Chapter 7 Confidence IntervalsTrue/False 1. The t distribution always has n degrees of freedom. Answer: False Difficulty: Easy 2. Assuming the same level of significance α, as the sample size increases, the value of tα/2 approaches the value of zα/2. Answer: True Difficulty: Medium 3. When constructing a confidence interval for a sample proportion, the t distribution is appropriate if the sample size is small. Answer: False Difficulty: Medium 4. When the population is normally distributed and the population standard deviation σ is unknown, then for any sample size n, the sampling distribution of X is based on the z distribution. Answer: False Difficulty: Medium (REF) 5. When the sample size and sample standard deviation remain the same, a 99% confidence interval for a population mean, µ will be narrower than the 95% confidence interval for µ . Answer: False Difficulty: Medium (REF) 6. When the level of confidence and sample standard deviation remain the same, a confidence interval for a population mean based on a sample of n = 100 will be narrower than a confidence interval for a population mean based on a sample of n = 50. Answer: True Difficulty: Medium 7. When the level of confidence and the sample size remain the same, a confidence interval for a population mean µ will be wider, when the sample standard deviation s is small than when s is large. Answer: False Difficulty: Medium Bowerman, Essentials of Business Statistics, 2/e 192 8. When the level of confidence and sample proportion p remain the same, a confidence interval for a population proportion p based on a sample of n = 100 will be wider than a confidence interval for p based on a sample of n = 400. Answer: True Difficulty: Medium 9. When the level of confidence and sample size remain the same, a confidence interval for a ˆ ˆ ˆ ˆ population proportion p will be narrower when p (1 − p) is larger than when p (1 − p) is smaller. Answer: False Difficulty: Medium (REF) 10. When solving for the sample size needed to compute a 95% confidence interval for a ˆ population proportion “p”, having a given error bound “B”, we choose a value of p that makes ˆ ˆ p (1 − p) as small as reasonably possible. Answer: False Difficulty: Medium (REF) 11. When determining the sample size n, if the value found for n is 79.2, we would choose to sample 79 observations. Answer: False Difficulty: Medium (REF) Multiple Choice 12. The t distribution approaches the _______________ as the sample size ___________. A) Binomial, increases B) Binomial, decreases C) Z, decreases D) Z, increases Answer: D Difficulty: Medium (REF) 13. The width of a confidence interval will be: A) Narrower for 99% confidence than 95% confidence. B) Wider for a sample size of 100 than for a sample size of 50. C) Narrower for 90% confidence than 95% confidence. D) Wider when the sample standard deviation (s) is small than when s is large. Answer: C Difficulty: Medium 193 Bowerman, Essentials of Business Statistics, 2/e 14. As standard deviation increases, samples size _____________ to achieve a specified level of confidence. A) Increases B) Decreases C) Remains the same Answer: A Difficulty: Medium 15. When determining the sample size, if the value found is not an integer initially, the next highest integer value will ____________ be chosen. A) Always B) Sometimes C) Never Answer: A Difficulty: Medium 16. When constructing a confidence interval for a population mean, if a population is normally distributed and a small sample is taken, then the distribution of X is based on _____ distribution. A) z B) t C) Neither D) Both A and B Answer: B Difficulty: Medium 17. A confidence interval increases in width as A) The level of confidence increases B) n decreases C) s increases D) All of the above Answer: D Difficulty: Medium (REF) 18. The width of a confidence interval will be: A) Narrower for 98% confidence than for 90% confidence. B) Wider for a sample size of 64 than for a sample size of 36. C) Wider for a 99% confidence than for 95% confidence D) Narrower for sample size of 25 than for a sample size of 36. E) None of the above Answer: C Difficulty: Medium Bowerman, Essentials of Business Statistics, 2/e 194 A Research and Development Laboratory researcher for a paint company is measuring the level a certain chemical contained in a certain type of paint.19. When the level of confidence and sample standard deviation remain the same. each can of paint contains 10% of the chemical. 2/e . µ will be _________________ the 95% confidence interval for µ . Essentials of Business Statistics. When the sample size and sample standard deviation remain the same. B) Interval estimates are less accurate than point estimates. If the paint contains too much of this chemical. How many cans of paint should the sample contain if the researcher wants to be 98% certain of being within 1% of the true proportion of this chemical? A) 4887 B) 1107 C) 26 D) 645 Answer: A Difficulty: Medium (AS) 20. a confidence interval for a population mean µ will be ________________. a 99% confidence interval for a population mean. Which of the following is an advantage of confidence interval estimate over a point estimate for a population parameter? A) Interval estimates are more precise than point estimates. C) Interval estimates are both more accurate and more precise than point estimates. they will be the same 195 Bowerman. Answer: D Difficulty: Medium 21. A) Wider than B) Narrower than C) Equal to Answer: B Difficulty: Medium 23. When the level of confidence and the sample size remain the same. A) Wider B) Narrower C) Neither A nor B. On the average. when the sample standard deviation s is small than when s is large. the quality of the paint will be compromised. A) Wider than B) Narrower than C) Equal to Answer: A Difficulty: Medium (REF) 22. a confidence interval for a population mean based on a sample of n = 100 will be ______________ a confidence interval for a population mean based on a sample of n = 50. D) Interval estimates take into account the fact that the statistic being used to estimate the population parameter is a random variable. 