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

Answer:

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

( 0.77024, 0.81776)

Step-by-step explanation:

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' = $\frac{883}{1112} = 0.7940$

                           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

$(p-z_{\alpha } \sqrt{\frac{pq}{n} } ,p + z_{\alpha } \sqrt{\frac{pq}{n} } )$

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

$(0.7940-1.96\sqrt{\frac{0.7940 X 0.206}{1112} } ,0.7940 + 1.96 \sqrt{\frac{0.7940 X 0.206}{1112} } )$

(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)