[R] Simple General Statistics and R question (with 3 line example) - get z value from pairwise.wilcox.test

[JP]

Thanks once again Peter, I understand your points -- I fiddled and
googled and read some more and found an eas(ier) route:

library(coin)
t <- wilcoxsign_test(a ~ b, alternative = "two.sided", distribution =
exact())  # This is equivalent to paired wilcox.test
pval <- pvalue(t)
sweet.zscore <- statistic(t) # signed and everything

# correct for multiple testing if you are doing a number of the above...

I would never have got here without your guidance - so kudos to you.
JP


On 5 May 2011 14:52, peter dalgaard <pdalgd at gmail.com> wrote:
>
> On May 5, 2011, at 10:58 , JP wrote:
>
>> On 4 May 2011 15:32, peter dalgaard <pdalgd at gmail.com> wrote:
>>
>>>
>>> On May 4, 2011, at 15:11 , JP wrote:
>>>
>>>> Peter thanks for the fantastically simple and understandable explanation...
>>>>
>>>> To sum it up... to find the z values of a number of pairwise wilcox
>>>> tests do the following:
>>>>
>>>> # pairwise tests with bonferroni correction
>>>> x <- pairwise.wilcox.test(a, b, alternative="two.sided",
>>>> p.adj="bonferroni", exact=F, paired=T)
>>>
>>>
>>> You probably don't want the bonferroni correction there. Rather p.adj="none". You generally correct the p values for multiple testing, not the test statistics.
>>>
>>
>> Oh, I see thanks... of course since I have 5 groups (samples) and 10
>> comparisons I still have to correct when quoting p values...
>>
>>
>>> (My sentiment would be to pick apart the stats:::wilcox.test.default function and clone the computation of Z from it, but presumably backtracking from the p value is a useful expedient.)
>>>
>>
>> Should this be so onerous for the user [read non-statistician] ?
>
>
> My main reservation is that if the p value gets very small or close to 1 (for 1-sided tests), then it might not be possible to reconstruct the underlying Z (qnorm(pnorm(Z)) breaks down around Z = -37 and Z=+8). If you get sensible values, there's probably not a problem.
>
>
>>
>>>> # what is the data structure we got back
>>>> is.matrix(x$p.value)
>>>> # p vals
>>>> x$p.value
>>>> # z.scores for each
>>>> z.score <- qnorm(x$p.value / 2)
>>>>
>>>
>>> Hmm, you're not actually getting a signed z out of this, you might want to try alternative="greater" and drop the division by 2 inside qnorm(). (If the signs come out inverted, I meant "less" not "greater"...)
>>>
>>
>> But I need a two sided test (changing the alternative would change the
>> hypothesis!)...  do I still do this?
>
> For the z's, yes.
>
> Hmm, the asymmetry of qnorm(pnorm(...)) indicates that you might want to be a bit more careful.
>
>> All my z values are negative....
>>
>> Is this correct?
>
> No, that's the problem. The sign of Z should depend on whether the signed ranks are predominantly positive or negative. If you do a two-sided p-value and divide by two, then by definition you get something less than .5 and qnorm of that will be negative. So if the 2-sided p is 0.04, the one-sided p will be either 0.02 or 0.98, depending on the alternative and on whether the V statistic is above or below its expectation. Notice that this corresponds to Z statistics of opposite sign:
>
>> qnorm(.02)
> [1] -2.053749
>> qnorm(.98)
> [1] 2.053749
>
> --
> Peter Dalgaard
> 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
>
>
>