A previous study gave $(0.58,0: 66)$ as a confidence interval for $p$. Find the minimum sample size required to ensure that your estimate $(\widehat{p})$ will be within 0.1 of $p$ with $99.5 \%$ confidence.

IMPORTANT: Use a critical value that you found with your calculator (not from a table), and round it to $\underline{3}$ places after the decimal point before you plug it into a formula and perform your calculations. Do not round-off any other intermediate results.

Step 1 ：First, calculate the sample proportion \(\widehat{p}\) and the margin of error \(E\) from the given confidence interval. The sample proportion \(\widehat{p}\) is the average of the lower and upper bounds of the confidence interval, and the margin of error \(E\) is half the difference between the upper and lower bounds of the confidence interval. In this case, \(\widehat{p} = \frac{0.58 + 0.66}{2} = 0.62\) and \(E = \frac{0.66 - 0.58}{2} = 0.04\).

Step 2 ：Next, find the Z-score corresponding to 99.5% confidence. The Z-score is a measure of how many standard deviations an element is from the mean. In this case, the Z-score corresponding to 99.5% confidence is approximately 2.576.

Step 3 ：Finally, rearrange the formula for the margin of error to solve for the sample size \(n\). The formula for the margin of error is \(E = Z \sqrt{\frac{\widehat{p}(1 - \widehat{p})}{n}}\), so solving for \(n\) gives \(n = \frac{Z^2 \widehat{p}(1 - \widehat{p})}{E^2}\). Plugging in the values we have, \(n = \frac{(2.576)^2 * 0.62 * (1 - 0.62)}{(0.04)^2}\), which rounds up to 977.

Step 4 ：Final Answer: The minimum sample size required to ensure that the estimate will be within 0.1 of the population proportion with 99.5% confidence is \(\boxed{977}\).