Chapter 9: Inferences Based on Two SamplesCHAPTER 9 Section 9.1 1. a. E X Y E X E Y 4.1 4.5 .4 , irrespective of sample sizes. b. V X Y V X V Y X Y c. 2. 12 22 1.8 2 2.0 2 .0724 , and the s.d. of m n 100 100 .0724 .2691 . A normal curve with mean and s.d. as given in a and b (because m = n = 100, the CLT implies that both X and Y have approximately normal distributions, so X Y does also). The shape is not necessarily that of a normal curve when m = n = 10, because the CLT cannot be invoked. So if the two lifetime population distributions are not normal, the distribution of X Y will typically be quite complicated. The test statistic value is xy z z 1.96 . We compute s12 s 22 , and H0 will be rejected if either z 1.96 or m n z 42,500 40,400 1900 2200 45 45 2 2 2100 4.85 433.33 . Since 4.85 > 1.96, reject H0 and conclude that the two brands differ with respect to true average tread lives. 3. The test statistic value is z 2.33 . We compute z z x y 5000 s12 s 22 m n , and H0 will be rejected at level .01 if 42,500 36,800 5000 2200 2 1500 45 45 2 700 1.76 396.93 , which is not > 2.33, so we don’t reject H0 and conclude that the true average life for radials does not exceed that for economy brand by significantly more than 500. 23 Chapter 9: Inferences Based on Two Samples 4. a. From Exercise 2, the C.I. is x y 1.96 s12 s 22 2100 1.96 433.33 2100 849.33 m n 1250.67,2949.33 . In the context of this problem situation, the interval is moderately wide (a consequence of the standard deviations being large), so the information about 1 and 2 is not as precise as might be desirable. b. From Exercise 3, the upper bound is 5700 1.645 396.93 5700 652.95 6352.95 . 5. a. Ha says that the average calorie output for sufferers is more than 1 cal/cm 2/min below that 12 22 .2 2 .4 2 .1414 , so m n 10 10 .64 2.05 1 z 2.90 . At level .01, H0 is rejected if z 2.33 ; since .1414 for non-sufferers. –2.90 < -2.33, reject H0. b. P 2.90 .0019 c. 1.2 1 1 2.33 1 .92 .8212 .1414 d. mn .2 2.33 1.28 .2 2 2 65.15 , so use 66. 24 Chapter 9: Inferences Based on Two Samples 6. a. H0 should be rejected if z 2.33 . Since z 18.12 16.87 2.56 1.96 40 32 3.53 2.33 , H0 should be rejected at level .01. b. 1 0 1 2.33 .50 .3085 .3539 c. 2.56 1.96 1 1.96 .1169 .0529 n 37.06 , so use 2 40 n n 1.645 1.28 n = 38. d. Since n = 32 is not a large sample, it would no longer be appropriate to use the large sample z test of Section 9.1. A small sample t procedure should be used (Section 9.2), and the appropriate conclusion would follow. Note, however, that the test statistic of 3.53 would not change, and thus it shouldn’t come as a surprise that we would still reject H 0 at the .01 significance level. 7. 1 Parameter of interest: 1 2 the true difference of means for males and 2 females on the Boredom Proneness Rating. Let 1 men’s average and 2 women’s average. H0: 1 2 0 3 H a: 1 2 0 4 5 6 7 z x y o x y 0 s12 s 22 m n s12 s 22 m n RR: z 1.645 z 10.40 9.26 0 1.83 4.83 2 4.68 2 97 148 Reject H0. The data indicates the average Boredom Proneness Rating is higher for males than for females. 25 s12 s 22 16 1. It appears that there could be a difference. the true average length of stays for patients given the treatment. No.0 129 129 6 28.6 10 2 2 1. and the p-value = 2[P(z > 1. we would expect most of the data to be within 2 standard deviations of the mean. and the distribution should be symmetric. d.1 19. There is very compelling evidence that the mean tensile strength of the 1078 grade exceeds that of the 1064 grade by more than 10. H0: 1 2 0 .588 .8 60 . The requested information can be provided by a 95% confidence interval for 1 2 : x y 1. The distribution is positively skewed.2 . a.7 6. b. The p value is larger than any reasonable . m n point estimate x y 19. With a normal distribution. H0: 1 2 10 Ha: 1 2 10 z x y o x y 10 s12 s 22 m n s12 s 22 m n RR: p value z 107.14 5. the p-value = 28.15. We will calculate a 95% confidence interval for μ.210 16. b.96 9.9 13.210 For a lower-tailed test.0.1271) = . so reject H0.44 .9 1.9 10. Let 1 1064 grade average and 2 1078 grade average.9 13. 26 39. 19.2 1. 2 sd’s above the mean is 98.96 10. c.7 2 39.1 15.1.21.Chapter 9: Inferences Based on Two Samples 8.8 60 60 2 6.2542.14)] = 2( .6 123. 1 2 3 4 5 6 7 Parameter of interest: 1 2 the true difference of mean tensile strength of the 1064 grade and the 1078 grade wire rod. There is no significant difference.3 2.57 0 . a. which is less than any . z 19.Ha: 1 2 0 .57 . but the distribution stops at zero on the left.412.9 9. so we do not reject H0.96 . 03 4 2. with Var ˆ 4Var X Var Y 2 1 m 2 2 n 2 2 Var ˆ .0025 . Ha: 0 . H 0 . so use n = 50.I.33 1.05 . so 2 . d = .33. 1 1 2 o 1 .31 N/mm2.6 59. The hypotheses are H0: 1 2 5 and Ha: 1 2 5 . Standard error = s . The test statistic is then ˆ (since o 0 ).0025 2.23 and -6.23. Because –1. so the use of the high purity steel cannot be justified. The appropriate hypotheses are H0: 0 vs.645 n . With 99% confidence we may say that the true difference between the average 7-day and 28-day strengths is between -11. .08 .9104 8.89 3.2 2 0. so 3. 27 . Since z 65.99. At level .2272 cannot be rejected in favor of Ha at this level.Chapter 9: Inferences Based on Two Samples a.9236 15.58 .41 higher than the corresponding average for female workers. The C. 1 2 .96 0.35 .6. so 0.97 and ˆ z 4. H0 is not rejected. .33.05 . b. With ˆ ˆ 2 2. .01.04. so normal is more than twice schizophrenic) The estimator of is 4 . z / 2 1. We are 95% confident that the true average blood lead level for male workers is between 0. is the square root of ˆ 2 X Y .96 . 49. and the test is one-tailed. .41 .2.08 .05 > -2.31 .46 m n 11.38 .0016 14.99 and 2.2891 .58 s12 s 22 8.5 3.53 .9236 . Substitution yields n m n x y z / 2 SE1 2 SE 2 2 5.69 6.05 .001. 13.08 . 11.3 43 45 2 2 . is x y 2.3 2 .77 2.8 5 2. and ˆ is obtained by replacing each i with S i . where 2 1 2 . 2272 X Y z /2 s12 s 22 .8 1.97 1. 12. ( 0 is equivalent to 2 1 2 . and H0 is rejected if z 2. H0 should be rejected if z 3.77 2.05. Using . z xy s12 s 22 n n .Chapter 9: Inferences Based on Two Samples a.27 18 . z = 1. decreases. 2 2 37. 16. the closeness of x and y suggests that there is essentially no difference between true average fracture toughness for type I and type I steels. so z As either m or n increases.694 .411 5 10 62 10 9 9 52 10 2 2 156 2 2 5 10 62 15 9 22 10 2 2 14 156 2 2 2 10 9 62 15 14 28 . z increases. The pvalue by itself would not have conveyed this message.2 2 .411 2 7. b. 52 10 2 2 106 2 c. Section 9.694 1. For n = 100.1586 .7 21 . z = 2. 