2/e 196 .Answer: B Difficulty: Medium ˆ 24. population standard deviation σ is unknown. and the sample size is n = 15. When the sample size and the sample proportion p remain the same.5 as reasonably possible ˆ ˆ D) p (1 − p) as close to . A) Wider than B) Narrower than C) Equal to Answer: B Difficulty: Medium ˆ 25. When the level of confidence and sample proportion p remain the same. When the level of confidence and sample size remain the same. we choose a value of p that makes: ˆ ˆ A) p (1 − p) as small as reasonably possible ˆ ˆ B) p (1 − p) as large as reasonably possible ˆ ˆ C) p (1 − p) as close to . When solving for the sample size needed to compute a 95% confidence interval for a ˆ population proportion “p”. Essentials of Business Statistics. they will be the same Answer: A Difficulty: Medium 27. having a given error bound “B”. A) Wider than B) Narrower than C) Equal to Answer: A Difficulty: Medium 26. the confidence interval for the population mean µ is based on the: A) z (normal) distribution B) t distribution C) Binomial distribution D) Poisson Distribution E) None of the above Answer: B Difficulty: Medium 28. a 90% confidence interval for a population proportion p will be ______________ the 99% confidence interval for p. a confidence interval for a ˆ ˆ ˆ ˆ population proportion p will be ______________ when p (1 − p) is larger than when p (1 − p) is smaller. A) Wider B) Narrower C) Neither A nor B. When the population is normally distributed.25 as reasonably possible E) Both B and D are correct Bowerman. a confidence interval for a population proportion p based on a sample of n = 100 will be ______________ a confidence interval for p based on a sample of n = 400. 9442 to 3.8355 to 3.4420 to 3.5065 to 3.0558 Answer: D Difficulty: Hard 197 Bowerman. A) Always B) Sometimes C) Never Answer: A Difficulty: Medium 32.Answer: E Difficulty: Medium (REF) 29. as the sample size increases. the value of tα /2 _____ approaches the value of Z a / 2 . When a confidence interval for a population proportion is constructed for a sample size n = ˆ 30 and the value of p = .8160 E) 2.09. Essentials of Business Statistics.4.1645 B) 2. In a manufacturing process a random sample of 9 bolts manufactured has a mean length of 3 inches with a variance of .4935 C) 2.8140 to 3. There is little difference between the values of tα/2 and zα/2 when the sample: A) size is small B) size is large C) mean is small D) mean is large E) standard deviation is small Answer: B Difficulty: Medium 31. 2/e . the interval is based on the: A) z distribution B) t distribution C) exponential distribution D) Poisson distribution E) None of the above Answer: A Difficulty: Medium 30.5580 D) 2. What is the 90% confidence interval for the true mean length of the bolt? A) 2. Assuming the same level of significance α . 692 C) 2.33.2392 B) .308 to 3.902 to 3. the auditing staff randomly selected 400 customer accounts and found that 80 of these accounts were delinquent.2515 E) . Essentials of Business Statistics.804 to 3.1714 to .3 inches. For this quarter.098 Answer: C Difficulty: Medium 35.902 to 3. The internal auditing staff of a local manufacturing company performs a sample audit each quarter to estimate the proportion of accounts that are delinquent (more than 90 days overdue). In a manufacturing process a random sample of 9 bolts manufactured has a mean length of 3 inches with a standard deviation of .231 D) 2.814 to 3.884 to 3.871 to 3.129 D) 2.2008 C) .3 inches.412 to 3.1671 to . In a manufacturing process a random sample of 36 bolts manufactured has a mean length of 3 inches with a standard deviation of . What is the 99% confidence interval for the true mean length of the bolt? A) 2.186 Answer: C Difficulty: Hard 34.098 B) 2.772 E) 2. What is the 95% confidence interval for the true mean length of the bolt? A) 2.770 to 3.228 to 3. What is the 95% confidence interval for the proportion of all delinquent customer accounts at this manufacturing company? A) .1608 to .117 C) 2.1485 to .196 B) 2.588 E) 2.1992 to .2286 Answer: A Difficulty: Hard Bowerman.2329 D) . 2/e 198 . the sample standard deviation is . A) 1842 B) 1548 C) 897 D) 632 E) 1267 Answer: B Difficulty: Hard 37.3 inches. The historical records show that over the past 8 years 70 percent of the accounts have been current. we are interested in measuring the average length of a certain type of bolt. Determine the sample size needed in order to be 99% confident that the sample proportion of the current customer accounts is within . Essentials of Business Statistics. we are interested in measuring the average length of a certain type of bolt. the width of the confidence interval ______________. The internal auditing staff of a local manufacturing company performs a sample audit each quarter to estimate the proportion of accounts that are current (between 0 and 60 days after billing). How many bolts should be sampled in order to make us 95% confident that the sample mean bolt length is within . Answer: increases Difficulty: Medium 199 Bowerman.36. In the construction of a confidence interval. as the confidence level required in estimating the mean increases. How many bolts should be sampled in order to make us 95% confident that the sample mean bolt length is within . In a manufacturing process. Based on a preliminary sample of 9 bolts. Past data indicates that the standard deviation is .02 inches of the true mean bolt length? A) 864.03 of the true proportion of all current accounts for this company.36 B) 80 C) 1470 D) 3989 E) 1197 Answer: E Difficulty: Hard Fill-in-the-Blank 39. 