Re: standard deviation and N-1

From: Herman Rubin (hrubin_at_odds.stat.purdue.edu)
Date: 07/10/04 

Date: 9 Jul 2004 20:27:03 -0500

In article <Pine.GSO.4.05.10407072147070.8955-100000@hercules.acsu.buffalo.edu>,
Victoria Florsheim <vf2@buffalo.edu> wrote:
>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?

>So far, I found this in my book (By Yates, Moore, McCabe):
>Why do we average by dividing by n-1 rather than n? Because the sum of
>deviations is always zero, the last deviation can be found once we know
>the other n-1. So we are not averaging n unrelated numbers. Only n-1 of
>the squared deviations can cary freely, and we average by dividing by the
>total by n-1. The n-1 is called the degrees of freedom of the variance or
>standard deviation.

>I sort of understand that, but could someone explain in simpler terms and
>expand on that? I'm still a little puzzled as to why n-1.

>Thanks.

That is an excuse, rather than an explanation. It was not
arrived at except from the following straightforward
mathematical statements; the first can be proved with
algebra only, and the second requires calculus and some
linear algebra.

1. If one computes the expected value of the sum of
squares of the deviations from the mean of n independent,
or just uncorrelated, random variables with the same
mean and variance, it is n-1 times the common variance.

2. If in addition the random variables are independent
and normally distributed, the sum of squares can be
expressed as the sum of squares of n-1 independent
normal variables with mean 0 and the common variance. 

-- 
This address is for information only.  I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu         Phone: (765)494-6054   FAX: (765)494-0558

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