Welcome to our community

Be a part of something great, join today!

[SOLVED] Probability of the error of type 2

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,003
Hey!! :eek:

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? (Wondering)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,684
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 $102.56$.
Hey mathmari !!

If we fail to reject the null hypothesis, we keep the null hypothesis don't we?
Isn't that the case if we find a sample mean less than $102.56$? (Wondering)

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?
I believe you have calculated the so called Power instead of the Type II error. (Worried)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,003
If we fail to reject the null hypothesis, we keep the null hypothesis don't we?
Isn't that the case if we find a sample mean less than $102.56$? (Wondering)
The error of type II is to accept the null hypothesis although it is wrong, isn't it?

I got stuck right now. We found the critival $\overline{X}$-value to be $102.56$ which is greater than $100$ and so we would accept the hypothesis $H_1$, or not? (Wondering)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,684
The error of type II is to accept the null hypothesis although it is wrong, isn't it?
Correct. (Nod)

I got stuck right now. We found the critival $\overline{X}$-value to be $102.56$ which is greater than $100$ and so we would accept the hypothesis $H_1$, or not?
Not quite.

Let's denote the critical $\overline{X}$-value as $\overline{X_0}^c$ to avoid confusion.
Note that $\overline{X_0}^c$ is calculated based on the null hypothesis for a certain significance level and sample size.
Now if we take a sample $x$ and its mean $\overline x$ is greater than $\overline{X_0}^c$, then we accept the alternative hypothesis, don't we?
And if $\overline x$ is less than $\overline{X_0}^c$, then we keep the null hypothesis, don't we? (Thinking)

The Type II Error is the probability that a sample follows the alternative distribution, but has a mean so close to the null hypothesis that we keep the null hypothesis even though it is wrong.
In our case:
$$\beta = \text{Type II Error} = P(\overline X < \overline{X_0}^c \mid \mu = \mu_1 = 102)$$
where $\overline{X_0}^c$ is calculated based on the null hypothesis with $\mu=\mu_0=100$. (Thinking)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,003
Let's denote the critical $\overline{X}$-value as $\overline{X}^c$ to avoid confusion.
Now if we take a sample $x$ and its mean $\overline x$ is greater than $\overline{X}^c$, then we accept the alternative hypothesis, don't we?
And if $\overline x$ is less than $\overline{X}^c$, then we keep the null hypothesis, don't we? (Thinking)

The Type II Error is the probability that a sample follows the alternative distribution, but has a mean so close to the null hypothesis that we keep the null hypothesis even though it is wrong.
In our case:
$$\beta = \text{Type II Error} = P(\overline X < \overline X^c \mid \mu = \mu_1 = 102)$$
where $\overline X^c$ is calculated based on the null hypothesis with $\mu=\mu_0=100$. (Thinking)

Ah ok!! So do we have the following? (Wondering)

\begin{equation*}\beta = P(\overline X < \overline X^c \mid \mu = \mu_1 = 102)=P\left (Z<\frac{102.56-102}{2}\right )=P\left (Z<\frac{0.56}{2}\right )=P\left (Z<0.28\right )=0.6103\end{equation*}
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,684
Ah ok!! So do we have the following?

\begin{equation*}\beta = P(\overline X < \overline X^c \mid \mu = \mu_1 = 102)=P\left (Z<\frac{102.56-102}{2}\right )=P\left (Z<\frac{0.56}{2}\right )=P\left (Z<0.28\right )=0.6103\end{equation*}
Yep. (Nod)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,003