Question

15) Fill in the missing highlighted entries in the partially completed one-way ANOVA table. What are the numerator and denominator degrees of freedom? At alpha=0.01, the critical value for this test is 4.67. What is your decision about this hypothesis test: do you reject the null hypothesis or not? Based on that decision, what is your interpretation: do the data support the claim, or not? Source of Variation Sum of Squares 32.96 Degrees of Freedom Mean Squared Error 4 Factor (between) Error (within) Total 3.48 21

Answer 1

Sol: we are given one-way ANOVA table. Level of significance L = 0.01 The Critlcal value for this test F= 4.67 source sum Degrees Mean of of of f-ratto Squared Variation Squares freedom Error 8.94 Factor (bet) 32.96 4 Error (within 59-16 17 3.48 Total 92.12 21 2-3687 We know that Error degrees of freedom - Total degrees of freedom - Faclor degrees of freedom → Error of = 21-4 = 14 → mean square Value - Sum of squares Respective degrees of freedom Sum of Squares (Error) - Mean Squared X Error of = 3.48 x17 = 59.16 Total Sum of squares factor Sum of sqs + Error Suem of sqs. = 32.96 + 59.16 = 92-12 + Mean Squares due to factor - to factor sum of squares Factor of = 32.46 = 8-24. - F-ratio= Factor Mean Squares Error Mean Squares = = 2.3678. = 2.368

- f-ratio = Factor Mean Squares - N F distribution Error Mean Squares with (4, 17) degrees of freedom. Numerator degrees of freedom =4. Denominator degrees of freedom = 17. Decision : Since Calculated F (F-ratio is less than the critical F. + 2.368 < 4.64 . . we fail to reject the sull hypothesis. are At we have equat. =0.01 Significance level, evidence to conclude that Mean effects of all factors are all factors equal -