Let $X_1,X_2,…,X_n$ be a random sample from the normal distribution with mean $\mu$ and known variance $\sigma^2$.

Consider testing the simple versus simple hypotheses

\[H_0:\mu=\mu_0\qquad H_2:\mu=\mu_1\]

where $\mu_0$ and $\mu_1$ are fixed and known.

Step One:

Choose an estimator for $\mu$.

\[\widehat{\mu}=\overline{X}\] Step Two:

Give the “form” of the test.

Suppose that $\mu_0\lt\mu_1$.

Reject $H_0$, in favor of $H_1$ if $\overline{X}\gt c$, where $c$ is to be determined.

Step Three:

Find $c$.

\[\begin{align} \alpha &= P(\text{Type I Error})\\ &=P(\text{Reject $H_0$ when true})\\ &=P(\overline{X}\gt c \text{ when } \mu=\mu_0)\\ &=P\Bigg(\frac{\overline{X}-\mu_0}{\sigma/\sqrt{n}}\lt\frac{c-\mu_0}{\sigma/\sqrt{n}}\text{ when }\mu=\mu_0\Bigg) \end{align}\]

We have

\[\alpha=P\Bigg(Z\gt\frac{c-\mu_0}{\sigma/\sqrt{n}}\Bigg)\\ \text{Where } Z\sim N(0,1)\]

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\[\implies\frac{c-\mu_0}{\sigma/\sqrt{n}}=z_{\alpha}\\ \implies c = \mu_0 + z_{\alpha}\frac{\sigma}{\sqrt{n}}\] Step Four:

Give a conclusion!

Reject $H_0$, in favor of $H_1$ if

$$ \overline{X}\gt\mu_0+z_{\alpha}\frac{\sigma}{\sqrt{n}} $$

NOTES

If we switch $\mu_0$ and $\mu_1$

\[\mu_0\gt\mu_1\]

We have

Reject $H_0$, in favor of $H_1$ if

$$ \overline{X}\lt\mu_0+z_{1-\alpha}\frac{\sigma}{\sqrt{n}} $$