2 k Experiments, Incomplete block designs for 2 k experiments, fractional 2 k experiments

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2k Experiments,Incomplete block designs for 2k

experiments, fractional 2k experiments

Factorial Experiments

• Dependent variable y

• k Categorical independent variables A, B, C, … (the Factors)

• Let– a = the number of categories of A– b = the number of categories of B– c = the number of categories of C– etc.

• t = abc... Treatment combinations

The Completely Randomized Design

• We form the set of all treatment combinations – the set of all combinations of the k factors

• Total number of treatment combinations– t = abc….

• In the completely randomized design n experimental units (test animals , test plots, etc. are randomly assigned to each treatment combination.– Total number of experimental units N = nt=nabc..

The ANOVA Table three factor experiment

Source SS df

A SSA a-1

B SSB b-1

C SSC c-1

AB SSAB (a-1)(b-1)

AC SSAC (a-1)(c-1)

BC SSBC (b-1)(c-1)

ABC SSABC (a-1)(b-1)(c-1)

Error SSError abc(n-1)

• If the number of factors, k, is large then it may be appropriate to keep the number of levels of each factor low (2 or 3) to keep the number of treatment combinations, t, small.

t = 2k if a = b =c = ... =2 or

t = 3k if a = b =c = ... =3

• The experimental designs are called 2k and 3k designs

The ANOVA Table 23 experiment

Source Sum of Squares d.f.

A SSA 1

B SSB 1

C SSC 1

AB SSAB 1

AC SSAC 1

BC SSBC 1

ABC SSABC 1

Error SSError 23(n – 1)

Notation for treatment combinations

for 2k experiments There are several methods for indicating treatment combinations in a 2k experiment and 3k experiment.

1.A sequence of small letters representing the factors with subscripts (0,1 for 2k experiment and 0, 1, 2 for a 3k experiment)

2.A sequence of k digits (0,1 for 2k experiment and 0, 1, 2 for a 3k experiment.

3.A third way of representing treatment combinations for 2k experiment is by representing each treatment combination by a sequence of small letters. If a factor is at its high level, it’s letter is present. If a factor is at its low level, it’s letter is not present.

The 8 treatment combinations in a 23 experiment

1.(a0, b0, c0), (a1, b0, c0), (a0, b1, c0), (a0, b0, c1),

(a1, b1, c0), (a1, b0, c1), (a0, b1, c1), (a1, b1, c1)

2.000, 100, 010, 001, 110, 101, 011, 111

3.1, a, b, c, ab, ac, bc, abc

In the last way of representing the treatment combinations, a more natural ordering is:

1, a, b, ab, c, ac, bc, abc

Using this ordering the 16 treatment combinations in a 24 experiment

1, a, b, ab, c, ac, bc, abc, d, da, db, dab, dc, dac, dbc, dabc

Notation for Linear contrasts treatment

combinations in a 2k experiments The linear contrast for 1 d.f. representing the Main effect of A

LA = (1 + b + c + bc) – (a + ab +ac + abc)

= comparison of the treatment combinations when A is at its low level with treatment combinations when A is at its high level.

Note: LA = (1 - a) (1 + b) (1 + c)

also

LB = (1 + a) (1 - b) (1 + c)

= (1 + a + c + ac) – (b + ab +bc + abc)

LC = (1 + a) (1 + b) (1 - c)

= (1 + a + b + ab) – (c + ca +cb + abc)

The linear contrast for 1 d.f. representing the interaction AB

LAB = (1 - a) (1 - b) (1 + c)

= (1 + ab + c + abc) – (a + b +ac + bc)

= comparison of the treatment combinations where A and B are both at a high level or both at a low level with treatment combinations either A is at its high level and B is at a low level or B is at its high level and A is at a low level.

LAC = (1 - a) (1 + b) (1 - c)

= (1 + ac + b + abc) – (a + c +ab + bc)

LBC = (1 + a) (1 - b) (1 - c)

= (1 + bc + a + abc) – (b + c +ac + ab)

The linear contrast for 1 d.f. representing the interaction ABC

LABC = (1 - a) (1 - b) (1 - c)

= (1 + ab + ac + bc) – (a + b + c + abc)

In general Linear contrasts are of the form:

L = (1 ± a)(1 ± b)(1 ± c) etc

We use minus (-) if the factor is present in the effect and plus (+) if the factor is not present.

