I was reading this article ,article link here, about the Central Limit Theorem, CLT, and how it can be used to determine if a cohort of interest is significantly different than the population (I might have phrased this poorly because I don't know the math terms). Using the CLT, I can calculate the cohort of interest's z-score and look at its p-value and either accept or reject the null hypothesis.
Basically, they do lots of random sampling of the population and use the sample means to determine the population mean. Then, they have a cohort which they're interested in and use the formula below to get the cohorts z-score in order to determine if they can reject the null hypothesis.
$M = \text{sample mean}$
$\mu = \text{population mean}$
${\sigma = \text{population standard deviation}}$
$n = \text{sample size}$
In my problem I have a population and I have a smaller cohort of people, which I will call cohort A, that I'm interested in analyzing. The population size is about 200,000 and cohort A is about 5,500. I am trying to test the hypothesis that cohort A is significantly more active than the population. Cohort A does not belong in the population. Also, in cohort A there are extreme outliers which are greatly shifting the mean. I want to use the median instead, to avoid the influence of the few extreme outliers, and read in this post, post link here, that if I apply the same techniques in the CLT and use the median instead, and a large n, I will get a normal distribution just like in the CLT (or at least that was my interpretation given my not so awesome math skills). If this holds true, can I use the formula in the image above and replace the means with medians in order to calculate cohort A's z-score so I can determine if I should accept or reject a null hypothesis that cohort A is significantly more physically active?