Question

A smart phone manufacturer is interested in constructing a 95% confidence interval for the proportion of smart phones that break before the warranty expires. 80 of the 1669 randomly selected smart phones broke before the warranty expired. Round answers to 4 decimal places where possible. a. With 95% confidence the proportion of all smart phones that break before the warranty expires is between and . b. If many groups of 1669 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:

A smart phone manufacturer is interested in constructing a 95% confidence interval for the proportion of smart phones that break before the warranty expires.

Sample size = n= 1669

x = number of smart phones that break before the warranty expires = 80

Thus sample proportion is:

Formula:

where

We need to find z_c value for c=95% confidence level.

Find Area = ( 1 + c ) / 2 = ( 1 + 0.95) /2 = 1.95 / 2 = 0.9750

Look in z table for Area = 0.9750 or its closest area and find z value.

Area = 0.9750 corresponds to 1.9 and 0.06 , thus z critical value = 1.96

That is : Z_c = 1.96

Thus

Thus

Correct interpretation is option a)

With 95% confidence the proportion of all smart phones that break before the warranty expires is between 0.0377 and 0.0581.