Provide Information Regarding Statistics & Econometrics : Quarter Fraction of a 2^k Factorial Experiment Lecture

Quarter Fraction of a 2^k Factorial Experiment

Lecture - 38

When the fraction index is equal to 2 that’s p = 2 of 2^k factorial experiment, then it is called quarter or 1/4 fraction factorial experiment.

The ¼ replicate of factorial is obtained by blocking 2 generating contrast. The generalize interaction of those 2 generators is also the defining contrast. Thus, if P and Q represent the generators chosen by the generalized interaction is PQ, the four fraction can be obtained as:

The defining relation of this design would be:

I = P = Q = PQ

The aliases of any effect are obtained by multiplying each contrast in the identity mod 2 with the effect. Since there are three contrasts in the defining relation. Each effect have 3 aliases.

For example, in factorial experiment with the defining relation:

I = ABE = CDE = ABCD

The alias of A is

A X ABE = BE

A X CDE =ACDE

A X ABCD = BCD

A = BE =ACDE = BCD

The defining contrast should involve contrasts so that are no important effect is aliased with any other important effect.

Analysis

To obtain a treatment combination of 2^(K-2) factorial experiment, first write down the treatment combination for 2^(K-2) using “+” and “-” sign for a full factorial experiment with (k – 2) factors. The additional factors are then identified with these contrasts in the defining relation having these factors.

Consider the construction of 2^(K-2) design with 5 factors i.e. A, B, C, D, E each with 2 levels. To obtain 2^(K-2) treatment combinations for the factors, first such that in each of the defining contrast, there must be one of these interacted letters. The rest of the procedure is the same as before.

The effect is obtained by:

The sum of square of an effect is given by;

Example:

write 1/4 replicate of a 2^5 factorial experiment outlines the pairs of aliases and the first two columns of ANOVA.

Solution: 2^5 factorial experiment consists 5 factors A, B, C, D, E and each factor have 2 levels. The replicate of a 2^5 is 2^(5-2) treatment combination. Let the two generators are ABCD and ABE, then one generalized interaction is

ABCD X ABE = CDE

The defining relation is

I = ABCD = ABE = CDE

The alias structure of the design is:

A X I

A = BCD = BE = ACDE

B = ACD = AE = BCDE

C = ABD = ABCE = DE

AC = BD = BCE =ADE

BC = AD = ACE = BDE

ABC = D = CE =A BDE

The eight-treatment combination for a 1/4 fraction of 2^5 factorial experiment are obtained by first writing eight combinations for 2^(5-2) = 2^3 factorial in AB and C. the rest of the factors D and E are identified with D = ABC and.AB = E The design is obtained as:

ANOVA Table for 2^(5-2) factorial experiment:

SV	df
A	1
B	1
C	1
D	1
E	1
Error	2
Total	7

Example:

Construct 1/4 replicate of a 2^6 factorial experiment by choosing the four factors interaction as defining contrast. Write down the pairs of aliases and outline ANOVA?

Solution:

The 2^6 factorial experiment consists 6 factors A, B, C, D, E, F and each factor have 2 levels. The 1/4 replicate of a is 2^6 = 2^4 16 treatment combination. Let the two generators are ABCD and CDEF, then one generalized interaction is

ABCD X CDEF = ABEF

So, the defining relation is

I = ABCD = CDEF = ABEF

The alias structure of the other design would be:

TC	P	Q	PQ
TC	ABCE	BCDF	ADEF
A	BCF	ABCDF	DEF
B	ACE	CDE	ABDEF
AB	CE	ACDE	BDEF
C	ABE	BDF	ACDEF
AC	BE	ABDF	CDEF
BC	AE	DF	ABCDEF
ABC	E	ADF	BDCEF
D	ABCDE	BCF	AEF
AD	BCDE	ABCF	EF
BD	ACDE	CF	ABEF
CD	ABDE	BF	ACEF
ACD	BDE	ABF	CEF
BCD	ADE	F	ABCEF
ABCD	DE	AF	BCEF

The sixteen treatment combinations of 1/4 fraction of 2^6 = 2^4 = 16 factorial are obtained by first writing sixteen combinations for factorial in A, B, C, D.

