Chapter 10
Statistical Inferences About Two Populations
LEARNING OBJECTIVES
The general focus of Chapter 10 is on testing hypotheses and constructing confidence intervals about
parameters from two populations, thereby enabling you to
1.
2.
3.
4.
5.
Test hypotheses and construct confidence intervals about the difference in two population means
using the z statistic.
Test hypotheses and establish confidence intervals about the difference in two population means
using the t statistic.
Test hypotheses and construct confidence intervals about the difference in two related populations
when the differences are normally distributed.
Test hypotheses and construct confidence intervals about the difference in two population
proportions.
Test hypotheses and construct confidence intervals about two population variances.
CHAPTER OUTLINE
10.1 Hypothesis Testing and Confidence Intervals about the Difference in Two Means using the z
Statistic: Population Variances Known
Hypothesis Testing
Confidence Intervals
Using the Computer to Test Hypotheses about the Difference in Two Population Means
Using the z Test
10.2 Hypothesis Testing and Confidence Intervals about the Difference in Two Means:
Independent Samples and Population Variances Unknown
Hypothesis Testing
Using the Computer to Test Hypotheses and Construct Confidence Intervals about the
Difference in Two Population Means Using the t Test
Confidence Intervals
10.3 Statistical Inferences For Two Related Populations
Hypothesis Testing
Using the Computer to Make Statistical Inferences about Two Related Populations
Confidence Intervals
10.4 Statistical Inferences About Two Population Proportions, p1 – p 2
Hypothesis Testing
Confidence Intervals
Using the Computer to Analyze the Difference in Two Proportions
10.5 Testing Hypotheses About Two Population Variances
Using the Computer to Test Hypotheses about Two Population Variances
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KEY WORDS
dependent samples
F distribution
F value
independent samples
matched-pairs test
related measures
STUDY QUESTIONS
1.
A researcher wants to estimate the difference in the means of two populations. A random sample of
40 items from the first population results in a sample mean of 433 with a population standard
deviation of 112. A random sample of 50 items from the second population results in a sample
mean of 467 with a population standard deviation of 120. From this information, a point estimate of
the difference of population means can be computed as _______________.
2.
Using the information from question 1, the researcher can compute a 95% confidence interval to
estimate the difference in population means. The resulting confidence interval is
_________________________.
3.
A random sample of 32 items is taken from a population which has a population variance of 93.
The resulting sample mean is 45.6. A random sample of 37 items is taken from a population which
has a population variance of 88. The resulting sample mean is 49.4. Using this information, a 98%
confidence interval can be computed to estimate the difference in means of these two populations.
The resulting interval is _________________________.
4.
A researcher desires to estimate the difference in means of two populations. To accomplish this,
he/she takes a random sample of 85 items from the first population. The sample yields a mean of
168 with a variance of 783. A random sample of 70 items is taken from the second population
yielding a mean of 161 with a population variance of 780. A 94% confidence interval is computed
to estimate the difference in population means. The resulting confidence interval is
_______________.
5.
Is there a difference in the average number years of experience of assembly line employees between
company A and company B? A researcher wants to conduct a statistical test to answer this question.
He is likely to be conducting a _______________-tailed test.
6.
The researcher who is conducting the test to determine if there is a difference in the average number
of years of experience of assembly line workers between companies A and B is using an alpha of
.10. The critical value of z for this problem is __________.
7.
Suppose the researcher conducting an experiment to compare the ages of workers at two
companies. The researcher randomly samples forty-five assembly-line workers from company A
and discovers that the sample average is 7.1 years with a population standard deviation of 2.3.
Fifty-two assembly-line workers from company B are randomly selected resulting in a sample
average of 6.2 years and a population standard deviation of 2.7. The observed z value for this
problem is _______________.
8.
Using an alpha of .10 and the critical values determined in questions 6 and 7, the decision is to
_______________ the null hypothesis.
Chapter 10: Statistical Inferences About Two Populations
9.
179
A researcher has a theory that the mean for population A is less than the mean for population B. To
test this, she randomly samples thirty-eight items from population A and determines that the sample
average is 38.4 with a population variance of 50.5. She randomly samples thirty-two items from
population B and determines that the sample average is 44.3 with a population variance of 48.6.
Alpha is .05. She is going to conduct a _______________-tailed test.
10.
Using the information from question 9, the critical z value is _______________.
11.
Using the information from question 9, the observed value of z is _______________.
12.
Using the results determined in question 10 and 11, the decision is to _______________ the null
hypothesis.
13.
A researcher is interested in testing to determine if the mean of population one is greater than the
mean of population two. He uses the following hypotheses to test this theory:
Ho: µ1 – µ2 = 0
Ha: µ1 – µ2 > 0
He randomly selects a sample of 8 items from population one resulting in a mean of 14.7 and a
standard deviation of 3.4. He randomly selects a sample of 12 items from population two resulting
in a mean of 11.5 and a standard deviation 2.9. He is using an alpha value of .10 to conduct this
test. The degrees of freedom for this problem are _____________. It is assumed that these values
are normally distributed in both populations.
14.
The critical table t value used to conduct the hypothesis test in question 13 is _______________.
15.
The t value calculated from the sample data is ______.
16.
Based on the observed t value obtained in question 15 and the critical table t value in question 14,
the researcher should _______________ the null hypothesis.
17.
What is the difference in the means of two populations? A researcher wishes to determine this by
