company one year from now is the same. (
8 points
)
a.
What is the sample proportion for the two regions?
Region 1
Region 2
Sample Size
95
115
Yes
72
80
Proportion
0.7579
0.6957
b.
Compute the test statistic z used to test the hypothesis.
z
=
0.7579
−
0.6957
√
0.7238
(
1
−
0.7238
)
(
1
95
+
1
115
)
¿
1.0034
c.
Can the firm conclude that the proportion of employees within the two regions who
plan to stay with the company one year from now is not the same? Use the p-value
approach and α = 0.05 to test the hypothesis stated above.
P
−
Value
=
0.3153

BUSI1013_Unit 9
4
The p-value is greater than 0.05, therefore we do NOT reject the null
hypothesis.
The firm could NOT conclude the hypothesis.
d.
Construct a 95% confidence interval for the difference of the population proportion for
the two regions.
(
0.7579
−
0.6957
)
±
1.96
√
0.7579
(
1
−
0.7579
)
95
+
0.6957
(
1
−
0.6957
)
115
0.0622
±
0.1204
= (-0.0582, 0.183)
3.
Use the data in and description of
BUSI1013-Case.xls
(from Unit 1 Exercise Question 6) to answer
this question. (4 Points for each part;
8 points total
)
a.
Perform a statistical test to see whether the proportion of members who would
recommend the Credit Union to their family and friends have increased as a result of the
pilot.
1-
Null hypothesis:
There is no change in proportion of members who
would recommend the Credit Union to their family and friends have
increased as a result of the pilot.
p1-p2=0
2-
Alternative hypothesis:
The proportion of members who would
recommend the Credit Union to their family and friends has increased
as a result of the pilot.
p2-p1>0
Number of people before pilot = 290; Number of people who said yes before
pilot = 185
Number of people after pilot = 290; Number of people who said yes after pilot
= 262
P
1
=
X
1
n
1
=
185
290
=
0.6379,
P
2
=
X
2
n
2
=
262
290
=
0.9034
Pooled estimate of p is
X
1
+
X
2
n
1
+
n
2
=
185
+
262
290
+
290
=
0.7707
The test statistic z:

BUSI1013_Unit 9
5
Z
=
P
2
−
P
1
√
P
(
1
−
P
)(
1
n
1
+
1
n
2
)
=
0.9034
−
0.6379
√
0.7707
(
1
−
0.7707
)(
1
290
+
1
290
)
=
7.61
The p-value corresponding to the test statistic is = 0.0000
The p-value is less than 0.0005, so therefore we reject the null hypothesis.

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