10 ch ken black solution

1. 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 the z statistic.

2. Test hypotheses and establish confidence intervals about the difference in two
population means using the t statistic.

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.
Conceptually, this is not radically different than the z test for a single sample mean shown
initially in Chapter 7. In analyzing the differences in two sample means where the
population variances are unknown, if it can be assumed that the populations are normally
distributed, a t test for independent samples can be used. There are two different

2. Chapter 10: Statistical Inferences About Two Populations 2
formulas given in the chapter to conduct this t test. One version 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 10.14 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 (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 Constr uct Confidence
Intervals about the Difference in Two Popu lation Means Using the t
Test
Confidence Intervals

10.3 Statistical Inferences For Two Related Populations
Hypothesis Testing
Using the Computer to Make Statistical Inference s about Two Related
Populations
Confidence Intervals

3. Chapter 10: Statistical Inferences About Two Populations 3

10.4 Statistical Inferences About Two Population Proportions, p
1 – p2
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 Independent Samples
F Distribution Matched-Pairs Test
F Value Related Measures





SOLUTIONS TO PROBLEMS IN CHAPTER 10


10.1 Sample 1
Sample 2

x 1 = 51.3 x2 = 53.2

s1
2 = 52 s2
2 = 60

n1 = 32 n2 = 32

a) H
o: µ1 - µ2 = 0
H
a: µ1 - µ2 < 0

For one-tail test, a = .10
z.10 = -1.28


z =
32
60
32
52
)0()2.533.51()()(
2
2
2
1
2
1
21
21
+
--
=
+
---
nn
xx
ss
mm
= -1.02

Since the observed z = -1.02 > z
c = -1.645, the decision is to fail to reject the null
hypothesis
.

4. Chapter 10: Statistical Inferences About Two Populations 4

b) Critical value method:

z
c =
2
2
2
1
2
1
21
21
)()(
nn
xx
c
ss
mm
+
---


-1.645 =
32
60
32
52
)0()(
21
+
--
cxx


(
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

n
1 = 32 n 2 = 31

x1 = 70.4 x2 = 68.7

s1 = 5.76 s2 = 6.1

For a 90% C.I., z
.05 = 1.645


2
2
2
1
2
1
21
)(
nn
zxx
ss
+±-

(70.4) – 68.7) +
1.645
31
1.6
32
76.5
22
+

1.7 ± 2.465


-.76 <
µ
1 - µ
2 < 4.16

5. Chapter 10: Statistical Inferences About Two Populations 5

10.3 a) Sample 1 Sample 2

x1 = 88.23 x2 = 81.2

s1
2 = 22.74 s2
2 = 26.65
n
1 = 30 n 2 = 30

H
o: µ 1 - µ2 = 0
H
a: µ 1 - µ2 ¹ 0

For two-tail test, use
a/2 = .01 z .01 = +
2.33

z =
30
65.26
30
74.22
)0()2.8123.88()()(
2
2
2
1
2
1
21
21
+
--
=
+
---
nn
xx
ss
mm
= 5.48

Since the observed z = 5.48 > z
.01 = 2.33, the decision is to reject the null
hypothesis.


b)
2
2
2
1
2
1
21
)(
nn
zxx
ss
+±-

(88.23 – 81.2) +
2.33
30
65.26
30
74.22
+

7.03 + 2.99


4.04 <
mmmm < 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 Food/Beverage

x1 = 1.96 x2 = 3.02

s1
2 = 1.0188 s2
2 = 0.9180

n1 = 50 n2 = 50

H
o: µ1 - µ2 = 0
H
a: µ1 - µ2 ¹ 0

For two-tail test, a/2 = .005
z.005 = ±2.575

6. Chapter 10: Statistical Inferences About Two Populations 6

z =
50
9180.0
50
0188.1
)0()02.396.1()()(
2
2
2
1
2
1
21
21
+
--
=
+
---
nn
xx
ss
mm
= -5.39

Since the observed z = -5.39 < z
c = -2.575, the decision is to reject the null
hypothesis.


10.5 A
B

n
1 = 40 n 2 = 37

x1 = 5.3 x2 = 6.5

s1
2 = 1.99 s2
2 = 2.36

For a 95% C.I., z
.025 = 1.96


2
2
2
1
2
1
21
)(
nn
zxx
ss
+±-

(5.3 – 6.5) +
1.96
37
36.2
40
99.1
+


-1.2 ± .66 -1.86 <
mmmm < -.54

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 n2 = 41

x1 = 1.84 x2 = 1.99

s1 = .38 s2 = .51

for a 98% C.I.,
z.01 = 2.33


2
2
2
1
2
1
21
)(
nn
zxx
ss
+±-

7. Chapter 10: Statistical Inferences About Two Populations 7
(1.84 - 1.99) ± 2.33
41
51.
35
38.
22
+

-.15 ± .2384

-.3884 <
µ
1 - µ
2 < .0884

Point Estimate = -.15


Hypothesis Test:

1) H
o: µ1 - µ2 = 0
H
a: µ1 - µ2 ¹ 0

2) z =
2
2
2
1
2
1
21
21
)()(
nn
xx
ss
mm
+
---


3) a = .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 =
41
)51(.
35
)38(.
)0()99.184.1(
22
+
--
= -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.

8. Chapter 10: Statistical Inferences About Two Populations 8

10.7 1994 2001

x
1 = 190 x
2 = 198

s1 = 18.50 s2 = 15.60
n
1 = 51 n 2 = 47 a = .01

H
0: m1 - m2 = 0
H
a: m1 - m2 < 0
For a one-tailed test, z
.01 = -2.33

z =
47
)60.15(
51
)50.18(
)0()198190()()(
22
2
2
2
1
2
1
21
21
+
--
=
+
---
nn
xx
ss
mm
= -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

n
1 = 31 n 2 = 31

x1 = 2.64 x2 = 2.36

s1
2 = .03 s2
2 = .015

For a 99% C.I., z
.005 = 2.575


2
2
2
1
2
1
21
)(
nn
zxx
ss
+±-

(2.64-2.36) ± 2.575
31
015.
31
03.
+


.28 ± .10 .18 <
mmmm < .38

Between $ .18 and $ .38 difference with Seattle being more expensive.

