Assume that a sample is used to estimate a population proportion p. Find the $99.9 \%$ confidence interval for a sample of size 241 with $62 \%$ successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

Step 1 ：We are given a sample size of 241 with 62% successes and we are asked to find the 99.9% confidence interval for the population proportion. The formula for a confidence interval is given by \(\hat{p} \pm Z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\), where \(\hat{p}\) is the sample proportion, Z is the Z-score corresponding to the desired level of confidence, and n is the sample size.

Step 2 ：In this case, \(\hat{p} = 0.62\), n = 241, and we need to find the Z-score for a 99.9% confidence level.

Step 3 ：The Z-score for a 99.9% confidence level can be found using a standard normal distribution table or a Z-score calculator. The Z-score that corresponds to a 99.9% confidence level is approximately 3.291.

Step 4 ：We can now substitute these values into the formula to find the confidence interval.

Step 5 ：Calculating the standard error (SE) using the formula \(SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\), we get SE = 0.03126648942551515.

Step 6 ：Substituting the values into the confidence interval formula, we get the lower limit of the confidence interval as \(CI_{lower} = \hat{p} - Z \times SE = 0.5171019833006296\) and the upper limit as \(CI_{upper} = \hat{p} + Z \times SE = 0.7228980166993704\).

Step 7 ：Rounding to three decimal places, we get the 99.9% confidence interval for the population proportion as \([0.517, 0.723]\).

Step 8 ：\(\boxed{0.517 \leq p \leq 0.723}\). We can be 99.9% confident that the true population proportion lies within this interval.