taking random samples of size 14 from each population and computing a 90% confidence interval.
The sample from the first population produces a mean of 780 with a standard deviation of 245. The
sample from the second population produces a mean of 890 with a standard deviation of 256. The
point estimate for the difference in the means of these two populations is _______________.
Assume that the values are normally distributed in each population.
18.
The table t value used to construct the confidence interval for the problem in question 17 is
__________.
19.
The confidence interval constructed for the problem in question 17 is __________.
20.
The matched-pairs t test deals with _______________ samples.
21.
A researcher wants to conduct a before/after study on 13 subjects to determine if a treatment results
in higher scores. The hypotheses are:
Ho: D = 0
Ha: D < 0
Scores are obtained on the subjects both before and after the treatment. After subtracting the after
scores from the before scores, the resulting value of d is –2.85 with a Sd of 1.01. The degrees of
freedom for this test are _______________. Assume that the data are normally distributed in the
population.
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Solutions Manual and Study Guide
22.
The critical table t value for the problem in question 21 is _______________ if  = .01.
23.
The observed t value for the problem in question 21 is _______________.
24.
For the problem in question 21 based on the critical table t value obtained in question 22 and the
observed t value obtained in question 23, the decision should be to _______________ the null
hypothesis.
25.
A researcher is conducting a matched-pairs study. She gathers data on each pair in the study
resulting in:
Pair
1
2
3
4
5
6
7
8
Group 1
10
13
11
14
12
12
10
8
Group 2
12
14
15
14
11
15
16
10
Assuming that the data are normally distributed in the population, the computed value of d
is _______________.
26.
The value of Sd for the problem in question 25 is _______________.
27.
The degrees of freedom for the problem in question 25 is _______________.
28.
The observed value of t for the problem in question 25 is _______________.
29.
A researcher desires to estimate the difference between two related populations. He gathers pairs of
data from the populations. The data are below:
Pair
1
2
3
4
5
6
Group 1
360
345
355
325
340
365
Group 2
280
290
300
270
300
310
It is assumed that the data are normally distributed in the population. Using this data, the value of
d is _______________.
30.
For the problem in 29, the value of Sd is __________.
31.
The point estimate for the population difference for the problem in question 29 is
_______________.
32.
The researcher conducting the study for the problem in question 29 wants to use a 95% level of
confidence. The table t value for this confidence interval is _______________.
33.
The confidence interval computed for the problem in question 29 is _______________.
Chapter 10: Statistical Inferences About Two Populations
181
34.
A researcher is interested in estimating the difference in two populations proportions. A sample of
1000 from each population results in sample proportions of .61 and .64. The point estimate of the
difference in the population proportions is _______________.
35.
Using the data from question 34, the researcher computes a 90% confidence interval to estimate the
difference in population proportions. The resulting confidence interval is
_________________________.
36.
A random sample of 400 items from a population shows that 110 of the sample items possess a
given characteristic. A random sample of 550 items from a second population resulted in 154 of the
sample items possessing the characteristic. Using this data, a 99% confidence interval is constructed
to estimate the difference in population proportions which possess the given characteristic. The
resulting confidence interval is _________________________.
37.
A researcher desires to estimate the difference in proportions of two populations. To accomplish
this, he/she samples 338 and 332 items respectively from each population. The resulting sample
proportions are .71 and .68 respectively. Using this data, a 90% confidence interval can be
computed to estimate the difference in population proportions. The resulting confidence interval is
_________________________.
38.
A statistician is being asked to test a new theory that the proportion of population A possessing a
given characteristic is greater than the proportion of population B possessing the characteristic. A
random sample of 625 from population A has been taken and it is determined that 463 possess the
characteristic. A random sample of 704 taken from population B results in 428 possessing the
characteristic. The alternative hypothesis for this problem is _______________.
39.
The observed value of z for question 38 is _______________.
40.
Suppose alpha is .10. The critical value of z for question 38 is _______________.
41.
Based on the results of question 39 and 40, the decision for the problem in question 38 is to
_______________ the null hypothesis.
42.
In testing hypotheses about two population variances, use the _______________________
distribution.
43.
Suppose we want to test the following hypothesis:
H0: 12 = 22 and Ha: 12 > 22
A sample of 9 items from population one yielded a sample standard deviation of 8.6. A sample of 8
items from population two yielded a sample standard deviation of 6.9. If alpha is .05, the critical F
value is ___________________________.
44.
The observed F value for question 45 is _______________________. The resulting decision is
_______________________.
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ANSWERS TO STUDY QUESTIONS
1. –34
23. –10.17
2. –82.07 < 1 – 2 < 14.07
24. Reject
3. –9.16 < 1 – 2 < 1.56
25. –2.125
4. –1.48 < 1 – 2 < 15.48
26. 2.232
5. Two
27. 7
6. + 1.645
28. –2.69
7. 1.77
29. 56.67
8. Reject
30. 12.91
9. One
31. 56.67
10. –1.645
32. 2.571
11. –3.50
33. 43.12 < D < 70.22
12. Reject
34. –.03
13. 18
35. –.066 < p1 – p2 < .006
14. 1.33
36. –.081 < p1 – p2 < .071
15. 2.26
37. –.0285 < p1 – p2 < .0885
16. Reject
38. pA – pB > 0
17. –110
39. 5.14
18. 1.706
40. 1.28
19. –271.56 < 1 – 2 < 51.56
41. Reject
20. Related
42. F
21. 12
43. 3.73
22. –2.681
44. 1.55, Fail to Reject the Null Hypothesis
Chapter 10: Statistical Inferences About Two Populations
183
SOLUTIONS TO ODD-NUMBERED PROBLEMS IN CHAPTER 10
10.1
a) Ho:
Ha:
Sample 1
Sample 2
x 1 = 51.3
12 = 52
x 2 = 53.2
22 = 60
n1 = 32
n2 = 32
µ1 – µ2 = 0
µ1 – µ2 < 0
For one-tail test,  = .10
z =
z.10 = –1.28
( x 1  x 2 )  ( 1   2 )
1
2
n1

