Chapter 7
Confidence 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 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,  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
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8. When the level of confidence and sample proportion p̂ 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 pˆ (1  pˆ ) is larger than when pˆ (1  pˆ ) 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 p̂ that makes
pˆ (1  pˆ ) 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
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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
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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
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24. When the sample size and the sample proportion p̂ 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 p̂ 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 pˆ (1  pˆ ) is larger than when pˆ (1  pˆ ) 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 p̂ that makes:
A) pˆ (1  pˆ ) as small as reasonably possible
B) pˆ (1  pˆ ) as large as reasonably possible
C) pˆ (1  pˆ ) as close to .5 as reasonably possible
D) pˆ (1  pˆ ) as close to .25 as reasonably possible
E) Both B and D are correct
Answer: E Difficulty: Medium (REF)
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29. When a confidence interval for a population proportion is constructed for a sample size n =
30 and the value of p̂ = .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 Z a / 2 .
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
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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
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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
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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
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Essay
48. A random sample of size 30 from a normal population yields X = 32.8 and s = 4.51.
Construct a 95 percent confidence interval for  .
Answer: (31.19, 34.41)
 4.51 
32.8  1.96 
  31.19 to 34.41
 30 
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)
 .0312 
1.0  3.355 
  .965 to 1.035
 9 
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)
 2.4 
18.6  3.499 
  15.63 to 21.57
 8
Difficulty: Medium
51. A sample of 12 items yields X = 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)
 1.5 
48.5  1.796 
  47.722 to 49.278
 12 
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)
 5.1 
21.6  1.96 
  20.6 to 22.6
 100 
Difficulty: Medium
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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)
(.42)(.58)
.42  1.96
 .389 to .451
1000 - 1
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)
114
pˆ 
 .19
600
(.19)(.81)
.19  1.96
 .159 to .221
600 - 1
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)
37
pˆ 
 .3083
120
(.3083)(.6 917)
.3083  1.96
 .225 to .391
120 - 1
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
2
 1.645(7) 
n 
  34
2


Difficulty: Hard
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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
(1.96)2 (.5)(.5)
n
 2401
(.02)2
Difficulty: Medium (AS)
58. What sample size is needed to estimate the proportion of highway speeders within 5 percent
using a 90 percent confidence level?
Answer: 271
(1.645)2 (.5)(.5)
n
 271
(.05)2
Difficulty: Medium (AS)
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
(1.96) 2 (1900)
n
 19
(20) 2
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)
 1.8 
31.7  1.645 
  31.49 to 31.91
 200 
Difficulty: Medium
61. Ten items of 100 are defective. Develop a 95 percent confidence interval for the population
proportion of defectives.
Answer: (.0412, .1588)
10
pˆ 
 .10
100
(.10)(.90)
.10  1.96
 .0412 to .1588
100
Difficulty: Hard
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62. What is a 95 percent confidence interval for  when n = 10, X =35.6, and s = 13.0?
Assume population normality.
Answer: (26.3, 44.9)
 13 
35.6  2.262 
  26.3 to 44.9
 10 
Difficulty: Medium
63. What is a 95 percent confidence interval for  when n = 10, X =34.1, and s=3.0? Assume
population normality.
Answer: (32.0, 36.2)
 3 
34.1  2.262 
  31.95 to 36.25
 10 
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)
 3.08 
63.8  1.96 
  63.43 to 64.17
 265 
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)
 24.592 
172  2.228 
  155.24 to 188.76
 11 
Difficulty: Hard
66. Find the 99 percent confidence interval for p when p̂ =.2, and n = 100.
Answer: (.096, .304)
(.2)(.8)
.2  2.575
 .096 to .304
100 - 1
Difficulty: Medium
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67. Find a 95 percent confidence interval for p when p̂ =.25 and n = 400.
Answer: (.208, .292)
(.25)(.75)
.25  1.96
 .208 to .292
400 - 1
Difficulty: Medium
68. Find a 99 percent confidence interval for p when p̂ = .51 and n = 1,000.
Answer: (.469, .551)
(.51)(.49)
.51  2.575
 .469 to .551
1000 - 1
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)
(.60)(.40)
.60  1.96
 .552 to .648
400 - 1
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
(2.575)2 (.5)(.5)
n
 664
(.05) 2
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
(1.645) 2 (.2)(.8)
n
 1083
.02
Difficulty: Medium
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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
(1.645) 2 (.4)(.6)
n
 260
(.05) 2
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)
180
pˆ 
 .2769
650
(.2769)(.7 231)
.2769  1.96
 .243 to .311
650 - 1
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
(1.645) 2 (.85)(.15)
n
 553
(.025) 2
Difficulty: Medium
75. Find a 99 percent confidence interval for  if X = 98.6, s = 2, and n = 5. Assume that the
sample is randomly selected from a normally distributed population.
Answer: (94.48, 102.72)
 2 
98.6  4.604 
  94.48 to 102.72
 5
Difficulty: Medium
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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)
9
pˆ 
 .15
60
(.15)(.85)
.15  1.96
 .06 to .24
60 - 1
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: X = 75 lbs., and s = 10 lbs.
78. Calculate a 90 percent confidence interval for  .
Answer: (72.67, 77.33)
 10 
75  1.645 
  72.67 to 77.33
 50 
Difficulty: Medium
79. Calculate a 95 percent confidence interval for  .
Answer: (72.23, 77.77)
 10 
75  1.96 
  72.23 to 77.77
 50 
Difficulty: Medium
80. Calculate a 99 percent confidence interval for  .
Answer: (71.36, 78.64)
 10 
75  2.575 
  71.36 to 78.64
 50 
Difficulty: Medium
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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 p̂ =.52.
83. What is the standard deviation of the population proportion?
Answer: .0112
(.52)(.48)
 
