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 / n
with µ = population mean, the summation being over all the x in the
(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 with
the sample variance:
s^2 = ?(x - m)^2 / n (Eq. 1)
where m = sample mean (usually written as x bar), and the sum is over all
the 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 all
the possible sample variances equals the true population variance if you
use the (n-1) version).
See also http://mathworld.wolfram.com/SampleVariance.html
I have a set of data with an (allegedly) known population mean µ but an
unknown ?^2. I wish to estimate ?^2. Under what circumstances should I
prefer Eq.1 over Eq.2?
Or should I ignore the sample mean altogether, and plug the known population
mean 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,