I measured a feature on 2 populations.
I want to make sure I have a meaningful difference between my populations.
The student's t-test requires a normal distribution, but the feature I measured follows an exponential distribution:
In the graph above you can see the distribution of the feature for both of my populations (one in orange, one in blue).
I found the likelihood-ratio test, but I don't understand how to apply it... I want an equivalent of the student's t-test but for exponential distributions.
EDIT:
I want to test if the mean is significantly different.
CLT looks promising.
EDIT 2:
So following Alberto Sinigaglia advice, I tried to use CLT.
Here is what I did:
For each of my populations, I sampled 50 values at random and computed the mean of the sample. I did this 1000 times for each population. So I end up with 1000 means for each population. Here is the distribution of the means:
It does not look like a normal distribution to me.
I'll put an idea for a two-sided hypothesis test here, but a study of power/optimality of the test is more work. We use the exponential pdf parametrisation $f_{\theta}(x)=\theta^{-1}e^{-x/\theta}$. The Lebesgue density of an IID exponential sample of length $n$ is, for $x \in \mathbb{R}^n$, $$\frac{dP_\theta}{d\lambda}(x)=\prod_{k\leq n}\theta^{-1} e^{-x_k/\theta}\mathbf{1}_{[0,\infty)}(x_k)=\theta^{-n}e^{-\theta^{-1} \sum_{k\leq n}x_k}\mathbf{1}_{[0,\infty)}(x_{(1)})$$ where $x_{(1)}=\min(x_1,x_2,...,x_n)$. The density describes an exponential family in the parameter $\theta\in \mathbb{R}^+$ with sufficient and complete statistic $T(X)=\sum_{k\leq n}X_k$. Consider that $T(X)\sim \textrm{Gamma}(n,\theta)$ in $(k,\textrm{scale})$ parametrization. Consider the samples $X^1,X^2$ of lengths $n^1,n^2$ drawn from exponentials with parameters $\theta^1,\theta^2$. We have $$\frac{\theta_2T(X^1)}{\theta_1T(X^2)}\sim \textrm{Betaprime}(n^1,n^2)$$ This quantity is pivotal in the ratio $\theta_2/\theta_1$. We can derive a confidence interval: $$\bigg\{c_1\leq \frac{\theta_2T(X^1)}{\theta_1T(X^2)}\leq c_2 \bigg\}\implies \bigg\{\frac{T(X_2)}{T(X_1)}c_1\leq \frac{\theta_2}{\theta_1}\leq \frac{T(X_2)}{T(X_1)}c_2\bigg\}$$ we can choose $c_1,c_2$ to be respectively the $\alpha/2,1-\alpha/2$ percentiles of the Beta prime distribution with parameters $n^1,n^2$. If $1$ is inside the confidence interval, we do not reject the null hypothesis that $\theta_1=\theta_2$. I ran some simulations and it looks like it works and the distribution of the pivotal quantity matches, feedback appreciated.