2
$\begingroup$

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

enter image description here

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!

$\endgroup$
  • 1
    $\begingroup$ On a different issue I don't see how you can get a UMP test when you are using a normal approximation. $\endgroup$ – Michael Chernick Apr 16 '17 at 19:16
  • 1
    $\begingroup$ 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) $\endgroup$ – Ben Bolker Apr 16 '17 at 20:06
  • 1
    $\begingroup$ @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. $\endgroup$ – Glen_b Apr 20 '17 at 7:51
  • $\begingroup$ Okay Glen I see your point. But the choice of the critical value is not exactly what would be used for the UMP test. $\endgroup$ – Michael Chernick Apr 20 '17 at 7:57
2
$\begingroup$

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 $T-E(T)=T+1/\beta_0$.

There's no typo, it's doing the right thing there.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.