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York University

Administrative Studies

ADMS 2320

Marwan Hassan

Winter

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CHAPTER 13
INFERENCE ABOUT COMPARING
TWO POPULATIONS
MULTIPLE CHOICE QUESTIONS
In the following multiple-choice questions, please circle the correct answer.
TEST QUESTIONS
76. In testing the hypotheses
H 0 μ 1 μ 2 25
H 1 μ1− μ 2 25
two random samples from two normal populations produced the following
results:
n = 42 x = 215 s = 15
1 1 1
n2= 32 x2= 180 s2= 20
What conclusion can we draw at the 5% significance level?
ANSWER:
Rejection region: t >0.05,72.667
Test statistic: t = 2.459
Conclusion: Reject the null hypotheses
QUESTIONS 77 AND 78 ARE BASED ON THE FOLLOWING INFORMATION:
In random samples of 40 from each of two normal populations, the following
statistics were
227 228 Chapter Thirteen
obtained:
x1= 76 s1= 8
x = 72 s = 6.5
2 2
77. Test at the 5% significance level to determine whether we can infer that the
two population means differ.
ANSWER:
H 0 μ 1 μ =20
H 1 μ 1 μ ≠20
Rejection region: |t| > t0.025,88.987
Test Statistic: t = 2.454
Conclusion: Reject the null hypothesis. Yes
78. Estimate with 95% confidence the difference between the two population
means.
ANSWER:
4 ±3.238= (0.762,0.7238 )
79. In testing the hypotheses
H 0 μ D = 5
H 1 μ D 5
two random samples from two normal populations produced the following
statistics:
nD = 20 xD = 9 s D = 7.5
What conclusion can we draw at the 1% significance level?
ANSWER:
Rejection region: t > t = 2.539
0.01,19
Test statistic: t = 2.385
Conclusion: Don’t reject the null hypothesis
QUESTIONS 80 AND 81 ARE BASED ON THE FOLLOWING INFORMATION:
In testing the hypotheses
H 0 p 1 p =20
H 1 p 1 p <20 , Inference About Comparing Two Populations 229
we found the following statistics:
n1= 400 , x1= 105 ,
n = 500 , x = 140
2 2
80. What conclusion can we draw at the 10% significance level?
ANSWER:
Rejection region: z < −z = −1.28
0.10
Test statistic: z = -0.587
Conclusion: Don’t reject the null hypothesis
81. Estimate with 90% confidence the difference between the two population
proportions.
ANSWER:
-0.0175 0.049 = (-0.0665, 0.0315)
±
QUESTIONS 82 AND 83 ARE BASED ON THE FOLLOWING INFORMATION:
In testing the hypotheses
H 0 μ 1 μ =20
H 1 μ 1 μ 2 0,
the following statistics were obtained:
n = 10 x = 58.6 s = 13.45
1 1 1
n2= 10 x2= 64.6 s = 21.15
82. Test the following hypotheses at the 5% level of significance
ANSWER:
Rejection region: t < −t0.05,18−1.734
Test statistic: t = -1.086
Conclusion: Don’t reject the null hypothesis
83. Estimate with 95% confidence the difference between the two population
means. 230 Chapter Thirteen
ANSWER:
-6.0 ± 11.6074
QUESTIONS 84 THROUGH 86 ARE BASED ON THE FOLLOWING INFORMATION:
In testing the hypotheses
H : p − p = 0.10
0 1 2
H : p − p < 0.10 ,
1 1 2
we found the following statistics.
n1= 400 x =1208
n 2 250 x =2115
84. What conclusion can we draw at the 10% significance level?
ANSWER:
Rejection region: z < −z0.10 −1.28
Test statistic: z = -0.9926
Conclusion: Don’t reject the null hypothesis
85. What is the p-value of the test in Question 84?
ANSWER:
p-value = 0.1611
86. Estimate with 90% confidence the difference between the two population
proportions.
ANSWER:
0.06 ± 0.0662 = (-0.0062 , 0.1262)
87. Random samples from two normal populations produced the following
statistics:
n = 10 2
1 s1= 32
n = 15 s = 22
2 2
Is there enough evidence at the 5% significance level to infer that the
variance of population 1 is larger than the variance of population 2?
