To address this hypothesis test, let’s proceed through each part of the problem, step by step.

a. State the Null and Alternative Hypotheses

• Null Hypothesis (H₀): The new study strategy has no effect on study hours, so the average study time remains the same as before, $\mu = 8.5$ hours.
• Alternative Hypothesis (H₁): The new study strategy has increased the average study time, so the average study time is now greater than before, $\mu > 8.5$ hours.

This is a one-tailed test since we are specifically testing for an increase in study hours.

b. Indicate the degrees of freedom (df), alpha level (α), and critical t-value

1. Degrees of freedom (df): $df = n - 1 = 30 - 1 = 29$.
2. Alpha level (α): Typically, α = 0.05 for hypothesis tests, unless specified otherwise.
3. Critical t-value: Using a one-tailed test with $\alpha = 0.05$ and $df = 29$, we look up or calculate the critical t-value, which is approximately $t_{\text{crit}} = 1.699$.

c. Calculate the t-statistic

To calculate the t-statistic, we need the standard deviation and the standard error of the sample mean.

1. Standard deviation (s): $s = \sqrt{\frac{SS}{n - 1}} = \sqrt{\frac{504}{29}} = \sqrt{17.3793} \approx 4.17$

2. Standard error (SE): $SE = \frac{s}{\sqrt{n}} = \frac{4.17}{\sqrt{30}} \approx \frac{4.17}{5.477} \approx 0.7614$

3. t-statistic: $t = \frac{M - \mu}{SE} = \frac{10.2 - 8.5}{0.7614} \approx \frac{1.7}{0.7614} \approx 2.23$

d. Interpret the Findings

• Decision regarding the null hypothesis: Since our calculated $t$-value of 2.23 is greater than the critical $t$-value of 1.699, we reject the null hypothesis.
• APA Style Interpretation: The results indicate that students who adopted the new study strategy spent significantly more time studying (M = 10.2, SD = 4.17) compared to the previous average of 8.5 hours, $t(29) = 2.23$, $p < .05$. This suggests that the new study strategy led to an increase in study hours.

e. Calculate Cohen’s d and r²

1. Cohen’s d: $d = \frac{M - \mu}{s} = \frac{10.2 - 8.5}{4.17} \approx \frac{1.7}{4.17} \approx 0.41$ Interpretation: A Cohen’s $d$ of 0.41 indicates a medium effect size, suggesting a moderate impact of the new study strategy on study hours.

2. $r^2$ (proportion of variance explained): $r^2 = \frac{t^2}{t^2 + df} = \frac{(2.23)^2}{(2.23)^2 + 29} = \frac{4.9729}{4.9729 + 29} \approx \frac{4.9729}{33.9729} \approx 0.146$ Interpretation: An $r^2$ of 0.146 indicates that about 14.6% of the variance in study hours can be explained by the adoption of the new study strategy, suggesting a moderate effect.

Summary

• Conclusion: The new study strategy significantly increased study hours among students, with a medium effect size (Cohen's $d = 0.41$) and accounting for 14.6% of the variance in study hours.

Would you like a more detailed breakdown of any step? Here are some related questions to explore:

1. How is Cohen's $d$ different from the $t$-statistic?
2. Why is the standard error essential in hypothesis testing?
3. How does the $r^2$ value help in interpreting the results?
4. What would a two-tailed test imply for this study?
5. How does sample size influence the outcome of a hypothesis test?

Tip: When reporting statistics in APA style, remember to include values such as $M$, $SD$, $t$, and $p$-value to provide context for your findings.