Really sorry. My formula is a rearrangement of the confidence interval formula shown below for ascertaining the maximum error.
E = Z(a/2) x SD/SQRT N The issue is you want to solve for N, but you have no standard deviation value. The formula then translates into n = (Z(a/2)*SD)/E)^2 Note: ^2 stands for squared. You have only the confidence interval, let's say 95% and E of 1%. Lets say that you want to find out how many people in the US have fake drivers licenses using these numbers. How large (N) must your sample be? Do you have to hypothesize a US population to solve this, ie. 300 m or can you solve it another way. It was suggested you can express the SD as a fraction of the E. ie. E = SD/2. "Randy Poe" <[EMAIL PROTECTED]> wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... > John Jackson wrote: > > > the forumla I was using was n = (Z?/e)^2 and attempting to express .05 as a > > fraction of a std dev. > > I think you posted that before, and it's still getting > garbled. We see a Z followed by a question mark, and > have no idea what was actually intended. > > - Randy ================================================================= Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =================================================================
