experiment has a single factor with five groups and two values in each group. In

Question

experiment has a single factor with five groups and two values in each group. In determining the among-group variation, there are 4 degrees of freedom. In determining the within-group variation, there are 5 degrees of freedom. In determining the total variation, there are 9 degrees of freedom Also, note that SSA = 1 44. SSW-45, SST = 1 89, MSA = 36 MSW= 9, and FSTATIT 4. Complete parts (a) through (d) Click here to view page 1 of the F table, Click here to view page 2 of the F table Click here to vew page 3 of the F table. Click here to view page 4 of the F table a. Construct the ANOVA summary table and fill in all values in the table Degrees cf Freedom Sum of Mean Square (Variance) Source mong groups ithin groups otal

Explanation / Answer

from above information below is ANOVA table:

b) for (4,5) degree of freedom and 0.05 level critical value of F =5.1922

c) Decision rule: reject null hypothesis if test statsitic F>5.1922

d)as test statsitic is not greater then crtiical value we can not reject null hypothesis

we do not have sufficient evidence to conclude that all four groups have different population means

source df SS MS F among groups 4 144 36 4 with in groups 5 45 9 total 9 189

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