CHAPTER 20
CONFIDENCE INTERVAL ESTIMATION
MULTIPLE CHOICE QUESTIONS
In the following multiple-choice questions, please circle the correct answer.
1.
The confidence interval for a proportion is based on the assumption of a large
sample size. A rule of thumb for checking the validity of this assumption is if
npL , n(1  pL ), npU , and n(1  pU ) are all greater than what value?
a. 0
b. n
c. 2
d. 3
e. 5
ANSWER:
2.
e
When the samples we want to compare are paired in some natural way, such as
pretest/posttest for each person or husband/wife pairs, a more appropriate form of
analysis is to not compare two separate variables, but their
.
a. difference
b. sum
c. ratio
d. total
e. product
ANSWER: a
3.
4.
Confidence intervals are a function of which of the following three things?
a. The population, the sample, and the standard deviation
b. The sample, the variable of interest, and the degrees of freedom
c. The data in the sample, the confidence level, and the sample size
d. The sampling distribution, the confidence level, and the degrees of freedom
e. The mean, the median, and the mode
ANSWER: c
The chi-square and F distributions are used primarily to make inferences about
population ___________.
a. means
b. variances
c. medians
d. modes
e. proportions
ANSWER: b
5.
If you increase the confidence level, the confidence interval
.
a. decreases
b. increases
c. stays the same
d. may increase or decrease, depending on the sample data
ANSWER: b
6.
A random sample allows us to use:
a. the rules of probabilities
b. the rules of large numbers
c. the laws of parameters
d. the laws of distributions
e. the laws of gravity
ANSWER: a
7.
Suppose there are 500 accounts in a population. You sample 50 of them and find
a sample total of $5,000. What would be your estimate for the population total?
a. $5,000
b. $50,000
c. $250,000
d. $2,500,000
e. None of the above
ANSWER: b
8.
9.
Suppose there are 400 accounts in a population. You sample 50 of them and find
a sample mean of $500. What would be your estimate for the population total?
a. $5,000
b. $50,000
c. $250,000
d. $2,500,000
e. None of the above
ANSWER: b
When we replace  with the sample standard deviation (s), we introduce a new
source of variability and the sampling distribution becomes the
.
a. t distribution
b. F distribution
c. chi-square distribution
d. robust distribution
ANSWER: a
10.
Another commonly used random mechanism, besides a simple random sample, is
called:
a. interval estimation
b. a random hypothesis test
c. a randomized experiment
d. a nuisance sample
ANSWER: c
11.
If the odds of a horse winning a race are 2 to 1, then the probability of this horse
winning the race is
.
a. 1/4
b. 1/3
c. 1/2
d. 2/3
e. 2/10
ANSWER:
12.
d
There are, generally speaking, two types of statistical inference. They are:
a. sample estimation and population estimation
b. confidence interval estimation and hypothesis testing
c. interval estimation for a mean and interval estimation for a proportion
d. independent sample estimation and dependent sample estimation
e. none of the above
ANSWER: b
13.
The t distribution has
a. n
b. 2
c. 10
d. n – 1
e. trillion
ANSWER:
14.
degrees of freedom.
d
If you are constructing a confidence interval for a single mean, the confidence
interval will
with an increase in the sample size.
a. decreases
b. increases
c. stays the same
d. may increase or decrease, depending on the sample data
ANSWER: a
15.
As the sample size increases, the t distribution becomes more similar to the
__ distribution.
a.
b.
c.
d.
e.
normal
exponential
F
chi-square
binomial
ANSWER:
16.
a
A parameter, such as  , is sometimes referred to as a ________ parameter,
because many times we need its value even though it is not the parameter of
primary interest.
a. special
b. random
c. nuisance
d. independent
e. dependent
ANSWER: c
17.
When you calculate the sample size for a proportion, you use an estimate for the
population proportion ( pest ). A conservative value for n can be obtained by using
pest =
.
a. 0.0
b. 0.05
c. 0.10
d. 0.50
e. 1.00
ANSWER:
d
QUESTIONS 18 THROUGH 23 ARE BASED ON THE FOLLOWING INFORMATION:
The following values have been calculated using the TDIST and TINV functions in
Excel. These values come from a t distribution with 15 degrees of freedom.
These values represent the probability to the right of the given positive values.
Value
1.00
1.20
1.40
t probability
0.1636
0.1209
0.0872
These values represent the t value for a given probability.
Probability
0.20
0.10
0.05
t value
1.3178
1.7109
2.0639
18.
What is the probability of a t-value smaller 1.00?
a. 0.1209
b. 0.1636
c. 0.8364
d. 0.8791
ANSWER:
19.
