Question

According to the Oxnard College Student Success Committee report in the previous year, we believe that 18% of students struggle in their classes because they don't spend more than 8 hours studying independently outside of a 4-unit class. For this year, you would like to obtain a new sample to estimate the proportiton of all Oxnard students who struggle in their classes because they don't study enough outside of the classrooms. You would like to be 99% confident that your estimate is within 5% of the true population proportion. How large of a sample size is required

Answer 1

Answer:

A sample size of 392 is required.

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:

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

18% of students struggle in their classes because they don't spend more than 8 hours studying independently outside of a 4-unit class.

This means that $\pi = 0.18$

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 $Z = 2.575$.

You would like to be 99% confident that your estimate is within 5% of the true population proportion. How large of a sample size is required?

A sample size of n is required.

n is found when M = 0.05. So

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

$0.05 = 2.575\sqrt{\frac{0.18*0.82}{n}}$

$0.05\sqrt{n} = 2.575\sqrt{0.18*0.82}$

$\sqrt{n} = \frac{2.575\sqrt{0.18*0.82}}{0.05}$

$(\sqrt{n})^2 = (\frac{2.575\sqrt{0.18*0.82}}{0.05})^2$

$n = 391.5$

Rounding up:

A sample size of 392 is required.