Download CHAPTER 10 SECTION 2: INTRODUCTION TO ESTIMATION

CHAPTER 10 SECTION 2: INTRODUCTION TO ESTIMATION
MULTIPLE CHOICE
47. The term 1   refers to:
a. the probability that a confidence interval does not contain the population parameter.
b. the confidence level.
c. the level of unbiasedness.
d. the level of consistency.
ANS: B
PTS: 1
REF: SECTION 10.2
48. The letter  in the formula for constructing a confidence interval estimate of the population mean is:
a. the level of confidence.
b. the probability that a particular confidence interval will contain the population mean.
c. the area in the lower tail of the sampling distribution of the sample mean.
d. None of these choices.
ANS: D
PTS: 1
REF: SECTION 10.2
49. Which of the following is an incorrect statement about a 90% confidence interval?
a. If we repeatedly draw samples of the same size from the same population, 90% of the
resulting confidence intervals will include .
b. There is a 90% probability that the population mean  will lie between the lower
confidence limit (LCL) and the upper confidence limit (UCL).
c. We are 90% confident that our sample mean equals the population mean .
d. 90% of the population values will lie within the confidence interval.
ANS: A
PTS: 1
REF: SECTION 10.2
50. The width of a confidence interval estimate of the population mean increases when the:
a. level of confidence increases
b. sample size decreases
c. value of the population standard deviation increases
d. All of these choices are true.
ANS: D
PTS: 1
REF: SECTION 10.2
51. If the confidence level is reduced, the confidence interval:
a. widens.
b. remains the same.
c. narrows.
d. disappears.
ANS: C
REF: SECTION 10.2
value for a 95% confidence interval estimate for a population mean  is
52. The
a.
b.
c.
d.
PTS: 1
0.95
0.025
1.65
1.96
ANS: D
PTS: 1
REF: SECTION 10.2
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53. In developing an interval estimate for a population mean, the population standard deviation  was
assumed to be 10. The interval estimate was 50.92  2.14. Had  equaled 20, the interval estimate
would be
a. 60.92  2.14
b. 50.92  12.14
c. 101.84  4.28
d. 50.92  4.28
ANS: D
PTS: 1
REF: SECTION 10.2
54. In developing an interval estimate for a population mean, a sample of 50 observations was used. The
interval estimate was 19.76  1.32. Had the sample size been 200 instead of 50, the interval estimate
would have been:
a. 19.76  .33
b. 19.76  .66
c. 19.76  5.28
d. None of these choices.
ANS: B
PTS: 1
REF: SECTION 10.2
55. After constructing a confidence interval estimate for a population mean, you believe that the interval is
useless because it is too wide. In order to correct this problem, you need to:
a. increase the population standard deviation.
b. increase the sample size.
c. increase the level of confidence.
d. increase the sample mean.
ANS: B
PTS: 1
REF: SECTION 10.2
56. Suppose an interval estimate for the population mean was 62.84 to 69.46. The population standard
deviation was assumed to be 6.50, and a sample of 100 observations was used. The mean of the sample
was:
a. 56.34
b. 62.96
c. 6.62
d. 66.15
ANS: D
PTS: 1
REF: SECTION 10.2
57. A confidence interval is defined as:
a. a point estimate plus or minus a specific confidence level.
b. a lower and upper confidence limit associated with a specific level of confidence.
c. an interval that has a 95% probability of containing the population parameter.
d. a lower and upper confidence limit that has a 95% probability of containing the population
parameter.
ANS: B
PTS: 1
REF: SECTION 10.2
58. Which of the following conditions does not allow you to use the formula
to estimate
?
a. Population is normally distributed and the population variance is known.
b. Population is not normally distributed but n is large; population variance is known.
c. Population has any distribution and n is any size.
d. All of these choices allow you to use the formula.
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ANS: C
PTS: 1
REF: SECTION 10.2
59. In the formula
, the  / 2 refers to:
a. the probability that the sample mean will not equal the population mean.
b. the probability that the confidence interval will not contain the population mean.
c. the level of confidence.
d. None of these choices.
