🚀 go-pugleaf

In article <1992Aug14.231916.23479@magnus.acs.ohio-state.edu> regeorge@magnus.acs.ohio-state.edu (Robert E George) writes:
>
>More intuitively, if we take a very sample, we are less likely to get
>extreme values and so our notion of what the population variance is (note
>that I am not proposing some particular estimator) will be unrealistic.
>For instance, I give an exam to two students. Their scores are 67 and
>71. I think, "Gee, there's not a lot of variability in these scores."
>But then 8 more students take the exam:
>     60 78  100 38  50 88  99 39
>
>and it now is clear there *is* more variability in these scores.
>
>But let me reiterate that T will *always* have a negative bias for the
>population variance whatever the sample size is

This is wrong.  If this set of 8 values is the "population", sure, for
your particular sample, the predicted variance happens to be less than
the population variance.  However, the mean variance of all samples
from this population will equal the population variance (i.e., the
sample variance is an unbiased estimator).  Note that most samples of
size 2 from this population will have much larger variances than the
one you picked, and many will have variances larger than the parent
population (i.e., you are just as likely to get [100 38] as [67 71]).

The reason the N-weighted variance is biased, is because it measures
variability around the sample mean rather than the population mean,
and this estimate absorbs some of the variability in the sample.  The
N-weighted variance of the sample around the population mean, and the
N-1 weighted variance of the sample around the sample mean, are both
unbiased estimates of the variance in the population.

        - David Hinds
          dhinds@allegro.stanford.edu