A survey 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?

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

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 [tex]n = 1100, \pi = \frac{870}{1100} = 0.7838[/tex]

95% confidence interval

So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]z = 1.96[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.7838 - 1.96\sqrt{\frac{0.7838*0.2162}{1100}} = 0.7595[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.7838 + 1.96\sqrt{\frac{0.7838*0.2162}{1100}} = 0.8081[/tex]

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