One-Way ANOVA
Use one-way ANOVA to test whether there are significant differences in mean plant growth across three types of fertilizer.
Problem Scenario
An agricultural researcher tests three fertilizers on tomato plants. Each fertilizer is applied to 5 plants, and the growth (in cm) after 4 weeks is measured. Fertilizer A: 20, 22, 19, 24, 21. Fertilizer B: 28, 30, 27, 26, 29. Fertilizer C: 24, 25, 23, 26, 22. At alpha = 0.05, is there a significant difference in mean growth among the three fertilizers?
Given Data
Requirements
- Calculate the group means and the grand mean
- Compute SSB (between-groups), SSW (within-groups), and the F-statistic
- Determine the p-value and state the conclusion
Solution
Step 1:
State the hypotheses. H_0: mu_A = mu_B = mu_C (all three population means are equal). H_a: At least one population mean is different.
Step 2:
Calculate the grand mean: (Sum of all 15 values) / 15 = (21.2*5 + 28.0*5 + 24.0*5) / 15 = (106 + 140 + 120) / 15 = 366 / 15 = 24.4.
Step 3:
Calculate SSB (Between-Groups Sum of Squares): SSB = n * Sum[(group mean - grand mean)^2] = 5[(21.2 - 24.4)^2 + (28.0 - 24.4)^2 + (24.0 - 24.4)^2] = 5[(-3.2)^2 + (3.6)^2 + (-0.4)^2] = 5[10.24 + 12.96 + 0.16] = 5(23.36) = 116.8.
Step 4:
Calculate SSW (Within-Groups Sum of Squares): For A: (20-21.2)^2 + (22-21.2)^2 + (19-21.2)^2 + (24-21.2)^2 + (21-21.2)^2 = 1.44 + 0.64 + 4.84 + 7.84 + 0.04 = 14.8. For B: (28-28)^2 + (30-28)^2 + (27-28)^2 + (26-28)^2 + (29-28)^2 = 0 + 4 + 1 + 4 + 1 = 10.0. For C: (24-24)^2 + (25-24)^2 + (23-24)^2 + (26-24)^2 + (22-24)^2 = 0 + 1 + 1 + 4 + 4 = 10.0. SSW = 14.8 + 10.0 + 10.0 = 34.8.
Step 5:
Calculate mean squares and F-statistic. MSB = SSB / (k - 1) = 116.8 / 2 = 58.4. MSW = SSW / (N - k) = 34.8 / 12 = 2.9. F = MSB / MSW = 58.4 / 2.9 = 20.14.
Step 6:
Find the p-value. With F = 20.14, df_between = 2, and df_within = 12, the p-value < 0.001 (the critical value F(0.05, 2, 12) = 3.885). Since 20.14 >> 3.885, we reject H_0. There is a significant difference in mean growth among the three fertilizers.
Final Answer
F(2, 12) = 20.14, p < 0.001. We reject H_0 at the 0.05 level. There is a statistically significant difference in mean tomato plant growth among the three fertilizers. Fertilizer B produced the highest mean growth (28.0 cm), followed by C (24.0 cm) and A (21.2 cm).
Key Takeaways
- ✓One-way ANOVA tests whether at least one group mean differs from the others. If significant, follow up with post-hoc tests (e.g., Tukey HSD) to identify which specific pairs differ.
- ✓ANOVA assumes normality within groups, equal variances across groups (homoscedasticity), and independent observations.
- ✓A large F-statistic indicates that the between-group variability is much greater than the within-group variability, suggesting real differences among group means.
Common Errors to Avoid
- ✗Concluding which specific groups differ based on ANOVA alone. ANOVA only tells you that at least one group differs; you need post-hoc tests to identify which pairs are significantly different.
- ✗Confusing the between-group and within-group degrees of freedom. df_between = k - 1 and df_within = N - k, where k is the number of groups and N is the total number of observations.
- ✗Failing to check the equal variance assumption. Use Levene's test or compare the largest and smallest group standard deviations; if they differ by more than a factor of 2, consider Welch's ANOVA.
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Common questions about this problem type
Running multiple t-tests inflates the family-wise Type I error rate. With 3 groups, you would need 3 pairwise t-tests, and the probability of at least one false positive rises above alpha. ANOVA controls the overall Type I error rate at alpha by using a single omnibus F-test.
Tukey's HSD (Honestly Significant Difference) is the most common choice when comparing all pairs of means. It controls the family-wise error rate while maintaining good statistical power. Other options include Bonferroni correction (more conservative) and Scheffe's method (more flexible for complex contrasts).