A study randomized a sample of 40 participants evenly into 4 different treatments. The sum of squares due to treatments is 500. The sum of squares due to error is 350. At 1% significance what is the conclusion of the analysis of variance?

Question

A study randomized a sample of 40 participants evenly into 4 different treatments.  The sum of squares due to treatments is 500.  The sum of squares due to error is 350.  At 1% significance what is the conclusion of the analysis of variance?

BlackTom AI · Accepted Answer

We start by identifying the components of the ANOVA table and the relevant test statistic.
Option A states: 'Reject the null hypothesis of equal means, at least one of the treatments has an effect.' This would be the correct conclusion if the computed F statistic exceeds the critical value at the 1% significance level.
Option B states: 'All four assumptions of the regression are satisfied and the model is valid.' While ANOVA relies on assumptions like independence, normality, and equal variances, the statement about all four regression assumptions being satisfied is not the direct conclusion of the F-test; moreover, ANOVA assumptions are not usually framed as four specific regression assumptions, making this claim not directly appropriate as the conclusion of the test.
Option C states: 'the MSTR and MSE are statistically equivalent.' If MSTR and MSE were statistically equivalent, the F statistic would be near 1, not a large value; this would indicate no evidence against the null. This option mischaracterizes the purpose of the F-test and the meaning of MS terms.
Option D states: 'Fail to reject the null hypothesis of equal means, the treatments have no effect.' This would be chosen if the F statistic were not significant at 1%; however, the result we compute shows a large F value, which would lead to rejection, not failure to reject.
Now, let's compute the actual F statistic to determine which option aligns with the data: the sum of squares due to treatments (SST) is 500 with 3 numerator degrees of freedom (k-1 where k=4), so MS_T = 500/3 ≈ 166.67. The sum of squares due to error (SSE) is 350 with 36 error degrees of freedom, so MS_E = 350/36 ≈ 9.72. The F statistic is F = MS_T / MS_E ≈ 166.67 / 9.72 ≈ 17.15.
To decide at 1% significance, we compare to the critical F value for (df1=3, df2=36) at alpha=0.01. The critical value F_{0.99}(3,36) is approximately 5.2. Since 17.15 > 5.2, we reject the null hypothesis of equal treatment means at the 1% level.
Putting it together:
- Option A is consistent with rejecting the null because there is strong evidence that at least one treatment differs in mean.
- Option B is not the direct conclusion of the ANOVA F-test and misstates the framework.
- Option C is incorrect because it implies equivalence of MSTR and MSE, which would not yield a large F.
- Option D would be true only if the F-test were not significant, which is not the case here.
In summary, the data lead to rejecting the null hypothesis and concluding that at least one treatment has a different mean from the others, corresponding to Option A.

A study randomized a sample of 40 participants evenly into 4 different treatments. The sum of squares due to treatments is 500. The sum of squares due to error is 350. At 1% significance what is the conclusion of the analysis of variance?

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