Question

The National Collegiate Athletic Association (NCAA) requires colleges to report the graduation rate of their athletes. At one large university, 69% of all students who entered in 1994 graduated within six years. Ninety-six (96) of the 152 students who entered with athletic scholarships graduated. Construct the 99% confidence interval for the proportion of athletes who graduate.

Answer 1

The 99% confidence interval for the proportion of athletes who graduate is (0.5309, 0.7323).

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`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:

`n = 152, \pi = (96)/(152) = 0.6316`

99% confidence level

So
`\alpha = 0.01`, z is the value of Z that has a pvalue of
`1 - (0.01)/(2) = 0.995`, so
`Z = 2.575`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.6316 - 2.575\sqrt{(0.6316*0.3684)/(152)} = 0.5309`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.6316 + 2.575\sqrt{(0.6316*0.3684)/(152)} = 0.7323`

The 99% confidence interval for the proportion of athletes who graduate is (0.5309, 0.7323).