I can say that Prof. Burrill's comments fit my thoughts exactly.

Plus, grain size measurements have a number of considerations that must be
included in the interpretation.  First, they cannot go below 0, and can reach
considerable sizes relative to a median.  An average grain size characterizes
a material only insofar as the processing of the sampled material has been
uniform.  If you are looking at a section that partly experienced critical
strain followed by annealing, then you could have some regions with huge
grains, relative to the surrounding ones.  A section through a bend region
can do this, easily.

Secondly, you can only measure the grains, in a typical cross section, in the
cross section you took, not the maximum cross section of the grain(s).  Thus,
the larger diameters may be more characteristic of the size than some of the
smaller diameters.  I believe there are ways for working out this effect, for
more or less uniform grain clusters.  But I doubt that it works for highly
variable grain sizes.

Thirdly, you don't say the specific historic conditions for the samples, but
if you are working with something like brass, especially a 'clean' one, then
recrystallization can do you some very clear grains, and the grain growth
which follows will make you some relatively huge ones.

In sum, the distribution of the grains, and as well the spread (standard
deviation) are significant to the metallurgical interpretation of your
samples, in addition to the 'average' grain size.

Assuming that your measuring/processing method for grain size takes into
account the effect of section size on the estimate of grain size, and that
you still have a wildly skewed distribution, I would suggest that you try the
log transform Don suggested.  It's worked for me on occasion.

However, I have never attempted to evaluate grain sizes in such a complete
analytic manner.

Cheers,
Jay

Donald Burrill wrote:

> On Sun, 13 Oct 2002, Maleck Kcelam wrote:
>
> > Dear sir or madam, I was making an experiment and I have a small
> > problem to write my final results. I've counted grain sizes in a metal
> > sample using a specific software and I obtained the following data:
>   [re-formatted for compactness -- DFB]
>
>  measure    mean  variance  st.dev.   C.V.   min.   max.     N
> >  1         3.6    22.3     4.7     1.32    1.1    88.5   4376
> >  2         4.3    18.3     4.3     0.98    1.5    96.9   4151
>
> > So, basically, I have three questions:
> >
> > 1)  Usually I would have written (average +/- standard deviation) but
> > the standard deviation is superior than my average ! How can I
> > represent my final result or what should I say because I did several
> > other measurements and they are quite the same?
>
> Evidently the distribution of your measurements is highly skewed;  it
> follows that the sample mean cannot be trusted as a measure of
> expectation, and the sample variance is highly inflated by the few very
> large values (maximum about 90, minimum about 1 to 1.5, mean about 4);
> which is why st.dev. > average for both data sets.
>
> Display your distributions, either as dotplots or frequency tables (or
> possibly stem-&-leaf diagrams, but these are likely to be cumbersome with
> such large sample sizes).  Investigate the very large values:  are they
> real?  Do they belong with the bulk of your observations?  If "yes" to
> both questions, consider
>   (a) reporting order statistics instead of average and s.d. (median,
> quartiles, possible 10th and 90th percentiles) so as to show the shape of
> the distribution you're trying to summarize (and if you like to report
> results in diagrams, a pair of box plots, one for each data set, would be
> informative to your readers);
>   (b) taking logarithms of your data values, displaying those
> distributions, and using them if they're (nearly) symmetrical (in which
> case mean & s.d. of the log values would be reasonable summary
> statistics) (and in that case the antilog of the mean logarithm is the
> geometric mean of the original data).
>
> > 2) Can I express my result as (average +/- coefficient of variation)?
>
> This does not appear to make any intuitive sense.  Why would one?
>
> > 3) I need to represent this measures in one single number, so, how
> > would I unite the two measurements?  Which formula should I use?
>
> If you report order statistics, all you can do is combine the two data
> sets into one (N = 8527) and find the median etc. of the combined data.
>
> If logarithms give you reasonable distributions, the usual "single number"
> would be the mean of the combined sample:
>   (N1*(average1) + N2*(average2)) / (N1 + N2).
>
> >       (mean average of  MEASURE 1+ MEASURE 2) +/- (????)
>
> The "+/- (????)" does not look to me like a "single number";  looks more
> like two numbers, an average of some kind +/- a measure of uncertainty.
>  If you end up using means and s.d.s (either because you cast out the very
> high values as not properly belonging to your data, or because you used a
> transformation (e.g., log) that made the distribution nearly symmetric),
> it might be appropriate to use a pooled standard deviation:
>
>    pooled variance = (var1*(N1-1) + var2*(N2-1)) / (N1 + N2 - 2),
>    pooled s.d. = square root of pooled variance.
>
> Not everyone would agree that this is appropriate, however;  and knowing
> nothing about the field in which you're operating, I cannot offer useful
> opinion on this point.
>
> > Could you possibly help me with this problem? Thank you in advance and
> > I hope to hearing from you soon.
>
>  -----------------------------------------------------------------------
>  Donald F. Burrill                                            [EMAIL PROTECTED]
>  56 Sebbins Pond Drive, Bedford, NH 03110                 (603) 626-0816
>  [Old address:  184 Nashua Road, Bedford, NH 03110       (603) 471-7128]

--
Jay Warner
Principal Scientist
Warner Consulting, Inc.
4444 North Green Bay Road
Racine, WI 53404-1216
USA

Ph: (262) 634-9100
FAX: (262) 681-1133
email: [EMAIL PROTECTED]
web: http://www.a2q.com

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