test for equality of population means at a 10% level of significance Independent random samples, each of size n (i.e. observe {$X_i$}$_{i=1,...n}$ and {$Y_j$}$_{j=1,...,n}$), are to be taken from 2 exponential distributions with unknown means $\theta_1$ and $\theta_2$ respectfully.
a) Derive a likelihood ratio test for $H_0: \theta_1 = \theta_2$ against $H_1: \theta_1 \not = \theta_2$ and in doing so show that the quotient $Q=\frac{\sum_{i=1}^n X_i}{\sum_{j=1}^n Y_j} = \frac{\over X}{\over Y}$ is an appropriate test statistic to use.
$$\text{I think I got this part correct but I will show you what I did.}$$
$$ \text{Please correct any mistakes. My real question is about part (b)}$$
When $H_0$ true ($\theta \in \omega$)
$$\theta_1 = \theta_2 = \theta$$
$$\hat \theta_{MLE} = \frac{\bar X + \bar Y}{2} \text{ (with proper derivation)}$$
When $H_1$true ($\theta \in Ω$)
$$\theta_1 \not = \theta_2$$
$$\hat \theta_{1 MLE} = \bar X \text{ and } \hat \theta_{2 MLE} = \bar Y \text{ (with proper derivations)}$$
Now to work out $\lambda$
$$\lambda = \frac{L(\hat \omega)}{L(\hat Ω)} = \frac{[\prod _{i=1}^n \frac{2}{\bar X + \bar Y}e^{-\frac{2X_i}{\bar X + \bar Y}}][\prod _{j=1}^n \frac{2}{\bar X + \bar Y}e^{-\frac{2Y_i}{\bar X + \bar Y}}]}{[\prod _{i=1}^n \frac{1}{\bar X}e^{-\frac{X_i}{\bar X}}][\prod _{j=1}^n \frac{1}{\bar Y}e^{-\frac{Y_i}{\bar Y}}]} = ... = [\frac{2 \sqrt Q}{1+Q}]^{2n} \leq K_\alpha$$
b) Two samples, each of size n=3, are obtained from independent exponential distributions yielding: 0.13, 0.15, 3.30 and 1.29, 2.03, 2.31, respectively. Test for equality of population means at the 10% level of significance.
Hint: if $W$ follows Gamma$(\alpha, \theta)$ then $\frac{2W}{\theta}$ follows $\chi^2_{2 \alpha}$ 
Now I did a sketch of say $f(Q) = [\frac{2 \sqrt Q}{1+Q}]^{2n}$ and see that it has a max where $Q=1$ and tends to zero as Q gets larger so:
$$\lambda \leq K_\alpha \text{ when }\frac{\bar X}{\bar Y} \leq c_1 \text{ or } \frac{\bar X}{\bar Y} \geq c_2$$
So the thing I am most confused with is the whole $\frac{\bar X}{\bar Y}$ thing because if it was just $\bar X$ for example then $\sum_{i=1}^n X_i$ of an exponential distribution is a Gamma distribution (using moment generating functions or some other method) but I can't see how to proceed here with this Q.
 A: Part (a) looks good to me. For part (b), I think what you need to remember is that the LRT selects the constants $c_1$, $c_2$ so that
$$
\alpha = P\left( \lambda \leq K_\alpha ; H_0 \text{ is true} \right) = P \left( \frac{\overline X}{\overline Y} \leq c_1 \text{ or } \frac{\overline X}{\overline Y} \geq c_2  ; H_0 \text{ is true} \right)
$$
So when $H_0$ is true, $\theta_1 = \theta_2$. We have $\overline{X} \sim \text{Gamma}(n,\frac{\theta_1}{n})$ and $\overline{Y} \sim \text{Gamma}(n,\frac{\theta_1}{n})$. It follows that $$\frac{\overline{X}}{\overline{Y}} = \frac{ \frac{2n \overline{X}}{\theta_1}}{ \frac{2n \overline{Y}}{\theta_1}} = \frac{U_1}{U_2}$$
is the ratio of independent two chi-squared random variables with $2n$ degrees of freedom. With one more trick, we have $$\frac{\overline{X}}{\overline{Y}} = \frac{U_1/(2n)}{U_2/(2n)}$$
which is the definition of an $F(2n,2n)$ random variable.
So then one choice for our test is via the quantiles $c_1 = F_{1-\alpha/2}(2n,2n)$ and $c_2 = F_{\alpha/2}(2n,2n)$.
