Solutions for Cpt 9 Problems
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
σ





=
=
+
+
= 1.73
9.2
H
0
:
1
=
2
H
1
:
1
≠
2
Decision rule: If
Z
< – 2.58 or
Z
> 2.58, reject
H
0
.
Test statistic: = 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
Z
=
(
X
1
–
X
2
) – (
1
–
2
)
1
2
n
1
+
2
2
n
2
=
(72 – 66) – 0
20
2
40
+
10
2
50
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





=
=
=
+
+
(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

+
=

a
+
1
2
3.7296<
12.2704

<
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View Full Document9.10
(a)
0
:
I
II
H
μ
=
Mean waiting time of Bank 1 and Bank 2 is the same.
1
:
I
II
H
Mean waiting time of Bank 1 and Bank 2 is different.
PHStat output:
t Test for Differences in Two Means
Data
Hypothesized Difference
0
Level of Significance
0.05
Population 1 Sample
Sample Size
15
Sample Mean
4.286667
Sample Standard Deviation
1.637985
Population 2 Sample
Sample Size
15
Sample Mean
7.114667
Sample Standard Deviation
2.082189
Intermediate Calculations
Population 1 Sample Degrees of Freedom
14
Population 2 Sample Degrees of Freedom
14
Total Degrees of Freedom
28
Pooled Variance
3.509254
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 Spring '08
 Judd
 Statistics, Normal Distribution, Standard Deviation, Null hypothesis, Statistical hypothesis testing

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