[R] Wilcoxon Test and Mean Ratios

[avinash barnwal]

Hi,

http://en.wikipedia.org/wiki/Wilcoxon_signed-rank_test

We can clearly see that null hypothesis is median different or not.
One way of proving non difference is P(X>Y) = P(X<Y) where X and Y are
ordered.

On 9/20/12, peter dalgaard <pdalgd at gmail.com> wrote:
>
> On Sep 20, 2012, at 02:43 , Thomas Lumley wrote:
>
>> On Thu, Sep 20, 2012 at 5:46 AM, Mohamed Radhouane Aniba
>> <aradwen at gmail.com> wrote:
>>> Hello All,
>>>
>>> I am writing to ask your opinion on how to interpret this case. I have
>>> two vectors "a" and "b" that I am trying to compare.
>>>
>>> The wilcoxon test is giving me a pvalue of 5.139217e-303 of a over b with
>>> the alternative "greater". Now if I make a summary on each of them I have
>>> the following
>>>
>>>> summary(a)
>>>     Min.   1st Qu.    Median      Mean   3rd Qu.      Max.
>>> 0.0000000 0.0001411 0.0002381 0.0002671 0.0003623 0.0012910
>>>> summary(c)
>>>     Min.   1st Qu.    Median      Mean   3rd Qu.      Max.
>>> 0.0000000 0.0000000 0.0000000 0.0004947 0.0002972 1.0000000
>>>
>>> The mean ratio is then around 0.5399031 which naively goes in opposite
>>> direction of the wilcoxon test ( I was expecting to find a ratio >> 1)
>>>
>>
>> There's nothing conceptually strange about the Wilcoxon test showing a
>> difference in the opposite direction to the difference in means.  It's
>> probably easiest to think about this in terms of the Mann-Whitney
>> version of the same test, which is based on the proportion of pairs of
>> one observation from each group where the `a' observation is higher.
>> Your 'c' vector has a lot more zeros, so a randomly chosen observation
>> from 'c' is likely to be smaller than one from 'a', but the non-zero
>> observations seem to be larger, so the mean of 'c' is higher.
>>
>> The Wilcoxon test probably isn't very useful in a setting like this,
>> since its results really make sense only under 'stochastic ordering',
>> where the shift is in the same direction across the whole
>> distribution.
>>
>>  -thomas
>
> I was sure I had seen a definition where X was "larger than" Y if P(X>Y) >
> P(Y<X), but that's obviously not the normal definition. Anyways, it is worth
> emphasizing that that is what the Wilcoxon test tests for, not whether the
> means differ, nor whether the medians do. As a counterexample of the latter,
> try
>
> x <- rep(0:1, c(60,40))
> y <- rep(0:1, c(80,20))
> wilcox.test(x,y)
> median(x)
> median(y)
>
> (and the "location shift" reference in wilcox.test output is a bit of a red
> herring.)
>
> --
> Peter Dalgaard, Professor,
> Center for Statistics, Copenhagen Business School
> Solbjerg Plads 3, 2000 Frederiksberg, Denmark
> Phone: (+45)38153501
> Email: pd.mes at cbs.dk  Priv: PDalgd at gmail.com
>
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>


-- 
Avinash Barnwal, M.Sc.
Statistics and Informatics
Department of Mathematics
IIT Kharagpur