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 \(\boxed{(0.442, 0.658)}\).