# Why do we assign fractional ranks in rank tests?

Both Mann-Whitney U test and Wilcoxon signed rank test follow this procedure:

1. sort the data
2. assign ranks, Ties receive a rank equal to the average of the ranks they span
3. compute the statistic using a sum the ranks
4. compare statistic against critical values and make verdict

For me the 3rd step seems like an unnecessary complication. If we intend to sum the ranks anyway, there is no difference if we sum 2 and 3 or 2.5 and 2.5.

Am I missing something?

• What happens when the $2$ comes from one group and the $3$ comes from the other group you are comparing it to? – whuber Mar 27 '14 at 18:50

When ties are internal to a group, of course the result of assigning the average rank makes no difference compared to say breaking ties at random just as you point out, but it does matter when there are ties across groups.

When there are ties across groups, how you deal with it will matter and there are several choices.

The one you mention - giving the average of the ranks to all tied values - is common, but not the only way it is done.

That approach has the disadvantage that you no longer have a set of integers from 1 to n, so it mucks up the distribution of the ranks under the null. Even under a normal approximation, it affects the variance of the distribution, though the calculation of the adjusted variance is for many of the common procedures not so onerous; if ties are not heavy it will often still be a very good approximation.

Another approach is simply to break ties at random ... which has the virtue of simplicity, but means two people may come to different conclusions on the same data

A third approach is to break ties in all possible ways (or sample from the set of all possible ways if there's too many to do otherwise), and then combine the p-values in some way (if all p-values are on the same side of the significance level, there's no difficulty); taking the average is usually what is done.

I think the best approach is to go back to the permutation distribution. You lose teh convenience of tables, but it's relatively easy to either enumerate (in small samples) or sample from (in larger samples) the permutation distribution (of the ranks); this is the "right" answer, really, and not hard to do with suitable software (it's rather easy in R in many of the common cases)

• +1 Nice summary. I believe that for many conventional rank-based tests (a) the expectation of the permutation distribution of the sampling statistic can be computed by averaging ties and (b) variances, confidence intervals, and p-values are computed according to the permutation distribution anyway. That's what's going on with formulas that tell you how to adjust the test statistics in terms of the numbers of ties and the sizes of the tied batches within each group. – whuber Mar 28 '14 at 15:37
• From what I've seen, the variance formulas apply to the permutation distribution, but it happens when the p-value is being obtained from a normal approximation (which is not that good under very heavy ties). See Conover's Practical Nonparametric Statistics, for example, where if I recall such variance formulas only appear in relation to such distributional approximations. If you're getting the whole permutation distribution, why calculate the variance at all? – Glen_b -Reinstate Monica Mar 28 '14 at 23:25
• Often exact critical values for the permutation distribution are offered in tables indexed by the two group sizes, but (obviously) only for small group sizes. The variance (which assumes only the permutation distribution for its calculation) is computed in order to find critical values for a Normal approximation to the sampling distribution of the test statistic, as you write, and it is true that this approximation can be poor with many ties. – whuber Mar 29 '14 at 20:27