1 2 o increases (the numerator is 1 2 o 2 o decreases. so z 1 positive). n increases also.43 17 .41 and p-value = 21 1. and since z is the numerator of n .41 .21 17.2 17.84 18.01 21. As decreases.44 2 24. a. so decreases. The very small difference in sample averages has been magnified by the large sample sizes – statistical rather than practical significance. n For n = 400.018 .0046. From a practical point of view.83 and p-value = . b. 1 9 29 .1265 =.0005) .0351125 . so we reject H0 and conclude that there is a difference in the densities of the two brick types. Using . The relevant hypotheses are H0: 1 2 0 vs. Ha: 1 2 0 .3329 15.007 . .6 1.40. We want a 95% confidence interval for 1 2 . For the given hypotheses.3 10 5. Because the interval is so wide. t .9 2.8241 2 5 115 .6.386 2 3. 3. 52 12 2 2 624 2 2 2 5 12 62 24 11 23 12.8 6 .6 2.07569 2 .007 20.46 . so use d.025.262 3.f. 18. since –1.2168 2 4.262 .764.82 2.01.95 .20 .9 2. .17)] < 2(. we don’t reject H0.164 2 6 p value .395 .07569 2 7 1.2405 2 .164 2 6 2 2 .0351125 2 . 20. 19. and 2 the corresponding value for CTS subjects.07569 .f. it does not appear that precise information is available.80 . so the interval is 13.78 6.Chapter 9: Inferences Based on Two Samples d. the rejection region is .73 21. 21.17 leads to a p-value of 2[ P(t > 6.0351125 .96 5.098 With H0: 1 2 0 vs. Let 1 the true average gap detection threshold for normal subjects. is t 4.2168 4.05 26 .032 6 9. which is less than most reasonable ' s .20 > -2.53 . = 9. and the test statistic 4 22. Ha: 1 2 0 . We will reject H0 if 5 t t .764. we will reject H0 if . or 15.84 26.2405 2 .001. the test statistic and the d.7 129.71 2.164 6 . and the test statistic t d.8241 2 4.f.240 2 5 5 t 2 2 6. We have insufficient evidence to claim that the true average gap detection threshold for CTS subjects exceeds that for normal subjects. we fail to reject H0. Since –2.46 is not 2.602 . 30 .15 2.Chapter 9: Inferences Based on Two Samples t t .602 .01. Since we are concerned about any possible difference between the two means.50833 StDev : 0.2442 0 and conclude that there does appear to be a difference between the two population average strengths.99 .012 5 11 2 2.670 P-Value: 1. 31 .93 3.530330 N: 24 Anderson-Darling Normality Test A-Squared: 0.20 .20 23.145 . it is plausible that both samples have been selected from normal population distributions.33 . We test H 0 : 1 2 0 vs. Normal Probability Plot for Poor Quality Fabric .3 1. a two-sided test is appropriate. a.999 .5 2.444206 N: 24 Av erage: 1. 22.80 .1201 2 2 4.50 .58 2 12 4.58 2 12 2 1. H a : 1 2 0 .0 2. we see that both plots illustrate sufficient linearity.000 Using Minitab to generate normal probability plots.159 . which we round down to 14.999 .20 .582 12 4.012 12 3.13 1. so .95 .58750 StDev : 0. which is 2.3964 14.396 P-Value: 0.8 1. The test statistic is t 19.1632 11 we reject H0 if t t .01 . 23.0039 .0 2.344 Anderson-Darling Normality Test A-Squared: -10.025. Therefore.05 .50 .001 .001 0.05 .01 .5 P: H: Av erage: 1.145 . so we reject H 1.8 1.3 1.Chapter 9: Inferences Based on Two Samples Let 1 the true average strength for wire-brushing preparation and let 2 the average strength for hand-chisel preparation.95 Probability Probability Normal Probability Plot for High Quality Fabric .80 . We need the degrees of freedom to find the rejection region: 1.99 .14 2. at the 5% significance level. 05 (not specified in the problem). 32 .4 + 1.7 relevant test statistic is t = 1.83 seconds b.5 extensibility (%) The comparative boxplot does not suggest a difference between average extensibility for the two types of fabrics. Ha: μ1 – μ2 > 0.05 4 1.76 4 64 49 ≈ 112. The true average time spent by blackbirds at the experimental location is not statistically significantly higher than at the natural location.0433265 2 . The 13.76 2 ) 2 2.05 2 1.671(2. Hence. There is insufficient evidence to claim that the true average extensibility differs for the two types of fabrics. Rounding to t = 1. which we round down to 10.Chapter 9: Inferences Based on Two Samples b.082. and using significance level . With degrees of freedom . so we statistic is t .228 .5 2. 24.5 1. which is not 2.38 . The test . Let μ1 and μ2 represent the true average time spent by blackbirds at the experimental and natural locations.025.37.00017906 10.05. we reject H0 if t t . Comparative Box Plot for High Quality and Poor Quality Fabric Poor Quality High Quality 0.65-1SE = 13.05 2 1.9. We wish to test H0: μ1 – μ2 = 0 v.5 . H a : 1 2 0 . c.0433265 cannot reject H0.4 and df = 120.05) = 16.228 in absolute value. we fail to reject the null hypothesis.10 2. the tabulated P-value is very roughly . respectively.08 . We test H 0 : 1 2 0 vs. with estimated df = 2.76 2 ( 2. a.4 9. 95% upper confidence bound: x + t. 3 1.306 and the desired interval is 40.5 28 731. The t-value 2.5 .8 2 3..7) ± (2. an upper tailed test is appropriate. For a 95% interval. The approximate degrees of freedom for this estimate are 2 11.5 88.9 2. i. which results in the interval 3.6.8 2.8 2 2 27 7 .96 sec.32 6 8.5 2 28 2 2 5 . but this number is very close to 54 – of course for this large number of df.00) 1.2 2.506 . Since we are interested in whether the potential drop is higher for alloy connections.175 7 t .32 6 8 2 2 2 .131 .59 8. we can no longer conclude that the means are different if we use a 95% confidence interval. We have sufficient evidence to claim that the true average potential drop for alloy connections is higher than that for EC connections.607 6. 95% CI for silvereyes’ average time – blackbirds’ average time at the natural location: (38.9 12. is t = 3. therefore. Because 0 does not lie inside this interval.306 5. so we reject H0.01.4674 2 18.3 21.269. that the two population means are not equal. we can conclude that the population means are not 33 . We calculate the degrees of freedom 5 .31. or about 54 30 (normally we would round down to 53.6362. so 11. Let 1 the true average potential drop for alloy connections and let 2 the true average potential drop for EC connections.0008 .0004 is less than the significance level of . Using the SAS output provided. the t value increases to about 2. using either 53 or 54 won’t make much difference in the critical t value) so the desired confidence interval is 91. we can be reasonably certain that the true difference 1 2 is not 0 and.Chapter 9: Inferences Based on Two Samples c.00 is based on estimated df = 55.76 2 5.5 2 28 731.0004 .68 5. Since this interval does contain 0. which we round down to 8.2 3.8 2 31 2 53. Our p- value of . Because 0 is not contained in this interval. the corresponding df is 37.32 8.931 .32 6 8 5 893. We test H 0 : 1 2 0 vs.4 – 9. there is strong evidence that 1 2 is not 0. H a : 1 2 0 .025.e.5.83 101.4 2. 