9. Chapter 10: Statistical Inferences About Two Populations 9

10.9 Canon Pioneer

x1 = 5.8 x2 = 5.0

s1 = 1.7 s2 = 1.4
n
1 = 36 n 2 = 45

H
o: µ 1 - µ2 = 0
H
a: µ 1 - µ2 ¹ 0

For two-tail test, a/2 = .025 z
.025 = ±1.96

z =
45
)4.1(
36
)7.1(
)0()0.58.5()()(
2
2
2
2
1
2
1
21
21
+
--
=
+
---
nn
xx
ss
mm
= 2.27

Since the observed z = 2.27 > z
c = 1.96, the decision is to reject the null hypothesis.


10.10 A
B

x1 = 8.05 x2 = 7.26

s1 = 1.36 s2 = 1.06
n
1 = 50 n 2 = 38

H
o: µ 1 - µ2 = 0
H
a: µ 1 - µ2 > 0

For one-tail test, a = .10 z
.10 = 1.28

z =
38
)06.1(
50
)36.1(
)0()26.705.8()()(
22
2
2
2
1
2
1
21
21
+
--
=
+
---
nn
xx
ss
mm
= 3.06

Since the observed z = 3.06 > z
c = 1.28, the decision is to reject the null
hypothesis.

10. Chapter 10: Statistical Inferences About Two Populations 10

10.11 H
o: µ1 - µ2 = 0 a = .01
H
a: µ1 - µ2 < 0 df = 8 + 11 - 2 = 17

Sample 1
Sample 2

n
1 = 8 n 2 = 11

x1 = 24.56 x2 = 26.42

s1
2 = 12.4 s2
2 = 15.8

For one-tail test, a = .01 Critical
t.01,19 = -2.567


t =
2121
2
2
21
2
1
21
2
1
11
2
)1()1(
)()(
nnnn
nsns
xx
+
-+
-+-
---
mm
=
11
1
8
1
2118
)10(8.15)7(4.12
)0()42.2656.24(
+
-+
+
--
= -1.05


Since the observed t = -1.05 > t
.01,19 = -2.567, the decision is to fail to reject the
null hypothesis
.


10.12 a) H
o: µ1 - µ2 = 0 a =.10
H
a: µ1 - µ2 ¹ 0 df = 20 + 20 - 2 = 38

Sample 1
Sample 2
n
1 = 20 n 2 = 20

x1 = 118 x2 = 113
s
1 = 23.9 s 2 = 21.6

For two-tail test, a/2 = .05 Critical t
.05,38 = 1.697 (used df=30)

t =
2121
2
2
21
2
1
21
2
1
11
2
)1()1(
)()(
nnnn
nsns
xx
+
-+
-+-
---
mm
=

t =
20
1
20
1
22020
)19()6.21()19()9.23(
)0()42.2656.24(
22
+
-+
+
--
= 0.69

Since the observed t = 0.69 < t .05,38 = 1.697, the decision is to fail to reject
the null hypothesis
.

11. Chapter 10: Statistical Inferences About Two Populations 11

b)
2121
2
2
21
2
1
21
11
2
)1()1(
)(
nnnn
nsns
txx +
-+
-+-
±-
=

(118 – 113) +
1.697
20
1
20
1
22020
)19()6.21()19()9.23(
22
+
-+
+


5 +
12.224

-7.224 < mmmm1 - mmmm2 < 17.224

10.13 H
o: µ1 - µ2 = 0 a = .05
H
a: µ1 - µ2 > 0 df = n 1 + n2 - 2 = 10 + 10 - 2 = 18

Sample 1
Sample 2

n
1 = 10 n 2 = 10

x1 = 45.38 x2 = 40.49
s
1 = 2.357 s 2 = 2.355

For one-tail test, a = .05 Critical t
.05,18 = 1.734

t =
2121
2
2
21
2
1
21
2
1
11
2
)1()1(
)()(
nnnn
nsns
xx
+
-+
-+-
---
mm
=

t =
10
1
10
1
21010
)9()355.2()9()357.2(
)0()49.4038.45(
22
+
-+
+
--
= 4.64

Since the observed t = 4.64 > t
.05,18 = 1.734, the decision is to reject the
null hypothesis.


10.14 H
o: µ1 - µ2 = 0 a =.01
H
a: µ1 - µ2 ¹ 0 df = 18 + 18 - 2 = 34

Sample 1
Sample 2

n
1 = 18 n 2 = 18

x1 = 5.333 x2 = 9.444
s
1
2 = 12 s 2
2 = 2.026

12. Chapter 10: Statistical Inferences About Two Populations 12
For two-tail test, a/2 = .005 Critical t .005,34 = ±2.457 (used df=30)

t =
2121
2
2
21
2
1
21
2
1
11
2
)1()1(
)()(
nnnn
nsns
xx
+
-+
-+-
---
mm
=

t =
18
1
18
1
21818
17)026.2()17(12
)0()444.9333.5(
+
-+
+
--
= -4.66

Since the observed t = -4.66 < t
.005,34 = -2.457

Reject the null hypothesis


b)
2121
2
2
21
2
1
21
11
2
)1()1(
)(
nnnn
nsns
txx +
-+
-+-
±-
=

(5.333 – 9.444) +
2.457
18
1
18
1
21818
)17)(026.2()17)(12(
+
-+
+


-4.111 + 2.1689


-6.2799 <
mmmm1 - mmmm2 < -1.9421

10.15 Peoria Evansville


n1 = 21 n2 = 26

1
x = 86,900
2
x = 84,000

s1 = 2,300 s2 = 1,750 df = 21 + 26 – 2

90% level of confidence, a/2 = .05
t .05,45 = 1.684 (used df = 40)


2121
2
2
21
2
1
21
11
2
)1()1(
)(
nnnn
nsns
txx +
-+
-+-
±-
=

(86,900 – 84,000) +
1.684
26
1
21
1
22621
)25()1750()20()2300(
22
+
-+
+
=
2,900 +
994.62


1905.38 <
mmmm1 - mmmm2 < 3894.62

13. Chapter 10: Statistical Inferences About Two Populations 13

10.16 H
o: µ1 - µ2 = 0 a = .05
H
a: µ1 - µ2 ¹ 0!= 0 df = 21 + 26 - 2 = 45

Peoria
Evansville

n
1 = 21 n 2 = 26

x1 = $86,900 x2 = $84,000
s
1 = $2,300 s 2 = $1,750

For two-tail test, a/2 = .025
Critical t
.025,45 = ± 2.021 (used df=40)

t =
2121
2
2
21
2
1
21
2
1
11
2
)1()1(
)()(
nnnn
nsns
xx
+
-+
-+-
---
mm
=

t =
26
1
21
1
22621
)25()750,1()20()300,2(
)0()000,84900,86(
22
+
-+
+
--
= 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) H
o: µ1 - µ2 = 0
H
a: µ1 - µ2 > 0

2) t =
2121
2
2
21
2
1
21
2
1
11
2
)1()1(
)()(
nnnn
nsns
xx
+
-+
-+-
---
mm


3) a = .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
Dallas

n
1 = 8 n 2 = 9

x1 = 47 x2 = 44
s
1 = 3 s 2 = 3

14. Chapter 10: Statistical Inferences About Two Populations 14

6) t =
9
1
8
1
15
)3(8)3(7
)0()4447(
22
+
+
--
= 2.06

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 n
m = 22 n no = 20

xm = 112 xno = 122
s
m = 11 s
no = 12

df = n
m + nno - 2 = 22 + 20 - 2 = 40

For a 98% Confidence Interval, a/2 = .01 and t
.01,40 = 2.423


2121
2
2
21
2
1
21
11
2
)1()1(
)(
nnnn
nsns
txx +
-+
-+-
±-
=

(112 – 122) +
2.423
20
1
22
1
22022
)19()12()21()11(
22
+
-+
+


-10 ± 8.63

-$18.63 <
µ1 - µ2 < -$1.37

Point Estimate = -$10



10.19 H
o: µ1 - µ2 = 0
H
a: µ1 - µ2 ¹ 0

df = n
1 + n2 - 2 = 11 + 11 - 2 = 20

Toronto
Mexico City

n
1 = 11 n 2 = 11

x1 = $67,381.82 x2 = $63,481.82
s
1 = $2,067.28 s 2 = $1,594.25

For a two-tail test, a/2 = .005 Critical t
.005,20 = ±2.845

15. Chapter 10: Statistical Inferences About Two Populations 15
t =
2121
2
2
21
2
1
21
2
1
11
2
)1()1(
)()(
nnnn
nsns
xx
+
-+
-+-
---
mm
=

t =
11
1
11
1
21111
)10()25.594,1()10()28.067,2(
)0()82.481,6382.381,67(
22
+
-+
+
--
= 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

n
1 = 11 n 2 = 11

x1 = $67,381.82 x2 = $63,481.82
s
1 = $2,067.28 s 2 = $1,594.25

df = n
1 + n2 - 2 = 11 + 11 - 2 = 20
For a 95% Level of Confidence, a/2 = .025 and t
.025,20 = 2.086


2121
2
2
21
2
1
21
11
2
)1()1(
)(
nnnn
nsns
txx +
-+
-+-
±-
=

($67,381.82 - $63,481.82) ± (2.086)
11
1
11
1
21111
)10()25.594,1()10()28.067,2(
22
+
-+
+


3,900 ± 1,641.9

2,258.1 <
µ1 - µ2 < 5,541.9

16. Chapter 10: Statistical Inferences About Two Populations 16
10.21 H o: D = 0
H
a: D > 0

Sample 1
Sample 2 d
38 22 16
27 28 -1
30 21 9
41 38 3
36 38 -2
38 26 12
33 19 14
35 31 4
44 35 9

n = 9 d=7.11 sd=6.45 a = .01

df =
n - 1 = 9 - 1 = 8

For one-tail test and a = .01, the critical
t.01,8 = ±2.896


t =
9
45.6
011.7-
=
-
n
s
Dd
d
= 3.31

Since the observed t = 3.31 > t
.01,8 = 2.896, the decision is to reject the null
hypothesis
.


10.22 H
o: D = 0
H
a: D ¹ 0

Before
After d
107 102 5
99 98 1
110 100 10
113 108 5
96 89 7
98 101 -3
100 99 1
102 102 0
107 105 2
109 110 -1
104 102 2
99 96 3
101 100 1

17. Chapter 10: Statistical Inferences About Two Populations 17

n = 13 d = 2.5385 s d=3.4789 a = .05
df = n - 1 = 13 - 1 = 12

For a two-tail test and a/2 = .025 Critical t
.025,12 = ±2.179

t =
13
4789.3
05385.2-
=
-
n
s
Dd
d
= 2.63

Since the observed t = 2.63 > t
.025,12 = 2.179, the decision is to reject the null
hypothesis
.




