IE341: Introduction to Design of Experiments

Last term we talked about testing the difference between two independent means. For means from a normal population, the test statistic is

where the denominator is the estimated standard deviation of the difference between two independent means. This denominator represents the random variation to be expected with two different samples. Only if the difference between the sample means is much greater than the expected random variation do we declare the means different.

B

B

A

A

BA

diff

BA

ns

ns

XX

s

XXt

22

We also covered the case where the two means are not independent, and what we must do to account for the fact that they are dependent.

And finally, we talked about the difference between two variances, where we used the F ratio. The F distribution is a ratio of two chi-square variables. So if s2

1 and s22 possess independent chi-s

quare distributions with v1 and v2 df, respectively, then

has the F distribution with v1 and v2 df.

2

22

1

21

vs

vs

F

All of this is valuable if we are testing only two means. But what if we want to test to see if there is a difference among three means, or four, or ten?

What if we want to know whether fertilizer A or fertilizer B or fertilizer C is best? In this case, fertilizer is called a factor, which is the condition under test.

A, B, C, the three types of fertilizer under test, are called levels of the factor fertilizer.

Or what if we want to know if treatment A or treatment B or treatment C or treatment D is best? In this case, treatment is called a factor.

A,B,C,D, the four types of treatment under test, are called levels of the factor treatment.

It should be noted that the factor may be quantitative or qualitative.

Enter the analysis of variance!

ANOVA, as it is usually called, is a way to test the differences between means in such situations.

Previously, we tested single-factor experiments with only two treatment levels. These experiments are called single-factor because there is only one factor under test. Single-factor experiments are more commonly called one-way experiments.

Now we move to single-factor experiments with more than

two treatment levels.

Let’s start with some notation.

Yij = ith observation in the jth level N = total number of experimental observations

= the grand mean of all N experimental observations

= the mean of the observations in the jth level

nj = number of observations in the jth level; the nj are called replicates.

Replication of the design refers to using more than one experimental unit for each level.

If there are the same number n replicates for each treatment, the design is said to be balanced.

Y

jY

N

YY

N

iij

1

j

n

iij

j n

YY

j

1

Designs are more powerful if they are balanced, but balance is not always possible.

Suppose you are doing an experiment and the equipment breaks down on one of the tests. Now, not by design but by circumstance, you have unequal numbers of replicates for the levels.

In all the formulas, we used nj as the number of replicates in treatment j, not n, so there is no problem.

Notation continued

= the effect of the jth level

J = number of treatment levels

eij = the “error” associated with the ith observation in the jth level, assumed to be independent normally distributed random

variables with mean = 0 and variance = σ2, which are constant for all levels of the factor.

j YY jj

For all experiments, randomization is critical. So to draw any conclusions from the experiment, we must require that the treatments be applied in random order.

We must also assign the experimental units to the treatments randomly.

If all this randomization occurs, the design is called a completely randomized design.

ANOVA begins with a linear statistical model

ijjij eYY

This model is for a one-way or single-factor ANOVA. The goal of the model is to test hypotheses about the treatment effects and to estimate them.

If the treatments have been selected by the experimenter, the model is called a fixed-effects model. In this case, the conclusions will apply only to the treatments under consideration.

Another type of model is the random effects model or components of variance model.

In this situation, the treatments used are a random sample from large population of treatments. Here the τ i are random variables and we are interested in their variability, not in the differences among the means being tested.

First, we will talk about fixed effects, completely randomized, balanced models.

In the model we showed earlier, the τ j are defined as deviations from the grand mean so

It follows that the mean of the jth treatment is

01

J

jj

jj YY

Now the hypothesis under test is: Ho: μ1= μ2 = μ3 = … μJ

Ha: μ j≠ μk for at least one j,k pair

The test procedure is ANOVA, which is a decomposition of the total sum of squares into its components parts according to the model.

The total SS is

and ANOVA is about dividing it into its component parts.

