TheoremProved

Exact significance level of the two-sided z-test

Statement

Let $X_1, \ldots, X_n$ be independent, identically distributed as $N(\mu_0, \sigma^2)$ with $\sigma$ known, and let $Z = \frac{\bar{X} - \mu_0}{\sigma / \sqrt{n}}$ . Let $z_{\alpha/2}$ be the value with $P(Z > z_{\alpha/2}) = \alpha/2$ for the standard normal distribution. Then the test that rejects $H_0: \mu = \mu_0$ when $|Z| > z_{\alpha/2}$ satisfies $P(\text{reject } H_0 \mid H_0) = \alpha$ exactly.

Why is it true?

The rejection region is built directly from the exact distribution of $Z$ under $H_0$ , so its probability under $H_0$ can be computed exactly rather than merely bounded, giving a test whose false-positive rate is controlled precisely at the chosen level $\alpha$ , not just approximately.

Proof sketch

Step 1 (distribution of $Z$ under $H_0$ ): since $X_1, \ldots, X_n$ are i.i.d. $N(\mu_0, \sigma^2)$ , the sample mean $\bar{X}$ is distributed as $N(\mu_0, \sigma^2/n)$ , so standardizing gives $Z = \frac{\bar{X} - \mu_0}{\sigma/\sqrt{n}} \sim N(0, 1)$ exactly.

Step 2 (probability of the rejection event): the rejection event is $|Z| > z_{\alpha/2}$ , which splits into the two disjoint events $Z > z_{\alpha/2}$ and $Z < -z_{\alpha/2}$ , so $P(|Z| > z_{\alpha/2}) = P(Z > z_{\alpha/2}) + P(Z < -z_{\alpha/2})$ .

Step 3 (use symmetry of the standard normal): the standard normal density is symmetric about $0$ , so $P(Z < -z_{\alpha/2}) = P(Z > z_{\alpha/2})$ ; by the definition of $z_{\alpha/2}$ , each of these two probabilities equals $\alpha/2$ .

Step 4 (add the two pieces): substituting into Step 2, $P(|Z| > z_{\alpha/2}) = \alpha/2 + \alpha/2 = \alpha$ , and since this probability was computed exactly under $H_0$ (using Step 1), $P(\text{reject } H_0 \mid H_0) = \alpha$ exactly, as claimed.

Topics that use this theorem

Hypothesis testing

Step-by-step proofs

No step-by-step proof yet for this theorem.