# How to transform t-statistic into a statistic using median and IQR?

I want to calculate similarity between various samples, but I'm limited to only knowing the IQR and medians. I came across a similar problem here but the top answer states that the newly proposed statistic follows the same distribution, but doesn't explain why it does:

$$t = \frac{\text{Mean}_1 - \text{Mean}_2}{\sqrt{\dfrac{s_1^2}{N_1}+\dfrac{s_2^2}{N_2}}}$$

$$u = \frac{\text{Median}_1 - \text{Median}_2}{\sqrt{\pi/2}\,\sqrt{J_1+J_2}}$$ where $$J_1=\frac{IQR_1^2}{1.82 N_1}, \ \ J_2 = \frac{IQR_2^2}{1.82 N_2}$$

I understand that essentially, we're comparing average values against variability. Specifically, I don't understand the following parts which should make these distributions similar:

1. Why do we divide the difference in medians by the square root of $$\pi/2$$ ?
2. Why do we divide the IQR by 1.82?
• Do you mean have only the medians and IRQ, not the sample size also ? Commented May 30, 2023 at 14:06
• The distributional assertion in the referenced answer is incorrect. It should be interpreted as having approximately the same distribution for large sample sizes.
– whuber
Commented May 30, 2023 at 15:48

The answer that you cited says: This is because the variances of sample medians are $$\pi/2$$ times the variances of sample means, and the IQRs are 1.82 times the standard deviations.

Here are more details:

$$t$$ distributed variables with $$d$$ degrees of freedom are defined as the ratio of a standard normal (N(0,1)) distributed variable $$X$$ and an independent variable $$S$$, for which $$d\cdot S^2$$ has $$\chi^2$$ distribution with $$d$$ degrees of freedom.

If sample 1 and sample 2 both come from an $$N(\mu,\sigma^2)$$ distribution, then the numerator $$(\mathrm{Mean}_1 - \mathrm{Mean}_2)$$ has a normal distribution with mean 0 and variance $$\tau^2 = \sigma^2(1/N_1 +1/N_2)$$. If you divide $$(\mathrm{Mean}_1 - \mathrm{Mean}_2)$$ by $$\tau = \sqrt{\sigma^2(1/N_1 +1/N_2)}$$, then you get an $$N(0,1)$$ distributed variable.

If you divide the denominator, $$\sqrt{S_1^2/N_1 + S_2^2/N_2}$$, by $$\tau$$, then (denominator)$$^2\cdot d$$ has a $$\chi^2$$ distribution with $$d=N_1+N_2-2$$ degrees of freedom. Therefore, $$T = \frac{\mathrm{Mean}_1 - \mathrm{Mean}_2}{\sqrt{S_1^2/N_1 + S_2^2/N_2}} = \frac{(\mathrm{Mean}_1 - \mathrm{Mean}_2)/\tau}{\sqrt{S_1^2/N_1 + S_2^2/N_2}/\tau}$$ follows a $$t$$-distribution with $$d=N_1+N_2-2$$ degrees of freedom.

Now, if you replace sample means by sample medians, the numerator of your $$t$$-variable spreads more. It is approximately $$N(0, \tau^2\cdot\pi/2)$$ distributed, but by dividing with $$\sqrt{\pi/2}$$, you are approximately back at the initial $$N(0,\tau^2)$$ distribution for the original numerator, $$\mathrm{Mean}_1 - \mathrm{Mean}_2$$.

Equally, the mean IQR in a standard normal is 1.349. IQR scales with the standard deviation for general normal distributions, $$N(\mu,\sigma^2)$$. Thus, $$\mathrm{IQR}/1.349$$ is an estimator for $$\sigma$$ and $$\mathrm{IQR}^2/1.349^2=\mathrm{IQR}^2/1.82$$ is an estimator for $$\sigma^2$$. Replacing $$S^2$$ with $$\mathrm{IQR}^2/1.82$$ gets you approximately back to the initial denominator.

• +1 for including "approximately"!
– whuber
Commented May 30, 2023 at 15:49