Question

xperiment was performed and you are given the following ANOVA table:

Source of Variation	Sum of Squares	df	Mean Square	F
Factor A: Programs	31	3	*	*
Factor B: Sex	11	1	*	*
Interaction	1.5	3	*	*
Error	4.5	4	*
Total	48	11

What advice would you give managers about the influence of sex and interaction? Use a .05 level of significance.

Answer 1

Solution:
Mean square can be calculated as =Sum of square/DF
Mean Square for Program A = 31/3 = 10.333
Mean square for Program B = 11/1 = 11
Mean square for interaction = 1.5/3 = 0.5
Mean square Error = 4.5/4 = 1.125
Fstat can be calculated as = Mean square/mean square Error
Fstat for Program A = 10.333/1.125 = 9.1851
Fstat for Program B= 11/1.125 = 9.7777
Fstat for Interaction = 0.5/1.125 = 0.4444
So p-value for Sex at F = 9.7777 and Df for numerator =1 and DF for denomenator = 4 is 0.035 this is significant because it is less than 0.05
And p-value for interaction at F= 0.444, DF for numerator =3 and Df for numerator = 4 is 0.72, it is not significant as this is greater than 0.05.
So we can say that there is a significance difference in sex but not significance difference in interaction.