Chapter 10 – Comparisons Involving Means, Experimental Design, and Analysis of Variance
Multiple Choice
1. To compute an interval estimate for the difference between the means of two populations, the t distribution
a.
is restricted to small sample situations
b.
is not restricted to small sample situations
c.
can be applied when the populations have equal means
d.
None of these alternatives is correct.
2. When developing an interval estimate for the difference between two sample means, with sample sizes of n1 and n2,
a.
n1 must be equal to n2
b.
n1 must be smaller than n2
c.
n1 must be larger than n2
d.
n1 and n2 can be of different sizes,
3. To construct an interval estimate for the difference between the means of two populations when the standard deviations
of the two populations are unknown, we must use a t distribution with (let n1 be the size of sample 1 and n2 the size of
sample 2)
a.
b.
c.
d.
4. When each data value in one sample is matched with a corresponding data value in another sample, the samples are
known as
a.
corresponding samples
b.
matched samples
c.
independent samples
d.
None of these alternatives is correct.
5. Independent simple random samples are taken to test the difference between the means of two populations whose
variances are not known. The sample sizes are n1 = 32 and n2 = 40. The correct distribution to use is the
a.
binomial distribution
b.
t distribution with 72 degrees of freedom
c.
t distribution with 71 degrees of freedom
d.
t distribution with 70 degrees of freedom
Chapter 10 – Comparisons Involving Means, Experimental Design, and Analysis of Variance
d
1
6. Independent simple random samples are taken to test the difference between the means of two populations whose
standard deviations are not known. The sample sizes are n1 = 25 and n2 = 35. The correct distribution to use is the
a.
Poisson distribution
b.
t distribution with 60 degrees of freedom
c.
t distribution with 59 degrees of freedom
d.
t distribution with 58 degrees of freedom
d
1
7. If two independent large samples are taken from two populations, the sampling distribution of the difference between
the two sample means
a.
can be approximated by a Poisson distribution
b.
will have a variance of one
c.
can be approximated by a normal distribution
d.
will have a mean of one
c
1
8. The standard error of is the
a.
variance of
b.
variance of the sampling distribution of
c.
standard deviation of the sampling distribution of
d.
difference between the two means
c
1
Exhibit 10-1
Salary information regarding male and female employees of a large company is shown below.
Male
Female
Sample Size
64
36
Sample Mean Salary (in $1,000)
44
41
Population Variance
128
72
9. Refer to Exhibit 10–1. The point estimate of the difference between the means of the two populations is
a.
–28
b.
3
c.
4
d.
-4
b
10. Refer to Exhibit 10–1. The standard error for the difference between the two means is
a.
4
b.
7.46
c.
4.24
d.
2.0
d
1
11. Refer to Exhibit 10–1. At 95% confidence, the margin of error is
a.
1.96
b.
1.645
c.
3.920
d.
2.000
c
1
12. Refer to Exhibit 10–1. The 95% confidence interval for the difference between the means of the two populations is
a.
0 to 6.92
b.
-2 to 2
c.
-1.96 to 1.96
d.
-0.92 to 6.92
d
1
13. Refer to Exhibit 10–1. If you are interested in testing whether or not the average salary of males is significantly greater
than that of females, the test statistic is
a.
2.0
b.
1.5
c.
1.96
d.
1.645
b
1
14. Refer to Exhibit 10–1. The p-value is
a.
0.0668
b.
0.0334
c.
1.336
d.
1.96
a
1
15. Refer to Exhibit 10–1. At 95% confidence, the conclusion is the
a.
average salary of males is significantly greater than females
1
Chapter 10 – Comparisons Involving Means, Experimental Design, and Analysis of Variance
b.
average salary of males is significantly lower than females
c.
salaries of males and females are equal
d.
None of these alternatives is correct.
d
1
Exhibit 10-2
The following information was obtained from matched samples.
The daily production rates for a sample of workers before and after a training program are shown below.
Worker
Before
After
1
20
22
2
25
23
3
27
27
4
23
20
5
22
25
6
20
19
7
17
18
16. Refer to Exhibit 10–2. The point estimate for the difference between the means of the two populations is
a.
-1
b.
-2
c.
0
d.
1
c
1
17. Refer to Exhibit 10–2. The null hypothesis to be tested is H0: μd = 0. The test statistic is
a.
-1.96
b.
1.96
c.
0
d.
1.645
c
1
18. Refer to Exhibit 10–2. Based on the results of question 18, the
a.
null hypothesis should be rejected
b.
null hypothesis should not be rejected
c.
alternative hypothesis should be accepted
d.
