Question

Consumer Reports (January 2005) indicates that profit margins on extended warranties are much greater than on the purchase of most products. In this exercise we consider a major electronics retailer that wishes to increase the proportion of customers who buy extended warranties on digital cameras. Historically, 20 percent of digital camera customers have purchased the retailer’s extended warranty. To increase this percentage, the retailer has decided to offer a new warranty that is less expensive and more comprehensive. Suppose that three months after starting to offer the new warranty, a random sample of 527 customer sales invoices shows that 132 out of 527 digital camera customers purchased the new warranty. Find a 95 percent confidence interval for the proportion of all digital camera customers who have purchased the new warranty. Are we 95 percent confident that this proportion exceeds .20? (Round your answers to 3 decimal places.) The 95 percent confidence interval is [, ]. , the entire interval above .20.

Answer 1

proportion of all digital camera customers who have purchased the new warranty=p=x/n=132/527=0.2505

SE(p)=sqrt(p(1-p)/n)=sqrt(0.2505*(1-0.2505)/527))=0.0189

(1-alpha)*100% confidence interval for proportion (P)=sample proportion (p) ±z(alpha/2)*SE(p)

95% confidence interval for P=p±z(0.1/2)*SE(p)=0.0.2505±1.96*0.0189=0.2505±0.0370=(0.2135,0.2875)

since the proportion P=0..20 doesnot belong to this interval (0.2135,0.2875), so we are 95 percent confident that this proportion exceeds 0.20.