Minimum Sample Size for Confidence IntervalDate: 09/21/2000 at 11:31:54 From: Steve Subject: Sample size for estimate for population proportion Let's say I have a population of 400 people. I need to find out if a value for each person is 1 or 0. Sampling each person is very time- consuming, however, so I don't want to have to sample any more than necessary. I need to find out how many people I need to sample to say that the proportion correct (1 = correct) in the sample is representative of the proportion correct in the population. I am trying to make a case that if the randomly drawn sample I choose is done right, I am "whatever" amount confident that the entire population is correct at the same rate. Thanks, Steve Date: 09/21/2000 at 15:55:56 From: Doctor Anthony Subject: Re: Sample size for estimate for population proportion Since we are talking about the means of samples we can by the Central Limit theorem use a normal distribution. If p = proportion of 1's, 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 = ------------- sqrt(ps.qs/n) where qs = 1 - ps. 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 say 10% we require: ps + 1.645 sqrt(ps.qs/n) = ps + 0.1 so 1.645 sqrt(ps.qs/n) = 0.1 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 qs = 1 - ps = 1/2 So we can put ps.qs = 0.5 x 0.5 = 0.25 In the worst case 1.645*sqrt(0.25/n) = 0.1 squaring 2.706 (0.25/n) = 0.01 n = 2.706 x 0.25/0.01 = 67.65 So take a sample of size 67.65 to get a 90% confidence limit with a maximum of 10% margin of error. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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