Chapter 10: Statistical Inferences About Two Populations
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 or in matched-pairs experiments from one population
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.
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.
In this chapter, the approach as to whether to use a z statistic or a t statistic for
analyzing the differences in two sample means is the same as that used in chapters 8 and
9. When the population variances are known, the z statistic can be used. However, if the
population variances are unknown and sample variances are being used, then the t test is
the appropriate statistic for the analysis. It is always an assumption underlying the use of
the t statistic that the populations are normally distributed. If sample sizes are small and
the population variances are known, the z statistic can be used if the populations are
normally distributed.
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Chapter 10: Statistical Inferences About Two Populations
In conducting a t test for the difference of two means from independent
populations, there are two different formulas given in the chapter. 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. In doing hand calculations, it is generally easier to use the “pooled”
variance formula because the degrees of freedom formula for the unequal variance
formula is quite complex. However, it is good to expose students to both formulas since
computer software packages often give you the option of using the “pooled” that assumes
equal population variances or the formula for unequal variances.
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.
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Chapter 10: Statistical Inferences About Two Populations
CHAPTER OUTLINE
10.1 Hypothesis Testing and Confidence Intervals about the Difference in Two Means
using the z Statistic (Population Variances Known)
Hypothesis Testing
Confidence Intervals
Using the Computer to Test Hypotheses about the Difference in Two
Population Means Using the z Test
10.2 Hypothesis Testing and Confidence Intervals about the Difference in Two Means:
Independent Samples and Population Variances Unknown
Hypothesis Testing
Using the Computer to Test Hypotheses and Construct Confidence
Intervals about the Difference in Two Population Means Using the
t Test
Confidence Intervals
10.3
Statistical Inferences for Two Related Populations
Hypothesis Testing
Using the Computer to Make Statistical Inferences about Two Related
Populations
Confidence Intervals
10.4
Statistical Inferences about Two Population Proportions, p1- p2
Hypothesis Testing
Confidence Intervals
10.5
Testing Hypotheses about Two Population Variances
Using the Computer to Test Hypotheses about Two Population Variances
KEY TERMS
Correlated t test
Dependent Samples
F Distribution
F Value
Independent Samples
Matched-Pairs Test
t test for related Measures
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SOLUTIONS TO PROBLEMS IN CHAPTER 10
10.1
Sample 1
x 1 = 51.3
σ12 = 52
n1 = 31
a)
Ho:
Ha:
Sample 2
x 2 = 53.2
σ22 = 60
n2 = 32
µ1 - µ2 = 0
µ1 - µ2 < 0
For one-tail test,  = .10
z =
z.10 = -1.28
( x1  x 2 )  ( 1   2 )
1
2
n1

2
2

(51.3  53.2)  (0)
n2
52 60

31 32
= -1.01
Since the observed z = -1.01 > zc = -1.28, the decision is to fail to reject
the null hypothesis.
b)
Critical value method:
zc =
( x 1  x 2 ) c  ( 1   2 )
 12
n1
-1.28 =

 22
n2
( x1  x 2 ) c  (0)
52 60

31 32
( x 1 - x 2)c = -2.41
c)
The area for z = -1.01 using Table A.5 is .3438.
The p-value is .5000 - .3438 = .1562
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Chapter 10: Statistical Inferences About Two Populations
10.2
Sample 1
n1 = 32
x 1 = 70.4
1 = 5.76
Sample 2
n2 = 31
x 2 = 68.7
2 = 6.1
For a 90% C.I.,
z.05 = 1.645
( x1  x 2 )  z
 12
n1

 22
n2
(70.4) – 68.7) + 1.645
5.762 6.12

32
31
1.7 ± 2.46
-.76 < µ1 - µ2 < 4.16
10.3
a)
Sample 1
Sample 2
x 1 = 88.23
12 = 22.74
n1 = 30
x 2 = 81.2
22 = 26.65
n2 = 30
Ho: µ1 - µ2 = 0
Ha: µ1 - µ2  0
For two-tail test, use /2 = .01
z =
( x 1  x 2 )  ( 1   2 )
1
2
n1

2
2
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.
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 12
b) ( x 1  x 2 )  z
n1

 22
(88.23 – 81.2) + 2.33
n2
22.74 26.65

30
30
7.03 + 2.99
4.04 <  < 10.02
This supports the decision made in a) to reject the null hypothesis because
zero is not in the interval.
10.4
Computers/electronics
Food/Beverage
x 1 = 1.96
12 = 1.0188
n1 = 50
Ho:
Ha:
x 2 = 3.02
22 = .9180
n2 = 50
µ1 - µ2 = 0
µ1 - µ2  0
For two-tail test, /2 = .005
z =
( x1  x 2 )  ( 1   2 )
1
2
n1