2/e . In a manufacturing process.02 inches of the true mean bolt length? A) 25 B) 49 C) 423 D) 601 E) 1225 Answer: D Difficulty: Hard 38.25 inches. A confidence interval for the population mean is an interval constructed around the _____. As the sample size n increases. decreasing the margin of error. As the significance level. Answer: decreases Difficulty: Medium 45. Answer: decreases Difficulty: Medium 41. and based on a sample the average weight of a sample of 20 boxes is 16 ounces. If everything else is held constant. Answer: Sample mean Difficulty: Medium Bowerman. Answer: Z Difficulty: Medium 42. the width of the confidence interval _______________. When establishing the confidence interval for the average weight of a cereal box. the width of the confidence interval _______________. Answer: decreases Difficulty: Medium 46. The appropriate test statistics to use is ________. 2/e 200 . Answer: decreases Difficulty: Medium 43. assume that the population standard deviation is known to be 2 ounces. Essentials of Business Statistics. As the margin of error decreases. As the standard deviation.40. α increases. As the stated confidence level decreases. the width of the confidence interval _______________. (σ ) decreases. Answer: decreases Difficulty: Medium 44. the width of the confidence interval _______________. Answer: increases Difficulty: Medium 47. the width of the confidence interval _______________. __________ the required sample size. construct a 99 percent confidence interval for µ . A random sample of size 30 from a normal population yields X = 32.6.5 grams and s = 1.1 and a mean of 21.035 9 Difficulty: Hard 50. Answer: (31. . A sample set of weights in pounds are 1. Assuming a normal distribution.63. Assume the population of weights are normally distributed.97. 2/e .51 32. Find a 99 percent confidence interval for the mean population weight. A sample of 12 items yields X = 48.6.278) 1. Answer: (. 1.96 = 20.01. . .19 to 34.8 and s = 4.5 48.03.4 grams.Essay 48. Construct a 95 percent confidence interval for µ .01.6 ± 1.965.5 ± 1.51. . 22. Answer: (15.04.0 ± 3. Essentials of Business Statistics. Answer: (20. construct a 90 percent confidence interval for the population mean weight. Construct a 95 percent confidence interval for µ .03.722.6 100 Difficulty: Medium 201 Bowerman.99. Answer: (47.96 = 31.4 18.41 30 Difficulty: Medium 49.41) 4.6 ± 3.1 21.95.8 ± 1.499 = 15.6) 5.57 8 Difficulty: Medium 51.97. 49.355 = . A sample of 8 items has an average fat content of 18. 21.278 12 Difficulty: Medium 52.0312 1.6 grams and a standard deviation of 2. 1.19. Assuming a normal distribution.63 to 21.035) . and 1. 1.722 to 49.6 to 22. A sample of 100 items has a standard deviation of 5. 1.57) 2.965 to 1. 34.5 grams.796 = 47. .58) →.391) 37 = .6 917) .96 600 .451 1000 .42)(.221 Difficulty: Hard (AS) 55. Answer: 34 1.451) .389 to . Construct a 95 percent confidence interval for the proportion of those people opposed to the tax increase.96 →.1 ˆ p= →.19 ±1. Answer: (.96 (. .391 120 . Answer: (. Find a 95 percent confidence interval for the actual success proportion of the procedure.53. What sample size is needed to obtain a 90 percent confidence interval for the mean protein content of meat if the estimate is to be within 2 pounds of the true mean value? Assume that the variance is 49 pounds.3083)(.221) 114 = . Answer: (. The success rate of a procedure is 37 per 120 cases in a sample. Construct a 95 percent confidence interval for the percentage of trucks that had bad signal lights.225. Essentials of Business Statistics.1 ˆ p= Difficulty: Medium 56.645(7) n = = 34 2 2 Difficulty: Hard Bowerman.19)(. 2/e 202 .159.159 to .81) . .389. Of a random sample of 600 trucks at a bridge. 420 are opposed to the tax increase.1 Difficulty: Medium (AS) 54.000 people.225 to . In a survey of 1.3083 120 (.42 ±1. 114 had bad signal lights.19 600 (.3083 ±1. 57.96 → .10)(.1588 100 Difficulty: Hard 203 Bowerman.8 31.91) 1.49.401 (1. Answer: (.90) . Develop a 95 percent confidence interval for the population proportion of defectives.8. What sample size is needed to obtain a 95 percent confidence interval for the proportion of fat in meat that is within 2 percent of the true value? Answer: 2.02) 2 Difficulty: Medium (AS) 58. What sample size is needed to estimate with 95 percent confidence the mean intake of calcium within 20 units of the true mean if the intake is normal with a variance of 1900 units? Answer: 19 (1.10 ± 1. Form a 90 percent confidence interval for the population mean.96) 2 (1900) n= ≅ 19 (20) 2 Difficulty: Medium 60. The mean is 31.645 ⇒ 31.645)2 (.05) 2 Difficulty: Medium (AS) 59.49 to 31. 31.10 100 (.91 200 Difficulty: Medium 61.5) n= ≅ 271 (.5) n= = 2401 (.7 and the standard deviation is 1.96) 2 (. Essentials of Business Statistics. Answer: (31. A sample of 200 observations is taken. . 2/e . What sample size is needed to estimate the proportion of highway speeders within 5 percent using a 90 percent confidence level? Answer: 271 (1.7 ± 1.0412 to .5)(. Ten items of 100 are defective.1588) 10 p= = .0412.5)(. X =35. 162.6 ± 2. Assume population normality.1 ± 2.24. 161. 44.08 63.2 ± 2.8 and s = 3.262 = 31. In a study of 265 subjects.575 (.3.43 to 64.262 = 26. What is a 95 percent confidence interval for µ when n = 10. What is a 95 percent confidence for µ ? Answer: (63. and s = 13.2)(.8 ± 1.0? Assume population normality. 182.0.304 100 .0? Assume population normality. and s=3.096.3 to 44.592 172 ± 2.2.1.8) → . 154. Given the following test scores. Find the 99 percent confidence interval for p when p =.08.9 10 Difficulty: Medium 63. 2/e 204 . Answer: (26.228 = 155. 166.76) 24. Answer: (155.9) 13 35. 158. find a 95 percent confidence interval for the population mean: 148.76 11 Difficulty: Hard ˆ 66.62.96 = 63. 160. 195.95 to 36. 64.25 10 Difficulty: Medium 64. 36.17) 3.096 to . X =34.24 to 188.6. 188. .2) 3 34.17 265 Difficulty: Medium 65. 236. What is a 95 percent confidence interval for µ when n = 10. the average score on the examination was 63. Answer: (32. and n = 100.304) . 