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It looks like you've uploaded an image related to a hypothesis test involving ANOVA (Analysis of Variance) for test scores with different caffeine levels.

Here’s a breakdown of how to approach the problem:

### 1. **Hypothesis Testing (ANOVA)**:
   - **Null Hypothesis (H₀)**: The population mean test scores for all three caffeine treatments (Water, Black Tea, Coffee) are equal.
   - **Alternative Hypothesis (H₁)**: At least one population mean is different from the others.

### 2. **ANOVA Table**:
   - You need to calculate the **degrees of freedom (df)** and **mean squares (MS)** for both "Between" and "Within" groups.
   - Formula for \( MS = \frac{SS}{df} \).
   - \( F \)-ratio is calculated as \( F = \frac{MS_{Between}}{MS_{Within}} \).

### 3. **Calculating F-statistic**:
   - Once \( F \) is calculated, you compare it against a critical value from the F-distribution to see if you can reject the null hypothesis.

### 4. **Conclusion**:
   - Using the significance level \( \alpha = 0.01 \), you check if the calculated \( F \)-statistic exceeds the critical \( F \)-value. Based on this, you conclude whether to reject or fail to reject the null hypothesis.

Let me help by calculating the necessary steps. Would you like to proceed with those?

A study is conducted to test the impact of different caffeine contents on students' test-taking abilities using ANOVA. The test involves three groups of students, each assigned to a drink (Water, Black Tea, Coffee) with different caffeine levels. The study aims to determine if the population mean test scores are the same across the three groups, with calculations for the ANOVA table, F-ratio, and conclusions based on the significance level α = 0.01.

Explore how to use ANOVA for hypothesis testing on the impact of different caffeine levels (Water, Black Tea, Coffee) on students' test scores. This problem covers the calculation of F-ratio, degrees of freedom, and comparison with the critical F-value to determine if caffeine affects test performance.

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