A sample of size $100$ is taken from a normal population with unknown mean µ and known variance $36$. An investigator wishes to test the hypotheses Null hypothesis: $µ=65$, Alternative hypothesis: $µ>65$. He decides on the following criteria:
Accept the Null hypothesis if the sample mean $x$ bar is smaller or equal to $66.5$.
Reject the Null hypothesis if $x$ bar is greater than $66.5$.
The probability that he makes a Type I error is $0.0062097...$
If he uses $µ=67.9$ for the alternative hypothesis, the probability that he makes a Type II error is $0.0098153...$
On which critical value should he decide for the sample mean if he wants P(Type I error)=P(Type II error)?
The test statistic for the hypothesis test $H_0:\mu =65$ versus $H_a:\mu>65$ is the $z-$score $Z=\frac{\bar{X}-65}{\sqrt{36/100}}$. We will reject $H_0$ if and only if $Z>z_{\alpha}$ if and only if $\bar{X}>z_{\alpha}\sqrt{36/100}+65$. Otherwise we won't reject $H_0$. So if $\mu=\mu_a \neq 65$ is the true population mean, then $$P(\text{Type II Error})=P(\bar{X} \leq z_{\alpha}\sqrt{36/100}+65|\mu=\mu_a)=P\bigg(Z<z_{\alpha}+\frac{65-\mu_a}{\sqrt{36/100}}\bigg)=\phi\Bigg(z_{\alpha}+\frac{65-\mu_a}{\sqrt{36/100}}\Bigg)$$ where $\phi(x)=\int_{-\infty}^{x}\frac{1}{\sqrt{2\pi}}e^{-t^2/2}dt$ and $Z\sim N(0,1)$. Similarly to before, $$\phi\Bigg(z_{\alpha}+\frac{65-\mu_a}{\sqrt{36/100}}\Bigg)=\alpha \iff z_{\alpha}+\frac{65-\mu_a}{\sqrt{36/100}}=-z_{\alpha}\iff 6z_{\alpha}=5\mu_a-325$$ Finally, if $\mu_a=67.9$, then $z_{\alpha}=29/12$ which induces the decision rule $$\text{Reject } H_0 \iff \bar{X}\in (66.45,\infty)$$ $$\text{Don't Reject } H_0 \iff \bar{X}\in (-\infty, 66.45]$$