PDA

View Full Version : Sampling

jaben
11-28-2005, 11:34 AM
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
11-28-2005, 11:41 AM
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
11-28-2005, 12:02 PM
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
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!!

JohnM
11-28-2005, 12:07 PM
You used the formula for estimating a proportion - you need to use the formula for estimating a mean.

For a 98% confidence, you need to use z = 2.326

jaben
11-28-2005, 12:14 PM
Thanks for the quick reply. Ok a few last questions. Did I do that correctly for the \$10 ( i put it .10^2) and was I supposed to do anything with the 500 standard deviation?
Thanks!

JohnM
11-28-2005, 12:20 PM
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.