Chapter 10: Statistical Inferences About Two Populations

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.
related populations when the differences are normally distributed.
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

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 Construct Confidence
Intervals about the Difference in Two Population Means Using the t
Test
10.3 Statistical Inferences For Two Related Populations
Hypothesis Testing
Using the Computer to Make Statistical Inferences about Two Related
Populations

10.4 Statistical Inferences About Two Population Proportions, p1 – p2
Hypothesis Testing
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 x 2 = 53.2
σ1
2
= 52 σ2
2
= 60
n1 = 32 n2 = 32
a) Ho: µ1 - µ2 = 0
Ha: µ1 - µ2 < 0
For one-tail test, α = .10 z.10 = -1.28
z =
32
60
32
52
)0()2.533.51()()(
2
2
2
1
2
1
2121
+
−−
=
+
−−−
nn
xx
σσ
µµ
= -1.02
Since the observed z = -1.02 > zc = -1.645, the decision is to fail to reject the null
hypothesis.

b) Critical value method:
zc =
2
2
2
1
2
1
2121 )()(
nn
xx c
σσ
µµ
+
−−−
-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
n1 = 32 n2 = 31
x 1 = 70.4 x 2 = 68.7
σ1 = 5.76 σ2 = 6.1
For a 90% C.I., z.05 = 1.645
2
2
2
1
2
1
21 )(
nn
zxx
σσ
+±−
(70.4) – 68.7) + 1.645
31
1.6
32
76.5 22
+
1.7 ± 2.465
-.76 < µ1 - µ2 < 4.16

10.3 a) Sample 1 Sample 2
x 1 = 88.23 x 2 = 81.2
σ1
2
= 22.74 σ2
2
= 26.65
n1 = 30 n2 = 30
Ho: µ1 - µ2 = 0
Ha: µ1 - µ2 ≠ 0
For two-tail test, use α/2 = .01 z.01 = + 2.33
z =
30
65.26
30
74.22
)0()2.8123.88()()(
2
2
2
1
2
1
2121
+
−−
=
+
−−−
nn
xx
σσ
µµ
= 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
σσ
+±−
(88.23 – 81.2) + 2.33
30
65.26
30
74.22
+
7.03 + 2.99
4.04 < µµµµ < 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
x 1 = 1.96 x 2 = 3.02
σ1
2
= 1.0188 σ2
2
= 0.9180
n1 = 50 n2 = 50
Ho: µ1 - µ2 = 0
Ha: µ1 - µ2 ≠ 0
For two-tail test, α/2 = .005 z.005 = ±2.575

z =
50
9180.0
50
0188.1
)0()02.396.1()()(
2
2
2
1
2
1
2121
+
−−
=
+
−−−
nn
xx
σσ
µµ
= -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 n2 = 37
x 1 = 5.3 x 2 = 6.5
σ1
2
= 1.99 σ2
2
= 2.36
For a 95% C.I., z.025 = 1.96
2
2
2
1
2
1
21 )(
nn
zxx
σσ
+±−
(5.3 – 6.5) + 1.96
37
36.2
40
99.1
+
-1.2 ± .66 -1.86 < µµµµ < -.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
x 1 = 1.84 x 2 = 1.99
σ1 = .38 σ2 = .51
for a 98% C.I., z.01 = 2.33
2
2
2
1
2
1
21 )(
nn
zxx
σσ
+±−

(1.84 - 1.99) ± 2.33
41
51.
35
38. 22
+
-.15 ± .2384
-.3884 < µ1 - µ2 < .0884
Point Estimate = -.15
Hypothesis Test:
1) Ho: µ1 - µ2 = 0
Ha: µ1 - µ2 ≠ 0
2) z =
2
2
2
1
2
1
2121 )()(
nn
xx
σσ
µµ
+
−−−
3) α = .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.

10.7 1994 2001
x 1 = 190 x 2 = 198
σ1 = 18.50 σ2 = 15.60
n1 = 51 n2 = 47 α = .01
H0: µ1 - µ2 = 0
Ha: µ1 - µ2 < 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
2121
+
−−
=
+
−−−
nn
xx
σσ
µµ
= -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 n2 = 31
x 1 = 2.64 x 2 = 2.36
σ1
2
= .03 σ2
2
= .015
For a 99% C.I., z.005 = 2.575
2
2
2
1
2
1
21 )(
nn
zxx
σσ
+±−
(2.64-2.36) ± 2.575
31
015.
31
03.
+
.28 ± .10 .18 < µµµµ < .38
Between $ .18 and $ .38 difference with Seattle being more expensive.

