Question

You want to obtain a sample to estimate a population proportion. At this point in time, you have no reasonable preliminary estimation for the population proportion. You would like to be 80% confident that you estimate is within 2.5% of the true population proportion. How large of a sample size is required?

Answer 1

`n=(0.5(1-0.5))/(((0.025)/(1.28))^2)=655.36`

And rounded up we have that n=656

Explanation:

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 80% of confidence, our significance level would be given by
`\alpha=1-0.80=0.20` and
`\alpha/2 =0.10`. And the critical value would be given by:

`z_(\alpha/2)=\pm 1.28`

Solution to the problem

The margin of error for the proportion interval is given by this formula:

`ME=z_(\alpha/2)\sqrt{(\hat p (1-\hat p))/(n)}` (a)

Since we don't have prior info for the proportion of interest we can use
`\hat p=0.5` as estimator. And on this case we have that
`ME =\pm 0.025` and we are interested in order to find the value of n, if we solve n from equation (a) we got:

`n=(\hat p (1-\hat p))/(((ME)/(z))^2)` (b)

And replacing into equation (b) the values from part a we got:

`n=(0.5(1-0.5))/(((0.025)/(1.28))^2)=655.36`

And rounded up we have that n=656