week 5 introduction to factorial designspeople.stat.sfu.ca/~jtg3/unit4.pdf · outline basic...

Week 5Introduction to Factorial Designs

Joslin Goh

Simon Fraser University

Week 5Introduction to Factorial Designs – p. 1/35

Outline

Basic concepts

The two-factor factorial design

The general factorial design


Basic concepts

Factorial designs: Designs in which all possible combinations of the levels of thefactors appear.

Example: The yield of a chemical process is being studied. The two mostimportant variables are thought to be the pressure and the temperature. Three

levels of each factor are selected, and a factorial experiment with two replicates isperformed. The yield data follow:

Pressure(psi)

Temperature(0C) 200 215 230

150 90.4 90.7 90.2

90.2 90.6 90.4

160 90.1 90.5 89.9

90.3 90.6 90.1

170 90.5 90.8 90.4

90.7 90.9 90.1


Advantages

Advantage over one-factor-at-a-time design

Less runs

Allows the interactions to be detected and estimated


Simple model

yijk = µij + ǫijk ,

whereyijk is the k-th observation for the ij-th treatmentµij is the mean response given by the i-th level of A and thej-th level of factor B.


Simple Maths

µ =1

ab

a∑

i=1

b∑

j=1

µij

µi. =1

b

b∑

j=1

µij

µ.j =1

a

a∑

i=1

µij

αi = µi. − µ , βj = µ.j − µ


Main Effect

Main effect: The change in the average response produced by a change in the

level of the factor.

Example: Consider two factors A and B each at two levels.

A B Response

+ + 52

+ - 40

- + 30

- - 20


Interaction Effect

Interaction effect of AB: The average difference in two A effects, one at low

level of factor B and the other at high level of factor B

Example: Consider two factors A and B each at two levels.

A B Response

+ + 12

+ - 50

- + 40

- - 20

INT(AB) = 12−40−(50−20)2

= −28


Blank page for more notes

You might want to have extra papers and a calculator?


Two Factor Fractional Design

General arrangement for a two-factor factorial design

Factor B

1 2 · · · b

Factor A

1 y111, y112 y121, y122 · · · y1b1, y1b2

· · · , y11n · · · , y12n · · · · · · , y1bn2 y211, y212 y221, y222 · · · y2b1, y2b2

· · · , y21n · · · , y22n · · · · · · , y2bn... · · · · · · · · · · · ·

· · · · · · · · · · · ·i yi11, yi12 yi21, yi22 · · · yib1, yib2

· · · , yi1n · · · , yi2n · · · · · · , yibn... · · · · · · · · · · · ·

· · · · · · · · · · · ·a ya11, ya12 ya21, ya22 · · · yab1, yab2

· · · , ya1n · · · , ya2n · · · · · · , yabn


Randomization

How is this a randomized design? Or is it even arandomized design?

How about blocking?


Model

yijk = µ+ τi + βj + (τβ)ij + ǫijk

where

i = 1, . . . , a; j = 1, . . . , b; k = 1, . . . , n

µ: the overall mean

τi: i-th treatment effect of factor A,∑

i τi = 0

βj : j-th treatment effect of factor B,∑

j βj = 0

(τβ)ij : interaction effect between τi and βj ,∑

i(τβ)ij =∑

j(τβ)ij = 0

ǫijk: experimental error iid∼ N(0, σ2)


Estimation

Define:

yi.. =b

∑

j=1

n∑

k=1

yijk yi.. =yi..

bn

y.j. =a

∑

i=1

n∑

k=1

yijk y.j. =y.j.

an

yij. =n∑

k=1

yijk yij. =yij.

n

y... =

a∑

i=1

b∑

j=1

n∑

k=1

yijk y... =y...

abn


To obtain the estimate,

Normal equations:

µ : abnµ+ bna∑

i=1

τi + anb∑

j=1

βj + na∑

i=1

b∑

j=1

ˆ(τβ)ij = y...

τi : bnµ+ bnτi + n

b∑

j=1

βj + n

b∑

j=1

ˆ(τβ)ij = yi..

βj : anµ+ n

a∑

i=1

τi + anβj + n

a∑

i=1

ˆ(τβ)ij = y.j.

(τβ)ij : nµ+ nτi + nβj + n ˆ(τβ)ij = yij.

Constraints:∑ai=1 τi = 0

∑ai=1

ˆ(τβ)ij = 0 j = 1, . . . , b∑b

j=1 βj = 0∑b

j=1ˆ(τβ)ij = 0 i = 1, . . . , a


The estimates are

µ = y...

τi = yi.. − y...

βj = y.j. − y...

ˆ(τβ)ij = yij. − yi.. − y.j. + y...

yijk = µ+ τi + βj + ˆ(τβ)ij = yij.


Hypothesis test

Three hypothesis are of interest. They are.

