Chapter 10: Statistical Inferences About Two Populations
1
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.
Test hypotheses and construct confidence intervals about the difference in two
population means using data from large samples.
2.
Test hypotheses and establish confidence intervals about the difference in two
population means using data from small samples when the population variances
are unknown and the populations are normally distributed.
3.
Test hypotheses and construct confidence intervals about the difference in two
related populations when the differences are normally distributed.
4.
Test hypotheses and construct confidence intervals about the difference in two
population proportions.
5.
Test hypotheses and construct confidence intervals about two population
variances when the two populations are normally distributed.
CHAPTER TEACHING STRATEGY
The major emphasis of chapter 10 is on analyzing data from two samples. The
student should be ready to deal with this topic given that he/she has tested hypotheses and
computed confidence intervals in previous chapters on single sample data.
The z test for analyzing the differences in two sample means is presented here
with emphasis on large samples. Conceptually, this is not radically different than the z
test for a single sample mean shown initially in Chapter 7. In this test, we can use either
sample or population variances since sample size is large. If sample sizes are small but
Chapter 10: Statistical Inferences About Two Populations
2
the population variances are known and the populations are normally distributed, this z
test can still be used. However, in many cases, the population variances are unknown. If
this is the case and sample sizes are small and it can be assumed that the populations are
normally distributed, a t test for independent samples can be used. There are two
different formulas given in the chapter to conduct this test. One version of this test uses a
"pooled" estimate of the population variance and assumes that the population variances
are equal. The other version does not assume equal population variances and is simpler
to compute. However, the degrees of freedom formula for this version is quite complex.
A t test is also included for related (non independent) samples. It is important that
the student be able to recognize when two samples are related and when they are
independent. The first portion of section 10.3 addresses this issue. To underscore the
potential difference in the outcome of the two techniques, it is sometimes valuable to
analyze some related measures data with both techniques and demonstrate that the results
and conclusions are usually quite different. You can have your students work problems
like this using both techniques to help them understand the differences between the two
tests (independent and dependent t tests) and the different outcomes they will obtain.
A z test of proportions for two samples is presented here along with an F test for
two population variances. This is a good place to introduce the student to the F
distribution in preparation for analysis of variance in Chapter 11. The student will begin
to understand that the F values have two different degrees of freedom. The F distribution
tables are upper tailed only. For this reason, formula 10.16 is given in the chapter to be
used to compute lower tailed F values for two-tailed tests.
CHAPTER OUTLINE
10.1 Hypothesis Testing and Confidence Intervals about the Difference in Two Means
using the z Statistic
Hypothesis Testing
Confidence Intervals
Using the Computer to Test Hypotheses and Construct Confidence
Intervals about the Difference in Two Population Means Using the
z Test
10.2 Hypothesis Testing and Confidence Intervals about the Difference in Two Means:
Small 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
Chapter 10: Statistical Inferences About Two Populations
3
Hypothesis Testing
Using the Computer to Make Statistical Inferences about Two Related
Populations
Confidence Intervals
10.4
Statistical Inferences About Two Population Proportions
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
KEY TERMS
Dependent Samples
F Distribution
F Value
Independent Samples
Matched-Pairs Data
Matched-Pairs Test
Paired Data
Related Measures
SOLUTIONS TO PROBLEMS IN CHAPTER 10
10.1
a) Ho:
Ha:
Sample 1
Sample 2
x 1 = 51.3
s12 = 52
n1 = 32
x 2 = 53.2
s22 = 60
n2 = 32
µ1 - µ2 = 0
µ1 - µ2 < 0
For one-tail test,  = .10
z =
( x 1  x 2 )  ( 1   2 )
2
2
s1
s
 2
n1
n2
z.10 = -1.28

(51.3  53.2)  (0)
52 60

32 32
= -1.02
Since the observed z = -1.02 > zc = -1.645, the decision is to fail to reject the null
hypothesis.
Chapter 10: Statistical Inferences About Two Populations
b) Critical value method:
zc =
( x 1  x 2 ) c  ( 1   2 )
2
2
s1
s
 2
n1
n2
-1.645 =
( 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
10.2
Sample 1
Sample 2
n1 = 32
x 1 = 70.4
s1 = 5.76
n2 = 31
x 2 = 68.7
s2 = 6.1
For a 90% C.I.,
z.05 = 1.645
2
2
s
s
( x1  x 2 )  z 1  2
n1
n2
(70.4) – 68.7) + 1.645
1.7 ± 2.465
-.76 < µ1 - µ2 < 4.16
5.762 6.12

