Question

A church group is interested in estimating the true proportion of Americans who attend weekly religious services. How large a sample size is required to estimate the true proportion to within 3% with 95% confidence

Answer 1

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

We can use an estimatior for the population proportion as
`\hat p=0.5` since we don't know previous info. And replacing into equation (b) the values from part a we got:

`n=(0.5(1-0.5))/(((0.03)/(1.96))^2)=1067.11`

And rounded up we have that n=1068

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 the confidence level is 99%, our significance level would be given by
`\alpha=1-0.95=0.05` and
`\alpha/2 =0.005`. And the critical value would be given by:

`z_(\alpha/2)=-1.96, z_(1-\alpha/2)=1.96`

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)

And on this case we have that
`ME =\pm 0.03` 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)

We can use an estimatior for the population proportion as
`\hat p=0.5` since we don't know previous info. And replacing into equation (b) the values from part a we got:

`n=(0.5(1-0.5))/(((0.03)/(1.96))^2)=1067.11`

And rounded up we have that n=1068