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

Answer:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$   (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:

$n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11$  

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 $\alpha=1-0.95=0.05$ 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:  

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$    (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:  

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$   (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:

$n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11$  

And rounded up we have that n=1068