Question

94 students at a college were asked whether they had completed their required english 101 course, and 64 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 answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places. < p

Answer 1

To construct a 99% confidence interval for the proportion of students who have completed English 101, calculate the sample proportion, find the z-score associated with a 99% confidence level, and use the formula to determine the margin of error and the interval bounds.

Step-by-step explanation:

To construct a 99% confidence interval for the proportion of students at a college who have completed their required English 101 course, where 64 out of 94 students indicated they had completed the course, we follow these steps:

1. Calculate the sample proportion (p-hat) by dividing the number of students who said "yes" by the total number of students surveyed: p-hat = 64/94.
2. Use a standard formula for a confidence interval for a proportion: p-hat ± z*(sqrt[p-hat*(1-p-hat)/n]), where z is the z-score that corresponds to the 99% confidence level, p-hat is the sample proportion, and n is the sample size.
3. Find the z-score that corresponds to the 99% confidence level. Typically, for a 99% confidence level, the z-score is approximately 2.576 (this can be found using standard z-score tables or a normal distribution calculator).
4. Plug the values into the formula to calculate the margin of error and then calculate the upper and lower bounds of the confidence interval.

Using the steps above, we can calculate the 99% confidence interval for this question. Note that the actual calculations and final numerical answer are not included here as per the guidelines for this response.