Question

A survey of an urban university showed that 750 or 1,100 students sampled attended a home football game during the season. Using the 90% level of confidence, what is the confidence interval for the proportion of students attending a football game?

Answer 1

The 90% confidence interval for the proportion of students attending a football game is (0.6587, 0.7049).

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:

A survey of an urban university showed that 750 or 1,100 students sampled attended a home football game during the season. This means that
`n = 1100, \pi = (750)/(1100) = 0.6818`

90% confidence interval

So
`\alpha = 0.1`, z is the value of Z that has a pvalue of
`1 - (0.1)/(2) = 0.95`, so
`Z = 1.645`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.6818 - 1.645\sqrt{(0.6818*0.3182)/(1100)} = 0.6587`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.6818 + 1.645\sqrt{(0.6818*0.3182)/(1100)} = 0.7049`

The 90% confidence interval for the proportion of students attending a football game is (0.6587, 0.7049).