32
31
4
Chapter 10: Statistical Inferences About Two Populations
10.3 a) Sample 1
Sample 2
x 1 = 88.23
s12 = 22.74
n1 = 30
Ho:
Ha:
x 2 = 81.2
s22 = 26.65
n2 = 30
µ1 - µ2 = 0
µ1 - µ2  0
For two-tail test, use /2 = .01
z =
5
( x 1  x 2 )  ( 1   2 )
2
2
s1
s
 2
n1
n2

z.01 = + 2.33
(88.23  81.2)  (0)
22.74 26.65

30
30
= 5.48
Since the observed z = 5.48 > z.01 = 2.33, the decision is to reject the null
hypothesis.
2
b) ( x 1  x 2 )  z
2
s1
s
 2
n1
n2
(88.23 – 81.2) + 2.33
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.
10.4
Computers/electronics
x 1 = 1.96
s12 = 1.018776
n1 = 50
Ho:
Ha:
Food/Beverage
x 2 = 3.02
s22 = .917959
n2 = 50
µ1 - µ2 = 0
µ1 - µ2  0
For two-tail test, /2 = .005
z.005 = ±2.575
Chapter 10: Statistical Inferences About Two Populations
z =
( x 1  x 2 )  ( 1   2 )
2
2

s1
s
 2
n1
n2
(1.96  3.02)  (0)
1.018776 0.917959

50
50
6
= -5.39
Since the observed z = -5.39 < zc = -2.575, the decision is to reject the null
hypothesis.
10.5
A
B
n1 = 40
x 1 = 5.3
s12 = 1.99
For a 95% C.I.,
z.025 = 1.96
2
( x1  x 2 )  z
n2 = 37
x 2 = 6.5
s22 = 2.36
2
s1
s
 2
n1
n2
(5.3 – 6.5) + 1.96
1.99 2.36

40
37
-1.86 <  < -.54
-1.2 ± .66
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.
10.6
Managers
Specialty
n1 = 35
x 1 = 1.84
S1 = .38
n2 = 41
x 2 = 1.99
S2 = .51
for a 98% C.I.,
z.01 = 2.33
2
2
s
s
( x1  x 2 )  z 1  2
n1
n2
Chapter 10: Statistical Inferences About Two Populations
(1.84 - 1.99) ± 2.33
7
.382 .512

35
41
-.15 ± .2384
-.3884 < µ1 - µ2 < .0884
Point Estimate = -.15
Hypothesis Test:
1) Ho: µ1 - µ2 = 0
Ha: µ1 - µ2  0
2) z =
( x 1  x 2 )  ( 1   2 )
 12
n1

 22
n2
3)  = .02
4) For a two-tailed test, z.01 = + 2.33. If the observed z value is greater than 2.33
or less than -2.33, then the decision will be to reject the null hypothesis.
5) Data given above
6) z =
(1.84  1.99)  (0)
(.38) 2 (.51) 2

35
41
= -1.47
7) Since z = -1.47 > z.01 = -2.33, the decision is to fail to reject the null
hypothesis.
8) There is no significant difference in the hourly rates
of the two groups.
Chapter 10: Statistical Inferences About Two Populations
10.7
1992
1999
x 1 = 190
s1 = 18.50
n1 = 51
x 2 = 198
s2 = 15.60
n2 = 47
H0: 1 - 2 = 0
Ha: 1 - 2 < 0
For a one-tailed test,
z =
 = .01
z.01 = -2.33
( x 1  x 2 )  ( 1   2 )
2
8
2
(190  198)  (0)

(18.50) 2 (15.60) 2

51
47
s1
s
 2
n1
n2
= -2.32
Since the observed z = -2.32 > z.01 = -2.33, the decision is to fail to reject the null
hypothesis.
10.8
Seattle
Atlanta
n1 = 31
x 1 = 2.64
s12 = .03
n2 = 31
x 2 = 2.36
s22 = .015
For a 99% C.I.,
z.005 = 2.575
2
( x1  x 2 )  z
(2.64-2.36) ± 2.575
.28 ± .10
2
s1
s
 2
n1
n2
.03 .015

31
31
.18 <  < .38
Between $ .18 and $ .38 difference with Seattle being more expensive.
Chapter 10: Statistical Inferences About Two Populations
10.9
Canon
Pioneer
x 1 = 5.8
s1 = 1.7
n1 = 36
x 2 = 5.0
s2 = 1.4
n2 = 45
Ho:
Ha:
µ1 - µ2 = 0
µ1 - µ2  0
For two-tail test, /2 = .025
z =
9
( x 1  x 2 )  ( 1   2 )
2
2

s1
s
 2
n1
n2
z.025 = ±1.96
(5.8  5.0)  (0)
(1.7) 2 (1.4)

36
45
= 2.27
Since the observed z = 2.27 > zc = 1.96, the decision is to reject the null hypothesis.
10.10
Ho:
Ha:
A
B
x 1 = 8.05
s1 = 1.36
n1 = 50
x 2 = 7.26
s2 = 1.06
n2 = 38
µ1 - µ2 = 0
µ1 - µ2 > 0
For one-tail test,  = .10
z =
( x 1  x 2 )  ( 1   2 )
2
2
s1
s
 2
n1
n2
z.10 = 1.28

(8.05  7.26)  (0)
(1.36) 2 (1.06) 2

50
38
= 3.06
Since the observed z = 3.06 > zc = 1.28, the decision is to reject the null
hypothesis.
Chapter 10: Statistical Inferences About Two Populations
 = .01
df = 8 + 11 - 2 = 17
10.11 Ho: µ1 - µ2 = 0
Ha: µ1 - µ2 < 0
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=
10
( 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
df = 20 + 20 - 2 = 38
10.12 a) Ho: µ1 - µ2 = 0
Ha: µ1 - µ2  0
Sample 1
n1 = 20
x 1 = 118
s1 = 23.9
Sample 2
n2 = 20
x 2 = 113
s2 = 21.6
For two-tail test,
t =
t =
/2 = .05
Critical t.05,38 = 1.697 (used df=30)
( 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)
(23.9) 2 (19)  (21.6) 2 (19) 1
1

