How to test statistical difference given only median and IQR? 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?
 A: 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.
A: 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.