 .0112
2000 - 1
Difficulty: Medium
84. Calculate a 90 percent confidence interval for p.
Answer: (.502, .538)
(.52)(.48)
.52  1.645
 .502 to .538
2000 - 1
Difficulty: Medium
85. Calculate a 95 percent confidence interval for p.
Answer: (.498, .542)
(.52)(.48)
.52  1.96
 .498 to .542
2000 - 1
Difficulty: Medium
86. Calculate a 99 percent confidence interval for p.
Answer: (.491, .549)
(.52)(.48)
.52  2.575
 .491 to .549
2000 - 1
Difficulty: Medium
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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 p̂ = .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 X =1.2
and s = 0.6. Calculate a 90 percent confidence interval for  .
Answer: (.852, 1.548)
.6
1.2  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 X = 1.2
and s = .6. Calculate a 95 percent confidence interval for  .
Answer: (.771, 1.629)
.6
1.2  2.262
 .771 to 1.629
10
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)
.36
1.2  3.25
 .583 to 1.817
10
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
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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)
 1.6 
4.25 (1.96) 
  (3.86 to 4.64)
 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.
93. Determine the probability that the lifetime for one drive will exceed 940 hours.
Answer: .6915
940  1000
60
Z

 .5
120
120
P ( Z  .5)  .1915  .5  .6915
Difficulty: Medium
94. Determine the probability that the lifetime for 9 disk drives will exceed 940 hours.
Answer: .9332
940  1000 60
Z

 1.5
120
40
9
P( Z  1.5)  .4332  .5  .9332
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)
 35 
238.4  2.33
  227.96 to 248.84
 61 
Difficulty: Medium
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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)
 (.4)(.6) 
80
  .343 to .457
 1.645

200
200
1


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)
640
(.8)(.2)
 2.05
800
800 - 1
.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
(2.33) 2 (.6)(.4)
n
 1448
.032
Difficulty: Medium
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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
(2.17) 2 (.06) 2
n
 43
(.02) 2
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)
120
(.12)(.88)
 1.96
1000
1000 - 1
(.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
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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) 2
B) 13
C) 136
D) 165
E) 166
Answer: E
Difficulty: 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: D
Difficulty: 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) 109
B) 76
C) 75
D) 62
E) 9
Answer: B
Difficulty: Hard
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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: E
Difficulty: 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: A
Difficulty: 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: D
Difficulty: Medium
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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:
Mean
Std Dev
N
18-24
4.17
0.75
241
AGE GROUP
25-44
4.04
0.81
768
45-64
4.31
0.82
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: B
Difficulty: 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: D
Difficulty: 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) 130
B) 65
C) 33
D) 17
E) 12
Answer: C
Difficulty: Medium (AS)
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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: C
Difficulty: 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: E
Difficulty: 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: A
Difficulty: Medium
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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 nonheart 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: A
Difficulty: 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 nonheart 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) 3
B) 7
C) 32
D) 48
E) 96
Answer: D
Difficulty: 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: E
Difficulty: Medium
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217
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) 97
B) 96
C) 73
D) 62
E) 10
Answer: A
Difficulty: 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: D
Difficulty: 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) 126
B) 124
C) 107
D) 12
E) 6
Answer: A
Difficulty: Medium
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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: 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. 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) 25
B) 124
C) 194
D) 618
E) 619
Answer: A
Difficulty: 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: B
Difficulty: Medium
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219
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: B
Difficulty: 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: E
Difficulty: 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: B
Difficulty: Hard (AS)
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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) 437
B) 33
C) 9
D) 6
E) 3
Answer: A
Difficulty: 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: C
Difficulty: 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) 4,239
B) 283
C) 212
D) 131
E) 66
Answer: A
Difficulty: Medium
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221
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) 677
B) 107
C) 35
D) 27
E) 11
Answer: B
Difficulty: 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: A
Difficulty: 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) 14
B) 38
C) 129
D) 1,371
E) 1,777
Answer: A
Difficulty: Medium (AS)
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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: C
Difficulty: 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: D
Difficulty: 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: B
Difficulty: Hard
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223
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: 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. 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) 4148
B) 2838
C) 1913
D) 744
E) 54
Answer: B
Difficulty: 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: D
Difficulty: Hard
224
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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: B
Difficulty: 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) 18
B) 307
C) 1717
D) 2000
E) 2965
Answer: C
Difficulty: Medium (AS)
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225