This is a paired t-test problem. The goal is to test if an anger-management course effectively reduces the number of times participants lose their temper, using data from before and after the course. Let's go through the steps for calculating the test statistic.

Step-by-step process:

1. Differences ($d = \text{After} - \text{Before}$): We compute the difference for each participant between the number of temper losses before and after the course.

2. Mean of differences ($\overline{d}$): After finding the differences, calculate their mean.

3. Standard deviation of differences ($s_d$): Compute the standard deviation of the differences.

4. Test statistic $t$: The formula for the paired t-test is: $t = \frac{\overline{d}}{s_d / \sqrt{n}}$ where:

   • $\overline{d}$ is the mean of the differences
   • $s_d$ is the standard deviation of the differences
   • $n$ is the number of participants (12 in this case)

Data from the table:

Before:

$[8, 8, 8, 3, 7, 6, 4, 7, 3, 12, 8, 4]$

After:

$[3, 4, 3, 3, 6, 4, 4, 6, 3, 11, 4, 5]$

I'll calculate the necessary values next.The calculations give the following results:

• Mean of the differences ($\overline{d}$): 1.833
• Standard deviation of the differences ($s_d$): 2.125
• Test statistic $t$: 2.989

Thus, the test statistic is approximately 2.989.

Would you like more details on this solution or have any further questions? Here are some related questions:

1. How do we interpret the test statistic in the context of this problem?
2. What is the next step to determine if the anger-management course was effective?
3. How is the paired t-test different from other t-tests?
4. What are the assumptions of the paired t-test?
5. How can you calculate the p-value for this test statistic?

Tip: When conducting a hypothesis test, always clearly state the null and alternative hypotheses before starting calculations.