2
2

(51.3  53.2)  (0)
= –1.02
52 60

32 32
n2
Since the observed z = –1.02 > zc = –1.645, the decision is to fail to reject the null hypothesis.
b) Critical value method:
zc =
( x 1  x 2 ) c  ( 1   2 )
 12
n1
–1.645 =

 22
n2
( x1  x 2 ) c  (0)
52 60

32 32
( x 1 – x 2)c = –3.08
c) The area for z = –1.02 using Table A.5 is .3461.
The p-value is .5000 – .3461 = .1539
184
10.3
Solutions Manual and Study Guide
a)
Sample 1
Sample 2
x 1 = 88.23
12 = 22.74
x 2 = 81.2
22 = 26.65
n1 = 30
n2 = 30
µ1 – µ2 = 0
µ1 – µ2  0
Ho:
Ha:
For two-tail test, use /2 = .01 z.01 = + 2.33
z =
( x 1  x 2 )  ( 1   2 )
 12
n1

 22
n2

(88.23  81.2)  (0)
= 5.48
22.74 26.65

30
30
Since the observed z = 5.48 > z.01 = 2.33, the decision is to reject the null hypothesis.
b)
( x1  x 2 )  z
 12
n1

(88.23 – 81.2) + 2.33
 22
n2
22.74 26.65

30
30
7.03 + 2.99
4.04 <  < 10.02
This supports the decision made in a) to reject the null hypothesis because zero is not in
the interval.
10.5
A
n1 = 40
x 1 = 5.3
12 = 1.99
B
n2 = 37
x 2 = 6.5
22 = 2.36
For a 95% C.I., z.025 = 1.96
( x1  x 2 )  z
 12
n1
(5.3 – 6.5) + 1.96
–1.2 ± .66

 22
n2
1.99 2.36

40
37
–1.86 <  < –.54
Chapter 10: Statistical Inferences About Two Populations
185
The results indicate that we are 95% confident that, on average, Plumber B does between 0.54
and 1.86 more jobs per day than Plumber A. Since zero does not lie in this interval, we are
confident that there is a difference between Plumber A and Plumber B.
10.7
1994
2001
x 1 = 190
1 = 18.50
x 2 = 198
2 = 15.60
n1 = 51
n2 = 47
H0: 1 – 2 = 0
Ha: 1 – 2 < 0
For a one-tailed test,
z =
z.01 = –2.33
( x 1  x 2 )  ( 1   2 )
 12
n1

