Given Variance, Check Mean

If we know the population variance $\sigma^2$ (e.g. from historical data), and we want to know if the mean of the new data meets a standard $\mu_0$ .

The main statistic to test is

z=\frac{\bar{x}-\mu_0}{\sigma/\sqrt{n}}

Hypothesis	$p$ -value	Decision Rule, Reject $H_0$ if …
$H_0:\mu=\mu_0$ vs. $H_1:\mu>\mu_0$	$\mathbb{P}(Z\gt z)$	$z>z_{\alpha}$
$H_0:\mu\le\mu_0$ vs. $H_1:\mu>\mu_0$	$\mathbb{P}(Z\gt z)$	$z>z_{\alpha}$
$H_0:\mu\ge\mu_0$ vs. $H_1:\mu\lt\mu_0$	$\mathbb{P}(Z\lt z)$	$z\lt -z_{\alpha}$
$H_0:\mu=\mu_0$ vs. $H_1:\mu\ne\mu_0$	$\mathbb{P}(\vert Z\vert \gt \vert z\vert)$	$\vert z\vert>z_{ \alpha /2}$

Assessing the Power of a Test

Let $H_0:\mu=\mu_0, H_1:\mu\gt\mu_0$ , fix $\mu^\ast\gt\mu_0$ , we have

\begin{aligned}\beta(\mu^\ast) &= \mathbb{P}(\bar{x} - \bar{x}_c | \mu^\ast) \\&=\mathbb{P}\Big( \frac{\bar{x}-\mu^\ast}{\sigma / \sqrt{n}} \lt \frac{\bar{x}_c - \mu^\ast}{\sigma/\sqrt{n}} \Big\vert \mu^\ast \Big)\\&=\mathbb{P} \Big( Z\lt \frac{\bar{x}_c-\mu^\ast}{\sigma/\sqrt{n}} \Big)\\&= \Phi\Big( \frac{\bar{x}_c-\mu^\ast}{\sigma/\sqrt{n}} \Big)\end{aligned}

Thus power $\pi(\mu^\ast)=1-\beta(\mu^\ast)=\Phi(\frac{\mu^\ast - \bar{x}_c}{\sigma/\sqrt{n}})$

Confidence Interval

Confidence interval covers $\mu$ with $1-\alpha$ is

[\bar{x}-z_{\alpha/2}\frac{\sigma}{\sqrt{n}}, \bar{x}+z_{\alpha/2}\frac{\sigma}{\sqrt{n}}]