170.43. Essentials of Business Statistics.1 Difficulty: Medium Bowerman. Answer: (. Find a 95 percent confidence interval for p when p =. Essentials of Business Statistics.648 Difficulty: Medium 70. An approximation is 20 percent. . In a survey of 400 people.60)(. We want to estimate with 99 percent confidence the percentage of buyers of cars who are under 30 years of age.96 (.75) 400 .469.02 2 Difficulty: Medium 205 Bowerman.05) 2 Difficulty: Hard 71.25 ±1.292 Difficulty: Medium ˆ 68.2)(.60 ±1.551 1000 . They want the estimate to be at the 90 percent confidence level and within 2 percent of the actual proportion.5) n= ≅ 664 (.551) .1 → .1 Difficulty: Medium 69. A cable TV company wants to estimate the percentage of cable boxes in use during an evening hour.575) 2 (.40) 400 . What sample size is needed? Answer: 1. Find a 95 percent confidence interval for the true proportion favoring new laws.8) ≅1083 .575 (. . Answer: (.083 n= (1.469 to .65) .49) → . 60 percent favor new zoning laws. Answer: (. 2/e .ˆ 67. What sample size is needed? Answer: 664 (2.25 and n = 400. A margin of error of 5 percentage points is desired.1 → .55.51 ± 2.000.292) .645) (.552 to .208.51 and n = 1.25)(.96 (.51)(. Find a 99 percent confidence interval for p when p = .5)(. . Answer: (.208 to . 6 ± 4.2769 ±1.645) 2 (. How large of a sample is needed? Answer: 553 (1. Assume that the sample is randomly selected from a normally distributed population.10 and within . An insurance company estimates 40 percent of its claims have errors.311 650 . and n = 5. s = 2.4)(.604 ⇒ 94.7 231) .243.6. 102.72) 2 98. 2/e 206 . Find a 99 percent confidence interval for µ if X = 98.1 ˆ p= Difficulty: Medium 74. In a randomly selected group of 650 automobile deaths.85. Construct a 95 percent confidence interval for the true proportion of all automobile accidents caused by alcohol.2769)(.025) 2 Difficulty: Medium 75. . 180 were alcohol related. You want to estimate the proportion of customers who are satisfied with their supermarket at α = .2769 650 (.15) n= ≅ 553 (.025 of the true value.645) 2 (.48 to 102. It has been estimated that p =. What sample size is needed if they wish to be within 5 percent of the actual? Answer: 260 (1. Answer: (.85)(.6) n= ≅ 260 (. Answer: (94.243 to .72 5 Difficulty: Medium Bowerman.72.48.96 → . The insurance company wants to estimate with 90 percent confidence the proportion of claims with errors.311) 180 = .05) 2 Difficulty: Hard 73. Essentials of Business Statistics. 23.575 ⇒ 71. 78. Essentials of Business Statistics.06 to .77) 10 75 ± 1. 77. A sample of 50 units is taken from a batch.15)(. Calculate a 90 percent confidence interval for µ .15 ±1.8) 700 ±1.33 50 Difficulty: Medium 79.64 50 Difficulty: Medium 81.36 to 78.64) 10 75 ± 2. 723.7)(.3) = 676.76. 2/e .2. Calculate a 95 percent confidence interval for µ . Calculate a 99 percent confidence interval for µ .. Find a 95 percent confidence interval for the proportion of vouchers with errors. The sample yielded the following results: X = 75 lbs.24 .15 60 (.96 ⇒ 72.000 cases. and s = 10 lbs.96 Difficulty: Medium 77. The 95% confidence interval for the average weight of a product is from 72.24) ˆ p= 9 = . Find a 90 percent confidence interval for the total number of successes per 1.23 to 77.77 50 Difficulty: Medium 80. . Answer: (676.06. Answer: (72.67. Answer: (71.2 to 723. The customer service manager for the XYZ Fastener Manufacturing Company examined 60 vouchers and found nine vouchers containing errors.8 Difficulty: Hard Use the following information to answer questions 78-80: The weight of a product is measured in pounds. Answer: (72. Answer: (.1 → . to 77.85) 60 .645 1000(. The success rate of a surgical procedure is 70 per 100 cases examined in a sample. 77.77 207 Bowerman. 78.67 to 77.23 lbs.36.645 ⇒ 72.33) 10 75 ± 1. 52. .575 (. . Difficulty: Medium Use the following information to answer questions 83-86: ˆ A sample of 2.502 to .1 Difficulty: Medium 86.0112 σρ = (.1 Difficulty: Medium 84. (proportion of customers satisfied Bowerman. Calculate a 95 percent confidence interval for p.542 2000 . Calculate a 90 percent confidence interval for p.498.549 2000 .52)(.000 people yielded p =.0112 2000 .1 Difficulty: Medium 87.60. 83.lbs.542) . because 77 is inside the interval.48) → . Answer: (. Answer: (. Calculate a 99 percent confidence interval for p.52)(. A sample of 2.52 ±1.48) = .52 ± 2.645 (.498 to . The 99% confidence interval for the average weight of a product is from 71.491.36 lbs.48) → . Essentials of Business Statistics. Can we conclude that µ = 77 using a 95 percent confidence interval? Answer: Yes. because 71 is outside the interval. Answer: No.538) .000 customers yielded p = .538 2000 . Answer: (. Customers of a company were surveyed as to whether they were satisfied with the service ˆ they received.1 Difficulty: Medium 85. to 78.52)(. Difficulty: Medium 82. .48) → . What is the standard deviation of the population proportion? Answer: .52 ±1. Can we conclude that µ is equal to 71 using a 99 percent confidence interval? Briefly explain.64 lbs. 2/e 208 .491 to .549) .52)(.502.96 (. 629. 1.with service in the sample) What is the standard deviation of the number of customers satisfied with service? Answer: 21.6. Calculate a 99 percent confidence interval for µ .262 ⇒ .6. 5.817 10 Difficulty: Medium 91.2 and the variance is .6 1. 1.91 σ = 2000(. Difficulty: Medium 88.852.2 and s = . and X =1.6)(. can we conclude that µ = . Essentials of Business Statistics.771. If the 95% confidence interval for a mean is from . Answer: (.36.2 and s = 0.771 to 1.2 ± 2. 2/e .833 ⇒ . and X = 1. Calculate a 95 percent confidence interval for µ . using a 95 percent confidence interval? Answer: No.25 ⇒ .5 is outside the interval. A random sample of size 10 is taken from a population assumed to be normal.2 ± 3.629) .629 10 Difficulty: Medium 90.548 10 Difficulty: Medium 89. Answer: (. Calculate a 90 percent confidence interval for µ .2 ± 1.771 to 1.583.548) .6 1. because .36 1.583 to 1. Difficulty: Medium 209 Bowerman. 