🚀 go-pugleaf

In article <seX2yRq00Uh785H2EB@andrew.cmu.edu>, dr3u+@andrew.cmu.edu (Daniel Read) writes:
> 
> Can someone explain why calculating the Standard Deviation (SD),
> for small samples, with (n-1) in the denominator is better than
> doing so with (n) in the denominator?  


I'm totally confused by some of the answers to this question.
Using n-1 in the denominator gives an unbiased estimate of the variance.

E (sum[(xi-xbar)**2]) = E (sum[(xi-mu)**2]) - E(sum[(mu-xbar)**2])
                      = E (sum[(xi-mu)**2]) - nE[(mu-xbar)**2]
                      = n sigma**2          - n sigma**2 / n
                      = (n-1) sigma**2

Of course, there's nothing magic about unbaisedness.  You might prefer a
maximum likelihood estimate (n in the denominator) or a minimum mse estimator
(n+1 in the denominator).  The reason n-1 is used is that you have to use
*something* and n-1 makes the cleanest formulas.

For example, Lindgren and Friedman, Pisani, and Purves define their sd with an
n in the denominator.  This means that their formula for CIs involving the t
distribution becomes

                  xbar +- c s / sqrt(n-1)
instead of the usual
                  xbar +- c s / sqrt(n)

(Note:  All of this assumes a whole bunch of qualifications and conditions that
I haven't stated explicitly, beginning with the fact that the variance should
exist!)