TC	A	B	C	D	E = ABC
1	-1	-1	-1	-1	-1
a	1	-1	-1	-1	1
b	-1	1	-1	-1	1
ab	1	1	-1	-1	-1
c	-1	-1	1	-1	1
ac	1	-1	1	-1	-1
bc	-1	1	1	-1	-1
abc	1	1	1	-1	1
d	-1	-1	-1	1	-1
ad	1	-1	-1	1	1
bd	-1	1	-1	1	1
abd	1	1	-1	1	-1
cd	-1	-1	1	1	1
acd	1	-1	1	1	-1
bcd	-1	1	1	1	-1
abcd	1	1	1	1	1

ANOVA Table:

SV	DF	SS	MS
A	1
B	1
C	1
D	1
E	1
F	1
Error	9
Total	15

Example: Test the significance at 5% of ¼ of 2^6 factorial experiment using the data below:

TC	A	B	C	D	Yield
1	-1	-1	-1	-1	6
2	1	-1	-1	-1	10
3	-1	1	-1	-1	32
4	1	1	-1	-1	60
5	-1	-1	1	-1	4
6	1	-1	1	-1	15
7	-1	1	1	-1	26
8	1	1	1	-1	60
9	-1	-1	-1	1	8
10	1	-1	-1	1	12
11	-1	1	-1	1	34
12	1	1	-1	1	60
13	-1	-1	1	1	16
14	1	-1	1	1	5
15	-1	1	1	1	37
16	1	1	1	1	52

Solution:

The ¼ of 2^6 = 2^4 = 16 treatment combinations

ANOVA Table:

TC	df	SS	MS	F	Ftab
A	1	770.0625	770.0625	8.65	5.11
B	1	5076.5625	5076.5625	57.03
C	1	3.0625	3.0625	0.033
D	1	7.5625	7.5625	0.078
E	1	0.5625	0.5625	0.006
F	1	0.5625	0.5625	0.006
Error	9	801.0625	89
Total	15

Example:

Construct 1/4 replicate of a 2^7 factorial experiment by choosing the four factors interaction as defining contrast. Write down the pairs of aliases and outline ANOVA?

Solution:

The 2^7 factorial experiment consists 7 factors A, B, C, D, E, F, G and each factor have 2 levels. The 1/4 replicate of a is 2^7 = 2^5 = 32 treatment combination. Let the two generators areLet the two generators P = ABDE and Q = BCDFG and one generalized interaction PQ = ACEFG

TC	P = ABDE	Q = BCDFG	PQ = ACEFG
ABD	E	ACFG	BCDEFG
BCD	ACE	F	ABEF
ABCD	CE	AF	BDEF
ACE	BD	ABFG	FG
ABCE	CE	ADEFG	BFG
ABCDE	CE	G	BDFG

We contract 4 fractions as:

ANOVA Table for 2^(7-2) factorial experiment

SV	df
A	1
B	1
C	1
D	1
E	1
F	1
G	1
Error	24
Total	31

1 / 8 Fraction of a 2^k Factorial Experiment

When the fraction index is equal to 3 that’s 2^k of factorial experiment, then it is called quarter or 1/8 fraction factorial experiment.

The 1/8 replicate of 2^k factorial is obtained by blocking 3 generating contrasts. The generalize interaction of those 3 generators is also the defining contrast.

Example: 1 / 8 replicate of 2^k factorial experiment.

The treatment combination is

The 3 generator ABCE, BCDF, ACDG

The three generators are E = ABC, F = BCD, G = ACD

I = ABCD = BCDF = ACDG

The aliases can obtained in the same manner discussed in earlier section.

The 8 fractions is obtained as: