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I am evaluating the levels of thyroid hormones and thyroid antibodies in 3 groups namely: pre-eclamptic pregnant women, non-pre-eclamptic pregnant women, and apparently healthy non-pregnant women. Because the data generated is not uniformly distributed, I decided to carry out Kruskal-Wallis Test. I want to report the median but I am confused about which one to report. Is it the one with less than/= to or the one with >?

Answers are highly appreciated in advance.

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  • $\begingroup$ I think you mean 'not normally distributed'. // Not sure what your last sentence mean. if you have 3 groups, report 3 medians. Maybe null hypothesis is $H_0: \eta_1 = \eta_2 = \eta_3$ and alternative is $H:a{ \mathrm{not all} \eta_i \mathrm{equal}.$ // Some people prefer to speak of 'locations' generically, instead of 'medians' specifically. The original K-W test was for medians, but modern software implementations may refer to locations. $\endgroup$
    – BruceET
    Commented Apr 23, 2020 at 23:35
  • $\begingroup$ As far as I can see, the original Kruskal-Wallis test (as presented in their 1952 paper) was not decribed as being "for medians" (nor is a test of medians in general). Neither the introduction nor section 4 on interpretation of the test present it that way from what I can see with a quick read. $\endgroup$
    – Glen_b
    Commented Apr 25, 2020 at 3:53
  • $\begingroup$ Thank you all for the answers so far. I appreciate. $\endgroup$
    – user282452
    Commented Apr 26, 2020 at 17:26

1 Answer 1

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Example using data simulated in R.

set.seed(1234)
x1 = rgamma(15, 3, .6)
x2 = rgamma(15, 3, .6) + .5
x3 = rgamma(15, 3, .6) + 2
x = c(x1,x2,x3);  g = rep(1:3, each=15)

In R, the K-W test output does not mention $H_0,$ which is rejected for my example, with P-value 0.025. You can use two-sample Wilcoxon tests to see which pairs of medians differ significantly.

kruskal.test(x ~ g)

        Kruskal-Wallis rank sum test

data:  x by g
 Kruskal-Wallis chi-squared = 7.4025, df = 2, p-value = 0.02469

Boxplots show sample medians:

 boxplot(x ~ g, col="skyblue2", pch=20)

enter image description here

From the simulation, we know that the population medians are $\eta_1 = 4.457, \eta_2 = 4.857, \eta_3 = 6.457.$

qgamma(.5, 3, .6)
[1] 4.456767

here are the sample medians:

median(x1);  median(x2);  median(x3)
[1] 4.215075
[1] 4.568092
[1] 6.14665

According to a 2-sample Wilcoxon test, the sample medians $H_1$ and $H_3$ differ significantly. If you do several such ad hoc tests you need to use some method to help avoid false discovery (such as Bonferroni).

wilcox.test(x1, x2)$p.val
[1] 0.6235943
wilcox.test(x1, x3)$p.val
[1] 0.01855373
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    $\begingroup$ Good answer. +1. My only comment is that I would recommend using a post-hoc test designed for the K-W test rather than use pairwise Wilcoxon-Mann-Whitney tests. Pairwise W-M-W tests in particular can lead to funky results that are difficult to interpret in aggregate. Appropriate post-hoc tests include Dunn test (1964), Conover test, and Nemenyi test. In R, these tests are available in the PMCMRplus package and others. $\endgroup$
    – Sal Mangiafico
    Commented Feb 11, 2022 at 10:45

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