What is the probability of a t-value larger than 1.20?
a. 0.0872
b. 0.1209
c. 0.1636
d. 0.2000
ANSWER:
20.
b
What would be the t-value where 0.05 of the values are in the upper tail?
a. +1.000
b. +1.318
c. +1.711
d. +2.064
ANSWER:
22.
b
What is the probability of a t-value between –1.40 and +1.40?
a. 0.7582
b. 0.8256
c. 0.9128
d. 0.9500
ANSWER:
21.
c
c
What would be the t-values where 0.10 of the values are in both tails (sum of both
tails)?
a. –1.000, +1.000
b. –1.318, +1.318
c. –1.711, +1.711
d. –2.064, +2.064
ANSWER: c
23.
What would be the t-values where 0.95 of the values would fall within this
interval?
a.
b.
c.
d.
–1.000, +1.000
–1.318, +1.315
–1.711, +1.711
–2.064, +2.064
ANSWER:
d
QUESTIONS 24 THROUGH 29 ARE BASED ON THE FOLLOWING INFORMATION:
The following values have been calculated using the TDIST and TINV functions in
Excel. These values come from a t distribution with 15 degrees of freedom.
These values represent the probability to the right of the given positive values.
Value
0.95
1.15
1.20
t probability
0.1786
0.1341
0.1244
These values represent the t value for a given probability.
Probability
0.20
0.15
0.10
24.
What is the probability of a t-value smaller than 1.20?
a. 0.8756
b. 0.8659
c. 0.1341
d. 0.1244
ANSWER:
25.
b
What is the probability of a t-value between –0.95 and +0.95?
a. 0.1786
b. 0.3572
c. 0.6428
d. 0.8214
ANSWER:
27.
a
What is the probability of a t-value larger than 1.15?
a. 0.1786
b. 0.1341
c. 0.1244
d. 0.1500
ANSWER:
26.
t value
1.341
1.517
1.753
c
What would be the t-value where 0.075 of the values are in the upper tail?
a. +1.000
b. +1.341
c. +1.517
d. +1.753
ANSWER:
28.
c
What would be the t-values where 0.80 of the values would fall within this
interval?
a. –1.000, +1.000
b. –1.341, +1.341
c. –1.517, +1.517
d. –1.753, +1.753
ANSWER: b
29.
What would be the t-values where 0.10 of the values are in both tails (sum of both
tails)?
a. –1.000, +1.000
b. –1.341, +1.341
c. –1.517, +1.517
d. –1.753, +1.753
ANSWER: d
TEST QUESTIONS
30.
You are told that a random sample of 150 people from Iowa has been given
cholesterol tests, and 60 of these people had levels over the “safe” count of 200.
Construct a 95% confidence interval for the population proportion of people in
Iowa with cholesterol levels over 200.
ANSWER:
n  150, pˆ  60 /150  .40
pˆ  Z pˆ (1  pˆ ) / n  0.40  1.96 (.40)(.60) /150  0.40  0.0784
Lower limit = 0.3216, and upper limit = 0.4784
31.
You are trying to estimate the average amount a family spends on food during a
year. In the past, the standard deviation of the amount a family has spent on food
during a year has been approximately $1200. If you want to be 99% sure that you
have estimated average family food expenditures within $60, how many families
do you need to survey?
ANSWER:
 est . =1200, z-multiple = 2.575, B = 60 . The sample size for a mean is given by
 z  multiple   est   2.575 1200 
n

 
B
60


 
2
32.
2
2653
You have been assigned to determine whether more people prefer Coke to Pepsi.
Assume that roughly half the population prefers Coke and half prefers Pepsi.
How large a sample would you need to take to ensure that you could
estimate, with 95% confidence, the proportion of people preferring Coke within
3% of the actual value?
ANSWER:
pest . = 0.50, z-multiple = 1.96, B = 0.03. The sample size for a proportion is given
by
 z  multiple 
 1.96 
n
 pest . (1  pest . )  
 (0.50)(0.50) 1068
B


 0.03 
2
2
QUESTIONS 33 THROUGH 35 ARE BASED ON THE FOLLOWING INFORMATION:
A marketing research consultant hired by Coca-Cola is interested in determining the
proportion of customers who favor Coke over other soft drinks. A random sample of 400
consumers was selected from the market under investigation and showed that 53%
favored Coca-Cola over other brands.
33.
Compute a 95% confidence interval for the true proportion of people who favor
Coke. Do the results of this poll convince you that a majority of people favors
Coke?
ANSWER:
0.53  0.0489 = (0.4811, 0.5789).
Since confidence interval ranges from 48% to 57.9%, it is difficult to conclude
that a majority of people favors Coke. It could be below 50%.
34.