20  20  2
20 20
=
= 0.69
Since the observed t = 0.69 < t.05,38 = 1.697, the decision is to fail to reject
the null hypothesis.
Chapter 10: Statistical Inferences About Two Populations
11
s (n  1)  s 2 (n2  1) 1
1

b) ( x1  x 2 )  t 1 1
n1  n2  2
n1 n2
=
2
(118 – 113) + 1.697
2
(23.9) 2 (19)  (21.6) 2 (19) 1
1

20  20  2
20 20
5 + 12.224
-7.224 < 1 - 2 < 17.224
 = .05
df = n1 + n2 - 2 = 10 + 10 - 2 = 18
10.13 Ho: µ1 - µ2 = 0
Ha: µ1 - µ2 > 0
Sample 1
Sample 2
n1 = 10
x 1 = 45.38
s1 = 2.357
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.
10.14 Ho: µ1 - µ2 = 0
Ha: µ1 - µ2  0
 =.01
df = 18 + 18 - 2 = 34
Sample 1
Sample 2
n1 = 18
x 1 = 5.333
s12 = 12
n2 = 18
x 2 = 9.444
s22 = 2.026
Chapter 10: Statistical Inferences About Two Populations
For two-tail test, /2 = .005
t =
t =
Critical t.005,34 = ±2.457 (used df=30)
( x1  x 2 )  ( 1   2 )
s1 (n1  1)  s 2 (n2  1) 1
1

n1  n2  2
n1 n2
2
12
2
=
(5.333  9.444)  (0)
= -4.66
12(17)  (2.026)17 1 1

18  18  2
18 18
Since the observed t = -4.66 < t.005,34 = -2.457
Reject the null hypothesis
s (n  1)  s 2 (n2  1) 1
1
( x1  x 2 )  t 1 1

n1  n2  2
n1 n2
2
b)
2
(5.333 – 9.444) + 2.457
=
(12)(17)  (2.026)(17) 1 1

18  18  2
18 18
-4.111 + 2.1689
-6.2799 < 1 - 2 < -1.9421
10.15 Peoria
Evansville
n1 = 21
x1 = 86,900
s1 = 2,300
n2 = 26
x 2 = 84,000
s2 = 1,750
90% level of confidence, /2 = .05
t .05,45 = 1.684 (used df = 40)
s1 (n1  1)  s 2 (n2  1) 1
1

n1  n2  2
n1 n2
2
( x1  x 2 )  t
df = 21 + 26 – 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
Chapter 10: Statistical Inferences About Two Populations
13
 = .05
df = 21 + 26 - 2 = 45
10.16 Ho: µ1 - µ2 = 0
Ha: µ1 - µ2  0!= 0
Peoria
Evansville
n1 = 21
x 1 = $86,900
s1 = $2,300
n2 = 26
x 2 = $84,000
s2 = $1,750
For two-tail test, /2 = .025
Critical t.025,45 = ± 2.021 (used df=40)
t =
t =
( x1  x 2 )  ( 1   2 )
s1 (n1  1)  s 2 (n2  1) 1
1

n1  n2  2
n1 n2
2
2
=
(86,900  84,000)  (0)
(2,300) 2 (20)  (1,750) 2 (25) 1
1

21  26  2
21 26
= 4.91
Since the observed t = 4.91 > t.025,45 = 2.021, the decision is to reject the null
hypothesis.
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
Dallas
n2 = 9
x 2 = 44
s2 = 3
Chapter 10: Statistical Inferences About Two Populations
6) t =
(47  44)  (0)
7(3) 2  8(3) 2
15
14
= 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.
10.18
nm = 22
x m = 112
sm = 11
nno = 20
x no = 122
sno = 12
df = nm + nno - 2 = 22 + 20 - 2 = 40
For a 98% Confidence Interval, /2 = .01 and
s (n  1)  s 2 (n2  1) 1
1
( x1  x 2 )  t 1 1

n1  n2  2
n1 n2
2
(112 – 122) + 2.423
t.01,40 = 2.423
2
=
(11)2 (21)  (12) 2 (19) 1
1

22  20  2
22 20
-10 ± 8.63
-$18.63 < µ1 - µ2 < -$1.37
Point Estimate = -$10
10.19 Ho: µ1 - µ2 = 0
Ha: µ1 - µ2  0
df = n1 + n2 - 2 = 11 + 11 - 2 = 20
Toronto
Mexico City
n1 = 11
x 1 = $67,381.82
s1 = $2,067.28
n2 = 11
x 2 = $63,481.82
s2 = $1,594.25
For a two-tail test, /2 = .005
Critical t.005,20 = ±2.845
Chapter 10: Statistical Inferences About Two Populations
t =
t =
( x1  x 2 )  ( 1   2 )
s1 (n1  1)  s 2 (n2  1) 1
1

n1  n2  2
n1 n2
2
2
15
=
(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.
10.20
Toronto
Mexico City
n1 = 11
x 1 = $67,381.82
s1 = $2,067.28
n2 = 11
x 2 = $63,481.82
s2 = $1,594.25
df = n1 + n2 - 2 = 11 + 11 - 2 = 20
For a 95% Level of Confidence, /2 = .025 and t.025,20 = 2.086
s (n  1)  s 2 (n2  1) 1
1
( x1  x 2 )  t 1 1

n1  n2  2
n1 n2
2
2
($67,381.82 - $63,481.82) ± (2.086)
3,900 ± 1,641.9
2,258.1 < µ1 - µ2 < 5,541.9
=
(2,067.28) 2 (10)  (1,594.25) 2 (10) 1 1

