Question

A political action committee is interested in finding out what proportion of voters will

support an environmental initiative. Obtain a sample size that will ensure a margin of

error of at most 0.09 for a 95% confidence interval. Similar initiatives in the past have

gotten 93% support.

Answer 1

Answer:

$n=\frac{0.93(1-0.93)}{(\frac{0.09}{1.96})^2}=30.875$  

And rounded up we have that n=31

Step-by-step explanation:

First we need to find the critical value for the margin of error desired. The confidence level is 0.95 so then the significance is $\alpha=1-0.95=0.05$ and the critical value for this case would be:

$z_{\alpha/2} =\pm 1.96$

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 $ME =\pm 0.09$ 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)  

The estimated proportion for this case is $\hat p =0.93$. And replacing into equation (b) the values from part a we got:

$n=\frac{0.93(1-0.93)}{(\frac{0.09}{1.96})^2}=30.875$  

And rounded up we have that n=31