10.23 n = 22
d= 40.56 s d = 26.58

For a 98% Level of Confidence, a/2 = .01, and df = n - 1 = 22 - 1 = 21

t
.01,21 = 2.518


n
s
td
d
±

40.56 ± (2.518)
22
58.26


40.56 ± 14.27


26.29 <
D < 54.83



10.24 Before After d
32 40 -8
28 25 3
35 36 -1
32 32 0
26 29 -3
25 31 -6
37 39 -2
16 30 -14
35 31 4

18. Chapter 10: Statistical Inferences About Two Populations 18

n = 9 d= -3 s d = 5.6347 a = .025
df = n - 1 = 9 - 1 = 8

For 90% level of confidence and a/2 = .025, t
.05,8 = 1.86

t =
n
s
td
d
±


t = -3 +
(1.86)

9
6347.5
= -3 ± 3.49


-0.49 <
D < 6.49


10.25 City Cost Resale d

Atlanta 20427 25163 -4736
Boston 27255 24625 2630
Des Moines 22115 12600 9515
Kansas City 23256 24588 -1332
Louisville 21887 19267 2620
Portland 24255 20150 4105
Raleigh-Durham 19852 22500 -2648
Reno 23624 16667 6957
Ridgewood 25885 26875 - 990
San Francisco 28999 35333 -6334
Tulsa 20836 16292 4544

d = 1302.82 s
d = 4938.22 n = 11, df = 10


a = .01 a/2 = .005 t
.005,10= 3.169


n
s
td
d
± = 1302.82 +
3.169
11
22.4938
= 1302.82 + 4718.42


-3415.6 <
D < 6021.2

19. Chapter 10: Statistical Inferences About Two Populations 19
10.26 Ho: D = 0
H
a: D < 0

Before
After d
2 4 -2
4 5 -1
1 3 -2
3 3 0
4 3 1
2 5 -3
2 6 -4
3 4 -1
1 5 -4

n = 9 d=-1.778 s d=1.716 a = .05 df = n - 1 = 9 - 1 = 8

For a one-tail test and a = .05, the critical t
.05,8 = -1.86


t = 9
716.1
0778.1--
=
-
n
s
Dd
d
= -3.11

Since the observed t = -3.11 < t
.05,8 = -1.86, the decision is to reject the null
hypothesis
.


10.27 Before
After d
255 197 58
230 225 5
290 215 75
242 215 27
300 240 60
250 235 15
215 190 25
230 240 -10
225 200 25
219 203 16
236 223 13

n = 11 d= 28.09 s d=25.813 df = n - 1 = 11 - 1 = 10

For a 98% level of confidence and a/2=.01, t
.01,10 = 2.764

20. Chapter 10: Statistical Inferences About Two Populations 20

n
s
td
d
±

28.09 ± (2.764)

11
813.25
= 28.09 ± 21.51


6.58 <
D < 49.60



10.28 H
0: D = 0
H
a: D > 0 n = 27 df = 27 – 1 = 26
d
= 3.17 sd = 5

Since
a = .01, the critical t.01,26 = 2.479


t =
27
5
071.3-
=
-
n
s
Dd
d
= 3.86

Since the observed t = 3.86 > t
.01,26 = 2.479, the decision is to reject the null
hypothesis
.

10.29 n = 21
d= 75 s d=30 df = 21 - 1 = 20

For a 90% confidence level, a/2=.05 and t
.05,20 = 1.725


n
s
td
d
±

75 +
1.725
21
30
= 75 ± 11.29


63.71 <
D < 86.29


10.30 H
o: D = 0
H
a: D ¹ 0


n = 15
d= -2.85 sd = 1.9 a = .01 df = 15 - 1 = 14

For a two-tail test,
a/2 = .005 and the critical t.005,14 = +
2.977

21. Chapter 10: Statistical Inferences About Two Populations 21
t =
15
9.1
085.2--
=
-
n
s
Dd
d
= -5.81

Since the observed t = -5.81 < t
.005,14 = -2.977, the decision is to reject the null
hypothesis
.



10.31 a) Sample 1
Sample 2

n
1 = 368 n 2 = 405
x
1 = 175 x 2 = 182


368
175
ˆ
1
1
1
==
n
x
p = .476
405
182
ˆ
2
2
2
==
n
x
p = .449


773
357
405368
182175
21
21
=
+
+
=
+
+
=
nn
xx
p = .462


H
o: p1 - p2 = 0
H
a: p1 - p2 ¹ 0

For two-tail,
a/2 = .025 and z .025 = ±1.96








+
--
=








+×
---
=
405
1
368
1
)538)(.462(.
)0()449.476(.
11
)()ˆˆ(
1
2121nn
qp
pppp
z =
0.75

Since the observed z = 0.75 < z
c = 1.96, the decision is to fail to reject the null
hypothesis
.

b) Sample 1
Sample 2


p
ˆ1 = .38 pˆ2 = .25
n
1 = 649 n 2 = 558


558649
)25(.558)38(.649ˆˆ
21
2211 +
+
=
+
+
=
nn
pnpn
p
= .32

22. Chapter 10: Statistical Inferences About Two Populations 22

H
o: p 1 - p2 = 0
H
a: p 1 - p2 > 0

For a one-tail test and a = .10, z
.10 = 1.28








+
--
=








+×
---
=
558
1
649
1
)68)(.32(.
)0()25.38(.
11
)()ˆˆ(
1
2121nn
qp
pppp
z =
4.83

Since the observed z = 4.83 > z
c = 1.28, the decision is to reject the null
hypothesis
.

