112
Chapter 9: TwoSample Tests with Numerical Data
CHAPTER 9
9.1
1
2
1
2
2
2
2
2
1
2
1
2
(
)
(
)
(72
66)
0
20
10
40
50
X
X
Z
n
n
μ
μ
σ
σ





=
=
+
+
= 1.73
9.2
H
0
:
μ
1
=
μ
2
H
1
:
μ
1
≠
μ
2
Decision rule: If
Z
< – 2.58 or
Z
> 2.58, reject
H
0
.
Test statistic:
Z
=
(
X
1
–
X
2
) – (
μ
1
–
μ
2
)
σ
1
2
n
1
+
σ
2
2
n
2
=
(72 – 66) – 0
20
2
40
+
10
2
50
= 1.73
Decision: Since
Z
calc
= 1.73 is between the critical bounds of
Z
=
±
2.58, do not reject
H
0
.
There is inadequate evidence to conclude the two population means are different.
9.3
p
value = 2(1.0 – 0.9582) = 0.0836
9.4
(a)
2
2
2
2
2
1
1
2
2
1
2
(
1)
(
1)
(7) 4
(14) 5
(
1)
(
1)
7
14
p
n
S
n
S
S
n
n

+

+
�
�
�
�
=
=

+

+
= 22
1
2
1
2
2
1
2
(
)
(
)
(42
34)
0
3.8959
1
1
1
1
22
8
15
p
X
X
t
S
n
n
μ
μ





=
=
=
�
�
�
�
+
+
�
�
�
�
�
�
�
�
(b)
df
= (
n
1
– 1) + (
n
2
– 1) = 7 + 14 = 21
(c)
Decision rule:
df
= 21. If
t
> 2.5177, reject
H
0
.
(d)
Decision: Since
t
calc
= 3.8959 is above the critical bound of
t
= 2.5177, reject
H
0
.
There is enough evidence to conclude that the first population mean is larger than the
second population mean.
(e)
We are sampling from two independent normal distributions having equal variances.
(f)
(
29
(
29
2
1
2
1
2
1
1
1
1
42
34
2.0796
22
8
15
p
X
X
t
S
n
n
�
�
�
�

+
=

q
+
�
�
�
�
�
�
�
�
1
2
3.7296<
12.2704
μ
μ

<
112
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113
Chapter 9: TwoSample Tests with Numerical Data
9.5
(a)
H
0
:
μ
1
=
μ
2
Light bulbs produced by machine 1 have the same average life
expectancy as light bulbs produced by machine 2.
H
1
:
μ
1
≠
μ
2
Light bulbs produced by machine 1 have a different average life
expectancy as light bulbs produced by machine 2.
Decision rule: If
Z
< – 1.96 or
Z
> 1.96, reject
H
0
.
Test statistic:
Z
=
(
X
1
–
X
2
) – (
μ
1
–
μ
2
)
σ
1
2
n
1
+
σ
2
2
n
2
=
(375 – 362) – 0
110
2
25
+
125
2
25
= 0.39
Decision: Since
Z
calc
= 0.39 is between the critical bounds of
±
1.96, do not reject
H
0
.
There is not enough evidence to conclude that light bulbs produced by machine 1
have a different average life expectancy from light bulbs produced by machine 2.
(b)
p
value = 2(1.0 – 0.6517) = 0.6966
The probability of obtaining samples whose means differ by 13 hours or more when
the null hypothesis is true is 0.6966.
9.6
(a)
H
0
:
μ
1
≤
μ
2
where Populations: 1 = new machine, 2 = old machine
The average breaking strength of parts produced by the new machine is not greater
than the average breaking strength of parts produced by the old machine.
H
1
:
μ
1
μ
2
The average breaking strength of parts produced by the new machine is greater
than the average breaking strength of parts produced by the old machine.
Decision rule: If
Z
> 2.33, reject
H
0
.
Test statistic:
Z
=
(
X
1
–
X
2
) – (
μ
1
–
μ
2
)
σ
1
2
n
1
+
σ
2
2
n
2
=
(72 – 65) – 0
9
2
100
+
10
2
100
= 5.20
Decision: Since
Z
calc
= 5.20 is above the critical bound of 2.33, reject
H
0
. There is
enough evidence to conclude that the average breaking strength of parts produced by
the new machine is greater than the average breaking strength of parts produced by
the old machine.
(b)
p
value = virtually zero. The probability of obtaining samples whose means differ by
7 or more units of strength when the null hypothesis is true is virtually zero.
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 Spring '08
 Harris
 Normal Distribution, Variance, Statistical hypothesis testing

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