On May 4, 2011, at 11:03 , JP wrote:
> On 3 May 2011 20:50, peter dalgaard <[email protected]> wrote:
>>
>> On Apr 28, 2011, at 15:18 , JP wrote:
>>
>>>
>>>
>>> I have found that when doing a wilcoxon signed ranked test you should
>>> report:
>>>
>>> - The median value (and not the mean or sd, presumably because of the
>>> underlying potential non normal distribution)
>>> - The Z score (or value)
>>> - r
>>> - p value
>>>
>>
>> ...printed on 40g/m^2 acid free paper with a pencil of 3B softness?
>>
>> Seriously, with nonparametrics, the p value is the only thing of real
>> interest, the other stuff is just attempting to check on authors doing their
>> calculations properly. The median difference is of some interest, but it is
>> not actually what is being tested, and in heavily tied data, it could even
>> be zero with a highly significant p-value. The Z score can in principle be
>> extracted from the p value (qnorm(p/2), basically) but it's obviously
>> unstable in the extreme cases. What is r? The correlation? Pearson, not
>> Spearman?
>>
>
> Thanks for this Peter - a couple of more questions:
>
> a <- rnorm(500)
> b <- runif(500, min=0, max=1)
> x <- wilcox.test(a, b, alternative="two.sided", exact=T, paired=T)
> x$statistic
>
> V
> 31835
>
> What is V? (is that the value Z of the test statistic)?
No. It's the sum of the positive ranks:
r <- rank(abs(x))
STATISTIC <- sum(r[x > 0])
names(STATISTIC) <- "V"
(where x is actually x-y in the paired case)
Subtract the expected value of V (sum(1:500)/2 == 62625) in your case, and
divide by the standard deviation (sqrt(500*501*1001/24)=3232.327) and you get
Z=-9.54. The slight discrepancy is likely due to your use of exact=T (so your p
value is not actually computed from Z).
>
> z.score <- qnorm(x$p.value/2)
> [1] -9.805352
>
> But what does this zscore show in practice?
That your test statistic is approx. 10 standard deviations away from its mean,
if the null hypothesis were to be true.
>
> The d.f. are suggested to be reported here:
> http://staff.bath.ac.uk/pssiw/stats2/page2/page3/page3.html
>
Some software replaces the asymptotic normal distribution of the rank sums with
the t-distribution with the same df as would be used in an ordinary t test.
However, since there is no such thing as an independent variance estimate in
the Wilcoxon test, it is hard to see how that should be an improvement. I have
it down to "coding by non-statistician".
> And r is mentioned here
> http://huberb.people.cofc.edu/Guide/Reporting_Statistics%20in%20Psychology.pdfs
>
>
Aha, so it's supposed to be the effect size. On the referenced site they
suggest to use r=Z/sqrt(N). (They even do so for the independent samples
version, which looks wrong to me).
>
>>> My questions are:
>>>
>>> - Are the above enough/correct values to report (some places even
>>> quote W and df) ?
>>
>> df is silly, and/or blatantly wrong...
>>
>>> What else would you suggest?
>>> - How do I calculate the Z score and r for the above example?
>>> - How do I get each statistic from the pairwise.wilcox.test call?
>>>
>>> Many Thanks
>>> JP
>>>
>>> ______________________________________________
>>> [email protected] 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.
>>
>> --
>> Peter Dalgaard
>> Center for Statistics, Copenhagen Business School
>> Solbjerg Plads 3, 2000 Frederiksberg, Denmark
>> Phone: (+45)38153501
>> Email: [email protected] Priv: [email protected]
>>
>>
--
Peter Dalgaard
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: [email protected] Priv: [email protected]
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