10.32 a) n
1 = 85 n 2 = 90 p
ˆ1 = .75 pˆ2 = .67

For a 90% Confidence Level, z
.05 = 1.645


2
22
1
11
21
ˆˆˆˆ
)ˆˆ(
n
qp
n
qp
zpp +±-

(.75 - .67) ± 1.645
90
)33)(.67(.
85
)25)(.75(.
+ = .08 ± .11


-.03 <
p
1 - p
2 < .19


b)
n1 = 1100 n2 = 1300 p
ˆ1 = .19 pˆ2 = .17

For a 95% Confidence Level, a/2 = .025 and
z.025 = 1.96


2
22
1
11
21
ˆˆˆˆ
)ˆˆ(
n
qp
n
qp
zpp +±-

(.19 - .17) +
1.96
1300
)83)(.17(.
1100
)81)(.19(.
+
= .02 ± .03


-.01 <
p
1 - p
2 < .05

23. Chapter 10: Statistical Inferences About Two Populations 23
c) n 1 = 430 n 2 = 399 x 1 = 275 x 2 = 275


430
275
ˆ
1
1
1
==
n
x
p = .64
399
275
ˆ
2
2
2
==
n
x
p = .69

For an 85% Confidence Level, a/2 = .075 and z
.075 = 1.44


2
22
1
11
21
ˆˆˆˆ
)ˆˆ(
n
qp
n
qp
zpp +±-


(.64 - .69) + 1.44
399
)31)(.69(.
430
)36)(.64(.
+
= -.05 ± .047

-.097 <
p1 - p2 < -.003


d) n
1 = 1500 n 2 = 1500 x 1 = 1050 x 2 = 1100


1500
1050
ˆ
1
1
1
==
n
x
p = .70
1500
1100
ˆ
2
2
2
==
n
x
p = .733

For an 80% Confidence Level, a/2 = .10 and z
.10 = 1.28


2
22
1
11
21
ˆˆˆˆ
)ˆˆ(
n
qp
n
qp
zpp
+±-

(.70 - .733) ± 1.28
1500
)267)(.733(.
1500
)30)(.70(.
+
= -.033 ± .02

-.053 <
p
1 - p
2 < -.013

10.33 H
0: pm - pw = 0
H
a: pm - pw < 0 n m = 374 n w = 481 p
ˆ
m = .59
pˆ
w = .70

For a one-tailed test and a = .05, z
.05 = -1.645


481374
)70(.481)59(.374ˆˆ
+
+
=
+
+
=
wm
wwmmnn
pnpn
p
= .652

24. Chapter 10: Statistical Inferences About Two Populations 24







+
--
=








+×
---
=
481
1
374
1
)348)(.652(.
)0()70.59(.
11
)()ˆˆ(
1
2121nn
qp
pppp
z =
-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
1
ˆp = .24
2
ˆp = .35

For a 90% Confidence Level,
a/2 = .05 and z.05 = +
1.645


2
22
1
11
21
ˆˆˆˆ
)ˆˆ(
n
qp
n
qp
zpp +±-

(.24 - .35) +
1.645
176
)65)(.35(.
210
)76)(.24(.
+
= -.11 +
.0765


-.1865 <
p1 – p2 < -.0335


10.35 Computer Firms Banks


p
ˆ1 = .48 pˆ2 = .56
n
1 = 56 n 2 = 89


8956
)56(.89)48(.56ˆˆ
21
2211 +
+
=
+
+
=
nn
pnpn
p
= .529

H
o: p 1 - p2 = 0
H
a: p 1 - p2 ¹ 0

For two-tail test, a/2 = .10 and z
c = ±1.28







+
--
=








+×
---
=
89
1
56
1
)471)(.529(.
)0()56.48(.
11
)()ˆˆ(
1
2121nn
qp
pppp
z =
-0.94

Since the observed z = -0.94 > z
c = -1.28, the decision is to fail to reject the null
hypothesis
.

25. Chapter 10: Statistical Inferences About Two Populations 25
10.36 A B

n
1 = 35 n 2 = 35
x
1 = 5 x 2 = 7


35
5
ˆ
1
1
1
==
n
x
p = .14
35
7
ˆ
2
2
2
==
n
x
p = .20

For a 98% Confidence Level, a/2 = .01 and z
.01 = 2.33


2
22
1
11
21
ˆˆˆˆ
)ˆˆ(
n
qp
n
qp
zpp +±-


(.14 - .20) ± 2.33
35
)80)(.20(.
35
)86)(.14(.
+
= -.06 ± .21

-.27 <
p1 - p2 < .15

10.37 H
0: p1 – p2 = 0
H
a: p1 – p2 ¹ 0

a = .10
p
ˆ
1
= .09 pˆ
2
= .06 n
1 = 780 n 2 = 915

For a two-tailed test, a/2 = .05 and z
.05 = +
1.645


915780
)06(.915)09(.780ˆˆ
21
2211 +
+
=
+
+
=
nn
pnpn
p
= .0738








+
--
=








+×
---
=
915
1
780
1
)9262)(.0738(.
)0()06.09(.
11
)()ˆˆ(
1
2121nn
qp
pppp
Z =
2.35

Since the observed z = 2.35 > z
.05 = 1.645, the decision is to reject the null
hypothesis
.

26. Chapter 10: Statistical Inferences About Two Populations 26
10.38 n 1 = 850 n 2 = 910 pˆ
1
= .60
p
ˆ2 = .52

For a 95% Confidence Level, a/2 = .025 and z
.025 = +
1.96


2
22
1
11
21
ˆˆˆˆ
)ˆˆ(
n
qp
n
qp
zpp +±-

(.60 - .52) +
1.96
910
)48)(.52(.
850
)40)(.60(.
+
= .08 +
.046


.034 <
p1 – p2 < .126

10.39 H
0: s1
2 = s 2
2 a = .01 n 1 = 10 s 1
2 = 562
H
a: s1
2 < s2
2 n 2 = 12 s 2
2 = 1013

df
num = 12 - 1 = 11 dfdenom = 10 - 1 = 9

Table F
.01,10,9 = 5.26

F =
562
1013
2
1
2
2
=
s
s
= 1.80

Since the observed F = 1.80 < F
.01,10,9 = 5.26, the decision is to fail to reject the
null hypothesis
.

10.40 H
0: s1
2 = s2
2 a = .05 n 1 = 5 S 1 = 4.68
H
a: s1
2 ¹ s2
2 n 2 = 19 S 2 = 2.78

df
num = 5 - 1 = 4 dfdenom = 19 - 1 = 18

The critical table F values are: F
.025,4,18 = 3.61 F
.95,18,4 = .277

F =
2
2
2
2
2
1
)78.2(
)68.4(
=
s
s
=
2.83

Since the observed F = 2.83 < F
.025,4,18 = 3.61, the decision is to fail to reject the
null hypothesis
.

