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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)?

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  • $\begingroup$ By critical value do you mean the level of significance $\alpha$? The level of significance represents the probability of a Type I Error which you can't exactly choose in your scenario since your rejection region has already been established. As you already said, $$\alpha=\mathbb{P}(\text{Reject } H_0|H_0 \text{ True})=\mathbb{P}(\bar{X}>66.5|\mu=65)=\mathbb{P}(Z>2.5)\approx 0.62\text{%}$$ $\endgroup$
    – user801306
    Commented Sep 15, 2020 at 18:07
  • $\begingroup$ Hello Matthew, thank you for your answer. By critical value, I mean the value of the test statistic, in this case, X bar, which separates the acceptance region from the rejection region. The 66.5 seen in the question above would be an example of a critical value, but that is not the correct answer to the question "On which critical value should he decide for the sample mean if he wants P(Type I error)=P(Type II error)?" since alpha and beta are not the same when x bar is 66.5 (see exact values above)... $\endgroup$ Commented Sep 15, 2020 at 18:21
  • $\begingroup$ Wait... so the decision rule the investigator is employing isn't constant and is subject to change? The way you worded it in the problem makes it seem like those are the rules the investigator is sticking to. $\endgroup$
    – user801306
    Commented Sep 15, 2020 at 18:36
  • $\begingroup$ The question may seem a little odd in its present form because originally it was divided into three parts. For simplicity, I provided the answers to the first two parts in the question. In the first two parts, we are asked to find the probability of the Type I and II errors when the investigator uses the original criteria seen next to the hypotheses (critical value is 66.5). In the last part he wants to know what critical value he should choose such that the probability of the two errors equal. Therefore, should we be solving for x bar and not Mu_a in your calculations below? $\endgroup$ Commented Sep 15, 2020 at 19:18
  • $\begingroup$ The probability of a Type II Error (which is usually denoted by $\beta$) depends on the value of $\mu_a$. If we want $\alpha = \beta$ are we to assume that $\mu_a=67.9$? $\endgroup$
    – user801306
    Commented Sep 15, 2020 at 19:24

2 Answers 2

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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 $$\text{Reject } H_0 \iff \bar{X}\in (66.45,\infty)$$ $$\text{Don't Reject } H_0 \iff \bar{X}\in (-\infty, 66.45]$$

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Suppose $\mu = \mu_a \neq 65$ is the true population mean. Then

$$P(\text{Type II Error})=P(\bar{X} <66.5|\mu = \mu_a)=P\Bigg(Z<\frac{66.5-\mu_a}{\sqrt{36/100}}\Bigg)=\phi\Bigg(\frac{66.5-\mu_a}{\sqrt{36/100}}\Bigg)$$ where $Z \sim N(0,1)$ and $\phi(x):=\int_{-\infty}^{x}\frac{1}{\sqrt{2\pi}}e^{-t^2/2}dt$. In order to find the value of $\mu_a$ so that we get $P(\text{Type I Error})=P(\text{Type II Error})$ we need to solve the equation $$\phi\bigg(\frac{66.5-\mu_a}{\sqrt{36/100}}\bigg)=0.0062097$$ for $\mu_a$. We get $$\frac{66.5-\mu_a}{\sqrt{36/100}}=\phi^{-1}(0.0062097)=-2.5\iff \mu_a=68$$

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