Finding all possible values of $\alpha$ such that there is a likelihood test of size exactly $\alpha$ Let $X\sim \mathrm{Bin}(2,\theta)$ and test $H_0:\theta =0.5$ against $H_1:\theta=0.75$. Find the possible values of $\alpha$ for which there is a likelihood ratio test of size exactly $\alpha$.
 A: Any reasonable test will reject this null hypothesis if $X$ is bigger than some critical number.  But $X$ can have only one of three values: $0$, $1$, and $2$.
There's the test that rejects $H_0$ precisely if $X=2$.  The probability of that, under the null hypothesis, is $0.5^2=0.25$.
There's the test that rejects $H_0$ precisely if $X\ge 1$.  The probability of that, under the null hypothesis, is $2(0.5)(1-0.5)+0.5^2=0.75$.
There's the test that rejects $H_0$ precisely if $X>2$, so that it never rejects $H_0$, and there's the test that rejects $H_0$ precisely if $X\ge0$, so that it always rejects $H_0$.  The probabilities of those are respectively $0$ and $1$.
So you have $0$, $0.25$, $0.75$, and $1$.
Now there's the question of showing that all of the above tests are likelihood-ratio tests, and that there are no other likelihood-ratio tests.  The likelihood ratio when $X=x$ is observed is
$$
\frac{\dbinom 2 x 0.75^x(1-0.75)^{2-x}}{\dbinom 2 x 0.5^x(1-0.5)^{2-x}} = \begin{cases} \text{something} & \text{if }x=0, \\   \text{something} & \text{if }x=1, \\  \text{something} & \text{if }x=2. \end{cases}
$$
You reject $H_0$ if this is too big.  How big is too big depends on $\alpha$.