Suppose 2,000 (not 400) people were polled and 53% favored Coke. Would you
now be convinced that a majority of people favor Coke? Why might your answer
be different than in Question 33?
ANSWER:
0.53  0.0219 = (0.5081, 0.5519).
In this case the 95% confidence interval is entirely above 50%, the data is now
more convincing than it was previously.
35.
How many people would have to be surveyed to be 95% confident that you can
estimate the fraction of people who favor Coca-Cola within 1%?
ANSWER:
9,569.43 or 9,570.
QUESTIONS 36 AND 37 ARE BASED ON THE FOLLOWING INFORMATION:
The employee benefits manager of a medium size business would like to estimate the
proportion of full-time employees who prefer adopting plan A of three available health
care plans in the coming annual enrollment period. A reliable frame of the company’s
employees and their tentative health care preferences are available. Using Excel, the
manager chose a random sample of size 50 from the frame. There were 17 employees in
the sample who preferred plan A.
36.
Construct a 99% confidence interval for the proportion of company employees
who prefer plan A. Assume that the population consists of the preferences of all
employees in the frame.
ANSWER:
n  50, pˆ  17 / 50  0.34
pˆ  Z pˆ (1  pˆ ) / n  0.34  2.575 (0.34)(0.66) / 50  0.34  0.1725
lower limit = 0.1675, upper limit = 0.5125
37.
Interpret the 99% confidence interval constructed in Question 36.
ANSWER:
We are 99% confident that the proportion of all employees who prefer plan A is
between 0.1675 and 0.5125.
QUESTIONS 38 THROUGH 40 ARE BASED ON THE FOLLOWING INFORMATION:
Q-Mart is interested in comparing its male and female customers. Q-Mart would like to
know if its female charge customers spend more money, on average, than its male charge
customers. They have collected random samples of 25 female customers and 22 male
customers. On average, women charge customers spend $102.23 and men charge
customers spend $86.46. Some information are shown below.
Summary statistics for two samples
Sample sizes
Sample means
Sample standard deviations
Female
25
102.23
93.393
Male
22
86.46
59.695
Confidence interval for difference between means
Sample mean difference
Pooled standard deviation
Std error of difference
38.
15.77
79.466
23.23
Using a t-value of 2.014, calculate a 95% confidence interval for the difference
between the average female purchase and the average male purchase. Would you
conclude that there is a significant difference between females and males in this
case? Explain.
ANSWER:
15.77  46.785 = (-31.015, 62.555). Since the range includes 0, there does not
appear to be a significant difference between the means of the two groups.
39.
What are the degrees of freedom for the t-statistic in this calculation? Explain
how you would calculate the degrees of freedom in this case.
ANSWER:
n1 + n2 – 2 = 45
40.
What is the assumption in this case that allows you to use the pooled standard
deviation for this confidence interval?
ANSWER:
In order to use the pooled standard deviation for this confidence interval, we must
assume that the two populations standard deviations are equal (  1   2 ).
QUESTIONS 41 AND 42 ARE BASED ON THE FOLLOWING INFORMATION:
A company employs two shifts of workers. Each shift produces a type of gasket where
the thickness is the critical dimension. The average thickness and the standard deviation
of thickness for shift 1, based on a random sample of 40 gaskets, are 10.85 mm and 0.16
mm, respectively. The similar figures for shift 2, based on a random sample of 30
gaskets, are 10.90 mm and 0.19 mm. Let 1  2 be the difference in thickness between
shifts 1 and 2, and assume that the population variances are equal.
41.
Construct a 95% confidence interval for 1  2 .
ANSWER:
n1  40, X1  10.85, s1  0.16
n2  30, X 2  10.90, s2  0.19
The pooled standard deviation is s p 
( X1  X 2 )  t s p
1
n1
(n1  1) s12  (n2  1) s22
= 0.1734
n1  n2  2
 n12  0.05  1.9955(0.1734)(0.2415)  0.05  0.0836
Lower limit = -0.1336, and upper limit = 0.0336.
42.
Based on your answer to Question 41, are you convinced that the gaskets from
shift 2 are, on average, wider than those from shift 1? Why or why not?
ANSWER:
The confidence interval extends from a negative number (indicating shift 2
thickness is larger) to a positive number (indicating shift 2 thickness is smaller).
So we are not absolutely sure which mean is greater.
QUESTIONS 43 AND 44 ARE BASED ON THE FOLLOWING INFORMATION:
A sample of 9 production managers with over 15 years of experience has an average
salary of $71,000 and a sample standard deviation of $18,000.
43.
You can be 95% confident that the mean salary for all production managers with
at least 15 years of experience is between what two numbers (the t-statistic with 8
degrees of freedom is 2.306)? What assumption are you making about the
distribution of salaries?
