Question

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 ( , )

Answer #1

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)

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 ( , )

138 students at a college were asked whether they had completed
their required English 101 course, and 109 students said "yes".
Construct the 90% 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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