26.3 8.83 18.306 11. 25.01 or so.06 2 = (17.95 . the test statistic.3. 27. and the p-value for our upper tailed test would be ½ (two-tailed p-value) = 1 2 .44 sec). when assuming unequal variances. 39. We test H 0 : 1 2 0 vs. Since 0 is not in this interval.24 26 there is no df = 37 row in Table A. We desire a 99% confidence interval. we reach exactly the same conclusion as before. Therefore.44 2 5 9 4.005.445 2 2 4. and the p-value from table A. is 34 .3 2 26 26 26 37.59 5 .023 . Let 1 the true average compression strength for strawberry drink and let 2 the true average compression strength for cola.2 2 4 .75 2 10 4.4 15 2. a.719 . A lower tailed test is appropriate.045.17 2 The degrees of freedom 22. Using 36 degrees of freedom (a more conservative choice).752 10 2. Calculating a confidence interval for 2 1 would change only the order of subtraction of the sample means.10) .10 so we reject H0 and conclude that the true average lean angle for older females is more than 10 degrees smaller than that of younger females. -6. The test statistic is t 44. which is < .08 2. H a : 1 2 10 . but the standard error calculation would give the same result as before. the population means are not equal. and the 99% C.3 . The data does suggest that the extra carbonation of cola results in a higher average compression strength.3).I.Chapter 9: Inferences Based on Two Samples equal. 36 2. We will test the hypotheses: H 0 : 1 2 10 vs.10 .5. t . 29. 1971. First we calculate the degrees of freedom: 2 2 . This p-value indicates strong support for the alternative hypothesis.8 is 3. H a : 1 2 0 . just the negatives of the endpoints of the original interval.36 25.4 2 15 2 14 14 29.75 2 10 2 2.8114 14 P(t 2.95 4 approx .2 2 4 . 30.4 2 29. so use df=25.44 2 5 4.5. which we would round down to 37.08 5. The p-value 77. the 95% interval estimate of 2 1 would be ( -31.3 2 26 26 2 2 .5 2. 28. The test statistic is x y 10 t 2. except that 2 . The 99% upper bound would be 9.434 .83 2. b.98.09 .32 .8 2.2 2 4.4 2.9473 7.719 9. We 26 26 are 99% confident that the true average load for carbon beams exceeds that for fiberglass beams by between 6.4 2.576 11.83 and 11. 35 .09 kN.4 42.Chapter 9: Inferences Based on Two Samples 33.6. The upper limit of the interval in part a does not give a 99% upper confidence bound.98 kN. meaning that the true average load for carbon beams exceeds that for fiberglass beams by at least 7. 2084 2 . a 95% confidence interval for mid range high range is 438.3 437.85 8.1183 . H a : 1 2 1 .818 ..79 2 14 13 2 1. 9 048.8) = .48 6.2084 1.Chapter 9: Inferences Based on Two Samples 31. b. . the standard 5% significance level).e.45 2. Comparative Box Plot for High Range and Mid Range 470 mid range 460 450 440 430 420 mid range high range The most notable feature of these boxplots is the larger amount of variation present in the mid-range data compared to the high-range data.g. Using df = 23. The test statistic is degrees of freedom 8.79 2 14 . 32..85 13 1. 36 With .83 . At α = . the p-value = P(t > 1.84. the interval spans zero) we would conclude that there is not sufficient evidence to suggest that the average value for mid-range and the average value for high-range differ. Otherwise.9. both look reasonably symmetric with no outliers present.069 6.05. We test H 0 : 1 2 1 vs. 15. Let 1 the true average proportional stress limit for red oak and let 2 the true average proportional stress limit for Douglas fir. there is sufficient evidence to claim that true average proportional stress limit for red oak exceeds that of Douglas fir by more than 1 MPa.28 2 10 13. We would reject H0 at significance levels greater than .65 1 t .28 2 10 2 1. a.12 17 2 Since plausible values for mid range high range are both positive and negative (i.046 (e.54 .69 7. 012 4 4 2 2 so t . b. Ha: μ1 – μ2 > 1. s 2 1.3. The pooled variance estimate is s 2p m 1 2 n 1 2 4 1 4 1 2 2 s1 s2 1. we fail to reject H0 at the 5% level.8 gives a one-sided P-value of . t . 6 2.571 . a pooled variance confidence interval for the difference between two means is x y t / 2 .447 1.64 . 34. Since our P-value > α = . Table A. 37 .025 .225 . Then the 1.20 .8 2 = 1. This interval is slightly wider.571 1. we need to calculate the degrees of freedom. so it does not support the conclusion that the two population means are different. The (5.05.70 2.012 4 4 3 . m n 2 s p 1 m 1n .025.1227 1 4 1 4 1.2252 1.094). but it still supports the same conclusion. We wish to test the hypotheses H0: μ1 – μ2 ≤ 1 v.34.571 .24.2 2 2.010 . respectively.447 and the desired interval is 13. c. 5 2. so s p 1.225 2 1.225 1. This interval contains 0. Following the usual format for most confidence intervals: statistic (critical value) (standard error).7938 .Chapter 9: Inferences Based on Two Samples 33.8) 1 t relevant test statistic is 3. Let μ1 and μ2 represent the true mean body mass decrease for the vegan diet and the control diet.012 4 4 .3971 5.8 3. We do not have statistically significant evidence that the true average weight loss for the vegan diet exceeds that for the control diet by more than 1 kg.3.098 (a computer will give the more accurate P-value of .9 12.90 . a.260 .3.943 . with estimated df = 60 using the 32 32 formula. y 12. s1 1. The sample means and standard deviations of the two samples are x 13. With df = m+n-1 = 6 for this interval.90 12.7 1.78 5 .20 2.010 m n 2 m n 2 4 4 2 4 4 2 1.2252 1.1227 .2 2. Using the two-sample t interval discussed earlier. 2 1.74 1. Rounding to t = 1.0686 3 interval is 13.33. we use the CI as follows: First. 25 0 11 .3 36.2 .021.998 6 t 7 Fail to reject H0.2) = . s D 11. First. The value of the test statistic is.5 5 1 1 2. and the p- value is P ( t < -2. There are two changes that must be made to the procedure we currently use. the equation used to compute the value of the t test statistic is: t x y sp 1 1 where sp is m n defined as in Exercise 34 above.Chapter 9: Inferences Based on Two Samples 35.021 > . we calculate sp as follows: 7 9 2 2 2.73 38 . then. The data does not indicate a significant mean difference in breaking load for the two fabric load conditions.544 .24 2. Assuming equal variances in the situation from Exercise 33.25 . 3 H0 : D 0 Ha : D 0 4 t 5 RR: t t . Section 9. 16 16 sp t 32. we fail to reject H0.01.8 40.8628 / 8 1. d 7.544 8 10 2.5 2.6 2.01. Second. 2 d D sD / n d 0 sD / n 7. The degrees of freedom = 16.8628 1 Parameter of Interest: D true average difference of breaking load for fabric in unabraded or abraded condition. 7 2. the degrees of freedom = m + n – 2. Since . 