MultiFactor Studies

1
Multi-Factor Studies

• Stat 701 Lecture
• E. Pena

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An Example with Two Factors

• To study effects of carbon content and tempering
  temperature on the strength of steel.
• Factor A Carbon Content
• Factor A Levels
• a1 Low Carbon level a2 High Carbon level
• Factor B Tempering Temperature
• Factor B Levels
• b1 Low temperature b2 High temperature
• This will be an example of a 22-factorial design

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Factor Level Combinations
Possible Treatment Combinations
Treatment 1 a1b1 Low Carbon and Low
Temperature Treatment 2 a1b2 Low Carbon and
High Temperature Treatment 3 a2b1 High Carbon
and Low Temperature Treatment 4 a2b2 High
Carbon and High Temperature
Possible Experimental Approaches

• Just use a completely randomized design with the
  four
• treatments above. What will be the disadvantages?
• Employ a factorial experiment where the two
  factors are delineated so their main effects and
  interaction effects could be ascertained.

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Why Multi-Factor Studies?
Advantages

• Efficiency and economy
• Informativeness
• Validity of findings

Disadvantages

• Could be costly if not properly designed.
• If interaction effects are strong, will be hard
  to determine or interpret main effects of
  factors.
• May need to employ fractional designs where only
  a subset of all possible treatments are included.

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Two-Factor Factorial Model(Balanced Design)
Let there be two factors A and B
Factor A has levels a1, a2, , aA
Factor B has levels b1, b2, , bB
Number of treatment combinations AB
In a balanced design, the same number
ofexperimental units per treatment combination
is allocated. Let k denote the number of eus
allocated to each treatment combination. Then n
ABk is the total number of eus in the study.
Note In allocating eus, the principle of
randomization should be observed.
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Model for Two-Factor Factorial Analysis
Yijl lth observation in treatment (ai,bj)
Yijl mij eijl
mij is the mean response of eus assigned (aibj)
eijl is the error component for the lth unit in
(aibj)

• Assumptions (fixed effects model)
• mij are fixed (but unknown) quantities
• Errors each have mean zero
• Errors are uncorrelated (independent) from each
  other
• Errors have equal variances
• Errors have normal distributions

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Interpretation of Parameters
Consider a two-factor study with Factor A having
2 levels and Factor B having 3 levels. For each
treatment combination, we have the (population)
mean response mij. We may summarize this in a
table of form
Factor A Main Effects ai mi. - m.. for
i1,2,,A Factor B Main Effects bj m.j - m..
for j1,2,,B Interaction Effects (ab)ij
mij - mi. - m.j m.. for i1,,A j1,..,B
Model mij m.. ai bj (ab)ij for
i1,,A j1,,B.
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Interpretation continued
Identities
Case 1 No interaction effects is when (ab)ij 0
for all i,j. What happens in this case? Pictorial
representation.
Case 2 Interaction effects are present.
Pictorial representation. Are main effects
meaningful? Strong and weak interactions.
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Two Examples
Example 1 A model without interaction effects.
Factor A Main Effects a1 4 - 5 -1 a2 6 -
5 1
Factor B Main Effects b1 5 - 5 0 b2 8 -
5 3 b3 2 - 5 -3
Interaction Effects (ab)11 4 - 4 - 5 5 0
all of them are zeros.
A Pictorial Representation of these Treatment
Means?
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Examples continued
Example 2 A model with interaction effects.
Factor A Main Effects a1 5 - 5 0 a2 5 -
5 0
Factor B Main Effects b1 5 - 5 0 b2 8 -
5 3 b3 2 - 5 -3
Interaction Effects (ab)11 4 - 5 - 5 5 -1
(ab)12 9 - 5 - 8 5 1 (ab)13 2 - 5 - 2
5 0 (ab)21 6 - 5 - 5 5 1 (ab)22 7 -
5 - 8 5 -1 (ab)23 2 - 5 - 2 5 0. Note
that the sum of these effects is 0.
Would appear that there are no differences in
Factor A levels!
Pictorial Representation?
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Example Drug Development for Hay Fever
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Tabular Presentation of Factorial Data with
Totals and Means
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Estimates of Main and Interaction Effects

• Factor A Main Effects
• A1 3.88 - 7.18 -3.3
• A2 7.83 - 7.18 0.65
• A3 9.83 - 7.18 2.65

• Factor B Main Effects
• B1 4.63 - 7.18 -2.55
• B2 7.93 - 7.18 0.75
• B3 8.98 - 7.18 1.80

Interaction Effects (A Level, B Level) (A1, B1)
2.475 - 3.88 - 4.63 7.18 1.145 (A1, B2) 4.6
- 3.88 - 7.93 7.18 -0.03 (A1, B3) 4.575 -
3.88 - 8.98 7.18 -1.105 (A2, B1) 5.45 - 7.83
- 4.63 7.18 0.17 (A2, B2) 8.925 - 7.83 -
7.93 7.18 0.345 (A2, B3) 9.125 - 7.83 - 8.98
7.18 -.505 (A3, B1) 5.975 - 9.83 - 4.63
7.18 -1.305 (A3, B2) 10.275 - 9.83 - 7.93
7.18 -.305 (A3, B3) 13.25 - 9.83 - 8.98 7.18
1.62
Estimates of Interaction effects far from zeros,
so indicative of interactions between A, B
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Three Dimensional Plot of the Estimated Cell Means
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Plot of the Estimated Cell Means in Two-Dimensions
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Treatment Means and Factor Level Means
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Results of Two-Factor Analysis using Minitab
Conclusions?
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Histogram of Residuals
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Normal Probability Plot
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Plot of Residuals vs Order
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Residuals vs Predicted Values
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SAS Program to Perform Two-Way Analysis with
Interaction Together with Analysis of Means
/ Hay Fever Drug Development / data hay input
relief FactorA FactorB RepNum
_at__at_ cards (Insert the data set here) proc
print proc anova class FactorA FactorB model
relief FactorA FactorB FactorAFactorB means
FactorA FactorB FactorAFactorB / tukey
bon run
The tukey and bon keywords are for performing
theTukey multiple comparisons procedure, while
bon is for the Bonferroni procedure.
One may also use the PROC GLM command above
instead of the PROC ANOVA command. The former
command is recommended for unbalanced designs.