Suppose that $X_1,...,X_n$ are i.i.d. data from a $N(\mu, 100)$ distribution. I am trying to find the rejection region for the likelihood ratio test for level $\alpha= 0.10$ of the test:
$H_0: \mu = 0$ versus $H_1: \mu= 1.5$
Finally, if $n = 25$, I would like to find the power of the test.
What confuses me about this question is that it is sort of non-standard.
My approach is to first write out the likelihood ratio:
$$\newcommand{\EXP}[1]{\exp\left[#1\right]} \frac{L(0)}{L(1.5)} = \EXP{\frac{\sum(X_i-.1.5)^2-\sum{X_i^2}}{2(100^2)}}$$.
Then, would my rejection region be:
$$\EXP{\frac{\sum(X_i-.1.5)^2-\sum{X_i^2}}{2(100^2)}} \leq k\text{ ?}$$
And finally, I am very confused about what the power of the test is, because it seems that I have already computed it in the process of writing out:
$$P\left(\EXP{\frac{\sum(X_i-.1.5)^2-\sum{X_i^2}}{2(100^2)}} \leq k\right)\leq \alpha = 0.10$$ and then, I would rearrange the above until I get $\sum{X_i}$ by itself, then apply the central limit theorem, then work backwards to find $k$.
However, I still dont know what they mean by the power of a test. I feel like it is staring right at me, but I am not sure what it is. Any help would be greatly appreciated!