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

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

[tex]n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11[/tex]

And rounded up we have that n=1068

Step-by-step 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 [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.005[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]

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

[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)

And on this case we have that [tex]ME =\pm 0.03[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:

[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)

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

[tex]n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11[/tex]

And rounded up we have that n=1068