Question 1177142: Application of the analysis of variance to a particular problem has resulted in the following ANOVA table. Complete the test at a significance level of 5%.
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Source | DF | SS | MS | F |
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Between | 2 | 457.1241 | 228.5621 | 52.74 |
within | 17 | 117.0165 | 4.3339 | |
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Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! **1. Hypotheses**
* **Null Hypothesis (H0):** The means of all the groups are equal.
* **Alternative Hypothesis (H1):** At least one group mean is different from the others.
**2. Significance Level**
* α = 0.05
**3. Degrees of Freedom**
* df(between) = 2 (from the ANOVA table)
* df(within) = 17 (from the ANOVA table)
**4. F-statistic**
* F = 52.74 (from the ANOVA table)
**5. Critical F-value**
* Using an F-distribution table or calculator with α = 0.05, df1 = 2, and df2 = 17, we find the critical F-value to be approximately 3.59.
**6. Decision**
* Since the calculated F-statistic (52.74) is greater than the critical F-value (3.59), we reject the null hypothesis.
**7. Conclusion**
At a 5% significance level, there is enough evidence to conclude that there is a statistically significant difference between the means of at least two of the groups.
**In simpler terms:** The ANOVA test shows that it's very unlikely that the observed differences between the group means are just due to random chance. Therefore, we can confidently say that there's a real difference in the average values of those groups.
**Further Analysis**
While ANOVA tells us that there's a difference, it doesn't tell us *which* groups are different. To find that out, you would need to perform post-hoc tests (like Tukey's HSD or Bonferroni correction) to compare specific pairs of group means.
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