Question

# 140 students at a college were asked whether they had completed their required English 101 course,...

140 students at a college were asked whether they had completed their required English 101 course, and 91 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 ( , )

Confidence interval for Population Proportion is given as below:

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

x = 91

n = 140

P = x/n = 91/140 = 0.65

Confidence level = 99%

Critical Z value = 2.5758

(by using z-table)

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Confidence Interval = 0.65 ± 2.5758* sqrt(0.65*(1 – 0.65)/140)

Confidence Interval = 0.65 ± 2.5758* 0.0403

Confidence Interval = 0.65 ± 0.1038

Lower limit = 0.65 ± 0.1038 =0.5462

Upper limit = 0.65 ± 0.1038 =0.7538

Confidence interval = (0.546, 0.754)