1
$\begingroup$

Wikipedia tells us: (link)

Tau-c (also called Stuart-Kendall Tau-c) is more suitable than Tau-b for the analysis of data based on non-square (i.e. rectangular) contingency tables.

What does $\tau_c$ try to solve? More specifically, is there a problem with applying Kendall's Tau-b on non-square contingency tables? When we compute $\tau_b(x,y)$, do we implictly assume that the number of levels of $x$ is the same as the number of levels of $y$? Or is it an issue with how Tau-b omits ties?

$\endgroup$
2
$\begingroup$

Largely it boils down to this: $\tau_b$ can't attain the values $\pm 1$ in a non-square table but $\tau_c$ can.

You can verify this with examples.

Consider the following tables

    I    II   III
A   30    0    0
B    0   30    0
C    0    0   30

    Ia   Ib   IIa  IIb   IIIa IIIb
A   15   15     0    0     0    0
B    0    0    15   15     0    0
C    0    0     0    0    15   15

Try computing the two measures on each. They're both 1 on the first table, and $\tau_c$ is 1 on both tables, but $\tau_b$ is only around 0.9 on the second table.

$\endgroup$
  • $\begingroup$ So is $\tau_c$ a more correct/general version of $\tau_b$? I wonder why $\frac{n_c-n_d}{n_c+n_d}$ isn't used instead of these formulas when we want to consider only non-tied pairs. For this case, it always scales correctly (i.e., attains $\pm1$). $\endgroup$ – Trisoloriansunscreen Jun 9 at 17:45
  • 1
    $\begingroup$ Indeed, I often do just the calculation you suggest, since it usually achieves my intent in each situation I want a Kendall-like measure of association. However, if I remember right, $\tau_c$ is not always identical to $\tau_b$ on square tables so it's not simply "a more correct version" of the same thing (however I think their p-values will be the same on square tables) $\endgroup$ – Glen_b Jun 10 at 0:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.