1.91. A random sample of size 10 is taken from a population assumed to be normal. and X = 1.852 to 1. Answer: (.817) . A random sample of size 10 is taken from a population assumed to be normal.4) = 21. Determine the 98% confidence interval.96 to 248.5) = .84 61 Difficulty: Medium Bowerman. On a particular day.6915 940 − 1000 60 Z= =− = − . the average pressure is 238.4 2.64) 64 Difficulty: Medium Use the following information to answer questions 93-94: The lifetime of a disk drive head is normally distributed with a population mean of 1000 hours and a standard deviation of 120 hours.5 120 120 P ( Z ≥ −. The test measures the force needed to break the rubber in pounds.4332 + .86 to 4. Determine the 95% confidence interval for the average number of strokes.33 = 227 .5) = .86 to 4.6 4. According to the sample results.1915 + .96 to 248. 93. Determine the probability that the lifetime for 9 disk drives will exceed 940 hours.64) 1. Essentials of Business Statistics.5 120 40 9 P ( Z ≥ −1. Answer: . A random sample of 61 pieces of rubber from different production batches is subjected to a stress test. 64 players completed the play on the 13th hole with average of 4. The quality control manager of a tire company wishes to estimate the tensile strength of a standard size of rubber used to make a class of radial tires. 2/e 210 .6 strokes. Determine the probability that the lifetime for one drive will exceed 940 hours. Answer: .96) = (3.6915 Difficulty: Medium 94.84) 35 238.25 strokes and standard deviation of 1.9332 940 − 1000 −60 Z= = = −1. Answer: (3.5 = . Answer: (227. A PGA (Professional Golf Association) tournament organizer is attempting to estimate the average number of strokes for the 13th hole on a given golf course.5 = .9332 Difficulty: Medium 95.25 m (1.4 pounds and the standard deviation is 35 pounds.92. 4) n= = 1448 . 2/e .8)(. An insurance analyst working for a car insurance company would like to determine the proportion of accident claims covered the company. A computer manufacturing company has sent a mail survey to 2800 of its randomly selected customers that have purchased the recently launched personal computer.8141) Difficulty: Medium 99. How large a sample should be taken to estimate the proportion of accident claims covered by the company if we want to be 98% confident that the sample percentage is within ±3% of the actual percentage of the accidents covered by the insurance company? Answer: 1448 (2.343 to .0613 (.33) 2 (. while 160 customers indicated they were not satisfied with their new computer.0613 240 Difficulty: Medium 97.6)(.05(.645 200 (.032 Difficulty: Medium 211 Bowerman.96 = .1 =.7589 to .8141) 640 ± 2.8 ± 2. 640 customers indicated that they were satisfied. Construct a 90% confidence interval estimate of the true proportion of claims covered by the insurance company. A random sample of 240 claims shows that the insurance company covered 90 accident claims while 150 claims were not covered. Construct a 96% confidence interval estimate of the true proportion of customers satisfied with their new computer. Answer: (. According to a preliminary estimate 60% of the claims are covered. An insurance analyst working for a car insurance company would like to determine the proportion of accident claims covered the company.343 to .96.6) 200 . A car insurance company would like to determine the proportion of accident claims covered by the company. Use a confidence interval of 95% and determine the margin of error. Answer: (. 800 customers responded to the survey.2) 800 . The survey asked the customers whether or not they were satisfied with the computer.0141) (.457 Difficulty: Medium 98. Answer: . Essentials of Business Statistics.05 800 (.625) 1.4)(. A random sample of 200 claims shows that the insurance company covered 80 accident claims while 120 claims were not covered.7859 to .375)(.457) 80 1.1 . how large a sample should be used? Answer: 43 n= (2.1 (.390 14. Essentials of Business Statistics. He asked one of his planners to check the timeliness of shipments for 1000 orders.17) 2 (.14) Difficulty: Hard Multiple Choice Use the following information to answer questions 102-103: A company is interested in estimating µ .02) 2 2 = 43 Difficulty: Medium 101.12)(.053 12.019 12. The sample mean is 12. the mean number of days of sick leave. The production manager for the XYZ manufacturing company is concerned that the customer orders are being shipped late.10 to . the mean number of days of sick leave taken by its employees. A) [10.348] E) [11.675] B) [10.100.10 to .2 days and the sample standard deviation is 10 days. In order to achieve a 97% confidence with a margin of error of .665] Answer: B Bowerman.381] D) [12. The planner randomly selected 1000 orders and found that 120 orders were shipped late.05 ounces.88) 1000 . Construct the 95% confidence interval for the proportion of orders shipped late. The firm’s statistician randomly selects 100 personnel files and notes the number of sick days taken by each employee. Calculate a 93% confidence interval for µ .96 1000 (.725 13. 