 = .01
 22
(190  198)  (0)

(18.50) 2 (15.60) 2

51
47
n2
= –2.32
Since the observed z = –2.32 > z.01 = –2.33, the decision is to fail to reject the null hypothesis.
10.9
Canon
x 1 = 5.8
1 = 1.7
n1 = 36
Ho:
Ha:
Pioneer
x 2 = 5.0
2 = 1.4
n2 = 45
µ1 – µ2 = 0
µ1 – µ2  0
For two-tail test, /2 = .025
z =
( x 1  x 2 )  ( 1   2 )
1
2
n1

2
2
n2
z.025 = ±1.96

(5.8  5.0)  (0)
2
= 2.27
(1.7)
(1.4)

36
45
Since the observed z = 2.27 > zc = 1.96, the decision is to reject the null hypothesis.
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Solutions Manual and Study Guide
10.11
Ho: µ1 – µ2 = 0
Ha: µ1 – µ2 < 0
 = .01
df = 8 + 11 – 2 = 17
Sample 1
Sample 2
n1 = 8
x 1 = 24.56
s12 = 12.4
n2 = 11
x 2 = 26.42
s22 = 15.8
For one-tail test,  = .01
Critical t.01,19 = –2.567
t=
( x1  x 2 )  ( 1   2 )
s1 (n1  1)  s 2 (n2  1) 1
1

n1  n2  2
n1 n2
2
2
=
(24.56  26.42)  (0)
= –1.05
12.4(7)  15.8(10) 1 1

8  11  2
8 11
Since the observed t = –1.05 > t.01,19 = –2.567, the decision is to fail to reject the null
hypothesis.
10.13
 = .05
Ho: µ1 – µ2 = 0
Ha: µ1 – µ2 > 0
df = n1 + n2 – 2 = 10 + 10 – 2 = 18
Sample 1
n1 = 10
x 1 = 45.38
s1 = 2.357
Sample 2
n2 = 10
x 2 = 40.49
s2 = 2.355
For one-tail test,  = .05
Critical t.05,18 = 1.734
t =
t =
( x1  x 2 )  ( 1   2 )
s1 (n1  1)  s 2 (n2  1) 1
1

n1  n2  2
n1 n2
2
2
(45.38  40.49)  (0)
(2.357) 2 (9)  (2.355) 2 (9) 1
1

10  10  2
10 10
=
= 4.64
Since the observed t = 4.64 > t.05,18 = 1.734, the decision is to reject the null hypothesis.
Chapter 10: Statistical Inferences About Two Populations
10.15
Peoria
n1 = 21
Evansville
n2 = 26
x1 = 86,900
x 2 = 84,000
s1 = 2,300
s2 = 1,750
90% level of confidence, /2 = .05
187
df = 21 + 26 – 2
t .05,45 = 1.684 (used df = 40)
s (n  1)  s 2 (n2  1) 1
1
( x1  x 2 )  t 1 1

=
n1  n2  2
n1 n2
2
2
(2300) 2 (20)  (1750) 2 (25) 1
1

(86,900 – 84,000) + 1.684
=
21  26  2
21 26
2,900 + 994.62
1905.38 < 1 – 2 < 3894.62
10.17
Let Boston be group 1
1) Ho: µ1 – µ2 = 0
Ha: µ1 – µ2 > 0
2)
t =
( x1  x 2 )  ( 1   2 )
s1 (n1  1)  s 2 (n2  1) 1
1

n1  n2  2
n1 n2
2
2
3)  = .01
4)
For a one-tailed test and df = 8 + 9 – 2 = 15, t.01,15 = 2.602. If the observed value of t is
greater than 2.602, the decision is to reject the null hypothesis.
5)
Boston
n1 = 8
x 1 = 47
s1 = 3
6)
t =
Dallas
n2 = 9
x 2 = 44
s2 = 3
(47  44)  (0)
7(3) 2  8(3) 2
15
= 2.06
1 1

8 9
7)
Since t = 2.06 < t.01,15 = 2.602, the decision is to fail to reject the null hypothesis.
8)
There is no significant difference in rental rates between Boston and Dallas.
188
10.19
Solutions Manual and Study Guide
Ho: µ1 – µ2 = 0
Ha: µ1 – µ2  0
df = n1 + n2 – 2 = 11 + 11 – 2 = 20
Toronto
n1 = 11
x 1 = $67,381.82
s1 = $2,067.28
Mexico City
n2 = 11
x 2 = $63,481.82
s2 = $1,594.25
For a two-tail test, /2 = .005 Critical t.005,20 = ±2.845
t =
t =
( x1  x 2 )  ( 1   2 )
s1 (n1  1)  s 2 (n2  1) 1
1

n1  n2  2
n1 n2
2
2
=
(67,381.82  63,481.82)  (0)
(2,067.28) 2 (10)  (1,594.25) 2 (10) 1 1

