Chapter 08Sampling Distributions and Estimation
True / False Questions1.The expected value of an unbiased
estimator is equal to the parameter whose value is being
estimated.TrueFalse
2.All estimators are biased since sampling errors always exist
to some extent.TrueFalse
3.An estimator must be unbiased if you are to use it for
statistical analysis.TrueFalse
4.The efficiency of an estimator depends on the variance of the
estimator's sampling distribution.TrueFalse
5.In comparing estimators, the more efficient estimator will
have a smaller standard error.TrueFalse
6.A 90 percent confidence interval will be wider than a 95
percent confidence interval, ceteris paribus.TrueFalse
7.In constructing a confidence interval for the mean, the z
distribution provides a result nearly identical to the t
distribution when n is large.TrueFalse
8.The Central Limit Theorem says that, if n exceeds 30, the
population will be normal.TrueFalse
9.The Central Limit Theorem says that a histogram of the sample
means will have a bell shape, even if the population is skewed and
the sample is small.TrueFalse
10.The confidence level refers to the procedure used to
construct the confidence interval, rather than to the particular
confidence interval we have constructed.TrueFalse
11.The Central Limit Theorem guarantees an approximately normal
sampling distribution when n is sufficiently large.TrueFalse
12.A sample of size 5 shows a mean of 45.2 and a sample standard
deviation of 6.4. The standard error of the sample mean is
approximately 2.86.TrueFalse
13.As n increases, the width of the confidence interval will
decrease, ceteris paribus.TrueFalse
14.As n increases, the standard error decreases.TrueFalse
15.A higher confidence level leads to a narrower confidence
interval, ceteris paribus.TrueFalse
16.When the sample standard deviation is used to construct a
confidence interval for the mean, we would use the Student's t
distribution instead of the normal distribution.TrueFalse
17.As long as the sample is more than one item, the standard
error of the sample mean will be smaller than the standard
deviation of the population.TrueFalse
18.For a sample size of 20, a 95 percent confidence interval
using the t distribution would be wider than one constructed using
the z distribution.TrueFalse
19.In constructing a confidence interval for a mean, the width
of the interval is dependent on the sample size, the confidence
level, and the population standard deviation.TrueFalse
20.In constructing confidence intervals, it is conservative to
use the z distribution when n 30.TrueFalse
21.The Central Limit Theorem can be applied to the sample
proportion.TrueFalse
22.The distribution of the sample proportion p = x/n is normal
when n 30.TrueFalse
23.The standard deviation of the sample proportion p = x/n
increases as n increases.TrueFalse
24.A 95 percent confidence interval constructed around p will be
wider than a 90 percent confidence interval.TrueFalse
25.The sample proportion is always the midpoint of a confidence
interval for the population proportion.TrueFalse
26.The standard error of the sample proportion is largest when =
.50.TrueFalse
27.The standard error of the sample proportion does not depend
on the confidence level.TrueFalse
28.To narrow the confidence interval for , we can either
increase n or decrease the level of confidence.TrueFalse
29.Ceteris paribus, the narrowest confidence interval for is
achieved when p = .50.TrueFalse
30.The statistic p = x/n may be assumed normally distributed
when np 10 and n(1 - p) 10.TrueFalse
31.The Student's t distribution is always symmetric and
bell-shaped, but its tails lie above the normal.TrueFalse
32.The confidence interval half-width when = .50 is called the
margin of error.TrueFalse
33.Based on the Rule of Three, if no events occur in n
independent trials we can set the upper 95 percent confidence bound
at 3/n.TrueFalse
34.The sample standard deviation s is halfway between the lower
and upper confidence limits for the population (i.e., the
confidence interval is symmetric around s).TrueFalse
35.In a sample size calculation, if the confidence level
decreases, the size of the sample needed will
increase.TrueFalse
36.To calculate the sample size needed for a survey to estimate
a proportion, the population standard deviation must be
known.TrueFalse
37.Assuming that = .50 is a quick and conservative approach to
use in a sample size calculation for a proportion.TrueFalse
38.To estimate the required sample size for a proportion, one
method is to take a small pilot sample to estimate and then apply
the sample size formula.TrueFalse
39.To estimate , you typically need a sample size equal to at
least 5 percent of your population.TrueFalse
40.To estimate a proportion with a 4 percent margin of error and
a 95 percent confidence level, the required sample size is over
800.TrueFalse
41.Approximately 95 percent of the population X values will lie
within the 95 percent confidence interval for the
mean.TrueFalse
42.A 99 percent confidence interval has more confidence but less
precision than a 95 percent confidence interval.TrueFalse
43.Sampling variation is not controllable by the
statistician.TrueFalse
44.The sample mean is not a random variable when the population
parameters are known.TrueFalse
45.The finite population correction factor (FPCF) can be ignored
if n = 7 and N = 700.TrueFalse
46.In constructing a confidence interval, the finite population
correction factor (FPCF) can be ignored if samples of 12 items are
drawn from a population of 300 items.TrueFalse
47.The finite population correction factor (FPCF) can be ignored
when the sample size is large relative to the population
size.TrueFalse
Multiple Choice Questions48.A sampling distribution describes
the distribution of:
A.a parameter.
B.a statistic.
C.either a parameter or a statistic.
D.neither a parameter nor a statistic.
49.As the sample size increases, the standard error of the
mean:
A.increases.
B.decreases.
C.may increase or decrease.
50.Which statement is most nearly correct, other things being
equal?
A.Doubling the sample size will cut the standard error of the
mean in half.
B.The standard error of the mean depends on the population
size.
C.Quadrupling the sample size roughly halves the standard error
of the mean.
D.The standard error of the mean depends on the confidence
level.
51.The width of a confidence interval for is not affected
by:
A.the sample size.
B.the confidence level.
C.the standard deviation.
D.the sample mean.
52.The Central Limit Theorem (CLT) implies that:
A.the population will be approximately normal if n 30.
B.repeated samples must be taken to obtain normality.
C.the distribution of the mean is approximately normal for large
n.
D.the mean follows the same distribution as the population.
53.The owner of Limp Pines Resort wanted to know the average age
of its clients. A random sample of 25 tourists is taken. It shows a
mean age of 46 years with a standard deviation of 5 years. The
width of a 98 percent CI for the true mean client age is
approximately:
A. 1.711 years.
B. 2.326 years.
C. 2.492 years.
D. 2.797 years.
54.In constructing a confidence interval for a mean with unknown
variance with a sample of 25 items, Bob used z instead of t. "Well,
at least my interval will be wider than necessary, so it was a
conservative error," said he. Is Bob's statement correct?
A.Yes.
B.No.
C.It depends on .
55.A random sample of 16 ATM transactions at the Last National
Bank of Flat Rock revealed a mean transaction time of 2.8 minutes
with a standard deviation of 1.2 minutes. The width (in minutes) of
the 95 percent confidence interval for the true mean transaction
time is:
A. 0.639
B. 0.588
C. 0.300
D. 2.131
56.We could narrow a 95 percent confidence interval by:
A.using 99 percent confidence.
B.using a larger sample.
C.raising the standard error.
57.The owner of Torpid Oaks B&B wanted to know the average
distance its guests had traveled. A random sample of 16 guests
showed a mean distance of 85 miles with a standard deviation of 32
miles. The 90 percent confidence interval (in miles) for the mean
is approximately:
A.(71.0, 99.0)
B.(71.8, 98.2)
C.(74.3, 95.7)
D.(68.7, 103.2)
58.A highway inspector needs an estimate of the mean weight of
trucks crossing a bridge on the interstate highway system. She
selects a random sample of 49 trucks and finds a mean of 15.8 tons
with a sample standard deviation of 3.85 tons. The 90 percent
confidence interval for the population mean is:
A.14.72 to 16.88 tons.
