Solved Example on Factorial 2 x 2 Experiment lecture 25

 Example - 2: 

The data in the following table are the form of  factorial experiment. Partition the treatment sums of squares into main effects and interaction effect. Interpret the result.

Solution: Let the treatment combination is considered as treatments and set up our hypotheses as:


i.   H0: τ1 = τ2 = τ3 = τ4 Vs. H1: τ1≠ τ2 ≠ τ3≠ τ4

     H01: Aj0   Vs.  H11: A ≠ 0  = 0, 1 

     H02: Bk 0  Vs.  H12: Bk   0  k 0, 1   

     H03 : (AB)jk = 0   Vs.  H03 : (AB)jk  0      

ii. The significance level; α = 0.05

iii. The test statistic:

 F = MST / MSE  ~ F (3,   16)

F1 = MSA / MSE ~ F (1, 16)

F2 = MSB / MSE ~ F (1, 16)

    F3 = MS(AB) / MSE ~ F (1, 16)

iv. Reject  H0, when F = > 3.29

     Reject  H01, when F1 = > 4.54

     Reject H02, when F2 = > 4.54
     
     Reject H03, when F3 = > 4.54

v.  Computation:








ANOVA Table:

vi. Remarks

     The F calculated value falling in the rejection region, it is concluded that the 4 treatments (2 treatments combination) are significant.

Further,

FA = 43 .82 Falling in the critical region, we have not accept . It is concluded that the levels of factor A has significant.

 FB = 0.18  Falling in the acceptance region, we have not sufficient evidence to reject . It is concluded that the levels of factor has non-significant.

FAB = 7.43 Falling in the critical region, we have not accept . It is concluded that factor A and B have interaction effect.

Example-3: 

An experiment is performed in order to investigate the effect of concentration of a reactant and the presence of the catalyst of the reaction time of a chemical process. Let the two levels of the reactant concentration be 15 % & 25 %. The catalyst has two levels high level denoting the presence of the catalyst and low level denoting the absence with three replicates. The data of the experiment are shown in the table below:

 

Treatment Combination

 

Replicate

I

II

III

A low, B low

28

25

27

A high, B low

36

32

32

A low, B high

18

19

23

A high, B high

31

30

29

Test the effects of reactant concentration and catalyst and their interaction.

Solution: We setup our hypotheses as:

i.   H0: τ1 = τ2 = τ3 = τ4 Vs. H1: τ1≠ τ2 ≠ τ3≠ τ4

     H01: Aj0   Vs.  H11: A ≠ 0  = 15 % ,  25 % 

     H02: Bk 0  Vs.  H12: Bk   0  k 0, 1   

     H03 : (AB)jk = 0   Vs.  H03 : (AB)jk  0      

ii. The significance level; α = 0.05

iii. The test statistic:

 F = MST / MSE  ~ F (3,   8)

F1 = MSA / MSE ~ F (1, 8)

F2 = MSB / MSE ~ F (1, 8)

    F3 = MS(AB) / MSE ~ F (1, 8)

iv. Reject  H0, when F = > 4.07

     Reject  H01, when F1 = > 5.12

     Reject H02, when F2 = > 5.12

     Reject H03, when F3 = > 5.12

v.  Computation:




 

ANOVA Table:




vi.   Remarks: 

The F calculated value falls in the rejection. It is concluded that the 4 factors combinations are significant. It is concluded from that  Factors A and B are significant but the interaction effect is insignificant.at 5 %.

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