Chapter 07

Chapter 7Confidence Intervals

True/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 is unknown, then for any sample size n, the sampling distribution of 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, will be narrower than the 95% confidence interval for . 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 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 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 population proportion p will be narrower when is larger than when is smaller. Answer: False Difficulty: Medium (REF)

10. When solving for the sample size needed to compute a 95% confidence interval for a population proportion “p”, having a given error bound “B”, we choose a value of that makes

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

Bowerman, Essentials of Business Statistics, 2/e193

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 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


19. A Research and Development Laboratory researcher for a paint company is measuring the level a certain chemical contained in a certain type of paint. If the paint contains too much of this chemical, the quality of the paint will be compromised. On the average, each can of paint contains 10% of the 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. 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. B) Interval estimates are less accurate than point estimates. C) Interval estimates are both more accurate and more precise than point estimates. D) Interval estimates take into account the fact that the statistic being used to estimate the population parameter is a random variable. Answer: D Difficulty: Medium

21. When the sample size and sample standard deviation remain the same, a 99% confidence interval for a population mean, will be _________________ the 95% confidence interval for . A) Wider than B) Narrower than C) Equal to Answer: A Difficulty: Medium (REF)

22. 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 ______________ a confidence interval for a population mean based on a sample of n = 50. 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 confidence interval for a population mean will be ________________, when the sample standard deviation s is small than when s is large. A) Wider B) Narrower C) Neither A nor B, they will be the same Answer: B Difficulty: Medium


24. When the sample size and the sample proportion remain the same, a 90% confidence interval for a population proportion p will be ______________ the 99% confidence interval for p. A) Wider than B) Narrower than C) Equal to Answer: B Difficulty: Medium

25. When the level of confidence and sample proportion remain the same, 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. A) Wider than B) Narrower than C) Equal to Answer: A Difficulty: Medium

26. When the level of confidence and sample size remain the same, a confidence interval for a population proportion p will be ______________ when is larger than when is smaller. A) Wider B) Narrower C) Neither A nor B, they will be the same Answer: A Difficulty: Medium

27. When the population is normally distributed, population standard deviation is unknown, and the sample size is n = 15; the confidence interval for the population mean 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. When solving for the sample size needed to compute a 95% confidence interval for a population proportion “p”, having a given error bound “B”, we choose a value of that makes: A) as small as reasonably possible B) as large as reasonably possible C) as close to .5 as reasonably possible D) as close to .25 as reasonably possible E) Both B and D are correct Answer: E Difficulty: Medium (REF)


29. When a confidence interval for a population proportion is constructed for a sample size n = 30 and the value of = .4, 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. 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. Assuming the same level of significance , as the sample size increases, the value of t/2 _____ approaches the value of .A) Always B) Sometimes C) Never Answer: A Difficulty: Medium

32. In a manufacturing process a random sample of 9 bolts manufactured has a mean length of 3 inches with a variance of .09. What is the 90% confidence interval for the true mean length of the bolt? A) 2.8355 to 3.1645 B) 2.5065 to 3.4935 C) 2.4420 to 3.5580 D) 2.8140 to 3.8160 E) 2.9442 to 3.0558 Answer: D Difficulty: Hard


33. In a manufacturing process a random sample of 9 bolts manufactured has a mean length of 3 inches with a standard deviation of .3 inches. What is the 95% confidence interval for the true mean length of the bolt? A) 2.804 to 3.196 B) 2.308 to 3.692 C) 2.770 to 3.231 D) 2.412 to 3.588 E) 2.814 to 3.186 Answer: C Difficulty: Hard

34. In a manufacturing process a random sample of 36 bolts manufactured has a mean length of 3 inches with a standard deviation of .3 inches. What is the 99% confidence interval for the true mean length of the bolt? A) 2.902 to 3.098 B) 2.884 to 3.117 C) 2.871 to 3.129 D) 2.228 to 3.772 E) 2.902 to 3.098 Answer: C Difficulty: Medium

35. 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). For this quarter, the auditing staff randomly selected 400 customer accounts and found that 80 of these accounts were delinquent. What is the 95% confidence interval for the proportion of all delinquent customer accounts at this manufacturing company? A) .1608 to .2392 B) .1992 to .2008 C) .1671 to .2329 D) .1485 to .2515 E) .1714 to .2286 Answer: A Difficulty: Hard


