Relation between Hypothesis Testing and Confidence Intervals

It may be recalled that rejecting a null hypothesis about $\theta$ ( $\theta= \theta_0$, say) at the 5% significance level is equivalent to saying that the value $\theta_0$ is not included in a 95% confidence interval for $\theta$. So we have a duality property here. We will illustrate this with an example.

Example 9..1
Consider the family of normal distributions with unknown mean $\mu$ and known variance $\sigma^2$. Let $z_{\alpha}$ be defined by $P(Z\geq z_{\alpha})=\alpha$. For a 2-sided alternative (using a 2-tailed test), the rejection region for a test of size $\alpha$ is

$$\left\{\overline{x}: \ \frac{\vert\overline{x}-\mu_0\vert}{\sigma/\sqrt{n}} \ > \ z_{\alpha/2} \right\}.$$

That is,

$$\left\{\overline{x}: \ \overline{x} > \mu_0 \, + \, \frac{z_... ...erline{x}<\mu_0- \frac{z_{\alpha/2}\sigma}{\sqrt{n}} \right\}.$$

This is the event {X $\in C(\theta)$} and it has probability $\alpha$. The complementary event,
{X $\notin$ C($\theta$)} has probability $1-\alpha$. The latter event can be written equivalently as

$$\left\{x: \mu_0- z_{\alpha/2}\sigma/\sqrt{n}<\overline{x}<\mu_0+z_{\alpha/2} \sigma/\sqrt{n} \right\}.$$

which is equivalent to

$$\overline{x}-z_{\alpha/2}\sigma/\sqrt{n} < \mu_0 < \overline{x} + x_{\alpha/2}\sigma/\sqrt{n}$$

and

$$( \overline{x} - z_{\alpha/2}\sigma/\sqrt{n}, \ \overline{x}+z_{\alpha/2}\sigma/\sqrt{n})$$

is a $100 (1-\alpha)\%$ confidence interval for $\mu$.