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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?

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  • $\begingroup$ 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. $\endgroup$
    – Michael M
    Commented May 29, 2020 at 5:14

1 Answer 1

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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.

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    $\begingroup$ very cool. Thank you very much! $\endgroup$ Commented May 30, 2020 at 7:12
  • $\begingroup$ Considering the heat around here today, I thought something very cool would be a good idea. $\endgroup$
    – BruceET
    Commented May 30, 2020 at 7:26
  • $\begingroup$ 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. $\endgroup$
    – BruceET
    Commented May 30, 2020 at 16:31

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