Can the t-test be used to test the difference between percentiles of 2 samples? [duplicate]

For example, instead of testing the difference in mean, I want to test the difference in the 75th percentile of the 2 groups.

Does the central limit theorem hold? and what would be the equation for the standard error?

• If you assume normality and the usual assumptions of the two-sample t test, then a difference in the mean implies a difference in any percentile. Commented May 29, 2020 at 5:14

Quoting from Theorem 7.5.1 (p243) of Bain & Englehardt (1992), except for notation:

Let $$X_1, X_2, \dots, X_n$$ be a random sample from a continuous distribution with PDF $$f(x)$$ that is continuous and nonzero at the $$p$$th percentile $$x_p,$$ for $$0 < p < 1.$$ If $$k/n \rightarrow p$$ (with $$k-np$$ bounded), then the $$k$$th order statistic $$X_{(k)}$$ is asymptotically normal with mean $$x_p$$ and variance $$c^2/n,$$ where $$c = p(1-p)/f(x_p).$$

So there is a CLT for the 75th percentile and the asymptotic variance is as specified in the theorem. Instead of requiring a finite population variance $$\sigma^2,$$ as in the CLT for means, the requirement is roughly that the quantile $$x_p$$ of the distribution be precisely defined, with $$f(x_p) > 0.$$

Suppose you have a sample of size $$n=625$$ from a population distributed as $$\mathsf{Norm}(\mu = 100, \sigma = 15),$$ with 75th percentile $$110.1173$$ and $$c^2/n = \frac{3/16}{0.0212n} = 0.6684.$$

f = dnorm(qnorm(.75, 100,15), 100,15); f
[1] 0.0211851
(3/16)/(625*f^2)
[1] 0.6684363


If we simulate $$m=100\,000$$ such samples, we see that the variance of the resulting $$m$$ 75th quantiles is in good agreement with the theoretical asymptotic variance.

q = replicate(10^5, as.numeric(
quantile(rnorm(625, 100,15),.75) ))
var(q)
[1] 0.6679301


I am not sure exactly what null and alternative hypotheses you will test and for what distribution. Of course, the 2-sample t test programmed into statistical software programs uses means and variances. I suppose power will be poorer using 75th percentiles than using means. Notice that the variance of a sample mean of $$n=625$$ observations from the normal distribution of my example above has variance $$\sigma^2/n = 15^2/625 = 0.360 < 0.668.$$

So I will leave the rest to you.

• very cool. Thank you very much! Commented May 30, 2020 at 7:12
• Considering the heat around here today, I thought something very cool would be a good idea. Commented May 30, 2020 at 7:26
• Maybe should have pointed out explicitly that the condition $0 < p < 1$ of the theorem excludes the max and the min. No CLT applies to max or min. They have 'extreme value' distributions. Commented May 30, 2020 at 16:31