Probability of the error of type 2

• MHB
• mathmari
In summary, the conversation discusses a test with a null hypothesis that the mean is less than or equal to 100 and an alternative hypothesis that the mean is greater than 100. The sample has a standard deviation of 20 and the real mean is 102. The significance level is 0.1 and the probability of error of type II is calculated to be 0.6103. The conversation also clarifies the difference between accepting the null hypothesis and keeping it.
mathmari
Gold Member
MHB
Hey!

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)

mathmari said:
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)

mathmari said:
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)

Klaas van Aarsen said:
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)

mathmari said:
The error of type II is to accept the null hypothesis although it is wrong, isn't it?

Correct. (Nod)

mathmari said:
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)

Klaas van Aarsen said:
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*}

mathmari said:
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)

Klaas van Aarsen said:
Yep. (Nod)

Great! Thanks a lot! (Mmm)

1. What is the probability of the error of type 2?

The probability of the error of type 2, also known as beta (β), is the likelihood of accepting a null hypothesis when it is actually false. In other words, it is the probability of failing to reject a false null hypothesis.

2. How is the probability of the error of type 2 calculated?

The probability of the error of type 2 is calculated by subtracting the power (1-α) from 1. This can also be represented as β = 1 - power.

3. What factors can affect the probability of the error of type 2?

The probability of the error of type 2 can be affected by the sample size, the size of the effect being measured, and the level of significance (α) chosen for the test. A larger sample size and a larger effect size can decrease the probability of type 2 error, while a smaller α can increase it.

4. How does the probability of the error of type 2 relate to the probability of the error of type 1?

The probability of the error of type 2 and the probability of the error of type 1 (α) are inversely related. This means that as the probability of type 2 error decreases, the probability of type 1 error increases, and vice versa.

5. Why is it important to consider the probability of the error of type 2 in statistical analysis?

It is important to consider the probability of the error of type 2 in statistical analysis because it allows us to understand the potential errors in our conclusions. If the probability of type 2 error is high, it means that there is a high chance of accepting a false null hypothesis, which can lead to incorrect conclusions and decisions.

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