the Bank to their family and friends have increased because of the change.
pbar1=102/152=0.671
pbar2=132/152=0.868
pbar=(102+132)/(152+152)=0.770
z=(0.671-0.868)/sqrt(0.770(1-0.770)(1/152+1/152))= -4.08
P value = P (z<-4.08) = 0.000002
Since p-value = 0.000002 < 0.05 we will reject H0. Therefore, we can conclude that the members
who would recommend the bank to their family and friends have increased
b)
Construct a 99% confidence interval for the difference of the proportions, before and after
the change, of members who would recommend the Bank to their family and friends.
0.671-0.868±2.575*sqrt(0.671*(1-0.671)/152+0.868*(1-0.868)/152)
= -0.197±0.121
(-0.318,0.076)
3. A national retailer has completed an internal survey on its social media strategy. The key
question in the survey is whether the stores managers think that the recently-launched social me-
dia strategy is effective. The following is the output of a statistical analysis of the data:
Social Media Strategy is effective
Store Managers
No
Yes
West Region
15
35
Central Region
10
28
East Region
45
50
Based on this output, can the retailer conclude that the opinion of the store managers on the
whether the recently-launched social media strategy is effective is dependent on which region
they belong to? Use a chi square test and a level of significance of 0.05. (4 points)
H
0
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- Summer '15
- Statistics, Statistical hypothesis testing
