A survey of an urban university (population of 25,450) showed that 750 of 1,100 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)[0.655, 0.705]
C)[0.644, 0.716]
D)[0.680, 0.681]

Question

Quizplus · Accepted Answer

To find the confidence interval for the proportion, we will use the formula:
sample proportion +/- z-score * standard error of the proportion
where sample proportion = 750/1100 = 0.682, z-score for 99% confidence level = 2.576 and standard error of proportion = sqrt((0.682*0.318)/1100) = 0.017
Plugging in the values, we get:
0.682 +/- 2.576*0.017
The confidence interval comes out to be [0.767, 0.814]. Therefore, the best choice is (A).

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