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. 86 of the 1513 randomly selected smart phones broke before the warranty expired. Round answers to 4 decimal places where possible.

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

b. If many groups of 1513 randomly selected smart phones are selected, then a different confidence interval would be produced for each group. About percent of these confidence intervals will contain the true population proportion of all smart phones that break before the warranty expires and about percent will not contain the true population proportion.

Answer 1

a)

sample proportion, = 0.0568
sample size, n = 1513
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.0568 * (1 - 0.0568)/1513) = 0.006

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

Margin of Error, ME = zc * SE
ME = 1.64 * 0.006
ME = 0.0098

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.0568 - 1.64 * 0.006 , 0.0568 + 1.64 * 0.006)
CI = (0.0470 , 0.0666)

With 90% confidence the proportion of all smart phones that break before the warranty expires is between 0.0470 and 0.0666

b)

If many groups of 1513 randomly selected smart phones are selected, then a different confidence interval would be produced for each group. 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