On Oct 6, 2009, at 1:51 PM, Kjetil Halvorsen wrote:
On Sat, Oct 3, 2009 at 2:45 PM, David Winsemius <dwinsem...@comcast.net
> wrote:
Do you have a citation for that statement? I cannot convince myself
that it
should be true.
OK. that took some time, since I have no nonparametrics book with me,
but it is a fairly
standard assumption the friedman.test shares with wilcox.test and
others. One online reference giving this is:
http://www.mathworks.com/access/helpdesk/help/toolbox/stats/index.html?/access/helpdesk/help/toolbox/stats/friedman.html&http://www.google.cl/search
?q=assumptions+of+friedman
+test&ie=utf-8&oe=utf-8&aq=t&rls=com.ubuntu:en-
US:unofficial&client=firefox-a
specifically:
"Friedman's test makes the following assumptions about the data in X:
*
All data come from populations having the same continuous
distribution, apart from possibly different locations due to column
and row effects.
*
All observations are mutually independent. "
This is also easy to investigate by simulation in R:
I did:
A[, 1] <- rnorm(100, 0, 1)
A[, 2] <- rnorm(100, 0, 5)
A[, 3] <- rnorm(100, 0, 500)
friedman.test(A)
Friedman rank sum test
data: A
Friedman chi-squared = 2.96, df = 2, p-value = 0.2276
which surprised me! This test seems to be somewhat robust against
variance heterogeneity ???, but that case is not included in the
usual theory.
I do not see that your citation implied that there would be a material
impact (especially toward false positive results which I take to be
the meaning of "not robust") from a violation of the equi-variance
assumption, ... only that equivariance was the basis of the
derivation of the statistical theory. The test might even be
conservative for all we know until the question has been subjected to
simulation studies. And then your simulation suggested not much of a
problem, which does not seem surprising to me given that a rank
transformation has been applied to the data. So I remain unconvinced.
--
Regards;
David.
Kjetil
After looking at the CRAN Task View, I would suggest the OP look at
rlm(MASS) or lmrob(robustbase).
--
David
On Oct 2, 2009, at 11:05 AM, Kjetil Halvorsen wrote:
On Fri, Oct 2, 2009 at 8:45 AM, David Winsemius <dwinsem...@comcast.net
>
wrote:
There are multiple routes to "robust" statistics, but the quick
answer to
this question is probably friedman.test
I don't think friedman.test is robust to variance heterogeneity.
It is
only robust to
non-normality.
Kjetil
I seem to remember a CRAN Task View on the area of Robust
Statistics.
--
David Winsemius
On Oct 2, 2009, at 3:05 AM, Maike Luhmann wrote:
Dear list members,
I am looking for an alternative function for a two-way ANOVA in
the case
of
variance heterogeneity. For one-way ANOVA, I found
oneway.test(), but I
didn't find anything alike for two-way ANOVA. Does anyone have a
suggestion?
Thank you!
Maike Luhmann
Freie Universität Berlin
,
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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