27. Chapter 10: Statistical Inferences About Two Populations 27

10.41 City 1 City 2

1.18 1.08
1.15 1.17
1.14 1.14
1.07 1.05
1.14 1.21
1.13 1.14
1.09 1.11
1.13 1.19
1.13 1.12
1.03 1.13

n
1 = 10 df1 = 9 n 2 = 10 df2 = 9

s
1
2 = .0018989 s 2
2 = .0023378

H
0: s1
2 = s2
2 a = .10 a/2 = .05
H
a: s1
2 ¹ s2
2

Upper tail critical F value = F
.05,9,9 = 3.18

Lower tail critical F value = F
.95,9,9 = 0.314

F =
0023378.
0018989.
2
2
2
1
=
s
s
= 0.81

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) H
0: s1
2 = s2
2
H
a: s1
2 ¹ s2
2

2) F =
2
2
2
1
s
s

3) a = .01

4) df
1 = 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

28. 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) s
1
2 = 393.4 s
2
2 = 702.7

6) F =
7.702
4.393
=
0.56

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 Chicag o.

10.43 H
0: s1
2 = s2
2 a = .05 n1 = 12 s1 = 7.52
H
a: s1
2 > s2
2 n2 = 15 s2 = 6.08

df
num = 12 - 1 = 11 dfdenom = 15 - 1 = 14

The critical table
F value is F.05,10,14 = 5.26

F =
2
2
2
2
2
1
)08.6(
)52.7(
=
s
s
=
1.53

Since the observed F = 1.53 < F
.05,10,14 = 2.60, the decision is to fail to reject the
null hypothesis
.

10.44 H
0: s1
2 = s2
2 a = .01 n 1 = 15 s 1
2 = 91.5
H
a: s1
2 ¹ s2
2 n 2 = 15 s 2
2 = 67.3

df
num = 15 - 1 = 14 dfdenom = 15 - 1 = 14

The critical table F values are: F
.005,12,14 = 4.43 F .995,14,12 = .226

F =
3.67
5.91
2
2
2
1
=
s
s
= 1.36

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.

29. Chapter 10: Statistical Inferences About Two Populations 29

10.45 H
o: µ 1 - µ2 = 0
H
a: µ 1 - µ2 ¹ 0

For a = .10 and a two-tailed test, a/2 = .05 and z
.05 = +
1.645

Sample 1 Sample 2


1x = 138.4
2
x = 142.5

s1 = 6.71 s2 = 8.92
n
1 = 48 n 2 = 39

z = 39
)92.8(
48
)71.6(
)0()5.1424.138()()(
2
2
2
2
1
2
1
21
21
+
--
=
+
---
nn
xx
ss
mm
= -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
Sample 2


1
x = 34.9
2
x = 27.6

s1
2
= 2.97 s2
2
= 3.50
n
1 = 34 n 2 = 31

For 98% Confidence Level, z
.01 = 2.33


2
2
2
1
2
1
21
)(
nn
zxx
ss
+±-

(34.9 – 27.6) +
2.33
31
50.3
34
97.2
+ = 7.3 + 1.04


6.26 <
mmmm1 - mmmm2 < 8.34

30. Chapter 10: Statistical Inferences About Two Populations 30
10.47 Ho: µ 1 - µ2 = 0
H
a: µ 1 - µ2 > 0

Sample 1
Sample 2


1x= 2.06
2x = 1.93
s
1
2
= .176 s 2
2
= .143
n
1 = 12 n 2 = 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 =
2121
2
2
21
2
1
21
2
1
11
2
)1()1(
)()(
nnnn
nsns
xx
+
-+
-+-
---
mm


t =
15
1
12
1
25
)14)(143(.)11)(176(.
)0()93.106.2(
+
+
--
= 0.85

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 Sample 2
x1 = 74.6 x2 = 70.9

s1
2
= 10.5 s2
2
= 11.4

n1 = 18 n2 = 19

For 95% confidence, a/2 = .025.
Using df = 18 + 19 - 2 = 35,
t
35,.025 = 2.042


2121
2
2
21
2
1
21
11
2
)1()1(
)(
nnnn
nsns
txx +
-+
-+-
±-


(74.6 – 70.9) +
2.042
20
1
20
1
22020
)19()6.21()19()9.23(
22
+
-+
+


3.7 +
2.22

1.48 < mmmm1 - mmmm2 < 5.92

31. Chapter 10: Statistical Inferences About Two Populations 31

10.49 H
o: D = 0 a = .01
H
a: D < 0

n = 21 df = 20
d = -1.16 s
d = 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 =
21
01.1
016.1--
=
-
n
s
Dd
d
= -5.26

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 d

1 47 63 -16
2 33 35 - 2
3 38 36 2
4 50 56 - 6
5 39 44 - 5
6 27 29 - 2
7 35 32 3
8 46 54 - 8
9 41 47 - 6

d = -4.44 s d = 5.703 df = 8

For a 99% Confidence Level, a/2 = .005 and t
8,.005 = 3.355


n
s
td
d
± = -4.44 +
3.355
9
703.5
= -4.44 + 6.38


-10.82 <
D < 1.94

10.51 H
o: p1 - p2 = 0 a = .05 a/2 = .025
H
a: p1 - p2 ¹ 0 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.

32. Chapter 10: Statistical Inferences About Two Populations 32

Sample 1 Sample 2

x
1 = 345 x 2 = 421
n
1 = 783 n 2 = 896


896783
421345
21
21 +
+
=
+
+
=
nn
xx
p
= .4562


783
345
ˆ
1
1
1
==
n
x
p = .4406
896
421
ˆ
2
2
2
==
n
x
p = .4699








+
--
=








+×
---
=
896
1
783
1
)5438)(.4562(.
)0()4699.4406(.
11
)()ˆˆ(
1
2121nn
qp
pppp
z =
-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

n
1 = 409 n 2 = 378

p
ˆ1 = .71 pˆ2 = .67

For a 99% Confidence Level,
a/2 = .005 and z .005 = 2.575


2
22
1
11
21
ˆˆˆˆ
)ˆˆ(
n
qp
n
qp
zpp +±-

(.71 - .67) +
2.575
378
)33)(.67(.
409
)29)(.71(.
+
= .04 ± .085


-.045 <
p
1 - p
2 < .125

10.53 H
0: s1
2 = s2
2 a = .05 n 1 = 8 s 1
2
= 46
H
a: s1
2 ¹ s2
2 n 2 = 10 S 2
2 = 37

df
num = 8 - 1 = 7 dfdenom = 10 - 1 = 9
The critical F values are: F
.025,7,9 = 4.20 F
.975,9,7 = .238

