Analysis of Variance 

The computations for an analysis-of-variance can be summarized in an anova table. 

 

Source of 

Variation 

Sum of
Squares
 

Degrees of 

Freedom 

Mean Square 

F-Ratio 

p-value 

`*`(Row, `*`(A)) 

SSA 

`+`(a, `-`(1)) 

MSA = `/`(`*`(SSA), `*`(`+`(a, `-`(1)))) 

f = `/`(`*`(MSA), `*`(MSE)) 



 

`*`(Column, `*`(B)) 

SSB 

`+`(b, `-`(1)) 

MSB = `/`(`*`(SSB), `*`(`+`(b, `-`(1)))) 

f = `/`(`*`(MSB), `*`(MSE)) 


 

`*`(Interaction, `*`(AB))
`*`(Interaction, `*`(AB))
 

 

SSAB 

(`+`(a, `-`(1)))(`+`(b, `-`(1))) 

MSAB = `/`(`*`(SSAB), `*`((`+`(a, `-`(1)))(`+`(b, `-`(1))))) 

f = `/`(`*`(MSAB), `*`(MSE)) 



 

Error 

SSE 

ab(`+`(c, `-`(1))) 

MSE = `/`(`*`(SSE), `*`(ab(`+`(c, `-`(1))))) 

 

 

Total 

SSTO 

`+`(abc, `-`(1)) 

 

 

 

 

The null hypothesis 

 

is rejected at the alpha-level of significance if the F-Ratio  

f[alpha] `+`(a, `-`(1)), ab(`+`(c, `-`(1))) 

 

The null hypothesis 

 

is rejected at the alpha-level of significance if the F-Ratio  

f[alpha] `+`(b, `-`(1)), ab(`+`(c, `-`(1))) 

 

The null hypothesis 

H[AB]; -1; gamma[ij] = 0 

is rejected at the alpha-level of significance if the F-Ratio  

f[alpha] (`+`(a, `-`(1)))(`+`(b, `-`(1))), ab(`+`(c, `-`(1))) 

 

 

 

Example 3 

An experiment was carried out to test the hardness of an alloy. Three operators made the experiment on the five machines available. Each of the operators carried out the experiment three times for each treatment combination. The data recorded are 

 

Machine 

Operator 

1 

2 

3 

4 

5 

A 

15 

16 

18 

6 

4 

5 

8 

5 

9 

4 

6 

5 

6 

8 

7 

B 

25 

20 

23 

15 

8 

17 

8 

9 

10 

14 

12 

8 

10 

8 

9 

C 

12 

17 

22 

13 

14 

12 

4 

13 

8 

9 

11 

12 

7 

9 

8 

 

Test the hypotheses 

a) H[A]; -1No row effect , b) H[B]; -1No column effect , c) H[AB]; -1No interaction 

Solution 

> restart:MathMaple:-ini():
 

> L:=[[[15,6,8,4,6],[16,4,5,6,8],[18,5,9,5,7]],[[25,15,8,14,10],[20,8,9,12,8],[23,17,10,8,9]],[[12,13,4,9,7],[17,14,13,11,9],[22,12,8,12,8]]];
 

[[[15, 6, 8, 4, 6], [16, 4, 5, 6, 8], [18, 5, 9, 5, 7]], [[25, 15, 8, 14, 10], [20, 8, 9, 12, 8], [23, 17, 10, 8, 9]], [[12, 13, 4, 9, 7], [17, 14, 13, 11, 9], [22, 12, 8, 12, 8]]]
[[[15, 6, 8, 4, 6], [16, 4, 5, 6, 8], [18, 5, 9, 5, 7]], [[25, 15, 8, 14, 10], [20, 8, 9, 12, 8], [23, 17, 10, 8, 9]], [[12, 13, 4, 9, 7], [17, 14, 13, 11, 9], [22, 12, 8, 12, 8]]]
 

The analysis of variance table is 

> TwoFactorAnova(L,Multiple);
 

Matrix(%id = 151011788)
 

a) 

The critical f-value is 

> f[0.05]=ProbTable([FRatio,2,30],p=0.95);
 

f[0.5e-1] = 3.315829501
 

reject H[A](no row effect). 

b) 

The critical f-value is 

> f[0.05]=ProbTable([FRatio,4,30],p=0.95);
 

f[0.5e-1] = 2.689627574
 

reject H[B](no column effect). 

c) 

The critical f-value is 

> f[0.05]=ProbTable([FRatio,8,30],p=0.95);
 

f[0.5e-1] = 2.266163274
 

accept H[AB](no interaction).