2/e 212 .14) 120 ±1. Answer: (. A statistical quality control process for cereal production measures the weight of a cereal box. 102.06) (.02 ounces. The population standard deviation is known to be .735 12.010] C) [12. A random sample of 64 days gives us a sample mean of 14. the mean number of cars passing the intersection? A) [12.438 15. After calculating the confidence interval.426] E) [14. The state highway department is studying traffic patterns on one of the busiest highways in the state. What is the 92% confidence interval estimate of μ.183 14. Essentials of Business Statistics. As part of the study.189 14.205 cars and a sample standard deviation of 1.010 cars. A random sample of 64 days gives us a sample mean of 14. The state highway department is studying traffic patterns on one of the busiest highways in the state.221] D) [13. How many personnel files would the director have to select in order to estimate μ to within 2 days with a 99% confidence interval? A) 2 B) 13 C) 136 D) 165 E) 166 Answer: E Difficulty: Hard 104. Given this precision. and needing to be 99% confident.984 14.227] Answer: D Difficulty: Medium 105.010 cars. the department needs to estimate the average number of vehicles that pass an intersection each day. the highway department officials determined that the precision is too low for their needs. As part of the study.205 cars and a sample standard deviation of 1.382] C) [12. how many days do they need to sample? A) 109 B) 76 C) 75 D) 62 E) 9 Answer: B Difficulty: Hard 213 Bowerman. the department needs to estimate the average number of vehicles that pass an intersection each day. 2/e .972] B) [14.103. They feel the precision should be 300 cars.028 14. Essentials of Business Statistics. Department of Health and Human Services collected sample data for 772 males between the ages of 18 and 24. In a study of factors affecting soldiers’ decisions to reenlist.78] D) [28.21] B) [62.24] B) [15. That sample group has a mean height of 69. A psychologist is collecting data on the time it takes to learn a certain task. 320 subjects were measured for an index of satisfaction and the sample mean is 28.8 inches. A) [63. Find the 99% confidence interval for the mean height of all males between the ages of 18 and 24.47 17.91] C) [69. The U.48] Answer: D Difficulty: Medium Bowerman. the sample mean is 16. Construct the 95% confidence interval for the mean time required by all adults to learn the task.75] D) [69.82 45.75 28.75] B) [27.00] Answer: A Difficulty: Hard 108.85 29.93] E) [69.56] D) [15.56 24.96 29.29 17.65 69.96] Answer: E Difficulty: Medium 107.49 76.40 minutes and the sample standard deviation is 4. A) [8.64] C) [11.60 29. Use the given sample data to construct the 98% confidence interval for the population mean.S. A) [27.47 69.19 76.00 minutes.8 and the sample standard deviation is 7.7 inches with a standard deviation of 2.12 48.106.44 69. For 50 randomly selected adult subjects.33] C) [16.85] E) [28. 2/e 214 .51] E) [17.3.24 16. 2/e . Unoccupied seats on flights cause airlines to lose revenue.82 677 Mean Std Dev N Construct a 98% confidence interval for the mean hours spent on personal tasks.20] Answer: D Difficulty: Medium (AS) 111. Calculate a 90% confidence interval for µ . Research has been conducted that studies the role that the age of workers has in determining the hours per month spent on personal tasks.17 0.81 768 45-64 4.34] B) [11.10] D) [2.92] E) [3.05] E) [11.96 4.95 4.98 4.6 seats and a standard deviation of 4.57 11.109.1 seats.86 18.15 12. Suppose a large airline wants to estimate its average number of unoccupied seats per flight over the past year. the mean number of unoccupied seats per flight during the past year.31 0. A) [3.1 seats.95] C) [11.11] C) [3. 225 flight records are randomly selected and the number of unoccupied seats is noted with a sample mean of 11.16 5. 225 flight records are randomly selected and the number of unoccupied seats is noted with a sample mean of 11. The data are: 18-24 4.12] B) [3.75 241 AGE GROUP 25-44 4. Suppose a large airline wants to estimate its average number of unoccupied seats per flight over the past year.25 11.13] Answer: B Difficulty: Hard 110. How many flights should we select if we wish to estimate μ to within 2 seats and be 95% confident? A) 130 B) 65 C) 33 D) 17 E) 12 Answer: C Difficulty: Medium (AS) 215 Bowerman.97 4.63] D) [11. Unoccupied seats on flights cause airlines to lose revenue.686 adults was observed for one month. A) [4.30 12. Essentials of Business Statistics.04 0. A sample of 1.6 seats and a standard deviation of 4. 979 $47.469 $68.08] B) [-6.653] Answer: E Difficulty: Hard 114.651] D) [32.00 with a standard deviation of $124.149.533 34.269 34.08 270.816 with a standard deviation of $12. Essentials of Business Statistics.163] B) [$46. An auditor was hired to verify the accuracy of a company’s new billing system. $34. Thirty-five (35) invoices produced since the system was installed were sampled.69 8.40] E) [-9. a random sample of 30 stocks was drawn and their closing prices on the last trading day of 1993 were observed with a mean of 33.65 and $31.514 34.112.66] Answer: C Difficulty: Hard 113. Estimate the average price of a share of stock in the portfolio at the end of 1993 with a 90% confidence interval.732 40.17 11. 1991.897] Answer: A Difficulty: Medium Bowerman.557. it was found that the average salary was $46. and 1992 the average prices of a share of stock in a money market portfolio were $34.783 $46. 2/e 216 . At the end of 1990.157 $47.040] C) [$46. In a random sample of 651 computer scientists who subscribed to a web-based daily news update.69] C) [-44.26 respectively. Construct a 97% confidence interval for the mean error per invoice.434] C) [32.583 and a standard deviation of 19. The average error on the invoices was $1. A) [-268. Calculate a 91% confidence interval for the mean salary of computer scientists.849] E) [$45.83. To investigate the average share price at the end of 1993.48] D) [-38. A) [$25.832 39.475] D) [$46.40 40.633] E) [32.592 $47.00. A) [27.48 46.334] B) [26. A random sample of 20 hospitals in one state had a mean LOS for nonheart patients in 2000 of 3.036] E) [5.144] D) [5.33] Answer: A Difficulty: Medium 116.36] B) [3.24 4.2 days.27 4.93] C) [3. A) [5. the average LOS for non-heart patient was 4. Ten cups of soup are brought with results of a mean of 5.38 4.2 days.8 days and a standard deviation of 1. A random sample of 20 hospitals in one state had a mean LOS for nonheart patients in 2000 of 3. How large a sample of hospitals would we need to be 99% confident that the sample mean is within 0.716 6.34 4.93 ounces and a standard deviation of 0.115. 