A survey of an urban university (population of 25,450) showed that 750 of 1100 students sampled attended a home football game during the season. Using the 99% level of confidence, what is the confidence interval for the proportion of students attending a football game?
A) [0.767, 0.814]
B) [0.6550, 0.7050]
C) [0.6659, 0.6941]
D) [0.6795, 0.6805]

Question

Quizplus · Accepted Answer

To find the confidence interval for population proportion, we can use the formula:

p̂ ± z*√[(p̂q̂)/n]

where p̂ is the sample proportion, q̂ = 1 - p̂, n is the sample size, and z is the z-score for the desired level of confidence.

In this case, we have:

p̂ = 750/1100 = 0.6818

q̂ = 1 - p̂ = 0.3182

n = 1100

We want a 99% level of confidence, which corresponds to a z-score of 2.576.

Plug these values into the formula:

0.6818 ± 2.576*√[(0.6818*0.3182)/1100]

Simplifying:

0.6818 ± 0.0161

So the confidence interval is [0.6659, 0.6941].

Therefore, the best choice is C.

A Survey of an Urban University (Population of 25,450) Showed