36. 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). The historical records show that over the past 8 years 70 percent of the accounts have been current. Determine the sample size needed in order to be 99% confident that the sample proportion of the current customer accounts is within .03 of the true proportion of all current accounts for this company. A) 1842 B) 1548 C) 897 D) 632 E) 1267 Answer: B Difficulty: Hard

37. In a manufacturing process, we are interested in measuring the average length of a certain type of bolt. Past data indicates that the standard deviation is .25 inches. How many bolts should be sampled in order to make us 95% confident that the sample mean bolt length is within .02 inches of the true mean bolt length? A) 25 B) 49 C) 423 D) 601 E) 1225 Answer: D Difficulty: Hard

38. In a manufacturing process, we are interested in measuring the average length of a certain type of bolt. Based on a preliminary sample of 9 bolts, the sample standard deviation is .3 inches. How many bolts should be sampled in order to make us 95% confident that the sample mean bolt length is within .02 inches of the true mean bolt length? A) 864.36 B) 80 C) 1470 D) 3989 E) 1197 Answer: E Difficulty: Hard

Fill-in-the-Blank

39. In the construction of a confidence interval, as the confidence level required in estimating the mean increases, the width of the confidence interval ______________. Answer: increases Difficulty: Medium


40. As the sample size n increases, the width of the confidence interval _______________. Answer: decreases Difficulty: Medium

41. When establishing the confidence interval for the average weight of a cereal box, assume that the population standard deviation is known to be 2 ounces, and based on a sample the average weight of a sample of 20 boxes is 16 ounces. The appropriate test statistics to use is ________. Answer: Z Difficulty: Medium

42. As the significance level, α increases, the width of the confidence interval _______________. Answer: decreases Difficulty: Medium

43. As the standard deviation, () decreases, the width of the confidence interval _______________. Answer: decreases Difficulty: Medium

44. As the stated confidence level decreases, the width of the confidence interval _______________. Answer: decreases Difficulty: Medium

45. As the margin of error decreases, the width of the confidence interval _______________. Answer: decreases Difficulty: Medium

46. If everything else is held constant, decreasing the margin of error, __________ the required sample size. Answer: increases Difficulty: Medium

47. A confidence interval for the population mean is an interval constructed around the _____. Answer: Sample mean Difficulty: Medium


Essay

48. A random sample of size 30 from a normal population yields = 32.8 and s = 4.51. Construct a 95 percent confidence interval for . Answer: (31.19, 34.41)

4.5132.8 1.96 31.19 34.41

30to

Difficulty: Medium

49. A sample set of weights in pounds are 1.01, .95, 1.03, 1.04, .97, .97, .99, 1.01, and 1.03. Assume the population of weights are normally distributed. Find a 99 percent confidence interval for the mean population weight. Answer: (.965, 1.035)

.03121.0 3.355 .965 1.035

9to

Difficulty: Hard

50. A sample of 8 items has an average fat content of 18.6 grams and a standard deviation of 2.4 grams. Assuming a normal distribution, construct a 99 percent confidence interval for . Answer: (15.63, 21.57)

2.418.6 3.499 15.63 21.57

8to

Difficulty: Medium

51. A sample of 12 items yields = 48.5 grams and s = 1.5 grams. Assuming a normal distribution, construct a 90 percent confidence interval for the population mean weight. Answer: (47.722, 49.278)

1.548.5 1.796 47.722 49.278

12to

Difficulty: Medium

52. A sample of 100 items has a standard deviation of 5.1 and a mean of 21.6. Construct a 95 percent confidence interval for . Answer: (20.6, 22.6)

5.121.6 1.96 20.6 22.6

100to

Difficulty: Medium


53. In a survey of 1,000 people, 420 are opposed to the tax increase. Construct a 95 percent confidence interval for the proportion of those people opposed to the tax increase. Answer: (.389, .451)

Difficulty: Medium (AS)

54. Of a random sample of 600 trucks at a bridge, 114 had bad signal lights. Construct a 95 percent confidence interval for the percentage of trucks that had bad signal lights. Answer: (.159, .221)

Difficulty: Hard (AS)

55. The success rate of a procedure is 37 per 120 cases in a sample. Find a 95 percent confidence interval for the actual success proportion of the procedure. Answer: (.225, .391)

Difficulty: Medium

56. 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. Answer: 34

Difficulty: Hard


57. 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: 2,401

2

2

(1.96) (.5)(.5)2401

(.02)n


58. What sample size is needed to estimate the proportion of highway speeders within 5 percent using a 90 percent confidence level? Answer: 271

