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To draw a conclusion from this hypothesis testing question, we will follow these steps:

### 1. **State the Hypotheses:**
   - **Null Hypothesis \( H_0 \):** The anger-management course has no effect on reducing the number of times participants lose their temper. Mathematically, \( H_0: \mu_d = 0 \), where \( \mu_d \) is the mean difference between the number of temper losses before and after the course.
   - **Alternative Hypothesis \( H_a \):** The anger-management course reduces the number of temper losses. Mathematically, \( H_a: \mu_d > 0 \).

### 2. **Calculate the Differences \( d \):**
   The differences \( d = \text{Before} - \text{After} \) for each participant are calculated as:

   \[
   d = \begin{bmatrix}
   12 - 10 = 2, & 4 - 5 = -1, & 11 - 9 = 2, & 4 - 3 = 1, & 5 - 5 = 0, & 7 - 6 = 1, & 3 - 3 = 0, \\
   4 - 11 = -7, & 10 - 9 = 1, & 8 - 12 = -4, & 11 - 4 = 7, & 8 - 4 = 4
   \end{bmatrix}
   \]

   So, the differences \( d \) are: \( [2, -1, 2, 1, 0, 1, 0, -7, 1, -4, 7, 4] \).

### 3. **Compute the Sample Statistics:**
   - **Mean of the differences \( \bar{d} \):**

   \[
   \bar{d} = \frac{2 + (-1) + 2 + 1 + 0 + 1 + 0 + (-7) + 1 + (-4) + 7 + 4}{12}
   \]

   Let's compute it.

### 4. **Perform the Hypothesis Test:**
   - Using a **paired t-test** for this data at the 0.05 level of significance.

Let me calculate the sample mean \( \bar{d} \), standard deviation, and t-statistic to conclude the test.### 3. **Test Results:**
- **Mean of the differences \( \bar{d} \):** 0.5
- **Standard deviation of the differences \( s_d \):** 3.55
- **Calculated t-statistic:** 0.487
- **Critical t-value (one-tailed, \( \alpha = 0.05 \)):** 1.796
- **p-value:** 0.318

### 4. **Conclusion:**
Since the calculated t-statistic (0.487) is less than the critical t-value (1.796) and the p-value (0.318) is greater than the significance level of 0.05, we **fail to reject the null hypothesis**.

### Interpretation:
There is not enough evidence to support the claim that the anger-management course significantly reduces the number of times participants lose their temper.

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Would you like to go over any specific details, or have questions about the steps? Here are some related questions to explore:

1. What assumptions are necessary for a paired t-test?
2. How is the p-value interpreted in hypothesis testing?
3. What are alternative methods to test this hypothesis, such as a non-parametric test?
4. What could affect the effectiveness of the anger-management course in this study?
5. How might increasing the sample size affect the results?

**Tip:** When conducting hypothesis tests, always check if the assumptions (like normality and independence) are satisfied for accurate results.

An anger-management course claims that, after completing its seminar, participants will lose their tempers less often. A random sample of 12 seminar participants is chosen, and these participants are asked to record the number of times that they lost their tempers in the two weeks prior to the course. After the course is over, the same participants are asked to record the number of times that they lost their tempers in the next two weeks. Using these data, test the claim at the 0.05 level of significance assuming that the population distribution of the paired differences is approximately normal. Let d = (participants after completing the anger-management course) – (participants before completing the anger-management course).

Test whether an anger-management course significantly reduces temper loss using a paired t-test at the 0.05 significance level. This analysis compares the number of temper losses before and after the course, calculates the mean difference, and uses the t-statistic for hypothesis testing. Includes detailed calculations and interpretation.

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