ANSWER:
2 2
H 0σ /1 = 2
2 2 1
H:1 /σ1> 2 Inference About Comparing Two Populations 231
Rejection region: F >F0.05,9,14.65
Test statistics: F = 1.455
Conclusion: Don’t reject the null hypothesis. No
QUESTIONS 88 THROUGH 90 ARE BASED ON THE FOLLOWING INFORMATION:
In testing the hypotheses
H 0 μ1− μ2= 0
H : μ − μ ≠ 0 ,
1 1 2
two random samples from two normal populations produced the following statistics:
n1= 50 x1= 35 s1= 28
n2= 50 x2= 30 s2= 10
88. What conclusion can we draw at the 5% significance level?
ANSWER:
There is reason to believe that the population variances are unequal.
Rejection region: |t| >0.025,61.0
Test statistic: t = 1.189
Conclusion: Don’t reject the null hypotheses
89. Estimate with 95% confidence the difference between the two population
means.
ANSWER:
5.0± 8.409 = (-3.409, 13.409)
90. Explain how to use the confidence interval in Question 89 for testing the
hypotheses.
ANSWER:
Since the hypothesized value μ 0 0 is included in the 95% confidence
interval, we fail to reject the null hypothesis at= 0.05.
QUESTIONS 91 THROUGH 93 ARE BASED ON THE FOLLOWING INFORMATION:
The following data were generated from a matched pairs experiment.
Pair: 1 2 3 4 5 6 7
Sample 1: 8 15 7 9 10 13 11
Sample 2: 12 18 8 9 12 11 10
91. Determine whether these data are sufficient to infer at the 10% significance
level that the two population means differ. 232 Chapter Thirteen
ANSWER:
H 0 μ D = 0
H : μ ≠ 0
1 D
Rejection region: |t| > t0.05,61.943
Test Statistics: t = -1.225
Conclusion: Don’t reject the null hypothesis. No
92. Estimate with 90% confidence the mean difference.
ANSWER:
–1.0 ± 1.587 = (-2.587, 0.587)
93. Briefly describe what the interval estimate in Question 92 tells you, and
explain how to use it to test the hypotheses.
ANSWER:
We estimate that the mean difference is between –2.587 and 0.587. Since
the hypothesized value 0 is included in the 90% interval estimate, we fail to
reject the null hypothesis at α = 0.10.
QUESTIONS 94 THROUGH 98 ARE BASED ON THE FOLLOWING INFORMATION:
In testing the hypotheses
H 0 p1− p 2 0.10
H : p − p ≠ 0.10
1 1 2
we found the following statistics.
n1= 150 x1= 72
n2= 175 x =270
94. What conclusion can we draw at the 5% significance level?
ANSWER:
Rejection region: |z| > z0.051.96
Test statistic: z = -0.36
Conclusion: Don’t reject the null hypothesis Inference About Comparing Two Populations 233
95. What is the p-value of the test?
ANSWER:
p-value = 0.7188 …(actually 0.7166)
96. Briefly explain how to use the p-value to test the hypotheses.
ANSWER:
Since p-value = 0.147 > α = 0.05, we fail to reject the null hypothesis
97. Estimate with 95% confidence the difference between the two population
proportions.
ANSWER:
0.08 ± 0.107 = (-0.027, 0.117)
98. Explain how to use the confidence interval in Question 97 to test the
hypotheses.
ANSWER:
We estimate that the difference between the population proportions lies
between –0.028 and 0.118. Since the hypothesized value 0 is included in the
95% interval estimate, we fail to reject the null hypothesis at = 0.05.
QUESTIONS 99 THROUGH 102 ARE BASED ON THE FOLLOWING INFORMATION:
Random samples from two normal populations produced the following statistics:
n1= 16 s1= 55
2
n2= 14 s2= 118
99. Is there enough evidence at the 10% significance level to infer that the two
population variances differ?
ANSWER:
2 2
H 0σ 1σ =2 1
H:σ /σ ≠ 2 1
1 1 2
Rejection region: F >F = 2.53 or F < F =1/ F ≈0.403
0.05,15,13 0.95,15,13 0.05,13,15
Test statistics: F = 0.466
Conclusion: Don’t reject the null hypothesis. No 234 Chapter Thirteen
100. Estimate with 90% confidence the ratio of the two population variances.
ANSWER:
2 2
LCL = ( s / s2 ) /F 0.05,15,131842
1
2 2 F
UCL = ( s1/ s2 ) . 0.05,13,15.1566
101. Briefly describe what the interval estimate in Question 100 tells you.
ANSWER:
We estimate that ( σ 1σ 2) lies between 0.1842 and 1.1566.
102. Briefly explain how to use the interval estimate in Question 100 to test the
hypotheses.
ANSWER:
Since the hypothesized value 1 is included in the 90% interval estimate, we
fail to reject
the null hypothesis at α = 0.10.
QUESTIONS 103 THROUGH 106 ARE BASED ON THE FOLLOWING INFORMATION:
In testing the hypotheses
H : p − p = 0
0 1 2
H1: p1− p2> 0 ,
we found the following statistics:
n 1 200 x,= 10 , n 2 400 , x2= 140
103. What conclusion can we draw at the 5% significance level?
ANSWER:
Rejection region: z > 0.05 1.645
Test statistic: z = 1.199
Conclusion: Don’t reject the null hypothesis
104. What is the p-value of the test?
ANSWER:
p-value = 0.1151
105. Briefly explain how to use the p-value for testing the hypotheses.
ANSWER:
Since p-value = 0.1151 > α = 0.05, we fail to reject the null hypothesis. Inference About Comparing Two Populations 235
106. Estimate with 95% confidence the difference between the two population
proportions.
ANSWER:
0.05 ± 0.0824 = (-0.0324, 0.1324)
107. In testing the hypotheses
H 0 p 1 p =20
H : p − p ≠ 0
1 1 2
we found the following statistics.
n1= 60 x1= 36
n2= 80 x2= 52
What conclusion can we draw at the 1% significance level?
ANSWER:
Rejection region: |z| > z0.0052.575
Test statistic: z = -0.61
Conclusion: Don’t reject the null hypothesis 236 Chapter Thirteen
QUESTIONS 108 THROUGH 110 ARE BASED ON THE FOLLOWING INFORMATION:
In order to test the hypotheses
H 0 μ1− μ2= 0
H : μ − μ ≠ 0
1 1 2 ,
we independently draw a random sample of 18 observations from a normal
population with standard deviation of 15, and another random sample of 12 from
a second normal population with standard deviation of 25.
108. If we set the level of significance at 10%, determine the power of the test
μ − μ = 5
when 1 2
ANSWER:
Power = 1 - β = 0.1631
109. Re-do Question 108 if the level of significance is reduced from 10% to 5%.
ANSWER:
Power = 1 - β = 0.095
110. Describe the effect of reducing the level of significance on the power of the
test.
ANSWER:
As the level of significance decreases from 10% to 5%, the power of the test
decreases from 0.1631 to 0.095.
QUESTIONS 111 THROUGH 113 ARE BASED ON THE FOLLOWING INFORMATION:
The general manager of a chain of fast food chicken restaurants wants to determine
how effective their promotional campaigns are. In these campaigns “20% off”
coupons are widely distributed. These coupons are only valid for one week. To
examine their effectiveness, the executive records the daily gross sales (in $1,000s)
in one restaurant during the campaign and during the week after the campaign
ends. The data is shown below. Can they infer at the 5% significance level that
sales increase during the campaign?
Day Sales During Sales After Campaign
Campaign
Sunday 18.1 16.6
Monday 10.0 8.8
Tuesday 9.1 8.6
Wednesday 8.4 8.3
Thursday 10.8 10.1
Friday 13.1 12.3
Saturday 20.8 18.9 Inference About Comparing Two Populations 237
ANSWER:
H 0 μ D = 0
H : μ > 0
1 D
Rejection region: t > 1.943
Test statistic: t = 4.111
Conclusion: Reject the null hypothesis. Yes
112. Estimate with 95% confidence the mean difference.
ANSWER:
0.957 ± 0.57 = (0.387, 1.527)
113. Briefly explain what the interval estimate in Question 112 tells you.
ANSWER:
We estimate that the daily sales during the campaign increase on average
between 0.387 and 1.527 thousand dollars
QUESTIONS 114 THROUGH 116 ARE BASED ON THE FOLLOWING INFORMATION:
A survey of 1,500 Canadians reveals that 945 believe that there is too much
violence on television. In a survey of 1,500 Americans, 810 believe that there is too
much television violence.
114. Can we infer at the 99% significance level that the proportion of Canadians
and Americans who believe that there is too much violence on television
differ?
ANSWER:
H 0 p 1 p 2 0
H 1 p1− p 2 0
Rejection region: |z| > z = 2.575
0.005
Test statistic: z = 5.0
Conclusion: Reject the null hypothesis. Yes
115. Estimate with 99% confidence the difference in the proportion of Canadians
and Americans who believe that there is too much violence on television.
