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I conducted a Wilcoxon rank sum test with continuity correction, which produces W = 7195.5, p-value = 0.07094. This lets me confidently reject the null hypothesis of no true location shift at a confidence level < 0.1

My question on this is relatively general:

Based on my descriptive graphs, i.e. boxplots, jitter plots, it is not immediately visible which of the two groups I am comparing has the higher/lower median/mean. Since I know that the Wilcoxon test compares pseudo-medians, I am not sure which way would be appropriate to answer the question which of the two groups of which I know are significantly different w.r.t. the outcome variable, is is greater than the other.

I see the following possibilities:

  1. Conduct a one-sided test and see which is significant.
  2. Look at the means/medians within the two groups and whichever is greater is the one which is significantly greater.

Are these two possibilities valid in order to answer my question? Possibility 2 can be problematic, if the mean of group A is bigger than the mean of group B, but this is different for the median.

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    $\begingroup$ It is perfectly possible to construct a dataset where the two groups have identical medians yet the Wilcoxon is significant. So it might well not be possible to see any difference in your plots. $\endgroup$ – mdewey Jun 1 '16 at 15:11
  • $\begingroup$ You really should not "confidently" (in a common sense) reject the null hypothesis of no true location shift "at a confidence level < 0.1". You can choose this confidence level, that's fine, but this is generally considered to be a very weak evidence (at best) so don't be "confident" (in a common sense) of your results. $\endgroup$ – amoeba says Reinstate Monica Jun 2 '16 at 11:26
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Since I know that the Wilcoxon test compares pseudo-medians

Not quite. The Wilcoxon signed rank test compares the one-sample Hodges-Lehmann statistic (median-of-within-sample-pairwise-averages, equivalently median of Walsh averages, or pseudomedian) to 0. But the rank-sum test compares the two-sample Hodges-Lehmann statistic (the median of between-sample pairwise differences as described in the second paragraph under "Definition" at the link) to zero -- it does not compare two one-sample pseudomedians.

Based on my descriptive graphs, i.e. boxplots, jitter plots, it is not immediately visible which of the two groups I am comparing has the higher/lower median/mean.

You seem to be assuming that the difference-in-mean and the difference-in-median will behave like the median-pairwise-difference (in the sense that if one is different the other two will be and in the same direction).

This will often be true but it is not necessarily the case.

A population (or indeed, a sample) can have any of the three be different in some given direction while one or both the others are not different or even arranged in the opposite direction.

I am not sure which way would be appropriate to answer the question which of the two groups of which I know are significantly different w.r.t. the outcome variable, is is greater than the other.

If you're asking "which group did the rank sum test think had higher location?" (i.e. what difference caused the rejection?), then compute the median pairwise difference. This is a simple calculation and most decent stats packages will offer the calculation (at least as an option relating to a confidence interval for the difference) with the rank sum test.

If you're asking "in what direction do the medians differ" or "in what direction do the means differ" then you won't necessarily have an answer consistent with the rank sum test -- if you care about one of those, test that instead (perhaps with a permutation test based on those particular statistics).

If you assume that the distributions are the same up to a possible location shift under the alternative, and if you assume population means exist (generally a quite reasonable assumption) then you've already made the necessary assumption to attribute the direction of difference to the direction the rank sum test looked at.

Conduct a one-sided test and see which is significant.

You could, if you do it at half the significance level, but it would seem to be a fairly involved way to go about finding what can be obtained via a simple sample calculation. If you don't have a convenient way to do the calculation otherwise, it should work just fine.

Look at the means/medians within the two groups and whichever is greater is the one which is significantly greater.

You could find mean and median both have group 1 greater but the rank sum rejected in the other direction. Or you could have mean greater in one direction and median greater in the other direction. Or you could have both group-means and both group-medians be equal even though the rank-sum test rejected.

So this would not be a generally good choice.


Here's an example in R. First, adding the confidence interval calculation gets the location difference estimate, then calculating it directly from the sample. This assumes there's already data (in x and y):

> wilcox.test(x,y,conf.int=TRUE,conf.level=0.9)

        Wilcoxon rank sum test

data:  x and y
W = 9, p-value = 0.05927
alternative hypothesis: true location shift is not equal to 0
90 percent confidence interval:
 -15.3239889  -0.5774458
sample estimates:
difference in location 
             -8.891949 

> median(outer(x,y,"-")) # calculate median of pairwise differences
[1] -8.891949

So this tells us that y tends to be larger (as measured by the rank-sum statistic) than x (since the x-y differences tend to be negative)


An aside on this bit:

This let's me confidently reject the null hypothesis of no true location shift at a confidence level < 0.1

There's a few thing wrong there. That doesn't really let us "confidently" do anything, and 0.1 would be your significance level, not a confidence level. If I wanted to speak with some sort of confidence about an effect, I'd tend to be looking at effect sizes, confidence intervals and I'd at least want some (before-seeing-the-data) sense of the power against an anticipated/useful effect-size.

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