Question

In a study of government financial aid for college​ students, it becomes necessary to estimate the percentage of​ full-time college students who earn a​ bachelor's degree in four years or less. Find the sample size needed to estimate that percentage. Use a 0.04 margin of error and use a confidence level of 99​%. Complete parts​ below.
​(a) Assume that nothing is known about the percentage to be estimated.
(b) Assume prior studies have shown that about 55​% of​ full-time students earn​ bachelor's degrees in four years or less.

Answer 1

Answer:

a) n = 1037.

b) n = 1026.

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 $1 - \frac{\alpha}{2}$

The margin of error is:

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

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 $Z = 2.575$.

​(a) Assume that nothing is known about the percentage to be estimated.

We need to find n when M = 0.04.

We dont know the percentage to be estimated, so we use $\pi = 0.5$, which is when we are going to need the largest sample size.

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

$0.04 = 2.575\sqrt{\frac{0.5*0.5}{n}}$

$0.04\sqrt{n} = 2.575*0.5$

$(\sqrt{n}) = \frac{2.575*0.5}{0.04}$

$(\sqrt{n})^{2} = (\frac{2.575*0.5}{0.04})^{2}$

$n = 1036.03$

Rounding up

n = 1037.

(b) Assume prior studies have shown that about 55​% of​ full-time students earn​ bachelor's degrees in four years or less.

$\pi = 0.55$

So

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

$0.04 = 2.575\sqrt{\frac{0.55*0.45}{n}}$

$0.04\sqrt{n} = 2.575*\sqrt{0.55*0.45}$

$(\sqrt{n}) = \frac{2.575*\sqrt{0.55*0.45}}{0.04}$

$(\sqrt{n})^{2} = (\frac{2.575*\sqrt{0.55*0.45}}{0.04})^{2}$

$n = 1025.7$

Rounding up

n = 1026.