```Date: Aug 31, 2011 1:06 AM
Author: Steven D'Aprano
Subject: Which sample variance should I choose?

The population variance is given by:?^2 = ?(x - µ)^2 / nwith µ = population mean, the summation being over all the x in thepopulation.(for brevity, I haven't attempted to show subscript-i on the x).If you don't have the entire population, you can estimate the variance withthe sample variance:s^2 = ?(x - m)^2 / n  (Eq. 1)where m = sample mean (usually written as x bar), and the sum is over allthe x in the sample. A second estimator is:s^2 = ?(x - m)^2 / (n-1)  (Eq. 2)which some people prefer because it is unbiased (that is, the average of allthe possible sample variances equals the true population variance if youuse the (n-1) version).See also http://mathworld.wolfram.com/SampleVariance.htmlI have a set of data with an (allegedly) known population mean µ but anunknown ?^2. I wish to estimate ?^2. Under what circumstances should Iprefer Eq.1 over Eq.2?Or should I ignore the sample mean altogether, and plug the known populationmean into one of the two equations? I.e.:s^2 = ?(x - µ)^2 / n      (Eq. 3)s^2 = ?(x - µ)^2 / (n-1)  (Eq. 4)Under what circumstances should I prefer each of these four estimators of?^2 and what are the pros and cons of each?Thanks in advance,-- Steven
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