Math Problem Statement

Draw a conclusion and interpret the decision.

Draw a conclusion and interpret the decision.

Solution

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=(afterbefore)d = (\text{after} - \text{before}). To draw a conclusion, we'll follow these steps:

Step 1: Hypotheses

  • Null hypothesis (H0H_0): There is no significant difference in temper loss before and after the anger-management course, meaning the average difference μd\mu_d between before and after is zero. H0:μd=0H_0: \mu_d = 0
  • Alternative hypothesis (H1H_1): The anger-management course reduces temper loss, so the average difference μd\mu_d is less than zero. H1:μd<0H_1: \mu_d < 0

Step 2: Data and Paired Differences

We will calculate the differences dd for each participant and use them to compute the test statistic for a paired tt-test. The differences are:

d=(1012),(54),(911),(95),(57),(63),(34),(1110),(98),(1211),(412),(48)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=2,1,2,4,2,3,1,1,1,1,8,4d = -2, 1, -2, 4, -2, 3, -1, 1, 1, 1, -8, -4

Step 3: Perform Paired tt-Test

Next, we calculate the mean d\overline{d}, the standard deviation sds_d, and use the paired tt-test formula to find the test statistic: t=dsd/nt = \frac{\overline{d}}{s_d / \sqrt{n}} where n=12n = 12 (the number of pairs).

Let me compute this step.### Step 4: Results of the Paired tt-Test

  • Mean of the differences d=0.67\overline{d} = -0.67
  • Standard deviation of the differences sd=3.26s_d = 3.26
  • Test statistic t=0.71t = -0.71
  • pp-value for the one-tailed test p=0.247p = 0.247

Step 5: Conclusion

At the 0.05 significance level, the pp-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.


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 tt-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 tt-test.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Paired t-Test
Significance Level

Formulas

Test Statistic: t = \frac{\overline{d}}{s_d / \sqrt{n}}
Null Hypothesis: H_0: \mu_d = 0
Alternative Hypothesis: H_1: \mu_d < 0

Theorems

Paired t-Test: A test used to compare the means of two related groups
p-value interpretation: If the p-value < alpha, reject H_0

Suitable Grade Level

College/Undergraduate