# 26.2 - Uniformly Most Powerful Tests

26.2 - Uniformly Most Powerful Tests

The Neyman Pearson Lemma is all well and good for deriving the best hypothesis tests for testing a simple null hypothesis against a simple alternative hypothesis, but the reality is that we typically are interested in testing a simple null hypothesis, such as $$H_0 \colon \mu = 10$$ against a composite alternative hypothesis, such as $$H_A \colon \mu > 10$$. The good news is that we can extend the Neyman Pearson Lemma to account for composite alternative hypotheses, providing we take into account each simple alternative specified in H_A. Doing so creates what is called a uniformly most powerful (or UMP) test.

Uniformly Most Powerful (UMP) test

A test defined by a critical region C of size $$\alpha$$ is a uniformly most powerful (UMP) test if it is a most powerful test against each simple alternative in the alternative hypothesis $$H_A$$. The critical region C is called a uniformly most powerful critical region of size $$\alpha$$.

Let's demonstrate by returning to the normal example from the previous page, but this time specifying a composite alternative hypothesis.

## Example 26-6

Suppose $$X_1, X_2, \colon, X_n$$ is a random sample from a normal population with mean $$\mu$$ and variance 16. Find the test with the best critical region, that is, find the uniformly most powerful test, with a sample size of $$n = 16$$ and a significance level $$\alpha$$ = 0.05 to test the simple null hypothesis $$H_0: \mu = 10$$ against the composite alternative hypothesis $$H_A: \mu > 10$$.

For each simple alternative in $$H_A , \mu = \mu_a$$, say, the ratio of the likelihood functions is:

$$\dfrac{L(10)}{L(\mu_\alpha)}= \dfrac{(32\pi)^{-16/2} exp \left[ -(1/32)\sum_{i=1}^{16}(x_i -10)^2 \right]}{(32\pi)^{-16/2} exp \left[ -(1/32)\sum_{i=1}^{16}(x_i -\mu_\alpha)^2 \right]} \le k$$

Simplifying, we get:

$$exp \left[ - \left(\dfrac{1}{32} \right) \left(\sum_{i=1}^{16}(x_i -10)^2 - \sum_{i=1}^{16}(x_i -\mu_\alpha)^2 \right) \right] \le k$$

And, simplifying yet more, we get:

Taking the natural logarithm of both sides of the inequality, collecting like terms, and multiplying through by 32, we get:

$$-2(\mu_\alpha - 10) \sum x_i +16 (\mu_{\alpha}^{2} - 10^2) \le 32 ln(k)$$

Moving the constant term on the left-side of the inequality to the right-side, and dividing through by $$-16(2(\mu_\alpha - 10))$$, we get:

$$\dfrac{1}{16} \sum x_i \ge - \dfrac{1}{16(2(\mu_\alpha - 10))}(32 ln(k) - 16(\mu_{\alpha}^{2} - 10^2)) = k^*$$

In summary, we have shown that the ratio of the likelihoods is small, that is:

$$\dfrac{L(10)}{L(\mu_\alpha)} \le k$$

if and only if:

$$\bar{x} \ge k^*$$

Therefore, the best critical region of size $$\alpha$$ for testing $$H_0: \mu = 10$$ against each simple alternative $$H_A \colon \mu = \mu_a$$, where $$\mu_a > 10$$, is given by:

$$C= \left\{ (x_1, x_1, ... , x_n): \bar{x} \ge k^* \right\}$$

where $$k^*$$ is selected such that the probability of committing a Type I error is $$\alpha$$, that is:

$$\alpha = P(\bar{X} \ge k^*) \text{ when } \mu = 10$$

Because the critical region C defines a test that is most powerful against each simple alternative $$\mu_a > 10$$, this is a uniformly most powerful test, and C is a uniformly most powerful critical region of size $$\alpha$$.

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