2

2

(1.645) (.5)(.5)271

(.05)n


59. 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

2

2

(1.96) (1900)19

(20)n

Difficulty: Medium

60. A sample of 200 observations is taken. The mean is 31.7 and the standard deviation is 1.8. Form a 90 percent confidence interval for the population mean. Answer: (31.49, 31.91)

1.831.7 1.645 31.49 31.91

200to

Difficulty: Medium

61. Ten items of 100 are defective. Develop a 95 percent confidence interval for the population proportion of defectives. Answer: (.0412, .1588)

10ˆ .10

100

(.10)(.90).10 1.96 .0412 .1588

100

p

to

Difficulty: Hard


62. What is a 95 percent confidence interval for when n = 10, =35.6, and s = 13.0? Assume population normality. Answer: (26.3, 44.9)

1335.6 2.262 26.3 44.9

10to

Difficulty: Medium

63. What is a 95 percent confidence interval for when n = 10, =34.1, and s=3.0? Assume population normality. Answer: (32.0, 36.2)

334.1 2.262 31.95 36.25

10to

Difficulty: Medium

64. In a study of 265 subjects, the average score on the examination was 63.8 and s = 3.08. What is a 95 percent confidence for ? Answer: (63.43, 64.17)

3.0863.8 1.96 63.43 64.17

265to

Difficulty: Medium

65. Given the following test scores, find a 95 percent confidence interval for the population mean: 148, 154, 158, 160, 161, 162, 166, 170, 182, 195, 236. Assume population normality. Answer: (155.24, 188.76)

24.592172 2.228 155.24 188.76

11to

Difficulty: Hard

66. Find the 99 percent confidence interval for p when =.2, and n = 100. Answer: (.096, .304)

Difficulty: Medium


67. Find a 95 percent confidence interval for p when =.25 and n = 400. Answer: (.208, .292)

Difficulty: Medium

68. Find a 99 percent confidence interval for p when = .51 and n = 1,000. Answer: (.469, .551)

Difficulty: Medium

69. In a survey of 400 people, 60 percent favor new zoning laws. Find a 95 percent confidence interval for the true proportion favoring new laws. Answer: (.55, .65)

Difficulty: Medium

70. We want to estimate with 99 percent confidence the percentage of buyers of cars who are under 30 years of age. A margin of error of 5 percentage points is desired. What sample size is needed? Answer: 664

2

2

(2.575) (.5)(.5)664

(.05)n

Difficulty: Hard

71. A cable TV company wants to estimate the percentage of cable boxes in use during an evening hour. An approximation is 20 percent. They want the estimate to be at the 90 percent confidence level and within 2 percent of the actual proportion. What sample size is needed? Answer: 1,083

Difficulty: Medium


72. An insurance company estimates 40 percent of its claims have errors. The insurance company wants to estimate with 90 percent confidence the proportion of claims with errors. What sample size is needed if they wish to be within 5 percent of the actual? Answer: 260

2

2

(1.645) (.4)(.6)260

(.05)n

Difficulty: Hard

73. In a randomly selected group of 650 automobile deaths, 180 were alcohol related. Construct a 95 percent confidence interval for the true proportion of all automobile accidents caused by alcohol. Answer: (.243, .311)

Difficulty: Medium

74. You want to estimate the proportion of customers who are satisfied with their supermarket at α = .10 and within .025 of the true value. It has been estimated that p =.85. How large of a sample is needed? Answer: 553

2

2

(1.645) (.85)(.15)553

(.025)n

Difficulty: Medium

75. Find a 99 percent confidence interval for if = 98.6, s = 2, and n = 5. Assume that the sample is randomly selected from a normally distributed population. Answer: (94.48, 102.72)

298.6 4.604 94.48 102.72

5to

Difficulty: Medium


76. The customer service manager for the XYZ Fastener Manufacturing Company examined 60 vouchers and found nine vouchers containing errors. Find a 95 percent confidence interval for the proportion of vouchers with errors. Answer: (.06, .24)

Difficulty: Medium

77. The success rate of a surgical procedure is 70 per 100 cases examined in a sample. Find a 90 percent confidence interval for the total number of successes per 1,000 cases. Answer: (676.2, 723.8)700 1.645 1000(.7)(.3) 676.2 to 723.8 Difficulty: Hard

Use the following information to answer questions 78-80:The weight of a product is measured in pounds. A sample of 50 units is taken from a batch. The sample yielded the following results: = 75 lbs., and s = 10 lbs.