11  11  2
11 11
Chapter 10: Statistical Inferences About Two Populations
10.21 Ho:
Ha:
16
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
16
-1
9
3
-2
12
14
4
9
d =7.11
sd=6.45
 = .01
df = n - 1 = 9 - 1 = 8
For one-tail test and  = .01,
t =
the critical t.01,8 = ±2.896
d  D 7.11  0
= 3.31

sd
6.45
n
9
Since the observed t = 3.31 > t.01,8 = 2.896, the decision is to reject the null
hypothesis.
10.22 Ho:
Ha:
Before
107
99
110
113
96
98
100
102
107
109
104
99
101
D=0
D0
After
102
98
100
108
89
101
99
102
105
110
102
96
100
d
5
1
10
5
7
-3
1
0
2
-1
2
3
1
Chapter 10: Statistical Inferences About Two Populations
n = 13
d = 2.5385
df = n - 1 = 13 - 1 = 12
sd=3.4789
For a two-tail test and /2 = .025
17
 = .05
Critical t.025,12 = ±2.179
d  D 2.5385  0
= 2.63

sd
3.4789
t =
13
n
Since the observed t = 2.63 > t.025,12 = 2.179, the decision is to reject the null
hypothesis.
d = 40.56
10.23 n = 22
sd = 26.58
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
10.24
Before
32
28
35
32
26
25
37
16
35
After
40
25
36
32
29
31
39
30
31
d
-8
3
-1
0
-3
-6
-2
-14
4
Chapter 10: Statistical Inferences About Two Populations
n=9
d = -3
df = n - 1 = 9 - 1 = 8
sd = 5.6347
For 90% level of confidence and /2 = .025,
t = d t
18
 = .025
t.05,8 = 1.86
sd
n
t = -3 + (1.86)
5.6347
= -3 ± 3.49
9
-0.49 < D < 6.49
10.25 City
Atlanta
Boston
Des Moines
Kansas City
Louisville
Portland
Raleigh-Durham
Reno
Ridgewood
San Francisco
Tulsa
d = 1302.82
 = .01
d t
sd
n
Cost
Resale
d
20427
27255
22115
23256
21887
24255
19852
23624
25885
28999
20836
25163
24625
12600
24588
19267
20150
22500
16667
26875
35333
16292
-4736
2630
9515
-1332
2620
4105
-2648
6957
- 990
-6334
4544
sd = 4938.22
/2 = .005
df = 10
t.005,10= 3.169
= 1302.82 + 3.169
-3415.6 < D < 6021.2
n = 11,
4938.22
= 1302.82 + 4718.42
11
Chapter 10: Statistical Inferences About Two Populations
10.26 Ho:
Ha:
19
D=0
D<0
Before
2
4
1
3
4
2
2
3
1
After
4
5
3
3
3
5
6
4
5
d =-1.778
n=9
d
-2
-1
-2
0
1
-3
-4
-1
-4
 = .05
sd=1.716
df = n - 1 = 9 - 1 = 8
For a one-tail test and  = .05, the critical t.05,8 = -1.86
t =
d  D  1.778  0

= -3.11
sd
1.716
n
9
Since the observed t = -3.11 < t.05,8 = -1.86, the decision is to reject the null
hypothesis.
10.27
Before
255
230
290
242
300
250
215
230
225
219
236
n = 11
After
197
225
215
215
240
235
190
240
200
203
223
d = 28.09
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
Chapter 10: Statistical Inferences About Two Populations
d t
20
sd
n
28.09 ± (2.764)
25.813
= 28.09 ± 21.51
11
6.58 < D < 49.60
10.28 H0: D = 0
Ha: D > 0
n = 27
df = 27 – 1 = 26
d = 3.17
sd = 5
Since  = .01, the critical t.01,26 = 2.479
t =
d  D 3.71  0

= 3.86
sd
5
27
n
Since the observed t = 3.86 > t.01,26 = 2.479, the decision is to reject the null
hypothesis.
d = 75
10.29 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
10.30 Ho:
Ha:
n = 15
D=0
D0
d = -2.85
sd = 1.9
 = .01
For a two-tail test, /2 = .005 and the critical t.005,14 = + 2.977
df = 15 - 1 = 14
Chapter 10: Statistical Inferences About Two Populations
21
d  D  2.85  0
= -5.81

sd
1.9
t =
n
15
Since the observed t = -5.81 < t.005,14 = -2.977, the decision is to reject the null
hypothesis.
10.31 a)
pˆ 1 
p
Sample 1
Sample 2
n1 = 368
x1 = 175
n2 = 405
x2 = 182
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)
p
Sample 1
Sample 2
p̂ 1 = .38
n1 = 649
p̂ 2 = .25
n2 = 558
n1 pˆ 1  n2 pˆ 2 649(.38)  558(.25)
= .32