B.14.90 to 16.70 tons.
C.14.69 to 16.91 tons.
D.14.88 to 16.72 tons.
59.To determine a 72 percent level of confidence for a
proportion, the value of z is approximately:
A. 1.65
B. 0.77
C. 1.08
D. 1.55
60.To estimate the average annual expenses of students on books
and class materials a sample of size 36 is taken. The sample mean
is $850 and the sample standard deviation is $54. A 99 percent
confidence interval for the population mean is:
A.$823.72 to $876.28
B.$832.36 to $867.64
C.$826.82 to $873.18
D.$825.48 to $874.52
61.In constructing a 95 percent confidence interval, if you
increase n to 4n, the width of your confidence interval will
(assuming other things remain the same) be:
A.about 25 percent of its former width.
B.about two times wider.
C.about 50 percent of its former width.
D.about four times wider.
62.Which of the following is not a characteristic of the t
distribution?
A.It is a continuous distribution.
B.It has a mean of 0.
C.It is a symmetric distribution.
D.It approaches z as degrees of freedom decrease.
63.Which statement is incorrect? Explain.
A.If p = .50 and n = 100, the standard error of the sample
proportion is .05.
B.In a sample size calculation for estimating , it is
conservative to assume = .50.
C.If n = 250 and p = .06, we cannot assume normality in a
confidence interval for .
64.What is the approximate width of a 90 percent confidence
interval for the true population proportion if there are 12
successes in a sample of 25?
A. .196
B. .164
C. .480
D. .206
65.A poll showed that 48 out of 120 randomly chosen graduates of
California medical schools last year intended to specialize in
family practice. What is the width of a 90 percent confidence
interval for the proportion that plan to specialize in family
practice?
A. .0447
B. .0736
C. .0876
D. .0894
66.What is the approximate width of an 80 percent confidence
interval for the true population proportion if there are 12
successes in a sample of 80?
A. .078
B. .066
C. .051
D. .094
67.A random sample of 160 commercial customers of PayMor Lumber
revealed that 32 had paid their accounts within a month of billing.
The 95 percent confidence interval for the true proportion of
customers who pay within a month would be:
A.0.148 to 0.252
B.0.138 to 0.262
C.0.144 to 0.256
D.0.153 to 0.247
68.A random sample of 160 commercial customers of PayMor Lumber
revealed that 32 had paid their accounts within a month of billing.
Can normality be assumed for the sample proportion?
A.Yes.
B.No.
C.Need more information to say.
69.The conservative sample size required for a 95 percent
confidence interval for with an error of 0.04 is:
A.271.
B.423.
C.385.
D.601.
70.Last week, 108 cars received parking violations in the main
university parking lot. Of these, 27 had unpaid parking tickets
from a previous violation. Assuming that last week was a random
sample of all parking violators, find the 95 percent confidence
interval for the percentage of parking violators that have prior
unpaid parking tickets.
A.18.1 to 31.9 percent.
B.16.8 to 33.2 percent.
C.15.3 to 34.7 percent.
D.19.5 to 30.5 percent.
71.In a random sample of 810 women employees, it is found that
81 would prefer working for a female boss. The width of the 95
percent confidence interval for the proportion of women who prefer
a female boss is:
A. .0288
B. .0105
C. .0207
D. .0196
72.Jolly Blue Giant Health Insurance (JBGHI) is concerned about
rising lab test costs and would like to know what proportion of the
positive lab tests for prostate cancer are actually proven correct
through subsequent biopsy. JBGHI demands a sample large enough to
ensure an error of 2 percent with 90 percent confidence. What is
the necessary sample size?
A.4,148
B.2,401
C.1,692
D.1,604
73.A university wants to estimate the average distance that
commuter students travel to get to class with an error of 3 miles
and 90 percent confidence. What sample size would be needed,
assuming that travel distances are normally distributed with a
range of X = 0 to X = 50 miles, using the Empirical Rule 3 to
estimate .
A.About 28 students
B.About 47 students
C.About 30 students
D.About 21 students
74.A financial institution wishes to estimate the mean balances
owed by its credit card customers. The population standard
deviation is $300. If a 99 percent confidence interval is used and
an interval of $75 is desired, how many cardholders should be
sampled?
A.3382
B.629
C.87
D.107
75.A company wants to estimate the time its trucks take to drive
from city A to city B. The standard deviation is known to be 12
minutes. What sample size is required in order that error will not
exceed 2 minutes, with 95 percent confidence?
A.12 observations
B.139 observations
C.36 observations
D.129 observations
76.In a large lecture class, the professor announced that the
scores on a recent exam were normally distributed with a range from
51 to 87. Using the Empirical Rule 3 to estimate , how many
students would you need to sample to estimate the true mean score
for the class with 90 percent confidence and an error of 2?
A.About 17 students
B.About 35 students
C.About 188 students
D.About 25 students
77.Using the conventional polling definition, find the margin of
error for a customer satisfaction survey of 225 customers who have
recently dined at Applebee's.
A. 5.0 percent
B. 4.2 percent
C. 7.1 percent
D. 6.5 percent
78.A marketing firm is asked to estimate the percent of existing
customers who would purchase a "digital upgrade" to their basic
cable TV service. The firm wants 99 percent confidence and an error
of 5 percent. What is the required sample size (to the next higher
integer)?
A.664
B.625
C.801
D.957
79.An airport traffic analyst wants to estimate the proportion
of daily takeoffs by small business jets (as opposed to commercial
passenger jets or other aircraft) with an error of 4 percent with
90 percent confidence. What sample size should the analyst use?
A.385
B.601
C.410
D.423
80.Ersatz Beneficial Insurance wants to estimate the cost of
damage to cars due to accidents. The standard deviation of the cost
is known to be $200. They want to estimate the mean cost using a 95
percent confidence interval within $10. What is the minimum sample
size n?
A.1083
B.4002
C.1537
D.2301
81.Professor York randomly surveyed 240 students at Oxnard
University and found that 150 of the students surveyed watch more
than 10 hours of television weekly. Develop a 95 percent confidence
interval to estimate the true proportion of students who watch more
than 10 hours of television each week. The confidence interval
is:
A..533 to .717
B..564 to .686
C..552 to .698
D..551 to .739
82.Professor York randomly surveyed 240 students at Oxnard
University and found that 150 of the students surveyed watch more
than 10 hours of television weekly. How many additional students
would Professor York have to sample to estimate the proportion of
all Oxnard University students who watch more than 10 hours of
television each week within 3 percent with 99 percent
confidence?
A.761
B.1001
C.1489
D.1728
83.The sample proportion is in the middle of the confidence
interval for the population proportion:
A.in any sample.
B.only if the samples are large.
C.only if is not too far from .50.
84.For a sample of size 16, the critical values of chi-square
for a 95 percent confidence interval for the population variance
are:
A.6.262, 27.49
B.6.908, 28.85
C.5.629, 26.12
D.7.261, 25.00
85.For a sample of size 11, the critical values of chi-square
for a 90 percent confidence interval for the population variance
are:
A.6.262, 27.49
B.6.908, 28.85
C.3.940, 18.31
D.3.247, 20.48
86.For a sample of size 18, the critical values of chi-square
for a 99 percent confidence interval for the population variance
are:
A.6.262, 27.49
B.5.697, 35.72
C.5.629, 26.12
D.7.261, 25.00
87.Which of the following statements is most nearly correct,
other things being equal?
A.Using Student's t instead of z makes a confidence interval
narrower.
B.The table values of z and t are about the same when the mean
is large.