33. 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 =
37
46
2
2
2
1
=
s
s
= 1.24

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

xt = $75,000 xw = $45,000
s
t = $22,000 s w = $15,500
n
t = 27 n w = 29

df = 27 + 29 - 2 = 54

For a 95% Confidence Level,
a/2 = .025 and t .025,40 = 2.021 (used df=40)


2121
2
2
21
2
1
21
11
2
)1()1(
)(
nnnn
nsns
txx +
-+
-+-
±-


(75,000 – 45,000) +
2.021
29
1
27
1
22927
)28()500,15()26()000,22(
22
+
-+
+

30,000 ± 10,220.73


19,779.27 <
µ
1 - µ
2 < 40,220.73

10.55 Morning Afternoon d
43 41 2
51 49 2
37 44 -7
24 32 -8
47 46 1
44 42 2
50 47 3
55 51 4
46 49 -3

n = 9 d = -0.444 sd =4.447 df = 9 - 1 = 8

For a 90% Confidence Level:
a/2 = .05 and t.05,8 = 1.86

34. Chapter 10: Statistical Inferences About Two Populations 34

n
s
td
d
±

-0.444 +
(1.86)

9
447.4
= -0.444 ± 2.757


-3.201 <
D < 2.313

10.56 Let group 1 be 1990

H
o: p1 - p2 = 0
H
a: p1 - p2 < 0 a = .05

The critical table
z value is: z.05 = -1.645


n1 = 1300 n2 = 1450
1
ˆp = .447
2
ˆp = .487


14501300
)1450)(487(.)1300)(447(.ˆˆ
21
2211 +
+
=
+
+
=
nn
pnpn
p
= .468








+
--
=








+×
---
=
1450
1
1300
1
)532)(.468(.
)0()487.447(.
11
)()ˆˆ(
1
2121nn
qp
pppp
z =
-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

n
1 = 16 n 2 = 14
x1 = 26,400 x2 = 25,800
s
1 = 1,200 s 2 = 1,050

H
0: s1
2 = s2
2
H
a: s1
2 ¹ s2
2

df
num = 16 – 1 = 15 df denom = 14 – 1 = 13

The critical F values are: F
.025,15,13 = 3.05 F .975,15,13 = 0.33

35. Chapter 10: Statistical Inferences About Two Populations 35
F =
500,102,1
000,440,1
2
2
2
1
=
s
s
= 1.31

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 H
0: s1
2 = s2
2 a = .01 n 1 = 8 n 2 = 7
H
a: s1
2 ¹ s2
2 S 1
2 = 72,909 S 2
2 = 129,569

df
num = 6 dfdenom = 7

The critical F values are: F
.005,6,7 = 9.16 F
.995,7,6 = .11

F =
909,72
569,129
2
2
2
1
=
s
s = 1.78

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

n
1 = 60 n 2 = 41

x1 = 631 x2 = 848

s1 = 100 s2 = 100

For a 95% Confidence Level,
a/2 = .025 and z.025 = 1.96


2
2
2
1
2
1
21
)(
nn
zxx
ss
+±-

(631 – 848) +
1.96
41
100
60
100
22
+ = -217 ± 39.7


-256.7 <
µ
1 - µ
2 < -177.3

36. Chapter 10: Statistical Inferences About Two Populations 36
10.60 Ho: µ 1 - µ2 = 0 a = .01
H
a: µ 1 - µ2 ¹ 0 df = 20 + 24 - 2 = 42

Detroit
Charlotte

n
1 = 20 n 2 = 24

x1 = 17.53 x2 = 14.89

s1 = 3.2 s2 = 2.7

For two-tail test, a/2 = .005 and the critical t
.005,42 = ±2.704 (used df=40)


t =
2121
2
2
21
2
1
21
2
1
11
2
)1()1(
)()(
nnnn
nsns
xx
+
-+
-+-
---
mm



t =
24
1
20
1
42
)23()7.2()19()2.3(
)0()89.1453.17(
22
+
+
--
= 2.97

Since the observed
t = 2.97 > t.005,42 = 2.704, the decision is to reject the null
hypothesis .


10.61 With Fertilizer
Without Fertilizer

x1 = 38.4 x2 = 23.1

s1 = 9.8 s2 = 7.4

n1 = 35 n2 = 35

H
o: µ1 - µ2 = 0
H
a: µ1 - µ2 > 0

For one-tail test, a = .01 and
z.01 = 2.33


z =
35
)4.7(
35
)8.9(
)0()1.234.38()()(
2
2
2
2
1
2
1
21
21
+
--
=
+
---
nn
xx
ss
mm
= 7.37

Since the observed
z = 7.37 > z.01 = 2.33, the decision is to reject the null
hypothesis .

37. Chapter 10: Statistical Inferences About Two Populations 37
10.62 Specialty Discount

n
1 = 350 n 2 = 500

p
ˆ1 = .75 pˆ2 = .52

For a 90% Confidence Level,
a/2 = .05 and z .05 = 1.645


2
22
1
11
21
ˆˆˆˆ
)ˆˆ(
n
qp
n
qp
zpp +±-


(.75 - .52) + 1.645
500
)48)(.52(.
350
)25)(.75(.
+
= .23 ± .053


.177 <
p1 - p2 < .283

10.63 H
0: s1
2 = s2
2 a = .05 n 1 = 27 s 1 = 22,000
H
a: s1
2 ¹ s2
2 n 2 = 29 s 2 = 15,500

df
num = 27 - 1 = 26 dfdenom = 29 - 1 = 28

The critical F values are: F
.025,24,28 = 2.17 F .975,28,24 = .46

F =
2
2
2
2
2
1
500,15
000,22
=s
s
= 2.01
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.

