Post on 03-Jan-2016
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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 CombinationsPossible Treatment Combinations:
Treatment 1: a1b1 = Low Carbon and Low TemperatureTreatment 2: a1b2 = Low Carbon and High TemperatureTreatment 3: a2b1 = High Carbon and Low TemperatureTreatment 4: a2b2 = High Carbon and High Temperature
Possible Experimental Approaches:
• Just use a completely randomized design with the fourtreatments 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 denotethe 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 shouldbe observed.
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Model for Two-Factor Factorial Analysis
Yijl = lth observation in treatment (ai,bj)
Yijl = ij + ijl
ij is the mean response of eu’s assigned (aibj)
ijl is the error component for the lth unit in (aibj)
Assumptions (fixed effects model):
• ij 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 ij. We may summarize this in a table of form:
TreatmentMeans
b1 b2 b3 Factor A LevelMeans
a1
11
12
13
1.
a2
21
22
23
2.
Factor B LevelMeans
.1
.2
.3
..
Factor A Main Effects: i = i. - .. for i=1,2,…,AFactor B Main Effects: j = .j - .. for j=1,2,…,BInteraction Effects: ()ij = ij - i. - .j + .. for i=1,…,A; j=1,..,B
Model: ij = .. + i + j + ()ij for i=1,…,A; j=1,…,B.
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Interpretation … continued
A
i
B
jij
A
iijj
B
jiji
AB
BjA
AiB
1 1..
1.
1.
1
,...,2,1,1
,...,2,1,1
Identities
Case 1: No interaction effects is when ()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 ExamplesExample 1: A model without interaction effects.
TreatmentMeans
b1 b2 b3 Factor A LevelMeans
a1
11 = 4 12 = 7
13 = 1 1. = 4
a2
21 = 6 22 = 9
23 = 3 2. = 6
Factor B LevelMeans
.1 = 5
.2 = 8 .3 = 2
.. = 5
Factor A Main Effects: 1 = 4 - 5 = -1; 2 = 6 - 5 = +1
Factor B Main Effects: 1 = 5 - 5 = 0; 2 = 8 - 5 = +3; 3 = 2 - 5 = -3
Interaction Effects: ()11 = 4 - 4 - 5 + 5 = 0; all of them are zeros.
A Pictorial Representation of these Treatment Means?
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Examples … continuedExample 2: A model with interaction effects.
TreatmentMeans
b1 b2 b3 Factor A LevelMeans
a1
11 = 4 12 = 9
13 = 2 1. = 5
a2
21 = 6 22 = 7
23 = 2 2. = 5
Factor B LevelMeans
.1 = 5
.2 = 8 .3 = 2
.. = 5
Factor A Main Effects: 1 = 5 - 5 = 0; 2 = 5 - 5 = 0
Factor B Main Effects: 1 = 5 - 5 = 0; 2 = 8 - 5 = +3; 3 = 2 - 5 = -3
Interaction Effects: ()11 = 4 - 5 - 5 + 5 = -1; ()12 = 9 - 5 - 8 + 5 = +1;()13 = 2 - 5 - 2 + 5 = 0; ()21 = 6 - 5 - 5 + 5 = +1; ()22 = 7 - 5 - 8 + 5 = -1; ()23 = 2 - 5 - 2 + 5 = 0. Note that the sum of these effects is 0.
Pictorial Representation?
Would appear that there are no differences in Factor A levels!