224 nanograms/m3.4239 ..4239 2. Then t . 38.32 2.025.55 = 4. which yields an IQR of 90. and a 95% confidence interval for the population mean difference between indoor and outdoor concentration is .95. d . This exercise calls for paired analysis. largely because of the amount of variation in the differences.037 . where d = (indoor value – outdoor value).00. Clearly.1 and the upper and lower quartiles are 88.Chapter 9: Inferences Based on Two Samples 37. We can 33 be highly confident.037 .3868 .32 s d 1 1n . The most significant feature of these boxplots is the fact that their locations (medians) are far apart. b.224.13715 . this is a wide prediction interval.2868 . This prediction interval means that the indoor concentration may exceed the outdoor concentration by as much as . The median of the “Normal” data is 46. that the true average concentration of hexavalent chromium outdoors exceeds the true average concentration indoors by between . First.5611 nanograms/m3.037 . A 95% prediction interval for the difference in concentration for the 34th house is d t. which yields an IQR of 49. for the 34th house.3758 .55 – 45.55 and 49. The median of the “High” data is 90.. Comparative Boxplots for Normal and High Strength Concrete Mix 90 80 70 60 50 40 High: Normal: 39 .55 = 2. These 33 differences are summarized as follows: n = 33. a.4239 2.3868 .3868 1 331 1. compute the difference between indoor and outdoor concentrations of hexavalent chromium for each of the 33 houses.55.40.2868 and . a.95 – 88.025.4239 .80 and the upper and lower quartiles are 45.55 and 90.3758 nanograms/m3 and that the outdoor concentration may exceed the indoor concentration by a much as 1. at the 95% confidence level.5611. s d . the corresponding P-value is around .018 < .025.23 2. A 95% upper confidence bound for μd = d + t.833(103. b.39. Taking differences in the order “Normal” – “High”.74 2. 41. This data is paired because the two measurements are taken for each of 15 test conditions.83 .9 ) = . and conclude that the data indicates that the higher level of illumination yields a decrease of more than 5 seconds in true average task completion time.178 / 9 corresponding p-value is P( t > 1.7)] = sD / n 228 / 14 2[.2 0 2. we conclude that the two population means are not equal. sd = 103.018. at the typical significance level of .845)/ No.05. c. i. We wish to test 105.. From the data. With d 7. with estimated df = 17 and P-value = . we reject H0 and conclude that true average TBBMC during postweaning does exceed the average during lactation by more than 25 grams.600 . we can reject H0 . There is strong evidence to support the claim that the true average difference between intake values measured by the two methods is not 0. at the 5% significance level. The test statistic is t = 103.047. b.047.6 t 1. A normal probability plot shows that the data could easily follow a normal distribution.46. Hence. H a : d 0 . We would reject H0 at any alpha level greater than .89 grams.34 42. we find d 42. a.145 .600 5 2.23 . far greater than 103. n = 105.7. 39.05.404 44.87 1. with test statistic t d 0 167. d = 105. the 1. Since . So. In turn.45. the new standard error is is roughly 235.23 2.7 25 the hypotheses H0: μd ≤ 25 v.009] = . Therefore.9sd/ 10 = 165.845.018. n = 10. 40 . Ha: μd > 25. If we pretend the two samples are independent.Chapter 9: Inferences Based on Two Samples b. we would reject H0. Because 42. There is a difference between them. With t .34 . Let μd = true mean difference in TBBMC. and s d 4.178 .7 + 1. we can conclude that the difference between the population means is not 0. We test H 0 : d 0 vs. and s d 4.39 4. at 9df.9 .845 / 10 2.329 (all using a computer). With degrees of freedom n – 1 = 8.145 15 0 is not contained in this interval.7 . a 95% confidence interval for the difference between the population means is 4. indicating that it is reasonable to assume that the differences follow a normal distribution. we have to work with the differences of the two samples. a. A normal probability plot of the 15 differences shows that the data follows (approximately) a straight line. H a : d 5 .e. the resulting t statistic is just t = 0. 40. postweaning minus lactation.63.14 2.845/ 10 . The two-tailed p-value is 2[ P( t > 2. 7.05. We test H 0 : d 5 vs. 11sd/ n = –0.81)/ 10 = (–2.028 (from Table 0 .8) Reject H0. this interval estimate includes the value zero. b.05. A 95% lower confidence bound for the population mean difference is: sd 23. the data follow a reasonably straight path.201(2.81. suggesting that true mean strength is not significantly different under these two conditions. sd = 2.63 1.6 13. A normal probability plot validates our use of the t distribution. 1. n = 12. Since n = 12.2 .025. 45. Although there is a “jump” in the middle of the Normal Probability plot.54 49.761 d t . c. 2 3 H0 : D 0 Ha : D 0 4 t 5 RR: P-value < .60 + 10.91 = 2858. d D sD / n d 0 sD / n 12. b.54.05 MPa).753 n We are 95% confident that the true mean difference between modulus of elasticity after 1 minute and after 4 weeks is at most 2858. We need to check the differences to see if the assumption of normality is plausible.14.73 ± 2. so there is no strong reason for doubting the normality of the population of differences. Let d exp osed un exp osed . A 95% upper confidence bound for μD sd is d t .54 = 49.7 0 2.645 2635.14 n 38. 15 We are 95% confident that the true mean difference between age at onset of Cushing’s disease symptoms and age at diagnosis is greater than -49. A 95% CI for μd is d ± t. with corresponding p-value . The normal probability plot below strongly substantiates that condition. 44. df = 9 6 t 7 A.15 508. a.54. In particular. From the data.60 1. 1 Parameter of interest: d denotes the true average difference of spatial ability in brothers exposed to DES and brothers not exposed to DES.63 222.Chapter 9: Inferences Based on Two Samples 42.18 38.73. a. 41 . 16 2635.05.52 MPa. d = –0. we must check that the differences are plausibly from a normal population.5 43.05. The data supports the idea that exposure to DES reduces spatial ability.60 10. A 95% upper confidence bound for the corresponding population mean difference is 38.14. Let μd = the true mean difference in strength between curing under moist conditions and laboratory drying conditions.14 . 0432 1 6.2 1.72 .18 2.150 4.0 . 180 300 180 .4 47.96 . so power = 1.0359 rejected. .96 .84 . y 4 21. H0 is rejected. while s1 = s2 = 8.33 .14 . H0 will be rejected if z z . 