2
2
n2

z.005 = ±2.575
(1.96  3.02)  (0)
1.0188 0.9180

50
50
= -5.39
Since the observed z = -5.39 < zc = -2.575, the decision is to reject the null
hypothesis.
10.5
A
n1 = 40
x 1 = 5.3
12 = 1.99
For a 95% C.I.,
B
n2 = 37
x 2 = 6.5
22 = 2.36
z.025 = 1.96
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Chapter 10: Statistical Inferences About Two Populations
( x1  x 2 )  z
 12
n1
(5.3 – 6.5) + 1.96

 22
n2
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
n1 = 35
x 1 = 1.84
1 = .38
Specialty
n2 = 41
x 2 = 1.99
2 = .51
for a 98% C.I.,
( x1  x 2 )  z
 12
n1
z.01 = 2.33

 22
n2
.382 .512

(1.84 - 1.99) ± 2.33
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
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Chapter 10: Statistical Inferences About Two Populations
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.
10.7
1999
2009
x 1 = 190
1 = 18.50
n1 = 51
x 2 = 198
2 = 15.60
n2 = 47
 = .01
H0: 1 - 2 = 0
Ha: 1 - 2 < 0
For a one-tailed test,
z =
z.01 = -2.33
( x 1  x 2 )  ( 1   2 )
1
2
n1

2
2

n2
(190  198)  (0)
(18.50) 2 (15.60) 2

51
47
= -2.32
Since the observed z = -2.32 > z.01 = -2.33, the decision is to fail to reject the null
hypothesis.
10.8
Vancouver
Montreal
n1 = 31
x 1 = 3.88
12 = .12
n2 = 31
x 2 = 5.59
22 = .06
For a 99% C.I.,
z.005 = 2.575
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Chapter 10: Statistical Inferences About Two Populations
 12
( x1  x 2 )  z
n1

 22
n2
(3.88 – 5.59) ± 2.575
.12 .06

31 31
−1.91 <  < −1.51
− 1.71 ± .20
Between $ 1.51 and $ 1.91 difference with Montreal being more expensive.
10.9
Manan
x 1 = 5.8
1 = 1.7
n1 = 36
Ho:
Ha:
Prairie
x 2 = 5.0
2 = 1.4
n2 = 45
µ1 - µ2 = 0
µ1 - µ2  0
For two-tail test, /2 = .025
z =
( x 1  x 2 )  ( 1   2 )
 12
n1

 22

n2
z.025 = ±1.96
(5.8  5.0)  (0)
(1.7) 2 (1.4) 2

36
45
= 2.27
Since the observed z = 2.27 > zc = 1.96, the decision is to reject the null
hypothesis.
10.10
A
x 1 = 8.05
1 = 1.36
n1 = 50
Ho:
Ha:
B
x 2 = 7.26
2 = 1.06
n2 = 38
µ1 - µ2 = 0
µ1 - µ2 > 0
For one-tail test,  = .10
z =
( x 1  x 2 )  ( 1   2 )
1
2
n1

2
2
n2
z.10 = 1.28

(8.05  7.26)  (0)
(1.36) 2 (1.06) 2

50
38
= 3.06
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Chapter 10: Statistical Inferences About Two Populations
Since the observed z = 3.06 > zc = 1.28, the decision is to reject the null
hypothesis.
 = .01
10.11 Ho: µ1 - µ2 = 0
Ha: µ1 - µ2 < 0
df = 8 + 11 - 2 = 17
Sample 1
n1 = 8
x 1 = 24.56
s12 = 12.4
Sample 2
n2 = 11
x 2 = 26.42
s22 = 15.8
For one-tail test,  = .01
Critical t.01,17 = -2.567
t=
( x1  x 2 )  ( 1   2 )
s1 (n1  1)  s 2 (n2  1) 1
1

n1  n2  2
n1 n2
2
2
=
(24.56  26.42)  (0)
= -1.05
12.4(7)  15.8(10) 1 1

8  11  2
8 11
Since the observed t = -1.05 > t.01,19 = -2.567, the decision is to fail to reject the
null hypothesis.
10.12 a)
Ho: µ1 - µ2 = 0
Ha: µ1 - µ2  0
 =.10
Sample 1
n1 = 20
x 1 = 118
s1 = 23.9
Sample 2
n2 = 20
x 2 = 113
s2 = 21.6
df = 20 + 20 - 2 = 38
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
(118  113)  (0)
(23.9) (19)  (21.6) 2 (19) 1
1

