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

`n=(0.93(1-0.93))/(((0.09)/(1.96))^2)=30.875`

And rounded up we have that n=31

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{(\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=(\hat p (1-\hat p))/(((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=(0.93(1-0.93))/(((0.09)/(1.96))^2)=30.875`

And rounded up we have that n=31