2/e .67 3.6 days. Construct a 99% confidence interval for the true machine-fill amount.824 6. The coffee/soup machine at the local bus station is supposed to fill cups with 6 ounces of soup. Health insurers and the federal government are both putting pressure on hospitals to shorten the average length of stay (LOS) of their patients.6 days.046] C) [5. Calculate a 95% confidence interval for the population mean LOS for non-heart patients in the state’s hospitals in 2000.13 ounces.8 days and a standard deviation of 1.064] Answer: E Difficulty: Medium 217 Bowerman. Essentials of Business Statistics.796 6. the average LOS for non-heart patient was 4.888 5. In 1996.972] B) [5.814 6.5 days of the population mean? A) 3 B) 7 C) 32 D) 48 E) 96 Answer: D Difficulty: Hard 117. A) [3. Health insurers and the federal government are both putting pressure on hospitals to shorten the average length of stay (LOS) of their patients.26] D) [3. In 1996.22] E) [3. 9964] C) [0.9860 1.996 ounces with a standard deviation of 0.0275] B) [0. The coffee/soup machine at the local bus station is supposed to fill cups with 6 ounces of soup.9942 0. A random sample of 25 pieces of candy produces a mean of 0. Ten cups of soup are brought with results of a mean of 5.9645 1. A random sample of 25 pieces of candy produces a mean of 0.03 ounces of the population mean? A) 97 B) 96 C) 73 D) 62 E) 10 Answer: A Difficulty: Hard 119.00 ounces. How large a sample of soups would we need to be 95% confident that the sample mean is within 0. Construct a 98% confidence interval for the mean weight of all such candy. A local company makes a candy that is supposed to weigh 1.0060] D) [0.9980] E) [0.004 ounces.9956 0. Essentials of Business Statistics.118.93 ounces and a standard deviation of 0. A) [0. How many pieces of candy must we sample if we want to be 99% confident that the sample mean is within 0.001 ounces of the true mean? A) 126 B) 124 C) 107 D) 12 E) 6 Answer: A Difficulty: Medium Bowerman.00 ounces.004 ounces.996 ounces with a standard deviation of 0.9978] Answer: D Difficulty: Hard 120.13 ounces. A local company makes a candy that is supposed to weigh 1.9940 0. 2/e 218 . Their mean score is 71. 2/e . to within 0.26 102.121.70] E) [25. A sample of five cars is tested and a mean of 28.16 30.44 77. A sample of five cars is tested and a mean of 28.7 miles.46 78. A sociologist develops a test designed to measure a person’s attitudes about disabled people.57] Answer: C Difficulty: Medium 122. An environmental group at a local college is conducting independent tests to determine the distance a particular make of automobile will travel while consuming only 1 gallon of gas. Essentials of Business Statistics.85 31.66 81.83 30.21 79. Construct the 99% confidence interval for the mean score of all subjects. find the 95% confidence interval for the mean distance traveled by all such cars using 1 gallon of gas.5 miles and be 99% confident? A) 25 B) 124 C) 194 D) 618 E) 619 Answer: A Difficulty: Medium 123.2 miles is obtained. An environmental group at a local college is conducting independent tests to determine the distance a particular make of automobile will travel while consuming only 1 gallon of gas.2 miles is obtained.94] C) [60. A) [26.96] E) [63. How many cars should the environmental group test if they wish to estimate μ.70] C) [24.14] B) [63.70 35.19] Answer: B Difficulty: Medium 219 Bowerman.55] D) [26. and 16 randomly selected subjects are given the test.24] B) [20.74] D) [64.2 with a standard deviation of 10.5. A) [40.70 29. Assuming that the standard deviation is 2. mean miles per 1 gallon. 513 32. A) [11.829 74.29 51.896] B) [11.67 50.626] D) [12.24] B) [47.487 69. the percentage of contamination removed from each soil sample was measured with a mean of 49.3% and a standard deviation of 1. Estimate the mean percentage of contamination removed at a 98% confidence level.007] E) [61.633] C) [12.13] Answer: B Difficulty: Hard (AS) Bowerman. Construct a 99% confidence interval for the true mean number of full-time employees at office furniture dealers. A federal bank examiner is interested in estimating the mean outstanding defaulted loans balance of all defaulted loans over the last three years.455 employees at 22 office furniture dealers in a major metropolitan area with a standard deviation of 18.317] Answer: E Difficulty: Medium 126.47 52. 2/e 220 . After 95 days.52.918 with a standard deviation of $16. Three contaminated soil samples were treated. Calculate a 90% confidence interval for the mean balance of defaulted loans over the past three years. There is an average of 22.299 96. Essentials of Business Statistics.93] C) [47.349] B) [39.838] Answer: B Difficulty: Medium 125.397] E) [20. A random sample of 20 defaulted loans yielded a mean of $67.277 33. An experiment was conducted to determine the effectiveness of a method designed to remove oil wastes found in soil.124.552.31] D) [47.329 78.284 32.5%. A) [48.72] E) [46.537] C) [57. A) [66.014 33.36 50.072 24.40.519 74.507] D) [61.88 50. 3% and a standard deviation of 1.5%.5 cm. An experiment was conducted to determine the effectiveness of a method designed to remove oil wastes found in soil. Three contaminated soil samples were treated.54 74. Essentials of Business Statistics. After 95 days.239 B) 283 C) 212 D) 131 E) 66 Answer: A Difficulty: Medium 221 Bowerman.30 74.70] C) [70.127. 2/e . On a standard IQ test. the standard deviation is 15.46] E) [71.10] Answer: C Difficulty: Medium 129.5 cm and 4. If we wished to narrow the boundary around μ for a 98% confidence interval to within 0.10 74. A botanist measures the heights of 16 seedlings and obtains a mean and standard deviation of 72.91 82. respectively. the percentage of contamination removed from each soil sample was measured with a mean of 49. A) [62. A) 437 B) 33 C) 9 D) 6 E) 3 Answer: A Difficulty: Hard 128. How many random IQ scores must be obtained if we want to find the true population mean (with an allowable error of 0.5) and we want 97% confidence in the results? A) 4.90 73.5%.09] B) [70. how many soil samples should be in our experiment. Find the 95% confidence interval for the mean height of seedlings in the population form which the sample was selected.90] D) [70. 