11  11  2
11 11
= 4.95
Since the observed t = 4.95 > t.005,20 = 2.845, the decision is to Reject the null hypothesis.
10.21
Ho:
Ha:
D=0
D>0
Sample 1
38
27
30
41
36
38
33
35
44
n=9
Sample 2
22
28
21
38
38
26
19
31
35
d =7.11
d
16
–1
9
3
–2
12
14
4
9
sd = 6.45
 = .01
df = n – 1 = 9 – 1 = 8
For one-tail test and  = .01,
the critical t.01,8 = ±2.896
Chapter 10: Statistical Inferences About Two Populations
t =
d  D 7.11  0
= 3.31

sd
6.45
9
n
Since the observed t = 3.31 > t.01,8 = 2.896, the decision is to reject the null hypothesis.
10.23
d = 40.56 sd = 26.58
n = 22
For a 98% Level of Confidence, /2 = .01, and df = n – 1 = 22 – 1 = 21
t.01,21 = 2.518
d t
sd
n
40.56 ± (2.518)
26.58
22
40.56 ± 14.27
26.29 < D < 54.83
10.25
City
Atlanta
Boston
Des Moines
Kansas City
Louisville
Portland
Raleigh-Durham
Reno
Ridgewood
San Francisco
Tulsa
d = 1302.82
 = .01
d t
Cost
20427
27255
22115
23256
21887
24255
19852
23624
25885
28999
20836
sd = 4938.22
/2 = .005
sd
n
Resale
25163
24625
12600
24588
19267
20150
22500
16667
26875
35333
16292
n = 11, df = 10
t.005,10= 3.169
= 1302.82 + 3.169
–3415.6 < D < 6021.2
d
–4736
2630
9515
–1332
2620
4105
–2648
6957
–990
–6334
4544
4938.22
= 1302.82 + 4718.42
11
189
190
10.27
Solutions Manual and Study Guide
Before
255
230
290
242
300
250
215
230
225
219
236
After
197
225
215
215
240
235
190
240
200
203
223
d = 28.09
n = 11
d
58
5
75
27
60
15
25
–10
25
16
13
sd=25.813
df = n – 1 = 11 – 1 = 10
For a 98% level of confidence and /2=.01, t.01,10 = 2.764
d t
sd
n
28.09 ± (2.764)
25.813
= 28.09 ± 21.51
11
6.58 < D < 49.60
10.29
d = 75
n = 21
sd=30
df = 21 – 1 = 20
For a 90% confidence level, /2=.05 and t.05,20 = 1.725
d t
sd
n
75 + 1.725
30
= 75 ± 11.29
21
63.71 < D < 86.29
Chapter 10: Statistical Inferences About Two Populations
10.31 a)
Sample 1
Sample 2
n1 = 368
x1 = 175
n2 = 405
x2 = 182
pˆ 1 
p
x1 175
= .476

n1 368
pˆ 2 
x2 182
= .449

n2 405
x1  x2 175  182 357
= .462


n1  n2 368  405 773
Ho: p1 – p2 = 0
Ha: p1 – p2  0
For two-tail, /2 = .025 and z.025 = ±1.96
z
( pˆ 1  pˆ 2 )  ( p1  p 2 )
1
1
p  q  
 n1 n 



(.476  .449)  (0)
1 
 1
(.462)(. 538)


 368 405 
= 0.75
Since the observed z = 0.75 < zc = 1.96, the decision is to fail to reject the null hypothesis.
b)
Sample 1
p̂ 1 = .38
n1 = 649
p
Ho:
Ha:
Sample 2
p̂ 2 = .25
n2 = 558
n1 pˆ 1  n2 pˆ 2 649(.38)  558(.25)
= .32

n1  n2
649  558
p1 – p2 = 0
p1 – p2 > 0
For a one-tail test and  = .10,
z
( pˆ 1  pˆ 2 )  ( p1  p 2 )
1
1
p  q  
 n1 n 



z.10 = 1.28
(.38  .25)  (0)
1 
 1
(.32)(. 68)


 649 558 
= 4.83
Since the observed z = 4.83 > zc = 1.28, the decision is to reject the null hypothesis.
191
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Solutions Manual and Study Guide
10.33
H0: pm – pw = 0
Ha: pm – pw < 0 nm = 374
nw = 481
p̂ = .59
m
p̂ = .70
w
For a one-tailed test and  = .05, z.05 = –1.645
p
z
nm pˆ m  nw pˆ w 374(.59)  481(.70)
= .652

nm  n w
374  481
( pˆ 1  pˆ 2 )  ( p1  p 2 )
1
1
p  q  
 n1 n 



(.59  .70)  (0)
= –3.35
1 
 1
(.652)(. 348)


