# Uniformly Most Powerful Test Gamma Distribution

In this worked-out solution, I'm convinced there is a typo:

In standarizing the variable, I understand how typically, we're supposed to subtract the mean from the variable in the numerator, so why are we adding (T+1/B_0)? Shouldn't it be subtraction? The rest makes sense. Thanks!

• On a different issue I don't see how you can get a UMP test when you are using a normal approximation. – Michael Chernick Apr 16 '17 at 19:16
• worth pointing out the source? Since these are from 2006 course notes you might have trouble giving feedback (but it does look like the professor is still around) – Ben Bolker Apr 16 '17 at 20:06
• @MichaelChernick The argument shows that a test based on $T$ is equivalent to a likelihood ratio test; the approximation comes only in calculating the critical value (i.e. the test is still most powerful, what's approximate is the significance level -- instead of a 5% test you get roughly 4.4% by using the normal approximation). Since $-T$ is itself gamma-distributed, one can compute an exact rejection rule -- "reject if $T\geq -0.8414/\beta_0$" (instead of the normal approximation's "reject if $T\geq -.8355/\beta_0$") but it's essentially beside the point to the main issue. – Glen_b Apr 20 '17 at 7:51
• Okay Glen I see your point. But the choice of the critical value is not exactly what would be used for the UMP test. – Michael Chernick Apr 20 '17 at 7:57

We can see from the information there that $E(X_i)=1/\beta_0$ (under the null, whence $E(\bar{X}_n) = 1/\beta_0$), and that $T=-\bar{X}_n$.
Consequently the expected value of $T$ is $-1/\beta_0$ and so $T-E(T)=T+1/\beta_0$.