In a research study conducted at a university, it was observed that students, on average, spent μ = 8.5 hours per week studying for a particular course last semester. To evaluate the effectiveness of a new study strategy introduced this semester, a sample of n = 30 students who adopted the strategy were selected. The average number of study hours for this sample was M = 10.2 hours with a Sum of Squares (SS) = 504. Can it be concluded that the new study strategy led to a significant increase in study hours compared to the previous semester? a. State the null & alternative hypotheses.  b. Indicate the df, the alpha, and the t critical value.  c. Calculate your t-statistic (include the standard deviation, standard error and t-value). d. Interpret the findings (include your decision regarding the null hypothesis and a description of the pattern of results with supporting statistics in APA style). e. Calculate the Cohen’s d and r-squared value. Indicate the size of the Cohen’s d and describe how much variance of hours studying can be accounted for by whether students adopted the new study strategy. 

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.

In a research study conducted at a university, it was observed that students, on average, spent μ = 8.5 hours per week studying for a particular course last semester. To evaluate the effectiveness of a new study strategy introduced this semester, a sample of n = 30 students who adopted the strategy were selected. The average number of study hours for this sample was M = 10.2 hours with a Sum of Squares (SS) = 504. Can it be concluded that the new study strategy led to a significant increase in study hours compared to the previous semester? a. State the null & alternative hypotheses. b. Indicate the df, the alpha, and the t critical value. c. Calculate your t-statistic (include the standard deviation, standard error and t-value). d. Interpret the findings (include your decision regarding the null hypothesis and a description of the pattern of results with supporting statistics in APA style). e. Calculate the Cohen’s d and r-squared value. Indicate the size of the Cohen’s d and describe how much variance of hours studying can be accounted for by whether students adopted the new study strategy.

Explore a hypothesis test evaluating if a new study strategy increased weekly study hours among university students. Learn about null and alternative hypotheses, degrees of freedom, t-statistics, effect sizes, and r² values, with results interpreted in APA style. Ideal for undergraduate statistics.

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