090 . What sample size should be used? A) 677 B) 107 C) 35 D) 27 E) 11 Answer: B Difficulty: Hard 131.244] Answer: A Difficulty: Medium (AS) 132. Despite the response from the restaurant chain. Recently. Of the 319 contacted. 29 indicated that they would not go back to the restaurant because of the potential for food poisoning Construct a 95% confidence interval for the true proportion of the market who still refuse to visit any of the restaurants in the chain three months after the event.196] D) [. The source was identified and corrective actions were taken to make sure that the food poisoning would not reoccur. a case of food poisoning was traced to a particular restaurant chain.118 .025 pounds of its true value with a 99% confidence.777 Answer: A Difficulty: Medium (AS) Bowerman. A) [. For quality control purposes. 29 indicated that they would not go back to the restaurant because of the potential for food poisoning. The source was identified and corrective actions were taken to make sure that the food poisoning would not reoccur. 02 of ρ. Recently. Of the 319 contacted. What sample size would be needed in order to be 99% confident that the sample proportion is within . 2/e 222 .000.240 .339] E) [.371 E) 1. many consumers refused to visit the restaurant for some time after the event. Despite the response from the restaurant chain.130.5 pounds ( σ = 0. When the machine is properly calibrated. Essentials of Business Statistics.1). A survey was conducted three months after the food poisoning occurred with a sample of 319 patrons contacted. the mean inflation pressure s 13.122] B) [. A manufacturer of dodge balls uses a machine to inflate its new balls to a pressure of 13. the manufacturer wishes to estimate the mean inflation pressure to within 0. many consumers refused to visit the restaurant for some time after the event.5 pounds. a case of food poisoning was traced to a particular restaurant chain. but uncontrollable factors can cause the pressure of individual dodge balls to vary. the true proportion of customers who refuse to go back to the restaurant? A) 14 B) 38 C) 129 D) 1.059 .091] C) [. A survey was conducted three months after the food poisoning occurred with a sample of 319 patrons contacted. 027] B) [0.076] E) [0.140 .003 0.015] C) [0. At a sobriety checkpoint the sheriff’s department screened 676 drivers and 6 were arrested for DWI.938] D) [0. A) [0. Construct the 90% confidence interval for the proportion of all adults who feel the same way.200] E) [. 175 respondents listed jewelry as being among the romantic gifts men want to receive.171] C) [.085] B) [.170] Answer: C Difficulty: Hard 134.950] Answer: D Difficulty: Hard 135.946] C) [. Find the 92% confidence interval for the true proportion of drivers who were DWI that evening.943] E) [.025 . Construct the 98% confidence interval for the proportion of all adults who include jewelry among their responses.000 0. A) [.920 .133.914 .917 . Essentials of Business Statistics.078 .030] Answer: B Difficulty: Hard 223 Bowerman.004 adults. A) [.050 . 2/e .006 0.000 0.220] D) [.169 .940] D) [. In a poll of 1. 93% indicated that restaurants and bars should refuse service to patrons who have had too much to drink. In a survey of 308 adults.108] B) [.902 0.910 .120 . How many samples would be needed to create a 99% confidence interval that is within 0.115] D) [.2034] B) [0.0039 0. To investigate the rate at which employees with cancer are fired or laid off. 2/e 224 .198] C) [.0371 0.276] Answer: D Difficulty: Medium 137. The Ohio Department of Agriculture tested 203 fuel samples across the state in 1999 for accuracy of the reported octane level.1361] D) [0. Seven (7) were either fired or laid off due to their illness. Find a 99% confidence interval for the true population proportion of premium grade fuel-quality failures. 14 out of 105 samples failed (they didn’t meet ASTM specification and the FTC Octane posting rule).0078 0.045 . The Ohio Department of Agriculture tested 203 fuel samples across the state in 1999 for accuracy of the reported octane level.136.0000 0. a telephone survey was taken of 100 cancer survivors who worked while undergoing treatment. A) [.068 .1400] Answer: D Difficulty: Hard Bowerman.0278 0. Construct a 90% confidence interval for the true percentage of all cancer patients who are fired or laid off due to their illness. 14 out of 105 samples failed (they didn’t meet ASTM specification and the FTC Octane posting rule).023 .221] B) [. Essentials of Business Statistics. A) [0. For premium grade.02 of the true proportion of premium grade fuel-quality failures? A) 4148 B) 2838 C) 1913 D) 744 E) 54 Answer: B Difficulty: Hard 138.1122] E) [0.1029] C) [0.219] E) [.100 .047 . For premium grade. A) [0. Construct a 95% confidence interval for the proportion definitely unwilling to pay fees. Essentials of Business Statistics.241] E) [0. 13.220 0.938 were definitely not willing to pay such fees. Of these.233] C) [0.212 0. 13.232] D) [0.938 were definitely not willing to pay such fees. In 1995.000 internet users were surveyed and asked about their willingness to pay fees for access to websites.139. 2.286 0. 2/e . How large a sample is necessary to estimate the proportion of interest to within 2% in a 95% confidence interval? A) 18 B) 307 C) 1717 D) 2000 E) 2965 Answer: C Difficulty: Medium (AS) 225 Bowerman. Of these.219 0. 2.214 0. In 1995.245] Answer: B Difficulty: Medium (AS) 140.000 internet users were surveyed and asked about their willingness to pay fees for access to websites.302] B) [0.