Dear Thomas,

Thank you very much for your answer! Can I use your function in my analysis?

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()?
This makes sometime people like me a bit confused!

Regards,
Cheba

2010/4/6 Thomas Lumley <tlumley@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
>>
>>        [[alternative HTML version deleted]]
>>
>> ______________________________________________
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>>
> Thomas Lumley                   Assoc. Professor, Biostatistics
> tlumley@u.washington.edu        University of Washington, Seattle
>
>

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