48. and pˆ 2 . Section 9.16. y1 6.2 1 .2875 .5 -5. pˆ . H0 is .2067 4.263 . and x 4 .2100 .5 0. y 2 15. and 1). so sp = 8.54 2.0421 .275 and .0 Differences 2.0432 .5 .96 . b. Because 4. -7. a.2875 .01 2. With d 1 and s d 0 (the dI’s are 1.300 .20 . With p pˆ z 30 80 210 .0421 .84 1.0 x1 .18 .150 .4167 . With p pˆ 2 z 63 .300 . ˆ 1 . ˆ1 H0 will be rejected if z 1.263 .33 . p .2100 . 42 . the proportion of those who repeat after inducement appears lower than those who repeat after no inducement.150 . and qˆ .5 5.0427 Since 4.96.96 and t = . and 300 75 63 75 .0 -2.9967 . x 2 .737 1 200 1 600 . x3 .0432 . The calculated test statistic is 200 600 800 .4167 .Chapter 9: Inferences Based on Two Samples Normal Probability Plot of Differences Normal 99 95 90 Percent 80 70 60 50 40 30 20 10 5 1 46. 1. 1.7125 3001 1801 . y 3 1.96 .737 . 0160.78.8) 2 (. b.8333)[529 1 563 1 ] = 0.1910 .433 > α = .176 . p-value = . 52.96 pˆ 1 pˆ 2 z / 2 pˆ1qˆ1 m ˆ ˆ p2nq2 224 171 395 395 126 140 266 266 395 .65) / 2 1. Let p1 and p2 denote the true incidence rates of GI problems for the olestra and control groups. 1210.4247 Fail to Reject H0. 1 3 H 0 : p1 p 2 0 H a : p1 p 2 0 4 z 5 Reject H0 if p-value < . p2 = true proportion of untreated bulbs that are marketable.2)(. The two-sided P-value is 2P(Z ≥ 0.39 .0774 . Ha: p1 – p2 ≠ 0.0934 . The pooled ˆ proportion is p statistic is z = 529(.05) 2 common sample size of m = n = 1211 would be required. Let . H 0 : p1 p 2 0 .05 .96 n (. Parameter of interest: p1 – p2 = true difference in proportions of those responding to two different survey covers. so a Let p1 = true proportion of irradiated bulbs that are marketable. The test statistic is z 43 pˆ 1 pˆ 2 pˆ qˆ 1 m 1 n ˆ1 .1667)(.158 (.28 (.85) (. and 180 . 1. The data do not suggest a statistically significant difference between the incidence rates of GI problems between the two groups. We wish to test H0: p1 – μ2 = 0 v.10 2 6 7 50.35)(1. The data does not indicate that plain cover surveys have a lower response rate.158) = . A 95% confidence interval is 224 395 126 266 1.05. With p 153 . respectively.1667. p2 = Picture. a.. hence we fail to reject the null hypothesis. from which the relevant test 529 563 . The hypotheses are H 0 : p1 p 2 0 vs.1708 .15)(. z pˆ 1 pˆ 2 pˆ qˆ m1 1n 104 207 109 213 213 207 420 420 2071 2131 . Let p1 = Plain.176) 563(. 266 51.Chapter 9: Inferences Based on Two Samples 49.78) = .850 . 44 . Taking the antilogs of the upper and lower bounds mx ny gives the confidence interval for itself.850 . 54.2 . z 180 360 .933 .0037 .661 .598 .244 1801 1801 . pˆ .43.360.1213 . The Z statistic is then . so Var X3 X2 X X2 n pˆ 2 pˆ 3 3 . 53. pˆ 2 pˆ 3 X2 X3 n n d.68 . ln ˆ .31 times more likely to suffer a heart attack than those who do.. a.68 . Then taking the antilogs of the two bounds gives the CI for to be 1. X1 X 3 X1 X 2 n X3 X2 .818 .2 0 . so reject H0 at any reasonable level.43 and 2.1213 . 034 104 11. H0 can be rejected but at level . thus we wish to test H 0 : p 3 p 2 versus H a : p3 p 2 .96 . n p p3 X3 X2 2 . so p1 + p3 > p1 + p2 becomes p3 > p2. 037 1. ln ˆ z / 2 b. At level . p2 = p3.037 104 . This suggests aspirin therapy may be effective in reducing the risk of a heart attack.001 H0 would not be rejected.756 . a.661 . so the CI for ln is 11.598 1.045 The p-value = 1 4.845 10. A 95% large sample confidence interval formula for ln is mx n y . b.836 .034 189 11.Chapter 9: Inferences Based on Two Samples pˆ 2 119 272 .31 . The estimator of (p1 + p3) – (p1 + p2) is c.756 .01. so P 1 2. We are 95% confident that people who do not take the aspirin treatment are between 1. ˆ 189 11. . The “after” success probability is p1 + p3 while the “before” probability is p1 + p2.2. and the standard deviation is 10. The computed value of Z is 200 150 200 150 2. Radiation appears to be beneficial.189 4. which is estimated by n n When H0 is true. 690 1.10.16 .95.5 4.64 2(.74 P F 4.6.95 . a.01.10.177 .5.5 . we can say that the upper-tailed p-value is between .14 .8 d.71 1 . 4 6.5. 56.550 .8.64 .9.690 .01.00 is less than F.07 .95 .95. and the 95% C.30 f.05.8 1 F. so L=. 5.82 . Since the given f value of 4. From Table A.5.212 .5. Section 9.177 F 4. 1 . 2.5 e.8. 45 .207 .5 57.05.12. F.Chapter 9: Inferences Based on Two Samples 55.05.05.10.106 .10 h. Since the given f of 2. F.33 and F.96 n=769. row 8.25 .94 . The two tailed p-value = 2 P F 5.10 5. column 5. 4.10. F.8 3.21 .96 .1 requires n n n Using p1 = q1 = p2 = q2 = .I.75 falls between F.01.74 P F .12 g.35. From column 8.7719 .01 . is 40 42 .99.271 F.64 P . pˆ 1 15 7 29 .5.02 .10.10 3.177 .25 .01) . 58.10 2.5 1 . the p-value > . F..52 . so P F 6. L 21.99.69 .5.01. F. b.550 .01. F. 1 c.8. F.01 and .16 . F. a. c.05. b. Since F. pˆ 2 .12 4. row 5.5. 0995 . For the hypotheses H 0 : 1 2 versus H a : 1 2 . So we can say .05.9.384 .10 1 . Since . 46 .5 F.05.95. 60.10. For a lower tailed test.10.384 is in neither rejection region. Since . 4 6.08 . however looking at both df = 30 and df = 40 columns.814 2.9 1 3.5 (but obviously closer to . = m – 1 = 10 – 1 = 9.10) = . we find a test statistic of f = 1. 59.2110 . = n – 1 = 8 – 1 = 7. 61. With numerator d. We wish to test H 0 : 1 2 vs. f 1.44 2 .001.10.f. we see that for denominator df = 20.40).95.5 F. we must first use formula 9.001< p-value < .5. F. We can say that the pvalue > .9 to find the critical values: 1 1 F. There is not sufficient evidence that the standard deviation of the strength distribution for fused specimens is smaller than that of not-fused specimens. Let 12 variance in weight gain for low-dose treatment. 62.f.72 F. of 35 in Table A.63 . 4 1 F.f.44) (40. so we cannot reject H0.01.5.10 .99.75 2 4. = m – 1 = 10 – 1 = 9. H a : 1 2 .05. we reject H0 if f F.2110.10 .9.00 or f F. which is obviously > . H a : 12 22 .05. The calculated test statistic is 2.9.7 . With numerator d.01 < p-value < .51 indicates the P-value is greater than 2(. 1. and denominator d.05. we test H 0 : 1 2 vs.22.05 if f F. The test statistic 2 is f f 2 2 2 s12 . and 22 variance in weight gain for control condition. and denominator d.85 2.20. we do not reject H0 and conclude that there is no significant difference between the two standard deviations.0995 < f = . and we reject H0 at level .5.22 < 1. H a : 1 2 . = n – 1 = 5 – 1 = 4.3 2 205. The calculated test statistic is f 277.f.f. so reject H0 at level .200 < . With 1 true standard deviation for not-fused specimens and 2 true standard deviation for fused specimens.01.90.01 and F.814 . The data does suggest that there is more variability in the low-dose weight gains. 22 2.08 .05 F. There is no column for numerator d. s 22 54 2 32 2 2. .10. F.19.3030 . At df = (47.10.Chapter 9: Inferences Based on Two Samples d.10. e.05).9 2 1. our f value is between F.01.275 .9. 