78. Calculate a 90 percent confidence interval for . Answer: (72.67, 77.33)

1075 1.645 72.67 77.33

50to

Difficulty: Medium


1075 1.96 72.23 77.77

50to

Difficulty: Medium


1075 2.575 71.36 78.64

50to

Difficulty: Medium

81. The 95% confidence interval for the average weight of a product is from 72.23 lbs. to 77.77


lbs. Can we conclude that = 77 using a 95 percent confidence interval? Answer: Yes, because 77 is inside the interval. Difficulty: Medium

82. The 99% confidence interval for the average weight of a product is from 71.36 lbs. to 78.64 lbs. Can we conclude that is equal to 71 using a 99 percent confidence interval? Briefly explain. Answer: No, because 71 is outside the interval. Difficulty: Medium

Use the following information to answer questions 83-86:A sample of 2,000 people yielded =.52.

83. What is the standard deviation of the population proportion? Answer: .0112

Difficulty: Medium

84. Calculate a 90 percent confidence interval for p. Answer: (.502, .538)

Difficulty: Medium


Difficulty: Medium


Difficulty: Medium

87. Customers of a company were surveyed as to whether they were satisfied with the service they received. A sample of 2,000 customers yielded = .60. (proportion of customers satisfied


with service in the sample) What is the standard deviation of the number of customers satisfied with service? Answer: 21.91

2000(.6)(.4) 21.91. Difficulty: Medium

88. A random sample of size 10 is taken from a population assumed to be normal, and =1.2 and s = 0.6. Calculate a 90 percent confidence interval for . Answer: (.852, 1.548)

.61.2 1.833 .852 to 1.548

10

Difficulty: Medium

89. A random sample of size 10 is taken from a population assumed to be normal, and = 1.2 and s = .6. Calculate a 95 percent confidence interval for . Answer: (.771, 1.629)

Difficulty: Medium

90. A random sample of size 10 is taken from a population assumed to be normal, and X = 1.2 and the variance is .36. Calculate a 99 percent confidence interval for . Answer: (.583, 1.817)

Difficulty: Medium

91. If the 95% confidence interval for a mean is from .771 to 1.629, can we conclude that = .5, using a 95 percent confidence interval? Answer: No, because .5 is outside the interval. Difficulty: Medium


92. 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. On a particular day, 64 players completed the play on the 13th hole with average of 4.25 strokes and standard deviation of 1.6 strokes. Determine the 95% confidence interval for the average number of strokes. Answer: (3.86 to 4.64)

1.64.25 (1.96) (3.86 4.64)

64to

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.

93. Determine the probability that the lifetime for one drive will exceed 940 hours. Answer: .6915

940 1000 60.5

120 120( .5) .1915 .5 .6915

Z

P Z

Difficulty: Medium

94. Determine the probability that the lifetime for 9 disk drives will exceed 940 hours. Answer: .9332

940 1000 601.5

120 409

( 1.5) .4332 .5 .9332

Z

P Z

Difficulty: Medium

95. 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. A random sample of 61 pieces of rubber from different production batches is subjected to a stress test. The test measures the force needed to break the rubber in pounds. According to the sample results, the average pressure is 238.4 pounds and the standard deviation is 35 pounds. Determine the 98% confidence interval. Answer: (227.96 to 248.84)

Difficulty: Medium


96. An insurance analyst working for a car insurance company would like to determine the proportion of accident claims covered the company. A random sample of 240 claims shows that the insurance company covered 90 accident claims while 150 claims were not covered. Use a confidence interval of 95% and determine the margin of error. Answer: .0613

(.375)(.625)1.96 .0613

240

Difficulty: Medium

97. An insurance analyst working for a car insurance company would like to determine the proportion of accident claims covered the company. A random sample of 200 claims shows that the insurance company covered 80 accident claims while 120 claims were not covered. Construct a 90% confidence interval estimate of the true proportion of claims covered by the insurance company. Answer: (.343 to .457)

Difficulty: Medium

98. A computer manufacturing company has sent a mail survey to 2800 of its randomly selected customers that have purchased the recently launched personal computer. The survey asked the customers whether or not they were satisfied with the computer. 800 customers responded to the survey. 640 customers indicated that they were satisfied, while 160 customers indicated they were not satisfied with their new computer. Construct a 96% confidence interval estimate of the true proportion of customers satisfied with their new computer. Answer: (.7589 to .8141)

