 The test statistic.6222 23 21 2 (3473. 4e 615 Exercise C. N 2 N3 2 The variance estimate is ˆ 2p = 22 3964. Exercise Solutions.649 The mean of W is given by E (W ) E( X1 X2 X3 X4 X5) E( X1 ) E( X 2 ) E( X 3 ) E( X 4 ) E( X 5 ) 1 2 3 4 5 The variance of W is given by var(W ) var( X 1 X2 X3 X4 X5 ) var( X 1 ) var( X 2 ) var( X 3 ) var( X 4 ) var( X 5 ) 2 1 2 2 2 3 2 4 2 5 To derive the second line for var(W ) we have used the result that cov( X i . we do not reject the null hypothesis that the means are equal.018 and t (iv) Since 2.458 P t(42) 0.458 1 21 0. is (X2 t ˆ 2 p X3) 1 N2 t N2 1 N3 where ˆ 2p = N3 2 ( N 2 1) ˆ 22 ( N 3 1) ˆ 32 . .018 < 0. Principles of Econometrics.05 level of significance and degrees of freedom 42. the rejection (iii) Using an 2.9 0. for i j . because the X i are independent.458 < 2.0392 20 2838.458 0. regions are t 2.018 .9) 2 The value of the test statistic is t (4847. X j ) 0. when H0 is true.018. a 95% interval estimator for i 1 is ˆ 2 w ) 2 i Ni Z (0.025) w . is also distributed normally. we need to replace it with an estimate ˆ w where ˆ 2w 5 i 1 ˆ i2 Ni The resulting 95% interval estimator ˆ Z (0.96 1835. we have used the result that cov( X i .5)2 and ˆ 31653 and an approximate 95% interval estimate is ˆ Z (0. we obtain ˆ 2w (1835. Exercise Solutions. it follows that the X i are normally distributed and that ˆ . Because w is unknown. For the Fulton Fish data.025) ˆ w 31653 1.11 (continued) (e) The mean for ˆ is given by E( ˆ ) E X1 X2 E X1 1 X3 X4 E X2 2 3 X5 E X3 4 E X4 E X5 5 The variance of ˆ is given by var( ˆ ) var X 1 X2 var X 1 X3 X4 var X 2 X5 var X 3 var X 4 2 1 2 2 2 3 2 4 2 5 N1 N2 N3 N4 N5 var X 5 In deriving this variance. Since the X i are distributed normally.025) ˆ w is an approximate one in large samples. which is a linear function of the X i . X j ) 0 because sales on different days are assumed independent. where 5 i 1 i 5 2 w and Hence. ˆ ~ N ( . 4e 616 Exercise C. 35251) .5 (28055.Appendix C. Principles of Econometrics.