On Aug 8, 2012, at 18:16 , BELLAY Juliette wrote:
> Dear list,
>
> I am facing a problem in my statistical analyses on R.
> My experiments are about plants, I record there growth after each cutting
> (every 3 weeks).
> 'BC' is for the plant, and '1' to '5' is the time of cutting and recording.
>
> The data and R script are :
>
> ""
> BourdCoup <- c(21, 7.2, 9.2, 0, 8.52, 14.7, 8.31, 6.2, 127.05, 115.2, 100.7,
> 24, 162.64, 136.8, 95.1, 78.93, 104.7, 21.2, 35.8, 21, 66, 103.4, 74.9, 43.8,
> 32.1, 14.9, 2.1, 14.4, 36.1, 35, 53.2, 28.9, 0,0,0,8.7,2.7,4.2,0,21.2)
>
> groupBC <- as.factor(rep(c("BC1","BC2","BC3","BC4","BC5"), times=1, each=8))
>
> BC1 <- BourdCoup[groupBC=="BC1"]
> BC2 <- BourdCoup[groupBC=="BC2"]
> BC3 <- BourdCoup[groupBC=="BC3"]
> BC4 <- BourdCoup[groupBC=="BC4"]
> BC5 <- BourdCoup[groupBC=="BC5"]
> ""
>
> I therefore realise a wilcoxon test for paired series, and compare them one
> by one.
>
> ""
>> wilcox.test(BC1,BC2, paired=T)
>
> Wilcoxon signed rank test
>
> data: BC1 and BC2
> V = 0, p-value = 0.007813
> alternative hypothesis: true location shift is not equal to 0
> ""
>
> The problem is that it always gives me the same p-value (0.007813), whatever
> vectors I try. And when I remove "paired=T" from the script, all p-values
> differ.
>
> Could you help me please ?
First of all, that's not actually true:
> wilcox.test(BC1, BC5, paired=TRUE)$p.value
[1] 0.3828125
> wilcox.test(BC1, BC4, paired=TRUE)$p.value
[1] 0.015625
If V=0, then the p-value is 2 * 2^(-8) = 1/128 = 0.0078125. This happens if all
differences have the same sign. And....
> outer(1:5,1:5, Vectorize(function(i,j) sum(sign(M[,i]-M[,j]))))
[,1] [,2] [,3] [,4] [,5]
[1,] 0 -8 -8 -6 4
[2,] 8 0 8 8 8
[3,] 8 -8 0 8 8
[4,] 6 -8 -8 0 8
[5,] -4 -8 -8 -8 0
... that is indeed what happens in 8 out of 10 paired comparisons.
--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd....@cbs.dk Priv: pda...@gmail.com
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