# What is it meant by the "rejection region" and "power" of a likelihood ratio test?

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.

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!

• I assume that your $H_0$ is $\mu = 0$, correct? Mar 27, 2014 at 11:59
• Sounds like routine bookwork. Please read the link and edit as appropriate. Mar 27, 2014 at 13:20
• You don't need the CLT in this instance because $\sum_i X_i$ is a normal random variable from the get go; no asymptotic approaches to normality are involved. Mar 27, 2014 at 13:44
• I think you are correct; the answer to the power question is indeed staring at us. It may be lurking in the ambiguity in the last statement: with respect to which distribution is the probability meant to be calculated? $H_0$ or $H_1$?
– whuber
Sep 19, 2014 at 18:34

As it is said in the comment. You don't need the CLT: you should already know the distribution of $\sum\limits_{i=1}^n X_{i}$ depending on which hypothesis $H_{0}$ or $H_{1}$ you assume.
• To find $k$, you need the distribution of $\sum\limits_{i=1}^n X_{i}$ under $H_{0}$ and do the manipulation as you said: isolate $\sum\limits_{i=1}^n X_{i}$ in the equation.
• To find the power of the test: I recall you the definition. $power=P(T\in \bar{A}|H_{1})$ where $T$ is your statistic of your test, here for instance $\sum\limits_{i=1}^n X_{i}$ and $\bar{A}$ is the rejected region. It is the probability of rejecting your initial hypothesis when this hypothesis is false. So how much your test can be usefull, powerfull. So according to the definition, this time you will need the distribution of $\sum\limits_{i=1}^n X_{i}$ not under $H_{0}$ but under $H_{1}$. Then again use $\alpha$ and a numerical table of it to find the probabiity.