TWO WAY ANOVA WITH REPLICATION

Also called a Factorial Experiment.Factorial Experiment is used to evaluate 2 or more factors simultaneously. Replication means an independent repeat of each factor combination. The purpose of factorial experiment is to examine:

1. The effect of factor A on the dependent variable, y.2. The effect of factor B on the dependent variable, y along with3. The effects of the interactions between different levels of the factors on the dependent variable, y. Interaction exists when the effect of a level for one factor depends on which level of the other factor is present. Advantages of Factorial Experiment over one factor at a time (one-way ANOVA) – more efficient & allow interactions to be detected.

The effect model for a factorial experiment can be written as:

: The response from the kth experimental unit receiving the

th level of factor and the th level of factor

: Overall mean

: An effect due to the th level of factor

ijk

i

j

y

i A j B

i A

: An effect due to the th level of factor

: An interaction effect of the th level of factor with jth level of factor

: A random error associated with the response from the th e

ij

ijk

j B

i A B

k

xperimental

unit receiving the th level of factor combined with jth level of factor

i A B

1

1

1ijk i j ijkij

i ,...,a

y j ,...,b

k ,...,r

There are three sets of hypothesis:

1. Factor A effect:

2. Factor B effect:

3. Interaction effect:

The results obtained in this analysis are summarized in the following ANOVA table:

Two way Factorial Treatment Structure

1A

2A

3A

1B 2B

111

11211

11n

y

yy

.

y

121

12212

12n

y

yy

.

y

211

21221

21n

y

yy

.

y

311

31231

31n

y

yy

.

y

221

22222

22n

y

yy

.

y

321

32232

32n

y

yy

.

y

1y 2y

1y

2y

3y

y

where

2 2

2 2

2 2

22

i

j

ij

ijk

y ySSA

br abr

y ySSB

ar abr

y ySSAB SSA SSB

r abr

ySST y

abrSSE SST SSA SSB SSAB

Example:

The two-way table gives data for a 2x2 factorial experiment with two

observations per factor – level combination.

Construct the ANOVA table for this experiment and do a complete analysis

at a level of significance 0.05.

Factor A

Factor B

Level 1 2

1 29.635.2

47.342.1

2 12.917.6

28.422.7

Solution:

Factor A

Factor B

Level 1 2

1 29.635.264.8

47.342.189.4

2 12.917.630.5

28.422.751.1

154.2

81.6

95.3 140.5 235.8

Solution:

1. Set up hypothesis

Factor A effect:

Factor B effect:

Interaction effect:

0 1 2

1

: 0

: at least one 0a

i

H

H

0 1 2

1

: 0

: at least one 0b

j

H

H

0

1

: 0 for all

: at least one 0

ij

ij

H i, j

H

22

22 2 2

2 2

2 2 2

235 829 6 35 2 22 7

2 2 2

972 715

154 2 81 6 235 8

4 8

658 845

ijk

i

ySST y

abr

.. . .

.

y ySSA

br abr

. . .

.

2 2

2 2 295 3 140 5 235 8

4 8

255 38

jy ySSB

ar abr

. . .

.

2 2

2 2 2 2 264 8 89 4 30 5 51 1 235 8 658 845 255 38

2 8

2

972 715 658 845 255 38 2

56 49

ijy ySSAB SSA SSB

r abr

. . . . .. .

SSE SST SSA SSB SSAB

. . .

.

2. Calculation (given the ANOVA table is as follows):

3. With = 0.05 we reject if :

Source of

Variation

SS df MS F

A 658.845 1 658.845 46.652

B 255.38 1 255.38 18.083

AB 2 1 2 0.1416

Error 56.49 4 14.1225

Total 972.715 7

0H

1 1

1 1

1 1 1

for effect of factor A

for effect of factor B

for effect of interaction

A ,a ,ab r

B ,b ,ab r

AB , a b ,ab r

F F

F F

F F

4. From ANOVA/table F, the critical and F effects are given as follow:

5. Factor A : since , thus we reject

We conclude that the difference level of A effect the response

Factor B : since , thus we reject

We conclude that the difference level of B effect the response

Interaction: since , thus we failed to reject

We conclude that no interaction between factor A and factor B.

0 05 1 41 1

0 05 1 41 1

0 05 1 71 1 1

46 652 and 7 71

18 083 and 7 71

0 1416 and 7 71

A . , ,,a ,ab r

B . , ,,b ,ab r

AB . , ,, a b ,ab r

F . F F .

F . F F .

F . F F .

0 05 1 446 652 > 7 71A . , ,F . F . 0H

0 05 1 418 083 > 7 71B . , ,F . F .

0 05 1 40 1416 7 71AB . , ,F . F .

0H

0H

Exercise:In a study to determine which are the important source

of variation in an industrial process, 3 measurements are taken on

yield for 3 operators chosen randomly and 4 batches a raw

materials chosen randomly. It was decided that a significance test should

be made at the 0.05 level of significance to determine if the variance

components due to batches, operators, and interaction are significant. In

addition, estimates of variance components are to be computed.

The data are as follows, with the response being percent by weight.

Batch

1 2 3 4

Operator

1 66.968.167.2

68.367.467.7

69.069.867.5

69.370.971.4

2 66.365.465.8

68.166.967.6

69.768.869.2

69.469.670.0

3 65.666.365.2

66.066.967.3

67.166.267.4

67.968.468.7

Perform the analysis of variance of this experiment at level of significance

0.05. State your conclusion