Assume that a sample is used to estimate a population proportion p. Find the $99 \%$ confidence interval for a sample of size 368 with 133 successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places. C.I. $=$

Step 1 ：Given a sample size of 368 with 133 successes, we want to find the 99% confidence interval for the population proportion p.

Step 2 ：We start by calculating the sample proportion, denoted as \(\hat{p}\), which is the ratio of successes to the sample size. In this case, \(\hat{p} = \frac{133}{368} = 0.361\).

Step 3 ：The Z-score for a 99% confidence interval is 2.576. This value is obtained from a standard normal distribution table.

Step 4 ：We then calculate the standard error (SE) using the formula \(SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\), where n is the sample size. Substituting the given values, we get \(SE = \sqrt{\frac{0.361(1-0.361)}{368}} = 0.025\).

Step 5 ：Finally, we calculate the confidence interval (CI) using the formula \(CI = \hat{p} \pm Z \times SE\). Substituting the calculated values, we get \(CI = 0.361 \pm 2.576 \times 0.025\).

Step 6 ：This gives us a lower limit of \(0.361 - 2.576 \times 0.025 = 0.297\) and an upper limit of \(0.361 + 2.576 \times 0.025 = 0.426\).

Step 7 ：\(\boxed{\text{Final Answer: The 99% confidence interval for a sample of size 368 with 133 successes is }(0.297, 0.426)}\)