# Margin of Error

#### statisticsStudent

I've been agonizing over this problem involving a sample mean and margin of error. I feel like the sample size I got as my answer is too small?

Question is:

How large a sample is needed to estimate a population proportion to within a margin of error of 4 % at the 98 % confidence level?

First, I calculated the critical value as 2.33 and then plugged the margin of error, .4, and 2.33 into the formula:

n = (z* divided by E)^2 x p-hat x q-hat or (2.33 divided by .4) ^2 x 1/2 x 1/2

I got the 1/2 since p-hat and q-hat are both unknowns so you have to use the most conservative estimate, or 1/2.

Anyway, my answer is 8.49 or sample size of 9

I feel like this is way too small of a sample size! What'd I do wrong?

#### statisticsStudent

update: I think it's supposed to be 849, not 8.49 (Giggle)

#### Danneedshelp

Since we have that $$\displaystyle (1-\alpha)=.98$$, $$\displaystyle \alpha$$ must equal 0.02 and $$\displaystyle \frac{\alpha}{2}=.01$$. So, the z value we are looking for is $$\displaystyle z_{\frac{\alpha}{2}}=z_{.01}=2.33$$. We then require that

$$\displaystyle 2.33\sqrt{\frac{pq}{n}}=.4$$.

Since the we do not know anything about $$\displaystyle p$$ we use $$\displaystyle p=.5$$.

I have not computed the value, but if you just solve for n you should find your answer. However, I think your answer is correct.