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

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' =

                           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

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

(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)

its the second choice.

step-by-step explanation:

the elements in the range will be 7 less than the elements in the domain.

answer: what is the question you're trying to ask?

step-by-step explanation: