Question

A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than examine the records for all students, the dean randomly selects 200 students and finds that 118 of them are receiving financial aid. If the dean wanted to estimate the proportion of all students receiving financial aid to within 1% with 90% reliability, how many students would need to be sampled

Answer 1

Answer:

6546 students would need to be sampled.

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 z-score that has a p-value of $1 - \frac{\alpha}{2}$.

The margin of error is:

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

The dean randomly selects 200 students and finds that 118 of them are receiving financial aid.

This means that $n = 200, \pi = \frac{118}{200} = 0.59$

90% confidence level

So $\alpha = 0.1$, z is the value of Z that has a p-value of $1 - \frac{0.1}{2} = 0.95$, so $Z = 1.645$.

If the dean wanted to estimate the proportion of all students receiving financial aid to within 1% with 90% reliability, how many students would need to be sampled?

n students would need to be sampled, and n is found when M = 0.01. So

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

$0.01 = 1.645\sqrt{\frac{0.59*0.41}{n}}$

$0.01\sqrt{n} = 1.645\sqrt{0.59*0.41}$

$\sqrt{n} = \frac{1.645\sqrt{0.59*0.41}}{0.01}$

$(\sqrt{n})^2 = (\frac{1.645\sqrt{0.59*0.41}}{0.01})^2$

$n = 6545.9$

Rounding up:

6546 students would need to be sampled.