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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, $S_x$ is distributed $\Gamma(n_x,\mu_x)$; similarly that for the sum of the $Y$s, $S_y$ is $\Gamma(n_y,\mu_y)$.

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.

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 $\Gamma(n_y,\mu_y)$.

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]

You can test equality of the mean parameters against the alternative that the mean parameters are unequal with a likelihood ratio test. (However, if the mean parameters do differ and the distribution is exponential, this is a scale shift, not a location shift.)

I believe that (but haven't checked) the LR test comes out to be equivalent to the following:

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.

[To check that this was in fact the same as the LR test in small samples one would need to show the LR statistic was monotonic in $\bar x/\bar y$.]

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 $\bar x/\bar y$):

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.

[To check that this was in fact the same as the LR test in small samples one would need to show the LR statistic was monotonic in $\bar x/\bar y$.]