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

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

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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 ntnabc..

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The ANOVA Table three factor experiment
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• 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

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The ANOVA Table 23 experiment
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Notation for treatment combinations for 2k
experiments

• There are several methods for indicating
  treatment combinations in a 2k experiment and 3k
  experiment.
• A sequence of small letters representing the
  factors with subscripts (0,1 for 2k experiment
  and 0, 1, 2 for a 3k experiment)
• A sequence of k digits (0,1 for 2k experiment and
  0, 1, 2 for a 3k experiment.
• 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, its
  letter is present. If a factor is at its low
  level, its letter is not present.

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• The 8 treatment combinations in a 23 experiment
• (a0, b0, c0), (a1, b0, c0), (a0, b1, c0), (a0,
  b0, c1),
• (a1, b1, c0), (a1, b0, c1), (a0, b1, c1), (a1,
  b1, c1)
• 000, 100, 010, 001, 110, 101, 011, 111
• 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

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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)

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• 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)

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• 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.

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

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 - - - - -
-
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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
• 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.

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Table Illustration of Yates Algorithm for a 23
factorial Design ( of replicates n 4)
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Yates Algorithm (continued)

• In the yield column enter the total yields for
  each treatment combination.
• Fill in as many columns headed by Roman numerals
  as there are factors in the experiment in the
  following way.
• Add successive pairs in the previous column. (1st
  2nd), (3rd 4th) etc
• Subtract successive pairs in the previous column
  (2nd - 1st), (4th - 3rd) etc
• To obtain entries in column II repeat steps 3a
  and 3b on the entries of column I.
• To obtain entries in column III repeat steps 3a
  and 3b on the entries of column II
• Continue in this way until as many columns have
  been filled as factors.

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Yates Algorithm (continued) - Computation of SSs

• Square the effect total (entry in last column).
• Divide the result by the number of observations
  n2k.

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Strategy for a single replication (n 1)
The ANOVA Table 23 experiment

• 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

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

• Pool together these degrees of freedom to
  estimate Error

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Randomized Block design for 2k experiments
Blocks
...
n
1
2
3
4
A Randomized Block Design for a 23 experiment
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The ANOVA Table 23 experiment in RB design
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Incomplete Block designs for 2k
experimentsConfounding
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A Randomized Block Design for a 23 experiment
Blocks
...
n
1
2
3
4
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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.
Blocks
1
2
Example a 23 experiment in blocks of size 4 (1
replication). The ABC interaction is confounded
with blocks
abc
1
a
bc
b
ac
c
ab
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Blocks
In this experiment the linear contrast
1
2
LABC (1 ab ac bc) (a b c abc)
In addition to measuring the ABC interaction it
is also subject to block to block differences.
abc
1
a
bc
b
ac
The ABC interaction it is said to be confounded
with block differences.
c
ab
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
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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
Blocks
1
2
abc
1
a
bc
b
ac
c
ab
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The ANOVA Table 23 experiment in incomplete
design with 2 blocks of size 4
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Confounding more than one interaction to
further reduce block size
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• Example contrasts for 23 experiment

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
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Example a 23 experiment in blocks of size 2 (1
replicate). BC and ABC interaction is confounded
in the four block.
Block 1
Block 2
Block 3
Block 4
1
a
ab
b
bc
abc
ac
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
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The ANOVA Table 23 experiment in incomplete
design with 4 blocks of size 2 (ABC, BC and hence
A confounded with blocks)
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
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• Rule (for determining additional contrasts that
  are confounded with block)
• Multiply the confounded interactions together.
• 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
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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.
Block 1
Block 2
Block 3
Block 4
1
b
a
ab
abc
ac
bc
c
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The ANOVA Table 23 experiment in incomplete
design with 4 blocks of size 2 (AC, BC and hence
AB confounded with blocks)
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
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Partial confounding
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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.
Replicate 1 BC confounded
Replicate 2 AC confounded
Replicate 3 AB confounded
Block 1
Block 2
Block 3
Block 4
Block 5
Block 6
1
ab
abc
bc
1
a
bc
c
b
a
c
b
abc
ac
1
ab
ab
ac
a
b
ac
c
abc
bc
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,
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The ANOVA Table
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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.
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The ANOVA Table
F0.05(1,11) 4.84 and F0.01(1,11) 9.65
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Fractional Factorials
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• 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
• Use only n 1 replicate and use higher order
  interactions to estimate error. It is very rare
  thqt the higher order interactions are
  significant
• 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

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• 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)

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Example 24 experiment, A, B, C, D - contrasts
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
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The treatments and contrasts of a ½ 24 24-1
experiment
Notice that some of the contrasts are
equivalent e.g. A and BCD, B and ACD, etc In
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 contrast e.g. (A)(ABCD)
A2BCD BCD
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• Aliased contrasts in a 24 -1 design with ABCD the
  defining contrast
• A with BCD
• B with ACD
• C with ABD
• D with ABC
• AB with CD
• AC with BD
• AD with BC
• If an effect is aliased with another effect you
  can either estimate one or the other but not both

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The ANOVA for a 24 -1 design with ABCD the
defining contrast
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Example ¼ 24 experiment
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
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The treatments and contrasts of a ¼ 24 24-2
experiment

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

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The ANOVA for a 24 -1 design with ABCD the
defining contrast
There may be better choices for the defining
contrasts The smaller fraction of a 2k design
becomes more appropriate as k increases.