20  20  2
20 20
2
=
= 0.69
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Chapter 10: Statistical Inferences About Two Populations
Since the observed t = 0.69 < t.05,38 = 1.697, the decision is to fail to reject
the null hypothesis.
s1 (n1  1)  s 2 (n2  1) 1
1

n1  n2  2
n1 n2
2
b)
( x1  x 2 )  t
(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
10.13
Ho: µ1 - µ2 = 0
Ha: µ1 - µ2 > 0
 = .05
Sample 1
n1 = 10
x 1 = 45.38
s1 = 2.357
Sample 2
n2 = 10
x 2 = 40.49
s2 = 2.355
For one-tail test,  = .05
Critical t.05,18 = 1.734
t =
t =
df = n1 + n2 - 2 = 10 + 10 - 2 = 18
( 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.
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Chapter 10: Statistical Inferences About Two Populations
10.14
 =.01
Ho: µ1 - µ2 = 0
Ha: µ1 - µ2  0
df = 18 + 18 - 2 = 34
Sample 1
n1 = 18
x 1 = 5.333
s12 = 12
Sample 2
n2 = 18
x 2 = 9.444
s22 = 2.026
For two-tail test, /2 = .005
t =
t =
Critical t.005,34 = ±2.75 (used df=30)
( x 1  x 2 )  ( 1   2 )
s1 (n1  1)  s 2 (n2  1) 1
1

n1  n2  2
n1 n2
2
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 = -2.75, reject the null hypothesis.
b)
For 98% confidence, t.01, 30 = 2.457
s (n  1)  s 2 (n2  1) 1
1
( x1  x 2 )  t 1 1

n1  n2  2
n1 n2
2
2
(5.333 – 9.444) + 2.457
=
(12)(17)  (2.026)(17) 1 1

18  18  2
18 18
-4.111 + 2.169
-6.280 < 1 - 2 < -1.942
10.15
Montreal
n1 = 21
x1 = 252,000
s1 = 4,900
Halifax
n2 = 26
x 2 = 243,000
s2 = 3,700
90% level of confidence, /2 = .05
df = 21 + 26 – 2 = 45
t .05,45 = 1.684 (used df = 40)
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Chapter 10: Statistical Inferences About Two Populations
s1 (n1  1)  s 2 (n2  1) 1
1

n1  n2  2
n1 n2
2
( x1  x 2 )  t
2
=
(4,900) 2 (20)  (3,700) 2 (25) 1
1

(252,000 – 243,000) + 1.684
=
21  26  2
21 26
9,000 + 2112.24
6887.76 < 1 - 2 < 11,112.24
Since zero does not fall into the 90% confidence interval, we can state at  = 0.10
that there is a difference in the mean prices.
 = .10
df = 12 + 12 - 2 = 22
10.16 Ho: µ1 - µ2 = 0
Ha: µ1 - µ2  0
Co-op
n1 = 12
x 1 = $15.645
s1 = $1.093
Interns
n2 = 12
x 2 = $15.439
s2 = $0.958
For two-tail test, /2 = .05
Critical t.05,22 = ± 1.717
t =
t =
( x 1  x 2 )  ( 1   2 )
s1 (n1  1)  s 2 (n2  1) 1
1

n1  n2  2
n1 n2
2
2
=
(15.645  15.439)  (0)
(1.093) 2 (11)  (0.958) 2 (11) 1
1

12  12  2
12 12
= 0.49
Since the observed t = 0.49 < t.05,22 = 1.717, the decision is to fail to reject the
null hypothesis.
90% Confidence Interval:
s1 (n1  1)  s 2 (n2  1) 1
1

n1  n2  2
n1 n2
2
( x1  x 2 )  t
t.05,22 = ± 1.717
2
=
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Chapter 10: Statistical Inferences About Two Populations
(1.093) 2 (11)  (0.958) 2 (11) 1
1

=
12  12  2
12 12
(15.645 – 15.439) + 1.717
0.206 + 0.7204
-0.5144 < 1 - 2 < 0.9264
10.17 Let Toronto be group 1
1)
Ho: µ1 - µ2 = 0
Ha: µ1 - µ2 > 0
2) t =
( x 1  x 2 )  ( 1   2 )
s1 (n1  1)  s 2 (n2  1) 1
1

n1  n2  2
n1 n2
2
2
3)  = .05
4) For a one-tailed test and df = 8 + 9 - 2 = 15, t.05,15 = 1.753. If the observed
value of t is greater than 1.753, the decision is to reject the null hypothesis.
5) Toronto
n1 = 8
x 1 = 47
s1 = 3
6) t =
Montreal
n2 = 9
x 2 = 44
s2 = 3
(47  44)  (0)
7(3) 2  8(3) 2
15
= 2.06
1 1

