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)[0.7671,0.8143]
B)[0.6550,0.7050]
C)[0.6464,0.7172]
D)[0.6805,0.6815]

Question

Quizplus · Accepted Answer

To find the confidence interval for the proportion of students attending a football game, we use the formula:
CI = p ± z*(sqrt((p*(1-p))/n))
where
p = sample proportion = 750/1100 = 0.6818
n = sample size = 1100
z = z-score for 99% level of confidence = 2.576 (from standard normal distribution table)

Substituting these values, we get:
CI = 0.6818 ± 2.576*(sqrt((0.6818*(1-0.6818))/1100))
CI = 0.6818 ± 0.0354
CI = (0.6464, 0.7172)

Therefore, the best choice is C, which gives the correct confidence interval for the proportion of students attending a football game.

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