.8 2.05(.0141)(.7859 to .8141)Difficulty: Medium

99. A car insurance company would like to determine the proportion of accident claims covered by the company. According to a preliminary estimate 60% of the claims are covered. 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

2

(2.33) (.6)(.4)1448

.03n

Difficulty: Medium


100. A statistical quality control process for cereal production measures the weight of a cereal box. The population standard deviation is known to be .05 ounces. In order to achieve a 97% confidence with a margin of error of .02 ounces, how large a sample should be used? Answer: 43

Difficulty: Medium

101. The production manager for the XYZ manufacturing company is concerned that the customer orders are being shipped late. He asked one of his planners to check the timeliness of shipments for 1000 orders. The planner randomly selected 1000 orders and found that 120 orders were shipped late. Construct the 95% confidence interval for the proportion of orders shipped late. Answer: (.10 to .14)

(.10 to .14)Difficulty: Hard

Multiple Choice

Use the following information to answer questions 102-103:A company is interested in estimating , the mean number of days of sick leave taken by its employees. The firm’s statistician randomly selects 100 personnel files and notes the number of sick days taken by each employee. The sample mean is 12.2 days and the sample standard deviation is 10 days.

102. Calculate a 93% confidence interval for , the mean number of days of sick leave.A) [10.725 13.675]B) [10.390 14.010]C) [12.019 12.381]D) [12.053 12.348]E) [11.735 12.665]Answer: B


103. 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) 2B) 13C) 136D) 165E) 166Answer: EDifficulty: Hard

104. The state highway department is studying traffic patterns on one of the busiest highways in the state. As part of the study, the department needs to estimate the average number of vehicles that pass an intersection each day. A random sample of 64 days gives us a sample mean of 14,205 cars and a sample standard deviation of 1,010 cars. What is the 92% confidence interval estimate of μ, the mean number of cars passing the intersection?A) [12,438 15,972]B) [14,028 14,382]C) [12,189 14,221]D) [13,984 14,426]E) [14,183 14,227]Answer: DDifficulty: Medium

105. The state highway department is studying traffic patterns on one of the busiest highways in the state. As part of the study, the department needs to estimate the average number of vehicles that pass an intersection each day. A random sample of 64 days gives us a sample mean of 14,205 cars and a sample standard deviation of 1,010 cars. After calculating the confidence interval, the highway department officials determined that the precision is too low for their needs. They feel the precision should be 300 cars. Given this precision, and needing to be 99% confident, how many days do they need to sample?A) 109B) 76C) 75D) 62E) 9Answer: BDifficulty: Hard


106. The U.S. Department of Health and Human Services collected sample data for 772 males between the ages of 18 and 24. That sample group has a mean height of 69.7 inches with a standard deviation of 2.8 inches. Find the 99% confidence interval for the mean height of all males between the ages of 18 and 24.A) [63.19 76.21]B) [62.49 76.91]C) [69.65 69.75]D) [69.47 69.93]E) [69.44 69.96]Answer: EDifficulty: Medium

107. In a study of factors affecting soldiers’ decisions to reenlist, 320 subjects were measured for an index of satisfaction and the sample mean is 28.8 and the sample standard deviation is 7.3. Use the given sample data to construct the 98% confidence interval for the population mean.A) [27.85 29.75]B) [27.96 29.64]C) [11.82 45.78]D) [28.75 28.85]E) [28.60 29.00]Answer: ADifficulty: Hard

108. A psychologist is collecting data on the time it takes to learn a certain task. For 50 randomly selected adult subjects, the sample mean is 16.40 minutes and the sample standard deviation is 4.00 minutes. Construct the 95% confidence interval for the mean time required by all adults to learn the task.A) [8.56 24.24]B) [15.47 17.33]C) [16.24 16.56]D) [15.29 17.51]E) [17.12 48.48]Answer: DDifficulty: Medium


109. Research has been conducted that studies the role that the age of workers has in determining the hours per month spent on personal tasks. A sample of 1,686 adults was observed for one month. The data are:

