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Hi, my name is Rosalie. I have a question about sample variance. To give an unbiased estimate of the population variance, the denominator of the sample variance should be (n1) instead of n. I tried to convince myself by comparing the population variance and sample variance (with denominator n and n1): Consider the population {1, 2, 3, 4, 5} Population mean mu = (1+2+3+4+5)/5=3 Population variance sigma =[(13)^{2} + (23)^{2}+ ··· +(53)^{2}]/5 = 2 Now, all the possible samples of size n = 4 drawn from the above population are considered.
It looks like the mean of the variance calculated by dividing by n (1.875) gives a better estimate of the population variance (2). Where did I go wrong? Thanks for your help. 

Hi Rosalie, The fact that the sample variance, calculated by dividing by n1, is an unbiased estimator for the population variance is true if the population is infinite or you sample with replacement. In your example if you sample with replacement then you get 5^{4} = 625 possible samples rather than the 5 you examined. Rather than doing an example that large I am going to repeat your calculation but using the population {1,2,3}and a sample of size 2.
Now, all the possible samples of size n = 2 drawn from the above population with replacement are considered.
Penny 