ANS: D
PTS: 1
REF: SECTION 10.2
60. Which of the following is not a part of the formula for constructing a confidence interval estimate of
the population mean?
a. A point estimate of the population mean.
b. The standard error of the sampling distribution of the sample mean.
c. The confidence level.
d. The value of the population mean.
ANS: D
PTS: 1
REF: SECTION 10.2
61. The larger the confidence level, the:
a. smaller the value of z / 2.
b. wider the confidence interval.
c. smaller the probability that the confidence interval will contain the population mean.
d. None of these choices.
ANS: B
PTS: 1
REF: SECTION 10.2
62. A 99% confidence interval estimate of the population mean  can be interpreted to mean:
a. if all possible sample are taken and confidence intervals created, 99% of them would
include the true population mean somewhere within their interval.
b. we have 99% confidence that we have selected a sample whose interval does include the
population mean.
c. we estimate that the population mean falls between the lower and upper confidence limits,
and this type of estimator is correct 99% of the time.
d. All of these choices are true.
ANS: D
PTS: 1
REF: SECTION 10.2
63. It is desired to estimate the average total compensation of CEOs in the service industry. Data were
randomly collected from 18 CEOs and 95% confidence interval was calculated to be ($2,190,000,
$4,720,000). Based on the interval above, do you believe the actual average total compensation of
CEOs in the service industry could be $3,000,000?
a. Yes, and I am sure of that.
b. Yes, and I am 95% confident of that.
c. No, and I am sure of that.
d. No, and I am 95% confident of that.
ANS: B
PTS: 1
REF: SECTION 10.2
64. Suppose a 95% confidence interval for  turns out to be (1,000, 2,100). What does it mean to be 95%
confident?
a. In repeated sampling, the population parameter would fall in the resulting interval 95% of
the time.
b. 95% of the observations in the entire population fall in the given interval.
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c. 95% of the observations in the sample fall in the given interval.
d. None of these choices.
ANS: A
PTS: 1
REF: SECTION 10.2
TRUE/FALSE
65. A confidence interval is an interval estimate for which there is a specified degree of certainty that the
actual value of the population parameter will fall within the interval.
ANS: T
PTS: 1
REF: SECTION 10.2
66. One can reduce the width of a confidence interval by taking a smaller sample size.
ANS: F
PTS: 1
REF: SECTION 10.2
67. The width of a 95% confidence interval is 0.95.
ANS: F
PTS: 1
REF: SECTION 10.2
68. The difference between the sample statistic and actual value of the population parameter is the
confidence level of the estimate.
ANS: F
PTS: 1
REF: SECTION 10.2
69. The term 1   refers to the probability that a confidence interval does not contain the population
parameter.
ANS: F
PTS: 1
REF: SECTION 10.2
70. In the formula
, the subscript  / 2 refers to the area in the lower tail or upper tail of the
sampling distribution of the sample mean.
ANS: T
PTS: 1
REF: SECTION 10.2
71. The larger the confidence level used in constructing a confidence interval estimate of the population
mean, the narrower the confidence interval.
ANS: F
PTS: 1
REF: SECTION 10.2
72. In order to construct a confidence interval estimate of the population mean, the value of the population
mean is needed.
ANS: F
PTS: 1
REF: SECTION 10.2
73. In developing an interval estimate for a population mean, the population standard deviation  was
assumed to be 8. The interval estimate was 50.0  2.50. Had  equaled 16, the interval estimate would
be 100  5.0.
ANS: F
PTS: 1
REF: SECTION 10.2
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74. A 95% confidence interval estimate for a population mean  is determined to be 75 to 85. If the
confidence level is reduced to 80%, the confidence interval for  becomes wider.
ANS: F
PTS: 1
REF: SECTION 10.2
75. When constructing confidence interval for a parameter, we generally set the confidence level 1  
close to 1 (usually between 0.90 and 0.99) because it is the probability that the interval includes the
actual value of the population parameter.
ANS: T
PTS: 1
REF: SECTION 10.2
76. Suppose that a 90% confidence interval for  is given by
90% confident that  falls between
and
.
ANS: F
PTS: 1
. This notation means that we are
REF: SECTION 10.2
77. We cannot interpret the confidence interval estimate of  as a probability statement about  because
the population mean is a fixed quantity.