n1  n2
649  558
Chapter 10: Statistical Inferences About Two Populations
Ho:
Ha:
p1 - p2 = 0
p1 - p2 > 0
For a one-tail test and  = .10,
z
22
( 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.
10.32 a) n1 = 85
n2 = 90
p̂ 1 = .75
For a 90% Confidence Level,
( pˆ 1  pˆ 2 )  z
p̂ 2 = .67
z.05 = 1.645
pˆ 1qˆ1 pˆ 2 qˆ 2

n1
n2
(.75 - .67) ± 1.645
(.75)(.25) (.67)(.33)
= .08 ± .11

85
90
-.03 < p1 - p2 < .19
b) n1 = 1100
n2 = 1300
p̂ 1 = .19
p̂ 2 = .17
For a 95% Confidence Level, /2 = .025 and z.025 = 1.96
( pˆ 1  pˆ 2 )  z
pˆ 1qˆ1 pˆ 2 qˆ 2

n1
n2
(.19 - .17) + 1.96
-.01 < p1 - p2 < .05
(.19)(.81) (.17)(.83)

= .02 ± .03
1100
1300
Chapter 10: Statistical Inferences About Two Populations
c) n1 = 430
n2 = 399
x1 275
= .64

n1 430
pˆ 1 
x1 = 275
pˆ 2 
x2 = 275
x2 275
= .69

n2 399
For an 85% Confidence Level, /2 = .075 and
( pˆ 1  pˆ 2 )  z
23
z.075 = 1.44
pˆ 1qˆ1 pˆ 2 qˆ 2

n1
n2
(.64 - .69) + 1.44
(.64)(.36) (.69)(.31)
= -.05 ± .047

430
399
-.097 < p1 - p2 < -.003
d) n1 = 1500
pˆ 1 
n2 = 1500
x1 1050
= .70

n1 1500
x1 = 1050
pˆ 2 
x2 = 1100
x2 1100
= .733

n2 1500
For an 80% Confidence Level, /2 = .10 and z.10 = 1.28
( pˆ 1  pˆ 2 )  z
pˆ 1qˆ1 pˆ 2 qˆ 2

n1
n2
(.70 - .733) ± 1.28
(.70)(.30) (.733)(.267)
= -.033 ± .02

1500
1500
-.053 < p1 - p2 < -.013
10.33 H0: pm - pw = 0
Ha: pm - pw < 0
nm = 374
For a one-tailed test and  = .05,
p
nw = 481
z.05 = -1.645
nm pˆ m  nw pˆ w 374(.59)  481(.70)
= .652

nm  n w
374  481
p̂
m
= .59
p̂
w
= .70
Chapter 10: Statistical Inferences About Two Populations
z
( pˆ 1  pˆ 2 )  ( p1  p 2 )
1
1
p  q  
 n1 n 



(.59  .70)  (0)
1 
 1
(.652)(. 348)


 374 481 
24
= -3.35
Since the observed z = -3.35 < z.05 = -1.645, the decision is to reject the null
hypothesis.
10.34 n1 = 210
n2 = 176
p̂1 = .24
p̂2 = .35
For a 90% Confidence Level, /2 = .05 and
( pˆ 1  pˆ 2 )  z
z.05 = + 1.645
pˆ 1qˆ1 pˆ 2 qˆ 2

n1
n2
(.24 - .35) + 1.645
(.24)(.76) (.35)(.65)
= -.11 + .0765

210
176
-.1865 < p1 – p2 < -.0335
10.35 Computer Firms
Banks
p̂ 1 = .48
n1 = 56
p
Ho:
Ha:
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
10.36
A
B
n1 = 35
x1 = 5
pˆ 1 
25
n2 = 35
x2 = 7
x1
5
= .14

n1 35
pˆ 2 
x2
7
= .20

n2 35
For a 98% Confidence Level, /2 = .01 and z.01 = 2.33
pˆ 1qˆ1 pˆ 2 qˆ 2

n1
n2
( pˆ 1  pˆ 2 )  z
(.14 - .20) ± 2.33
(.14)(.86) (.20)(.80)

35
35
= -.06 ± .21
-.27 < p1 - p2 < .15
10.37 H0: p1 – p2 = 0
Ha: p1 – p2  0
 = .10
p̂ = .09
1
p̂ = .06
2
For a two-tailed test, /2 = .05 and
p
Z
n1 = 780
n2 = 915
z.05 = + 1.645
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.
Chapter 10: Statistical Inferences About Two Populations
10.38 n1 = 850
n2 = 910
p̂ = .60
1
26
p̂ 2 = .52
For a 95% Confidence Level, /2 = .025 and z.025 = + 1.96
( pˆ 1  pˆ 2 )  z
pˆ 1qˆ1 pˆ 2 qˆ 2

n1
n2
(.60 - .52) + 1.96
(.60)(.40) (.52)(.48)
= .08 + .046

850
910
.034 < p1 – p2 < .126
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.
10.40 H0: 12 = 22
Ha: 12  22
dfnum = 5 - 1 = 4
 = .05
F =
2
s2
2