H0 : τ1 = τ2 = · · · = τa = 0

H1 : τi 6= 0 ∃ i

H0 : β1 = β2 = · · · = βb = 0

H1 : βj 6= 0 ∃ j

H0 : (τβ)ij = 0 for all i, j

H1 : ∃ i 6= j, (τβ)ij 6= 0


Analysis of variance

Define:

SSTotal =a∑

i=1

b∑

j=1

n∑

k=1

(yijk − y...)2

=∑

i

∑

j

∑

k

y2ijk −

y2...

abn

SSA = bna∑

i=1

(yi.. − y...)2

=1

bn

∑

i

y2i.. −

y2...

abn

SSB = anb∑

j=1

(y.j. − y...)2

=1

an

∑

j

y2.j. −

y2...

abn

SSAB = n

a∑

i=1

b∑

j=1

(yij. − yi.. − y.j. + y...)2=

1

n

∑

i

∑

j

y2ij. −

y2...

abn− SSA − SSB

SSError =a∑

i=1

b∑

j=1

n∑

k=1

(yijk − yij.)2

We have

SSTotal = SSA + SSB + SSAB + SSError


ANOVA Table

Source Sum of Squares D.O.F. Mean Squares F

A Treatments SSA a− 1 MSA = SSAa−1

MSA

MSError

B Treatments SSB b− 1 MSB = SSBb−1

MSB

MSError

Interaction SSAB (a− 1)(b− 1) MSAB = SSAB(a−1)(b−1)

MSAB

MSError

Error SSError ab(n− 1) MSError = SSErrorab(n−1)

MSA

MSError

H0∼ Fa−1,ab(n−1)

MSB

MSError

H0∼ Fb−1,ab(n−1)

MSAB

MSError

H0∼ F(a−1)(b−1),ab(n−1)


More blanks!


Multiple comparisons

Case 1: The interactions are not significant, we can compare µ+ τi and µ+ τj

by using the statistics

tij =yi.. − yj..√2MSError

nb

Fisher LSD method claims µ+ τi and µ+ τj are significantly different atlevel α if |tij | > tα/2,ab(n−1)

Tukey method claims µ+ τi and µ+ τj are significantly different at level α if|tij | >

qα,a,ab(n−1)√2


Case 2: The interactions are significant. We can make comparison at fixed level of theother factor.For example: Fix the temperature at level 2 (700F )

y12. = 57.25 (Material type 1)



3 V S. 1 | y32. − y12.√

2MSError/n| = |145.75− 57.25

√

2 ∗ 675.21/4| = 4.817 >

qα,3,27√2

= 2.479

3 V S. 2 | y32. − y22.√

2MSError/n| = |145.75− 119.75

√

2 ∗ 675.21/4| = 1.415 <

qα,3,27√2

= 2.479

2 V S. 1 | y22. − y12.√

2MSError/n| = |119.75− 57.25

√

2 ∗ 675.21/4| = 3.402 >

qα,3,27√2

= 2.479


Model adequacy checking

Residuals: rijk = yijk − yijk = yijk − yij.


Example

An example

An engineer is designing a battery for use in a device that will be subjected to some extremevariations in temperature. The only design parameter that he can select at this point is theplate material for the battery, and he has three possible choices. When the devices ismanufactured and is shipped to the filed, the engineer has no control over the temperatureextremes that the device will encounter, and he knows from experience that temperature willprobably affect the effective battery life. However, temperature can be controlled in theproduct development laboratory for the purpose of a test. The engineer decides to test allthree plate materials at three temperature levels – 15, 90, and 1250F because thesetemperature levels are consistent with the product end-use environment.

The engineer wants to answer the following questions:

a. What effects do material type and temperature have on the life of the battery?

b. Is there a choice of material that would give uniformly long life regardless oftemperature?


All combinations of plate material and temperature are tested. Four batteries aretested at each combination of plate material and temperature. All 36 tests are run

in random order.

Material TypeTemperature(0F )

15 70 125

1 130 155 34 40 20 70

74 180 80 75 82 58

2 150 188 136 122 25 70

159 126 106 115 58 45

3 138 110 174 120 96 104

168 160 150 139 82 60


Useful Plots

M1 M2 M3

5010

015

0

material

resp

onse

125F 15F 70F

5010

015

0

temperature

resp

onse

6080

100

140

battery$material

mea

n of

bat

tery

$res

pons

e

M1 M2 M3

battery$temperature

70F15F125F

6080

100

140

battery$temperature

mea

n of

bat

tery

$res

pons

e

125F 15F 70F

battery$material

M3M2M1


# Fit the model

> g <- lm(response ˜ material * temperature, battery)

> anova(g)

Analysis of Variance Table

Response: response

Df Sum Sq Mean Sq F value Pr(>F)

material 2 10684 5342 7.9114 0.001976 **temperature 2 39119 19559 28.9677 1.909e-07 ***material:temperature 4 9614 2403 3.5595 0.018611 *Residuals 27 18231 675

---

Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1

Note : You can also use the command

g <- lm(response ˜ material+temperature+material * temperature, battery)

Conclusion : Both factors and their interactions are statistically significant.


