Question

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 90% level of confidence, what is the confidence interval for the proportion of students attending a football game

Answer 1

The 90% level of confidence interval For the Population proportion

(0.6569 ,0.7031)

Explanation:

Explanation:-

Given sample size 'n' = 1100

Given sample proportion

`p^(-) =(x)/(n) = (750)/(1100) = 0.68`

The 90% level of confidence interval For the Population proportion is determined by

`(p^(-) - Z_{(\alpha )/(2) } \sqrt{(p(1-p))/(n) } ,p^(-) + Z_{(\alpha )/(2) } \sqrt{(p(1-p))/(n) })`

90% of level of significance Z-value

`Z_{(0.10)/(2) } = Z_(0.05) =1.645`

`(0.68 - 1.645 \sqrt{(0.68(1-0.68))/(1100) } ,(0.68 + 1.645 \sqrt{(0.68(1-0.68))/(1100)`

On calculation , we get

(0.68 -0.0231 ,0.68 +0.0231)

(0.6569 ,0.7031)

Final answer:-

The 90% level of confidence interval For the Population proportion

(0.6569 ,0.7031)