I have two samples of data, a baseline sample, and a treatment sample.
The hypothesis is that the treatment sample has a higher mean than the baseline sample.
Both samples are exponential in shape. Since the data is rather large, I only have the mean and the number of elements for each sample at the time I will be running the test.
How can I test that hypothesis? I'm guessing that it is super easy, and I've come across several references to using the F-Test, but I'm not sure how the parameters map.
Best Answer
You can test equality of the mean parameters against the alternative that the mean parameters are unequal with a likelihood ratio test (LR test). (However, if the mean parameters do differ and the distribution is exponential, this is a scale shift, not a location shift.)
For a one-tailed test (but only asymptotically in the two tailed case), I believe that the LR test comes out to be equivalent to the following (to show that this is in fact the same as the LR test for the one-tailed case one would need to show the LR statistic was monotonic in $\bar x/\bar y$):
Let's say we parameterize the $i$th observation in the first exponential as having pdf $1/\mu_x \exp(-x_i/\mu_x)$ and the $j$th observation in the second sample as having pdf $1/\mu_y \exp(-y_j/\mu_y)$ (over the obvious domains for the observations and parameters).
(To be clear, we're working in the mean-form not the rate-form here; this won't affect the outcome of the calculations.)
Since the distribution of $X_i$ is a special case of the gamma, $\Gamma(1,\mu_x)$, the distribution of the sum of $X$'s, $S_x=\sum_i X_i$ is distributed $\Gamma(n_x,\mu_x)$; similarly that for the sum of the $Y$s, $S_y$ is $\Gamma(n_y,\mu_y)$.
Because of the relationship between gamma distributions and chi-squared distributions, it turns out that $2/\mu_x S_x$ is distributed $\chi^2_{2n_x}$. The ratio of two chi-squares on their degrees of freedom is F. Hence the ratio, $\frac{\mu_y}{\mu_x}\frac{S_x/n_x}{S_y/n_y} \sim F_{2n_x,2n_y}$.
Under the null hypothesis of equality of means, then, $\bar x/\bar y \sim F_{2n_x,2n_y}$, and under the two sided alternative, the values might tend to be either smaller or larger than a value from the null distribution, so you need a two-tailed test.
Simulation to check that we didn't make some simple mistake in the algebra:
Here I simulated 1000 samples of size 30 for $X$ and 20 for $Y$ from an exponential distribution with the same mean, and computed the above ratio-of-means statistic.
Below is a histogram of the resulting distribution as well as a curve showing the $F$ distribution we computed under the null:
Example, with discussion of computation of two-tailed p-values:
To illustrate the calculation, here's two small samples from exponential distributions. The X-sample has 14 observations from a population with mean 10, the Y-sample has 17 observations from a population with mean 15:
The sample means are 12.082 and 16.077 respectively. The ratio of means is 0.7515
The area to the left is straightforward, since it's in the lower tail (calc in R):
We need the probability for the other tail. If the distribution was symmetric in the inverse, it would be straightforward to do this.
A common convention with the ratio of variances F-test (which is similarly two tailed) is simply to double the one-tailed p-value (effectively what is going on as here; that's also what seems to be done in R, for example); in this case it gives a p-value of 0.44.
However, if you do it with a formal rejection rule, by putting an area of $\alpha/2$ in each tail, you'd get critical values as described here. The p-value is then the largest $\alpha$ that would lead to rejection, which is equivalent to adding the one tailed p-value above to the one-tailed p-value in the other tail for the degrees of freedom interchanged. In the above example that gives a p-value of 0.43.
[Neither of these rules are "optimal" in small samples]