A smart phone manufacturer is interested in constructing a 99% confidence interval

Question

A smart phone manufacturer is interested in constructing a 99% confidence interval for the proportion of smart phones that break before the warranty expires. 91 of the 1578 randomly selected smart phones broke before the warranty expired. Round answers to 4 decimal places where possible.

a. With 99% confidence the proportion of all smart phones that break before the warranty expires is between ___ and ___ .

b. If many groups of 1578 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

99% CI for number of phones that break before warranty expires

91 of 1578 phones broke before warranty expired

99% CI is plus or minus about 3 standard deviations from the mean, but more precisely 2.576, not 3. 3 is for 99.7% CI

assuming a normal distribution,

and mean proportion = 91/1578 = .0577

mean number = 91 but you need the standard deviation, which isn't expressly given

use a binomial distribution, as there are two possibilities, either warranty expires or doesn't when the phone breaks down

the normal distribution approximates the binomial

mean = np = 1578(.0577) = 91 and variance = npq =np(1-p) = 1578(.0577)(.9423) =about 85.7493.

standard deviation = square root of variance = sqr(85.7493) = about 9.2601

CI = 91+/- 2.576(9.2601) = 23.8540

CI is the interval (67.1460, 114.8540)