ANSWER:
$71,000  $13,836 = ($57,164, $84,836). The assumption is that the population is
normal or near normal. This is particularly important since the sample size is so
small (9). However, the t distribution is rather robust to violations of normality.
44.
What sample size would be needed to ensure that we could estimate the true mean
salary of all production managers with more than 15 years of experience and have
only 5 chances in 100 of being off by more than $600?
ANSWER:
69.18 or 70
QUESTIONS 45 THROUGH 50 REQUIRE THE USE OF EXCEL:
45.
Compute P(1.50  t10  1.00), where t10 has a t-distribution with 10 degrees of
freedom.
ANSWER:
0.74730
46.
Compute P(1.50  t100  1.00), where t100 has a t-distribution with 100 degrees of
freedom.
ANSWER:
0.77176
47.
Compute P(1.50  Z  1.00), where Z is a standard normal random variable.
ANSWER:
0.77454
48.
Compare the result of Question 47 to the results obtained in Questions 45 and 46.
How do you explain the difference in these probabilities?
ANSWER:
The variance of t with a small degree of freedom is larger than a t with a large
degree of freedom, which is larger than for a Z. This explains why the “between”
probabilities in Questions 45, 46, and 47 increase.
49.
Find the 75th percentile of the t-distribution with 25 degrees of freedom.
ANSWER:
0.32217
50.
Find the 75th percentile of the t-distribution with 5 degrees of freedom.
ANSWER:
0.33672
QUESTIONS 51 and 52 ARE BASED ON THE FOLLOWING INFORMATION:
A sample of 40 country CD recordings of Willie Nelson has been examined. The average
playing time of these recordings is 51.3 minutes, and the standard deviation is 5.8
minutes.
51.
Construct a 95% confidence interval for the mean playing time of all Willie
Nelson recordings.
ANSWER:
n = 10, X = 54000, s = 15000
X  t (s / n)  51.3  2.0227(5.8/ 40)  51.3  1.855
Lower limit = 49.445, and upper limit = 53.155
52.
Interpret the confidence interval you constructed.
ANSWER:
We are 95% confident that the mean playing time of all Willie Nelson recordings
is between. 49.445 and 53.155 minutes.
QUESTIONS 53 AND 54 ARE BASED ON THE FOLLOWING INFORMATION:
A department store is interested in the average balance that is carried on its store’s credit
card. A sample of 40 accounts reveals an average balance of $1,250 and a standard
deviation of $350.
53.
Find a 95% confidence interval for the mean account balance on this store’s credit
card (the t-statistic with 39 degrees of freedom is 2.02).
ANSWER:
$1,250  $111.79 = ($1,138.21, $1,361.79).
54.
What sample size would be needed to ensure that we could estimate the true mean
account balance and have only 5 chances in 100 of being off by more than $100?
ANSWER:
49.98 or 50.
QUESTIONS 55 AND 56 ARE BASED ON THE FOLLOWING INFORMATION:
A market research consultant hired by Coke Classic Company is interested in
estimating the difference between the proportions of female and male customers who
favor Coke Classic over Pepsi Cola in Chicago. A random sample of 200 consumers
from the market under investigation showed the following frequency distribution.
Coke
Pepsi
55.
Male
72
58
130
Female
38
32
70
110
90
200
Construct a 95% confidence interval for the difference between the proportions of
male and female customers who prefer Coke Classic over Pepsi Cola.
ANSWER:
n1  number of males = 130, n 2 = number of females = 70
Pˆ1  proportion of males who favor Coke over Pepsi = 72/130 = 0.5538
Pˆ  proportion of females who favor Coke over Pepsi = 0.5429
2
SE ( pˆ1  pˆ 2 ) 
pˆ1 (1  pˆ1 ) pˆ 2 (1  pˆ 2 )

= 0.0738
n1
n2
( Pˆ1  Pˆ2 )  Z  SE ( Pˆ1  Pˆ2 )  0.0109  1.96(.0738)  0.0109  0.1446
Lower limit = -0.1337, and upper limit = 0.1555
56.
Interpret the constructed confidence interval.
ANSWER:
We are 95% confident that the population difference between these proportions is
between –13.37% and 15.55%.
QUESTIONS 57 THROUGH 60 ARE BASED ON THE FOLLOWING INFORMATION:
The percent defective for parts produced by a manufacturing process is targeted at 4%.
The process is monitored daily by taking samples of sizes n = 160 units. Suppose that
today’s sample contains 14 defectives.
57.
58.
Determine a 95% confidence interval for the proportion defective for the process
today.
ANSWER:
0.0875  0.0438 = (0.0437, 0.1313).
Based on your answer to Question 57, is it still reasonable to think the overall
proportion defective produced by today’s process is actually the targeted 4%?
