A smart phone manufacturer is interested in constructing a $90 \%$ confidence interval for the proportion of smart phones that break before the warranty expires. 90 of the 1571 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 1571 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
Answer
Final Answer: With 90% confidence the proportion of all smart phones that break before the warranty expires is between 0.0476 and 0.0669. So, the final answer is \(\boxed{[0.0476, 0.0669]}\).
Step 1 ：Given that the sample size (n) is 1571 and the number of successes (phones that broke before the warranty expired) is 90, we can calculate the sample proportion (\(\hat{p}\)) as the number of successes divided by the sample size, which is \(\hat{p} = \frac{90}{1571} = 0.0573\).
Step 2 ：The z-score (Z) corresponding to the desired confidence level of 90% is 1.645.
Step 3 ：We can calculate the standard error (SE) using the formula \(SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.0573(1-0.0573)}{1571}} = 0.0059\).
Step 4 ：Finally, we can calculate the confidence interval using the formula \(\hat{p} \pm Z_{\alpha/2} \times SE\). The lower limit of the confidence interval is \(0.0573 - 1.645 \times 0.0059 = 0.0476\) and the upper limit of the confidence interval is \(0.0573 + 1.645 \times 0.0059 = 0.0669\).
Step 5 ：Final Answer: With 90% confidence the proportion of all smart phones that break before the warranty expires is between 0.0476 and 0.0669. So, the final answer is \(\boxed{[0.0476, 0.0669]}\).