C.For a given confidence level, the z value is always smaller
then the t value.
D.Student's t is rarely used because it is more conservative to
use z.
88.The Central Limit Theorem (CLT):
A.applies only to samples from normal populations.
B.applies to any population.
C.applies best to populations that are skewed.
D.applies only when and are known.
89.In which situation may the sample proportion safely be
assumed to follow a normal distribution?
A.12 successes in a sample of 72 items
B.8 successes in a sample of 40 items
C.6 successes in a sample of 200 items
D.4 successes in a sample of 500 items
90.In which situation may the sample proportion safely be
assumed to follow a normal distribution?
A.n = 100, = .06
B.n = 250, = .02
C.n = 30, = .50
D.n = 500, = .01
91.If = 12, find the sample size to estimate the mean with an
error of 4 and 95 percent confidence (rounded to the next higher
integer).
A.75
B.35
C.58
D.113
92.If = 25, find the sample size to estimate the mean with an
error of 3 and 90 percent confidence (rounded to the next higher
integer).
A.426
B.512
C.267
D.188
93.Sampling error can be avoided:
A.by using an unbiased estimator.
B.by eliminating nonresponses (e.g., older people).
C.by no method under the statistician's control.
D.either by using an unbiased estimator or by eliminating
nonresponse.
94.A consistent estimator for the mean:
A.converges on the true parameter as the variance increases.
B.converges on the true parameter as the sample size
increases.
C.consistently follows a normal distribution.
D.is impossible to obtain using real sample data.
95.Concerning confidence intervals, which statement is most
nearly correct?
A.We should use z instead of t when n is large.
B.We use the Student's t distribution when is unknown.
C.We use the Student's t distribution to narrow the confidence
interval.
96.The standard error of the mean decreases when:
A.the sample size decreases.
B.the standard deviation increases.
C.the standard deviation decreases or n increases.
D.the population size decreases.
97.For a given sample size, the higher the confidence level,
the:
A.more accurate the point estimate.
B.smaller the standard error.
C.smaller the interval width.
D.greater the interval width.
98.A sample is taken and a confidence interval is constructed
for the mean of the distribution. At the center of the interval is
always which value?
A.The sample mean
B.The population mean
C.Neither nor since with a sample anything can happen
D.Both and as long as there are not too many outliers
99.If a normal population has parameters = 40 and = 8, then for
a sample size n = 4:
A.the standard error of the sample mean is approximately 2.
B.the standard error of the sample mean is approximately 4.
C.the standard error of the sample mean is approximately 8.
D.the standard error of the sample mean is approximately 10.
Short Answer Questions100.On the basis of a survey of 545
television viewers, a statistician has constructed a confidence
interval and estimated that the proportion of people who watched
the season premiere of Glee is between .16 and .24. What level of
confidence did the statistician use in constructing this interval?
Explain carefully, showing all steps in your reasoning.
101.Read the news story below. Using the 95 percent confidence
level, what sample size would be needed to estimate the true
proportion of stores selling cigarettes to minors with an error of
3 percent? Explain carefully, showing all steps in your
reasoning.
102.In a survey, 858 out of 2600 homeowners said they expected
good economic conditions to continue for the next 12 months.
Construct a 95 percent confidence interval for "good times" in the
next 12 months.
103.Fulsome University has 16,059 students. In a sample of 200
students, 12 were born outside the United States. Construct a 95
percent confidence interval for the true population proportion. How
large a sample is needed to estimate the true proportion of Fulsome
students who were born outside the United States with an error of
2.5 percent and 95 percent confidence? Show your work and explain
fully.
104.List differences and similarities between Student's t and
the standard normal distribution.
105.Why does pose a problem for sample size calculation for a
mean? How can be approximated when it is unknown?
Chapter 08 Sampling Distributions and Estimation Answer Key
True / False Questions1.The expected value of an unbiased
estimator is equal to the parameter whose value is being
estimated.TRUEAn unbiased estimator's expected value is the true
parameter value.
AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning
Objective: 08-02 Explain the desirable properties of
estimators.Topic: Estimators and Sampling Error
2.All estimators are biased since sampling errors always exist
to some extent.FALSESome estimators are systematically biased,
regardless of sampling error.
AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning
Objective: 08-02 Explain the desirable properties of
estimators.Topic: Estimators and Sampling Error
3.An estimator must be unbiased if you are to use it for
statistical analysis.FALSEAn estimator can be useful as long as its
bias is known.
AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning
Objective: 08-02 Explain the desirable properties of
estimators.Topic: Estimators and Sampling Error
4.The efficiency of an estimator depends on the variance of the
estimator's sampling distribution.TRUEEfficiency is measured by the
variance of the estimator's sampling distribution.
AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning
Objective: 08-02 Explain the desirable properties of
estimators.Topic: Estimators and Sampling Error
5.In comparing estimators, the more efficient estimator will
have a smaller standard error.TRUEEfficiency is measured by the
variance of the estimator's sampling distribution.
AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning
Objective: 08-02 Explain the desirable properties of
estimators.Topic: Estimators and Sampling Error
6.A 90 percent confidence interval will be wider than a 95
percent confidence interval, ceteris paribus.FALSEWe can make a
more precise statement about the true parameter if we are willing
to sacrifice some confidence. For example, z.025 = 1.960 (for 95
percent confidence) gives a wider interval than z.05 = 1.645 (for
90 percent confidence). The proffered statement would also hold
true for the Student's t distribution.
AACSB: AnalyticBlooms: UnderstandDifficulty: 1 EasyLearning
Objective: 08-05 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Mean () with
Known
7.In constructing a confidence interval for the mean, the z
distribution provides a result nearly identical to the t
distribution when n is large.TRUEStudent's t approaches z as sample
size increases.
AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning
Objective: 08-06 Know when to use Student's t instead of z to
estimate .Topic: Confidence Interval for a Mean () with Unknown
8.The Central Limit Theorem says that, if n exceeds 30, the
population will be normal.FALSEThe population cannot be
changed.
AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning
Objective: 08-03 State the Central Limit Theorem for a mean.Topic:
Sample Mean and the Central Limit Theorem
9.The Central Limit Theorem says that a histogram of the sample
means will have a bell shape, even if the population is skewed and
the sample is small.FALSEA large sample size may be required if the
population is skewed.
AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning
Objective: 08-03 State the Central Limit Theorem for a mean.Topic:
Sample Mean and the Central Limit Theorem
10.The confidence level refers to the procedure used to
construct the confidence interval, rather than to the particular
confidence interval we have constructed.TRUEA particular interval
either does or does not contain the true parameter.
AACSB: AnalyticBlooms: UnderstandDifficulty: 2 MediumLearning
Objective: 08-05 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Mean () with
Known
11.The Central Limit Theorem guarantees an approximately normal
sampling distribution when n is sufficiently large.TRUEYes,
although a large sample size may be required if the population is
skewed.
AACSB: AnalyticBlooms: UnderstandDifficulty: 2 MediumLearning
Objective: 08-03 State the Central Limit Theorem for a mean.Topic:
Sample Mean and the Central Limit Theorem
12.A sample of size 5 shows a mean of 45.2 and a sample standard
deviation of 6.4. The standard error of the sample mean is
approximately 2.86.TRUEThe standard error is the standard deviation
divided by the square root of the sample size.
AACSB: AnalyticBlooms: ApplyDifficulty: 1 EasyLearning
Objective: 08-04 Explain how sample size affects the standard
error.Topic: Confidence Interval for a Mean () with Unknown
13.As n increases, the width of the confidence interval will
decrease, ceteris paribus.TRUEThe standard error is the standard
deviation divided by the square root of the sample size.
