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