As their name says, Kendall's tau for correlation try to quantify if the two sets of data tend to vary in the same direction, whereas Kendall's W for concordance try to quantify if the two sets of data are in fact the same.

Obviously, concordance is a special case of correlation, but the contrary is wrong, you can have correlation without concordance.

As their name says, Kendall's tau for correlation try to quantify if the two sets of data tend to vary in the same direction, whereas Kendall's W for concordance try to quantify if the two sets of data are in fact the same.

Obviously, concordance is a special case of correlation, but the contrary is wrong, you can have correlation without concordance.

Thanks Emmanuel Curis that was a very crisp explanation.

Kendall’s coefficient of concordance W summarises the concordance between several ranked lists. If there are only two lists, then W = (1 + RS)/2 (RS = Spearman’s rank correlation). Its range is [0, 1].

Kendall’s rank correlation TAU applies only to pairs of ranked lists and measures the co-relation; its values are in the range [-1, 1]. For this pair of data, one may also apply Spearman’s rank correlation, which differs somehow.

Thus, for pairs of ranked data, one has (amongst others) the alternative between

*Kendall’s rank correlation and

*Spearman’s rank correlation (which is equal to 2W-1)

 If the variables are normally distributed, then there are reasons to prefer Kendall’s W over Spearman’s RS (efficiency and robustness).

If you have more than two ranked data lists to compare, there is no alternative as TAU can be calculated only for all pairs separately while W summarises the concordance in all lists.

Thank you very much Manfred Borovcnik. Your explanation completely clears my doubts. It's all very clear now, I had this question in mind what if the data set is more than a pair: and now I know W holds the key. Thanks again.

Thank you very much for this answers, i've had the same dilema all week long while finishing my dissertation! keep up the good work

The distinction made by Emmanuel Curis is misleading. Concordance does not mean "the same" in value. . For example, if Group A has values of (1, 2, 3, 4, 5) and Group B has values of (6, 7, 8, 9, 10) ---- across raters "a" to "e", say, and in that order ---- then they have perfect concordance (W = 1.0).

I think the best way to think about Kendall's W is that it is a standardized version --- or effect size for --- Friedman's test. The raters act as blocks, so that the values are treated as relative to the other values for that rater. Maybe it is helpful to look at a worked through example ( http://rcompanion.org/handbook/F_10.html )

Also, I think the Wikipedia article isn't too bad. ( https://en.wikipedia.org/wiki/Kendall%27s_W ).

______________________

* I deleted: "As Manfred Borovcnik points out, Kendall's W is related to correlation."

@ Salvatore: could you develop in what way « concordance » of two observations is not the idea that they are basically the same (I mean, not considering cosmetic changes in values like concordance between « red » [English] and « rouge » [French] or similar situations, like probably the example you give — without context defining what your values are, any interpretation is possible)?

Note that Manfred also says that Kendall's W is a concordance coefficient (even if related to Spearman's rank correlation coefficient), and that Kendall's tau is a correlation coefficient. And the wikipedia link you give also says Kendall's W is a concordance coefficient.

Emmanuel ,

The question is about Kendall's W. It's not about the English word "concordance" or about the concept of concordance in English. No one is disputing that Kendall's W is called "coefficient of concordance". It just matters how the statistic works out mathematically and how one would interpret it.

Consider the situation where you have two instructors and each is rated on an ordinal scale 1 to 9 by 6 raters.

If the data are, across raters (Example 1):

Instructor A = (1, 2, 2, 6, 6, 7)

Instructor B = (8, 8, 8, 9, 9, 9)

(I'll also attach some data which may be easier to see.)

In this case, Kendall's W is 1. That is, there is prefect concordance. This is because every rater rates Instructor B as higher than Instructor A.

When you said, "Kendall's tau for correlation try to quantify if the two sets of data tend to vary in the same direction, whereas Kendall's W for concordance try to quantify if the two sets of data are in fact the same", I took this to mean that in this case you would think that A and B are correlated, but that they would have weak concordance, because the values are not the same. I may have understood you incorrectly. In any case, I think what you wrote is likely to mislead the reader, and it is important to point this out since you have the highest-rated answer on this thread.

To think about the statistic more, Kendall's W treats the values as relative to the rater's other values. If the data are, across raters (Example 2):

Instructor A = (1, 2, 2, 8, 8, 8)

Instructor B = (2, 3, 4, 9, 9, 9)

Kendall's W is 1. Again, each rater rates B as higher than A. It is clear that raters "a", "b", and "c" have rated both Instructors as low. And raters "d", "e", "f" are the opposite. They have rated both Instructors as high. But the important fact for Kendall's W is that all raters rated B as higher than A.