None of these alternatives is correct.
b
1
Exhibit 10-3
A statistics teacher wants to see if there is any difference in the abilities of students enrolled in statistics today and those
enrolled five years ago. A sample of final examination scores from students enrolled today and from students enrolled five
Chapter 10 – Comparisons Involving Means, Experimental Design, and Analysis of Variance
years ago was taken. You are given the following information.
Today
Five Years Ago
82
88
σ2
112.5
54
n
45
36
19. Refer to Exhibit 10–3. The point estimate for the difference between the means of the two populations is
a.
58.5
b.
9
c.
-9
d.
-6
d
1
20. Refer to Exhibit 10–3. The standard error of is
a.
12.9
b.
9.3
c.
4
d.
2
d
1
21. Refer to Exhibit 10–3. The 95% confidence interval for the difference between the two population means is
a.
-9.92 to -2.08
b.
-3.92 to 3.92
c.
-13.84 to 1.84
d.
-24.228 to 12.23
a
1
22. Refer to Exhibit 10–3. The test statistic for the difference between the two population means is
a.
-.47
b.
-.65
c.
-1.5
d.
-3
d
1
23. Refer to Exhibit 10–3. The p-value for the difference between the two population means is
a.
.0014
b.
.0028
c.
.4986
d.
.9972
b
1
24. Refer to Exhibit 10–3. What is the conclusion that can be reached about the difference in the average final examination
scores between the two classes? (Use a .05 level of significance.)
a.
There is a statistically significant difference in the average final examination scores between the two classes.
b.
There is no statistically significant difference in the average final examination scores between the two classes.
c.
It is impossible to make a decision on the basis of the information given.
d.
There is a difference, but it is not significant.
a
1
Exhibit 10-4
The following information was obtained from independent random samples.
Assume normally distributed populations with equal variances.
Sample 1
Sample 2
Sample Mean
45
42
Sample Variance
85
90
Sample Size
10
12
25. Refer to Exhibit 10–4. The point estimate for the difference between the means of the two populations is
a.
0
b.
2
c.
3
d.
15
c
1
26. Refer to Exhibit 10–4. The standard error of is
a.
3.0
b.
4.0
c.
8.372
d.
19.48
b
1
27. Refer to Exhibit 10–4. The degrees of freedom for the t-distribution are
a.
22
b.
21
c.
20
d.
19
d
1
28. Refer to Exhibit 10–4. The 95% confidence interval for the difference between the two population means is
a.
-5.372 to 11.372
b.
-5 to 3
c.
-4.86 to 10.86
Chapter 10 – Comparisons Involving Means, Experimental Design, and Analysis of Variance
d.
-2.65 to 8.65
a
1
Exhibit 10-5
The following information was obtained from matched samples.
Individual
Method 1
Method 2
1
7
5
2
5
9
3
6
8
4
7
7
5
5
6
29. Refer to Exhibit 10–5. The point estimate for the difference between the means of the two populations is
a.
-1
b.
0
c.
1
d.
2
a
1
30. Refer to Exhibit 10–5. The 95% confidence interval for the difference between the two population means is
a.
-3.776 to 1.776
b.
-2.776 to 2.776
c.
-1.776 to 2.776
d.
0 to 3.776
a
1
31. Refer to Exhibit 10–5. If the null hypothesis is tested at the 5% level, the null hypothesis
a.
should be rejected
b.
should not be rejected
c.
should be revised
d.
None of these alternatives is correct.
b
1
32. Refer to Exhibit 10–5. The null hypothesis tested is H0: μd = 0. The test statistic for the difference between the two
population means is
a.
2
b.
0
c.
-1
d.
-2
c
1
Chapter 10 – Comparisons Involving Means, Experimental Design, and Analysis of Variance
Exhibit 10-6
The management of a department store is interested in estimating the difference between the mean credit purchases of
customers using the store’s credit card versus those customers using a national major credit card. You are given the
following information.
Store’s Card
Major Credit Card
Sample size
64
49
Sample mean
$140
$125
Population standard deviation
$10
$8
33. Refer to Exhibit 10–6. A point estimate for the difference between the mean purchases of the users of the two credit
cards is
a.
2
b.
18
c.
265
d.
15
d
1
34. Refer to Exhibit 10–6. At 95% confidence, the margin of error is
a.
1.694
b.
3.32
c.
1.96
d.
15
1
35. Refer to Exhibit 10–6. A 95% confidence interval estimate for the difference between the average purchases of the
customers using the two different credit cards is
a.
