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If I only know the medians and IQRs from samples (along with the number of samples of each population) drawn from two distributions, how to test the medians of the two distributions are different or not?

t-test is based on mean and sd. But I don't find a test based on median and IQR. Is there such a statistical test?

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  • $\begingroup$ "..along with number of samples of each population..." Did you mean sample size of each sample? Is your sample size large? $\endgroup$
    – Dayne
    Feb 16, 2022 at 4:17
  • $\begingroup$ Yes. As n1 and n2 in t-test en.wikipedia.org/wiki/…. They could be large or small, but not very small. $\endgroup$
    – user1424739
    Feb 16, 2022 at 4:24
  • $\begingroup$ The problem is that if you do not have large sample, you should know (or at least be able to make reasonable assumptions) about the underlying distribution of the population. And then derive the sampling distribution of an appropriate statistic which is a function of sample median. If the sample is large, however, and asymptotic distribution will make your life easy. See this for more. $\endgroup$
    – Dayne
    Feb 16, 2022 at 4:36
  • $\begingroup$ For non-parametric tests, see this: en.wikipedia.org/wiki/Median_test $\endgroup$
    – Dayne
    Feb 16, 2022 at 4:38
  • $\begingroup$ A solution can be separated into two cases: large samples and small samples. Since large sample case may be easier, a partial solution for this case may be offered first. $\endgroup$
    – user1424739
    Feb 16, 2022 at 4:57

2 Answers 2

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If the parent distributions are normal, then the $t$-statistic $$t = \frac{\text{Mean}_1 - \text{Mean}_2}{\sqrt{\dfrac{s_1^2}{N_1}+\dfrac{s_2^2}{N_2}}}$$ has the same distribution as $$u = \frac{\text{Median}_1 - \text{Median}_2}{\sqrt{\pi/2}\,\sqrt{J_1+J_2}}$$ where $$J_1=\frac{IQR_1^2}{1.82 N_1}, \ \ J_2 = \frac{IQR_2^2}{1.82 N_2}$$ This is because the variances of sample medians are $\pi/2$ times the variances of sample means, and the IQRs are $\sqrt{1.82^{\phantom'}}$ times the standard deviations.

So if the populations are normal or roughly normal, you can apply the $t$-test to $u$, using $$\nu=\frac{(J_1+J_2)^2}{\dfrac{J_1^2}{N_1-1} + \dfrac{J_2^2}{N_2 -1}}$$ degrees of freedom.

This follows Welch's $t$-test, and when the variances or sample sizes are equal, the formulas can also be simplified to follow Student’s $t$-test.

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    $\begingroup$ This isn't quite correct: $u$ does not have a Student t distribution, nor does it have the same distribution as $t.$ It's a plausible approximation, but -- especially for small $N_i$ -- needs to be checked. $\endgroup$
    – whuber ♦
    May 30 at 15:48
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Non-parametric tests, like Mann-Whitney, assign numeric ranks to all the observations in the groups, then test whether the sums of ranks in the groups are different. So, AFAIK, we cannot conduct a non-parametric test just by knowing the median+IQR.

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  • $\begingroup$ The first part of the answer is not asked in the question. I am only interested in a solution using medians and IQRs. If such a solution is not available, I will need a proof that it is not possible to create a test using just medians and IQRs. $\endgroup$
    – user1424739
    Feb 16, 2022 at 4:56
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    $\begingroup$ This answer is logically flawed: just because some nonparametric tests use ranks and not all ranks are available does not imply a non-parametric test is impossible. Indeed, there is an obvious such test lying on the table: compare the statistics you do have. That is analyzed at stats.stackexchange.com/questions/86931. $\endgroup$
    – whuber ♦
    Feb 17, 2022 at 17:12
  • $\begingroup$ Thanks, @whuber. I had not read such non-rank methods in the mainstream statistical books. As I understood, such tests can be used under some minor assumptions that these robust statistics or 5-letter summaries are from F distribution, and the sample sizes are known, right? $\endgroup$
    – maaniB
    Feb 19, 2022 at 10:13
  • $\begingroup$ This is unrelated to the F distribution. $\endgroup$
    – whuber ♦
    Feb 19, 2022 at 18:51

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