10.9 Canon Pioneer
x 1 = 5.8 x 2 = 5.0
σ1 = 1.7 σ2 = 1.4
n1 = 36 n2 = 45
Ho: µ1 - µ2 = 0
Ha: µ1 - µ2 ≠ 0
For two-tail test, α/2 = .025 z.025 = ±1.96
z =
45
)4.1(
36
)7.1(
)0()0.58.5()()(
2
2
2
2
1
2
1
2121
+
−−
=
+
−−−
nn
xx
σσ
µµ
= 2.27
Since the observed z = 2.27 > zc = 1.96, the decision is to reject the null hypothesis.
10.10 A B
x 1 = 8.05 x 2 = 7.26
σ1 = 1.36 σ2 = 1.06
n1 = 50 n2 = 38
Ho: µ1 - µ2 = 0
Ha: µ1 - µ2 > 0
For one-tail test, α = .10 z.10 = 1.28
z =
38
)06.1(
50
)36.1(
)0()26.705.8()()(
22
2
2
2
1
2
1
2121
+
−−
=
+
−−−
nn
xx
σσ
µµ
= 3.06
Since the observed z = 3.06 > zc = 1.28, the decision is to reject the null
hypothesis.

10.11 Ho: µ1 - µ2 = 0 α = .01
Ha: µ1 - µ2 < 0 df = 8 + 11 - 2 = 17
Sample 1 Sample 2
n1 = 8 n2 = 11
x 1 = 24.56 x 2 = 26.42
s1
2
= 12.4 s2
2
= 15.8
For one-tail test, α = .01 Critical t.01,19 = -2.567
t =
2121
2
2
21
2
1
2121
11
2
)1()1(
)()(
nnnn
nsns
xx
+
−+
−+−
−−− µµ
=
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) Ho: µ1 - µ2 = 0 α =.10
Ha: µ1 - µ2 ≠ 0 df = 20 + 20 - 2 = 38
Sample 1 Sample 2
n1 = 20 n2 = 20
x 1 = 118 x 2 = 113
s1 = 23.9 s2 = 21.6
For two-tail test, α/2 = .05 Critical t.05,38 = 1.697 (used df=30)
t =
2121
2
2
21
2
1
2121
11
2
)1()1(
)()(
nnnn
nsns
xx
+
−+
−+−
−−− µµ
=
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.

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 < µµµµ1 - µµµµ2 < 17.224
10.13 Ho: µ1 - µ2 = 0 α = .05
Ha: µ1 - µ2 > 0 df = n1 + n2 - 2 = 10 + 10 - 2 = 18
Sample 1 Sample 2
n1 = 10 n2 = 10
x 1 = 45.38 x 2 = 40.49
s1 = 2.357 s2 = 2.355
For one-tail test, α = .05 Critical t.05,18 = 1.734
t =
2121
2
2
21
2
1
2121
11
2
)1()1(
)()(
nnnn
nsns
xx
+
−+
−+−
−−− µµ
=
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 Ho: µ1 - µ2 = 0 α =.01
Ha: µ1 - µ2 ≠ 0 df = 18 + 18 - 2 = 34
Sample 1 Sample 2
n1 = 18 n2 = 18
x 1 = 5.333 x 2 = 9.444
s1
2
= 12 s2
2
= 2.026

For two-tail test, α/2 = .005 Critical t.005,34 = ±2.457 (used df=30)
t =
2121
2
2
21
2
1
2121
11
2
)1()1(
)()(
nnnn
nsns
xx
+
−+
−+−
−−− µµ
=
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 < µµµµ1 - µµµµ2 < -1.9421
10.15 Peoria Evansville
n1 = 21 n2 = 26
1x = 86,900 2x = 84,000
s1 = 2,300 s2 = 1,750 df = 21 + 26 – 2
90% level of confidence, α/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 < µµµµ1 - µµµµ2 < 3894.62

10.16 Ho: µ1 - µ2 = 0 α = .05
Ha: µ1 - µ2 ≠ 0!= 0 df = 21 + 26 - 2 = 45
Peoria Evansville
n1 = 21 n2 = 26
x 1 = $86,900 x 2 = $84,000
s1 = $2,300 s2 = $1,750
For two-tail test, α/2 = .025
Critical t.025,45 = ± 2.021 (used df=40)
t =
2121
2
2
21
2
1
2121
11
2
)1()1(
)()(
nnnn
nsns
xx
+
−+
−+−
−−− µµ
=
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) Ho: µ1 - µ2 = 0
Ha: µ1 - µ2 > 0
2) t =
2121
2
2
21
2
1
2121
11
2
)1()1(
)()(
nnnn
nsns
xx
+
−+
−+−
−−− µµ
3) α = .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
n1 = 8 n2 = 9
x 1 = 47 x 2 = 44
s1 = 3 s2 = 3

