On Thu, 6 Apr 2000 [EMAIL PROTECTED] wrote:
> In response to Donald Burrill I would like to clarify what I said in my
> original post. I have a vector A with 102 elements. The mean of these
> elements is 50. These elements all are close to 50. The variance is
> ~90 which is small given the magnitude of the mean. The maximum and
> minimum values are also fairly close to 50.
If A is approximately Gaussian, with N = 102, one would not be surprised
if the maximum and minimum values were about 2 standard deviations from
the mean -- that is, about 50 plus or minus about 20, or from 30 to 70.
(Does this agree with what you mean by "fairly close to 50"?)
> There are not any clearly outlying observations. The distribution is
> symmetric about the mean and tight to the mean as well.
> Clearly, it is in some sense near to the mean and not very variable.
This may be clear to you; it is not so clear to me. Perhaps this is a
function of the graphical display you mention below.
> The other vector B has same n=102. It is from a different population
> ... Vector B has mean 0.3. It has variance that is 9.
> The distribution is symmetric (more or less) about the mean but the
> values are very spread out and not tight around the mean.
> Clearly the _variance_ of vector A is larger than the variance of B
> BUT I think I am right to say that A is less variable than B.
> My decision to say this is based on their graphical representations.
This would seem to imply that the two distributions were displayed in
separate graphical representations, and to two quite different scales.
If they had been displayed to the same scale (which might be unreasonable
to attempt without a VERY large piece of paper, given the disparity in
means and variances) vector B would have appeared much more densely
packed (i.e., visually less variable) than vector A. After all, B's
standard deviation is 3, about one-third that of A.
On the other hand, in some disciplines it is not uncommon to refer to the
"coefficient of variation" (CV) as a measure of dispersion. (I never
used it much myself, because I could not justify it as intelligible, let
alone interpretable, in social-science contexts.) It is defined thus:
CV = s/m
wehre s = standard deviation and m = mean. For your data,
vector A has a CV of about 9.5/50, or about 0.2;
vector B has a CV of 3/0.3, or about 10.
In this sense, it would be entirely appropriate to say that the A data
are less variable than the B data, and by nearly two orders of magnitude.
What I don't know, of course, is whether the CV is an appropriate
measure of dispersion for the kinds of data you have. Notice that it is
a very difficult thing to deal with if the mean of a distribution is in
any danger of being very near to zero: if the mean be zero, CV is
infinite.
> Maybe it is wrong to think there is a way to quantify this.
No: trust your intuition. Most things people notice _can_ be
quantified, but it isn't always clear how best to do so, and one may
encounter disagreements on the propriety of various ways of quantifying
something. That doesn't mean that it shouldn't be done, nor that it
isn't worth doing: it may mean that whatever one is trying to quantify
is something really interesting.
But one thing you have to address is what it _means_ if the
variability of data set A is less than that of data set B. What
consequences ensue from such a declaration? And what consequences ensue
if the declaration should turn out to be wrong? (As a result of further
research, perhaps.) Presumably such a result has _some_ meaning for you,
or you would not have raised the question in the first place.
> Maybe this isn't very much an issue of statistics.
It's more, I think, an issue in the interface (or intersection) between
statistics and your substantive discipline. Most interesting questions
are like that, I think.
-- DFB.
------------------------------------------------------------------------
Donald F. Burrill [EMAIL PROTECTED]
348 Hyde Hall, Plymouth State College, [EMAIL PROTECTED]
MSC #29, Plymouth, NH 03264 603-535-2597
184 Nashua Road, Bedford, NH 03110 603-471-7128
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