S1 = 4.68
S2 = 2.78
dfdenom = 19 - 1 = 18
The critical table F values are:
s1
n1 = 5
n2 = 19
F.025,4,18 = 3.61
F.95,18,4 = .277
(4.68) 2
= 2.83
(2.78) 2
Since the observed F = 2.83 < F.025,4,18 = 3.61, the decision is to fail to reject the
null hypothesis.
Chapter 10: Statistical Inferences About Two Populations
10.41
City 1
27
City 2
1.18
1.15
1.14
1.07
1.14
1.13
1.09
1.13
1.13
1.03
1.08
1.17
1.14
1.05
1.21
1.14
1.11
1.19
1.12
1.13
n1 = 10
df1 = 9
n2 = 10
df2 = 9
s12 = .0018989
s22 = .0023378
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.
10.42 Let Houston = group 1 and Chicago = group 2
1) H0: 12 = 22
Ha: 12  22
s1
2
s2
3)  = .01
2
2) F =
4) df1 = 12
df2 = 10 This is a two-tailed test
The critical table F values are:
F.005,12,10 = 5.66
F.995,10,12 = .177
Chapter 10: Statistical Inferences About Two Populations
28
If the observed value is greater than 5.66 or less than .177, the decision will be
to reject the null hypothesis.
5) s12 = 393.4
6) F =
s22 = 702.7
393.4
= 0.56
702.7
7) Since F = 0.56 is greater than .177 and less than 5.66,
the decision is to fail to reject the null hypothesis.
8) There is no significant difference in the variances of
number of days between Houston and Chicago.
10.43 H0: 12 = 22
Ha: 12 > 22
dfnum = 12 - 1 = 11
 = .05
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.
10.44 H0: 12 = 22
Ha: 12  22
 = .01
dfnum = 15 - 1 = 14
n1 = 15
n2 = 15
dfdenom = 15 - 1 = 14
The critical table F values are:
F =
s1
2
s2
2

s12 = 91.5
s22 = 67.3
F.005,12,14 = 4.43 F.995,14,12 = .226
91.5
= 1.36
67.3
Since the observed F = 1.36 < F.005,12,14 = 4.43 and > F.995,14,12 = .226, the decision
is to fail to reject the null hypothesis.
Chapter 10: Statistical Inferences About Two Populations
10.45 Ho:
Ha:
29
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
s1 = 6.71
n1 = 48
x 2 = 142.5
s2 = 8.92
n2 = 39
z =
( x 1  x 2 )  ( 1   2 )
2
2

s1
s
 2
n1
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.
10.46 Sample 1
x1 = 34.9
s12 = 2.97
n1 = 34
Sample 2
x 2 = 27.6
s22 = 3.50
n2 = 31
For 98% Confidence Level,
2
( x1  x 2 )  z
z.01 = 2.33
2
s1
s
 2
n1
n2
(34.9 – 27.6) + 2.33
2.97 3.50

34
31
6.26 < 1 - 2 < 8.34
= 7.3 + 1.04
Chapter 10: Statistical Inferences About Two Populations
10.47 Ho:
Ha:
30
µ1 - µ2 = 0
µ1 - µ2 > 0
Sample 1
Sample 2
x1 = 2.06
s12 = .176
n1 = 12
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.
t =
t =
( x1  x 2 )  ( 1   2 )
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
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.
10.48
Sample 1
x 1 = 74.6
s12 = 10.5
n1 = 18
Sample 2
x 2 = 70.9
s22 = 11.4
n2 = 19
For 95% confidence, /2 = .025.
Using df = 18 + 19 - 2 = 35, t35,.025 = 2.042
s (n  1)  s 2 (n2  1) 1
1
( x1  x 2 )  t 1 1

n1  n2  2
n1 n2
2
(74.6 – 70.9) + 2.042
3.7 + 2.22
1.48 < 1 - 2 < 5.92
2
(23.9) 2 (19)  (21.6) 2 (19) 1
1

20  20  2
20 20
Chapter 10: Statistical Inferences About Two Populations
 = .01
10.49 Ho: D = 0
Ha: D < 0
n = 21
31
df = 20
d = -1.16
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
21
n
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.
10.50
Respondent
Before
After
1
2
3
4
5
6
7
8
9
47
33
38
50
39
27
35
46
41
63
35
36
56
44
29
32
54
47
d = -4.44
sd = 5.703
d
-16
-2
2
-6
-5
-2
3
-8
-6
df = 8
For a 99% Confidence Level, /2 = .005 and t8,.005 = 3.355
d t
sd
n
= -4.44 + 3.355
5.703
= -4.44 + 6.38
9
-10.82 < D < 1.94
10.51 Ho:
Ha:
p1 - p2 = 0
p1 - p2  0
 = .05
/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.
Chapter 10: Statistical Inferences About Two Populations
Sample 1
Sample 2
x1 = 345
n1 = 783
x2 = 421
n2 = 896
x1  x2 345  421
= .4562

n1  n2 783  896
p
pˆ 1 
z
32
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.
10.52 Sample 1
Sample 2
n1 = 409
p̂ 1 = .71
n2 = 378
p̂ 2 = .67
For a 99% Confidence Level, /2 = .005 and
( pˆ 1  pˆ 2 )  z
z.005 = 2.575
pˆ 1qˆ1 pˆ 2 qˆ 2

n1
n2
(.71 - .67) + 2.575
(.71)(.29) (.67)(.33)

= .04 ± .085
409
378
-.045 < p1 - p2 < .125
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
Chapter 10: Statistical Inferences About Two Populations
33
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.
10.54
Term
Whole Life
x t = $75,000
st = $22,000
nt = 27
x w = $45,000
sw = $15,500
nw = 29
df = 27 + 29 - 2 = 54
For a 95% Confidence Level, /2 = .025 and
s1 (n1  1)  s 2 (n2  1) 1
1

n1  n2  2
n1 n2
2
( x1  x 2 )  t
t.025,40 = 2.021 (used df=40)
2
(75,000 – 45,000) + 2.021
(22,000) 2 (26)  (15,500) 2 (28) 1
1

27  29  2
27 29
30,000 ± 10,220.73
19,779.27 < µ1 - µ2 < 40,220.73
10.55
Morning
43
51
37
24
47
44
50
55
46
n=9
Afternoon
41
49
44
32
46
42
47
51
49
d = -0.444
For a 90% Confidence Level:
d
2
2
-7
-8
1
2
3
4
-3
sd =4.447
/2 = .05 and t.05,8 = 1.86
df = 9 - 1 = 8
Chapter 10: Statistical Inferences About Two Populations
d t
34
sd
n
-0.444 + (1.86)
4.447
= -0.444 ± 2.757
9
-3.201 < D < 2.313
10.56 Let group 1 be 1990
Ho:
Ha:
p1 - p2 = 0
p1 - p2 < 0
 = .05
The critical table z value is:
n1 = 1300
p
z
n2 = 1450
z.05 = -1.645
p̂1 = .447
p̂2 = .487
n1 pˆ 1  n2 pˆ 2 (.447)(1300)  (.487)(1450)
= .468

n1  n2
1300  1450
( pˆ 1  pˆ 2 )  ( p1  p 2 )
1
1
p  q  
 n1 n 



(.447  .487)  (0)
1 
 1
(.468)(. 532)


 1300 1450 
= -2.10
Since the observed z = -3.73 is less than z.05 = -1.645, the decision is to reject the
null hypothesis. 1997 has a significantly higher proportion.
10.57
Accounting
Data Entry
n1 = 16
x 1 = 26,400
s1 = 1,200
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
Chapter 10: Statistical Inferences About Two Populations
F =
s1
2
s2
2

35
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.
10.58 H0: 12 = 22
Ha: 12  22
dfnum = 6
 = .01
n1 = 8
n2 = 7
S12 = 72,909
S22 = 142,512
dfdenom = 7
The critical F values are:
F =
s1
2
s2
2

F.005,6,7 = 9.16
F.995,7,6 = .11
142,512
= 1.95
72,909
Since F = 1.95 < F.005,6,7 = 9.16 but also > F.995,7,6 = .11, the decision is to fail to
reject the null hypothesis. There is no difference in the variances of the shifts.
10.59
Men
Women
n1 = 60
x 1 = 631
s1 = 100
n2 = 41
x 2 = 848
s2 = 100
For a 95% Confidence Level, /2 = .025 and z.025 = 1.96
2
( x1  x 2 )  z
2
s1
s
 2
n1
n2
1002 1002

(631 – 848) + 1.96
= -217 ± 39.7
60
41
-256.7 < µ1 - µ2 < -177.3
Chapter 10: Statistical Inferences About Two Populations
10.60 Ho:
Ha:
36
 = .01
df = 20 + 24 - 2 = 42
µ1 - µ2 = 0
µ1 - µ2  0
Detroit
Charlotte
n1 = 20
x 1 = 17.53
s1 = 3.2
n2 = 24
x 2 = 14.89
s2 = 2.7
For two-tail test, /2 = .005 and the critical t.005,42 = ±2.704 (used df=40)
t =
t =
( x1  x 2 )  ( 1   2 )
s1 (n1  1)  s 2 (n2  1) 1
1

n1  n2  2
n1 n2
2
2
(17.53  14.89)  (0)
(3.2) (19)  (2.7) 2 (23) 1
1

42
20 24
2
= 2.97
Since the observed t = 2.97 > t.005,42 = 2.704, the decision is to reject the null
hypothesis.
10.61
Ho:
Ha:
With Fertilizer
Without Fertilizer
x 1 = 38.4
s1 = 9.8
n1 = 35
x 2 = 23.1
s2 = 7.4
n2 = 35
µ1 - µ2 = 0
µ1 - µ2 > 0
For one-tail test,  = .01 and z.01 = 2.33
z =
( x 1  x 2 )  ( 1   2 )
2
2
s1
s
 2
n1
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.
Chapter 10: Statistical Inferences About Two Populations
10.62
Specialty
Discount
n1 = 350
p̂ 1 = .75
n2 = 500
p̂ 2 = .52
For a 90% Confidence Level, /2 = .05 and
( pˆ 1  pˆ 2 )  z
37
z.05 = 1.645
pˆ 1qˆ1 pˆ 2 qˆ 2

n1
n2
(.75 - .52) + 1.645
(.75)(.25) (.52)(.48)