The sign of coefficients of each treatment for each contrast (LA, LB, LAB, LC, LAC, LBC, LABC) is illustrated in the table below:

I A B AB C AC BC ABC

1 + + + + + + + +a + - + - + - + -b + + - - + + - -ab + - - + + - - +c + + + + - - - -

ac + - + - - + - +bc + + - - - - + +abc + - - + - + + -

Treat

ment

contrast

For the main effects (LA, LB, LC) the sign is negative (-) if the letter is present in the treatment, positive (+) if the letter is not present. The interactions are products of the main effects:

+ × + = + - × + = - + × - = - - × - = +

Yates Algorithm

This is a method for computing the Linear contrasts of the effects and their sum of squares (S.S.) The algorithm is illustrated with the Table on the next slide

The algorithm is as follows:

1.Treatments are listed in the standard order (1, a, b, ab, c, ac, bc, abc etc.) i.e. Starting with 1, then adding one letter at a time followed by all combinations with letters that have been previously added.

TreatmentTotal yield I II III Effect SS

1 121 302 663 1362 Total

a 181 361 699 408 A 5202.00

b 104 296 213 166 B 861.12

ab 257 403 195 188 AB 1104.50

c 123 60 59 36 C 40.50

ac 173 153 107 -18 AC 10.12

ab 129 50 93 48 AB 72.00

abc 274 145 95 2 ABC 0.12

Table: Illustration of Yates Algorithm for a 23 factorial Design (# of replicates n = 4)

Yates Algorithm (continued)

2. In the yield column enter the total yields for each treatment combination.

3. Fill in as many columns headed by Roman numerals as there are factors in the experiment in the following way.

a. Add successive pairs in the previous column. (1st +2nd), (3rd + 4th) etc

b. Subtract successive pairs in the previous column (2nd - 1st), (4th - 3rd) etc

4. To obtain entries in column II repeat steps 3a and 3b on the entries of column I.

5. To obtain entries in column III repeat steps 3a and 3b on the entries of column II

6. Continue in this way until as many columns have been filled as factors.

Yates Algorithm (continued) - Computation of SS’s

7. Square the effect total (entry in last column).

8. Divide the result by the number of observations n2k.

Strategy for a single replication (n = 1)

If n = 1 then there is 0 df for estimating error. In practice the higher order interactions are not usually present. One makes this assumption and pools together these degrees of freedom to estimate Error

The ANOVA Table 23 experimentSource Sum of Squares d.f.

A SSA 1

B SSB 1

C SSC 1

AB SSAB 1

AC SSAC 1

BC SSBC 1

ABC SSABC 1

Error SSError 23(n – 1)

In a 7 factor experiment (each at two levels) there are 27 =128 treatments.

There are:

EffectsMain 71

7

nsinteractiofactor -2 ,212

7


7


7


7


7

ninteractiofactor -7 ,17

7

Pool together these degrees of freedom to estimate Error

ANOVA table:

Source d.f.

Main Effects 7

2-factor interactions 21





7-factor interaction 1

Randomized Block design for 2k experiments

1

a

b

ab

c

ac

bc

abc

1

a

b

ab

c

ac

bc

abc

1

a

b

ab

c

ac

bc

abc

1

a

b

ab

c

ac

bc

abc

1

a

b

ab

c

ac

bc

abc

Blocks1 2 3 4 n...

A Randomized Block Design for a 23 experiment

The ANOVA Table 23 experiment in RB design


Blocks SSBlocks n - 1

A SSA 1

B SSB 1

C SSC 1

AB SSAB 1

AC SSAC 1

BC SSBC 1

ABC SSABC 1

Error SSError (23 – 1)(n – 1)

Incomplete Block designs for 2k experiments

Confounding

1

a

b

ab

c

ac

bc

abc

1

a

b

ab

c

ac

bc

abc

1

a

b

ab

c

ac

bc

abc

1

a

b

ab

c

ac

bc

abc

1

a

b

ab

c

ac

bc

abc

Blocks1 2 3 4 n...