AACSB: AnalyticBlooms: UnderstandDifficulty: 1 EasyLearning
Objective: 08-05 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Mean () with
Known
14.As n increases, the standard error decreases.TRUEThe standard
error is the standard deviation divided by the square root of the
sample size.
AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning
Objective: 08-04 Explain how sample size affects the standard
error.Topic: Sample Mean and the Central Limit Theorem
15.A higher confidence level leads to a narrower confidence
interval, ceteris paribus.FALSEHigher confidence requires more
uncertainty (a wider interval). For example, z.025 = 1.960 (for 95
percent confidence) gives a wider interval than z.05 = 1.645 (for
90 percent confidence). The proffered statement would also hold
true for the Student's t distribution.
AACSB: AnalyticBlooms: UnderstandDifficulty: 2 MediumLearning
Objective: 08-05 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Mean () with
Known
16.When the sample standard deviation is used to construct a
confidence interval for the mean, we would use the Student's t
distribution instead of the normal distribution.TRUEWe should use t
when the population variance is unknown.
AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning
Objective: 08-06 Know when to use Student's t instead of z to
estimate .Topic: Confidence Interval for a Mean () with Unknown
17.As long as the sample is more than one item, the standard
error of the sample mean will be smaller than the standard
deviation of the population.TRUEThe standard error is the standard
deviation divided by the square root of the sample size.
AACSB: AnalyticBlooms: UnderstandDifficulty: 1 EasyLearning
Objective: 08-04 Explain how sample size affects the standard
error.Topic: Sample Mean and the Central Limit Theorem
18.For a sample size of 20, a 95 percent confidence interval
using the t distribution would be wider than one constructed using
the z distribution.TRUEStudent's t is always larger than z for the
same level of confidence.
AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning
Objective: 08-06 Know when to use Student's t instead of z to
estimate .Topic: Confidence Interval for a Mean () with Unknown
19.In constructing a confidence interval for a mean, the width
of the interval is dependent on the sample size, the confidence
level, and the population standard deviation.TRUEThe confidence
interval depends on all of these.
AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning
Objective: 08-05 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Mean () with
Known
20.In constructing confidence intervals, it is conservative to
use the z distribution when n 30.FALSEWhile t and z may be similar
for large samples, it is more conservative to use t.
AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning
Objective: 08-06 Know when to use Student's t instead of z to
estimate .Topic: Confidence Interval for a Mean () with Unknown
21.The Central Limit Theorem can be applied to the sample
proportion.TRUEWe are sampling a Bernoulli population, but the CLT
still applies.
AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning
Objective: 08-07 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Proportion ()
22.The distribution of the sample proportion p = x/n is normal
when n 30.FALSEWe want at least 10 successes and 10 failures to
assume that p is normally distributed.
AACSB: AnalyticBlooms: UnderstandDifficulty: 1 EasyLearning
Objective: 08-07 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Proportion ()
23.The standard deviation of the sample proportion p = x/n
increases as n increases.FALSEThe proffered statement is backwards
because n is in the denominator of [p(1 - p)/n]1/2.
AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning
Objective: 08-07 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Proportion ()
24.A 95 percent confidence interval constructed around p will be
wider than a 90 percent confidence interval.TRUEHigher confidence
requires more uncertainty (a wider interval).
AACSB: AnalyticBlooms: UnderstandDifficulty: 1 EasyLearning
Objective: 08-07 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Proportion ()
25.The sample proportion is always the midpoint of a confidence
interval for the population proportion.TRUEThe interval is p z[p(1
- p)/n]1/2.
AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning
Objective: 08-07 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Proportion ()
26.The standard error of the sample proportion is largest when =
.50.TRUEThe value of [(1 - )/n]1/2 is smaller for any value less
than = .50.
AACSB: AnalyticBlooms: UnderstandDifficulty: 2 MediumLearning
Objective: 08-07 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Proportion ()
27.The standard error of the sample proportion does not depend
on the confidence level.TRUEThe standard error of p is [(1 -
)/n]1/2.
AACSB: AnalyticBlooms: UnderstandDifficulty: 2 MediumLearning
Objective: 08-07 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Proportion ()
28.To narrow the confidence interval for , we can either
increase n or decrease the level of confidence.TRUEThe interval is
p z[p(1 - p)/n]1/2.
AACSB: AnalyticBlooms: UnderstandDifficulty: 2 MediumLearning
Objective: 08-07 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Proportion ()
29.Ceteris paribus, the narrowest confidence interval for is
achieved when p = .50.FALSEThe value of [p(1 - p)/n]1/2 is smaller
for any value less than = .50.
AACSB: AnalyticBlooms: UnderstandDifficulty: 2 MediumLearning
Objective: 08-07 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Proportion ()
30.The statistic p = x/n may be assumed normally distributed
when np 10 and n(1 - p) 10.TRUEWe want at least 10 successes and 10
failures in the sample to assume normality of p.
AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning
Objective: 08-07 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Proportion ()
31.The Student's t distribution is always symmetric and
bell-shaped, but its tails lie above the normal.TRUEStudent's t
resembles a normal, but its PDF is above the normal PDF in the
tails.
AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning
Objective: 08-06 Know when to use Student's t instead of z to
estimate .Topic: Confidence Interval for a Mean () with Unknown
32.The confidence interval half-width when = .50 is called the
margin of error.TRUEPollsters use this definition.
AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning
Objective: 08-07 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Proportion ()
33.Based on the Rule of Three, if no events occur in n
independent trials we can set the upper 95 percent confidence bound
at 3/n.TRUEWe need a special rule because when p = 0 we can't apply
the usual formula p z[p(1 - p)/n]1/2.
AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning
Objective: 08-07 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Proportion ()
34.The sample standard deviation s is halfway between the lower
and upper confidence limits for the population (i.e., the
confidence interval is symmetric around s).FALSEThe chi-square
distribution is not symmetric.
AACSB: AnalyticBlooms: ApplyDifficulty: 3 HardLearning
Objective: 08-10 Construct a confidence interval for a variance
(optional).Topic: Confidence Interval for a Population Variance, 2
(Optional)
35.In a sample size calculation, if the confidence level
decreases, the size of the sample needed will increase.FALSEReduced
confidence allows a smaller sample.
AACSB: AnalyticBlooms: UnderstandDifficulty: 2 MediumLearning
Objective: 08-09 Calculate sample size to estimate a mean or
proportion.Topic: Sample Size Determination for a Mean
36.To calculate the sample size needed for a survey to estimate
a proportion, the population standard deviation must be
known.FALSEFor a proportion, the sample size formula requires not
.
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-09 Calculate sample size to estimate a mean or
proportion.Topic: Sample Size Determination for a Proportion
37.Assuming that = .50 is a quick and conservative approach to
use in a sample size calculation for a proportion.TRUEAssuming that
= .50 is quick and safe (but may give a larger sample than is
needed).
AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning
Objective: 08-09 Calculate sample size to estimate a mean or
proportion.Topic: Sample Size Determination for a Proportion
38.To estimate the required sample size for a proportion, one
method is to take a small pilot sample to estimate and then apply
the sample size formula.TRUEThis is a common method, but assuming
that = .50 is quicker and safer.
AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning
Objective: 08-09 Calculate sample size to estimate a mean or
proportion.Topic: Sample Size Determination for a Proportion
39.To estimate , you typically need a sample size equal to at
least 5 percent of your population.FALSEThe sample size n bears no
necessary relation to N.
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-09 Calculate sample size to estimate a mean or
proportion.Topic: Sample Size Determination for a Proportion
40.To estimate a proportion with a 4 percent margin of error and
a 95 percent confidence level, the required sample size is over
800.FALSEn = (z/E)2()(1 - ) = (1.96/.04)2(.50)(1 - .50) =
600.25.