 374 481 
Since the observed z = –3.35 < z.05 = –1.645, the decision is to reject the null hypothesis.
10.35 Computer Firms
p̂ 1 = .48
n1 = 56
p
Ho:
Ha:
Banks
p̂ 2 = .56
n2 = 89
n1 pˆ 1  n2 pˆ 2 56(.48)  89(.56)
= .529

n1  n2
56  89
p1 – p2 = 0
p1 – p2  0
For two-tail test, /2 = .10 and zc = ±1.28
z
( pˆ 1  pˆ 2 )  ( p1  p 2 )
1
1
p  q  
 n1 n 



(.48  .56)  (0)
1 
 1
(.529)(. 471)  
 56 89 
=
–0.94
Since the observed z = –0.94 > zc = –1.28, the decision is to fail to reject the null hypothesis.
Chapter 10: Statistical Inferences About Two Populations
10.37
193
H0: p1 – p2 = 0
Ha: p1 – p2  0
 = .10
p̂ = .09
p̂ = .06
1
n1 = 780
2
n2 = 915
For a two-tailed test, /2 = .05 and z.05 = + 1.645
p
Z
n1 pˆ 1  n2 pˆ 2 780(.09)  915(.06)
= .0738

n1  n2
780  915
( pˆ 1  pˆ 2 )  ( p1  p 2 )
1
1
p  q  
 n1 n 



(.09  .06)  (0)
1 
 1
(.0738)(. 9262)


 780 915 
= 2.35
Since the observed z = 2.35 > z.05 = 1.645, the decision is to reject the null hypothesis.
10.39
H0: 12 = 22
Ha: 12 < 22
 = .01
dfnum = 12 – 1 = 11
n1 = 10
n2 = 12
s12 = 562
s22 = 1013
dfdenom = 10 – 1 = 9
Table F.01,10,9 = 5.26
F =
s2
2
s1
2

1013
= 1.80
562
Since the observed F = 1.80 < F.01,10,9 = 5.26, the decision is to fail to reject the null hypothesis.
10.41 City 1
1.18
1.15
1.14
1.07
1.14
1.13
1.09
1.13
1.13
1.03
n1 = 10
City 2
1.08
1.17
1.14
1.05
1.21
1.14
1.11
1.19
1.12
1.13
df1 = 9
s12 = .0018989
n2 = 10
df2 = 9
s22 = .0023378
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Solutions Manual and Study Guide
H0: 12 = 22
Ha: 12  22
 = .10 /2 = .05
Upper tail critical F value = F.05,9,9 = 3.18
Lower tail critical F value = F.95,9,9 = 0.314
F =
s1
2
s2
2

.0018989
= 0.81
.0023378
Since the observed F = 0.81 is greater than the lower tail critical value of 0.314 and less than the
upper tail critical value of 3.18, the decision is to fail to reject the null hypothesis.
10.43
H0: 12 = 22
Ha: 12 > 22
 = .05
dfnum = 12 – 1 = 11
n1 = 12
n2 = 15
s1 = 7.52
s2 = 6.08
dfdenom = 15 – 1 = 14
The critical table F value is F.05,10,14 = 5.26
F=
s1
2
s2
2
(7.52) 2
= 1.53

(6.08) 2
Since the observed F = 1.53 < F.05,10,14 = 2.60, the decision is to fail to reject the null hypothesis.
10.45
Ho:
Ha:
µ1 – µ2 = 0
µ1 – µ2  0
For  = .10 and a two-tailed test, /2 = .05 and z.05 = + 1.645
Sample 1
Sample 2
x1 = 138.4
1 = 6.71
x 2 = 142.5
2 = 8.92
n1 = 48
n2 = 39
z =
( x 1  x 2 )  ( 1   2 )
1
2
n1

2
2
n2

(138.4  142.5)  (0)
(6.71) 2 (8.92)