Explain your reasoning.
ANSWER:
No, since 4% falls outside of this range.
59.
The confidence interval in Question 57 is based on the assumption of a large
sample size. Is this sample size sufficiently large in this example? Explain how
you arrived at your answer.
ANSWER:
Yes. Because npL , n(1  pL ), npU , and n(1  pU ) are all greater than 5.0.
60.
How many units would have to be sampled to be 95% confident that you can
estimate the fraction of defective parts within 2% (using the information from
today’s sample)?
ANSWER:
766.40 or 767.
QUESTIONS 61 AND 62 ARE BASED ON THE FOLLOWING INFORMATION:
Auditors of Independent Bank are interested in comparing the reported value of all 1775
customer saving account balances with their own findings regarding the actual value of
such assets. Rather than reviewing the records of each savings account at the bank, the
auditors randomly selected a sample of 100 savings account balances from the frame.
The sample mean and sample standard deviations were $505.75 and 360.95, respectively.
61.
Construct a 90% confidence interval for the total value of all savings account
balances within this bank. Assume that the population consists of all savings
account balances in the frame.
ANSWER:
N  1775, n  100, X  505.75, s  360.95
NX  1.6604( Ns / n )  1775(505.75)  1.6604(1775  360.95 / 100)
 897706.25 106379.55
= ($719,326.70, $1,004,085.8)
62.
Interpret the 90% confidence interval constructed in Question 61.
ANSWER:
We are 90% confident that the total balance of all 1775 savings account balances
within the bank are between $791,327 and $1,004,086.
QUESTIONS 63 AND 64 ARE BASED ON THE FOLLOWING INFORMATION:
A real estate agent has collected a random sample of 40 houses that were recently sold in
Grand Rapids, Michigan. She is interested in comparing the appraised value and recent
selling price (in thousands of dollars) of the houses in this particular market. The values
of these two variables for each of the 40 randomly selected houses are shown below.
House
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Value
140.93
132.42
118.30
122.14
149.82
128.91
134.61
121.99
150.50
142.87
155.55
128.50
143.36
119.65
122.57
145.27
149.73
147.70
117.53
140.13
Price
140.24
129.89
121.14
111.23
145.14
139.01
129.34
113.61
141.05
152.90
157.79
135.57
151.99
120.53
118.64
149.51
146.86
143.88
118.52
146.07
House
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
Value
136.57
130.44
118.13
130.98
131.33
141.10
117.87
160.58
151.10
120.15
133.17
140.16
124.56
127.97
101.93
131.47
121.27
143.55
136.89
106.11
Price
135.35
121.54
132.98
147.53
128.49
141.93
123.55
162.03
157.39
114.55
139.54
149.92
122.08
136.51
109.41
127.29
120.45
151.96
132.54
114.33
63.
Using the sample data, generate a 95% confidence interval for the mean
difference between the appraised values and selling prices of the houses sold in
Grand Rapids.
ANSWER:
We applied the paired sample analysis using n  40, X D  1.612, sD  6.794 ,
where: D = Difference = Appraised value – selling price.
X D  t (sD / n)  1.612  2.0227(6.794 / 40)  1.612  2.173
Lower limit = -3.785, and Upper limit = 0.561 (in thousands of dollars)
64.
Interpret the constructed confidence interval for the real estate agent.
ANSWER:
We are 95% confident that the actual mean difference between the appraised
values and selling prices of all the houses sold in Grand Rapids is between -$3785
and $561.
QUESTIONS 65 THROUGH 69 REQUIRE THE USE OF EXCEL:
65.
Compute P(t15  2.0), where t15 has a t-distribution with 15 degrees of freedom.
ANSWER:
0.03197
66.
Compute P(t150  2.0), where t150 has a t-distribution with 150 degrees of freedom.
ANSWER:
0.02365
67.
How do you explain the difference between the results obtained in Questions 65
and 66?
ANSWER:
The smaller the degrees of freedom, the higher the variance of t, and so the larger
the tail probabilities are.
68.
Compute P( Z  2.0), where Z is a standard normal random variable.
ANSWER:
0.02275
69.
Compare the results of Question 68 to the results obtained in Questions 65 and 66.
How do you explain the difference in these probabilities?
ANSWER:
First, the variance of t with a small degree of freedom is larger than a t with a
large degree of freedom, which is larger than for a Z. This explains why the
probabilities in Questions 65, 66, and 68 increases. Second, when the sample size
is large, the degrees of freedom of t are large; and that the t distribution and the
standard normal distribution are practically indistinguishable. This explains why
the probabilities in Questions 66 and 68 are close.
