Provide Information Regarding Statistics & Econometrics : Confounding in more than Two Blocks Lecture

CONFOUNDING

More than Two Blocks

Lecture - 35

In 2^k factorial experiment with small k like k = 2, 3, 4 confounding in two blocks is reasonable but high value of k like k = 6, 7, ... confounding in two blocks is not a wise. The block size is again significantly increase and the use of confounding technique is in vein. The 2^k factorial experiment for large k, it is required to confounded in more than two blocks. In such situation more than one higher order interaction is confounded.

Let us consider 2^5 factorial experiment is divided into 4 blocks, we confounded 2 interactions ABC and ADE. The four blocks are obtained by simultaneously solving the two equations,

X1 + X2 + X3 = 0, 1 (mod 2) and X1 + X4 + X5 = 0, 1 (mod 2).

The block 1 consist the treatment combination which satisfy X1 + X2 + X3 = 0 (mod 2) and

X1 + X4 + X5 = 0, (mod 2).

The Block 2 consist the treatment combination which satisfy X1 + X2 + X3 = 0 (mod 2) and

X1 + X4 + X5 = 1 (mod 2).

The Block 3 consist the treatment combination which satisfy X1 + X2 + X3 = 1 (mod 2) and

X1 + X4 + X5 = 0 (mod 2).

The Block 4 consist the treatment combination which satisfy X1 + X2 + X3 = 1 (mod 2) and

X1 + X4 + X5 = 1 (mod 2).

As we confounded ABC and ADE in the above 4 blocks, which are originally chosen for confounding are called generator. When we confounded two interactions (i.e. ABC and ADE) there must be another interaction confounded in blocks. This is called generalized interaction. The generalized interaction may be obtained from the multiplication of ABC and ADE.

In general, to divide 2^k factorial experiment into 2^r blocks; chose generators, their

( 2^q -q - 1) generalized interactions are obtained from all possible multiplication of the generators modulus 2.

Example:

To confound 2^6 factorial experiment in 8 blocks, we have 2^3 = 8 blocks. Here q = 3 generators.

There generalized interaction is (2^q - q - 1) = (2^3 - 3 - 1) = 4 generalized interactions.

Let the generators are ABC, ADE, BDF. The four generalized interactions are obtained as:

So, there are 3 generators ABC, ADE, BDF while 4 generalized interactions are BCDE, ACDF, ABEF and CEF.

The generalized interactions like BCDE, ACDF, and ABEF cannot be chosen as generators.

Example:

Confound 2^6 factorial experiment in 8 blocks.

Solution:

The number of blocks = 8 which is equivalent to 2^3, so, q = 3

The number of generalized interaction = (2^q - q - 1)

The number of generalized interaction = (2^3 - 3 - 1)

The number of generalized interaction = 4

Let the generators are ABC, ADE, BDF. The four generalized interactions are obtained as:

So, there are 3 generators ABC, ADE, BDF while 4 generalized interactions are BCDE, ACDF, ABEF and CEF.

The generalized interactions like BCDE, ACDF, and ABEF cannot be chosen as generators.

Example:

Confound the two generators ABC and ADE in 2^5 factorial experiment in 4 blocks. Find the number of generalized interaction. also determine the generalized interaction. Highlight the procedure to obtained the four blocks.

Solution:

The number of blocks = 4 which is equivalent to 2^2, so, q = 2

The number of generalized interaction = (2^q - q - 1)

The number of generalized interaction = (2^2 - 2 - 1)

The number of generalized interaction = 1

The generalized interaction is obtained by the product of generators.

ABC X ADE = A^2 BCDE

ABC X ADE = BCDE

The four blocks are obtained by solving simultaneously the following two equations:

X1 + X2 + X3 = 0, 1 (mod 2)

X1 + X4 + X5 = 0, 1 (mod 2)

Block I consist the treatment combinations which satisfy the equation:

X1 + X2 + X3 = 0 (mod 2) & X1 + X4 + X5 = 0 (mod 2)

Block II consist the treatment combinations which satisfy the equation:

X1 + X2 + X3 = 1 (mod 2) & X1 + X4 + X5 = 0 (mod 2)

Block III consist the treatment combinations which satisfy the equation:

X1 + X2 + X3 = 0 (mod 2) & X1 + X4 + X5 = 1 (mod 2)

Block IV consist the treatment combinations which satisfy the equation:

X1 + X2 + X3 = 1 (mod 2) & X1 + X4 + X5 = 1 (mod 2)

The four blocks can be obtained as:

ANOVA Table:

Question: In 2^5 factorial experiment placed in 4 blocks of size 8. The interaction ADE and BCDE are confounded. Layout the design and outline the ANOVA table.

Solution: In 2^5 factorial experiment (A, B, C, D, and E) in 4 blocks,

The number of blocks = 4

The number of blocks = 2^2

So, q = 2 (No. of Generators)

that's

ADE and BCDE

The number of generalized interactions = 2^q - q -1

The number of generalized interactions= 2^2 - 2 - 1

ADE X BCDE = ABCD^2 E^2

ADE X BCDE = ABC

The four blocks are obtained by solving simultaneously the following two equations:

X1 + X4 + X5 = 0, 1 (mod 2)

X2 + X3 + X4 + X5 = 0, 1 (mod 2)

Block I consist the treatment combinations which satisfy the equation:

X1 + X4 + X5 = 0 (mod 2) & X2 + X3 + X4 + X5 = 0 (mod 2)

Block II consist the treatment combinations which satisfy the equation:

X1 + X4 + X5 = 1 (mod 2) & X2 + X3 + X4 + X5 = 0 (mod 2)

Block III consist the treatment combinations which satisfy the equation:

X1 + X4 + X5 = 0 (mod 2) & X2 + X3 + X4 + X5 = 1 (mod 2)

Block IV consist the treatment combinations which satisfy the equation:

X1 + X4 + X5 = 1 (mod 2) & X2 + X3 + X4 + X5 = 1 (mod 2)

ANOVA Table:

SV	df
Block	3
Main effect	5
Two factors interaction	10
Three factors interaction	8 4 1	Pooled Error = 13
Four factors interaction
Five factors interaction
Total	31