8 9
7) Since t = 2.06 > t.05,15 = 1.753, the decision is to reject the null
hypothesis.
8) The rental rates in Toronto are significantly higher than in Montreal.
10.18
nE = 22
x E = 112
sE = 11
nQ = 20
x Q = 122
sQ = 12
df = nE + nQ - 2 = 22 + 20 - 2 = 40
For a 98% Confidence Interval, /2 = .01 and
t.01,40 = 2.423
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Chapter 10: Statistical Inferences About Two Populations
s E (n E  1)  sQ (nQ  1)
2
(x E  xQ )  t
2
n E 1  nQ  2
(112 – 122) + 2.423
1
1

n E nQ
=
(11)2 (21)  (12) 2 (19) 1
1

22  20  2
22 20
-10 ± 8.60
-$18.60 < µ1 - µ2 < -$1.40
Point Estimate = -$10
10.19
Ho: µ1 - µ2 = 0
Ha: µ1 - µ2  0
df = n1 + n2 - 2 = 11 + 11 - 2 = 20
Perth
n1 = 11
x 1 = $67,381.82
s1 = $2,067.28
Mexico City
n2 = 11
x 2 = $63,390.91
s2 = $1,526.08
For a two-tail test, /2 = .025
t =
t =
Critical t.025,20 = ±2.086
( x 1  x 2 )  ( 1   2 )
s1 (n1  1)  s 2 (n2  1) 1
1

n1  n2  2
n1 n2
2
2
=
(67,381.82  63,390.91)  (0)
(2,067.28) 2 (10)  (1,526.08) 2 (10) 1 1

11  11  2
11 11
= 5.15
Since the observed t = 5.15 > t.025,20 = 2.086, the decision is to reject the null
hypothesis.
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Chapter 10: Statistical Inferences About Two Populations
10.20
Ho: µ1 - µ2 = 0
Ha: µ1 - µ2 > 0
df = n1 + n2 - 2 = 9 + 10 - 2 = 17
Men
n1 = 9
x 1 = $110.92
s1 = $28.79
Women
n2 = 10
x 2 = $75.48
s2 = $30.51
This is a one-tail test,  = .01
t =
t =
Critical t.01,17 = 2.567
( x 1  x 2 )  ( 1   2 )
s1 (n1  1)  s 2 (n2  1) 1
1

n1  n2  2
n1 n2
2
2
(110.92  75.48)  (0)
(28.79) 2 (8)  (30.51) 2 (9) 1 1

9  10  2
9 10
=
= 2.60
Since the observed t = 2.60 > t.01,17 = 2.567, the decision is to reject the null
hypothesis.
10.21 Ho: D = 0
Ha: D > 0
Sample 1
38
27
30
41
36
38
33
35
44
n=9
Sample 2
22
28
21
38
38
26
19
31
35
d =7.11
d
16
-1
9
3
-2
12
14
4
9
sd=6.45
 = .01
df = n - 1 = 9 - 1 = 8
For one-tail test and  = .01,
the critical t.01,8 = ±2.896
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t =
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:
D=0
D0
Before
107
99
110
113
96
98
100
102
107
109
104
99
101
After
102
98
100
108
89
101
99
102
105
110
102
96
100
n = 13
d = 2.5385
df = n - 1 = 13 - 1 = 12
d
5
1
10
5
7
-3
1
0
2
-1
2
3
1
sd=3.4789
For a two-tail test and /2 = .025
t =
 = .05
Critical t.025,12 = ±2.179
d  D 2.5385  0

= 2.63
sd
3.4789
n
13
Since the observed t = 2.63 > t.025,12 = 2.179, the decision is to reject the null
hypothesis.
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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
n=9
d = -3
df = n - 1 = 9 - 1 = 8
d
-8
3
-1
0
-3
-6
-2
-14
4
sd = 5.6347
For 90% level of confidence and /2 = .05,
t = d t
 = .025
t.05,8 = 1.86
sd
n
t = -3 + (1.86)
5.6347
= -3 ± 3.49
9
-6.49 < D < 0.49
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10.25 City
Cost
Resale
d
Calgary
Edmonton
Fredericton
Halifax
London, ON
Montreal
Regina
Toronto
Vancouver
Victoria
Winnipeg
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
d = 1302.82
sd = 4938.22
n = 11,
df = 10
 = .01
d t
/2 = .005
sd
n
t.005,10= 3.169
= 1302.82 + 3.169
4938.22
= 1302.82 + 4718.42
11
-3415.6 < D < 6021.2
10.26 Ho:
Ha:
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
sd=1.716
 = .05
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
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Since the observed t = -3.11 < t.05,8 = -1.86, the decision is to reject the null
hypothesis.
10.27
Before
6.59
5.95
7.50
6.26
7.76
6.46
5.56
5.95
5.82
5.66
6.10
After
5.09
5.82
5.56
5.56
6.21
6.08
4.91
6.21
5.17
5.25
5.77
d = 0.7255
n = 11
d
1.50
0.13
1.94
0.70
1.55
0.38
0.65
-0.26
0.65
0.41
0.33
sd=0.6684
df = n - 1 = 11 - 1 = 10
For a 98% level of confidence and /2=.01, t.01,10 = 2.764
d t
sd
n
0.7255 ± (2.764)
0.6684
11
= 0.7255 ± 0.5570
0.1685 < D < 1.2825
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
n
27
Since the observed t = 3.86 > t.01,26 = 2.479, the decision is to reject the null
hypothesis.
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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:
D=0
D0
d = 1.32
n = 15
sd = 2.08
 = .01
df = 15 - 1 = 14
For a two-tail test, /2 = .005 and the critical t.005,14 = + 2.977
d  D 1.32  0