49 to 64
b.
11.68 to 18.32
c.
125 to 140
d.
8 to 10
b
1
Exhibit 10-7
In order to estimate the difference between the average hourly wages of employees of two branches of a department store,
the following data have been gathered.
Downtown Store
North Mall Store
Sample size
25
20
Sample mean
$9
$8
Sample standard deviation
$2
$1
36. Refer to Exhibit 10–7. A point estimate for the difference between the two sample means is
a.
1
b.
2
Chapter 10 – Comparisons Involving Means, Experimental Design, and Analysis of Variance
c.
3
d.
4
a
1
37. Refer to Exhibit 10–7. A 95% interval estimate for the difference between the two population means is
a.
0.078 to 1.922
b.
1.922 to 2.078
c.
1.09 to 4.078
d.
1.078 to 2.922
a
1
Exhibit 10-8
In order to determine whether or not there is a significant difference between the hourly wages of two companies, the
following data have been accumulated.
Company A
Company B
Sample size
80
60
Sample mean
$6.75
$6.25
Population standard deviation
$1.00
$0.95
38. Refer to Exhibit 10–8. A point estimate for the difference between the two sample means is
a.
20
b.
0.50
c.
0.25
d.
1.00
b
1
39. Refer to Exhibit 10–8. The test statistic is
a.
0.098
b.
1.645
c.
2.75
d.
3.01
d
1
40. Refer to Exhibit 10–8. The p-value is
a.
0.0013
b.
0.0026
c.
0.0042
d.
0.0084
b
1
41. Refer to Exhibit 10–8. The null hypothesis
Chapter 10 – Comparisons Involving Means, Experimental Design, and Analysis of Variance
a.
should be rejected
b.
should not be rejected
c.
should be revised
d.
None of these alternatives is correct.
a
1
Exhibit 10-9
Two major automobile manufacturers have produced compact cars with the same size engines. We are interested in
determining whether or not there is a significant difference in the MPG (miles per gallon) of the two brands of
automobiles. A random sample of eight cars from each manufacturer is selected, and eight drivers are selected to drive
each automobile for a specified distance. The following data show the results of the test.
Driver
Manufacturer A
Manufacturer B
1
32
28
2
27
22
3
26
27
4
26
24
5
25
24
6
29
25
7
31
28
8
25
27
42. Refer to Exhibit 10–9. The mean for the differences is
a.
0.50
b.
1.5
c.
2.0
d.
2.5
c
1
43. Refer to Exhibit 10–9. The test statistic is
a.
1.645
b.
1.96
c.
2.096
d.
2.256
d
1
44. Refer to Exhibit 10–9. At 90% confidence the null hypothesis
a.
should not be rejected
b.
should be rejected
c.
should be revised
d.
None of these alternatives is correct.
b
1
Exhibit 10–10
Chapter 10 – Comparisons Involving Means, Experimental Design, and Analysis of Variance
In order to determine whether or not there is a significant difference between the hourly wages of two companies, the
following data have been accumulated.
Company 1
Company 2
n1 = 80
n2 = 60
= $10.80
= = $10.00
= $2.00
= $1.50
45. Refer to Exhibit 10–10. The null hypothesis for this test is
a.
μ1 – μ2 ≠ 0
b.
μ1 – μ2 > 0
c.
μ1 – μ2 < 0
d.
μ1 – μ2 = 0
46. Refer to Exhibit 10–10. The point estimate of the difference between the means is
a.
20
b.
0.8
c.
0.50
d.
–20
47. Refer to Exhibit 10–10. The test statistic has a value of
a.
1.96
b.
1.645
c.
0.80
d.
2.7
48. Refer to Exhibit 10–10. The p-value is
a.
0.0035
b.
0.007
c.
0.4965
d.
1.96
49. In an analysis of variance problem if SST = 120 and SSTR = 80, then SSE is
a.
200
b.
40
c.
80
Chapter 10 – Comparisons Involving Means, Experimental Design, and Analysis of Variance
d.
120
50. In the analysis of variance procedure (ANOVA), factor refers to
a.
the dependent variable
b.
the independent variable
c.
different levels of a treatment
d.
the critical value of F
51. In an analysis of variance problem involving 3 treatments and 10 observations per treatment, SSE = 399.6. The MSE
for this situation is
a.
133.2
b.
13.32
c.
14.8
d.
30.0
c
52. When an analysis of variance is performed on samples drawn from k populations, the mean square between treatments
(MSTR) is
a.
SSTR/nT
b.
SSTR/(nT – 1)
c.
SSTR/k
d.
