Solved – How to obtain p-values for a given Kendall’s Tau value (not calculated from data)

kendall-taup-valuer

I want to obtain p-values for Parker's Tau-U (Parker et al., 2011).
which is a measure based on Kendall's Tau.

Tau-U is related to Kendall's Tau in such a way that it's possible to translate
a Tau-U value to a Kendall's Tau value.

This means that I can use the Kendall() function in the Kendall package in R
to compute p-values for Tau-U. The only problem is that the Kendall() function only accepts two equal length data vectors for which it computes Kendall's Tau. However, I want to input a Kendall's Tau value directly (the Tau-U value expressed as a Kendall's Tau value) and then obtain a p-value for it.

Does any know if there are any R-functions that will allow me to input a Kendall's Tau value directly? There is other (non-free) software available that will do this for me (e.g. StatsDirect) but I need to find a way in R because I need the p-values for implementation in a permutation test (based on my own code).

Thanks in advance!

Regards.
Bart

Best Answer

My initial answer was incorrect as pointed out by Silverfish. After some review of the tau variations and looking at the paper the test statistic is calculated independently from the tau value. As such, it doesn't appear to be possible to directly use the tau value to calculate the p-value. You will need to calculate the statistic from the data. You will therefore need to review the paper to see how the calculation of the statistic differs from the other tau variations.

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Solved – Which correlation should be used for non-normal data: Spearman’s rho versus Kendall’s tau versus Kendal’s tau-b

Pearson's correlation doesn't assume normality, so you should use it. You really don't need Kendall's tau in your example.

In your analysis, you should start off with a simple plot. Like this:

grass  <- c(0,0,0,8,0,0,0,2,6,0,0,1,0,0,0,8,0,7,15)
height <- c(0,0,0,16,0,0,0,2,6,0,0,1,0,0,0,15,0,7,15)
plot(grass, height)

This is clearly linear and monotonic positive, and so it was pointless for you to do a normality test.

Both Pearson and Spearman give similar results:

cor(grass, height) # 0.9300721
cor(grass, height, method='spearman') # 0.9947245

In this example, it's not really not that important to do a correlation test because you know the p-value will be very small and your result will be significant. But let's do it anyway:

cor(grass, height) # 8.227e-09
cor.test(grass, height, method='spearman', exact=FALSE) # 2.2e-16

In the first line, we did a test for Pearson's correlation. Your very small p-value give you confidence that your correlation is not zero (not very surprising).

In the second line, we have a test for Spearman's correlation. Note that your zeros made the default exact test impossible, so you'll need to set exact to FALSE to approximate your test statistic with t approximation. Again, this is very significant.

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Solved – Which correlation should be used for non-normal data: Spearman’s rho versus Kendall’s tau versus Kendal’s tau-b

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