= 2.46
sd
2.08
t =
n
15
Since the observed t = 2.46 < t.005,14 = 2.977, we fail to reject the null
hypothesis.
10.31 a)
Sample 1
n1 = 368
x1 = 175
pˆ 1 
p
Sample 2
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
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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
Ho:
Ha:
Sample 1
p̂ 1 = .38
n1 = 649
Sample 2
p̂ 2 = .25
n2 = 558
n1 pˆ 1  n2 pˆ 2 649(.38)  558(.25)
= .32

n1  n2
649  558
p1 - p2 = 0
p1 - p2 > 0
For a one-tail test and  = .10,
z
( pˆ 1  pˆ 2 )  ( p1  p 2 )
1
1 
p  q  
 n1 n2 

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
p̂ 2 = .67
For a 90% Confidence Level, z.05 = 1.645
( pˆ 1  pˆ 2 )  z
pˆ 1qˆ1 pˆ 2 qˆ 2

n1
n2
(.75 - .67) ± 1.645
(.75)(.25) (.67)(.33)
= .08 ± .11

85
90
-.03 < p1 - p2 < .19
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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
(.19)(.81) (.17)(.83)
= .02 ± .03

1100
1300
-.01 < p1 - p2 < .05
c)
n1 = 430
pˆ 1 
n2 = 399
x1 275
= .64

n1 430
x1 = 275
pˆ 2 
x2 = 275
x2 275
= .69

n2 399
For an 85% Confidence Level, /2 = .075 and
( pˆ 1  pˆ 2 )  z
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
x1 1050
= .70

n1 1500
x2 = 1100
pˆ 2 
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
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(.70 - .733) ± 1.28
(.70)(.30) (.733)(.267)
= -.033 ± .021

1500
1500
-.054 < p1 - p2 < -.012
10.33 H0: pm - pw = 0
Ha: pm - pw < 0
nm = 374
For a one-tailed test and  = .05,
p
z
nw = 481
p̂
m
= .59
p̂
w
= .70
z.05 = -1.645
nm pˆ m  nw pˆ w 374(.59)  481(.70)
= .652

nm  n w
374  481
( pˆ 1  pˆ 2 )  ( p1  p 2 )
1
1 
p  q  
 n1 n2 

(.59  .70)  (0)
1 
 1
(.652)(. 348)


 374 481 
= -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
For a 90% Confidence Level, /2 = .05 and
( pˆ 1  pˆ 2 )  z
p̂2 = .35
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
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10.35 Computer Firms
p̂ 1 = .48
n1 = 56
Banks
p̂ 2 = .56
n2 = 89
n1 pˆ 1  n2 pˆ 2 56(.48)  89(.56)
= .529

n1  n2
56  89
p
Ho:
Ha:
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.
10.36
A
n1 = 35
x1 = 5
pˆ 1 
B
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 1  pˆ 2 )  z
pˆ 1qˆ1 pˆ 2 qˆ 2

n1
n2
(.14 - .20) ± 2.33
(.14)(.86) (.20)(.80)

35
35
= -.06 ± .21
-.27 < p1 - p2 < .15
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10.37 H0: p1 – p2 = 0
Ha: p1 – p2  0
 = .10
p̂ = .09
p̂ = .06
1
For a two-tailed test, /2 = .05 and
p
Z
n1 = 780
2
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.
10.38 n1 = 850
n2 = 910
p̂ = .60
p̂ 2 = .52
1
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 = 10 - 1 = 9
Table F.99, 9,11 
n1 = 10
n2 = 12
s12 = 562
s22 = 1013
dfdenom = 12 - 1 = 11
1
F.01,11, 9

