0
$\begingroup$

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}

To:

\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?

$\endgroup$
0
$\begingroup$

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 $(\beta)$.

$\endgroup$
  • $\begingroup$ Makes perfect sense. Thank you for a clear and succinct answer. $\endgroup$ – this_name_here Nov 30 '18 at 23:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.