# Why Kendall's Tau-c?

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?

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.

• 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 $\pm1$). – Trisoloriansunscreen Jun 9 at 17:45
• 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) – Glen_b Jun 10 at 0:39