Question

A recent study at Winthrop University was done to determine the ratio of democrats to republicans. Use the information in each problem to come up with an appropriate sample size to accurately estimate the mean. Let a republican represent a success r. If no preliminary sample is taken, how large should the sample be to be 90% sure that the estimate is within .03 of the population proportion?

Answer 1

Answer:

A sample of 752 should be taken.

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 of:

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

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 $Z = 1.645$.

No preliminary sample is taken

This means that $\pi = 0.5$

How large should the sample be to be 90% sure that the estimate is within .03 of the population proportion?

This is n for which M = 0.03. So

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

$0.03 = 1.645\sqrt{\frac{0.5*0.5}{n}}$

$0.03\sqrt{n} = 1.645*0.5$

$\sqrt{n} = \frac{1.645*0.5}{0.03}$

$(\sqrt{n})^2 = (\frac{1.645*0.5}{0.03})^2$

$n = 751.7$

Rounding up:

A sample of 752 should be taken.