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)= G2/npq= 152.72/60; (2)= Σ x2= 12+ 1.92+……+3.52; (3)= (Σ Ai2)/nq= (82.42 + 70.32)/30; (4)=(ΣRj2)/np= (12.02+20.72+……+47.02)/12; (5)=[Σ (ARij2)]/n= (6.02+11.52+…..+22.02)/6; (6)=(Σ Hk2)/q= (15.12+14.82+ ……11.82)/5

 

Source of variation

Computational formula    

Sum of square  

df

MS

F

(probability)  

 

Between Hives

(6)-(1)

3.84

(pn-1)=11

0.35

 

 

Group (Product)

(3)-(1)

2.44

(p-1)=1

2.44

17.48

(p=0.0019)

Hives within

groups

(6)-(3)

1.40

p(n-1)=10

0.14

 

 

 

Within Hives

(2)-(6)

70.23

pn(q-1)=48

 

 

 

Repeated

(4)-(1)

66.92

(q-1)=4

16.73

274.70

(p<0.0001)

Interaction Group x R

(5)-(3)-(4)+(1)

0.88

(p-1)(q-1)=4

0.22

3.61

(p=0.0132)

R x Hives within groups  

(2)-(5)-(6)+(3)

2.44

p(n-1)(q-1)=40

0.061