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- Apr 14, 2013

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We have data of a sample of $100$ people from a population with standard deviation $\sigma=20$.

We consider the following test: \begin{align*}H_0 : \ \mu\leq 100 \\ H_1 : \ \mu>100\end{align*}

The real mean is $\mu=102$ and the significance level is $\alpha=0.1$.

I want to calculate the probability of the error of type 2.

I have done the following:

The statistic function is: \begin{equation*}Z=\frac{\overline{X}-\mu}{\sigma_{\overline{X}}}=\frac{\overline{X}-\mu}{\frac{\sigma}{\sqrt{n}}}=\frac{\overline{X}-100}{\frac{20}{\sqrt{100}}}=\frac{\overline{X}-100}{\frac{20}{10}}=\frac{\overline{X}-100}{2}\end{equation*}

where $\overline{X}$ is the estimation of $\mu$.

For the significance level $\alpha=0.1$ the critical value is $Z_{c} = 1.28$ and the region of rejection of $Η_0$ is $R = \{Z\mid Z > 1.28\}$.

The critical value $Z_{c}$ corresponds to a critical value $\overline{X}_{c}$ such that \begin{equation*}P\left (Z>1.28\right )=P\left (\overline{X}>\overline{X}_c \mid \mu=100 , \sigma_{\overline{X}}=2\right ) =1-\alpha=0.9\end{equation*}

We can find the value of $\overline{X}_c$ solving th following equation: \begin{equation*}Z_c=1.28 \Rightarrow \frac{\overline{X}_c-100}{2}=1.28 \Rightarrow \overline{X}_c-100=2.56 \Rightarrow \overline{X}_c=102.56\end{equation*}

So incorrectly we fail to reject the null hypothesis if we take a sample mean greater than $102.56$.

The probability to take a sample mean greater than $102.56$ given $\mu=102$ and $\sigma_{\overline{X}}=2$, i.e. the probability of error of type II is \begin{align*}P\left (\overline{X}>102.56\mid \mu=102, \sigma_{\overline{X}}=2\right )&=P\left (Z>\frac{102.56-102}{2}\right )=P\left (Z>\frac{0.56}{2}\right )=P\left (Z>0.28\right )\\ & =1-P\left (Z\leq 0.28\right )=1-0.6103=0.3897\end{align*}

Is everything correct?