Question

A smart phone manufacturer is interested in constructing a 90% confidence interval for the proportion of smart phones that break before the warranty expires. 83 of the 1543 randomly selected smart phones broke before the warranty expired.

a. With 90% confidence the proportion of all smart phones that break before the warranty expires is between answer ____ and answer ______.

b. If many groups of 1543 randomly selected smart phones are selected, then a different confidence interval would be produced for each group. About answer _____ percent of these confidence intervals will contain the true population proportion of all smart phones that break before the warranty expires and about answer ______ percent will not contain the true population proportion.

Answer 1

a)
sample proportion, = 0.0538
sample size, n = 1543
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.0538 * (1 - 0.0538)/1543) = 0.0057

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.64

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.0538 - 1.64 * 0.0057 , 0.0538 + 1.64 * 0.0057)
CI = (0.0445 , 0.0631)

With 90% confidence the proportion of all smart phones that break before the warranty expires is between 0.0445 and 0.0631

b)
About 90 percent of these confidence intervals will contain the true population proportion of all smart phones that break before the warranty expires and about 10 percent will not contain the true population proportion.