Question

You want to obtain a sample to estimate a population proportion. At this point in time, you have no reasonable estimate for the population proportion. You would like to be 98% confident that you esimate is within 0.1% of the true population proportion. How large of a sample size is required?

Answer 1

Answer:

A sample size of at least 1,353,733 is required.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , 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 .

The margin of error is:

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

98% confidence level

So $\alpha = 0.02$, z is the value of Z that has a pvalue of $1 - \frac{0.02}{2} = 0.99$, so $Z = 2.327$.  

You would like to be 98% confident that you esimate is within 0.1% of the true population proportion. How large of a sample size is required?

We need a sample size of at least n.

n is found when M = 0.001.

Since we don't have an estimate for the proportion, we use the worst case scenario, that is $\pi = 0.5$

So

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

$0.001 = 2.327\sqrt{\frac{0.5*0.5}{n}}$

$0.001\sqrt{n} = 2.327*0.5$

$\sqrt{n} = \frac{2.327*0.5}{0.001}$

$(\sqrt{n})^{2} = (\frac{2.327*0.5}{0.001})^{2}$

$n = 1353732.25$

Rounding up

A sample size of at least 1,353,733 is required.