ANS: T
PTS: 1
REF: SECTION 10.2
78. The width of the confidence interval estimate of the population mean  is a function of only two
quantities: the population standard deviation  and the sample size n.
ANS: F
PTS: 1
REF: SECTION 10.2
79. Doubling the population standard deviation  has the effect of doubling the width of the confidence
interval estimate of .
ANS: T
PTS: 1
REF: SECTION 10.2
80. Increasing the value of 1   narrows a confidence interval.
ANS: F
PTS: 1
REF: SECTION 10.2
81. Suppose that a 95% confidence interval for  is given by
. This notation means that, if we
repeatedly draw samples of the same size from the same population, 95% of the values of will be
such that  would lie somewhere between
and
.
ANS: T
PTS: 1
REF: SECTION 10.2
82. When constructing confidence interval estimate of , doubling the sample size n decreases the width
of the interval by half.
ANS: F
PTS: 1
REF: SECTION 10.2
83. In this chapter you need four values to construct the confidence interval estimate of . They are the
sample mean, the sample size, the population standard deviation, and the confidence level.
ANS: T
PTS: 1
REF: SECTION 10.2
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84. Given a mean of 2.1 and a standard deviation of 0.7, a 90% confidence interval will be 2.1  0.7.
ANS: F
PTS: 1
REF: SECTION 10.2
85. Suppose a sample size of 5 has mean 9.60. If the population variance is 5 and the population is
normally distributed, the lower limit for a 92% confidence interval is 7.85.
ANS: T
PTS: 1
REF: SECTION 10.2
86. Other things being equal, as the confidence level increases, the width of the confidence interval
increases.
ANS: T
PTS: 1
REF: SECTION 10.2
87. Other things being equal, the confidence interval for the mean will be wider for 99% confidence than
for 95% confidence.
ANS: T
PTS: 1
REF: SECTION 10.2
88. The lower limit of the 90% confidence interval for , where n = 64, = 70, and  = 20, is 65.89.
ANS: T
PTS: 1
REF: SECTION 10.2
COMPLETION
89. When estimating the population mean using a confidence interval, the sample mean is in the
____________________ of the interval.
ANS: middle
PTS: 1
REF: SECTION 10.2
90. A confidence interval is a probability statement about the ____________________ mean.
ANS: sample
PTS: 1
REF: SECTION 10.2
91. The confidence ____________________ is the chance that the interval contains the parameter, over
repeated sampling.
ANS: level
PTS: 1
REF: SECTION 10.2
92. In the formula for a confidence interval for , 1   is called the ____________________.
ANS: confidence level
PTS: 1
REF: SECTION 10.2
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93. The smallest number in a confidence interval is abbreviated by ____________________.
ANS: LCL
PTS: 1
REF: SECTION 10.2
94. It is ____________________ to say that a 95% confidence interval means there is a 95% chance that
the parameter lies in that confidence interval.
ANS:
wrong
incorrect
PTS: 1
REF: SECTION 10.2
95. The population mean is a(n) ____________________ but ____________________ quantity.
ANS:
fixed; unknown
unknown; fixed
PTS: 1
REF: SECTION 10.2
96. If you increase the confidence level, the confidence interval becomes ____________________.
ANS:
wider
bigger
larger
PTS: 1
REF: SECTION 10.2
97. We do not have control over the population standard deviation when forming a confidence interval,
but we can control the ____________________ and the ____________________.
ANS:
confidence level; sample size
sample size; confidence level
PTS: 1
REF: SECTION 10.2
98. The confidence level that is considered to be the standard amongst statisticians is
____________________%.
ANS: 95
PTS: 1
REF: SECTION 10.2
99. If the population values are very close to each other, the width of the confidence interval is
____________________ than if the population values are very far apart.
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copied, or distributed without the prior consent of the publisher.
ANS:
narrower
smaller
PTS: 1
REF: SECTION 10.2
100. Increasing the sample size fourfold decreases the width of the confidence interval by
____________________.
ANS:
half
one-half
50%
PTS: 1
REF: SECTION 10.2
SHORT ANSWER
101. A survey of 100 retailers revealed that the mean after-tax profit was $80,000. Assume the population
standard deviation is $15,000:
a.
b.
Determine the 95% confidence interval estimate of the mean after-tax profit for all
retailers.