1
F.01,10, 9

1
 .190
5.26
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F =
s1
2
s2
2

562
= .5548
1013
Since the observed F = .5548 >F.99,9,11 = .190, the decision is to fail to reject the
null hypothesis.
10.40 H0: 12 = 22
Ha: 12  22
 = .05
dfnum = 5 - 1 = 4
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..975,18,4 = .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.
10.41
City 1
City 2
1.029
1.020
1.017
1.002
0.996
1.017
1.014
1.014
1.014
0.983
n1 = 10
0.999
1.026
1.017
1.008
0.990
1.038
1.017
1.032
1.011
1.014
df1 = 9
n2 = 10
df2 = 9
s12 = 0.0001769
s22 = 0.0002104
H0: 12 = 22
Ha: 12  22
 = .01 /2 = .005
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Upper tail critical F value = F.005,9,9 = 6.54
Lower tail critical F value = F.995,9,9 = 0.153
2
s
.0001769
F = 12 
= 0.8408
.0002104
s2
Since the observed F = 0.8408 is greater than the lower tail critical value of 0.153
and less than the upper tail critical value of 6.54, the decision is to fail
to reject the null hypothesis.
10.42 Let Thunder Bay = group 1 and Moncton = group 2
1) H0: 12 = 22
Ha: 12  22
2
s1
2) F =
2
s2
3)  = .01
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
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 Thunder Bay and Moncton.
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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 = 2.60
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.
10.45 Ho:
Ha:
µ1 - µ2 = 0
µ1 - µ2  0
For  = .10 and a two-tailed test, /2 = .05 and z.05 = + 1.645
Sample 1
x 1 = 138.4
1 = 6.71
n1 = 48
Sample 2
x 2 = 142.5
2 = 8.92
n2 = 39
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Chapter 10: Statistical Inferences About Two Populations
z =
( x 1  x 2 )  ( 1   2 )
 12
n1

 22
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
x 1 = 34.9
12 = 2.97
n1 = 34
Sample 2
x 2 = 27.6
22 = 3.50
n2 = 31
For 98% Confidence Level, z.01 = 2.33
( x1  x 2 )  z
 12
n1

 22
n2
2.97 3.50

34
31
(34.9 – 27.6) + 2.33
= 7.3 + 1.04
6.26 < 1 - 2 < 8.34
10.47 Ho: µ1 - µ2 = 0
Ha: µ1 - µ2 > 0
Sample 1
x 1= 2.06
s12 = .176
n1 = 12
Sample 2
x 2 = 1.93
s22 = .143
n2 = 15
 = .05
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 =
( x 1  x 2 )  ( 1   2 )
s1 (n1  1)  s 2 (n2  1) 1
1

n1  n2  2
n1 n2
2
2
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(2.06  1.93)  (0)
t =
(.176)(11)  (.143)(14) 1 1

25
12 15
= 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
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, t30,.025 = 2.042
s1 (n1  1)  s 2 (n2  1) 1
1

n1  n2  2
n1 n2
2
( x1  x 2 )  t
2
(10.5)(17)  (11.4)(18) 1 1

18  19  2
18 19
(74.6 – 70.9) + 2.042
3.7 + 2.22
1.48 < 1 - 2 < 5.92
 = .01
10.49 Ho: D = 0
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 =
d  D  1.16  0

= -5.26
sd
1.01
n
21
Since the observed value of t = -5.26 is less than the critical t value of -2.528, the
decision is to reject the null hypothesis. The population difference is less than
zero.
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10.50
Respondent
1
2
3
4
5
6
7
8
9
Before
47
33
38
50
39
27
35
46
41
After
63
35
36
56
44
29
32
54
47
d
-16
-2
2
-6
-5
-2
3
-8
-6
sd = 5.703
df = 8
d = -4.44
For a 99% Confidence Level, /2 = .005 and t8,.005 = 3.355
d t
sd
= -4.44 + 3.355
n
5.703
= -4.44 + 6.38
9
-10.82 < D < 1.94
 = .05
10.51 Ho: p1 - p2 = 0
Ha: p1 - p2  0
/2 = .025
z.025 = + 1.96
If the observed value of z is greater than 1.96 or less than -1.96, then the decision
will be to reject the null hypothesis.
Sample 1
x1 = 345
n1 = 783
x1  x2 345  421
= .4562

n1  n2 783  896
p
pˆ 1 
z
Sample 2
x2 = 421
n2 = 896
x1 345
= .4406

n1 783
( pˆ 1  pˆ 2 )  ( p1  p 2 )
1
1
p  q  
 n1 n 


pˆ 2 

x2 421
= .4699

n2 896
(.4406  .4699)  (0)
1 
 1
(.4562)(. 5438)


 783 896 
= -1.20
Since the observed value of z = -1.20 is greater than -1.96, the decision is to fail to
reject the null hypothesis. There is no significant difference.
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Chapter 10: Statistical Inferences About Two Populations
10.52 Sample 1
n1 = 409
p̂ 1 = .71
Sample 2
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
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
x t = $75,000
st = $22,000
nt = 27
Whole Life
x w = $45,000
sw = $15,500
nw = 29
df = 27 + 29 - 2 = 54
For a 95% Confidence Level, /2 = .025 and
t.025,50 = 2.009 (used df=50)
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Chapter 10: Statistical Inferences About Two Populations
s1 (n1  1)  s 2 (n2  1) 1
1

