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.

Respuesta :

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 [tex]\pi[/tex], and a confidence level of [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:

[tex]n = 152, \pi = \frac{96}{152} = 0.6316[/tex]

99% confidence level

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

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6316 - 2.575\sqrt{\frac{0.6316*0.3684}{152}} = 0.5309[/tex]

The upper limit of this interval is:

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

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