Question

A bookstore claims that 60% of its visitors are return customers. You sample 200 random customers entering the store and find that 130 of them had previously visited and bought items from the bookstore. Estimate the proportion of return customers entering the store at the 95% confidence level.

Answer 1

Given that,

Point estimate = sample proportion = = x / n = 130 / 200 = 0.65

Z_/2 = 1.96

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 * (((0.65 * 0.35) / 200)

= 0.066

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.65 - 0.066 < p < 0.65 + 0.066

0.584 < p < 0.716

(0.584 , 0.716)