Question

We are interested in conducting a study to determine the percentage of voters of a state would vote for the incumbent governor. What is the minimum sample size needed to estimate the population proportion with a margin of error of .05 or less at 95% confidenc

Answer 1

`n=(0.5(1-0.5))/(((0.05)/(1.96))^2)=384.16`

And rounded up we have that n=385

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 our interval is at 95% of confidence, our significance level would be given by
`\alpha=1-0.95=0.05` and
`\alpha/2 =0.025`. 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 0.05` 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=0.5`. And replacing into equation (b) the values from part a we got:

`n=(0.5(1-0.5))/(((0.05)/(1.96))^2)=384.16`

And rounded up we have that n=385