Question

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?

Answer 1

the answer is [0.767, 0.815] :)

Answer 2

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

Step-by-step explanation:

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

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

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
`n = 1100, \pi = (870)/(1100) = 0.7838`

95% confidence interval

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (0.05)/(2) = 0.975`, so
`z = 1.96`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.7838 - 1.96\sqrt{(0.7838*0.2162)/(1100)} = 0.7595`

The upper limit of this interval is:

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

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