🚀 go-pugleaf

In article <1992Aug15.000049.29790@leland.Stanford.EDU> dhinds@leland.Stanford.
EDU (David Hinds) writes:
>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]).
[deletions]
  The 8 values are not necessarily the whole population (which may
well be finite).
   For any given range of values, the probability that a RS of size
8 will be contained in that range is no greater than the probability
that a RS of size 2 will be contained in that range.

My intent (as I made clear) wasn't that we think of a particular
estimator, just that we
look at the data and try to see how variable it is.

Robert George
(speaking only for myself)