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 nm = 22 nno = 20
x m = 112 x no = 122
sm = 11 sno = 12
df = nm + nno - 2 = 22 + 20 - 2 = 40
For a 98% Confidence Interval, α/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 Ho: µ1 - µ2 = 0
Ha: µ1 - µ2 ≠ 0
df = n1 + n2 - 2 = 11 + 11 - 2 = 20
Toronto Mexico City
n1 = 11 n2 = 11
x 1 = $67,381.82 x 2 = $63,481.82
s1 = $2,067.28 s2 = $1,594.25
For a two-tail test, α/2 = .005 Critical t.005,20 = ±2.845

t =
2121
2
2
21
2
1
2121
11
2
)1()1(
)()(
nnnn
nsns
xx
+
−+
−+−
−−− µµ
=
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
n1 = 11 n2 = 11
x 1 = $67,381.82 x 2 = $63,481.82
s1 = $2,067.28 s2 = $1,594.25
df = n1 + n2 - 2 = 11 + 11 - 2 = 20
For a 95% Level of Confidence, α/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

10.21 Ho: D = 0
Ha: 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 α = .01
df = n - 1 = 9 - 1 = 8
For one-tail test and α = .01, the critical t.01,8 = ±2.896
t =
9
45.6
011.7 −
=
−
n
s
Dd
d
= 3.31
hypothesis.
10.22 Ho: D = 0
Ha: D ≠ 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

n = 13 d = 2.5385 sd=3.4789 α = .05
df = n - 1 = 13 - 1 = 12
For a two-tail test and α/2 = .025 Critical t.025,12 = ±2.179
t =
13
4789.3
05385.2 −
=
−
n
s
Dd
d
= 2.63
hypothesis.
10.23 n = 22 d = 40.56 sd = 26.58
For a 98% Level of Confidence, α/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

n = 9 d = -3 sd = 5.6347 α = .025
df = n - 1 = 9 - 1 = 8
For 90% level of confidence and α/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 sd = 4938.22 n = 11, df = 10
α = .01 α/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

10.26 Ho: D = 0
Ha: 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 sd=1.716 α = .05 df = n - 1 = 9 - 1 = 8
For a one-tail test and α = .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.
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 sd=25.813 df = n - 1 = 11 - 1 = 10
For a 98% level of confidence and α/2=.01, t.01,10 = 2.764

n
s
td d
±
28.09 ± (2.764)
11
813.25
= 28.09 ± 21.51
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 α = .01, the critical t.01,26 = 2.479
t =
27
5
071.3 −
=
−
n
s
Dd
d
= 3.86
hypothesis.
10.29 n = 21 d = 75 sd=30 df = 21 - 1 = 20
For a 90% confidence level, α/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 Ho: D = 0
Ha: D ≠ 0
n = 15 d = -2.85 sd = 1.9 α = .01 df = 15 - 1 = 14
For a two-tail test, α/2 = .005 and the critical t.005,14 = + 2.977

t =
15
9.1
085.2 −−
=
−
n
s
Dd
d
= -5.81
hypothesis.
10.31 a) Sample 1 Sample 2
n1 = 368 n2 = 405
x1 = 175 x2 = 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
Ho: p1 - p2 = 0
Ha: p1 - p2 ≠ 0
For two-tail, α/2 = .025 and z.025 = ±1.96






+
−−
=








+⋅
−−−
=
405
1
368
1
)538)(.462(.
)0()449.476(.
11
)()ˆˆ(
1
2121
nn
qp
pppp
z = 0.75
Since the observed z = 0.75 < zc = 1.96, the decision is to fail to reject the null
hypothesis.
b) Sample 1 Sample 2
pˆ 1 = .38 pˆ 2 = .25
n1 = 649 n2 = 558
558649
)25(.558)38(.649ˆˆ
21
2211
+
+
=
+
+
=
nn
pnpn
p = .32

