# Sampling

#### jaben

##### New Member
Hi, I was looking at your examples for determining the minimum sample size required to estimate a population. Ok with Z^2, you put 1.96, and that is a 95% confidence level. Why is it 1.96? Secondly, on the ((p*q), you have put the .51*.49 and I understand that, but what if you didn't know how many voted for the particular candidate in the last election? What would you put for p*q? Thanks
*Sorry I posted this on the examples page by accident.

#### JohnM

##### TS Contributor
95% of the area under the normal curve is within z=-1.96 and z=1.96.

If you have absolutely no idea what p or q was, assume the worst case and let p=.50 and q=.50. This will give you the largest standard error of the proportion and will result in the largest sample size. It is a conservative approach.

#### jaben

##### New Member
Ok I am going to post the whole problem, then post my calculations. Can you tell me if this is correct?
Estimate the mean dollars that each cardholder will spend each month. It would like to be within plus or minus $10 of the true mean with a 98% confidence level. The standard deviation is thought to be$500. HOw many card holders should be sampled.
Ok I did this: n=((.50*.50)/.10^2)*.196^2
n=(.25/.01)*3.8416
n= 25*3.8416 =96.04
Thanks!

#### JohnM

##### TS Contributor
Look at the formula for estimating a mean - it's different than the one you used. You also need a value of z for the desired level of power.