48
39
= –2.38
Since the observed value of z = –2.38 is less than the critical value of z = –1.645, the decision is
to reject the null hypothesis. There is a significant difference in the means of the two
populations.
Chapter 10: Statistical Inferences About Two Populations
10.47
195
µ1 – µ2 = 0
µ1 – µ2 > 0
Ho:
Ha:
Sample 1
x1 = 2.06
s12 = .176
n1 = 12
Sample 2
x 2 = 1.93
s22 = .143
n2 = 15
This is a one-tailed test with df = 12 + 15 – 2 = 25. The critical value is
t.05,25 = 1.708. If the observed value is greater than 1.708, the decision will be to reject the null
hypothesis.
( x1  x 2 )  ( 1   2 )
t =
s1 (n1  1)  s 2 (n2  1) 1
1

n1  n2  2
n1 n2
2
2
(2.06  1.93)  (0)
= 0.85
(.176)(11)  (.143)(14) 1
1

25
12 15
t =
Since the observed value of t = 0.85 is less than the critical value of t = 1.708, the decision is to
fail to reject the null hypothesis. The mean for population one is not significantly greater than
the mean for population two.
10.49
 = .01
Ho: D = 0
Ha: D < 0
n = 21
d = –1.16
df = 20
sd = 1.01
The critical t.01,20 = –2.528. If the observed t is less than –2.528, then the decision will be to reject
the null hypothesis.
t =
d  D  1.16  0

= –5.26
sd
1.01
n
21
Since the observed value of t = –5.26 is less than the critical t value of –2.528, the decision is to
reject the null hypothesis. The population difference is less than zero.
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Solutions Manual and Study Guide
10.51
Ho:
Ha:
 = .05
p1 – p2 = 0
p1 – p2  0
/2 = .025
z.025 = + 1.96
If the observed value of z is greater than 1.96 or less than –1.96, then the decision will be to reject
the null hypothesis.
Sample 1
x1 = 345
n1 = 783
x1  x2 345  421
= .4562

n1  n2 783  896
p
pˆ 1 
z
Sample 2
x2 = 421
n2 = 896
x1 345
= .4406

n1 783
( pˆ 1  pˆ 2 )  ( p1  p 2 )
1
1
p  q  
 n1 n 


pˆ 2 

x2 421
= .4699

n2 896
(.4406  .4699)  (0)
1 
 1
(.4562)(. 5438)


 783 896 
= –1.20
Since the observed value of z = –1.20 is greater than –1.96, the decision is to fail to reject the
null hypothesis. There is no significant difference in the population proportions.
10.53
H0: 12 = 22
Ha: 12  22
 = .05
n1 = 8
n2 = 10
dfnum = 8 – 1 = 7 dfdenom = 10 – 1 = 9
The critical F values are:
F.025,7,9 = 4.20
s12 = 46
S22 = 37
F.975,9,7 = .238
If the observed value of F is greater than 4.20 or less than .238, then the decision will be to reject
the null hypothesis.
F =
s1
2
s2
2

46
= 1.24
37
Since the observed F = 1.24 is less than F.025,7,9 =4.20 and greater than
F.975,9,7 = .238, the decision is to fail to reject the null hypothesis. There is no significant
difference in the variances of the two populations.
Chapter 10: Statistical Inferences About Two Populations
10.55
Morning
43
51
37
24
47
44
50
55
46
Afternoon
41
49
44
32
46
42
47
51
49
d = –0.444
n=9
For a 90% Confidence Level:
d t
197
d
2
2
–7
–8
1
2
3
4
–3
sd =4.447
df = 9 – 1 = 8
/2 = .05 and t.05,8 = 1.86
sd
n
–0.444 + (1.86)
4.447
= –0.444 ± 2.757
9
–3.201 < D < 2.313
10.57
Accounting
n1 = 16
x 1 = 26,400
s1 = 1,200
Data Entry
n2 = 14
x 2 = 25,800
s2 = 1,050
H0: 12 = 22
Ha: 12  22
dfnum = 16 – 1 = 15
dfdenom = 14 – 1 = 13
The critical F values are:
F.025,15,13 = 3.05 F.975,15,13 = 0.33
F =
s1
2
s2
2

1,440,000
= 1.31
1,102,500
Since the observed F = 1.31 is less than F.025,15,13 = 3.05 and greater than F.975,15,13 = 0.33,
the decision is to fail to reject the null hypothesis.
198
10.59
Solutions Manual and Study Guide
Men
n1 = 60
Women
n2 = 41
x 1 = 631
1 = 100
x 2 = 848
2 = 100
For a 95% Confidence Level, /2 = .025 and z.025 = 1.96
( x1  x 2 )  z
 12

n1
 22
n2
1002 1002

= –217 ± 39.7
60
41
(631 – 848) + 1.96
–256.7 < µ1 – µ2 < –177.3
10.61
With Fertilizer
Without Fertilizer
x 1 = 38.4
1 = 9.8
x 2 = 23.1
2 = 7.4
n1 = 35
n2 = 35
µ1 – µ2 = 0
µ1 – µ2 > 0
Ho:
Ha:
For one-tail test,  = .01 and z.01 = 2.33
z =
( x 1  x 2 )  ( 1   2 )
1
2
n1

