Question

A researcher would like to estimate the population proportion of adults living in a certain town who have at least a high school education. No more information is available about its value. How large of a sample size is needed to estimate it to within 0.15 of the true value with 99% confidence?

Answer 1

Answer:

We would need a sample of size at least 74.

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 $Z = 2.575$.

How large of a sample size is needed to estimate it to within 0.15 of the true value with 99% confidence?

We need a sample of size at least n

n is found when $M = 0.15$

We don't know the proportion, so we estimate $\pi = 0.5$, which is the case for which we are going to need the largest sample size.

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

$0.15 = 2.575\sqrt{\frac{0.5*0.5}{n}}$

$0.15\sqrt{n} = 2.575*0.5$

$\sqrt{n} = \frac{2.575*0.5}{0.15}$

$(\sqrt{n})^{2} = (\frac{2.575*0.5}{0.15})^{2}$

$n = 73.67$

Rounding up

We would need a sample of size at least 74.