# Cramér's $V$ for one variable

In Wikipedia's summary on Cramér's $V$, it is mentioned that it can (also) be used in the one-dimensional case as measure of concentration.

1. What is the formula in this special case?

No reference, link or explanation is offered at the article*, so this is guesswork, but I believe this is the intent/reasoning:

*(and as such it should probably be removed from Wikipedia)

1) Cramér's $V$ is a measure of association; it's a function of $\phi$ which is also a measure of association; they cannot really mean that there's a measure of association here.

2) However, Cramér's $V$ can be defined in terms of the statistic for the chi-square test of independence:

$V = \sqrt{ \frac{\chi^2/n}{\min(k - 1,r-1)}}=\sqrt{ \frac{\chi^2}{n(k - 1)}}$ (when $r=1$)

If you place the goodness-of-fit chi-square value in that formula in place of the usual independence chi-square, while you could not (to my mind) reasonably call that Cramer's $V$ (since it doesn't measure association), it would be a perfectly good measure of concentration. It's 0 when the categories are perfectly evenly represented and it's 1 when all the values are in a single category.

• Thanks, Glen. Would you think the probabilities behind the goodness-of-fit chisquared are always to be chosen uniform in such application? – Michael M Aug 5 '14 at 6:43
• If they weren't, it wouldn't be a measure of concentration in the same sense. – Glen_b Aug 5 '14 at 7:36
• +1, it couldn't reasonably be called a measure of association, but I do think it can still be thought of as a measure of the relevant type of effect size for such analyses. Test statistics & p-values conflate magnitude & N to give some sense of the clarity of the result; this formula seems to effectively disentangle the magnitude from N as other measures of effect size do (eg, $d$ & $r$). I think you can also use it to measure the deviation from non-uniform null distributions & it would be doing the same thing, it's just the word "concentration" would no longer characterize it well. – gung Aug 9 '14 at 18:25
• @gung +1 I'd agree with everything you have there. – Glen_b Aug 9 '14 at 20:56