On Jun 24, 2010, at 6:42 PM, Joris Meys wrote:
On Fri, Jun 25, 2010 at 12:17 AM, David Winsemius
<dwinsem...@comcast.net> wrote:
On Jun 24, 2010, at 6:09 PM, Joris Meys wrote:
I do agree that one should not trust solely on sources like
wikipedia
and graphpad, although they contain a lot of valuable information.
This said, it is not too difficult to illustrate why, in the case of
the one-sample signed rank test,
That is a key point. I was assuming that you were using the paired
sample
version of the WSRT and I may have been misleading the OP. For the
one-sample situation, the assumption of symmetry is needed but for
the
paired sampling version of the test, the location shift becomes the
tested
hypothesis, and no assumptions about the form of the hypothesis are
made
except that they be the same.
I believe you mean the form of the distributions. The assumption that
the distributions of both samples are the same (or similar, it is a
robust test) implies that the differences x_i - y_i are more or less
symmetrically distributed. Key point here that we're not talking about
the distribution of the populations/samples (as done in the OP) but
about the distribution of the difference. I may not have been clear
enough on that one.
What I meant about different hypotheses was that in the single sample
case the H0 was mean (or median) = mu_pop and in the paired two sample
the H0 was mean(distr_A_i - distr_B_1) =0. And yes, I did miss the
OP's point. My apologies.
--
David.
Cheers
Joris
Any consideration of median or mean (which
will be the same in the case of symmetric distributions) gets lost
in the
paired test case.
--
David.
the differences should be not to far
away from symmetrical. It just needs some reflection on how the
statistic is calculated. If you have an asymmetrical distribution,
you
have a lot of small differences with a negative sign and a lot of
large differences with a positive sign if you test against the
median
or mean. Hence the sum of ranks for one side will be higher than for
the other, leading eventually to a significant result.
An extreme example :
set.seed(100)
y <- rnorm(100,1,2)^2
wilcox.test(y,mu=median(y))
Wilcoxon signed rank test with continuity correction
data: y
V = 3240.5, p-value = 0.01396
alternative hypothesis: true location is not equal to 1.829867
wilcox.test(y,mu=mean(y))
Wilcoxon signed rank test with continuity correction
data: y
V = 1763, p-value = 0.008837
alternative hypothesis: true location is not equal to 5.137409
Which brings us to the question what location is actually tested in
the wilcoxon test. For the measure of location to be the mean (or
median), one has to assume that the distribution of the
differences is
rather symmetrical, which implies your data has to be distributed
somewhat symmetrical. The test is robust against violations of this
-implicit- assumption, but in more extreme cases skewness does
matter.
Cheers
Joris
On Thu, Jun 24, 2010 at 7:40 PM, David Winsemius <dwinsem...@comcast.net
>
wrote:
You are being misled. Simply finding a statement on a statistics
software
website, even one as reputable as Graphpad (???), does not mean
that it
is
necessarily true. My understanding (confirmed reviewing
"Nonparametric
statistical methods for complete and censored data" by M. M. Desu,
Damaraju
Raghavarao, is that the Wilcoxon signed-rank test does not
require that
the
underlying distributions be symmetric. The above quotation is
highly
inaccurate.
--
David.
--
Joris Meys
Statistical consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
tel : +32 9 264 59 87
joris.m...@ugent.be
-------------------------------
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php
--
Joris Meys
Statistical consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
tel : +32 9 264 59 87
joris.m...@ugent.be
-------------------------------
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php
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