Explain why you can use the confidence interval formula here, even though the population
is not necessarily normal.
ANS:
a.
b.
LCL = $77,060 and UCL = $82,940.
By the Central Limit Theorem, as long as n is large (and 100 is considered large), the
sample means have an approximate normal distribution.
PTS: 1
REF: SECTION 10.2
102. The temperature readings for 20 winter days (degrees Fahrenheit) in Grand Rapids, Michigan, are
normally distributed with a mean of 5.5 degrees and a standard deviation of 1.5 degrees. Determine the
90% confidence interval estimate for the winter mean temperature.
ANS:
LCL = 4.9 degrees and UCL = 6.05 degrees.
PTS: 1
REF: SECTION 10.2
103. A sample of 49 measurements of tensile strength for roof hangers are calculated to have a mean of
2.45 and a standard deviation of 0.25. (Units are Newton's per square meter.)
a. Determine the 95% confidence interval for mean tensile strength for all hangers.
b. Interpret this confidence interval.
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copied, or distributed without the prior consent of the publisher.
ANS:
a. LCL = 2.38 and UCL = 2.52 (Newton's per square meter).
b. I estimate the mean tensile strength for all roof hangars to be between 2.38 and 2.52
Newton's per square meter, based on my sample. We know that 95% of the time this
method will contain the true population mean.
PTS: 1
REF: SECTION 10.2
104. A random sample of 10 waitresses in Iowa City, Iowa, revealed the following hourly earnings,
including tips: 19, 18, 15, 16, 18, 17, 16, 18, 20, and 14. (Units are dollars.)
a. If the hourly earnings are normally distributed with a standard deviation of $4.50, estimate
with 95% confidence the mean hourly earnings for all waitresses in Iowa City.
b. Interpret your confidence interval.
ANS:
a. LCL = $14.31 and UCL = $19.89.
b. Based on my sample, I estimate that the mean hourly earnings is between $14.31 and
$19.89. I am 95% confident that my sampling process yields a correct result.
PTS: 1
REF: SECTION 10.2
Time Spent on Internet
Suppose that the amount of time teenagers spend on the internet per week is normally distributed with
a standard deviation of 1.5 hours. A sample of 100 teenagers is selected at random, and the sample
mean computed as 6.5 hours.
105. {Time Spent on Internet Narrative} Determine the 95% confidence interval estimate of the population
mean.
ANS:
LCL = 6.21 and UCL = 6.79 (hours per week)
PTS: 1
REF: SECTION 10.2
106. {Time Spent on Internet Narrative} Interpret the 95% confidence interval for this situation.
ANS:
Based on our sample, we estimate the average time teenagers spend on the internet is between 6.21
hours and 6.79 hours per week. We are 95% in our sampling procedure that yields our results.
PTS: 1
REF: SECTION 10.2
107. {Time Spent on Internet Narrative} Determine and interpret the 99% confidence interval estimate of
the population mean.
ANS:
Based on our sample we estimate the average time teenagers spent on the internet LCL = 6.11 and
UCL = 6.89 (hours per week). Repeated samples correctly estimate this average 99% of the time.
PTS: 1
REF: SECTION 10.2
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108. {Time Spent on Internet Narrative} Determine and interpret the 90% confidence interval estimate of
the population mean.
ANS:
LCL = 6.25 and UCL = 6.75 (hours per week). We are 90% confident in the process that leads us to
our results over repeated samples.
PTS: 1
REF: SECTION 10.2
109. {Time Spent on Internet Narrative} Determine the 95% confidence interval estimate of the population
mean if the sample size is changed to 300.
ANS:
LCL = 6.33 and UCL = 6.67 (hours per week)
PTS: 1
REF: SECTION 10.2
110. {Time Spent on Internet Narrative} Determine the 95% confidence interval estimate of the population
mean if the sample size is changed to 36.
ANS:
LCL = 6.01 and UCL = 6.99 (hours per week)
PTS: 1
REF: SECTION 10.2
111. {Time Spent on Internet Narrative} Determine the 95% confidence interval estimate of the population
mean if the population standard deviation is changed to 2.
