There is a specific problem associated w/this formula: "I need to do a spot check of our inventory of CDs to ascertain which ones are genuine CDs from the factory and which ones are counterfeit CDs "burned" by a forger. I have recommeded that we take a random sample by SKU numbers and I was told it was OK as long as whatever estimate I produced, my error of estimate would be no more than 5% with a 99% confidence level. I am wondering how I should go about determining how many CDs, of which there are 10,000, to randomly look at."
the forumla I was using was n = (Z?/e)^2 and attempting to express .05 as a fraction of a std dev. "Glen Barnett" <[EMAIL PROTECTED]> wrote in message 9oug3c$su1$[EMAIL PROTECTED]">news:9oug3c$su1$[EMAIL PROTECTED]... > > John Jackson <[EMAIL PROTECTED]> wrote in message > MGns7.49824$[EMAIL PROTECTED]">news:MGns7.49824$[EMAIL PROTECTED]... > > re: the formula: > > > > n = (Z?/e)2 > > This formula hasn't come over at all well. Please note that newsgroups > work in ascii. What's it supposed to look like? What's it a formula for? > > > could you express E as a % of a standard deviation . > > What's E? The above formula doesn't have a (capital) E. > > What is Z? n? e? > > > In other words does a .02 error translate into .02/1 standard deviations, > > assuming you are dealing w/a normal distribution? > > ? How does this relate to the formula above? > > Glen > ================================================================= Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =================================================================
