1. Multi-Factor Studies
Stat 701 Lecture
E. Pena

2. 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

3. 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.

4. 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.

5. 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 eu’s allocated to each treatment combination.
Then n = ABk is the total number of eu’s in the study.
Note: In allocating eu’s, the principle of randomization should
be observed.

6. Model for Two-Factor Factorial Analysis
Yijl = lth observation in treatment (ai,bj)
Yijl = mij + eijl
mij is the mean response of eu’s 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

7. 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 i=1,2,…,A
Factor B Main Effects: bj = m.j - m.. for j=1,2,…,B
Interaction Effects: (ab)ij = mij - mi. - m.j + m.. for i=1,…,A; j=1,..,B
Model: mij = m.. + ai + bj + (ab)ij for i=1,…,A; j=1,…,B.

8. 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.

9. Two Examples Example 1: A model without interaction effects.
Factor A Main Effects: a1 = = -1; a2 = = +1
Factor B Main Effects: b1 = = 0; b2 = = +3; b3 = = -3
Interaction Effects: (ab)11 = = 0; all of them are zeros.
A Pictorial Representation of these Treatment Means?

10. Examples … continued Example 2: A model with interaction effects.
Factor A Main Effects: a1 = = 0; a2 = = 0
Factor B Main Effects: b1 = = 0; b2 = = +3; b3 = = -3
Interaction Effects: (ab)11 = = -1; (ab)12 = = +1;
(ab)13 = = 0; (ab)21 = = +1; (ab)22 = = -1; (ab)23 = = 0. Note that the sum of these effects is 0.
Would appear that there are no differences in Factor A levels!
Pictorial Representation?

11. Example: Drug Development for Hay Fever

12. Tabular Presentation of Factorial Data
with Totals and Means

13. Estimates of Main and Interaction Effects
Factor A Main Effects
A1: = -3.3
A2: = 0.65
A3: = 2.65
Factor B Main Effects
B1: = -2.55
B2: = 0.75
B3: = 1.80
Interaction Effects (A Level, B Level)
(A1, B1): = 1.145
(A1, B2): = -0.03
(A1, B3): =
(A2, B1): = 0.17
(A2, B2): = 0.345
(A2, B3): = -.505
(A3, B1): =
(A3, B2): = -.305
(A3, B3): = 1.62
Estimates of
Interaction effects
far from zeros, so
indicative of
interactions
between A, B

14. Three Dimensional Plot of the Estimated
Cell Means

15. Plot of the Estimated Cell Means in
Two-Dimensions

16. Treatment Means and Factor Level Means

17. Results of Two-Factor Analysis using Minitab
Conclusions?

18. Histogram of Residuals

19. Normal Probability Plot

20. Plot of Residuals vs Order

21. Residuals vs Predicted Values

22. 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
cards;
(Insert the data set here) ;
proc print;
proc anova;
class FactorA FactorB;
model relief = FactorA FactorB FactorA*FactorB;
means FactorA FactorB FactorA*FactorB / 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.