pbar11021520671 pbar21321520868 z0671 0868sqrt07701

Pbar11021520671 pbar21321520868 z0671 0868sqrt07701

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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
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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

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