(This article was first published on

A colleague recently contacted us with the following question: "My outcome is skewed-- how can I compare medians across multiple categories?" **SAS and R**, and kindly contributed to R-bloggers)What they were asking for was a generalization of the Wilcoxon rank-sum test (also known as the Mann-Whitney-Wilcoxon test, among other monikers) to more than two groups. For the record, the answer is that the Kruskal-Wallis test is the generalization of the Wilcoxon, the one-way ANOVA to the Wilcoxon's t-test.

But this question is based on a false premise: that the the Wilcoxon rank-sum test is used to compare medians. The premise is based on a misunderstanding of the null hypothesis of the test. The actual null hypothesis is that there is a 50% probability that a random value from one population exceeds an random value from the other population. The practical value of this is hard to see, and thus in many places, including textbooks, the null hypothesis is presented as "the two populations have equal medians". The actual null hypothesis

**can**be expressed as the latter median hypothesis, but only under the additional assumption that the shapes of the distributions are identical in each group.

In other words, our interpretation of the test as comparing the medians of the distributions requires the location-shift-

**only**alternative to be the case. Since this is rarely true, and never assessed, we should probably use extreme caution in using, and especially in interpreting, the Wilcoxon rank-sum test.

To demonstrate this issue, we present a simple simulation, showing a case of two samples with equal medians but very different shapes. In one group, the values are exponential with mean = 1 and therefore median log 2, in the other they are normal with mean and median = log 2 and variance = 1. We generate 10,000 observations and show that the Wilcoxon rank-sum test rejects the null hypothesis. If we interpret the null incorrectly as applying to the medians, we will be misled. If our interest actually centered on the medians for some reasons, an appropriate test that would not be sensitive to the shape of the distribution could be found in a quantile regression. Another, of course, would be the median test. We show that this test does not reject the null hypothesis of equal medians, even with the large sample size.

We leave aside the deeper questions of whether comparing medians is a useful substitute for comparing means, or whether means should not be compared when distributions are skewed.

**R**

In R, we simulate two separate vectors of data, then feed them directly to the

`wilcox.test()`function (section 2.4.2).

This shows a very small p-value, denoting the fact

y1 = rexp(10000)

y2 = rnorm(10000) + log(2)

wilcox.test(y1,y2)

**not**that the medians are unequal but that one or the other of these distributions generally has larger values. Hint: it might be the one with the long tail and no negative values.

To set up the quantile regression, we put the observations into a single vector and make a group indicator vector. Then we load the quantreg package and use the

Wilcoxon rank sum test with continuity correction

data: y1 and y2

W = 55318328, p-value < 2.2e-16

alternative hypothesis: true location shift is not equal to 0

`rq()`function (section 4.4.4) to fit the median regression.

This shows we would fail to reject the null of equal medians.

y = c(y1,y2)

c = rep(1:2, each = 10000)

library(quantreg)

summary(rq(y~as.factor(c)))

Call: rq(formula = y ~ as.factor(c))

tau: [1] 0.5

Coefficients:

Value Std. Error t value Pr(>|t|)

(Intercept) 0.68840 0.01067 64.53047 0.00000

as.factor(c)2 -0.00167 0.01692 -0.09844 0.92159

Warning message:

In rq.fit.br(c, y, tau = tau, ...) : Solution may be nonunique

**SAS**

In SAS, we make a data set with a group indicator and use it to generate data conditionally.

The Wilcoxon rank-sum test is in

data wtest;

do i = 1 to 20000;

c = (i gt 10000);

if c eq 0 then y = ranexp(0);

else y = normal(0) + log(2);

output;

end;

run;

`proc npar1way`(section 2.4.2).

As with R, we find a very small p-value.

proc npar1way data = wtest wilcoxon;

class c;

var y;

run;

The median regression is in

The NPAR1WAY Procedure

Wilcoxon Scores (Rank Sums) for Variable y

Classified by Variable c

Sum of Expected Std Dev Mean

c N Scores Under H0 Under H0 Score

0 10000 106061416 100005000 408258.497 10606.1416

1 10000 93948584 100005000 408258.497 9394.8584

Wilcoxon Two-Sample Test

Statistic 106061416.0000

Normal Approximation

Z 14.8348

One-Sided Pr > Z <.0001

Two-Sided Pr > |Z| <.0001

t Approximation

One-Sided Pr > Z <.0001

Two-Sided Pr > |Z| <.0001

Z includes a continuity correction of 0.5.

Kruskal-Wallis Test

Chi-Square 220.0700

DF 1

Pr > Chi-Square <.0001

`proc quantreg`(section 4.4.4). As in R, we fail to reject the null of equal medians.

proc quantreg data =wtest;

class c;

model y = c;

run;

The QUANTREG Procedure

Quantile and Objective Function

Predicted Value at Mean 0.6909

Parameter Estimates

Standard 95% Confidence

Parameter DF Estimate Error Limits t Value

Intercept 1 0.6909 0.0125 0.6663 0.7155 55.06

c 0 1 0.0165 0.0160 -0.0148 0.0479 1.03

c 1 0 0.0000 0.0000 0.0000 0.0000 .

Parameter Estimates

Parameter Pr > |t|

Intercept <.0001

c 0 0.3016

c 1 .

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