Question

2. Provide a complete ANOVA table from the following information for one-way design of 3 groups with 5r) within-group repetitions. You may use pf (t, df1, df2) function in R when computing p-value Group 1 5 40 10 Group 25 50 10 Group 35 60 10

Answer 1

ANOVA Table... d.f. 2 12 14 Source of Variation Sum of Squares Variance 49.99930.0000 Between Groups: 1000.0000 Within Groups: 120.0017 Total: 1120.0017 10.0001

p value <-1-pf(49.993,2,12)

This command in R will give the p value =1.514*e^-6~0

p value <0.05 indicates that the null hypothesis can be rejected at 5% significance level which again implies that atleast one treatment differs from the others. So a post hoc analysis is done to determine which are the means differing from one another

Tukey HSD Post-hoc Test Group 1 vs Group 2: Diff=10.0000, Group 1 vs Group 3: Diff-20.0000 , Group 2 vs Group 3: Diffz10.0000, 95%CI-4.6642 to 15.3358, p=0.0008 95%CI-14.6642 to 25.3358, p-0. ΟΘΘΘ 95%CI-4.6642 to 15.3358, p:0.0008

The p value for all the pairs are less than 0.05. The above results indicates that all the groups significantly differ from one another.