A survey of an urban university showed that 750 of 1100 stud

A survey of an urban university 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?

[0.767, 0.814]

[0.6550, 0.7050]

[0.6456, 0.7179

[0.6795, 0.6805]

Solution

Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=750
Sample Size(n)=1100
Sample proportion = x/n =0.6818
Confidence Interval = [ 0.6818 ±Z a/2 ( Sqrt ( 0.6818*0.3182) /1100)]
= [ 0.6818 - 2.576* Sqrt(0.0002) , 0.6818 + 2.58* Sqrt(0.0002) ]
= [ 0.6456,0.7179] [ANSWER]