4. We test f H 0 : 12 22 vs. H0 is not rejected.Chapter 9: Inferences Based on Two Samples Hence. The data does not suggest a significant difference in the two population variances. 47 . the C.05.m 1. Supplementary Exercises 65.22) = 2( . The p-value for a two- 9 tailed test is approximately 2P(T > 3.m 1.6 50 241 50 3. 1. And. 48 . 9. We are confident that the ratio of the standard deviation of triacetate porosity distribution to that of the cotton porosity distribution is at most 8.18 .59 2 3.9. The approximate d. S 2 / 2 P F1 / 2. The data indicates a significant difference.524 .Chapter 9: Inferences Based on Two Samples 63.28 and F. This small of a p-value gives strong support for the alternative hypothesis.108 .15.95. is . we are assuming the data were randomly sampled from their respective populations.28 is (. n 1 parentheses is clearly equivalent to 2 .003) = . n 1 22 S 22 F / 2. I.m 1.023.10. we are assuming here that compression strengths for both fixed and floating test platens are normally distributed.1 2 9 15.99). 1.9 s12 2 is (. as always.006.05. which we round down to 15. The test statistic is t x y s12 s 22 m n 807 757 27 2 412 10 10 241 2 72.41).n 1 1 .3 074. 22 12 1 .f. so 1 s 22 F.3. for 64. The set of inequalities inside the S2 / 2 2 S 2 F1 / 2. m = n = 4. 1 3. S12 1 S12 22 2 2 Substituting the sample values s1 and s 2 yields the confidence interval for 2 .m 1. H a : 1 2 0 .79 2 8. Due to the small sample sizes (10 each). 3 9.n 1 12 12 F / 2.160 and s2 = .22 15. and 1 taking the square root of each endpoint yields the confidence interval for we need F. Then with s1 = .10 . 3.9 2 168. and for A 95% upper bound for 2 is 1 2 . We test H 0 : 1 2 0 vs. Chapter 9: Inferences Based on Two Samples 66. Comparative Boxplot of Tree Density Between Fertilizer Plots and Control Plots 1400 Fertiliz 1300 1200 1100 1000 Fertiliz Control Although the median of the fertilizer plot is higher than that of the control plots. Since this interval contains 0. and pˆ 2 . (d.673 . and since we have a lower tailed test. 0 is a plausible value for the difference. and a twotailed p-value of . the p-value will be > . H a : 1 2 0 yields a t value of -. At this point we notice that since pˆ 1 pˆ 2 . H 0 : p1 p 2 0 . 49 . 75 66 . p2 = true proportion of returned questionnaires that included an incentive. c. We fail to reject H0. We would fail to reject H0. A test of H 0 : 1 2 0 vs. This data does not suggest that including an incentive increases the likelihood of a response. 67.682 . = 13).f. With 95% confidence we can say that the true average difference between the tree density of the fertilizer plots and that of the control plots is somewhere between –144 and 120.20. a. The hypotheses are H 0 : p1 p 2 0 vs.85. the data does not indicate a significant difference in the means.5. The test statistic is pˆ 1 z pˆ 1 pˆ 2 pˆ qˆ m1 1n . the fertilizer plot data appears negatively skewed. b. Let p1 = true proportion of returned questionnaires that included no incentive. while the opposite is true for the control plot data. which further supports the conclusion based on the p-value. the 110 98 numerator of the z statistic will be > 0. n = 11. A 95% confidence interval for the difference between the true average strengths for the two types of joints is 80. For a 90% n1 n 2 1.1517. The . we cannot say that there is a significant difference between them. k = 95%.77.3 1691. so 473.9 63. Furthermore. Summary quantities are m = 24. 70.09.23 1. 50 9.23 3.77 . That is.37 . and n = 10. y 101. x 103.66 . the strength of a single joint with a side coating would be at least 51.35 609. With s 2p 13.9 473.) b.73 .025.14 43.60.025.23 20. 10 with a confidence level of 95%.59 2 10 5.83. d.23 t . A 95% lower prediction bound for the strength of a single joint with a side coating is x t . the confidence interval is 2. s2 = 3.70 1 24 111 2. we find s12 s 22 1082.96 2 10 .Chapter 9: Inferences Based on Two Samples 68. Equating this value to the expression on the right of the 2 x1 x 2 95% confidence interval formula.70 .03 .74.96 interval. That is. c. 69. That is.18 and 5.3.45 59.3 .833 63.55 2.18.23 3. We use the pooled t interval based on 24 + 11 – 2 = 33 d. The tolerance critical value is obtained from Table A.645 552.35 .025. a two-sided tolerance interval for capturing at least 95% of the strength values of joints with side coating is x (tolerance critical value)s.68 and s p 3.025. the associated z value is 1.379 5.28. For a confidence level of 95%.645.3 908. 95% confidence requires t.9 s 1 1n 63. s1 = 3. We are confident that the difference between true average dry densities for the two sampling methods is between -. we can be highly confident that at least 95% of all joints with side coatings have strength values between 43.3 1082.96 n is x t .95 63. with a confidence level of 95%.55 2.03 3.96 1 101 63.78 .5.6 299.46 51.33 2.23 1.23 11.3 1.9 609. the average strength of joints with a side coating is at least 59. a.11 . the interval is 63.28 . The center of any confidence interval for 1 2 is always x1 x 2 .96 63. Because the interval contains 0. so the 90% confidence interval is then 609.6 552. half of the width of this interval is 2 1691.f.78 (Note: this bound is valid only if the distribution of joint strength is normal. Thus. A 95% lower confidence bound for the true average strength of joints with a side coating s 5.833 5..09 and 83.37.9 .6 with 95% confidence.6 . The large sample 99% confidence interval for 1 2 is 3975.14.33 lb-in.15 2. we have insufficient evidence to claim that the population mean penetration differs for the two types of bearings.131 3.57 17.120 .85 . the confidence interval spans zero.85 4.025. 71.1340 .61. then. we can say that the true average strength for joints without side coating exceeds that of joints with side coating by between 10.18 18.25 .7. m = n = 40.33 . The bound on the error of estimation is quite large.Chapter 9: Inferences Based on Two Samples 91.25. We would have 17 4.0 .61 10. First compute the difference between the amount of cone penetration for commutator and pinion bearings for each of the 17 motors./in.16 2. The interval is .11 and 25.0 2. 17. With 95% confidence.12 293. y 2795.43 22. s1 = 245.109681 35. Then t .1.05 . These 17 differences are summarized as follows: n = 17. 245.0 1560 1020. s2 = 293.131 . with very high confidence.109681 2 35.11.18 . x 3975. and t . the value of 1 2 is not 0.f.72 2.105216 2 approximate degrees of freedom is 91.105216 2 9 15.7 2 40 40 The value 0 is not contained in this interval so we can state that. This exercise calls for a paired analysis.025 .0 2795.0 . 