Asurvey of an urban university (population of 25,450) showed that 870 of 1,100 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 95% confidence interval for the proportion of students supporting the fee increase is (0.7595, 0.8081).

Explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence interval , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

For this problem, we have that:

870 of 1,100 students sampled supported a fee increase to fund improvements to the student recreation center. So

95% confidence interval

So , z is the value of Z that has a pvalue of , so .

The lower limit of this interval is:

The upper limit of this interval is:

The 95% confidence interval for the proportion of students supporting the fee increase is (0.7595, 0.8081).