SOLUTION: A Two-Way ANOVA Summary Table is shown below. Source SS d.f. MS F Factor A 16.63 2 8.31 4.89 Factor B 5.13 2 2.57 1.51 Interaction A × B 17.88 4 4.47 2.63 Within (error) 6

Question 1193825: A Two-Way ANOVA Summary Table is shown below.
Source SS d.f. MS F
Factor A 16.63 2 8.31 4.89
Factor B 5.13 2 2.57 1.51
Interaction A × B 17.88 4 4.47 2.63
Within (error) 61.20 36 1.7
Total 100.844 44
The critical value for the F test using = 0.05, d.f.N. = 2, and d.f.D = 36 is 3.26. Then,
A.only the null hypothesis concerning Factor A should be rejected.
B.only the null hypothesis concerning Factor B should be rejected.
C.all of the null hypotheses should be accepted.
Answer by parmen(42) (Show Source): You can put this solution on YOUR website!
To answer this question, we compare the **F values** for each factor and interaction term to the critical value (\( F_{critical} = 3.26 \)) provided for \( \alpha = 0.05 \), \( \text{d.f.}_{\text{numerator}} = 2 \), and \( \text{d.f.}_{\text{denominator}} = 36 \).
### Step 1: Analyze \( F \)-statistics
1. **Factor A**:
- \( F_A = 4.89 \)
- \( F_A > 3.26 \): **Reject the null hypothesis** for Factor A.
2. **Factor B**:
- \( F_B = 1.51 \)
- \( F_B < 3.26 \): **Do not reject the null hypothesis** for Factor B.
3. **Interaction A × B**:
- \( F_{A \times B} = 2.63 \)
- \( F_{A \times B} < 3.26 \): **Do not reject the null hypothesis** for the interaction term.
### Step 2: Conclusion
- The null hypothesis for **Factor A** is rejected because its \( F \)-statistic is greater than the critical value.
- The null hypotheses for **Factor B** and the **interaction term** are not rejected because their \( F \)-statistics are less than the critical value.
### Final Answer:
**A. Only the null hypothesis concerning Factor A should be rejected.**
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