On Wed, 7 Apr 2010, cheba meier wrote:
Dear Thomas,
Thank you very much for your answer! Can I use your function in my analysis?
Sure.
The wilcox.test() tests the differences in average ranks (H0: F(X)=F(Y)),
why then do people in many studies compare t.test() with wilcox.test()?
The Wilcoxon test is slightly more complicated than difference in average ranks.
The null hypothesis is actually that P(X>Y)=0.5, in the sense that if this
hypothesis is true the test will not reject even for arbitrarily large sample
sizes. However, because the test has truly weird behaviour when comparing
arbitrary distributions, it's really not useful unless you have good reason to
expect stochastic ordering, that is, the entire distribution is shifted up or down
(though not necessarily by the same amount).
One reason that people compare t-tests and Wilcoxon tests is that a lot of the
elegant statistical theory for two-sample testing is for location-shift
alternatives, that is, for situations where you know the two distributions are
the same shape but don't know if one of them is shifted up or down relative to
the other. If you happen to know that the distributions are the same shape
then all the tests are asking the same scientific question, so comparing them
makes sense.
Now, in reality we never know that we have a location-shift alternative. It's
still reasonable to hope that some of the lessons learned from studying
location-shift alternatives still apply (some do, some don't), even if you do
understand that reality is more complex. On the other hand, some people either
don't know that reality is more complex or want to pretend it isn't.
A few years ago I gathered up ten introductory statistics and biostatistics
textbooks that I and colleagues had to hand, to see what they had to say about
the Wilcoxon test. If I recall correctly, eight of them said something that
was both untrue and importantly misleading. Of the remaining two, one didn't
mention the Wilcoxon test.
-thomas
This makes sometime people like me a bit confused!
Regards,
Cheba
2010/4/6 Thomas Lumley <tlum...@u.washington.edu>
None of them.
- mood.test() looks promising until you read the help page and
see that it does not do Mood's test for equality of quantiles,
it does Mood's test for equality of scale parameters.
- wilcox.test() is not a test for equal medians
- ks.test() is not a test for equal medians.
Mood's test for the median involves dichotomizing the data at
the pooled median and then doing Fisher's exact test to see if
the binary variable has the same mean in the two samples.
median.test<-function(x,y){
z<-c(x,y)
g <- rep(1:2, c(length(x),length(y)))
m<-median(z)
fisher.test(z<m,g)$p.value
}
Like most exact tests, it is quite conservative at small sample
sizes.
-thomas
On Tue, 6 Apr 2010, cheba meier wrote:
Dear all,
What is the right test to test whether the median of two groups
are
statistically significant? Is it the wilcox.test, mood.test or
the ks.test?
In the text book I have got there is explanation for the
Wilcoxon (Mann
Whitney) test which tests ob the two variable are from the same
population
and also ks.test!
Regards,
Cheba
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Thomas Lumley Assoc. Professor, Biostatistics
tlum...@u.washington.edu University of Washington, Seattle
Thomas Lumley Assoc. Professor, Biostatistics
tlum...@u.washington.edu University of Washington, Seattle
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