350
500
= .23 ± .053
.177 < p1 - p2 < .283
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
10.64
Name Brand
54
55
59
53
54
61
51
53
n=8
Store Brand
49
50
52
51
50
56
47
49
d = 4.5
d
5
5
7
2
4
5
4
4
sd=1.414
df = 8 - 1 = 7
For a 90% Confidence Level, /2 = .05 and
d t
38
t.05,7 = 1.895
sd
n
4.5 + 1.895
1.414
= 4.5 ± .947
8
3.553 < D < 5.447
10.65 Ho:
Ha:
 = .01
df = 23 + 19 - 2 = 40
µ1 - µ2 = 0
µ1 - µ2 < 0
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
t =
t =
( x1  x 2 )  ( 1   2 )
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
Chapter 10: Statistical Inferences About Two Populations
39
Since the observed t = -2.44 < t.01,40 = -2.423, the decision is to reject the null
hypothesis.
10.66
Wednesday
71
56
75
68
74
n=5
Friday
53
47
52
55
58
d = 15.8
sd = 5.263
d
18
9
23
13
16
df = 5 - 1 = 4
 = .05
Ho: D = 0
Ha: D > 0
For one-tail test,  = .05 and the critical t.05,4 = 2.132
d  D 15.8  0
= 6.71

sd
5.263
t =
n
5
Since the observed t = 6.71 > t.05,4 = 2.132, the decision is to reject the null
hypothesis.
10.67 Ho: P1 - P2 = 0
Ha: P1 - P2  0
pˆ 1 
p
 = .05
Machine 1
Machine 2
x1 = 38
n1 = 191
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
Chapter 10: Statistical Inferences About Two Populations
z
( pˆ 1  pˆ 2 )  ( p1  p 2 )
1
1
p  q  
 n1 n 



(.199  .104)  (0)
1 
 1
(.15)(. 85)


 191 202 
40
= 2.64
Since the observed z = 2.64 > zc = 1.96, the decision is to reject the null
hypothesis.
10.68 Construction
Telephone Repair
n1 = 338
x1 = 297
pˆ 1 
n2 = 281
x2 = 192
x1 297
= .879

n1 338
pˆ 2 
x2 192
= .683

n2 281
For a 90% Confidence Level, /2 = .05 and z.05 = 1.645
( pˆ 1  pˆ 2 )  z
pˆ 1qˆ1 pˆ 2 qˆ 2

n1
n2
(.879)(.121) (.683)(.317)
= .196 ± .054

338
281
(.879 - .683) + 1.645
.142 < p1 - p2 < .250
10.69
Aerospace
Automobile
n1 = 33
x 1 = 12.4
s1 = 2.9
n2 = 35
x 2 = 4.6
s2 = 1.8
For a 99% Confidence Level, /2 = .005 and z.005 = 2.575
2
2
s
s
( x1  x 2 )  z 1  2
n1
n2
(12.4 – 4.6) + 2.575
6.28 < µ1 - µ2 < 9.32
(2.9) 2 (1.8) 2

33
35
= 7.8 ± 1.52
Chapter 10: Statistical Inferences About Two Populations
10.70
Ho:
Ha:
Discount
Specialty
x 1 = $47.20
s1 = $12.45
n1 = 60
x 2 = $27.40
s2 = $9.82
n2 = 40
µ1 - µ2 = 0
µ1 - µ2  0
 = .01
41
For two-tail test, /2 = .005 and zc = ±2.575
( x 1  x 2 )  ( 1   2 )
z =
2
2

(47.20  27.40)  (0)
s1
s
 2
n1
n2
(12.45) 2 (9.82) 2

60
40
= 8.86
Since the observed z = 8.86 > zc = 2.575, the decision is to reject the null
hypothesis.
10.71
Before
12
7
10
16
8
n=5
After
8
3
8
9
5
d = 4.0
sd = 1.8708
d
4
4
2
7
3
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
10.72 Ho:
Ha:
42
 = .01
df = 10 + 6 - 2 = 14
µ1 - µ2 = 0
µ1 - µ2  0
A
B
n1 = 10
x 1 = 18.3
s12 = 17.122
n2 = 6
x 2 = 9.667
s22 = 7.467
For two-tail test, /2 = .005 and the critical t.005,14 = ±2.977
( x1  x 2 )  ( 1   2 )
t =
t =
s1 (n1  1)  s 2 (n2  1) 1
1

n1  n2  2
n1 n2
2
2
(18.3  9.667)  (0)
=
(17.122)(9)  (7.467)(5) 1 1

14
10 6
4.52
Since the observed t = 4.52 > t.005,14 = 2.977, the decision is to reject the null
hypothesis.
10.73 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 - µ2 = 0
µ1 - µ2 > 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.
10.74 H0: D = 0
Ha: D  0
This is a related measures before and after study. Fourteen people were involved
in the study. Before the treatment, the sample mean was 4.357 and after the
Chapter 10: Statistical Inferences About Two Populations
43
treatment, the mean was 5.214. The higher number after the treatment indicates
that subjects were more likely to “blow the whistle” after having been through the
treatment. The observed t value was –3.12 which was more extreme than twotailed table t value of + 2.16 causing the researcher to reject the null hypothesis.
This is underscored by a p-value of .0081 which is less than  = .05. The study
concludes that there is a significantly higher likelihood of “blowing the whistle”
after the treatment.
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 .
10.76 A test of differences of the variances of the populations of the two machines is
being computed. The hypotheses are:
H0: 12 = 22
Ha: 12  22
Twenty-six pipes were measured for sample one and twenty-six pipes were
measured for sample two. The observed F = 1.79 is not significant at  = .05 for
a two-tailed test since the associated p-value is .0758. There is no significant
difference in the variance of pipe lengths for pipes produced by machine A versus
machine B.