51 .18 2. 72. Because of this. d 4..120 to say that the population mean difference has not been precisely estimated. which is equivalent to concluding that the population means are not equal. where d = (commutator value – pinion value).72 7. so we use 15 9 d.58 1180 . and the 95% confidence interval for the population mean difference between penetration for the commutator armature bearing and penetration for the pinion bearing is: 35. Also. s d 35. 1 18.3309 101. Then the hypotheses are H 0 : 1 2 0 vs. the results are different. H0 is rejected and we conclude that the average tear length for Brand A is larger than that of Brand B. No. The test statistic is t 31. so we would reject H0. s1 = 14. the two-sample t test is appropriate.8.252 120. which we round down to 27. m = 16. The data supports the article’s authors’ claim. the two-tailed p-value 2 .5 .25 . so we use 18. which is less than the specified significance level. H a : 1 2 0 . x 74.8. Assuming both populations have normal distributions.2. From Table A.1 120.84 .0 .0 .64 .97 13 2.0 61. Assuming both populations have normal distributions. m = 11. the p-value = P( t > 2. a. From Table A.007. The relevant hypotheses are H 0 : 1 2 0 vs.9207 freedom 31.5 2 16 14 2 2 approximate d.086083 2 .66 . Let 1 denote the true mean headability rating for aluminum killed steel specimens and 2 denote the true mean headability rating for silicon killed steel. Since we can assume that the distributions from which the samples were taken are normal.047203 . H a : 1 2 0 .Chapter 9: Inferences Based on Two Samples 73.012 . obviously.6) = .086083 57. The test statistic is t .6 . y 61. is .014 .3309 2 101.252 2 18.5 2 27. 75. 14. H a : 1 2 0 .006 .05. y 129.03888 .047203 2 29 . 74.f. s2 = 12. Let 1 denote the true average tear length for Brand A and let 2 denote the true average tear length for Brand B. 52 .5. The approximate degrees of freedom .9207 2 10 14 2. The two-tailed p-value 29 2 . n = 14.5 2 16 14 15 The test statistic is t 74.8 2 12.7. The relevant hypotheses are H 0 : 1 2 0 vs. x 98. The approximate degrees of 18. At a significance level of .028 .66 2.8 2 12. we use the two-sample t test. n = 15. so the 2 14. so we use 57.03888 2 . the two-sample t test is appropriate.0 14. s2 = 39.8 2 16 1214. s1 = 14.1 .2 .1. 05. = m + n – 2 = 8 + 9 – 2 = 15. this data does not show that the heated average strength exceeds the average strength for the control population.024 and s d .Chapter 9: Inferences Based on Two Samples b. 78.556 . There is insufficient evidence that the true average strength for males exceeds that for females by more than 25N. .004) = .0305 .02. 76. the test statistic changes to t 31. From Table A.073.742 1 8 19 3.06.252 .9 4.49 . we fail to reject H0. a.05. The mean of a lognormal distribution is e / 2 . Therefore. At significance level . The relevant hypotheses are H 0 : 1 2 0 (which is equivalent to saying 1 2 0 ) versus H a : 1 2 0 (which is the same as saying 1 2 0 ). the lower tailed p- 5 value associated with t = -1. where d denotes the difference between the population average control strength minus the population average heated strength.8 ) = . H a : 1 2 25 . -.742 .8. The test statistic is s 2p t x *y * sp 1 m 1n 18. Assuming both populations from which these samples were taken are normally 53 .e.278 .8 is P( t < -1. where and are the parameters of the lognormal distribution (i. The test statistic is t . For the hypotheses H 0 : 1 2 25 vs.0 11.0 . the p-value P t .76 1. The pooled variance is m 1 2 n 1 2 8 1 9 1 2 2 s1 s2 4. H0 is rejected and we conclude that there is a difference between 1 and 2 . The sample mean and standard deviation of the differences are d . H0 should not be rejected. No. Since the p-value is greater than any sensible choice of .8 . The pooled t test is based on d. At significance level . The relevant hypotheses are H 0 : d 0 vs. then 1 2 would imply that 1 2 . and -.1 25 120. This is paired data.0305 1. b. which is equivalent to saying that there is a difference between 1 and 2 . With degrees of freedom 18. H a : d 0 . 0. Let 1 denote the true average ratio for young men and 2 denote the true average ratio for elderly men. then even if 1 2 . From Table A. So when 1 2 . the p-value associated with t = 3.0 is 2P( t > 3.024 .05. = 5 – 1 = 4. with d.6 .6 m n 2 m n 2 8 9 2 89 2 22. so the paired t test is employed.04 3. The observed differences (control – heated) are: -. the two means 1 and 2 (given by the formula above) would not be equal.008. the mean and standard deviation of ln(x)).7. 77.01.f..0 ) = 2(.f.8) = P( t > 1. However. when 2 1 2 . so s p 4.0 4. 97)/ 6 = 325. We wish to test H0: μ37.5 .71 .Chapter 9: Inferences Based on Two Samples distributed.09) ± t.09).9 34. = 20 and the p-value is P(t > 13 12 7.08 2 6 6 approximate P-value is .276) = (289. The relevant estimated df = 9.81. We are 95% confident that the true average breaking force in a dry medium at 37° is between 289.73 170.5) ≈ 0. 362.43). thus a t-test is not appropriate for this small a sample size.70.03 N and 362.09 N more than the true average breaking force in a wet medium at 37°. = 54 .05 .22 2 .dry – μ37.73 ± t.dry – μ22.58. c. (The plot for poor visibility isn’t as bad. The normal probability plot below indicates the data for good visibility does not follow a normal distribution. the relevant hypotheses are H 0 : 1 2 0 vs. H a : 1 2 0 .97 2 = (–30.60) 100 = 2. we reject H0.dry – μ22.025. b.f.97 2 41.025. We are 95% confident that the true average 6 6 breaking force in a dry medium at 37° is between 30. The estimated df = 9 again.43 N.73 – 306. A 95% CI for μ37.wet = (325.47 6.dry = 325.dry > 0. and the 34. 79.5(34. The value of the test statistic is 7. Since the p-value is .03.dry = 0 v.571(14. The relevant test statistic is t (325. A 95% CI for μ37. we reject H0 and conclude that true average force in a dry medium at 37° is indeed more than 100 N greater than the average at 22°. We have sufficient evidence to claim that the true average ratio for young men exceeds that for elderly men. The d.015.) 99 95 90 80 Percent 70 60 50 40 30 20 10 5 1 -1 0 1 Good 2 3 80.97 2 39.73 ± 2. Hence.81 N less and 70.28 2 t 7. a. Ha: μ37. 88 .88 t 1.181 14 12 With the pooled method. Pooled: The degrees of freedom are m n 2 14 12 2 24 and the pooled variance is 13 11 2 2 . However.757 1.79 2 14 15. We obtain the differences by subtracting intake value from expenditure value.52 1. 83. 