ANS:
LCL = 6.11 and UCL = 6.89 (hours per week)
PTS: 1
REF: SECTION 10.2
112. {Time Spent on Internet Narrative} Determine the 95% confidence interval estimate of the population
mean if the population standard deviation is changed to 1.2.
ANS:
LCL = 6.26 and UCL = 6.74 (hours per week)
PTS: 1
REF: SECTION 10.2
113. {Time Spent on Internet Narrative} Determine the 95% confidence interval estimate of the population
mean if the sample mean is changed to 5.0 hours.
ANS:
LCL = 4.71 and UCL = 5.29 (hours per week)
PTS: 1
REF: SECTION 10.2
114. {Time Spent on Internet Narrative} Determine the 95% confidence interval estimate of the population
mean if the sample mean is changed to 8.5 hours.
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copied, or distributed without the prior consent of the publisher.
ANS:
LCL = 8.21 and UCL = 8.79 (hours per week)
PTS: 1
REF: SECTION 10.2
115. {Time Spent on Internet Narrative} What happens to the width of a confidence interval in each of the
following situations?
a. Confidence level increases
b. Confidence level decreases
c. Sample size increases
d. Sample size decreases
e. Population standard deviation increases
f. Population standard deviation decreases
g. Sample mean increases
h. Sample mean decreases
ANS:
a. Widens
b. Narrows
c. Narrows
d. Widens
e. Widens
f. Narrows
g. No change
h. No change
PTS: 1
REF: SECTION 10.2
116. A random sample of 10 university students was surveyed to determine the amount of time spent
weekly using a personal computer. The times are: 13, 14, 5, 6, 8, 10, 7, 12, 15, and 3. If the times are
normally distributed with a standard deviation of 5.2 hours, estimate with 90% confidence the mean
weekly time spent using a personal computer by all university students.
ANS:
LCL = 6.59 and UCL = 12.01 hours.
PTS: 1
REF: SECTION 10.2
117. A financial analyst wanted to determine the mean annual return on mutual funds. A random sample of
60 returns shows a mean of 12%. If the population standard deviation is assumed to be 4%, estimate
with 95% confidence the mean annual return on all mutual funds.
ANS:
LCL = 11% and UCL = 13%
PTS: 1
REF: SECTION 10.2
118. An economist is interested in studying the incomes of consumers in a particular region. The population
standard deviation is known to be $1,000. A random sample of 50 individuals resulted in an average
income of $15,000. What is the upper end point in a 99% confidence interval for the average income?
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ANS:
UCL = $15,364.87
PTS: 1
REF: SECTION 10.2
119. An economist is interested in studying the incomes of consumers in a particular region. The population
standard deviation is known to be $1,000. A random sample of 50 individuals resulted in an average
income of $15,000. What is the width of the 99% confidence interval?
ANS:
Width = $729.74
PTS: 1
REF: SECTION 10.2
120. A quality control engineer is interested in the mean length of sheet insulation being cut automatically
by machine. It is known that the standard deviation in the cutting length is 0.15 feet. A sample of 60
cut sheets yields a mean length of 12.15 feet. This sample will be used to obtain a 99% confidence
interval for the mean length cut by machine.
a.
b.
What is the z / 2 value to use in obtaining the confidence interval?
Develop the 99% confidence interval for .
ANS:
a.
b.
z / 2 = 2.58
LCL = 12.10 and UCL = 12.20 (feet)
PTS: 1
REF: SECTION 10.2
Hotel Rooms Rented
A hotel chain wants to estimate the average number of rooms rented daily in each month. The
population of rooms rented daily is assumed to be normally distributed for each with a standard
deviation of 24 rooms.
121. {Hotel Rooms Rented Narrative} During January, a sample of 16 days has a sample mean of 48
rooms. This information is used to calculate an interval estimate for the population mean to be from 40
to 56 rooms. What is the level of confidence of this interval?
ANS:
81.64%
PTS: 1
REF: SECTION 10.2
122. {Hotel Rooms Rented Narrative} During February, a sample of 25 days has a sample mean of 37
rooms. Use this information to calculate a 92% confidence interval for the population mean.
ANS:
LCL = 28.6 and UCL = 45.4 (rooms)
PTS: 1
REF: SECTION 10.2
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copied, or distributed without the prior consent of the publisher.