Statistical tests for a population mean of a random variable with an unknown distribution

I have some data whose distribution is a priori unknown. A Q-Q plot shows that the distribution has fatter tails than a normal curve, and when I look at some samples I have available, the excess kurtosis is fairly large and fluctuates a lot, which is the behavior I would expect if the distribution did not have a well-defined fourth moment. For a particular hypothesis I want to test, I have 55 relevant pieces of data, and I want to test the hypothesis that their mean has a certain value. (Actually I'm using the median, but that's a side issue.)

The following is what I did, but it seems kind of ad hoc and arbitrary. I considered the possibility of a Student's $$t$$-distribution as the underlying distribution for my data. I played around with the number of degrees of freedom $$\nu$$ and looked at Q-Q plots for several data sets I had available, and in general $$\nu=4$$ seemed like the best fit -- significantly better than a normal, and better in most cases than $$\nu=2$$ or 3. So then I reasoned that if it were a Student's $$t$$ with $$\nu=4$$, it would have a finite variance, and therefore the central limit theorem would hold. Based on this, I expect the mean of 55 observations to be well approximated by a normal distribution, and then I can use standard tests for that situation. Taking into account the factor of $$\sqrt{\pi/2}$$ because I'm using the median, this is the equivalent of a 3.1 $$\sigma$$ effect, which is a high level of significance.

Is there a better way of going about this? What I did seems weak to me in two ways:

(1) It was pretty arbitrary to use Student's $$t$$-distribution.

(2) I would like to have some way of estimating whether the normal approximation is good enough for $$n=55$$ with a significance of 3.1 $$\sigma$$. The central limit theorem is a statement about what happens as $$n$$ approaches infinity, and I would expect that the approximation would become better more quickly when we were at, say, 1 $$\sigma$$, and would become better more slowly when we were at 5 $$\sigma$$. That is, for the mean of a finite number of random variables drawn from a distribution with fatter tails than a normal, I would expect the tails for the mean's distribution would asymptotically get completely different from those of the normal, when you got far enough out in the tails.

Issue #2 seems like it could be addressed simply by doing a little Monte Carlo simulation, but this presupposes that I know the distribution, so it doesn't address issue #1.

I agree that the choice of a t distribution to model your data may be arbitrary. But from what you say, the sample must be roughly symmetrical.

To test for a population median using roughly symmetrical data of unknown distribution, a nonparametric Wilcoxon signed rank test seems appropriate. Suppose you have $$n = 55$$ observations as in the vector x below, with numerical and graphical summaries from R, as shown below:

summary(x);  sd(x)
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
6.520   9.359   9.820   9.736  10.266  13.318
[1] 1.154808   # sample standard deviation

boxplot(x, horizontal=T, col="skyblue2", pch=20)


histogram(x, prob=T, col="skyblue2")


A (distinctly nonlinear) normal probability plot suggests that data were sampled from a heavy-tailed non-normal population.

qqnorm(x, datax="T");  qqline(x, datax=T, col="green")


Here are results from such a one-sample Wilcoxon test:

wilcox.test(x, mu = 9.5)   \$ mu is hypothetical median

Wilcoxon signed rank test with continuity correction

data:  x
V = 1058, p-value = 0.016
alternative hypothesis: true location is not equal to 9.5


The null hypothesis $$H_0: \eta = 9.5$$ is rejected at level 5% in favor of the alternative $$H_a: \eta \ne 9.5.$$

Note: The data, sampled in R as below, are from a Laplace (also called 'double exponential') distribution with median $$\eta = 10.$$

set.seed(2021)
x = rexp(55) - rexp(55) + 10

• Perfect, this was exactly the information I was looking for, thanks!
– user122182
Apr 3, 2021 at 23:50