How Many Samples?Date: 03/30/99 at 12:59:04 From: Greg Carboni Subject: Sample Size Hi Dr. Math, I have a somewhat tricky question about sample size determination that I have not found anywhere in the literature. Let's assume we have a continuous process with about 8 critical parameters. As a result of the variation in flows and overall heat removal we have a quality characteristic with a standard deviation of 0.36 units. I can back out method variation and determine a sample size at 95% confidence interval of 9. [B = Za/2 * (s/Sqrt(n))] However, I don't infer anything valuable from that. Nine samples in how long a continuous run? If I am taking the critical parameters and saving the data every 2 minutes, this more than satisfies the minimum sample size, but I still don't know what this represents. Since there is gradual error that builds up in field instruments, I would never be assured that the estimated/predicted value (derived theoretically from the reading of the instruments) was real. So, here lies the problem to which I referred. For a continuous operation, how does one determine the number of samples necessary so the mean of the samples represents the mean of the whole? Also, how does one determine the number of feedback samples necessary to determine the validity of predicted values taken from high sample rates of critical flows? I know this is somewhat outside your usual area of reply, but I thought you might have some experience (or know someone who does) or might have encountered this type of issue in the past. Any help you can provide is greatly appreciated. GAC Date: 03/30/99 at 15:20:20 From: Doctor Anthony Subject: Re: Sample Size Without knowing all the details of the information you are dealing with, the best I can do is show you an example of how we find sample size to get confidence limits to within a required margin of error. Imagine you are responsible for estimating the percentage of TV households that are tuned to the Tonight Show with Jay Leno on a particular night. You want to have 90% confidence that your sample percentage has a margin of error of not more than 2%. Assuming that nothing is known about the proportion of households tuned in to television after 11 P.M., how many TV households must be surveyed? If p = proportion tuned to the TV show, then if we take a random sample of size n, and ps is the unbiased estimator of p we get (using the normal approximation to the binomial), ps - p z = --------------- where qs = 1 - ps. sqrt(ps.qs/n) and with 90% confidence limits we have ps-p Prob[-1.645 < ------------- < 1.645] = 0.90 sqrt(ps.qs/n) Prob[ps - 1.645 sqrt(ps.qs/n) < p < ps + 1.645 sqrt(ps.qs/n)] = 0.90 Now if we are to have a margin of error of not more than 2% we require ps + 1.645 sqrt(ps.qs/n) = ps + 0.02 so 1.645 sqrt(ps.qs/n) = 0.02 since we don't know the value of ps, we must assume the worst case with ps.qs a maximum. That is ps(1 - ps) = maximum ps - ps^2 = maximum differentiating 1 - 2.ps = 0 so ps = 1/2 and therefore qs = 1/2 also. So we can put ps.qs = 0.5 x 0.5 = 0.25 In the worst case 1.645 sqrt(0.25/n) = 0.02 squaring 2.706 (0.25/n) = 0.0004 n = 2.706 x 0.25/0.0004 = 1691.25 So take a sample of size 1692 to get a 90% confidence limit with a maximum of 2% margin of error. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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