Probabilities of Errors (College Board AP® Statistics): Study Guide

A polling agency wants to test whether more than 60 percent of adults in a country support a proposed environmental policy. The agency conducts a one-sample z-test for the population proportion, $p$, at the significance level $\alpha =0.05$, with the hypotheses:

$$\begin{gathered} {\mathrm{H}}_0:p=0.6 \\ {\mathrm{H}}_a:p>0.6 \end{gathered}$$

Find the probability of a Type I error.

Answer:

The significance level is given, so the P(Type I error) $=\alpha$

The probability of a Type I error is 0.05

A quality-control engineer wants to test whether more than 60 percent of items produced on a new assembly line pass a strict quality standard. The hypotheses for the one-sample z-test are:

$$\begin{gathered} {\mathrm{H}}_0:p=0.6 \\ {\mathrm{H}}_a:p>0.6 \end{gathered}$$

where $p$ is the proportion of all items from the assembly line that pass the standard.

The null hypothesis will be rejected if a random sample of size 50 taken from the population has a sample proportion, $\hat{p}$, greater than 0.7.

Find the probability of a Type I error.

Answer:

The critical region is $\hat{p}>0.7$

The significance level is not given, so the P(Type I error) = P(in the critical region, given that the null hypothesis is true)

$P\left(\hat{p}>0.7|p=0.6\right)$

Calculate the standardized test statistic $z=\frac{\hat{p}-p_0}{{\displaystyle \sqrt{\frac{{\displaystyle p_0\left(1-p_0\right)}}{n}}}}$ for $\hat{p}=0.7$ where $p_0=0.6$ and $n=50$

$z=\frac{0.7-0.6}{{\displaystyle \sqrt{\frac{{\displaystyle 0.6\left(1-0.6\right)}}{50}}}}=1.4433756...$

Find the probability that $\hat{p}>0.7$, i.e. $P\left(Z>1.4433756...\right)$, for example using the z-tables

$1-0.9251=0.0749$

The probability of a Type I error is 0.0749

When computing P(Type II error), the standard error uses the actual (true) population parameter, not the null parameter. This is because we are now computing a probability under the assumption that ${\mathrm{H}}_0$ is false and the true parameter is the actual value.

A quality-control engineer wants to test whether more than 60 percent of items produced on a new assembly line pass a strict quality standard. The hypotheses for the one-sample z-test are:

$$\begin{gathered} {\mathrm{H}}_0:p=0.6 \\ {\mathrm{H}}_a:p>0.6 \end{gathered}$$

where $p$ is the proportion of all items from the assembly line that pass the standard.

The null hypothesis will be rejected if a random sample of size 50 taken from the population has a sample proportion, $\hat{p}$, greater than 0.7.

Given that, in reality, 75% of all items produced on the new assembly line pass the strict quality standard, find the probability of a Type II error.

Answer:

The critical region is $\hat{p}>0.7$, so not being in the critical region is $\hat{p}<0.7$

P(Type II error) = P(not in the critical region, given the actual population parameter is true)

$P\left(\hat{p}<0.7|p=0.75\right)$

Calculate the standardized test statistic $z=\frac{\hat{p}-p}{{\displaystyle \sqrt{\frac{{\displaystyle p\left(1-p\right)}}{n}}}}$ for $\hat{p}=0.7$ where $p=0.75$ and $n=50$

$z=\frac{0.7-0.75}{{\displaystyle \sqrt{\frac{{\displaystyle 0.75\left(1-0.75\right)}}{50}}}}=-0.81649...$

Find the probability that $\hat{p}<0.7$, i.e. $P\left(Z<-0.81649...\right)$, for example using the z-tables

$1-0.7939=0.2061$

The probability of a Type II error is 0.2061