3n n b.035) = . We then wish 20 2 2.9 )] = 2( .070. Ha: 1 2 0 . a.197 3. there are more degrees of freedom. We wish to test H0: 1 2 versus Ha: 1 2 Unpooled: With H0: 1 2 0 vs. which yields n = 47. We are testing the hypotheses H0: μd = 0 vs Ha: μd ≠ 0.001.181 .48 9. That is.522 14 12 2 2 . we will use a paired t test.88 1. m = 141. Taking the derivative with respect to n and 400 n n 55 . we would reject the null hypothesis and conclude that there is a difference between average intake and expenditure.95 15 52 1.79 1. The test statistic is 24 24 .05 or . the first is m = 3n. Using either significance level .79 2 1. if we are willing to assume equal variances (which might or might not be valid here). We wish to find the n which minimizes 2 z / 2 n which minimizes 900 400 . we would not reject. 82.36 .79 2 1.046) . The p-value = 2[P( t > 1. Test statistic t 1.522 14 12 13 t .Chapter 9: Inferences Based on Two Samples 81. we will reject H0 if p value . and the test statistic 11 8. or equivalently. so s p 1.89 . and the p-value is smaller than with the unpooled method. the 400 n n 900 400 .3970 . Because of the nature of the data.465 1.092. With n denoting the second sample size. at significance level .88)] ≈ .58 900 400 . the pooled test is more capable of detecting a significant difference between the sample means.4869 = .01.8)] =2(.12 2 .008. 2 .88 with df = n – 1 = 6 leads to a p-value of 7 2[P(t>3.81 leads to a p-value of about 2[P(t15 > 1. 24 1 1 . 0043) .129 . The data for this analysis are precisely the differences utilized in parts a 4. Since 0.10.15) 0 ≈ 0. the test statistic is t = 56 .676.01). respectively.01). 6. Hence. n = 17.5μ2 v. on average.73) = .05. The estimated df = 20 17 19 (using software).129 911 1101 .871 . H a : p1 p 2 0 . Ha: μ1 – μ2 > 0. A 2. . qˆ . d = 0.603 2 = 3. 91 110 201 . d = 4.0086 .927 .066. The data does not suggest that the average tip after an introduction is more than 50% greater than the average tip without introduction.54. a.5 X 2 .83 < 1.63 2(.83. Let μ1 and μ2 represent the true mean differences in side-to-side AP translation for pitchers and position players.001. Plug in (1. Hence. . in pitchers than in position players. The test statistic is pˆ 1 z z pˆ 1 pˆ 2 pˆ qˆ m1 1n .03 and 6. We are 95% confident that the true mean difference between dominant and nondominant arm translation for position players is between 2.5(14.066 0. The hypotheses are 2 H 0 : p1 p 2 0 vs.120.955 2 1. and b.03 and 6.025. For the paired data on position players. .05. The two survival rates appear to differ.16 = 2. With 73 102 175 . Let p1 = true survival rate at 11 C . and the resulting 95% CI is (2. We are 95% confident that the true mean difference between dominant and nondominant arm translation for pitchers is between 2.233 3.676. 85.8975 conservative df estimate here is ν = 50 – 1 = 49. m = 400 – n = 240.233. whose estimated standard error equals s (ˆ) 12 2 s12 s2 (1.000 0 .03.49 ≈ 1.5) 2 2 .871 . 1. p2 = true survival rate at 30 C . 86. pˆ . and pˆ 2 . and t.5) 2 2 . so reject H0 at most reasonable levels (.802 . and sd = 1.0475 The p-value = 2 2. We wish to test the hypotheses H0: μ1 – μ2 = 0 v.603. 2 400 n 2 0 . or 5n 2 3200n 640. H0: θ ≤ 0 v. and the resulting 95% CI is (–0. Ha: μ1 > 1. whence 9n 2 4 400 n . we fail to reject H0 at the 5% significance level. even at the 1% level.025. we concur with the authors’ assessment that this difference is greater. and the corresponding P-value is P(t > 3. t. n = 19. Ha: θ > 0. b.10. using the hint. t.125 2.73.63 1. c.5μ2 – or. We want to test the hypothesis H0: μ1 ≤ 1.802 .10.10). using the fact that V (ˆ) n1 n2 n1 n2 the values provided to get a test statistic t = 22.18 = 2.101.927 . For the paired data on pitchers.63 . and sd = 3. Our point estimate of θ is ˆ X 1 1.955. This yields n = 160.Chapter 9: Inferences Based on Two Samples equating to 0 yields 900 400 n 84. 34 .0834 . and m = n = 8.8264.025. giving . . .0294. s1 2. With x 11 . so no separate assessment could be made.0901 . 88.68 .0079 name than for a black name. d = 1.Chapter 9: Inferences Based on Two Samples 87. A lower tailed test would be 34 46 appropriate. and t = .34 .05 . In the situation described.000 respectively. 100. 89.95. 1 2 1 would have little practical significance. If the i ' s referred to true average IQ’s resulting from two different conditions.0668 .21 . yet very large sample sizes would yield statistical significance in this situation.2 . so H0 is rejected . n 200 14. H 0 : p1 p 2 will be rejected at level z z. so 1. H 0 : 1 2 0 is tested against H a : 1 2 0 using the two-sample t test.131 or if t 2.0000 for n = 25.131 .9015.15 2. p .g.. It appears that a response is more likely for a white . rejecting H0 at level .96.0332 4. pˆ 2 167 2500 . and ˆ . and . 90. so the p-value 1. With z pˆ 1 250 2500 in favor of H a : p1 p 2 if either .645 . 2500.10 . and 10. so H0 is not rejected. 57 .79 . a randomized block design). y 9. so we would not judge the drug to be effective.05 if either t t .142 . the effect of carpeting would be mixed up with any effects due to the different types of hospitals. The experiment should have been designed so that a separate assessment could be obtained (e. The computed value of Z is z 34 46 1. . 1 2 10 . s 2 3.20 .142 n n 0 0 .645 14.05 1. sp = 2. 94 ± 1. The test statistic value is t 1402.318083 2 4.699 . 2. 92. H0 is not rejected.97 2 9.24 1419.24 1400 10.557 . With the obvious point estimates ˆ1 X . True average weight does not appear to exceed 1400. or A large-sample confidence interval for λ1 – λ2 is (ˆ ˆ ) z 1 2 /2 m n x y .94 ± .62 and y 2.29.1 . mn (X Y ) 0 X Y Z ˆ 2 Y . With 1 1 ˆ n m m n x 1.35 = (-1.f. First. With degrees of freedom = 29. 30 7.616 and y 2.56 .43 . a.39 4. 93. where λ can be estimated for the m n m n mX nY variance by the pooled estimate ˆ . respectively. t .1 < 1.-. With x 1. Let 1 and 2 denote the true average weights for operations 1 and 2. so we can reject H0 at level . we reject H0 if m t t . ˆ1 ˆ2 .63 10.5 . 29 1.Chapter 9: Inferences Based on Two Samples 91.3 . 57 2. Var X Y 1 2 1 1 under H0.025.699.3 and p-value = 2 5.318083 6. H 0 : 1 1400 will be tested against H a : 1 1400 using a one-sample t test with test statistic t x 1400 s1 .96(. The data indicates that there is a significant difference between the true mean weights of the packages for the two operations. we have a large-sample test statistic of X Y .30672 2 29 57.05. so we would certainly reject H 0 : 1 2 in favor of H a : 1 2 .39 7. ( x y ) z / 2 58 .011363 3. 17.0006 .59).00 Because 1.24 1. The relevant hypotheses are H 0 : 1 2 0 vs.011363 2 3.97 30 2.96 2 30 The d. z = -5. the 95% confidence interval m n for λ1 – λ2 is -.30672 17. so use df = 57. b.05. The value of the test statistic is t 1402.177) = -. H a : 1 2 0 .000 29 .