QUESTIONS 70 THROUGH 72 ARE BASED ON THE FOLLOWING INFORMATION:
Senior management of a consulting services firm is concerned about a growing decline in
the firm’s weekly number of billable hours. The firm expects each professional
employee to spend at least 40 hours per week on work. In an effort to understand this
problem better, management would like to estimate the standard deviation of the number
of hours their employees spend on work-related activities in a typical week. Rather than
reviewing the records of all the firm’s full-time employees, the management randomly
selected a sample of size 50 from the available frame. The sample mean and sample
standard deviations were 48.5 and 7.5 hours, respectively.
70.
Construct a 99% confidence interval for the standard deviation of the number of
hours this firm’s employees spend on work-related activities in a typical week.
ANSWER:
n  50, X  48.5, s  7.5
71.
Lower limit =
n  1s / 2 / 2  49(7.5) / 78.2305 =5.936
Upper limit =
n  1s / 12 / 2  49(7.5) / 27.2494  10.057
Interpret the 99% confidence interval constructed in Question 70.
ANSWER:
We are 99% confident that the population standard deviation is between 5.936
and 10.057.
72.
Given the target range of 40 to 60 hours of work per week, should senior
management be concerned about the number of hours their employees are
currently devoting to work? Explain why or why not.
ANSWER:
The best guess for the population mean is 48.5 hours per week, and about 95% of
all employees are within 2 standard deviations of this, where we are almost sure
(99% sure) that this standard deviation is between 5.9 and 10.1. But even if the
standard deviation is only 5.9, then 48.5 2 standard deviations will produce the
range 36.7 to 60.3. Maybe management should be concerned.
QUESTIONS 73 THROUGH 75 REQUIRE THE USE OF EXCEL:
73.
Compute P(t20  0.95), where t20 has a t-distribution with 20 degrees of
freedom.
ANSWER:
Because of the symmetry of the t distribution, this left-hand tail probability can be
calculated exactly like right-hand tail. The answer is 0.17673.
74.
Compute P(t2  0.95), where t2 has a t-distribution with 2 degrees of freedom.
ANSWER:
Because of the symmetry of the t distribution, this left-hand tail probability can be
calculated exactly like right-hand tail. The answer is 0.22119.
75.
How do you explain the difference between the results obtained in Questions 73
and 74?
ANSWER:
The larger the degrees of freedom, the lower the variance of t, so the smaller the
tail probabilities are. This explains why the probability in Question 73 is smaller
than that in Question 74.
QUESTIONS 76 AND 77 ARE BASED ON THE FOLLOWING INFORMATION:
A sample of 10 quality control managers with over 15 years of experience has an average
salary of $54,000 and a standard deviation of $15,000.
76.
You can be 95% confident that the mean salary for all quality control managers
with at least 15 years of experience is between what two numbers? What
assumptions are you making about the distribution of salaries?
ANSWER:
n = 10, X = 54000, s = 15000
X  t (s / n)  54000  2.2622(15000 / 10)  54000  10730.557
Lower limit = 43,269.443, and upper limit = 64,730.557
We must assume that the population distribution of salaries is normal, especially
since the sample size is so small.
77.
What size sample would be needed to ensure that we could estimate the true mean
salary of all quality control managers with more than 15 years of experience and
have only 2 chances in 100 of being off by more than $800?
ANSWER:
 est . =15000, z-multiple = 2.326, B = 800
The approximate sample size required to produce a 98% confidence interval for
the mean is given by
 z  multiple   est   2.326 15000 
n

 
B
800


 
2
2
1903
QUESTIONS 78 THROUGH 80 ARE BASED ON THE FOLLOWING INFORMATION:
Q-Mart is interested in comparing customer who used its own charge card with those who
use other types of credit cards. Q-Mart would like to know if customers who use the QMart card spend more money per visit, on average, than customers who use some other
type of credit card. They have collected information on a random sample of 38 charge
customers and the data is presented below. On average, the person using a Q-Mart card
spends $192.81 per visit and customers using another type of card spend $104.47 per
visit. Use the information below to answer the following questions.
Summary statistics for two samples
Q-Mart
Sample sizes
13
Sample means
192.81
Sample standard deviations
115.243
Other Charges
25
104.47
71.139
Confidence interval for difference between means
Sample mean difference
Pooled standard deviation
Std error of difference
78.
88.34
88.323
30.201
Using a t-value of 2.023, calculate a 95% confidence interval for the difference
between the average Q-Mart charge and the average charge on another type of
credit card. Would you conclude that there is a significant difference between the
two types of customers in this case? Explain.
ANSWER:
88.34  61.0966 = +27.2434 – +149.4366. Since the range does not include 0,
there appears to be a significant difference between the means of the two groups.