Ho: p1 - p2 = 0
Ha: p1 - p2 > 0
For a one-tail test and α = .10, z.10 = 1.28






+
−−
=








+⋅
−−−
=
558
1
649
1
)68)(.32(.
)0()25.38(.
11
)()ˆˆ(
1
2121
nn
qp
pppp
z = 4.83
hypothesis.
10.32 a) n1 = 85 n2 = 90 pˆ 1 = .75 pˆ 2 = .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 < p1 - p2 < .19
b) n1 = 1100 n2 = 1300 pˆ 1 = .19 pˆ 2 = .17
For a 95% Confidence Level, α/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 < p1 - p2 < .05

c) n1 = 430 n2 = 399 x1 = 275 x2 = 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, α/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) n1 = 1500 n2 = 1500 x1 = 1050 x2 = 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, α/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 < p1 - p2 < -.013
10.33 H0: pm - pw = 0
Ha: pm - pw < 0 nm = 374 nw = 481 pˆ
m = .59 pˆ
w = .70
For a one-tailed test and α = .05, z.05 = -1.645
481374
)70(.481)59(.374ˆˆ
+
+
=
+
+
=
wm
wwmm
nn
pnpn
p = .652







+
−−
=








+⋅
−−−
=
481
1
374
1
)348)(.652(.
)0()70.59(.
11
)()ˆˆ(
1
2121
nn
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, α/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
pˆ 1 = .48 pˆ 2 = .56
n1 = 56 n2 = 89
8956
)56(.89)48(.56ˆˆ
21
2211
+
+
=
+
+
=
nn
pnpn
p = .529
Ho: p1 - p2 = 0
Ha: p1 - p2 ≠ 0
For two-tail test, α/2 = .10 and zc = ±1.28






+
−−
=








+⋅
−−−
=
89
1
56
1
)471)(.529(.
)0()56.48(.
11
)()ˆˆ(
1
2121
nn
qp
pppp
z = -0.94
Since the observed z = -0.94 > zc = -1.28, the decision is to fail to reject the null
hypothesis.

10.36 A B
n1 = 35 n2 = 35
x1 = 5 x2 = 7
35
5
ˆ
1
1
1 ==
n
x
p = .14
35
7
ˆ
2
2
2 ==
n
x
p = .20
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 H0: p1 – p2 = 0
Ha: p1 – p2 ≠ 0
α = .10 pˆ
1
= .09 pˆ
2
= .06 n1 = 780 n2 = 915
For a two-tailed test, α/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
2121
nn
qp
pppp
Z = 2.35
hypothesis.

10.38 n1 = 850 n2 = 910 pˆ
1
= .60 pˆ 2 = .52
For a 95% Confidence Level, α/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 H0: σ1
2
= σ2
2
α = .01 n1 = 10 s1
2
= 562
Ha: σ1
2
< σ2
2
n2 = 12 s2
2
= 1013
dfnum = 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 H0: σ1
2
= σ2
2
α = .05 n1 = 5 S1 = 4.68
Ha: σ1
2
≠ σ2
2
n2 = 19 S2 = 2.78
dfnum = 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
null hypothesis.

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
n1 = 10 df1 = 9 n2 = 10 df2 = 9
s1
2
= .0018989 s2
2
= .0023378
H0: σ1
2
= σ2
2
α = .10 α/2 = .05
Ha: σ1
2
≠ σ2
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) H0: σ1
2
= σ2
2
Ha: σ1
2
≠ σ2
2
2) F = 2
2
2
1
s
s
3) α = .01
4) df1 = 12 df2 = 10 This is a two-tailed test

If the observed value is greater than 5.66 or less than .177, the decision will be
to reject the null hypothesis.
5) s1
2
= 393.4 s2
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 Chicago.
10.43 H0: σ1
2
= σ2
2
α = .05 n1 = 12 s1 = 7.52
Ha: σ1
2
> σ2
2
n2 = 15 s2 = 6.08
dfnum = 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
null hypothesis.
10.44 H0: σ1
2
= σ2
2
α = .01 n1 = 15 s1
2
= 91.5
Ha: σ1
2
≠ σ2
2
n2 = 15 s2
2
= 67.3
dfnum = 15 - 1 = 14 dfdenom = 15 - 1 = 14
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.

10.45 Ho: µ1 - µ2 = 0
Ha: µ1 - µ2 ≠ 0
For α = .10 and a two-tailed test, α/2 = .05 and z.05 = + 1.645
Sample 1 Sample 2
1x = 138.4 2x = 142.5
σ1 = 6.71 σ2 = 8.92
n1 = 48 n2 = 39
z =
39
)92.8(
48
)71.6(
)0()5.1424.138()()(
2
2
2
2
1
2
1
2121
+
−−
=
+
−−−
nn
xx
σσ
µµ
= -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.
1x = 34.9 2x = 27.6
σ1
2
= 2.97 σ2
2
= 3.50
n1 = 34 n2 = 31
For 98% Confidence Level, z.01 = 2.33
2
2
2
1
2
1
21 )(
nn
zxx
σσ
+±−
(34.9 – 27.6) + 2.33
31
50.3
34
97.2
+ = 7.3 + 1.04
6.26 < µµµµ1 - µµµµ2 < 8.34

10.47 Ho: µ1 - µ2 = 0
Ha: µ1 - µ2 >0
Sample 1 Sample 2
1x = 2.06 2x = 1.93
s1
2
= .176 s2
2
= .143
n1 = 12 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 =
2121
2
2
21
2
1
2121
11
2
)1()1(
)()(
nnnn
nsns
xx
+
−+
−+−
−−− µµ
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.
x 1 = 74.6 x 2 = 70.9
s1
2
= 10.5 s2
2
= 11.4
n1 = 18 n2 = 19
For 95% confidence, α/2 = .025.
Using df = 18 + 19 - 2 = 35, t35,.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 < µµµµ1 - µµµµ2 < 5.92

10.49 Ho: D = 0 α = .01
Ha: D < 0
n = 21 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 =
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 sd = 5.703 df = 8
For a 99% Confidence Level, α/2 = .005 and t8,.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 Ho: p1 - p2 = 0 α = .05 α/2 = .025
Ha: p1 - p2 ≠ 0 z.025 = + 1.96
If the observed value of z is greater than 1.96 or less than -1.96, then the decision

Sample 1 Sample 2
x1 = 345 x2 = 421
n1 = 783 n2 = 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
2121
nn
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.
n1 = 409 n2 = 378
pˆ 1 = .71 pˆ 2 = .67
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 < p1 - p2 < .125
10.53 H0: σ1
2
= σ2
2
α = .05 n1 = 8 s1
2
= 46
Ha: σ1
2
≠ σ2
2
n2 = 10 S2
2
= 37
dfnum = 8 - 1 = 7 dfdenom = 10 - 1 = 9
The critical F values are: F.025,7,9 = 4.20 F.975,9,7 = .238

If the observed value of F is greater than 4.20 or less than .238, then the decision
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
x t = $75,000 x w = $45,000
st = $22,000 sw = $15,500
nt = 27 nw = 29
df = 27 + 29 - 2 = 54
For a 95% Confidence Level, α/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 < µ1 - µ2 < 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: α/2 = .05 and t.05,8 = 1.86

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
Ho: p1 - p2 = 0
Ha: p1 - p2 < 0 α = .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
2121
nn
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
n1 = 16 n2 = 14
x 1 = 26,400 x 2 = 25,800
s1 = 1,200 s2 = 1,050
H0: σ1
2
= σ2
2
Ha: σ1
2
≠ σ2
2
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 =
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 H0: σ1
2
= σ2
2
α = .01 n1 = 8 n2 = 7
Ha: σ1
2
≠ σ2
2
S1
2
= 72,909 S2
2
= 129,569
dfnum = 6 dfdenom = 7
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
n1 = 60 n2 = 41
x 1 = 631 x 2 = 848
σ1 = 100 σ2 = 100
2
2
2
1
2
1
21 )(
nn
zxx
σσ
+±−
(631 – 848) + 1.96
41
100
60
100 22
+ = -217 ± 39.7
-256.7 < µ1 - µ2 < -177.3

10.60 Ho: µ1 - µ2 = 0 α = .01
Ha: µ1 - µ2 ≠ 0 df = 20 + 24 - 2 = 42
Detroit Charlotte
n1 = 20 n2 = 24
x 1 = 17.53 x 2 = 14.89
s1 = 3.2 s2 = 2.7
For two-tail test, α/2 = .005 and the critical t.005,42 = ±2.704 (used df=40)
t =
2121
2
2
21
2
1
2121
11
2
)1()1(
)()(
nnnn
nsns
xx
+
−+
−+−
−−− µµ
t =
24
1
20
1
42
)23()7.2()19()2.3(
)0()89.1453.17(
22
+
+
−−
= 2.97
hypothesis.
10.61 With Fertilizer Without Fertilizer
x 1 = 38.4 x 2 = 23.1
σ1 = 9.8 σ2 = 7.4
n1 = 35 n2 = 35
Ho: µ1 - µ2 = 0
Ha: µ1 - µ2 > 0
For one-tail test, α = .01 and z.01 = 2.33
z =
35
)4.7(
35
)8.9(
)0()1.234.38()()(
2
2
2
2
1
2
1
2121
+
−−
=
+
−−−
nn
xx
σσ
µµ
= 7.37
hypothesis.

10.62 Specialty Discount
n1 = 350 n2 = 500
pˆ 1 = .75 pˆ 2 = .52
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 H0: σ1
2
= σ2
2
α = .05 n1 = 27 s1 = 22,000
Ha: σ1
2
≠ σ2
2
n2 = 29 s2 = 15,500
dfnum = 27 - 1 = 26 dfdenom = 29 - 1 = 28
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.

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, α/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 Ho: µ1 - µ2 = 0 α = .01
Ha: µ1 - µ2 < 0 df = 23 + 19 - 2 = 40
Wisconsin Tennessee
n1 = 23 n2 = 19
x 1 = 69.652 x 2 = 71.7368
s1
2
= 9.9644 s2
2
= 4.6491
For one-tail test, α = .01 and the critical t.01,40 = -2.423
t =
2121
2
2
21
2
1
2121
11
2
)1()1(
)()(
nnnn
nsns
xx
+
−+
−+−
−−− µµ
t =
19
1
23
1
40
)18)(6491.4()22)(9644.9(
)0()7368.71652.69(
+
+
−−
= -2.44

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
Ho: D = 0 α = .05
Ha: D > 0
For one-tail test, α = .05 and the critical t.05,4 = 2.132
t =
5
263.5
08.15 −
=
−
n
s
Dd
d
= 6.71
hypothesis.
10.67 Ho: P1 - P2 = 0 α = .05
Ha: P1 - P2 ≠ 0
Machine 1 Machine 2
x1 = 38 x2 = 21
n1 = 191 n2 = 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, α/2 = .025 and the critical z values are: z.025 = ±1.96







+
−−
=








+⋅
−−−
=
202
1
191
1
)85)(.15(.
)0()104.199(.
11
)()ˆˆ(
1
2121
nn
qp
pppp
z = 2.64
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
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
x 1 = 12.4 x 2 = 4.6
σ1 = 2.9 σ2 = 1.8
2
2
2
1
2
1
21 )(
nn
zxx
σσ
+±−
(12.4 – 4.6) + 2.575
35
)8.1(
33
)9.2( 22
+ = 7.8 ± 1.52
6.28 < µ1 - µ2 < 9.32

10.70 Discount Specialty
x 1 = $47.20 x 2 = $27.40
σ1 = $12.45 σ2 = $9.82
n1 = 60 n2 = 40
Ho: µ1 - µ2 = 0 α = .01
Ha: µ1 - µ2 ≠ 0
For two-tail test, α/2 = .005 and zc = ±2.575
z =
40
)82.9(
60
)45.12(
)0()40.2720.47()()(
22
2
2
2
1
2
1
2121
+
−−
=
+
−−−
nn
xx
σσ
µµ
= 8.86
hypothesis.
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
Ho: D = 0 α = .01
Ha: D > 0
For one-tail test, α = .01 and the critical t.01,4 = 3.747
t =
5
8708.1
00.4 −
=
−
n
s
Dd
d
= 4.78
hypothesis.

10.72 Ho: µ1 - µ2 = 0 α = .01
Ha: µ1 - µ2 ≠ 0 df = 10 + 6 - 2 = 14
A B
n1 = 10 n2 = 6
x 1 = 18.3 x 2 = 9.667
s1
2
= 17.122 s2
2
= 7.467
For two-tail test, α/2 = .005 and the critical t.005,14 = ±2.977
t =
2121
2
2
21
2
1
2121
11
2
)1()1(
)()(
nnnn
nsns
xx
+
−+
−+−
−−− µµ
t =
6
1
10
1
14
)5)(467.7()9)(122.17(
)0()667.93.18(
+
+
−−
= 4.52
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: µ1 - µ2 = 0
Ha: µ1 - µ2 > 0
n1 = 19 n2 = 23 x1 = 130.4 x 2 = 128.4
S1 = 12.9 S2 = 13.9 α = .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

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 α = .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 α.
10.76 A test of differences of the variances of the populations of the two machines is
being computed. The hypotheses are:
H0: σ1
2
= σ2
2
Ha: σ1
2
≠ σ2
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 α = .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.

Chapter 10: Statistical Inferences About Two Populations

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