Finally, Kendall's W is quite different than correlation . If the data are, across raters (Example 3):

Instructor A = c(1, 1, 1, 5, 5, 5)

Instructor B = c(6, 6, 9, 6, 6, 9))

Kendall's W is 1. Again, each rater rates B as higher than A. But rho and tau for correlation are 0.

My post was indeed unclear, sorry, because I had in mind quantitative data for which there is no ambiguity on what means « concordance » in terms of « values ».

For (ordinal or not) qualitative data, observed values are just labels of abstract classes, and concordance is (in my mind) on these classes, not on the observed values — in your two first examples, on « low » vs « high » for instance. But the definition is complex when the number of classes is different. So thanks for clarifying this point.

I do not agree with your last example, for which there anyway should not be any perfect concordance. According to the W = (1 + RS)/2 relationship given in previous post, if RS = 0 (no correlation), then W = 1/2. And that is indeed the value given by the W online calculator:

https://www.statstodo.com/KendallW_Pgm.php

May be some problems with handling ties?

That's also make sense basically: you cannot imagine concordance without correlation, as I said.

Hi, Emmanuel.

* * *

For the calculator you posted, it won't accept data for which there are only two subjects, so I don't know how you got the 1/2 value from it.

* * *

I don't think the ties are creating any problems here. You can change the data to

A = (1.1, 1.2, 1.3, 5.1, 5.2, 5.3)

B = (6.1, 6.2, 9.1, 6.3, 6.4, 9.2)

And W still equals 1.

* * *

In general, it shouldn't matter for these statistics if the data being used is represented as words or as numbers, as long as they can be ordered. That is, you can calculate Kendall's W with data represented by "Strongly disagree", "Disagree", "Neutral", "Agree", or "Strongly agree". If you represent these as 1, 2, 3, 4, 5, you will get the same answer. If you represent them by 1, 10, 20, 90, 106, you will get the same answer.

The exact relationship depends on the number of judges, and is according to the attached PDF, for k judges

W = ((k - 1) rS + 1) / k

where rS is the average of all rS between pairs of judges, which reduces for 2 judges to the formula given by Manfred Borovcnik .

I just realized that I interpreted your data as 2 raters on 6 objects, and note as 6 raters on 2 objets, which explains how I entered the data on the calculator.

Whatever k is, if rS = 0, W cannot be null, since it is equal to 1/k. But when you speak of correlation (tau or R) equal to 0, to which correlation do you refer? between raters (6*5/2 possibilities, which one?)? average of these possibilités (rS in the formula above)? or between A and B, in which case the comparison is meaningless, because you are changing the problem...

My point on correlation and concordance as a special case of concordance was for k = 2 raters, and correlation/concordance between raters

Of course, when k > 2, there is no clear equivalent of correlation between raters, since correlations are for pairs...

Great. That should clear up the relationship between rho and W.

Yes, the correlations in my examples were all between A and B (subjects, not raters).

Can anyone respond to my following question?

"Is it OK to use Spearman's rho for correlation (i.e. relationship and the strength of relationship between variables) and Kendall's W for hypothesis testing as well as to know the effect size?" (Kendall's W also provides Chi-square and degree of freedom in addition to pair comparison and statistical significance i.e. for Bonferroni Correction. Moreover, analysis of my data using Kendall's W provides more logical and better results that Jonckheere-Terpstra or Kruskal Wallis test. Please suggest? By the way, I am not a statistician but a social science student.

First, select the test the most suited to the question you're asking, then do the analysis. Not the other way round! So selecting W just because the *results* seems better is not good practice. You should instead think about the exact questions the three different tests are answering to, then select the closest to your own question.

If questions are close, the three methods should give similar results. If not, there's something special in your data that must be investigated.

You'd better stay consistent in your choices: if you think Spearman's rho is better (since you want to use if for correlation), then do the test and effect size based on Spearman's rho. I don't understand why the fact that W provides a chi-square with is degress of freedom would be of any additional value in this respect: Spearman's rho is already a kind of effect size by itself, and one can obtain p-values for testing it against 0.

Impossible to say more without the data and details. But you may have much better answers by consulting a local statistician and discuss with him your own problem!

@ArshadKhan ,

Please see:

https://www.researchgate.net/post/Is_it_OK_to_use_Spearmans_rho_for_correlation_ie_relationship_and_Kendalls_W_for_hypothesis