Answer Happy

In a completely randomized design, seven experimental units were used for each of the five levels of the factor. Consider the following ANOVA table. Source of Variation Treatments Error Total Sum Degrees of Squares of Freedom 380 110 490 # O Ho: H₁ H₂ H3 H4 Submit Answer (a) What hypotheses are implied in this problem? H5 Ha: M₁ = H₂ = H3 = H4 = H5 Need Help? Read It 4 Ho: M₁ = H₂ = μ3 = μ₁= 15 H₂: M₂ M₂ # H3 H4 H5 30 O Ho: At least two of the population means are equal. H: At least two of the population means are different. 34 O Ho: Not all the population means are equal. H₂H₂ = H₂ = H3 H4 H5 Ho: H₁ H₂ H3=14=15 H: Not all the population means are equal. Mean Square 95 3.67 F p-value (b) At the a = 0.05 level of significance, can we reject the null hypothesis in part (a)? Explain. Because the p-value sa = 0.05, we cannot reject Ho. O Because the p-value > a = 0.05, we can reject Ho O Because the p-value > a = 0.05, we cannot reject Ho O Because the p-value sa = 0.05, we can reject Ho 25.91 0.0000 View Previous Question Question 9 of 20 View Next Question Home My Assignments Copyright © 1998-2022 Cengage Learning, Inc. All Rights Reserved | TERMS OF USE PRIVACY