SS = variability of the differences among the J levels

SSε = pooled variability of the random error within levels

jn

i

J

jijTotal YYSS

1 1

2)(

2

1

)( YYnSS j

J

jj

2

1 1

)( j

J

j

n

iijerror YYSS

j

This is easy to see because

But the cross-product term vanishes because

)(2)(

)()(

1 11 1

2

1

2

2

1 11 1

2

jij

J

j

n

ij

J

j

n

ijij

J

jjj

J

j

n

ijijj

J

j

n

iij

YYYYYYYYn

YYYYYY

jj

jj

0)(1

j

n

iij YY

j

So SStotal = SS treatments + SS error

Most of the time, this is called

SStotal = SS between + SS within

Each of these terms becomes an MS (mean square) term when divided by the appropriate df.

JN

SS

df

SSMS

J

SS

df

SSMS

error

error

errorerror

treatments

treatments

treatmentstreatments

1

The df for SSerror = N-J because

and the df for SSbetween = J-1 because

there are J levels.

JN

SS

nnn

snsnsn

snYY

error

J

JJ

J

jjj

J

jj

n

iij

j

)1(...)1()1(

)1(...)1()1(

)1()(

21

2222

211

1

2

1

2

1

Now the expected values of each of these terms are E(MSerror) = σ2

E(MStreatments) =

11

2

2

J

nJ

jjj

Now if there are no differences among the treatment means, then for all j.

So we can test for differences with our old friend F

with J -1 and N -J df. Under Ho, both numerator and denominator are estimates

of σ2 so the result will not be significant.

Under Ha, the result should be significant because the numerator is estimating the treatment effects as well as σ2.

0j

error

treatments

MS

MSF

The results of an ANOVA are presented in an ANOVA table. For this one-way, fixed-effects, balanced model:

Source SS df MS p Model SSbetween J-1 MSbetween p

Error SSwithin N-J MSwithin

Total SStotal N-1

Let’s look at a simple example.

A product engineer is investigating the tensile strength of a synthetic fiber to make men’s shirts. He knows from prior experience that the strength is affected by the weight percent of cotton in the material. He also knows that the percent should range between 10% and 40% so that the shirts can receive permanent press treatment.

The engineer decides to test 5 levels: 15%, 20%, 25%, 30%, 35%and to have 5 replicates in this design.

His data are%

15 7 7 15 11 9 9.8

20 12 17 12 18 18 15.4

25 14 18 18 19 19 17.6

30 19 25 22 19 23 21.6

35 7 10 11 15 11 10.8

15.04

jY

Y

In this tensile strength example, the ANOVA table is

In this case, we would reject Ho and declare that there is an effect of the cotton weight percent.

Source SS df MS p Model 475.76 4 118.94 <0.01 Error 161.20 20 8.06Total 636.96 24

We can estimate the treatment parameters by subtracting the grand mean from the treatment means. In this example,

τ 1 = 9.80 – 15.04 = -5.24 τ 2 = 15.40 – 15.04 = +0.36 τ 3 = 17.60 – 15.04 = -2.56 τ 4 = 21.60 – 15.04 = +6.56 τ 5 = 10.80 – 15.04 = -4.24 Clearly, treatment 4 is the best because it

provides the greatest tensile strength.

Now you could have computed these values from the raw data yourself instead of doing the ANOVA. You would get the same results, but you wouldn’t know if treatment 4 was significantly better.

But if you did a scatter diagram of the original data, you would see that treatment 4 was best, with no analysis whatsoever.

In fact, you should always look at the original data to see if the results do make sense. A scatter diagram of the raw data usually tells as much as any analysis can.

Scatter plot of tensile strength data

0

5

10

15

20

25

30

10 15 20 25 30 35 40

weight percent cotton

tensi

le s

trength

How do you test the adequacy of the model?

The model assumes certain assumptions that must hold for the ANOVA to be useful. Most importantly, that the errors are distributed normally and independently.

The error for each observation, sometimes called the residual, isjijij YYe

A residual check is very important for testing for nonconstant variance. The residuals should be structureless, that is, they should have no pattern whatsoever, which, in this case, they do not.

Scatter plot of residuals vs. fitted values

- 5

- 4

- 3

- 2

- 1

0

1

2

3

4

5

6

9 12 15 18 21

fitted value

resi

dual

These residuals show no extreme differences in variation because they all have about the same spread.