AGE GROUP18-24 25-44 45-64

Mean 4.17 4.04 4.31Std Dev 0.75 0.81 0.82N 241 768 677

Construct a 98% confidence interval for the mean hours spent on personal tasks.A) [3.96 4.12]B) [3.97 4.11]C) [3.98 4.10]D) [2.16 5.92]E) [3.95 4.13]Answer: BDifficulty: Hard

110. Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoccupied seats per flight over the past year. 225 flight records are randomly selected and the number of unoccupied seats is noted with a sample mean of 11.6 seats and a standard deviation of 4.1 seats. Calculate a 90% confidence interval for , the mean number of unoccupied seats per flight during the past year.A) [4.86 18.34]B) [11.25 11.95]C) [11.57 11.63]D) [11.15 12.05]E) [11.30 12.20]Answer: DDifficulty: Medium (AS)

111. Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoccupied seats per flight over the past year. 225 flight records are randomly selected and the number of unoccupied seats is noted with a sample mean of 11.6 seats and a standard deviation of 4.1 seats. How many flights should we select if we wish to estimate μ to within 2 seats and be 95% confident?A) 130B) 65C) 33D) 17E) 12Answer: CDifficulty: Medium (AS)


112. An auditor was hired to verify the accuracy of a company’s new billing system. Thirty-five (35) invoices produced since the system was installed were sampled. The average error on the invoices was $1.00 with a standard deviation of $124.00. Construct a 97% confidence interval for the mean error per invoice.A) [-268.08 270.08]B) [-6.69 8.69]C) [-44.48 46.48]D) [-38.40 40.40]E) [-9.17 11.66]Answer: CDifficulty: Hard

113. In a random sample of 651 computer scientists who subscribed to a web-based daily news update, it was found that the average salary was $46,816 with a standard deviation of $12,557. Calculate a 91% confidence interval for the mean salary of computer scientists.A) [$25,469 $68,163]B) [$46,592 $47,040]C) [$46,157 $47,475]D) [$46,783 $46,849]E) [$45,979 $47,653]Answer: EDifficulty: Hard

114. At the end of 1990, 1991, and 1992 the average prices of a share of stock in a money market portfolio were $34.83, $34.65 and $31.26 respectively. To investigate the average share price at the end of 1993, 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 33.583 and a standard deviation of 19.149. Estimate the average price of a share of stock in the portfolio at the end of 1993 with a 90% confidence interval.A) [27.832 39.334]B) [26.732 40.434]C) [32.514 34.651]D) [32.533 34.633]E) [32.269 34.897]Answer: ADifficulty: Medium


115. Health insurers and the federal government are both putting pressure on hospitals to shorten the average length of stay (LOS) of their patients. In 1996, the average LOS for non-heart patient was 4.6 days. A random sample of 20 hospitals in one state had a mean LOS for non-heart patients in 2000 of 3.8 days and a standard deviation of 1.2 days. Calculate a 95% confidence interval for the population mean LOS for non-heart patients in the state’s hospitals in 2000.A) [3.24 4.36]B) [3.67 3.93]C) [3.34 4.26]D) [3.38 4.22]E) [3.27 4.33]Answer: ADifficulty: Medium

116. Health insurers and the federal government are both putting pressure on hospitals to shorten the average length of stay (LOS) of their patients. In 1996, the average LOS for non-heart patient was 4.6 days. A random sample of 20 hospitals in one state had a mean LOS for non-heart patients in 2000 of 3.8 days and a standard deviation of 1.2 days. How large a sample of hospitals would we need to be 99% confident that the sample mean is within 0.5 days of the population mean?A) 3B) 7C) 32D) 48E) 96Answer: DDifficulty: Hard

117. The coffee/soup machine at the local bus station is supposed to fill cups with 6 ounces of soup. Ten cups of soup are brought with results of a mean of 5.93 ounces and a standard deviation of 0.13 ounces. Construct a 99% confidence interval for the true machine-fill amount.A) [5.888 5.972]B) [5.814 6.046]C) [5.716 6.144]D) [5.824 6.036]E) [5.796 6.064]Answer: EDifficulty: Medium


118. The coffee/soup machine at the local bus station is supposed to fill cups with 6 ounces of soup. Ten cups of soup are brought with results of a mean of 5.93 ounces and a standard deviation of 0.13 ounces. How large a sample of soups would we need to be 95% confident that the sample mean is within 0.03 ounces of the population mean?A) 97B) 96C) 73D) 62E) 10Answer: ADifficulty: Hard