In this case, it appears as though the Q-Mart charge card holders spend more
money than those who use other types of charge cards.
79.
What are the degrees of freedom for the t-statistic in this calculation? Explain
how you would calculate the degrees of freedom in this case.
ANSWER:
n1 + n2 – 2 = 36
80.
What is the assumption in this case that allows you to use the pooled standard
deviation for this confidence interval?
ANSWER:
In order to use the pooled standard deviation for this confidence interval, we must
assume that the two populations standard deviations are equal (  1   2 ).
QUESTIONS 81 THROUGH 84 ARE BASED ON THE FOLLOWING INFORMATION:
The average annual household income levels of citizens of selected U.S. cities are shown
below.
City
Index
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
81.
Household
Income
$54,300
$61,800
$61,400
$50,800
$56,200
$48,300
$61,600
$63,200
$55,200
$58,000
$77,600
$47,600
$62,700
$46,200
$64,300
$56,000
$53,400
$56,800
$51,200
$59,000
City
Index
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
Household
Income
$53,500
$45,600
$70,100
$108,700
$46,400
$56,700
$59,100
$46,300
$52,900
$56,300
$67,300
$63,800
$70,600
$49,800
$51,300
$56,600
$49,600
$67,400
$53,700
$48,700
City
Index
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
Household
Income
$61,500
$53,000
$51,000
$55,600
$51,600
$57,200
$54,300
$51,500
$53,500
$61,800
$44,800
$57,400
$48,100
$52,700
$57,400
$65,500
$59,600
$62,000
$49,700
$54,400
Use Excel to obtain a simple random sample of size 10 from this frame.
ANSWER:
I used StatPro’s Generate Random Samples to generate a sample of size 10, then
used the VLOOKUP function to get the corresponding incomes. The following
sample is obtained:
City Index
50
14
4
56
48
49
8
11
38
52
Income
61,800
46,200
50,800
65.500
51,500
53,500
63,200
77,600
67,400
57,400
82.
Using the sample generated in Question 81, construct a 95% confidence interval
for the mean average annual household income level of citizens in the selected
U.S. cities. Assume that the population consists of all average annual household
income levels in the given frame.
ANSWER:
n  10, X  59, 490, s  9, 439.803
X  2.2622(s / n)  59, 490  2.2622(9, 439.803/ 10)  59, 490  6,752.9561
Lower limit = 52,737.0439, upper limit = 66,242.9561
83.
Interpret the 95% confidence interval constructed in Question 82.
ANSWER:
We are 95% confident that the average of all incomes is between $52,737 and
$66,243.
84.
Does the 95% confidence interval contain the actual population mean? If not,
explain why not. What proportion of many similarly constructed confidence
intervals should include the true population mean value?
ANSWER:
This confidence interval easily captures the true population mean of $57,043.
Approximately 95% of the confidence intervals constructed in this way should
contain the true population mean.
QUESTIONS 85 THROUGH 91 ARE BASED ON THE FOLLOWING INFORMATION:
The personnel department of a large corporation wants to estimate the family dental
expenses of its employees to determine the feasibility of providing a dental insurance
plan. A random sample of 12 employees reveals the following family dental expenses (in
dollars) for the year 2001.
115
370
250
85.
Construct a 90% confidence interval estimate of the mean family dental expenses
for all employees of this corporation.
ANSWER:
93
540
225
177
425
318
182
275
228
86.
What assumption about the population distribution must be made to answer
Question 85?
ANSWER:
The population of dental expenses must be approximately normally distributed.
87.
Interpret the 90% confidence interval constructed in question 85.
ANSWER:
We are 90% confident that the mean family dental expenses for all employees of
this corporation is between $199.26 and $333.74.
88.
Suppose you used a 95% confidence interval in Question 85. What would be your
answer to Question 85?
ANSWER:
89.
Suppose the fourth value were 593 instead of 93. What would be your answer to
Question 88? What effect does this change have on the confidence interval?
ANSWER:
The additional $500 in dental expenses, divided across the sample of 12, raises
the mean by $41.67 and increases the standard deviation by nearly $18.20. The
interval width increases over $23 in the process.
90.
Construct a 90% confidence interval estimate for the standard deviation of family
dental expenses for all employees of this corporation.
ANSWER:
91.
Interpret the 90% confidence interval constructed in question 90.
ANSWER:
We are 90% confident that the standard deviation for family dental expenses for
all employees of this corporation is between 110.61 and 229.38.
QUESTIONS 92 AND 93 ARE BASED ON THE FOLLOWING INFORMATION:
An automobile dealer wants to estimate the proportion of customers who still own the
cars they purchased six years ago. A random sample of 200 customers selected from the
automobile dealer’s records indicates that 88 still own cars that were purchased six years
earlier.