SSTR/(k – 1)
e.
None of these alternatives is correct.
53. In an analysis of variance where the total sample size for the experiment is nT and the number of populations is k, the
mean square within treatments is
a.
SSE/(nT – k)
b.
SSTR/(nT – k)
c.
SSE/(k – 1)
d.
SSE/k
a
54. The F ratio in a completely randomized ANOVA is the ratio of
a.
MSTR/MSE
b.
MST/MSE
c.
MSE/MSTR
Chapter 10 – Comparisons Involving Means, Experimental Design, and Analysis of Variance
d.
MSE/MST
a
55. The critical F value with 6 numerator and 60 denominator degrees of freedom at α = .05 is
a.
3.74
b.
2.25
c.
2.37
d.
1.96
56. The ANOVA procedure is a statistical approach for determining whether or not
a.
the means of two samples are equal
b.
the means of two or more samples are equal
c.
the means of more than two samples are equal
d.
the means of two or more populations are equal
57. The variable of interest in an ANOVA procedure is called
a.
a partition
b.
a treatment
c.
either a partition or a treatment
d.
a factor
58. An ANOVA procedure is applied to data obtained from 6 samples where each sample contains 20 observations. The
degrees of freedom for the critical value of F are
a.
6 numerator and 20 denominator degrees of freedom
b.
5 numerator and 20 denominator degrees of freedom
c.
5 numerator and 114 denominator degrees of freedom
d.
6 numerator and 20 denominator degrees of freedom
c
59. In the ANOVA, treatment refers to
a.
experimental units
b.
different levels of a factor
c.
a factor
d.
applying antibiotic to a wound
60. The mean square is the sum of squares divided by
a.
the total number of observations
b.
its corresponding degrees of freedom
c.
its corresponding degrees of freedom minus one
d.
None of these alternatives is correct.
61. In factorial designs, the response produced when the treatments of one factor interact with the treatments of another in
influencing the response variable is known as
a.
main effect
b.
replication
c.
interaction
d.
None of these alternatives is correct.
c
62. An experimental design where the experimental units are randomly assigned to the treatments is known as
a.
factor block design
b.
random factor design
c.
completely randomized design
d.
None of these alternatives is correct.
c
63. The required condition for using an ANOVA procedure on data from several populations is that the
a.
the selected samples are dependent on each other
b.
sampled populations are all uniform
c.
sampled populations have equal variances
d.
sampled populations have equal means
c
64. An ANOVA procedure is used for data that was obtained from four sample groups each comprised of five
observations. The degrees of freedom for the critical value of F are
a.
3 and 20
b.
3 and 16
c.
4 and 17
d.
3 and 19
65. In ANOVA, which of the following is not affected by whether or not the population means are equal?
a.
b.
between-samples estimate of 2
Chapter 10 – Comparisons Involving Means, Experimental Design, and Analysis of Variance
c.
within-samples estimate of 2
d.
None of these alternatives is correct.
c
1
66. A term that means the same as the term “variable” in an ANOVA procedure is
a.
factor
b.
treatment
c.
replication
d.
variance within
a
1
67. In order to determine whether or not the means of two populations are equal,
a.
a t test must be performed
b.
an analysis of variance must be performed
c.
either a t test or an analysis of variance can be performed
d.
a chi-square test must be performed
c
1
68. The process of allocating the total sum of squares and degrees of freedom is called
a.
factoring
b.
blocking
c.
replicating
d.
partitioning
d
1
69. In a completely randomized design involving three treatments, the following information is provided:
Treatment 1
Treatment 2
Treatment 3
Sample Size
5
10
5
Sample Mean
4
8
9
The overall mean for all the treatments is
a.
7.00
b.
6.67
c.
7.25
d.
4.89
c
1
Exhibit 10–11
SSTR = 6,750
H0: μ1=μ2=μ3=μ4
SSE = 8,000
Ha: at least one mean is different
70. Refer to Exhibit 10–11. The mean square between treatments (MSTR) equals
a.
400
b.
500
c.
1,687.5
d.
2,250
71. Refer to Exhibit 10–11. The mean square within treatments (MSE) equals
a.
400
b.
500
c.
1,687.5
d.
2,250
72. Refer to Exhibit 10–11. The test statistic to test the null hypothesis equals
a.
0.22
b.
0.84
c.
4.22
d.
4.5
73. Refer to Exhibit 10–11. The null hypothesis is to be tested at the 5% level of significance. The critical value from the
table is
a.
2.87
b.
3.24
c.
4.08
d.