119. A local company makes a candy that is supposed to weigh 1.00 ounces. A random sample of 25 pieces of candy produces a mean of 0.996 ounces with a standard deviation of 0.004 ounces. Construct a 98% confidence interval for the mean weight of all such candy.A) [0.9645 1.0275]B) [0.9956 0.9964]C) [0.9860 1.0060]D) [0.9940 0.9980]E) [0.9942 0.9978]Answer: DDifficulty: Hard

120. A local company makes a candy that is supposed to weigh 1.00 ounces. A random sample of 25 pieces of candy produces a mean of 0.996 ounces with a standard deviation of 0.004 ounces. 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) 126B) 124C) 107D) 12E) 6Answer: ADifficulty: Medium


121. 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. A sample of five cars is tested and a mean of 28.2 miles is obtained. Assuming that the standard deviation is 2.7 miles, find the 95% confidence interval for the mean distance traveled by all such cars using 1 gallon of gas.A) [26.16 30.24]B) [20.70 35.70]C) [24.85 31.55]D) [26.70 29.70]E) [25.83 30.57]Answer: CDifficulty: 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. A sample of five cars is tested and a mean of 28.2 miles is obtained. How many cars should the environmental group test if they wish to estimate μ, mean miles per 1 gallon, to within 0.5 miles and be 99% confident?A) 25B) 124C) 194D) 618E) 619Answer: ADifficulty: Medium

123. A sociologist develops a test designed to measure a person’s attitudes about disabled people, and 16 randomly selected subjects are given the test. Their mean score is 71.2 with a standard deviation of 10.5. Construct the 99% confidence interval for the mean score of all subjects.A) [40.26 102.14]B) [63.46 78.94]C) [60.66 81.74]D) [64.44 77.96]E) [63.21 79.19]Answer: BDifficulty: Medium


124. There is an average of 22.455 employees at 22 office furniture dealers in a major metropolitan area with a standard deviation of 18.52. Construct a 99% confidence interval for the true mean number of full-time employees at office furniture dealers.A) [11.014 33.896]B) [11.277 33.633]C) [12.284 32.626]D) [12.513 32.397]E) [20.072 24.838]Answer: BDifficulty: Medium

125. A federal bank examiner is interested in estimating the mean outstanding defaulted loans balance of all defaulted loans over the last three years. A random sample of 20 defaulted loans yielded a mean of $67,918 with a standard deviation of $16,552.40. Calculate a 90% confidence interval for the mean balance of defaulted loans over the past three years.A) [66,487 69,349]B) [39,299 96,537]C) [57,329 78,507]D) [61,829 74,007]E) [61,519 74,317]Answer: EDifficulty: Medium

126. 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. After 95 days, the percentage of contamination removed from each soil sample was measured with a mean of 49.3% and a standard deviation of 1.5%. Estimate the mean percentage of contamination removed at a 98% confidence level. A) [48.36 50.24]B) [47.67 50.93]C) [47.29 51.31]D) [47.88 50.72]E) [46.47 52.13]Answer: BDifficulty: Hard (AS)


127. 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. After 95 days, the percentage of contamination removed from each soil sample was measured with a mean of 49.3% and a standard deviation of 1.5%. If we wished to narrow the boundary around μ for a 98% confidence interval to within 0.5%, how many soil samples should be in our experiment.A) 437B) 33C) 9D) 6E) 3Answer: ADifficulty: Hard

128. A botanist measures the heights of 16 seedlings and obtains a mean and standard deviation of 72.5 cm and 4.5 cm, respectively. Find the 95% confidence interval for the mean height of seedlings in the population form which the sample was selected.A) [62.91 82.09]B) [70.30 74.70]C) [70.10 74.90]D) [70.54 74.46]E) [71.90 73.10]Answer: CDifficulty: Medium

129. On a standard IQ test, the standard deviation is 15. 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) 4,239B) 283C) 212D) 131E) 66Answer: ADifficulty: Medium


130. A manufacturer of dodge balls uses a machine to inflate its new balls to a pressure of 13.5 pounds ( = 0.1). When the machine is properly calibrated, the mean inflation pressure s 13.5 pounds, but uncontrollable factors can cause the pressure of individual dodge balls to vary. For quality control purposes, the manufacturer wishes to estimate the mean inflation pressure to within 0.025 pounds of its true value with a 99% confidence. What sample size should be used?A) 677B) 107C) 35D) 27E) 11Answer: BDifficulty: Hard

