8.2.2.1. Analysis of variance for numerical example
The test
calculations are reported in the Table 10. In this example, factor group
and factor repeated measures show a P
value via a Fisher less than the classical level of significance (0.05): Group
(p = 0.0019) and repeated measures R (p < 0.00001). These observed
probabilities do not permit one to accept a null hypothesis of equality between
the levels inside each factor. However the experimenter is not authorized to
conclude the main factors because the interaction between the factors is
significant (p = 0.0132). This statistical observation shows that the mean time
profiles are not parallel between both groups (control and test product). The
experimenter does analyse this interaction for instance with comparisons
between groups at each time of measure.
Table 10.
Analysis of variance for the example reported in Table 9. Formulae used:
(1)= G^{2}/npq= 152.7^{2}/60; (2)= Σ x^{2}= 1^{2}+
1.9^{2}+……+3.5^{2}; (3)= (Σ Ai^{2})/nq= (82.4^{2}
+ 70.3^{2})/30; (4)=(ΣRj^{2})/np= (12.0^{2}+20.7^{2}+……+47.0^{2})/12;
(5)=[Σ (ARij^{2})]/n= (6.0^{2}+11.5^{2}+…..+22.0^{2})/6;
(6)=(Σ Hk^{2})/q= (15.1^{2}+14.8^{2}+ ……11.8^{2})/5


Source of variation 
Computational formula 
Sum of square 
df 
MS 
F 
(probability) 


Between Hives 
(6)(1) 
3.84 
(pn1)=11 
0.35 


Group (Product) 
(3)(1) 
2.44 
(p1)=1 
2.44 
17.48 
(p=0.0019) 
Hives within groups 
(6)(3) 
1.40 
p(n1)=10 
0.14 




Within Hives 
(2)(6) 
70.23 
pn(q1)=48 



Repeated 
(4)(1) 
66.92 
(q1)=4 
16.73 
274.70 
(p<0.0001) 
Interaction Group x R 
(5)(3)(4)+(1) 
0.88 
(p1)(q1)=4 
0.22 
3.61 
(p=0.0132) 
R x Hives within groups 
(2)(5)(6)+(3) 
2.44 
p(n1)(q1)=40 
0.061 

