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

Answer:

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

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 level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

$n = 152, \pi = \frac{96}{152} = 0.6316$

99% confidence level

So $\alpha = 0.01$, z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$, so $Z = 2.575$.

The lower limit of this interval is:

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

The upper limit of this interval is:

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

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