2
2
n2

(38.4  23.1)  (0)
(9.8) 2 (7.4)

35
35
= 7.37
Since the observed z = 7.37 > z.01 = 2.33, the decision is to reject the null
hypothesis.
10.63
H0: 12 = 22
Ha: 12  22
dfnum = 27 – 1 = 26
 = .05
F =
2
s2
2

s1 = 22,000
s2 = 15,500
dfdenom = 29 – 1 = 28
The critical F values are:
s1
n1 = 27
n2 = 29
F.025,24,28 = 2.17
F.975,28,24 = .46
22,000 2
= 2.01
15,500 2
Since the observed F = 2.01 < F.025,24,28 = 2.17 and > than F.975,28,24 = .46, the
decision is to fail to reject the null hypothesis.
Chapter 10: Statistical Inferences About Two Populations
10.65
 = .01
Ho: µ1 – µ2 = 0
Ha: µ1 – µ2 < 0
df = 23 + 19 – 2 = 40
Wisconsin
Tennessee
n1 = 23
x 1 = 69.652
s12 = 9.9644
n2 = 19
x 2 = 71.7368
s22 = 4.6491
For one-tail test,  = .01 and the critical t.01,40 = –2.423
( x1  x 2 )  ( 1   2 )
t =
s1 (n1  1)  s 2 (n2  1) 1
1

n1  n2  2
n1 n2
2
2
(69.652  71.7368)  (0)
= –2.44
(9.9644)( 22)  (4.6491)(18) 1
1

40
23 19
t =
Since the observed t = –2.44 < t.01,40 = –2.423, the decision is to reject the
null hypothesis.
10.67
 = .05
Ho: P1 – P2 = 0
Ha: P1 – P2  0
Machine 1
x1 = 38
n1 = 191
pˆ 1 
p
Machine 2
x2 = 21
n2 = 202
x1 38
= .199

n1 191
pˆ 2 
x2
21
= .104

n2 202
n1 pˆ 1  n2 pˆ 2 (.199)(191)  (.104)(202)
= .15

n1  n2
191  202
For two-tail, /2 = .025 and the critical z values are: z.025 = ±1.96
z
( pˆ 1  pˆ 2 )  ( p1  p 2 )
1
1
p  q  
 n1 n 



(.199  .104)  (0)
1 
 1
(.15)(. 85)


 191 202 
= 2.64
Since the observed z = 2.64 > zc = 1.96, the decision is to reject the null hypothesis.
199
200
10.69
Solutions Manual and Study Guide
Aerospace
n1 = 33
Automobile
n2 = 35
x 1 = 12.4
1 = 2.9
x 2 = 4.6
2 = 1.8
For a 99% Confidence Level, /2 = .005 and z.005 = 2.575
( x1  x 2 )  z
 12
n1

 22
n2
(2.9) 2 (1.8) 2

33
35
(12.4 – 4.6) + 2.575
= 7.8 ± 1.52
6.28 < µ1 – µ2 < 9.32
10.71
Before
12
7
10
16
8
After
8
3
8
9
5
d = 4.0
n=5
d
4
4
2
7
3
sd = 1.8708
df = 5 – 1 = 4
 = .01
Ho: D = 0
Ha: D > 0
For one-tail test,  = .01 and the critical t.01,4 = 3.747
t =
d  D 4.0  0

= 4.78
sd
1.8708
n
5
Since the observed t = 4.78 > t.01,4 = 3.747, the decision is to reject the null hypothesis.
Chapter 10: Statistical Inferences About Two Populations
10.73
201
A t test was used to test to determine if Hong Kong has significantly
different rates than Bombay. Let group 1 be Hong Kong.
Ho:
Ha:
µ1 – µ 2 = 0
µ1 – µ 2 > 0
n1 = 19
S1 = 12.9
n2 = 23
S2 = 13.9
x 1 = 130.4
x 2 = 128.4
 = .01
t = 0.48 with a p-value of .634 which is not significant at of .05. There is
not enough evidence in these data to declare that there is a difference in the
average rental rates of the two cities.
10.75
The point estimates from the sample data indicate that in the northern city the market share is
.3108 and in the southern city the market share is .2701. The point estimate for the difference in
the two proportions of market share are .0407. Since the 99% confidence interval ranges from –
.0394 to +.1207 and zero is in the interval, any hypothesis testing decision based on this interval
would result in failure to reject the null hypothesis. Alpha is .01 with a two-tailed test. This is
underscored by a calculated z value of 1.31 which has an associated p-value of .191 which, of
course, is not significant for any of the usual values of .
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Solutions Manual and Study Guide