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Consider a population with the pdf $N(\theta,1)$ where $\theta$ is unknown and hypothesis $H_0:\theta=5.5$ and $H_1:\theta=8$.Suppose that $\bar{X}=\frac{1}{9}\sum_{i=1}^9 X_i.$Reject $H_0$ iff $\bar{X} >7.5$.Power function is given by $Q(\theta)=P_{\theta }(\cal R).$ $\beta=P{\text{(type-II error)}}=P({\text{Accepting } H_0}$ when $H_1 \text{is true}).$

Why then $Q(\theta_1)=1-\beta$ and $Q(\theta)=1-\Phi(22.5-3\theta)?$

In particular how arises $3$ in front of the $\theta$?

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  • $\begingroup$ What $\sum_{i=1}^\theta X_i$ stnd for? $\endgroup$ – Deep North Feb 4 '18 at 22:26
  • $\begingroup$ See the corrected question. A random sample $(X_1,...,X_9)$ is collected and denote $\bar{X}=\frac{1}{9}\sum_{i=1}^9 X_i$ $\endgroup$ – user2925716 Feb 5 '18 at 15:32
  • $\begingroup$ @DeepNorth Can you answer now, after the clarification? $\endgroup$ – user2925716 Feb 5 '18 at 20:24
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    $\begingroup$ Hint: Find the distribution of $\bar X$. (This will show you where the $3$ comes from.) $\endgroup$ – whuber Feb 5 '18 at 21:17
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The power is defined as

$\gamma(\theta_1)$ or $Q(\theta_1)=P_{\theta_1}[(X_1,X_2,...,X_n)\in C]$

$(X_1,X_2,...,X_n)$ is sample space and $C$ is critical region, $\theta_1$ is the parameter under the $H_1$.

Intuitively, you can think the decivision rule as

$$Q(\theta_1)=P_{\theta_1}(S<k) \text{ or } Q(\theta_1)=P_{\theta_1}(S>k)$$

where $S$ is a statstics from the sample and $k$ is determined by type I error $\alpha$ i.e. the significant level and

$$\alpha=P_{\theta_0}[S<k] \text{ or } \alpha=P_{\theta_0}[S>k]$$

From your information you can calculate both the power and type I error $\alpha$

$Power=Q(\theta_1)=P_{\theta_1}(\bar{X}>7.5)=P_{\theta_1}(\frac{\bar{X}-\theta_1}{1/3}>\frac{7.5-\theta_1}{1/3})=1-\Phi(22.5-3\theta_1)=0.9331928$

$\theta_1=8$ here.

Accordingly, you also can calculate your type I error $\alpha$

Note from whuber's hint.

$\bar{X}$ has a $ N(\theta,\frac{\sigma^2}{n}) $distribution

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  • $\begingroup$ What is $k$ here? How can it be evaluated?How can I see that $k$ is 7.5 in my case? $\endgroup$ – user2925716 Feb 6 '18 at 13:46
  • $\begingroup$ Yes, it is 7.5 in your case, usually, you have the significant level first, then you calculate $k$ by yourself, here they already gave you the vaule $k$ for the critical region. $\endgroup$ – Deep North Feb 7 '18 at 22:53
  • $\begingroup$ But how can I see that $7.5$ is the correct value for my critical region? $\endgroup$ – user2925716 Feb 8 '18 at 14:14
  • $\begingroup$ Because it says Rject $H_0$ iff $\bar{X}>7.5$ $\endgroup$ – Deep North Feb 8 '18 at 22:12
  • $\begingroup$ Yes. And how do I decide which one of the following shall I use: $$\alpha=P_{\theta_0}[S<k] \text{ or } \alpha=P_{\theta_0}[S>k] ?$$ I.e. when $\alpha$ is given. $\endgroup$ – user2925716 Feb 10 '18 at 14:29

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