In article <8ciqv6$i4v$[EMAIL PROTECTED]>, <[EMAIL PROTECTED]> wrote:
>... 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. There are not any clearly
>outlying observations. The distribution is symetric about the mean and
>tight to the mean as well. Clearly, it is in some sense near to the
>mean and not very variable.
>The other vector B has same n=102. It is
>from a different population and I know the means of these two vectors
>are not equal. Vector B has mean 0.3. It has variance that is 9. The
>distribution is symetric (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.
I don't see why you think this. It would make some sense if these
variables were constrained to be positive, since you could then say
that even though A has a higher variability in absolute terms, it has
a lower variability relative to the mean.
However, it is not meaningful to discuss relative values for
quantities such as, say, longitute, where the "zero" value is
arbitrary. Typically, zero is non-arbitrary because the values are
all positive, for instance, they might be weights. You say, however,
that the distribution for variable B has mean 0.3, is symmetrical, and
has variance 9, hence standard deviation 3. Clearly, there are many
negative values.
I think you cannot expect to receive useful advise regarding whatever
your REAL problem is if you don't tell us what the data actually is
about, and what your purpose in looking at it is.
Radford Neal
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