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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.