Chapter 10
TwoSample Hypothesis Tests
10.1
For each problem, the following formulas were used:
1
2
1
2
calc
c
1
2
c
c
1
2
combined number of successes
p
p
x
x
=
where
p
z
combined sample size
n
n
1
1
p (1  p )
+
n
n

+
=
=
+
a.
Standard error: .0987
Z
Test Statistic:
−2.43
p
value: 0.0075
Z
Critical: 2.3263
Decision is not close:
reject
H
0
b.
Standard error: .0884
Z
Test Statistic:
2.26
p
value: .0237
Z
Critical: +/ 1.645
Decision is not close: reject
H
0
c.
Standard error: .07033
Z
Test Statistic:
−1.7063
p
value: 0.0440
Z
Critical: 1.645
Decision is close:
reject
H
0
10.2
For each problem, the following formulas are used:
1
2
1
2
calc
c
1
2
c
c
1
2
combined number of successes
p
p
x
x
=
where
z
combined sample size
n
n
1
1
p (1  p )
+
n
n

+
=
=
+
a.
Standard error: .0555
Z
Test Statistic:
1.4825
p
value: 0.1382
Z
Critical: +/ 1.9600
Decision is not close:
fail to reject
H
0
b.
Standard error: .0618
Z
Test Statistic:
−2.162
p
value: .0153
Z
Critical: −2.3263
Decision is not close:
reject
H
0
c.
Standard error: .01526
Z
Test Statistic:
1.638
p
value: .0507
Z
Critical: 1.645
Decision is close:
fail to reject
H
0
83
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View Full Document10.3
a.
Define
π
1
= proportion of shoppers that paid by debit card in 1999. Define
π
2
= proportion of shoppers
that paid by debit card in 2004.
H
0
:
π
1
=
π
2
versus
H
1
:
π
1
<
π
2
. This is a lefttailed test. Reject the null hypothesis if
z
< −2.33.
b
.
z
= −2.28 so we fail to reject the null hypothesis (although the decision is close.) The sample does not
provide strong enough evidence to conclude that there is a difference in the two proportions.
c.
p
value = .0113.
d.
Normality is assumed since
n
1
p
1
> 10 and
n
2
p
2
> 10.
10.4
a.
Define
π
1
= proportion of loyal mayonnaise purchasers. Define
π
2
= proportion of loyal soap purchasers.
H
0
:
π
1
=
π
2
versus
H
1
:
π
1
≠
π
2
. This is a twotailed test. Reject the null hypothesis if
z
< −1.96 or
z
> 1.96.
z
= 1.725 therefore we fail to reject the null hypothesis. The sample evidence does not show a significant
difference in the two proportions.
b.
95% confidence interval: (−.015, .255). Yes, the interval does contain zero.
10.5
a.
Define
π
1
= proportion of respondents in first group (the group given the gift certificate.) Define
π
2
=
proportion of respondents in the second group.
H
0
:
π
1
=
π
2
versus
H
1
:
π
1
≠
π
2
. This is a twotailed test. Reject the null hypothesis if
z
< −1.96 or
z
> 1.96.
z
= 2.021 therefore we reject the null hypothesis. The sample shows a significant difference in response
rates.
b.
95% confidence interval: (.0013, .0787). No, the interval does not contain zero. We estimate that the
response rate for the group given the gift certificate is higher than the group that did not receive the gift
certificate.
10.6
a.
Define
π
1
= proportion of flights with contaminated water in August and September 2004. Define
π
2
=
proportion of flights with contaminated water in November and December 2004.
H
0
:
π
1
=
π
2
versus
H
1
:
π
1
<
π
2
. Reject the null hypothesis if
z
< −1.645.
z
= −1.1397 so we fail to reject
the null hypothesis. The level of contamination was not lower in the first sample.
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 Spring '08
 HEATHERADAMS
 Variance, Null hypothesis, Statistical hypothesis testing, ν

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