Consider the following ANOVA table. The null hypothesis for this ANOVA problem is A) μ₁ = μ₂ = μ₃ = μ₄. B) μ₁ = μ₂ = μ₃ = μ₄ = μ₅. C) μ₁ = μ₂ = μ₃ = μ₄ = μ₅ = μ₆. D) μ₁ = μ₂ = ... = μ₂₀.

Question

Quizplus · Accepted Answer

The null hypothesis of an ANOVA test is that there is no significant difference between the means of the groups being compared. In this ANOVA table, there is one factor (S) with two levels (S1 and S2), and one factor (U) with two levels (U1 and U2). The null hypothesis is therefore that the means of the 5 groups (B0, B11, B12, B13, B14) are all equal within each combination of the two levels of factor S and U.

Choice A is incorrect because it only includes the first 4 groups (B0, B11, B12, B13) and does not include the last group (B14).

Choice C is incorrect because it includes an additional group (B15) that is not present in the ANOVA table.

Choice D is incorrect because it includes 120 groups, which is not consistent with the levels of factors S and U in this problem.

Consider the Following ANOVA Table. ​ ​

Consider the Following ANOVA Table.