Mind you, I'm a biologist, so I only generally know what these 2 tests are.
Okay, say you have 2 different groups, control and treated. Each group/sample has 35 individuals (n=35). You know the sample mean (x-bar) and the sample standard deviation (s) but you do not know the population mean (mu) or the population standard deviation (sigma). One of the general rules, I thought, of using a t-test was that you do not know the population mean (sigma) so I would think you'd use the t-test, BUT isn't one of the tenets of the central limit theorem that if n>30 then you can assume the sample standard deviation (s) equals the population standard deviation (sigma) and would then therefore use the z-score? Also, the central limit theorem states that n must be greater than 30 to assume normal distribution but I've also seen that n>20 is enough. Which is correct?
Okay, say you have 2 different groups, control and treated. Each group/sample has 35 individuals (n=35). You know the sample mean (x-bar) and the sample standard deviation (s) but you do not know the population mean (mu) or the population standard deviation (sigma). One of the general rules, I thought, of using a t-test was that you do not know the population mean (sigma) so I would think you'd use the t-test, BUT isn't one of the tenets of the central limit theorem that if n>30 then you can assume the sample standard deviation (s) equals the population standard deviation (sigma) and would then therefore use the z-score? Also, the central limit theorem states that n must be greater than 30 to assume normal distribution but I've also seen that n>20 is enough. Which is correct?
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If you don't know the population standard deviation then you would use the t-distribution (student distribution) regardless the size of your sample.