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-09 Calculate sample size to estimate a mean or
proportion.Topic: Sample Size Determination for a Proportion
41.Approximately 95 percent of the population X values will lie
within the 95 percent confidence interval for the mean.FALSEThe
confidence interval is for the true mean, not for individual X
values.
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-05 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Mean () with
Known
42.A 99 percent confidence interval has more confidence but less
precision than a 95 percent confidence interval.TRUEThe higher
confidence level widens the interval so it is less precise.
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-05 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Mean () with
Known
43.Sampling variation is not controllable by the
statistician.TRUESampling variation is inevitable.
AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning
Objective: 08-01 Define sampling error; parameter; and
estimator.Topic: Sampling Variation
44.The sample mean is not a random variable when the population
parameters are known.FALSEThe sample mean is a random variable
regardless of what we know about the population.
AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning
Objective: 08-01 Define sampling error; parameter; and
estimator.Topic: Sampling Variation
45.The finite population correction factor (FPCF) can be ignored
if n = 7 and N = 700.TRUEThe FPCF has a negligible effect when the
sample is less than 5 percent of the population.
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-08 Construct confidence intervals for finite
populations.Topic: Estimating from Finite Populations
46.In constructing a confidence interval, the finite population
correction factor (FPCF) can be ignored if samples of 12 items are
drawn from a population of 300 items.TRUEThe FPCF has a negligible
effect when the sample is less than 5 percent of the
population.
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-08 Construct confidence intervals for finite
populations.Topic: Estimating from Finite Populations
47.The finite population correction factor (FPCF) can be ignored
when the sample size is large relative to the population
size.TRUEThe FPCF has a negligible effect when n is small relative
to N.
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-08 Construct confidence intervals for finite
populations.Topic: Estimating from Finite Populations
Multiple Choice Questions48.A sampling distribution describes
the distribution of:
A.a parameter.
B.a statistic.
C.either a parameter or a statistic.
D.neither a parameter nor a statistic.
A statistic has a sampling distribution.
AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning
Objective: 08-03 State the Central Limit Theorem for a mean.Topic:
Sample Mean and the Central Limit Theorem
49.As the sample size increases, the standard error of the
mean:
A.increases.
B.decreases.
C.may increase or decrease.
The standard error of the mean is /(n)1/2.
AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning
Objective: 08-03 State the Central Limit Theorem for a mean.Topic:
Sample Mean and the Central Limit Theorem
50.Which statement is most nearly correct, other things being
equal?
A.Doubling the sample size will cut the standard error of the
mean in half.
B.The standard error of the mean depends on the population
size.
C.Quadrupling the sample size roughly halves the standard error
of the mean.
D.The standard error of the mean depends on the confidence
level.
The standard error of the mean is /(n)1/2 so replacing n by 4n
would cut the SEM in half.
AACSB: AnalyticBlooms: UnderstandDifficulty: 2 MediumLearning
Objective: 08-04 Explain how sample size affects the standard
error.Topic: Sample Mean and the Central Limit Theorem
51.The width of a confidence interval for is not affected
by:
A.the sample size.
B.the confidence level.
C.the standard deviation.
D.the sample mean.
The mean is not used in calculating the width of the confidence
interval z/(n)1/2.
AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning
Objective: 08-05 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Mean () with
Known
52.The Central Limit Theorem (CLT) implies that:
A.the population will be approximately normal if n 30.
B.repeated samples must be taken to obtain normality.
C.the distribution of the mean is approximately normal for large
n.
D.the mean follows the same distribution as the population.
The sampling distribution of the mean is asymptotically normal
for any population.
AACSB: AnalyticBlooms: UnderstandDifficulty: 2 MediumLearning
Objective: 08-03 State the Central Limit Theorem for a mean.Topic:
Sample Mean and the Central Limit Theorem
53.The owner of Limp Pines Resort wanted to know the average age
of its clients. A random sample of 25 tourists is taken. It shows a
mean age of 46 years with a standard deviation of 5 years. The
width of a 98 percent CI for the true mean client age is
approximately:
A. 1.711 years.
B. 2.326 years.
C. 2.492 years.
D. 2.797 years.
The width is ts/(n)1/2 = (2.492)(5)/(25)1/2 = 2.492.
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-06 Know when to use Student's t instead of z to
estimate .Topic: Confidence Interval for a Mean () with Unknown
54.In constructing a confidence interval for a mean with unknown
variance with a sample of 25 items, Bob used z instead of t. "Well,
at least my interval will be wider than necessary, so it was a
conservative error," said he. Is Bob's statement correct?
A.Yes.
B.No.
C.It depends on .
z is always smaller than t (ceteris paribus) so the interval
would be narrower than is justified.
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-06 Know when to use Student's t instead of z to
estimate .Topic: Confidence Interval for a Mean () with Unknown
55.A random sample of 16 ATM transactions at the Last National
Bank of Flat Rock revealed a mean transaction time of 2.8 minutes
with a standard deviation of 1.2 minutes. The width (in minutes) of
the 95 percent confidence interval for the true mean transaction
time is:
A. 0.639
B. 0.588
C. 0.300
D. 2.131
The width is ts/(n)1/2 = (2.131)(1.2)/(16)1/2 = 0.639.
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-06 Know when to use Student's t instead of z to
estimate .Topic: Confidence Interval for a Mean () with Unknown
56.We could narrow a 95 percent confidence interval by:
A.using 99 percent confidence.
B.using a larger sample.
C.raising the standard error.
A larger sample would narrow the interval width z/(n)1/2.
AACSB: AnalyticBlooms: UnderstandDifficulty: 1 EasyLearning
Objective: 08-05 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Mean () with
Known
57.The owner of Torpid Oaks B&B wanted to know the average
distance its guests had traveled. A random sample of 16 guests
showed a mean distance of 85 miles with a standard deviation of 32
miles. The 90 percent confidence interval (in miles) for the mean
is approximately:
A.(71.0, 99.0)
B.(71.8, 98.2)
C.(74.3, 95.7)
D.(68.7, 103.2)
The interval is 85 ts/(n)1/2 or 85 (1.753)(32)/(16)1/2 with d.f
= 15 (don't use z).
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-06 Know when to use Student's t instead of z to
estimate .Topic: Confidence Interval for a Mean () with Unknown
58.A highway inspector needs an estimate of the mean weight of
trucks crossing a bridge on the interstate highway system. She
selects a random sample of 49 trucks and finds a mean of 15.8 tons
with a sample standard deviation of 3.85 tons. The 90 percent
confidence interval for the population mean is:
A.14.72 to 16.88 tons.
B.14.90 to 16.70 tons.
C.14.69 to 16.91 tons.
D.14.88 to 16.72 tons.
The interval is 15.8 ts/(n)1/2 or 15.8 (1.677)(3.85)/(49)1/2
using d.f. = 48 (don't use z).
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-06 Know when to use Student's t instead of z to
estimate .Topic: Confidence Interval for a Mean () with Unknown
59.To determine a 72 percent level of confidence for a
proportion, the value of z is approximately:
A. 1.65
B. 0.77
C. 1.08
D. 1.55
Look up the z value that puts 14 percent in each tail.
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-07 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Proportion ()
60.To estimate the average annual expenses of students on books
and class materials a sample of size 36 is taken. The sample mean
is $850 and the sample standard deviation is $54. A 99 percent
confidence interval for the population mean is:
A.$823.72 to $876.28
B.$832.36 to $867.64
C.$826.82 to $873.18
D.$825.48 to $874.52
The interval is 850 ts/(n)1/2 or 850 (2.724)(54)/(36)1/2 with
d.f = 35 (don't use z).
