140 students at a college were asked whether they had completed their required English 101 course, and 77 students said "yes". Construct the $99 \%$ confidence interval for the proportion of students at the college who have completed their required English 101 course.
Enter your answers as decimals (not percents) accurate to three decimal places.
The Confidence Interval is (
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Step 1 ：Given that 140 students at a college were asked whether they had completed their required English 101 course, and 77 students said 'yes'. We are asked to construct the 99% confidence interval for the proportion of students at the college who have completed their required English 101 course.

Step 2 ：To construct a confidence interval for a proportion, we can use the formula: \[\hat{p} \pm Z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\] 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 3 ：In this case, \(\hat{p} = \frac{77}{140}\), \(Z = 2.576\) (for a 99% confidence interval), and \(n = 140\).

Step 4 ：Substituting these values into the formula, we get the standard error (SE) as 0.04204589329360411.

Step 5 ：Then, we calculate the lower and upper bounds of the confidence interval as 0.44168977887567584 and 0.6583102211243242 respectively.

Step 6 ：Rounding these values to three decimal places, we get the final answer.

Step 7 ：Final Answer: The 99% confidence interval for the proportion of students at the college who have completed their required English 101 course is \(\boxed{(0.442, 0.658)}\).