To find the test statistic and the p-value for your hypothesis test, we can use the following steps:

1. Calculate the test statistic (t): The formula for the t-statistic when the population standard deviation is unknown is:

   $t = \frac{\bar{d} - \mu_d}{s_d / \sqrt{n}}$

   where:

   • $\bar{d} = -10.5$ (mean difference)
   • $\mu_d = 0$ (mean under the null hypothesis)
   • $s_d = 25.2$ (standard deviation of the differences)
   • $n = 39$ (sample size)

   Plugging in the values:

   $t = \frac{-10.5 - 0}{25.2 / \sqrt{39}} \approx \frac{-10.5}{4.020} \approx -2.610$

   Thus, the test statistic is:

   $\text{test statistic} = -2.610$

2. Calculate the p-value: Since this is a one-tailed test (Ha: μd < 0), we will find the p-value corresponding to the calculated t-value with $n - 1 = 38$ degrees of freedom. Using a t-distribution table or calculator:

   The p-value for $t = -2.610$ and 38 degrees of freedom is approximately:

   $\text{p-value} \approx 0.0056$

Summary of Results:

• Test Statistic: $t = -2.610$ (rounded to three decimal places)
• p-value: $0.0056$ (rounded to four decimal places)

Conclusions:

• The p-value is less than $0.10$.
• This test statistic leads to a decision to reject the null.
• As such, the final conclusion is that there is sufficient evidence to warrant rejection of the claim that the mean difference of post-test from pre-test is less than 0.

Would you like more details on any of these calculations or concepts? Here are some follow-up questions you might consider:

1. How does the choice of significance level affect the results?
2. What assumptions are made in this hypothesis testing?
3. How would you interpret the effect size in this context?
4. What other statistical tests could be applied to similar data?
5. How could you visualize the data to better understand the results?

One tip: Always check the assumptions of the test before drawing conclusions!