2^3 Factorial Experiment lecture - 26

 Introduction to 2^3 Factorial Experiment

lecture - 26

The 2^3 factorial experiment is the extension of 2^2 factorial experiment to three factors named A, B, C and each factor at two levels say a0, a1, b0, b1, c0, c1. In this experiment 3 factors each at two levels are analyzed.

The 2^3 factorial experiment is represented by the following linear statistical model:

 Yijkl = μ + αj + βk + γl + (αβ)jk + (αγ)jl + (βγ)kl + (αβ)γjkl + ϵijkl        i = 1, 2, ..., r    j, k,  l = 0, 1

The standard form of 2^3 factors combinations are:

1

a

b

ab

c

ac

bc

abc

Assume that the total response values are:

The response values can be arranged in a three-dimensional contingency table as:

The sums of squares of total, blocks and treatment can be computed as:

The sums of squares A, B and interaction effect can also be computed as:

ANOVA Table:

Where:

 Example: The following table gives treatment combination of 3 factors:

Replicate	a	b	c	abc	1	ab	ac	bc
1	1.9	1.6	2.1	3.8	1.3	3.2	2.8	3.2
2	3.0	2.7	3.0	4.8	2.2	4.1	3.9	4.1
3	4.0	3.8	2.1	5.9	4.2	5.2	5.1	5.0
4	1.9	1.5	2.0	3.9	1.1	3.4	3.0	3.0

Write statistical model for the above stated experiment and test the significance of factors effects and interaction effects.

Solution: The statistical model of the above experiment (2^3 Factorial Experiment ) is:

Yijkl = μ + Aj + Bk + Cl + (AB)jk +(AC)jl +(BC)kl + (ABC)jkl + ϵijkli = 1, 2, 3, 4. j, k,  l = 0, 1

We setup our hypotheses as:

i.  The null and alternative hypotheses can be stated as:

H01 : Aj = 0 Vs. H11 : Aj ≠ 0

H02 : Bk = 0 Vs. H12 : Bj ≠ 0

H03 : Cl = 0 Vs. H13 : Cl ≠ 0 

H04 : ABjk = 0 Vs. H14 : ABjk ≠ 0

H05 : ACjl = 0 Vs. H15 : ACjl ≠ 0

H06 : BCkl = 0 Vs. H16 : BCkl ≠ 0

H07 : ABCjkl = 0 Vs. H17 : ABCjkl ≠ 0

ii. The significance level; α = 0.05

iii. The test statistic:

iv. Reject all H0's, when F = > 4.32

v. Computation: using Contrast method

ANOVA Table:

vi.                    Remarks:

The effects pf factor A, B, C and their order interaction (ABC) are significant.

Example: The data of the following table are form 2 X 2 factorial experiment. Partition the treatment sum of squares into main effects and interaction counter parts. Interpret your results:

Solution: The data look like from homogeneous experimental units. So, we are using  CRD 2^2 factorial experiment.

.i. Statement of hypotheses:

H0 : T1 = T2 = T3 = T4 Vs. H1 : T1 ≠ T2 ≠ T3 ≠ T4

H01 : αj = 0  Vs. H11 : αj≠ 0

H02 : βk = 0 VS. H12 : βk ≠ 0

H03 : (αβ)jk = 0  Vs. H13 : (αβ)jk ≠ 0

ii. The significance level; α = 0.05

iii. The test statistic: 2^2 factorial experiment with CRD

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, if F => 3.24

     Reject H0i, if F = > 4.49 i = 1, 2, 3

v.  Computation:

ANOVA Table:

vi.                    Remarks:

H0 is significant.and H03 are significant.

• Read more:Yates method for 2^k Factorial Experiment

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:

${\mathrm{<span\; style="font-size:\; medium;">i.</span> \; H}}_0: {\tau }_1 = {\tau }_2 = {\tau }_3 = {\tau }_{\mathrm{4\; Vs. }}$H1: τ1≠ τ2 ≠ τ3≠ τ4

$$\begin{gathered} \text{ \\ H01: Aj= 0 Vs. } \end{gathered}$$ H11: Aj  ≠ 0  j = 0, 1 

$$\begin{gathered} \text{ \\ H02: Bk = 0 Vs. } \end{gathered}$$H12: Bk  ≠ 0  k = 0, 1   

$$\begin{gathered} \text{ \\ H03 : (AB)jk = 0 Vs. } \end{gathered}$$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:

${\mathrm{<span\; style="font-size:\; medium;">i.</span> \; H}}_0: {\tau }_1 = {\tau }_2 = {\tau }_3 = {\tau }_{\mathrm{4\; Vs. }}$H1: τ1≠ τ2 ≠ τ3≠ τ4

$$\begin{gathered} \text{ \\ H01: Aj= 0 Vs. } \end{gathered}$$ H11: Aj  ≠ 0  j = 15 % ,  25 % 

$$\begin{gathered} \text{ \\ H02: Bk = 0 Vs. } \end{gathered}$$H12: Bk  ≠ 0  k = 0, 1   

$$\begin{gathered} \text{ \\ H03 : (AB)jk = 0 Vs. } \end{gathered}$$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 %.

• Read: 2^3 Factorial Experiment