Hypothesis testing: Finding critical values

In a nutshell

Critical values for a hypothesis test are values that form the boundaries of a critical region. A critical region is the region where you reject the null hypothesis.

Definitions

Critical region	The region of the sample space where the probability of the test statistic is less than the significance level. If the test statistic lies in the critical region for a particular hypothesis test, then reject the null hypothesis.
critical value(s)	The highest or lowest value that lies in the critical region.
acceptance region	The region of the sample space that is not part of the critical region.

Critical value for a one-tailed test

A one-tailed test has only one critical region, and hence it only has one critical value.

procedure

1.	Assume the null hypothesis to be true.
2.	Call the critical value $c$ . The critical region is either $X\leq c$ or $X\geq c$ , depending on the alternative hypothesis.
3.	Test values of $c$ , working out $P(X\leq c)$ or $P(X\geq c)$ for each value.
4.	The correct value of $c$ is the lowest or highest possible $c$ , such that the value $P(X\leq c)$ or $P(X\geq c)$ is less than the significance level.

Example 1

A casino is suspected of using weighted dice. A particular die is tested for whether or not it is biased towards landing on $1$ . The die is to be rolled $30$ times and the number of times it lands on $1$ is recorded. Find the critical region using a $5\%$ level of significance.

First, write down the test statistic and hypotheses:

Let $X$ be the number of times the die lands on $1$ .

$X\sim B(30,p)$

$H_0:p=\dfrac{1}{6}$

$H_1:p\gt \dfrac{1}{6}$

Assume the null hypothesis to be true:

$p=\dfrac{1}{6}\Rightarrow X\sim B(30,\dfrac{1}{6})$

Write down the correct critical region:

Since the inequality sign of the alternative hypothesis is $\gt$ , the critical region is $X\geq c$ , for the critical value $c$ .

Test different values of $c$ until the correct value has been found:

You want the lowest value of $c$ such that $P(X \geq c) <0.05$ .

It's very tedious to calculate binomial probabilities of the form $P(X\ge c)$ , so rearrange the inequality to obtain a probability of the form $P(X\le \_)$ :

$P(X\ge c)<0.05\\1-P(X\le c-1)<0.05\\P(X\le c-1)>0.95$

So, keep testing various numbers until you find the smallest one that yields a probability greater than $0.95$ :

$P(X\le10)=0.9933$ .

This is too high, so go lower:

$P(X\le9)=0.9803\\P(X\le6)=0.7765\\P(X\le7)=0.8863\\P(X\le8)=0.9494$ 

Therefore, the smallest probability that satisfies the inequality is $P(X\le9)$ . This corresponds to $P(X\le c-1)$ . Comparing these two gives:

$c-1=9\\c=10$ 

The critical region using a $5\%$ level of significance is $\underline{X\ge 10}$ .

Note: To interpret this contextually, this means that if the die lands on $1$ ten times or more, then there is sufficient evidence to reject the null hypothesis. To the $5\%$ significance level, the die will have been concluded to be biased towards $1$ .

Critical value for a two-tailed test

A two-tailed test has two critical regions: one at each end.

procedure

1.	Assume the null hypothesis to be true.
2.	There will be two critical values: call them $c_1$ and $c_2$ . The critical region is therefore $X\le c_1$ and $X\ge c_2$ .
3.	Focus on each end of the critical region individually. Make sure that $c_1$ and $c_2$ are chosen such that both probabilities are less than half of the significance level.

Example 2

A casino is suspected of using weighted dice. A particular die is tested for whether or not it is biased towards or against landing on $1$ . The die is to be rolled $30$ times and the number of times it lands on $1$ is recorded. Find the critical region using a $5\%$ level of significance.

Write down the test statistic and hypotheses:

Let $X$ be the number of times the die lands on $1$ .

$X\sim B(30,p)$

$H_0:p=\dfrac{1}{6}$

$H_1:p\neq\dfrac{1}{6}$

Assume the null hypothesis to be true:

$p=\dfrac{1}{6}\Rightarrow X\sim B(30,\dfrac{1}{6})$

Write down and define the critical region:

Let the critical values be $c_1$ and $c_2$ . The critical region is $X\leq c_1$ and $X\ge c_2$ .

First, focus on the 'lower tail' - the region $X\le c_1$ :

You want the greatest value of $c_1$ such that $P(X\le c_1)<0.05\div2=0.025$ .

$P(X\le 1)=0.0295\\P(X\le0)=P(X=0)=0.0042$

Therefore, $c_2=0$

Then, focus on the 'upper tail' - the region $X\ge c_2$ :

Want the smallest value of $c_2$ such that $P(X\ge c_2)<0.025$ .

Rearranging this gives $P(X\le c-1)>0.0975$ .

$P(X\le9)=0.9803\\P(X\le8)=0.9494$

Therefore, $c_2-1=9\Rightarrow c_2=10$ .

Write down the critical region using the values of $c_1$ and $c_2$ :

The critical region is $\underline{X=0}$ and $\underline{X\ge10}$ .

Note: Again, this means that if the die lands on $1$ either zero times or ten or more times out of $30$ tosses, then it will be judged to be biased to the $5\%$ significance level.

Actual significance level

The actual significance level is the probability of the critical region.

The actual significance level is also the probability of incorrectly rejecting the null hypothesis.

Example 3

Find the probability of incorrectly rejecting the null hypothesis in the above example.

The probability of incorrectly rejecting the null hypothesis is the actual significance level, which is the probability of the critical region. This is calculated to be:

$P(X=0)+P(X\ge10)=0.0042+0.0197=0.0239$

The probability of incorrectly rejecting the null hypothesis is $\underline{0.0239}$ , or $\underline{2.39\%}$ .