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.
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 pvalue 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 pvalue 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.
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