Probabilities of Errors (College Board AP® Statistics)

Revision Note

Author

Mark Curtis

Expertise

Maths

Probability of a Type I error

How do I find the probability of a Type I error if a significance level is given?

If the significance level of a test, $α$ , is known, then
- the probability of a Type I error is equal to the significance level, $α$
This is because the probability of a Type I error is
- the probability of rejecting $H_{0}$ given that $H_{0}$ is true
- which is the same definition as a significance level, $α$
  - $α$ is the preset probability at which you decide to reject $H_{0}$ , assuming it was true

Worked Example

A z-test for the population proportion, $p$ , at the significance level $α = 0.05$ has the hypotheses:

$H_{0} : p = 0.6 H_{a} : p > 0.6$

Find the probability of a Type I error.

Answer:

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

The probability of a Type I error is 0.05

How do I find the probability of a Type I error if a critical region is given?

If a significance level is not given, you may be given a critical region instead
- This is the range of values for the test statistic for which the null hypothesis is rejected
- The probability of a Type I error is the probability of being in the critical region (rejecting H₀) given H₀was true

Worked Example

A z-test for the population proportion, $p$ , has the hypotheses:

$H_{0} : p = 0.6 H_{a} : p > 0.6$

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)

$P (\hat{p} > 0.7 given that p = 0.6)$

Calculate the standardized test statistic $z = \frac{\hat{p} - p_{0}}{\sqrt{\frac{p_{0} (1 - p_{0})}{n}}}$ for $\hat{p} = 0.7$ where $p_{0} = 0.6$ and $n = 50$

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

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

$1 - 0.9251 = 0.0749$

The probability of a Type I error is 0.0749

Probability of a Type II error

How do I find the probability of a Type II error if a critical region is given?

Recall that a critical region is the range of values for the test statistic for which the null hypothesis is rejected
The probability of a Type II error is the probability of not rejecting the null hypothesis, despite it being false in reality
- This is the probability of not being in the critical region (not rejecting H₀) given H₀was false
  - You need to be given the actual (true) population parameter to find this
  - For example, H₀ assumed $p = \frac{1}{2}$ but actually $p = \frac{1}{3}$
  - P(Type II error) = P(not in the critical region, given the actual population parameter is true)

How do I reduce the probability of a Type II error?

The probability of a Type II error is reduced when one of the following is changed (and the others are kept the same):
- The sample size, $n$ , increases
- The significance level, $α$ , increases
- The standard error of the hypothesis test decreases
- The actual (true) population parameter is farther from the null population parameter

Because $α$ is also the probability of a Type I error, the above means:
- You can reduce the probability of a Type II error by increasing the significance level
  - However this will increase the probability of a Type I error
- You can reduce the probability of a Type I error by reducing the significance level
  - However this will increase the probability of a Type II error
- The only way to reduce both probabilities is by increasing the size of the sample

Worked Example

A z-test for the population proportion, $p$ , has the hypotheses:

$H_{0} : p = 0.6 H_{a} : p > 0.6$

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, the population proportion is actually $p = 0.75$ , 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)

$P (\hat{p} < 0.7 given that p = 0.75)$

Calculate the standardized test statistic $z = \frac{\hat{p} - p_{0}}{\sqrt{\frac{p_{0} (1 - p_{0})}{n}}}$ for $\hat{p} = 0.7$ where $p_{0} = 0.75$ and $n = 50$

$z = \frac{0.7 - 0.75}{\sqrt{\frac{0.6 (1 - 0.6)}{50}}} = - 0.721687 . . .$

Find the probability that $\hat{p} < 0.7$ , i.e. $P (Z < - 0.721687 . . .)$ , for example using the z-tables

0.2358

The probability of a Type II error is 0.2358

You've read 0 of your 10 free revision notes

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Next topic

Did this page help you?

Probabilities of Errors (College Board AP® Statistics)

Revision Note

Probability of a Type I error

How do I find the probability of a Type I error if a significance level is given?

Worked Example

How do I find the probability of a Type I error if a critical region is given?

Worked Example

Probability of a Type II error

How do I find the probability of a Type II error if a critical region is given?

How do I reduce the probability of a Type II error?

How are the probabilities of a Type I error and a Type II error related?

Worked Example

You've read 0 of your 10 free revision notes

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Mark Curtis