To determine whether the review section significantly harms test performance at a significance level of $\alpha = 0.002$, we can perform a paired t-test because the data comes in pairs (pre-test and post-test scores for each student).

Step 1: Define the hypotheses

The null and alternative hypotheses are as follows:

• $H_0$: The review section does not significantly harm test performance, i.e., there is no difference between pre-test and post-test scores. Mathematically, this can be written as $\mu_d = 0$, where $\mu_d$ is the mean of the differences between paired scores.
• $H_1$: The review section significantly harms test performance, i.e., the post-test scores are lower than the pre-test scores. Mathematically, this can be written as $\mu_d > 0$ (because we are testing for harm, we expect pre-test scores to be higher than post-test scores).

Step 2: Calculate the differences

We subtract the post-test scores from the pre-test scores to get the differences:

$\text{Differences} = \text{Pre-test} - \text{Post-test} = [51-60, 59-59, 92-50, 87-67, 78-62, 44-56, 92-64, 68-56, 76-56, 76-55]$ $\text{Differences} = [-9, 0, 42, 20, 16, -12, 28, 12, 20, 21]$

Step 3: Calculate the test statistic

To compute the test statistic, we first need the mean and standard deviation of the differences.

• Mean of the differences ($\bar{d}$): $\bar{d} = \frac{\sum \text{Differences}}{n} = \frac{-9 + 0 + 42 + 20 + 16 - 12 + 28 + 12 + 20 + 21}{10} = \frac{138}{10} = 13.8$

• Standard deviation of the differences ($s_d$): $s_d = \sqrt{\frac{\sum (\text{Difference} - \bar{d})^2}{n-1}}$ First, we calculate the squared differences: $(-9 - 13.8)^2, (0 - 13.8)^2, (42 - 13.8)^2, \ldots$ Summing them up and then dividing by $n-1$ gives the standard deviation.

Let's compute these values:

$s_d = \sqrt{\frac{(-9 - 13.8)^2 + (0 - 13.8)^2 + \dots}{9}} = \sqrt{366.93} \approx 19.152$

Now, the test statistic is given by: $t = \frac{\bar{d}}{s_d / \sqrt{n}} = \frac{13.8}{19.152 / \sqrt{10}} = \frac{13.8}{6.056} \approx 2.278$

Step 4: Find the p-value

Using a t-distribution table (with $n-1 = 9$ degrees of freedom), we can find the p-value for $t = 2.278$.

Looking up the value or using a calculator for the t-distribution: $\text{p-value} \approx 0.0255$

Step 5: Compare the p-value with $\alpha = 0.002$

The p-value $0.0255$ is greater than the significance level $\alpha = 0.002$. Therefore, we fail to reject the null hypothesis.

Conclusion

There is not sufficient evidence to conclude that the review section significantly harms test performance with a significance level of $\alpha = 0.002$.

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Would you like further details or clarifications on any part of this? Here are 5 related questions that could expand on this topic:

1. How does changing the significance level affect the result of this test?
2. What are the assumptions behind using a paired t-test?
3. How can we compute the p-value using software like Excel or Python?
4. Can you explain what a Type I and Type II error would mean in this context?
5. What other statistical tests could be used for similar paired data?

Tip: Always ensure the assumptions of normality and pairing are met before applying a paired t-test.