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. 93 of the 1567 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 and .

b. If many groups of 1567 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

Solution :

Given that,

a) Point estimate = sample proportion = = x / n = 93 / 1567 = 0.0593

1 - = 1 - 0.0593 = 0.9407

Z/2 = Z0.05 = 1.645

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

= 1.645 (((0.0593 * 0.9407) / 1567)

= 0.0098

A 90% confidence interval for population proportion p is ,

± E

= 0.0593  ± 0.0098

= ( 0.0495, 0.0691 )

With 90% confidence the proportion of all smart phones that break before the warranty expires is between 0.0495 and 0.0691.

b. If many groups of 1567 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.