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Example: Drug Development for Hay FeverReliefTime FactorA FactorB RepNum
2.4 1 1 12.7 1 1 22.3 1 1 32.5 1 1 44.6 1 2 14.2 1 2 24.9 1 2 34.7 1 2 44.8 1 3 14.5 1 3 24.4 1 3 34.6 1 3 45.8 2 1 15.2 2 1 25.5 2 1 35.3 2 1 48.9 2 2 19.1 2 2 28.7 2 2 39.0 2 2 49.1 2 3 19.3 2 3 28.7 2 3 39.4 2 3 46.1 3 1 15.7 3 1 25.9 3 1 36.2 3 1 49.9 3 2 1
10.5 3 2 210.6 3 2 310.1 3 2 413.5 3 3 113.0 3 3 213.3 3 3 313.2 3 3 4
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Factor B1 2 3 A Level
Totals andMeans
1 2.4, 2.7, 2.3,2.5
9.92.475
4.6, 4.2, 4.9,4.7
18.44.6
4.8, 4.5, 4.4,4.6
18.34.575
46.63.88
2 5.8, 5.2, 5.5,5.3
21.85.45
8.9, 9.1, 8.7,9.0
35.78.925
9.1, 9.3, 8.7,9.4
36.59.125
94.07.83
3 6.1, 5.7, 5.9,6.2
23.95.975
9.9, 10.5,10.6, 10.1
41.110.275
13.5, 13.0,13.3, 13.2
53.013.25
118.09.83
Factor A
B LevelTotals and
Means
55.64.63
95.27.93
107.88.98
258.67.18
Tabular Presentation of Factorial Datawith Totals and Means
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Estimates of Main and Interaction EffectsFactor 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 effectsfar from zeros, so
indicative ofinteractions
between A, B
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3
21
2B
7
12
2
EstMean
13A
Three Dimensional Plot of the EstimatedCell Means
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1 2 3
2
7
12
B Level
Est
ima
te o
fC
ell
Me
an
Level A1
Level A2
Level A3
Plot of the Estimated Cell Means inTwo-Dimensions
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1 2 3 1 2 3 1 2 3
1 2 3
-3.4E-01
0.3443140
-1
0
1
FactorBFactorA
Interaction Effects
Effe
ct
1 2 3
456789
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7.039657.327017.18333
FactorA
Main Effects
Mea
n
1 2 3
456789
10
7.039657.327017.18333
FactorB
Mea
n
FactorA FactorBLevel Value Level Value
123
123
123
123
Two-way ANOM for ReliefTime by FactorA, FactorB
Treatment Means and Factor Level Means
Rows: FactorA Columns: FactorB
1 2 3 All
1 4 4 4 12 2.475 4.600 4.575 3.883 0.171 0.294 0.171 1.059
2 4 4 4 12 5.450 8.925 9.125 7.833 0.265 0.171 0.310 1.777
3 4 4 4 12 5.975 10.275 13.250 9.833 0.222 0.330 0.208 3.128
All 12 12 12 36 4.633 7.933 8.983 7.183 1.622 2.540 3.707 3.272
Cell Contents -- ReliefTi:N Mean StDev
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Two-way Analysis of Variance
Analysis of Variance for ReliefTiSource DF SS MS F PFactorA 2 220.0200 110.0100 1827.86 0.000FactorB 2 123.6600 61.8300 1027.33 0.000Interaction 4 29.4250 7.3562 122.23 0.000Error 27 1.6250 0.0602Total 35 374.7300
Individual 95% CIFactorA Mean ------+---------+---------+---------+-----1 3.883 (*)2 7.833 (*)3 9.833 (*) ------+---------+---------+---------+----- 4.500 6.000 7.500 9.000
Individual 95% CIFactorB Mean ---+---------+---------+---------+--------1 4.633 (-*)2 7.933 (*)3 8.983 (*) ---+---------+---------+---------+-------- 4.800 6.000 7.200 8.400
Results of Two-Factor Analysis using Minitab
Conclusions?
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Histogram of Residuals
-0.4 -0.3 -0.2 -0.1 0.0 0.1 0.2 0.3
0
1
2
3
4
5
6
Residual
Fre
qu
en
cy
Histogram of the Residuals(response is ReliefTi)
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Normal Probability Plot
-0.5 -0.4 -0.3 -0.2 -0.1 0.0 0.1 0.2 0.3 0.4
-2
-1
0
1
2
No
rma
l S
co
re
Residual
Normal Probability Plot of the Residuals(response is ReliefTi)
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Plot of Residuals vs Order
5 10 15 20 25 30 35
-0.5
-0.4
-0.3
-0.2
-0.1
0.0
0.1
0.2
0.3
0.4
Observation Order
Re
sid
ua
l
Residuals Versus the Order of the Data(response is ReliefTi)
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Residuals vs Predicted Values
2 7 12
-0.5
-0.4
-0.3
-0.2
-0.1
0.0
0.1
0.2
0.3
0.4
Fitted Value
Re
sid
ua
l
Residuals Versus the Fitted Values(response is ReliefTi)
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SAS Program to Perform Two-Way Analysis with InteractionTogether 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 PROCANOVA command. The former command is recommended for unbalanceddesigns.