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-05 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Mean () with
Unknown
61.In constructing a 95 percent confidence interval, if you
increase n to 4n, the width of your confidence interval will
(assuming other things remain the same) be:
A.about 25 percent of its former width.
B.about two times wider.
C.about 50 percent of its former width.
D.about four times wider.
The standard error of the mean is /(n)1/2 so replacing n by 4n
would cut the SEM in half.
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-04 Explain how sample size affects the standard
error.Topic: Confidence Interval for a Mean () with Known
62.Which of the following is not a characteristic of the t
distribution?
A.It is a continuous distribution.
B.It has a mean of 0.
C.It is a symmetric distribution.
D.It approaches z as degrees of freedom decrease.
It approaches z as degrees of freedom increase.
AACSB: AnalyticBlooms: ApplyDifficulty: 1 EasyLearning
Objective: 08-06 Know when to use Student's t instead of z to
estimate .Topic: Confidence Interval for a Mean () with Unknown
63.Which statement is incorrect? Explain.
A.If p = .50 and n = 100, the standard error of the sample
proportion is .05.
B.In a sample size calculation for estimating , it is
conservative to assume = .50.
C.If n = 250 and p = .06, we cannot assume normality in a
confidence interval for .
Normality of p may be assumed because np = 15 and n(1 - p) =
235.
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-07 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Proportion ()
64.What is the approximate width of a 90 percent confidence
interval for the true population proportion if there are 12
successes in a sample of 25?
A. .196
B. .164
C. .480
D. .206
The interval width is z[p(1 - p)/n]1/2 =
(1.645)[(.48)(.52)/25]1/2.
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-07 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Proportion ()
65.A poll showed that 48 out of 120 randomly chosen graduates of
California medical schools last year intended to specialize in
family practice. What is the width of a 90 percent confidence
interval for the proportion that plan to specialize in family
practice?
A. .0447
B. .0736
C. .0876
D. .0894
The interval width is z[p(1 - p)/n]1/2 =
(1.645)[(.40)(.60)/120]1/2.
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-07 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Proportion ()
66.What is the approximate width of an 80 percent confidence
interval for the true population proportion if there are 12
successes in a sample of 80?
A. .078
B. .066
C. .051
D. .094
The interval width is z[p(1 - p)/n]1/2 =
(1.282)[(.15)(.85)/80]1/2.
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-07 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Proportion ()
67.A random sample of 160 commercial customers of PayMor Lumber
revealed that 32 had paid their accounts within a month of billing.
The 95 percent confidence interval for the true proportion of
customers who pay within a month would be:
A.0.148 to 0.252
B.0.138 to 0.262
C.0.144 to 0.256
D.0.153 to 0.247
The interval is p z[p(1 - p)/n]1/2 = .20
(1.960)[(.20)(.80)/160]1/2.
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-07 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Proportion ()
68.A random sample of 160 commercial customers of PayMor Lumber
revealed that 32 had paid their accounts within a month of billing.
Can normality be assumed for the sample proportion?
A.Yes.
B.No.
C.Need more information to say.
Yes, because there were at least 10 "successes" and at least 10
"failures" in the sample.
AACSB: AnalyticBlooms: ApplyDifficulty: 1 EasyLearning
Objective: 08-07 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Proportion ()
69.The conservative sample size required for a 95 percent
confidence interval for with an error of 0.04 is:
A.271.
B.423.
C.385.
D.601.
n = (z/E)2()(1 - ) = (1.96/.04)2(.50)(1 - .50) = 600.25 (round
up).
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-09 Calculate sample size to estimate a mean or
proportion.Topic: Sample Size Determination for a Proportion
70.Last week, 108 cars received parking violations in the main
university parking lot. Of these, 27 had unpaid parking tickets
from a previous violation. Assuming that last week was a random
sample of all parking violators, find the 95 percent confidence
interval for the percentage of parking violators that have prior
unpaid parking tickets.
A.18.1 to 31.9 percent.
B.16.8 to 33.2 percent.
C.15.3 to 34.7 percent.
D.19.5 to 30.5 percent.
The interval is p z[p(1 - p)/n]1/2 = .25
(1.960)[(.25)(.75)/108]1/2.
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-07 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Proportion ()
71.In a random sample of 810 women employees, it is found that
81 would prefer working for a female boss. The width of the 95
percent confidence interval for the proportion of women who prefer
a female boss is:
A. .0288
B. .0105
C. .0207
D. .0196
The width is z[p(1 - p)/n]1/2 or (1.960)[(.10)(.90)/810]1/2 or
.0207.
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-07 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Proportion ()
72.Jolly Blue Giant Health Insurance (JBGHI) is concerned about
rising lab test costs and would like to know what proportion of the
positive lab tests for prostate cancer are actually proven correct
through subsequent biopsy. JBGHI demands a sample large enough to
ensure an error of 2 percent with 90 percent confidence. What is
the necessary sample size?
A.4,148
B.2,401
C.1,692
D.1,604
n = (z/E)2()(1 - ) = (1.645/.02)2(.50)(1 - .50) = 1691.3 (round
up).
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-09 Calculate sample size to estimate a mean or
proportion.Topic: Sample Size Determination for a Proportion
73.A university wants to estimate the average distance that
commuter students travel to get to class with an error of 3 miles
and 90 percent confidence. What sample size would be needed,
assuming that travel distances are normally distributed with a
range of X = 0 to X = 50 miles, using the Empirical Rule 3 to
estimate .
A.About 28 students
B.About 47 students
C.About 30 students
D.About 21 students
Using = (50 - 0)/6 = 8.333, we get n = [z/E]2 =
[(1.645)(8.333)/3]2 = 20.9 (round up).
AACSB: AnalyticBlooms: ApplyDifficulty: 3 HardLearning
Objective: 08-09 Calculate sample size to estimate a mean or
proportion.Topic: Sample Size Determination for a Mean
74.A financial institution wishes to estimate the mean balances
owed by its credit card customers. The population standard
deviation is $300. If a 99 percent confidence interval is used and
an interval of $75 is desired, how many cardholders should be
sampled?
A.3382
B.629
C.87
D.107
n = [z/E]2 = [(2.576)(300)/75]2 = 106.2 (round up).
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-09 Calculate sample size to estimate a mean or
proportion.Topic: Sample Size Determination for a Mean
75.A company wants to estimate the time its trucks take to drive
from city A to city B. The standard deviation is known to be 12
minutes. What sample size is required in order that error will not
exceed 2 minutes, with 95 percent confidence?
A.12 observations
B.139 observations
C.36 observations
D.129 observations
n = [z/E]2 = [(1.960)(12)/2]2 = 138.3 (round up).
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-09 Calculate sample size to estimate a mean or
proportion.Topic: Sample Size Determination for a Mean
76.In a large lecture class, the professor announced that the
scores on a recent exam were normally distributed with a range from
51 to 87. Using the Empirical Rule 3 to estimate , how many
students would you need to sample to estimate the true mean score
for the class with 90 percent confidence and an error of 2?
A.About 17 students
B.About 35 students
C.About 188 students
D.About 25 students
Using = (87 - 51)/6 = 6, we get n = [z/E]2 = [(1.645)(6)/2]2 =
24.35 (round up).
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-09 Calculate sample size to estimate a mean or
proportion.Topic: Sample Size Determination for a Mean
77.Using the conventional polling definition, find the margin of
error for a customer satisfaction survey of 225 customers who have
recently dined at Applebee's.
