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RE: st: Mean Test


From   "Newson, Roger B" <r.newson@imperial.ac.uk>
To   <statalist@hsphsun2.harvard.edu>
Subject   RE: st: Mean Test
Date   Sun, 22 Jan 2006 20:03:44 -0000

Yes, one-way anova does indeed assume equality of variances. And yes,
this will lead to bias in the standard errors, if that assumption is
incorrect.

As I see it, the issues are as follows. The population variability of
small groups may be estimated using only the sampling variability of
these same small groups, as we do if we use the Huber standard error (eg
the unequal-variance t-test). Alternatively, the population variability
of small groups may be estimated using also the sampling variability of
larger groups, as we do if we use the Wedderburn standard error (eg the
equal-variance t-test). The decision to use the Huber standard error or
the Wedderburn standard error should therefore be based on whether we
think the population variability of small groups is better estimated
using the sample variability of these same small groups or using the
sample variability of other, larger groups. Therefore, Huber variances
tend to have the advantage if the number of observations in the smallest
group is large, whereas Wedderburn variances are better if the number of
observations in the smallest group is small and variances only vary
slightly from group to group. Unfortunately, if ww have enough data to
answer that question, then we probably have enough observations, even in
the smallest group, to be able to use Huber variances without much loss
of power. Therefore, I personally tend to choose between Huber and
Wedderburn variances a priori, based on numbers of observations per
group, instead of doing heteroskedasticity tests. If I choose Wedderburn
variances, then I make it clear, in the Methods section of the report,
that I am making an assumption of homoskedasticity. I do not try to test
this assumption, because, in these circumstances, there are not enough
data in the smallest group to be able to test this assumption.

I hope this helps. The references I gave in my earlier posting will also
probably help.

Best wishes

Roger


Roger Newson
Lecturer in Medical Statistics
POSTAL ADDRESS:
Respiratory Epidemiology and Public Health Group
National Heart and Lung Institute at Imperial College London
St Mary's Campus
Norfolk Place
London W2 1PG
STREET ADDRESS:
Respiratory Epidemiology and Public Health Group
National Heart and Lung Institute at Imperial College London
47 Praed Street
Paddington
London W1 1NR
TELEPHONE: (+44) 020 7594 0939
FAX: (+44) 020 7594 0942
EMAIL: r.newson@imperial.ac.uk
WEBSITE: http://www.kcl-phs.org.uk/rogernewson/
Opinions expressed are those of the author, not of the institution.


-----Original Message-----
From: owner-statalist@hsphsun2.harvard.edu
[mailto:owner-statalist@hsphsun2.harvard.edu] On Behalf Of Rijo John
Sent: 22 January 2006 18:51
To: statalist@hsphsun2.harvard.edu
Subject: Re: st: Mean Test

Thank you Dr. Maarten. But isn't  One way ANOVA also assume the
equality of variance (of the populations) that we consider? (excuse me
for my ignorance). Hence in a case where variances are statistically
different would the method you suggest give the 'right' result?

And thank you for pointing out that p-value thing being incorrect.

Rijo.

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