A Randomized Block Design for a 23 experiment

Incomplete Block designs for 2k experiments

A Randomized Block Design for a 2k experiment requires blocks of size 2k. The ability to detect treatment differences depends on the magnitude of the within block variability. This can be reduced by decreasing the block size.

abc

a

b

c

1 2

1

bc

ac

ab

Blocks

Example: a 23 experiment in blocks of size 4 (1 replication). The ABC interaction is confounded with blocks

In addition to measuring the ABC interaction it is also subject to block to block differences.

abc

a

b

c

1 2

1

bc

ac

ab

Blocks In this experiment the linear contrast

LABC = (1 + ab + ac + bc) – (a + b + c + abc)

The ABC interaction it is said to be confounded with block differences.

The linear contrasts

LA = (1 + b + c + bc) – (a + ab +ac + abc)

LB = (1 + a + c + ac) – (b + ab +bc + abc)

LC = (1 + a + b + ab) – (c + ca +cb + abc

LAB = (1 + ab + c + abc) – (a + b +ac + bc)

LAC = (1 + ac + b + abc) – (a + c +ab + bc)

LBC = (1 + bc + a + abc) – (b + c +ac + ab)

are not subject to block to block differences

abc

a

b

c

1 2

1

bc

ac

ab

Blocks

To confound an interaction (e. g. ABC) consider the linear contrast associated with the interaction:

LABC = 1 + ab + ac + bc – a – b – c – abc

Assign treatments associated with positive (+) coefficients to one block and treatments associated with negative (-) coefficients to the other block

The ANOVA Table 23 experiment in incomplete design with 2 blocks of size 4


Blocks SSBlocks 1

A SSA 1

B SSB 1

C SSC 1

AB SSAB 1

AC SSAC 1

BC SSBC 1

Total SSTotal 7

Confounding more than one interaction

to further reduce block size

Example: contrasts for 23 experiment


1 + + + + + + + +a + - + - + - + -b + + - - + + - -ab + - - + + - - +c + + + + - - - -

ac + - + - - + - +bc + + - - - - + +abc + - - + - + + -

Treat

ment

contrast

If I want to confound ABC, one places the treatments associated with the positive sign (+) in one block and the treatments associated with the negative sign (-) in the other block.

If I want to confound both BC and ABC, one chooses the blocks using the sign categories (+,+) (+,-) (-,+) (-,-)

Comment: There will also be a third contrast that will also be confounded

1

bc

Block 1 Block 2 Block 3 Block 4

Example: a 23 experiment in blocks of size 2 (1 replicate). BC and ABC interaction is confounded in the four block.

a

abc

ab

ac

b

c

LABC = (1 + ab + ac + bc) – (a + b + c + abc) and

LBC = (1 + bc + a + abc) – (b + c +ac + ab)

are confounded with blocks

LA = (1 + b + c + bc) – (a + ab +ac + abc)

is also confounded with blocks

LB = (1 + a + c + ac) – (b + ab +bc + abc)

LC = (1 + a + b + ab) – (c + ca +cb + abc

LAB = (1 + ab + c + abc) – (a + b +ac + bc)

LAC = (1 + ac + b + abc) – (a + c +ab + bc)

are not subject to block to block differences

The ANOVA Table 23 experiment in incomplete design with 4 blocks of size 2 (ABC, BC and hence A confounded with blocks)


Blocks SSBlocks 3

B SSB 1

C SSC 1

AB SSAB 1

AC SSAC 1

Total SSTotal 7

There are no degrees of freedom for Error.Solution: Assume either one or both of the two factor interactions are not present and use those degrees of freedom to estimate error

Rule: (for determining additional contrasts that are confounded with block)1.“Multiply” the confounded interactions together.2.If a factor is raised to the power 2, delete it

Example:Suppose that ABC and BC is confounded, then so also is (ABC)(BC) = AB2C2 = A.A better choice would be to confound AC and BC, then the third contrast that would be confounded would be (AC)(BC) = ABC2 = AB


1 + + + + + + + +a + - + - + - + -b + + - - + + - -ab + - - + + - - +c + + + + - - - -

ac + - + - - + - +bc + + - - - - + +abc + - - + - + + -

Treat

ment

contrast

If I want to confound both AC and BC, one chooses the blocks using the sign categories (+,+) (+,-) (-,+) (-,-). As noted this would also confound (AC)(BC) = ABC2 = AB.