38. Chapter 10: Statistical Inferences About Two Populations 38
10.64 Name Brand Store Brand d
54 49 5
55 50 5
59 52 7
53 51 2
54 50 4
61 56 5
51 47 4
53 49 4

n = 8 d = 4.5 sd=1.414 df = 8 - 1 = 7

For a 90% Confidence Level, a/2 = .05 and
t.05,7 = 1.895


n
s
td
d
±

4.5 +
1.895
8
414.1
= 4.5 ± .947


3.553 <
D < 5.447

10.65 H
o: µ1 - µ2 = 0 a = .01
H
a: µ1 - µ2 < 0 df = 23 + 19 - 2 = 40

Wisconsin
Tennessee

n
1 = 23 n2 = 19

x1 = 69.652 x2 = 71.7368

s1
2 = 9.9644 s2
2 = 4.6491

For one-tail test,
a = .01 and the critical t.01,40 = -2.423


t =
2121
2
2
21
2
1
21
2
1
11
2
)1()1(
)()(
nnnn
nsns
xx
+
-+
-+-
---
mm





t =
19
1
23
1
40
)18)(6491.4()22)(9644.9(
)0()7368.71652.69(
+
+
--
= -2.44

39. 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 Friday d
71 53 18
56 47 9
75 52 23
68 55 13
74 58 16

n = 5 d = 15.8 sd = 5.263 df = 5 - 1 = 4

H
o: D = 0 a = .05
H
a: D > 0

For one-tail test, a = .05 and the critical
t.05,4 = 2.132


t =
5
263.5
08.15-
=
-
n
s
Dd
d
= 6.71

Since the observed t = 6.71 > t
.05,4 = 2.132, the decision is to reject the null
hypothesis
.



10.67 H
o: P1 - P2 = 0 a = .05
H
a: P1 - P2 ¹ 0

Machine 1
Machine 2

x
1 = 38 x 2 = 21
n
1 = 191 n 2 = 202


191
38
ˆ
1
1
1
==
n
x
p = .199
202
21
ˆ
2
2
2
==
n
x
p = .104


202191
)202)(104(.)191)(199(.ˆˆ
21
2211 +
+
=
+
+
=
nn
pnpn
p
= .15

For two-tail, a/2 = .025 and the critical z values are: z
.025 = ±1.96

40. Chapter 10: Statistical Inferences About Two Populations 40







+
--
=








+×
---
=
202
1
191
1
)85)(.15(.
)0()104.199(.
11
)()ˆˆ(
1
2121nn
qp
pppp
z =
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 n2 = 281

x1 = 297 x2 = 192


338
297
ˆ
1
1
1
==
n
x
p = .879
281
192
ˆ
2
2
2
==
n
x
p = .683

For a 90% Confidence Level,
a/2 = .05 and z.05 = 1.645


2
22
1
11
21
ˆˆˆˆ
)ˆˆ(
n
qp
n
qp
zpp +±-


(.879 - .683) + 1.645
281
)317)(.683(.
338
)121)(.879(.
+
= .196 ± .054


.142 <
p1 - p2 < .250



10.69 Aerospace
Automobile


n1 = 33 n2 = 35

x1 = 12.4 x2 = 4.6

s1 = 2.9 s2 = 1.8

For a 99% Confidence Level,
a/2 = .005 and z.005 = 2.575


2
2
2
1
2
1
21
)(
nn
zxx
ss
+±-
(12.4 – 4.6) +
2.575
35
)8.1(
33
)9.2(
22
+ = 7.8 ± 1.52


6.28 <
µ
1 - µ
2 < 9.32

41. Chapter 10: Statistical Inferences About Two Populations 41

10.70 Discount Specialty

x1 = $47.20 x2 = $27.40

s1 = $12.45 s2 = $9.82
n
1 = 60 n 2 = 40

H
o: µ 1 - µ2 = 0 a = .01
H
a: µ 1 - µ2 ¹ 0

For two-tail test, a/2 = .005 and z
c = ±2.575

z =
40
)82.9(
60
)45.12(
)0()40.2720.47()()(
22
2
2
2
1
2
1
21
21
+
--
=
+
---
nn
xx
ss
mm
= 8.86

Since the observed z = 8.86 > z
c = 2.575, the decision is to reject the null
hypothesis .


10.71 Before
After d
12 8 4
7 3 4
10 8 2
16 9 7
8 5 3

n = 5 d = 4.0 sd = 1.8708 df = 5 - 1 = 4

H
o: D = 0 a = .01
H
a: D > 0

For one-tail test, a = .01 and the critical
t.01,4 = 3.747


t =
5
8708.1
00.4-
=
-
n
s
Dd
d
= 4.78

Since the observed t = 4.78 > t
.01,4 = 3.747, the decision is to reject the null
hypothesis
.

42. Chapter 10: Statistical Inferences About Two Populations 42

10.72 H
o: µ 1 - µ2 = 0 a = .01
H
a: µ 1 - µ2 ¹ 0 df = 10 + 6 - 2 = 14

A
B

n
1 = 10 n 2 = 6

x1 = 18.3 x2 = 9.667

s1
2 = 17.122 s2
2 = 7.467

For two-tail test, a/2 = .005 and the critical
t.005,14 = ±2.977


t =
2121
2
2
21
2
1
21
2
1
11
2
)1()1(
)()(
nnnn
nsns
xx
+
-+
-+-
---
mm



t =
6
1
10
1
14
)5)(467.7()9)(122.17(
)0()667.93.18(
+
+
--
= 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.

H
o: µ1 - µ2 = 0
H
a: µ1 - µ2 > 0


n1 = 19 n2 = 23
x1 = 130.4 x2 = 128.4

S
1 = 12.9 S2 = 13.9 a = .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 H
0: D = 0
H
a: D ¹ 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

43. 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 two-
tailed 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 a = .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 a.

10.76 A test of differences of the variances of the populations of the two machines is
being computed. The hypotheses are:

H
0: s1
2 = s2
2
H
a: s1
2 ¹ s2
2

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 a = .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.