Thread: variance and standard deviation

Why do we average by divivding by n-1 rather than n in calculating the variance? the formula I am using is s^2=[ sum of (x- X)^2]/(n-1) where X= mean. Thank you very much for your help!

Originally Posted by 0123

Why do we average by divivding by n-1 rather than n in calculating the variance? the formula I am using is s^2=[ sum of (x- X)^2]/(n-1) where X= mean. Thank you very much for your help!

Because you are using s^2 to estimate the population variance, and dividing
by n results in a biased estimate while dividing by n-1 gives an unbiased
estimator.

RonL

could you be more clear? because I do not understand why it is unbiased in the case on n-1. Thanks again.

Originally Posted by 0123

could you be more clear? because I do not understand why it is unbiased in the case on n-1. Thanks again.

Take my word for it, the proof is longer than I am prepared to reproduce here.

I can illustrate it by example:

Code:

>x=normal(100,5);
>m=sum(x)/5;         ..100 means of samples of size 5
>v=sum(x^2)/5-m^2;   ..100 (1/n) variances of samples of size 5 
>vv=sum(s')/100      .. average of 100 (1/n) variance of sample of size 5
         0.80403236 
>vv*5/4              .. corrected to 1/(n-1) variances
         1.00504045 
>

Here we see the average variance of samples of size 5 from a population of
variance 1 is ~0.8, while the corrected average variance is ~1, as required.

RonL