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TOTAL 20 samples ANY MISSING COMBINATIONS?? I wonder that the sample mean = population mean; BUT sample variance not equal (3.49) to population variance (2.92)…. ANYTHING WRONG!! Standard deviation of Xbar = σ^{2}/ n (take square root) * square root of (Nn/ N1)??? What this means???
Hi Willy, You are not missing any combinations. There are 6 choose 3 which is 20 combinations. You calculated the mean of each of the 20 samples which is correct. This gives a new population of 20 numbers called the sampling distribution of Xbar. (I can't easily write a bar over the X so I call it Xbar.) You found the mean of this population and verified that it is the same as the mean of the original population of 6 items. For the variance calculations you need to calculate the population variance of this new population of 20 items. You have the mean (3.5) so calculate the variance as you did for the population of 6 items. That is
I called this V since I don't want to confuse it with σ^{2}, the variance of the original population. If the sampling had been with replacement you would find that V = σ^{2}/n = σ^{2}/3. Since the sampling is without replacement there is a correction factor, sometimes called the finite population correction factor, which is (N  n)/(n  1) where n is the size of the sample (n = 3) and N is the size of the population (N = 6). If you check the numbers in your example you will find that
. Harley
 


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