8.7
74. Refer to Exhibit 10–11. The null hypothesis
a.
should be rejected
b.
should not be rejected
c.
was designed incorrectly
d.
None of these alternatives is correct.
a
Exhibit 10–12
Chapter 10 – Comparisons Involving Means, Experimental Design, and Analysis of Variance
To test whether or not there is a difference between treatments A, B, and C, a sample of 12 observations has been
randomly assigned to the 3 treatments. You are given the results below.
Treatment
Observation
A
20
30
25
33
B
22
26
20
28
C
40
30
28
22
75. Refer to Exhibit 10–12. The null hypothesis for this ANOVA problem is
a.
μ1=μ2
b.
μ1=μ2=μ3
c.
μ1=μ2=μ3=μ4
d.
μ1=μ2= … =μ12
76. Refer to Exhibit 10–12. The mean square between treatments (MSTR) equals
a.
1.872
b.
5.86
c.
34
d.
36
77. Refer to Exhibit 10–12. The mean square within treatments (MSE) equals
a.
1.872
b.
5.86
c.
34
d.
36
c
78. Refer to Exhibit 10–12. The test statistic to test the null hypothesis equals
a.
0.944
b.
1.059
c.
3.13
d.
19.231
79. Refer to Exhibit 10–12. The null hypothesis is to be tested at the 1% level of significance. The critical value from the
table is
a.
4.26
b.
8.02
c.
16.69
Chapter 10 – Comparisons Involving Means, Experimental Design, and Analysis of Variance
d.
99.39
b
1
80. Refer to Exhibit 10–12. The null hypothesis
a.
should be rejected
b.
should not be rejected
c.
should be revised
d.
None of these alternatives is correct.
b
1
Exhibit 10–13
In a completely randomized experimental design involving five treatments, thirteen observations were recorded for each
of the five treatments. The following information is provided.
SSTR = 200 (Sum Square Between Treatments)
SST = 800 (Total Sum Square)
81. Refer to Exhibit 10–13. The sum of squares within treatments (SSE) is
a.
1,000
b.
600
c.
200
d.
1,600
1
82. Refer to Exhibit 10–13. The number of degrees of freedom corresponding to between treatments is
a.
60
b.
59
c.
5
d.
4
d
1
83. Refer to Exhibit 10–13. The number of degrees of freedom corresponding to within treatments is
a.
60
b.
59
c.
5
d.
4
a
1
84. Refer to Exhibit 10–13. The mean square between treatments (MSTR) is
a.
3.34
b.
10.00
Chapter 10 – Comparisons Involving Means, Experimental Design, and Analysis of Variance
c.
50.00
d.
12.00
c
85. Refer to Exhibit 10–13. The mean square within treatments (MSE) is
a.
50
b.
10
c.
200
d.
600
86. Refer to Exhibit 10–13. If at a 5% level of significance we want to determine whether or not the means of the five
populations are equal, the critical value of F is
a.
2.53
b.
19.48
c.
4.98
d.
39.48
a
87. Refer to Exhibit 10–13. The conclusion of the test is that the five means
a.
are equal
b.
may be equal
c.
are not equal
d.
None of these alternatives is correct.
c
Exhibit 10–14
Part of an ANOVA table is shown below.
Source of Variation
Sum of
Squares
Degrees of
Freedom
Mean
Square
F
Between Treatments
180
3
Within Treatments (Error)
Total
480
18
88. Refer to Exhibit 10–14. The mean square between treatments (MSTR) is
a.
20
b.
60
c.
300
d.
15
89. Refer to Exhibit 10–14. The mean square within treatments (MSE) is
a.
60
b.
15
c.
300
d.
20
d
1
90. Refer to Exhibit 10–14. If at a 5% level of significance, we want to determine whether or not the means of the
populations are equal, the critical value of F is
a.
2.53
b.
19.48
c.
3.29
d.
5.86
c
1
91. Refer to Exhibit 10–14. The conclusion of the test is that the means
a.
are equal to fifty
b.
may be equal
c.
are not equal
d.
None of these alternatives is correct.
b
1
Exhibit 10–15
Part of an ANOVA table is shown below.
Source of Variation
Sum of
Squares
Degrees of
Freedom
Mean
Square
F
Between Treatments
64
8
Within Treatments (Error)
2
Total
100
92. Refer to Exhibit 10–15. The number of degrees of freedom corresponding to between treatments is
a.
18
b.
2
c.
4
d.
3
c
1
93. Refer to Exhibit 10–15. The number of degrees of freedom corresponding to within treatments is
a.
22
1