A. 5.0 percent
B. 4.2 percent
C. 7.1 percent
D. 6.5 percent
The margin of error is z[(1 - )/n]1/2 or
(1.960)[(.50)(.50)/225]1/2 or .065.
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-09 Calculate sample size to estimate a mean or
proportion.Topic: Sample Size Determination for a Proportion
78.A marketing firm is asked to estimate the percent of existing
customers who would purchase a "digital upgrade" to their basic
cable TV service. The firm wants 99 percent confidence and an error
of 5 percent. What is the required sample size (to the next higher
integer)?
A.664
B.625
C.801
D.957
n = (z/E)2()(1 - ) = (2.576/.05)2(.50)(1 - .50) = 663.6 (round
up).
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-09 Calculate sample size to estimate a mean or
proportion.Topic: Sample Size Determination for a Proportion
79.An airport traffic analyst wants to estimate the proportion
of daily takeoffs by small business jets (as opposed to commercial
passenger jets or other aircraft) with an error of 4 percent with
90 percent confidence. What sample size should the analyst use?
A.385
B.601
C.410
D.423
n = (z/E)2()(1 - ) = (1.645/.04)2(.50)(1 - .50) = 422.8 (round
up).
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-09 Calculate sample size to estimate a mean or
proportion.Topic: Sample Size Determination for a Proportion
80.Ersatz Beneficial Insurance wants to estimate the cost of
damage to cars due to accidents. The standard deviation of the cost
is known to be $200. They want to estimate the mean cost using a 95
percent confidence interval within $10. What is the minimum sample
size n?
A.1083
B.4002
C.1537
D.2301
n = [z/E]2 = [(1.960)(200)/10]2 = 1536.6 (round up).
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-09 Calculate sample size to estimate a mean or
proportion.Topic: Sample Size Determination for a Mean
81.Professor York randomly surveyed 240 students at Oxnard
University and found that 150 of the students surveyed watch more
than 10 hours of television weekly. Develop a 95 percent confidence
interval to estimate the true proportion of students who watch more
than 10 hours of television each week. The confidence interval
is:
A..533 to .717
B..564 to .686
C..552 to .698
D..551 to .739
The interval is p z[p(1 - p)/n]1/2 = .625
(1.960)[(.625)(.375)/240]1/2.
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-07 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Proportion ()
82.Professor York randomly surveyed 240 students at Oxnard
University and found that 150 of the students surveyed watch more
than 10 hours of television weekly. How many additional students
would Professor York have to sample to estimate the proportion of
all Oxnard University students who watch more than 10 hours of
television each week within 3 percent with 99 percent
confidence?
A.761
B.1001
C.1489
D.1728
Using p = .625 we get n = (z/E)2()(1 - ) =
(2.576/.03)2(.625)(.375) = 1728.06 (round up).
AACSB: AnalyticBlooms: ApplyDifficulty: 3 HardLearning
Objective: 08-09 Calculate sample size to estimate a mean or
proportion.Topic: Sample Size Determination for a Proportion
83.The sample proportion is in the middle of the confidence
interval for the population proportion:
A.in any sample.
B.only if the samples are large.
C.only if is not too far from .50.
The interval is p z[p(1 - p)/n]1/2.
AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning
Objective: 08-07 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Proportion ()
84.For a sample of size 16, the critical values of chi-square
for a 95 percent confidence interval for the population variance
are:
A.6.262, 27.49
B.6.908, 28.85
C.5.629, 26.12
D.7.261, 25.00
Using d.f. = n - 1 = 15, we get 2L = 6.262 and 2U = 27.49 from
Appendix E.
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-10 Construct a confidence interval for a variance
(optional).Topic: Confidence Interval for a Population Variance, 2
(Optional)
85.For a sample of size 11, the critical values of chi-square
for a 90 percent confidence interval for the population variance
are:
A.6.262, 27.49
B.6.908, 28.85
C.3.940, 18.31
D.3.247, 20.48
d.f. = n - 1 = 10, we get 2L = 3.940 and 2U = 18.31 from
Appendix E.
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-10 Construct a confidence interval for a variance
(optional).Topic: Confidence Interval for a Population Variance, 2
(Optional)
86.For a sample of size 18, the critical values of chi-square
for a 99 percent confidence interval for the population variance
are:
A.6.262, 27.49
B.5.697, 35.72
C.5.629, 26.12
D.7.261, 25.00
d.f. = n - 1 = 17, we get 2L = 5.697 and 2U = 35.72 from
Appendix E.
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-10 Construct a confidence interval for a variance
(optional).Topic: Confidence Interval for a Population Variance, 2
(Optional)
87.Which of the following statements is most nearly correct,
other things being equal?
A.Using Student's t instead of z makes a confidence interval
narrower.
B.The table values of z and t are about the same when the mean
is large.
C.For a given confidence level, the z value is always smaller
then the t value.
D.Student's t is rarely used because it is more conservative to
use z.
As n increases, t approaches z, but t is always larger than
z.
AACSB: AnalyticBlooms: UnderstandDifficulty: 2 MediumLearning
Objective: 08-04 Explain how sample size affects the standard
error.Topic: Confidence Interval for a Mean () with Unknown
88.The Central Limit Theorem (CLT):
A.applies only to samples from normal populations.
B.applies to any population.
C.applies best to populations that are skewed.
D.applies only when and are known.
The appeal of the CLT is that is applies to populations of any
shape.
AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning
Objective: 08-03 State the Central Limit Theorem for a mean.Topic:
Sample Mean and the Central Limit Theorem
89.In which situation may the sample proportion safely be
assumed to follow a normal distribution?
A.12 successes in a sample of 72 items
B.8 successes in a sample of 40 items
C.6 successes in a sample of 200 items
D.4 successes in a sample of 500 items
We prefer at least 10 "successes" and at least 10 "failures" to
assume that p is normal.
AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning
Objective: 08-03 State the Central Limit Theorem for a mean.Topic:
Confidence Interval for a Proportion ()
90.In which situation may the sample proportion safely be
assumed to follow a normal distribution?
A.n = 100, = .06
B.n = 250, = .02
C.n = 30, = .50
D.n = 500, = .01
We want n 10 and n(1 - ) 10 to assume that p is normal.
AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning
Objective: 08-03 State the Central Limit Theorem for a mean.Topic:
Confidence Interval for a Proportion ()
91.If = 12, find the sample size to estimate the mean with an
error of 4 and 95 percent confidence (rounded to the next higher
integer).
A.75
B.35
C.58
D.113
n = [z/E]2 = [(1.960)(12)/4]2 = 34.6 (round up).
AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning
Objective: 08-03 State the Central Limit Theorem for a mean.Topic:
Sample Size Determination for a Mean
92.If = 25, find the sample size to estimate the mean with an
error of 3 and 90 percent confidence (rounded to the next higher
integer).
A.426
B.512
C.267
D.188
n = [z/E]2 = [(1.645)(25)/3]2 = 187.9 (round up).
AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning
Objective: 08-03 State the Central Limit Theorem for a mean.Topic:
Sample Size Determination for a Mean
93.Sampling error can be avoided:
A.by using an unbiased estimator.
B.by eliminating nonresponses (e.g., older people).
C.by no method under the statistician's control.
D.either by using an unbiased estimator or by eliminating
nonresponse.
Sampling error occurs in any random sample used to estimate an
unknown parameter.
AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning
Objective: 08-02 Explain the desirable properties of
estimators.Topic: Estimators and Sampling Error
94.A consistent estimator for the mean:
A.converges on the true parameter as the variance increases.
B.converges on the true parameter as the sample size
increases.
C.consistently follows a normal distribution.
D.is impossible to obtain using real sample data.
The variance becomes smaller and the estimator approaches the
parameter as n increases.
AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning
Objective: 08-02 Explain the desirable properties of
estimators.Topic: Estimators and Sampling Error
95.Concerning confidence intervals, which statement is most
nearly correct?
A.We should use z instead of t when n is large.
B.We use the Student's t distribution when is unknown.
C.We use the Student's t distribution to narrow the confidence
interval.
Student's t distribution widens the confidence interval when is
unknown.
AACSB: AnalyticBlooms: UnderstandDifficulty: 2 MediumLearning
Objective: 08-06 Know when to use Student's t instead of z to
estimate .Topic: Confidence Interval for a Mean () with Unknown
96.The standard error of the mean decreases when:
A.the sample size decreases.
B.the standard deviation increases.
C.the standard deviation decreases or n increases.
D.the population size decreases.
The standard error of the mean /(n1/2) depends on n and .
AACSB: AnalyticBlooms: UnderstandDifficulty: 2 MediumLearning
Objective: 08-04 Explain how sample size affects the standard
error.Topic: Confidence Interval for a Mean () with Known
97.For a given sample size, the higher the confidence level,
the:
A.more accurate the point estimate.
B.smaller the standard error.
C.smaller the interval width.
D.greater the interval width.
To have more confidence, we must widen the interval. For
example, z.025 = 1.960 (for 95 percent confidence) gives a wider
interval than z.05 = 1.645 (for 90 percent confidence). The
proffered statement would also be true for the Student's t
distribution.
AACSB: AnalyticBlooms: UnderstandDifficulty: 2 MediumLearning
Objective: 08-05 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Mean () with
Known
98.A sample is taken and a confidence interval is constructed
for the mean of the distribution. At the center of the interval is
always which value?
A.The sample mean
B.The population mean
C.Neither nor since with a sample anything can happen
D.Both and as long as there are not too many outliers
The confidence interval for the mean is symmetric around the
sample mean.
AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning
Objective: 08-05 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Mean () with
Known
99.If a normal population has parameters = 40 and = 8, then for
a sample size n = 4:
A.the standard error of the sample mean is approximately 2.
B.the standard error of the sample mean is approximately 4.
C.the standard error of the sample mean is approximately 8.
D.the standard error of the sample mean is approximately 10.
The standard error is /(n1/2) = (8)/(41/2) = 4.
AACSB: AnalyticBlooms: ApplyDifficulty: 1 EasyLearning
Objective: 08-04 Explain how sample size affects the standard
error.Topic: Confidence Interval for a Mean () with Known
Short Answer Questions100.On the basis of a survey of 545
television viewers, a statistician has constructed a confidence
interval and estimated that the proportion of people who watched
the season premiere of Glee is between .16 and .24. What level of
confidence did the statistician use in constructing this interval?
Explain carefully, showing all steps in your reasoning.
We solve to get z = 2.33, which corresponds approximately to a
98 percent confidence level.
Feedback: The confidence interval is
and the interval half-width is .04 so we set
and p = .20 (the midpoint of the interval) to solve for
= 2.33 which corresponds approximately to a 98 percent
confidence level.
AACSB: AnalyticBlooms: EvaluateDifficulty: 3 HardLearning
Objective: 08-07 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Proportion ()
101.Read the news story below. Using the 95 percent confidence
level, what sample size would be needed to estimate the true
proportion of stores selling cigarettes to minors with an error of
3 percent? Explain carefully, showing all steps in your
reasoning.
=
= 813.5, or 814 (rounded up), using the sample proportion
because it is available (instead of assuming that = .50).
Feedback:
=
= 813.5, or 814 (rounded up). We use the sample proportion
because it is available, instead of assuming that = .50.
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-09 Calculate sample size to estimate a mean or
proportion.Topic: Sample Size Determination for a Proportion
102.In a survey, 858 out of 2600 homeowners said they expected
good economic conditions to continue for the next 12 months.
Construct a 95 percent confidence interval for "good times" in the
next 12 months.
The confidence interval is .3119 < < .3481.
Feedback:
or
or
or .33 .0181, so the confidence interval is .3119 < <
.3481.
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-07 Construct a 90; 95; or 99 percent confidence
interval for .Topic: Confidence Interval for a Proportion ()
103.Fulsome University has 16,059 students. In a sample of 200
students, 12 were born outside the United States. Construct a 95
percent confidence interval for the true population proportion. How
large a sample is needed to estimate the true proportion of Fulsome
students who were born outside the United States with an error of
2.5 percent and 95 percent confidence? Show your work and explain
fully.
We have sampled less than 5 percent of the population, so the
FPCF is unnecessary (i.e., we can ignore the population size. The
95 percent confidence interval is p z.025[p(1 - p)/n]1/2 = .06
(1.960)[(.06)(.94)/200]1/2 or .06 .032914 or .027 < < .093.
To reduce the error to .025, the required sample size is
or
= 346.7, or n = 347 (rounded up). We can use the sample value
for p so we do not need to assume that = .50.
Feedback: The 95 percent confidence interval is p z.025[p(1 -
p)/n]1/2 = .06 (1.960)[(.06)(.94)/200]1/2 or .06 .032914 or .027
< < .093. To reduce the error to .025, the required sample
size is
or
= 346.7, or n = 347 (rounded up). We have a sample value for p
so we do not need to assume that = .50. If you did assume = .50,
you would get an unnecessarily large required sample since the
preliminary sample indicates that is not .50. The sample does not
exceed 5 percent of the population size, so the finite population
correction would make little difference.
AACSB: AnalyticBlooms: ApplyDifficulty: 3 HardLearning
Objective: 08-09 Calculate sample size to estimate a mean or
proportion.Topic: Sample Size Determination for a Proportion
104.List differences and similarities between Student's t and
the standard normal distribution.
Both are bell-shaped and symmetric, but the Student's t
distribution lies below the standard normal in the middle, and its
tails are above the standard normal.
Feedback: They are both bell-shaped and symmetric. However, the
Student's t distribution lies below the standard normal in the
middle, and its tails are above the standard normal ("thicker" or
"heavier" tails). Therefore, the value of Student's t for a given
tail area will always be greater than the corresponding z value. We
use the Student's t whenever the standard deviation is estimated
from a sample, which is to say, most of the time.
AACSB: AnalyticBlooms: ApplyDifficulty: 1 EasyLearning
Objective: 08-06 Know when to use Student's t instead of z to
estimate .Topic: Confidence Interval for a Mean () with Unknown
105.Why does pose a problem for sample size calculation for a
mean? How can be approximated when it is unknown?
Truehe formula for the sample size to estimate requires knowing
. But because is unknown (we are trying to estimate it), then
probably is unknown as well. There are several ways to estimate :
(1) take a small preliminary sample and calculate the sample
standard deviation s as an estimate of ; or (2) if the range is
known, we can estimate = Range/6 because from the Empirical Rule 3
contains almost all of the data in a normal distribution (a
sometimes doubtful assumption if there are outliers or a skewed
population); or (3) we might have some value for from prior
experience (e.g., a previous sample or historical data).
Feedback: The formula for the sample size to estimate requires
knowing . But because is unknown (we are trying to estimate it),
then probably is unknown as well. There are several ways to
estimate : (1) take a small preliminary sample and calculate the
sample standard deviation s as an estimate of ; or (2) if the range
is known, we can estimate = Range/6 because from the Empirical Rule
3 contains almost all of the data in a normal distribution (a
sometimes doubtful assumption if there are outliers or a skewed
population); or (3) we might have some value for from prior
experience (e.g., a previous sample or historical data).
AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning
Objective: 08-09 Calculate sample size to estimate a mean or
proportion.Topic: Sample Size Determination for a Mean