document.write( "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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document.write( "Source | DF | SS | MS | F |
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document.write( "Between | 2 | 457.1241 | 228.5621 | 52.74 |
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document.write( "within | 17 | 117.0165 | 4.3339 | |
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Algebra.Com's Answer #850514 by CPhill(1959)![]() ![]() You can put this solution on YOUR website! **1. Hypotheses**\r \n" ); document.write( "\n" ); document.write( "* **Null Hypothesis (H0):** The means of all the groups are equal. \n" ); document.write( "* **Alternative Hypothesis (H1):** At least one group mean is different from the others.\r \n" ); document.write( "\n" ); document.write( "**2. Significance Level**\r \n" ); document.write( "\n" ); document.write( "* α = 0.05\r \n" ); document.write( "\n" ); document.write( "**3. Degrees of Freedom**\r \n" ); document.write( "\n" ); document.write( "* df(between) = 2 (from the ANOVA table) \n" ); document.write( "* df(within) = 17 (from the ANOVA table)\r \n" ); document.write( "\n" ); document.write( "**4. F-statistic**\r \n" ); document.write( "\n" ); document.write( "* F = 52.74 (from the ANOVA table)\r \n" ); document.write( "\n" ); document.write( "**5. Critical F-value**\r \n" ); document.write( "\n" ); document.write( "* 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.\r \n" ); document.write( "\n" ); document.write( "**6. Decision**\r \n" ); document.write( "\n" ); document.write( "* Since the calculated F-statistic (52.74) is greater than the critical F-value (3.59), we reject the null hypothesis.\r \n" ); document.write( "\n" ); document.write( "**7. Conclusion**\r \n" ); document.write( "\n" ); document.write( "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.\r \n" ); document.write( "\n" ); document.write( "**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.\r \n" ); document.write( "\n" ); document.write( "**Further Analysis**\r \n" ); document.write( "\n" ); document.write( "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. \n" ); document.write( " |