[R] Weird differing results when using the Wilcoxon-test
Cedric Laczny
cedric.laczny at gmx.de
Wed Aug 18 11:55:52 CEST 2010
I was able to trace down the unexpected behavior to the following line
SIGMA <- sqrt((n.x * n.y/12) * ((n.x + n.y + 1) -
sum(NTIES^3 - NTIES)/((n.x + n.y) * (n.x + n.y -
1))))
My calculations of the Z-score for the normal approximation where based on
using the standard deviation for ranks _without_ ties. The above formula seems
to account for ties and thus, yields a slightly different z-score. However, the
data seems to include at most 1 tie (based on rnorm), so it would be the same
result as if it contained no tie (1^3 - 1 has the same result as 0^3 - 0,
obviously ;) ) and thus I would expect the result to be the same as when using
the formula for the standard deviation without ties.
Interestigly, this z-score is also different from that reported by wilcox_test
(exact calculation of p-value).
I have not been able to find this formula in any textbook nearby or at any
website. Therefore I am wondering where it does actually com from?
Best,
Cedric
On Tuesday, 17. August 2010 13:48:10 Cedric Laczny wrote:
> Hi,
>
> I became a little bit confused when working with the Wilcoxon test in R.
> As far as I understood, there are mainly two versions:
> 1) wilcox.test{stats}, which is the default and an approximation,
> especially, when ties are involved
> 2) wilcox_test{coin}, which does calculate the distribution _exactly_ even,
> with ties.
>
> I have the following scenario:
>
> #---BeginCode---
> # big example
> size = 60
> big1 = rnorm(size, 0, 1)
> big2 = rnorm(size, 0.5, 1
>
> g1f = rep(1, size)
> g2f = rep(2, size)
> big = c(big1, big2)
> data_frame = data.frame(big, gr=as.factor(c(g1f, g2f)))
>
> wilcox_approx = wilcox.test(big1, big2)
> wilcox_exact = wilcox_test(big ~ gr, data=data_frame, distribution="exact")
> #---EndCode---
>
> I found here http://www-stat.stanford.edu/~susan/courses/s141/hononpara.pdf
> that wilcox.test (at least for the signed rank test) relies on exact
> (p-)values until N = n1 + n2 = 50.
> I can reproduce this, when using e.g. size = 15. The p-values then are the
> same, as I would expect it, having read the info from the link.
>
> #---BeginCode---
> print(paste("Wilcox approx p-value:", wilcox_approx$p.value), quote=F)
> print(paste("Wilcox exact p-value:", pvalue(wilcox_exact)), quote=F)
> #---EndCode---
>
> That said, if I set e.g. size = 60, then the p-values of wilcox.test and
> wilcox_test differ, as expected.
>
> What's puzzling me particularly is the differing results when wanting to
> calculate the p-value manually, for bigger sample sizes.
>
> So, if we get the W-score from wilcox.test:
>
> #---BeginCode---
> W_big = wilcox.test(big1, big2))$statistic
> #---EndCode---
>
> and "convert" it to a Z-score, like this:
>
> #---BeginCode---
> mu_big = (size^2)/2
> sd_big = sqrt(size*size*(size + size + 1)/12)
> N = size + size
> sd_big_corr = sqrt( (size * size) / (N * (N - 1)) * (N^3 - N) / 12 )
>
> Z_big = (((W_big - mu_big)/sd_big)
> #---EndCode---
>
> The Z-Score (Z_big) is equal to the statistic of wilcox_test.
> So far so good. And now comes the main problem.
> When I follow the documentation correctly, the p-value for a given
> W-score/- statistic ist calculated using the normal-approximation with the
> Z-score. However, when I do that, I get a different result than what I
> would expect. Because I would expect the p-value of wilcox.test to be
> equal to
> 2*pnorm(Z_big), which is in fact _not_ equal. Please see:
>
> #---BeginCode---
> p_value_manual = 2 * pnorm(Z_big)
>
> print("--- Resulting pvalues --- ", quote=F)
> print(paste("Wilcox approx p-value:", wilcox_approx$p.value), quote=F)
> print(paste("Wilcox exact p-value:", pvalue(wilcox_exact)), quote=F)
> print(paste("P-value manual:", p_value_manual), quote=F)
> #---EndCode---
>
> So how is the calculation of the p-value performed in wilcox.test, when the
> sample sizes are big? Because this might explain why the value differs from
> that being calculated manually.
>
> Best regards,
>
> Cedric
>
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