n1  n2  2
n1 n2
2
( x1  x 2 )  t
2
(22,000) 2 (26)  (15,500) 2 (28) 1
1

(75,000 – 45,000) + 2.009
27  29  2
27 29
30,000 ± 10,160.11
19,839.89 < µ1 - µ2 < 40,160.11
10.55
Morning
43
51
37
24
47
44
50
55
46
Afternoon
41
49
44
32
46
42
47
51
49
n=9
d = -0.444
For a 90% Confidence Level:
d t
d
2
2
-7
-8
1
2
3
4
-3
sd =4.447
df = 9 - 1 = 8
/2 = .05 and t.05,8 = 1.86
sd
n
-0.444 + (1.86)
4.447
= -0.444 ± 2.757
9
-3.201 < D < 2.313
10.56 Marketing
n1 = 400
x1 = 220
Ho: p1 - p2 = 0
Ha: p1 - p2 > 0
Accountants
n2 = 450
x2 = 216
 = .01
The critical table z value is:
z.01 = 2.33
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Chapter 10: Statistical Inferences About Two Populations
p̂1 =
p
z
220
= .55
400
p̂2 =
216
= .48
450
x1  x2 220  216
= .513

n1  n2 400  450
( pˆ 1  pˆ 2 )  ( p1  p 2 )
1
1
p  q  
 n1 n 



(.55  .48)  (0)
1 
 1
(.513)(.487)


 400 450 
= 2.04
Since the observed z = 2.04 is less than z.01 = 2.33, the decision is to fail to reject
the null hypothesis. There is no significant difference between marketing
managers and accountants in the proportion who keep track of obligations “in
their head”.
10.57
Accounting
n1 = 16
x 1 = 26,400
s1 = 1,200
Data Entry
n2 = 14
x 2 = 25,800
s2 = 1,050
H0: 12 = 22
Ha: 12  22
 = .05
dfnum = 16 – 1 = 15
dfdenom = 14 – 1 = 13
The critical F values are:
F.025,15,13 = 3.05
F =
s1
2
s2
2

and 
F.975,15,13 = 0.33
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
Men
Women
n1 = 60
x 1 = 631
1 = 100
n2 = 41
x 2 = 848
2 = 100
For a 95% Confidence Level, /2 = .025 and z.025 = 1.96
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Chapter 10: Statistical Inferences About Two Populations
( x1  x 2 )  z
 12
n1

 22
n2
1002 1002

(631 – 848) + 1.96
= -217 ± 39.7
60
41
-256.7 < µ1 - µ2 < -177.3
10.59 Ho:
Ha:
 = .01
df = 20 + 24 - 2 = 42
µ1 - µ2 = 0
µ1 - µ2  0
Guelph
n1 = 20
x 1 = 7.97
s1 = 1.5
Sudbury
n2 = 24
x 2 = 6.7
s2 = 1.2
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
(7.97  6.7)  (0)
(1.5) (19)  (1.2) 2 (23) 1
1

42
20 24
2
= 3.12
Since the observed t = 3.12 > t.005,40 = 2.704, the decision is to reject the null
hypothesis.
10.60
With Fertilizer
x 1 = 97.5
1 = 24.9
n1 = 35
Ho:
Ha:
Without Fertilizer
x 2 = 58.7
2 = 18.8
n2 = 35
µ1 - µ2 = 0
µ1 - µ2 > 0
For one-tail test,  = .01 and z.01 = 2.33
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Chapter 10: Statistical Inferences About Two Populations
z =
( x 1  x 2 )  ( 1   2 )
 12
n1

 22

n2
(97.5  58.7)  (0)
(24.9) 2 (18.8) 2

35
35
= 7.36
Since the observed z = 7.36 > z.01 = 2.33, the decision is to reject the null
hypothesis.
10.61
Specialty
n1 = 350
p̂ 1 = .75
Discount
n2 = 500
p̂ 2 = .52
 = 0.01, /2 = .05 and z.05 = ±1.645
p
n1 pˆ 1  n2 pˆ 2 350(.75)  500(.52)

 .615
n1  n2
350  500
Ho: p1 - p2 = 0
Ha: p1 - p2  0
z
( pˆ 1  pˆ 2 )  ( p1  p 2 )
(.75  .52)  (0)

 6.78
1 
1
 1
1 
(.615)(.385)


p  q  
 350 500 
 n1 n2 
Since the observed z = 6.78 > zc = 1.645, the decision is to reject the null
hypothesis.
10.62 H0: 12 = 22
Ha: 12  22
dfnum = 6
 = .01
n1 = 8
n2 = 7
2
s1 = 72,909
s2 = 129,569
2
dfdenom = 7
The critical F values are:
F =
s1
2
s2
2

