Question

A survey of an urban university (population of 25,450) showed that 883 of 1,112 students sampled supported a fee increase to fund improvements to the student recreation center. Using the 95% level of confidence, what is the confidence interval for the proportion of students supporting the fee increase

Answer 1

The confidence interval for the proportion of students supporting the fee increase

( 0.77024, 0.81776)

Step-by-step explanation:

Step-by-step explanation:

Given data a survey of an urban university (population of 25,450) showed that 883 of 1,112 students sampled supported a fee increase to fund improvements to the student recreation center.

Given sample size 'n' = 1112

Sample proportion 'p' =
`(883)/(1112) = 0.7940`

q = 1 - p = 1- 0.7940 = 0.206

The 95% level of confidence intervals

The confidence interval for the proportion of students supporting the fee increase

`(p-z_(\alpha ) \sqrt{(pq)/(n) } ,p + z_(\alpha ) \sqrt{(pq)/(n) } )`

The Z-score at 95% level of significance =1.96

`(0.7940-1.96\sqrt{(0.7940 X 0.206)/(1112) } ,0.7940 + 1.96 \sqrt{(0.7940 X 0.206)/(1112) } )`

(0.7940-0.02376 , 0.7940+0.02376)

( 0.77024, 0.81776)

Conclusion:-

The confidence interval for the proportion of students supporting the fee increase

( 0.77024, 0.81776)