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Draw a conclusion and interpret the decision.

The problem describes a hypothesis test for a paired sample, as we're comparing the number of times participants lost their temper before and after an anger-management course. The differences between the two periods (before and after the course) are calculated as \( d = (\text{after} - \text{before}) \). To draw a conclusion, we'll follow these steps:

### Step 1: Hypotheses
- **Null hypothesis (\( H_0 \)):** There is no significant difference in temper loss before and after the anger-management course, meaning the average difference \( \mu_d \) between before and after is zero.
  \[
  H_0: \mu_d = 0
  \]
- **Alternative hypothesis (\( H_1 \)):** The anger-management course reduces temper loss, so the average difference \( \mu_d \) is less than zero.
  \[
  H_1: \mu_d < 0
  \]

### Step 2: Data and Paired Differences
We will calculate the differences \( d \) for each participant and use them to compute the test statistic for a paired \( t \)-test. The differences are:

\[
d = (10-12), (5-4), (9-11), (9-5), (5-7), (6-3), (3-4), (11-10), (9-8), (12-11), (4-12), (4-8)
\]

This simplifies to:
\[
d = -2, 1, -2, 4, -2, 3, -1, 1, 1, 1, -8, -4
\]

### Step 3: Perform Paired \( t \)-Test
Next, we calculate the mean \( \overline{d} \), the standard deviation \( s_d \), and use the paired \( t \)-test formula to find the test statistic:
\[
t = \frac{\overline{d}}{s_d / \sqrt{n}}
\]
where \( n = 12 \) (the number of pairs).

Let me compute this step.### Step 4: Results of the Paired \( t \)-Test
- Mean of the differences \( \overline{d} = -0.67 \)
- Standard deviation of the differences \( s_d = 3.26 \)
- Test statistic \( t = -0.71 \)
- \( p \)-value for the one-tailed test \( p = 0.247 \)

### Step 5: Conclusion
At the 0.05 significance level, the \( p \)-value of 0.247 is greater than 0.05. Therefore, **we fail to reject the null hypothesis**.

### Interpretation
There is insufficient evidence to support the claim that the anger-management course significantly reduces the number of times participants lose their temper. The reduction observed in the number of temper losses after the course might be due to random chance rather than a systematic effect of the course.

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Would you like further details or have any questions?

Here are some related questions to consider:
1. How can the power of this test be improved?
2. What are the assumptions of a paired \( t \)-test?
3. What is the difference between a one-tailed and two-tailed test?
4. How would the conclusion change if the significance level was 0.10 instead of 0.05?
5. Can a non-parametric test be used for this data?

**Tip:** Always check the normality assumption of the differences when performing a paired \( t \)-test.

This problem tests the effectiveness of an anger-management course using a paired t-test. We examine if participants lose their temper less frequently after completing the course by comparing data before and after the course. The significance level is 0.05.

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