131. Recently, a case of food poisoning was traced to a particular restaurant chain. The source was identified and corrective actions were taken to make sure that the food poisoning would not reoccur. Despite the response from the restaurant chain, many consumers refused to visit the restaurant for some time after the event. A survey was conducted three months after the food poisoning occurred with a sample of 319 patrons contacted. 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.A) [.059 .122]B) [.090 .091]C) [.000.196]D) [.240 .339]E) [.118 .244]Answer: ADifficulty: Medium (AS)

132. Recently, a case of food poisoning was traced to a particular restaurant chain. The source was identified and corrective actions were taken to make sure that the food poisoning would not reoccur. Despite the response from the restaurant chain, many consumers refused to visit the restaurant for some time after the event. A survey was conducted three months after the food poisoning occurred with a sample of 319 patrons contacted. Of the 319 contacted, 29 indicated that they would not go back to the restaurant because of the potential for food poisoning. What sample size would be needed in order to be 99% confident that the sample proportion is within .02 of ρ, the true proportion of customers who refuse to go back to the restaurant?A) 14B) 38C) 129D) 1,371E) 1,777Answer: ADifficulty: Medium (AS)


133. In a survey of 308 adults, 175 respondents listed jewelry as being among the romantic gifts men want to receive. Construct the 98% confidence interval for the proportion of all adults who include jewelry among their responses.A) [.025 .085]B) [.169 .171]C) [.120 .220]D) [.140 .200]E) [.050 .170]Answer: CDifficulty: Hard

134. In a poll of 1,004 adults, 93% indicated that restaurants and bars should refuse service to patrons who have had too much to drink. Construct the 90% confidence interval for the proportion of all adults who feel the same way.A) [.078 .108]B) [.914 .946]C) [.920 .940]D) [.917 .943]E) [.910 .950]Answer: DDifficulty: Hard

135. At a sobriety checkpoint the sheriff’s department screened 676 drivers and 6 were arrested for DWI. Find the 92% confidence interval for the true proportion of drivers who were DWI that evening.A) [0.000 0.027]B) [0.003 0.015]C) [0.902 0.938]D) [0.000 0.076]E) [0.006 0.030]Answer: BDifficulty: Hard


136. The Ohio Department of Agriculture tested 203 fuel samples across the state in 1999 for accuracy of the reported octane level. For premium grade, 14 out of 105 samples failed (they didn’t meet ASTM specification and the FTC Octane posting rule). Find a 99% confidence interval for the true population proportion of premium grade fuel-quality failures.A) [.045 .221]B) [.068 .198]C) [.023 .115]D) [.047 .219]E) [.100 .276]Answer: DDifficulty: Medium

137. The Ohio Department of Agriculture tested 203 fuel samples across the state in 1999 for accuracy of the reported octane level. For premium grade, 14 out of 105 samples failed (they didn’t meet ASTM specification and the FTC Octane posting rule). How many samples would be needed to create a 99% confidence interval that is within 0.02 of the true proportion of premium grade fuel-quality failures?A) 4148B) 2838C) 1913D) 744E) 54Answer: BDifficulty: Hard

138. To investigate the rate at which employees with cancer are fired or laid off, a telephone survey was taken of 100 cancer survivors who worked while undergoing treatment. Seven (7) were either fired or laid off due to their illness. Construct a 90% confidence interval for the true percentage of all cancer patients who are fired or laid off due to their illness.A) [0.0000 0.2034]B) [0.0371 0.1029]C) [0.0039 0.1361]D) [0.0278 0.1122]E) [0.0078 0.1400]Answer: DDifficulty: Hard


139. In 1995, 13,000 internet users were surveyed and asked about their willingness to pay fees for access to websites. Of these, 2,938 were definitely not willing to pay such fees. Construct a 95% confidence interval for the proportion definitely unwilling to pay fees.A) [0.286 0.302]B) [0.219 0.233]C) [0.220 0.232]D) [0.212 0.241]E) [0.214 0.245]Answer: BDifficulty: Medium (AS)

140. In 1995, 13,000 internet users were surveyed and asked about their willingness to pay fees for access to websites. Of these, 2,938 were definitely not willing to pay such fees. How large a sample is necessary to estimate the proportion of interest to within 2% in a 95% confidence interval?A) 18B) 307C) 1717D) 2000E) 2965Answer: CDifficulty: Medium (AS)


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