ANSWER:
0.09 ± 0.0461 = (0.0439, 0.1361) 238 Chapter Thirteen
116. Briefly explain what the interval estimate in Question 115 tells you.
ANSWER:
We estimate that the proportion of Canadians who believe that there is too
much violence on television is between 4.39% and 13.61% higher than the
proportion of Americans who share the same view.
QUESTIONS 117 THROUGH 120 ARE BASED ON THE FOLLOWING INFORMATION:
A statistician wants to test for the equality of means in two independent samples
drawn from normal populations. However, he will not perform the equal-variance t-
test of the difference between the population means if the condition necessary for
its use is not satisfied. The data follow:
Sample 1: 7 9 6 15 7 10 8 12
Sample 2: 2 25 9 15 10 18 5 22 27 3
117. Given the data above, can the statistician conclude at the 5% significance
level that the required condition is not satisfied?
ANSWER:
H0:σ 1σ 2 1
2 2 1
H:1 /1 ≠ 2
Rejection region: F >F 0.025,7,920 or F < F0.975,7,9F0.025,9,7207
Test statistics: F = 0.108
Conclusion: Reject the null hypothesis. Yes
118. Estimate with 95% confidence the ratio of the two population variances.
ANSWER:
LCL = ( 2 s2 ) /F = 0.0257
s1/ 2 0.025,7,9
2
UCL = ( s / s2) .F0.025,9,7.5214
1
119. Briefly describe what the interval estimate in Question 118 tells you.
ANSWER:
We estimate that ( σ 1σ 2) lies between 0.0257 and 0.5214.
120. Briefly explain how to use the interval estimate in Question 118 to test the
hypotheses.
ANSWER:
Since the hypothesized value 1 is not included in the 95% interval estimate,
we reject the null hypothesis at α = 0.05. Inference About Comparing Two Populations 239
121. The owner of a service station wants to determine if owners of new cars (two
years old or less) change their cars’ oil more frequently than owners of older
cars (more than two years old). From his records he takes a random sample of
ten new cars and ten older cars and determines the number of times the oil
was changed in the last 12 months. The data follow. Do these data allow the
service station owner to infer at the 10% significance level that new car
owners change their cars’ oil more frequently than older car owners?
Frequency of Oil Changes in Past 12 Months
New Car Owners Old Cars Owners
6 4
3 2
3 1
3 2
4 3
3 2
6 2
5 3
5 2
4 1
ANSWER:
H0: μ1− μ2= 0
H1: μ1− μ2> 0
Rejection region: t > t =1.33
0.10,18
Test statistic: t = 2.914
Conclusion: Reject the null hypothesis. Yes
122. Because of the rising costs of industrial accidents, many chemical, mining, and
manufacturing firms have instituted safety courses. Employees are
encouraged to take these courses designed to heighten safety awareness. A
company is trying to decide which one of two courses to institute. To help
make a decision eight employees take course 1 and another eight take course
2. Each employee takes a test, which is graded out of a possible 25. The safety
test results are shown below. Assume that the scores are normally distributed.
Do these data provide sufficient evidence at the 5% level of significance to
infer that the marks from course 1 are lower than those of course 2?
Course 1 14 21 17 14 17 19 20 16
Course 2 20 18 22 15 23 21 19 15
ANSWER:
H0: μ1− μ2= 0
H1: μ1− μ2< 0
Rejection region: t < -0.05,141.761
Test statistic: t = -1.336
Conclusion: Don’t reject the null hypothesis. No 240 Chapter Thirteen
QUESTIONS 123 THROUGH 125 ARE BASED ON THE FOLLOWING INFORMATION:
Ten functionally illiterate adults were given an experimental one-week crash course
in reading. Each of the ten adults was given a reading test prior to the course and
another test after the course. The results are shown below.
Adult 1 2 3 4 5 6 7 8 9 10
Score after course 48 42 43 34 50 30 43 38 41 38
Score before course 31 34 18 30 44 28 34 33 27 32
123. Is there enough evidence to infer at the 5% significance level that the
reading scores have improved?
ANSWER:
H0: μD = 0
H :μ > 0
1 D
Rejection region: t >t0.05,9.833
Test statistic: t = 3.836
Conclusion: Reject the null hypothesis. Yes
124. Estimate the mean improvement with 95% confidence.
ANSWER:
8.6 ± 5.072 = (3.528, 13.672)
125. Briefly describe what the interval estimate in Question 124 tells you.
ANSWER:
We estimate that the scores after taking the course improve on the average
between 3.528 and 13.672 points.
126. A politician regularly polls her constituency to gauge her level of support
among voters. This month, 652 o

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