*Sorry I posted this on the examples page by accident.

- Thread starter jaben
- Start date

*Sorry I posted this on the examples page by accident.

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.

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

But I am wondering if I need to do anything with the $500? And would the .196 be different since my confidence level is 98% instead of 95%?

Thanks again!!