92.
Construct a 95% confidence interval estimate of the population proportion of all
customers who still own the cars they purchased six years ago
ANSWER:
pˆ (1– pˆ )
0.44(0.56)
= 0.44  0.0688
 0.44  1.96 
n
200
Lower limit = 0.3712, and upper limit = 0.5088
pˆ  Z 
93.
How can the result in Question 92 be used by the automobile dealer to study
satisfaction with cars purchased at the dealership?
ANSWER:
The dealer can infer that the proportion of all customers who still own the cars they
purchased at the dealership 6 years earlier is somewhere between 03712 and 0.5088
with a 95% level of confidence.
TRUE / FALSE QUESTIONS
94.
The degrees of freedom for the t and chi-square distributions is a numerical
parameter of the distribution that defines the precise shape of the distribution.
ANSWER:
95.
When all possible samples of size n are drawn from any population, then the
sampling distribution of the sample mean X is approximately normal provided
that n is reasonably large.
ANSWER:
96.
F
The standard error of the sampling distribution of the sample proportion p̂ , when
the sample size n = 50 and the population proportion p = 0.25, is 0.00375.
ANSWER:
98.
T
The mean of the sampling distribution of the sample proportion p̂ , when the
sample size n = 100 and the population proportion p = 0.58, is 58.0.
ANSWER:
97.
T
F
In developing a confidence interval for the population standard deviation  , we
make use of the fact that the sampling distribution of the sample standard
deviation S is not the normal distribution or the t distribution, but rather a rightskewed distribution called the chi-square distribution, which (for this procedure)
has n – 1 degrees of freedom.
ANSWER:
T
99.
As a general rule, the normal distribution is used to approximate the sampling
distribution of the sample proportion p̂ only if the sample size n is greater than
30.
ANSWER:
100.
In general, the paired-sample procedure is appropriate when the samples are
naturally paired in some way and there is a reasonably large positive correlation
between the pairs. In this case, the paired-sample procedure makes more efficient
use of the data and generally results in narrower confidence intervals.
ANSWER:
101.
F
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.
ANSWER:
106.
T
If two random samples of sizes 30 and 35 are selected independently from two
populations whose means are 85 and 90, then the mean of the sampling
distribution of the sample mean difference, X 1  X 2 , equals 5.
ANSWER:
105.
F
If two random samples of size 40 each are selected independently from two
populations whose variances are 35 and 45, then the standard error of the
sampling distribution of the sample mean difference, X 1  X 2 , equals 1.4142.
ANSWER:
104.
F
If a random sample of size 250 is taken from a population, where it is known that
the population proportion p = 0.4, then the mean of the sampling distribution of
the sample proportion p̂ is 0.60.
ANSWER:
103.
T
If the standard error of the sampling distribution of the sample proportion p̂ is
0.0324 for samples of size 200, then the population proportion must be 0.30.
ANSWER:
102.
F
T
The 95% confidence interval for the population mean  , given that the sample
size n = 49 and the population standard deviation  = 7, is X  1.96 .
ANSWER:
T
107.
In order to construct a confidence interval estimate of the population mean  , the
value of  must be given.
ANSWER:
108.
The interval estimate 18.5  2.5 was developed for a population mean when the
sample standard deviation S was 7.5. Had S equaled 15, the interval estimate
would be 37  5.0.
ANSWER:
109.
F
The t-distribution and the standard normal distribution are practically
indistinguishable as the degrees of freedom increase.
ANSWER:
115.
T
The lower limit of the 95% confidence interval for the population proportion p,
given that n = 300; and p̂ = 0.10 is 0.1339.
ANSWER:
114.
F
The upper limit of the 90% confidence interval for the population proportion p,
given that n = 100; and p̂ = 0.20 is 0.2658.
ANSWER:
113.
T
In general, increasing the confidence level will narrow the confidence interval,
and decreasing the confidence level widens the interval.
ANSWER:
112.
T
A 90% confidence interval estimate for a population mean  is determined to be
72.8 to 79.6. If the confidence level is reduced to 80%, the confidence interval for
 becomes narrower.
ANSWER:
111.
F
We can form a confidence interval for the population total T by finding a
confidence interval for the population mean  in the usual way, and then
multiplying each end point of the confidence interval by the population size N.
ANSWER:
110.
F
T
In determining the sample size n for estimating the population proportion p, a
conservative value of n can be obtained by using 0.50 as an estimate of p.
ANSWER:
T
116.
In developing confidence interval for the difference between two population
means using two independent samples, we use the pooled estimate s p in
estimating the standard error of the sampling distribution of the sample mean
difference X 1  X 2 if the populations are normal with equal variances.
ANSWER:
T