Useful Plots

−2 −1 0 1 2

−60

−20

20

Normal Q−Q Plot

Theoretical Quantiles

Sam

ple

Qua

ntile

s

0 5 10 15 20 25 30 35

−60

−20

20

run order

Res

idua

ls60 80 100 120 140 160

−60

−20

20

fitted vlaue

Res

idua

ls

1.0 1.5 2.0 2.5 3.0

−60

−20

20c(materials)

g$re

s

1.0 1.5 2.0 2.5 3.0

−60

−20

20

c(temperatures)

g$re

s


Check for interaction

a. Model

b. ANOVA table

> g <- lm(response ˜ material+temperature, battery)

> anova(g)

Analysis of Variance Table

Response: response

Df Sum Sq Mean Sq F value Pr(>F)

material 2 10684 5342 5.9472 0.006515 **temperature 2 39119 19559 21.7759 1.239e-06 ***Residuals 31 27845 898

---

Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1

c. Fitted value yijk = yi.. + y.j. − y...


Interaction?

40 60 80 100 120 140 160

−20

−10

010

20

fitted

diffe

renc

e


A Reasonable Design

One observation per cell (n = 1)

yij = µ+ τi + βj + (τβ)ij + ǫij

Problem :

Possible solution :


ANOVA Table

For a two-factor model with one observation per cell

Source Sum of Squares D.O.F. Mean Square Expected Mean Square

A∑a

i=1

y2i.

b−

y2..

aba− 1 MSA σ2 +

b∑

i τ2i

a−1

B∑b

j=1

y2.j

a−

y2..

abb− 1 MSB σ2 +

a∑

j β2j

b−1

Error or AB Subtraction (a− 1)(b− 1) MSError σ2 +∑

i

∑j(τβ)

2ij

(a−1)(b−1)

Total∑a

i=1

∑bj=1 y

2ij −

y2..

abab− 1


Tukey test is used for determining whether interaction is present.

Assumption: (τβ)ij = γτiβj where γ is an unknown parameter.

Procedure: Define

SSN =[∑a

i=1

∑bj=1 yijyi.y.j − y..(SSA + SSB +

y2..

ab)]2

abSSASSB

SS∗Error = SSError − SSN

and compute

F =SSN

SS∗Error/[(a− 1)(b− 1)− 1]

If F > Fα,1,(a−1)(b−1)−1 ,


Example: Tukey test of no interaction

TemperaturePressure

25 30 35 40 45 yi.

100 5 4 6 3 5 23

125 3 1 4 2 3 13

150 1 1 3 1 2 8

y.j 9 6 13 6 10 44 = y..

SSTotal = 166 − 129.07 = 36.93

SSA =1

5[23

2+ 13

2+ 8

2] −

442

3 ∗ 5= 23.33

SSB =1

3[9

2+ 6

2+ 13

2+ 6

102] −

442

3 ∗ 5= 11.60

SSError = SSTotal − SSA − SSB = 2.00

SSN =[7236 − 44 ∗ (23.33 + 11.60 + 129.07)]2

3 ∗ 5 ∗ 23.33 ∗ 11.60= 0.0985

SS∗

Error = SSError − SSN = 1.9015


The General Factorial Design

Design :

Consider p factors. Suppose that ith factor has mi levels and each treatment hasn replicates. A factorial design has n

∏pi=1 mi runs whose order is completely

random.

Analysis :Model, estimation, hypothesis test, multiple comparisons, and residual analysis

are similar to those for a two-factor factorial design.


Source Sum of Squares D.O.F

A∑a

i=1 nbc(αi)2 a− 1

B∑b

j=1 nac(βj)2 b− 1

C∑c

k=1 nab(δk)2 c− 1

A×B∑a

i=1

∑bj=1 nc((αβ)ij)

2 (a− 1)(b− 1)

A× C∑a

i=1

∑ck=1 nb((αδ)ik)

2 (a− 1)(c− 1)

B × C∑b

j=1

∑ck=1 na((βδ)jk)

2 (b− 1)(c− 1)

A×B × C∑a

i=1

∑bj=1

∑ck=1 n(

ˆ(γ)ijk)2 (a− 1)(b− 1)(c− 1)

Error∑a

i=1

∑bj=1

∑ck=1

∑nl=1(yijkl − yijk.)

2 abc(n− 1)

Total∑a

i=1

∑bj=1

∑ck=1

∑nl=1(yijkl − y....)

2 abcn− 1


week 5 introduction to factorial designspeople.stat.sfu.ca/~jtg3/unit4.pdf · outline basic...

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