A smart phone manufacturer is interested in constructing a $99 \%$ confidence interval for the proportion of smart phones that break before the warranty expires. 82 of the 1594 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 1594 randomly selected smart phones are selected, then a different confidence interval would be produced for each group. About 99 percent of these confidence intervals will contain the true population proportion of all smart phones that break before the warranty expires and about 1 population proportion. percent will not contain the true

Step 1 ：Given that the sample size (n) is 1594 and the number of successes (phones that broke before the warranty expires) is 82.

Step 2 ：First, we calculate the sample proportion (p̂), which is the number of successes divided by the sample size. So, \(p̂ = \frac{x}{n} = \frac{82}{1594} = 0.0514\).

Step 3 ：Next, we calculate the standard error of the proportion, which is the square root of \((p̂ * (1 - p̂)) / n\). So, the standard error is \(\sqrt{\frac{p̂ * (1 - p̂)}{n}} = \sqrt{\frac{0.0514 * (1 - 0.0514)}{1594}} = 0.0055\).

Step 4 ：The confidence interval is then calculated as \(p̂ ± Z * \) standard error, where Z is the Z-score for the desired confidence level. For a 99% confidence level, the Z-score is approximately 2.576. So, the confidence interval is \(0.0514 ± 2.576 * 0.0055\).

Step 5 ：Calculating the above expression, we get the confidence interval as \((0.0372, 0.0657)\).

Step 6 ：\(\boxed{\text{Final Answer: With 99% confidence the proportion of all smart phones that break before the warranty expires is between 0.0372 and 0.0657.}}\)