Re: standard deviation and N-1

From: Dimitris (dimitris_at_dyn.unipg.it)
Date: 07/08/04 

Date: 8 Jul 2004 01:56:34 -0700

> In high school, I learned that the formula for standard deviation has n in
> the denominator, but in college the book has N-1 in the denominator. What
> is the reason for this?

Suppose sig^2_s the variance of your sample, and sig^2_t the true
variance
of the population.

Then it can be worked out that the sample variance, sig^2_s has an
average (over
an infinite amount of samples) equal to

sig^2_s = (N-1)/N sig^2_t

where N is the size of the sample and, as mentioned, sig^2_t is the
true
variance of the population.

You can work this out from the formula for sig^2_s, and from the fact
that
members within a sample are independent from each other.

Therefore, if, instead of sig^2_s, you take N/(N-1) sig^2_s, (using
the 'N-1' formula) you will have a quantity whose average over many
samples will be the true variance of the population.

(For a trivial case, suppose N=1. Then, obviously, sig^2_s will always
be 0 and this will not tell you a thing about the true variance of the
population.)

Bear in mind that this is exact only for *variances*, NOT for
*standard deviations* (their square roots). But, to a good
approximation, one can normally
carry it over to standard deviations.

Hope this helps,

Dimitris

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