1

abc

Block 1 Block 2 Block 3 Block 4

b

ac

a

bc

ab

c

The ANOVA Table 23 experiment in incomplete design with 4 blocks of size 2 (AC, BC and hence AB confounded with blocks)


Blocks SSBlocks 3

A SSA 1

B SSB 1

C SSC 1

ABC SSABC 1

Total SSTotal 7

There are no degrees of freedom for Error.Solution: Assume that the three factor interaction is not present and use this degrees of freedom to estimate error

Partial confounding

1

bc

abc

a

Block 1

ab

c

ac

b

Replicate 1BC confounded

Block 2

abc

b

1

ac

Block 3

bc

a

ab

c

Replicate 2AC confounded

Block 4

1

c

ab

abc

Block 5

a

b

ac

bc

Replicate 3AB confounded

Block 6

Example: a 23 experiment in blocks of size 4 (3 replicates). BC interaction is confounded in 1st replication. AC interaction is confounded in 2nd replication. AB interaction is confounded in 3rd replication.

The main effects (A, B and C) and the three factor interaction ABC can be estimated using all three replicates. The two factor interaction AB can be estimated using replicates 1 and 2, AC using replicates 1 and 3, BC using replicates 2 and 3,

The ANOVA Table


Reps SSBlocks 2

Blocks within Reps SSBlocks(Reps) 3

A SSA 1

B SSB 1

C SSC 1

AB SSAB 1 Reps I,II

AC SSAC 1 Reps I,III

BC SSBC 1 Reps II,III

ABC SSABC 1

Error SSError 11

Total SSTotal 23

Example: A chemist is interested in determining how purity (Y) of a chemical product, depends on agitation rate (A), base component concentration (B) and concnetration of reagent (C). He decides to use a 23 design. Only 4 runs can be done each day (Block) and he wanted to have 3 replications of the experiment.

Replicate 1BC confounded

Replicate 2AC confounded

Replicate 3AB confounded

day 1 day 2 day 3 day 4 day 5 day 6

1 25 ab 43 abc 39 bc 38 1 26 a 43

bc 34 c 30 b 29 a 37 c 32 b 34

abc 42 ac 40 1 27 ab 46 ab 52 ac 40

a 25 b 33 ac 40 c 34 abc 51 bc 36

The ANOVA Table

SourceSum of Squares d.f.

Mean Square F

Reps 111.00 2 55.50

Blocks within Reps 108.00 3 36.00

A 600.00 1 600.00 40.6**

B 253.50 1 253.50 17.2**

C 54.00 1 54.00 3.7(ns)

AB (Reps I,II) 6.25 1 6.25 < 1

AC (Reps I,III) 1.00 1 1.00 < 1

BC (Reps II,III) 6.25 1 6.25 < 1

ABC 13.50 1 13.50 < 1

Error 162.50 11 14.77

Total 1316.00 23

F0.05(1,11) = 4.84 and F0.01(1,11) = 9.65

Fractional Factorials

In a 2k experiment the number of experimental units required may be quite large even for moderate values of k.

For k = 7, 27 = 128 and n27 = 256 if n = 2.

