I'm working through Intro to Probability and Statistics... and I'm a little confused about how to calculate the probability of a type II error, particularly as it relates to a simple hypothesis test, and would appreciate some clarification on this.

To explain my confusion, in 8.4.3 of the book, he gives example 8.2.4:

Let $X1,X2,...,Xn$ be a random sample from a $N(\mu,\sigma^2)$ distribution, where $\mu$ is unknown but $\sigma$ is known. Design a level $\alpha$ test to choose between

$H_0: \mu=\mu_0$,

$H_1: \mu \neq \mu_0$.

Which I understand completely. However, in 8.2.5 (just below), he asks:

For the above example (Example 8.24), find $\beta$, the probability of type II error, as a function of $\mu$.

Providing the solution:

\begin{align} \beta (\mu) &=P(\textrm{type II error}) = P(\textrm{accept }H_0 \; | \; \mu) \\ &= P\left(\left|\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}} \right| \lt z_{\frac{\alpha}{2}}\; | \; \mu \right). \end{align}

If $X_i \sim N(\mu,\sigma^2)$, then $\overline{X} \sim N(\mu, \frac{\sigma^2}{n})$. Thus,

\begin{align} \beta (\mu)&=P\left(\left|\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}} \right| \lt z_{\frac{\alpha}{2}}\; | \; \mu \right)\\ &=P\left(\mu_0- z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}} \leq \overline{X} \leq \mu_0+ z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}}\right)\\ &=\Phi\left(z_{\frac{\alpha}{2}}+\frac{\mu_0-\mu}{\sigma / \sqrt{n}}\right)-\Phi\left(-z_{\frac{\alpha}{2}}+\frac{\mu_0-\mu}{\sigma / \sqrt{n}}\right). \end{align}

I follow this until the end, where he jumps from:

\begin{align} &=P\left(\mu_0- z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}} \leq \overline{X} \leq \mu_0+ z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}}\right)\\ \end{align}


\begin{align} &=\Phi\left(z_{\frac{\alpha}{2}}+\frac{\mu_0-\mu}{\sigma / \sqrt{n}}\right)-\Phi\left(-z_{\frac{\alpha}{2}}+\frac{\mu_0-\mu}{\sigma / \sqrt{n}}\right). \end{align}

Further, since the alternative hypothesis is simply $\mu_1 \neq 2$, how exactly would one go about calculating this probability? What value of $\mu$ would be used here?

Would someone be able to clarify this, and elaborate on how to calculate the probability of a type II error given a simple hypothesis, such as this?


1 Answer 1


In $\Pr\left(\mu_0- z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}} \leq \overline{X} \leq \mu_0+ z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}}\right)$, $\bar X\sim N(\mu,\frac {\sigma^2}n)$, instead of standard normal distribution. We need to transform $\bar X$ into standard normal distribution by subtracting the mean and dividing by standard deviation.

$\Pr\left(\mu_0- z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}} \leq \overline{X} \leq \mu_0+ z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}}\right)$ = $\Pr\left(z_{\frac{\alpha}{2}}+\frac{\mu_0-\mu}{\sigma / \sqrt{n}} \leq \frac{\overline{X}-\mu}{\sigma/\sqrt n} \leq -z_{\frac{\alpha}{2}}+\frac{\mu_0-\mu}{\sigma / \sqrt{n}}\right)$

Now $\frac{\overline{X}-\mu}{\sigma/\sqrt n} \sim N(0,1)$, so you can get \begin{align} &=\Phi\left(-z_{\frac{\alpha}{2}}+\frac{\mu_0-\mu}{\sigma / \sqrt{n}}\right)-\Phi\left(z_{\frac{\alpha}{2}}+\frac{\mu_0-\mu}{\sigma / \sqrt{n}}\right). \end{align} where $\Phi$ is CDF of standard normal distribution.

For the question aout the value of $\mu$, as $\beta(\mu)$ indicates that $\beta$ is the function of $\mu$. So when you plug in the different values of $\mu$, you get the different probability of type II error $(\beta)$.

  • $\begingroup$ Makes perfect sense. Thank you for a clear and succinct answer. $\endgroup$ Nov 30, 2018 at 23:19

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