Sample size for the Mann-Whitney U test

2000/09/28

[StATS]: Sample size for the Mann-Whitney U test (September 28, 2000)

Dear Professor Mean, I need to calculate the sample size for the Mann-Whitney U test. How do I do this? – Bewildered Bob

Dear Bewildered,

Just to clarify things, Professor Mean should note that the Mann-Whitney U test is sometimes called the Wilcoxon rank sum test. There are independent publications in the mid 1940’s from Mann and Whitney and from Wilcoxon that outline a nonparametric approach to comparing two independent groups. The two tests have different forms but you can show that the two forms are equivalent. Some statisticians in the art of compromise have named the test after all three people, but there is still a dispute about whether it should be called the Wilcoxon-Mann-Whitney test or the Mann-Whitney-Wilcoxon test. After that explanation, you are probably more bewildered than ever.

There are several approaches you can take for computing sample size. A simple approximate approach uses information about the Pitman Asymptotic Relative Efficiency (ARE) to estimate sample size. The ARE compares the efficiency of two competing test statistics. It represents the an asymptotic limit of the ratio of sample sizes needed to achieve equal power. The limit is evaluated as the sample size increases without bound and the alternative being considered shrinks towards the null value. To calculate an ARE requires some heavy duty mathematics, but someone else has already done that work for you.

So to estimate sample size, you compute the sample size needed for a two-sample t-test, and then adjust the sample size based on the ARE of the Mann-Whitney U relative to the t-test. The trick, though, is that the ARE depends on the underlying distribution. That is a bit surprising, but the Mann-Whitney U test statistic is distribution free only under the null hypothesis. When the alternative hypothesis is true, the distribution of the Mann-Whitney U depends quite a bit on the underlying distribution.

This leads you to a dilemma. You probably want to use the Mann-Whitney U test because you can’t assume with any confidence what the underlying distribution is. There are several approaches you can take.

First, you can select a distibution that you believe is reasonable, and estimate sample size using that distribution.

For example, you might assume a normal distribution. Under the normal distribution the ARE of the Mann-Whitney U relative to the t-test is 0.955. Estimate the sample size for a t-test and then divide that sample size by 0.955. You will get a slightly larger sample size, but consider the extra data points as insurance against a possible violation of the underlying assumptions.

If you assume a logistic distribution, the ARE is 1.097. The logistic distribution has a symmetric bell shaped curve like the normal distribution but it has a slightly greater weight in the tails of the curve. This means that the logistic distribution is slighlty more likely to produce outliers than the normal distribution. If you believe that the logistic distribution is a reasonable distribution for your data, then you could compute the sample size for a t-test and divide it by 1.097. Interestingly, this tells you that the Mann-Whitney U is slightly more efficient when your data has a slightly greater tendency to produce outliers. One intuitive explanation is that the use of ranking in the Mann-Whitney U test reduces the influence of outliers.

You could examine a worst case scenario, since the A.R.E. for the Mann-Whitney U is never less than 0.864. So estimate the sample size for a t-test and divide by 0.864. This is a sample size that is big enough no matter what the underlying distribution.

All of the calculations described above are approximate, since they rely on asymptotic results (remmeber what the A in ARE stands for!). You can also use this ARE approach to estimating sample size for other nonparametric procedures, such as the sign test, the Wilcoxon signed rank test (not be be confused with the Wilcoxon rank sum test), and the Spearman correlation test.

Update from my weblog (March 8, 2005)

If you plan to use a nonparametric test, compute the sample size required for a t test and add 15%.

This assumes a reasonably high number of subjects (at least a few dozen) and a distribution which is not really unusual. I had not heard this rule; the author cites pages 76-81 of Lehmann, Nonparametrics: Statistical Methods Based on Ranks [BookFinder4U link]. I don’t have this book, so I can only guess as to the basis for this formula.

This rule could be based, I suppose, on the lower bound for the Asymptotic Relative Efficiency (ARE) of the Mann Whitney U test versus the t-test, which is 0.864. This says that no matter what the distribution, the ARE of the Mann Whitney U test can never be worse than 0.864 for a reasonably broad class of probability distributions. Inverting that gives you an increase in the sample size by a factor of 1.157. The same statement would also apply for the Wilcoxon Signed Ranks test, which can never have an ARE less than 0.864 compared to the paired t-test.

Summary

Bewildered Bob does not know how to estimate an appropriate sample size for the Mann-Whitney U test, a nonparametric test for the comparison of two independent groups. Professor Mean explains that you can find an approximate size by estimating the sample size for a t-test and then adjusting this size size based on the Asymptotic Relative Efficiency of the Mann-Whitney U test.

A technical description of ARE appears in chapter 5 of Randles and Wolfe. An alternate formula for sample size for the Mann-Whitney U test appears on page 120 of Hollander and Wolfe. There are other published approaches to sample size, which Professor Mean will add to these pages when time permits.

1. **Nonparametric Statistical Methods, Second Edition.
**Hollander M, Wolfe D.
New York NY: John Wiley and Sons, Inc (1999).
ISBN: 0-471-19045-4.
2. Introduction to the Theory of Nonparametric Statistics.
Randles RH, Wolfe DA.**
**New York, John Wiley & Sons (1979).
ISBN: 0-471-04245-5.

This page was written by Steve Simon while working at Children’s Mercy Hospital. Although I do not hold the copyright for this material, I am reproducing it here as a service, as it is no longer available on the Children’s Mercy Hospital website. Need more information? I have a page with general help resources. You can also browse for pages similar to this one at Category: Sample size justification

justification](../category/SampleSizeJustification.html) this one at [Category: Sample size with general help resources. You can also browse for pages similar to Children’s Mercy Hospital website. Need more information? I have a page reproducing it here as a service, as it is no longer available on the Hospital. Although I do not hold the copyright for this material, I am This page was written by Steve Simon while working at Children’s Mercy