Question

A survey of an urban university showed that 22​% of 2800 randomly selected students supported a fee increase to fund improvements to the student recreation center. What is the 90% confidence interval estimate for the proportion of all students at this university in favor of the fee​ increase? (Round to the nearest tenth of a​ percent.)

A.19.0​%< p​ <25.0​%

B.20.7​%< p​ <23.3​%

C.21.1​%< p​ <22.9​%

D.20.1​%< p​ <23.9​%

E.None of the above

Answer 1

Given : sample size=p=2800

Sample proportion=p=0.22

q=1-p=1-0.22=0.78

Now , ; From standard normal distribution table

Therefore , the 90% confidence interval estimate for the proportion of all students at this university in favor of the fee​ increase is ,

Correct option (B)