Suppose that we are interested in testing the simple null

Chapter 10, Problem 111E

(choose chapter or problem)

Suppose that we are interested in testing the simple null hypothesis \(H_{0}: \theta=\theta_{0}\) versus the simple alternative hypothesis \(H_{a}: \theta=\theta_{a}\). According to the Neyman-Pearson lemma, the test that maximizes the power at \(\theta_{a}\) has a rejection region determined by

                         \(\frac{L\left(\theta_{0}\right)}{L\left(\theta_{a}\right)}<k\)

In the context of a likelihood ratio test, if we are interested in the simple  and , as stated, then \(\Omega_{0}=\left\{\theta_{0}\right\}, \Omega_{a}=\left\{\theta_{a}\right)\), and \(\Omega=\left\{\theta_{0}, \theta_{a}\right)\).

a Show that the likelihood ratio  is given by

                  \(\lambda=\frac{L\left(\theta_{0}\right)}{m\left(L\left(\theta_{0}\right), L\left(\theta_{a}\right)\right.}=\frac{1}{m\left(1, \frac{\left.L \theta_{0}\right)}{L\left(\theta_{0}\right)}\right\}}\)

b Argue that \(\lambda<k\) if and only if, for some constant ,

                   \(\frac{L\left(\theta_{0}\right)}{L\left(\theta_{a}\right)}<k^{\prime}\)

c What do the results in parts (a) and (b) imply about likelihood ratio tests when both the null and alternative hypotheses are simple?

Equation transcription:

Text transcription:

H_{0}: \theta=\theta{0}

H_{a}: \theta=\theta{a}

\theta{a}

\frac{L\left(\theta{0}\right)}{L\left(\theta{a}\right)}<k

\Omega{0}=\left\{\theta{0}\right\}, \Omega{a}=\left\{\theta{a}\right)

\Omega=\left\{\theta{0}, \theta{a}\right)

\lambda=\frac{L\left(\theta{0}\right)}{m\left(L\left(\theta{0}\right), L\left(\theta{a}\right)\right.}=\frac{1}{m\left(1, \frac{\left.L \theta_{0}\right)}{L\left(\theta{0}\right)}\right\}}

\lambda<k

\frac{L\left(\theta{0}\right)}{L\left(\theta{a}\right)}<k^{\prime}