Solution:1.Use only n = 1 replicate and use higher order interactions to estimate error. It is very rare thqt the higher order interactions are significant

2.An alternative solution is to use ½ a replicate, ¼ a replicate, 1/8 a replicate etc. (i.e. a fractional replicate)

2k – 1 = ½ 2k design, 2k – 2 = ¼ 2k design

In a fractional factorial design, some ot he effects (interactions or main effects) may not be estimable. However it may be assumed that these effects are not present (in particular the higher order interactions)

Example: 24 experiment, A, B, C, D - contrasts

I A B AB C AC BC ABC D AD BD ABD CD ACD BCD ABCD1 + + + + + + + + + + + + + + + +a + - + - + - + - + - + - + - + -b + + - - + + - - + + - - + + - -ab + - - + + - - + + - - + + - - +c + + + + - - - - + + + + - - - -

ac + - + - - + - + + - + - - + - +bc + + - - - - + + + + - - - - + +abc + - - + - + + - + - - + - + + -d + + + + + + + + - - - - - - - -ad + - + - + - + - - + - + - + - +bd + + - - + + - - - - + + - - + +

abd + - - + + - - + - + + - - + + -cd + + + + - - - - - - - - + + + +acd + - + - - + - + - + - + + - + -bcd + + - - - - + + - - + + + + - -

abcd + - - + - + + - - + + - + - - +

To construct a ½ replicate of this design in which the four factor interaction, ABCD, select only the treatment combinations where the coefficient is positive (+) for ABCD

The treatments and contrasts of a ½ 24 = 24-1 experiment

Notice that some of the contrasts are equivalente.g.

A and BCD, B and ACD, etcIn this case the two contrasts are said to be aliased. Note the defining contrast, ABCD is aliased with the constant term I. To determine aliased contrasts multiply the any effect by the effect of the defining contraste.g. (A)×(ABCD) = A2BCD = BCD

I A B AB C AC BC ABC D AD BD ABD CD ACD BCD ABCD1 + + + + + + + + + + + + + + + +

ab + - - + + - - + + - - + + - - +ac + - + - - + - + + - + - - + - +bc + + - - - - + + + + - - - - + +ad + - + - + - + - - + - + - + - +bd + + - - + + - - - - + + - - + +cd + + + + - - - - - - - - + + + +

abcd + - - + - + + - - + + - + - - +

Aliased contrasts in a 24 -1 design with ABCD the defining contrast

1.A with BCD2.B with ACD3.C with ABD4.D with ABC5.AB with CD6.AC with BD7.AD with BC

If an effect is aliased with another effect you can either estimate one or the other but not both

The ANOVA for a 24 -1 design with ABCD the defining contrast

Source df

A 1

B 1

C 1

D 1

AB 1

AC 1

AD 1

Total 7

Example: ¼ 24 experiment

I A B AB C AC BC ABC D AD BD ABD CD ACD BCD ABCD1 + + + + + + + + + + + + + + + +a + - + - + - + - + - + - + - + -b + + - - + + - - + + - - + + - -ab + - - + + - - + + - - + + - - +c + + + + - - - - + + + + - - - -

ac + - + - - + - + + - + - - + - +bc + + - - - - + + + + - - - - + +abc + - - + - + + - + - - + - + + -d + + + + + + + + - - - - - - - -ad + - + - + - + - - + - + - + - +bd + + - - + + - - - - + + - - + +

abd + - - + + - - + - + + - - + + -cd + + + + - - - - - - - - + + + +acd + - + - - + - + - + - + + - + -bcd + + - - - - + + - - + + + + - -

abcd + - - + - + + - - + + - + - - +

To construct a ¼ replicate of the 24 design. Choose two defining contrasts, AB and CD, say and select only the treatment combinations where the coefficient is positive (+) for both AB and CD

The treatments and contrasts of a ¼ 24 = 24-2 experiment

Aliased contrasts1.I and AC and BD and ABCD2.A and C and ABD and BCD3.B and ABC and D and ACD4.AB and BC and AD and CD

I A B AB C AC BC ABC D AD BD ABD CD ACD BCD ABCD

1 + + + + + + + + + + + + + + + +ac + - + - - + - + + - + - - + - +bd + + - - + + - - - - + + - - + +

abcd + - - + - + + - - + + - + - - +

The ANOVA for a 24 -1 design with ABCD the defining contrast

Source df

A 1

B 1

AB 1

Total 3

There may be better choices for the defining contrastsThe smaller fraction of a 2k design becomes more appropriate as k increases.

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2 k Experiments, Incomplete block designs for 2 k experiments, fractional 2 k experiments