F.005,6,7 = 9.16
F.995,7,6 = .11
72,909
= .56
129,569
Since F = .56 < 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.
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Chapter 10: Statistical Inferences About Two Populations
10.63
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
t.05,7 = 1.895
sd
n
4.5 + 1.895
1.414
= 4.5 ± .947
8
3.553 < D < 5.447
10.64 Ho:
Ha:
 = .01
µ1 - µ2 = 0
µ1 - µ2 < 0
df = 23 + 19 - 2 = 40
Alberta
n1 = 23
x 1 = 20.917
s12 = 3.104
Saskatchewan
n2 = 19
x 2 = 22.074
s22 = 1.422
For one-tail test,  = .01 and the critical t.01,40 = -2.423
t =
t =
( x 1  x 2 )  ( 1   2 )
s1 (n1  1)  s 2 (n2  1) 1
1

n1  n2  2
n1 n2
2
2
(20.917  22.074)  (0)
(3.104)( 22)  (1.422)(18) 1
1

40
23 19
= -2.44
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Chapter 10: Statistical Inferences About Two Populations
Since the observed t = -2.44 < t.01,40 = -2.423, the decision is to reject the null
hypothesis.
10.65
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.
 = .05
10.66 Ho: p1 - p2 = 0
Ha: p1 - p2  0
Machine 1
x1 = 38
n1 = 191
pˆ 1 
p
Machine 2
x2 = 21
n2 = 202
x1 38
= .199

n1 191
pˆ 2 
x2
21
= .104

n2 202
n1 pˆ 1  n2 pˆ 2 (.199)(191)  (.104)(202)
= .15

n1  n2
191  202
For two-tail, /2 = .025 and the critical z values are:
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357
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 
= 2.64
Since the observed z = 2.64 > zc = 1.96, the decision is to reject the null
hypothesis.
10.67 Construction
n1 = 338
x1 = 297
pˆ 1 
Telephone Repair
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 - .683) + 1.645
(.879)(.121) (.683)(.317)
= .196 ± .054

338
281
.142 < p1 - p2 < .250
10.68
Aerospace
n1 = 33
x 1 = 12.4
1 = 2.9
Automobile
n2 = 35
x 2 = 4.6
2 = 1.8
For a 99% Confidence Level, /2 = .005 and z.005 = 2.575
( x1  x 2 )  z
 12
n1

(12.4 – 4.6) + 2.575
 22
n2
(2.9) 2 (1.8) 2

33
35
= 7.8 ± 1.52
6.28 < µ1 - µ2 < 9.32
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Chapter 10: Statistical Inferences About Two Populations
10.69
Discount
x 1 = $47.20
1 = $12.45
n1 = 60
Specialty
x 2 = $27.40
2 = $9.82
n2 = 40
 = .01
Ho: µ1 - µ2 = 0
Ha: µ1 - µ2  0
For two-tail test, /2 = .005 and zc = ±2.575
( x 1  x 2 )  ( 1   2 )
z =
1
2
n1

2
2

(47.20  27.40)  (0)
(12.45) 2 (9.82) 2

60
40
n2
= 8.86
Since the observed z = 8.86 > zc = 2.575, the decision is to reject the null
hypothesis.
10.70
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.
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Chapter 10: Statistical Inferences About Two Populations
 = .01
10.71 Ho: µ1 - µ2 = 0
Ha: µ1 - µ2  0
df = 10 + 6 - 2 = 14
A
n1 = 10
x 1 = 18.3
s12 = 17.122
B___
n2 = 6
x 2 = 9.667
s22 = 7.467
For two-tail test, /2 = .005 and the critical t.005,14 = ±2.977
t =
t =
( x1  x 2 )  ( 1   2 )
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.
10.72 A t test was used to test to determine if Hong Kong has significantly different
rates than Mumbai. Let group 1 be Hong Kong.
Ho:
Ha:
µ1 - µ2 = 0
µ1 - µ2  0
x 1 = 130.4
n1 = 19
n2 = 23
x 2 = 128.4
s12 = 166.41
s22 = 193.21
98% C.I. and /2 = .01
t = 0.48 with a p-value of .634 which is greater than  = .02. There is not enough
evidence in these data to declare that there is a difference in the average rental
rates of the two cities.
10.73 H0: D = 0
Ha: D  0
This is a related measure 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 twotailed table t value of + 2.16 and as a result, the researcher rejects the null
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Chapter 10: Statistical Inferences About Two Populations
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.
10.74 The point estimates from the sample data indicate that in the Alberta city the
market share is .310782 and in the Ontario city the market share is .27013o. The
point estimate for the difference in the two proportions of market share is
.0406524. Since the 99% confidence interval ranges from -.0393623 to +.120667
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 an observed 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
.
10.75 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-eight pipes were
measured for sample two. The observed F = 2.0575 is significant at  = .05 for a
one-tailed test since the associated p-value is .0348. The variance of pipe lengths
for machine 1 is significantly greater than the variance of pipe lengths for
machine 2.
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