They also do not show the presence of any outlier. An outlier is a residual value that is vey much larger than any of the others. The presence of an outlier can seriously jeopardize the ANOVA, so if one is found, its cause should be carefully investigated.

A histogram of residuals shows the distribution is slightly skewed. Small departures from symmetry are of less concern than heavy tails.

Histogram of Residuals

1

2

3

- 6 - 4 - 2 0 2 4 6

Residual

Fre

quency

Another check is for normality. If we do a normal probability plot of the residuals, we can see whether normality holds.

Normal probability plot

0

10

20

30

40

50

60

70

80

90

100

-4 -2 0 2 4 6Residual

Norm

al pro

bability

A normal probability plot is made with ascending ordered residuals on the x-axis and their cumulative probability points, 100(k-.5)/n, on the y-axis. k is the order of the residual and n = number of residuals. There is no evidence of an outlier here.

The previous slide is not exactly a normal probability plot because the y-axis is not scaled properly. But it does gives a pretty good suggestion of linearity.

A plot of residuals vs run order is useful to detect correlation between the residuals, a violation of the independence assumption.

Runs of positive or of negative residuals indicates correlation. None is observed here.

Residuals vs Run Order

- 5

- 4

- 3

- 2

- 1

0

1

2

3

4

5

6

0 5 10 15 20 25 30

Run Order

Resi

duals

One of the goals of the analysis is to estimate the level means. If the results of the ANOVA shows that the factor is significant, we know that at least one of the means stands out from the rest. But which one or ones?

The procedures for making these mean comparisons are called multiple comparison methods. These methods use linear combinations called contrasts.

A contrast is a particular linear combination of level means, such as to test the difference between level 4 and level 5.

Or if one wished to test the average of levels 1 and 3 vs levels 4 and 5, he would use .

In general, where

54 YY

)()( 5431 YYYY

J

jjjYcC

1

01

J

jjc

An important case of contrasts is called orthogonal contrasts. Two contrasts in a design with coefficients cj and dj are orthogonal if

01

J

jjjj dcn

There are many ways to choose the orthogonal contrast coefficients for a set of levels. For example, if level 1 is a control and levels 2 and 3 are two real treatments, a logical choice is to compare the average of the two treatments with the control:

and then the two treatments against one another:

These two contrasts are orthogonal because

132 2YYY

132 0YYY

0)0*2()1*1()1*1(3

1

j

jdc

Only J-1 orthogonal contrasts may be chosen because the J levels have only J-1 df. So for only three levels, the contrasts chosen exhaust those available for this experiment.

Contrasts must be chosen before seeing the data so that experimenters aren’t tempted to contrast the levels with the greatest differences.

For the tensile strength experiment with 5 levels and thus 4 df, the 4 contrasts are:

C1= 0(5)(9.8)+0(5)(15.4)+0(5)(17.6)-1(5)(21.6)+1(5)(10.8) =-54 C2= +1(5)(9.8)+0(5)(15.4)+1(5)(17.6)-1(5)(21.6)-1(5)(10.8) =-25 C3= +1(5)(9.8)+0(5)(15.4)-1(5)(17.6)+0(5)(21.6)+0(5)(10.8) =-39 C4= -1(5)(9.8)+4(5)(15.4)-1(5)(17.6)-1(5)(21.6)-1(5)(10.8) = 9

These 4 contrasts completely partition the SStreatments. Then the SS for each contrast is formed:

J

jjj

J

jjjj

C

cn

Ycn

SS

1

2

2

1

So for the 4 contrasts we have:

81.0

)]1()1()1()4()1[(5

9

25.31)]0()0()1()0()1[(5

39

25.31)]1()1()1()0()1[(5

25

6.291)]1()1()0()0(0[(5

54

22222

2

22222

2

22222

2

2222)2

2

4

3

2

1

C

C

C

C

SS

SS

SS

SS

Now the revised ANOVA table is

Source SS df MS p Weight % 475.76 4 118.94 <0.001 C1 291.60 1 291.60 <0.001 C2 31.25 1 31.25 <0.06 C3 152.10 1 152.10 <0.001 C4 0.81 1 0.81 <0.76 Error 161.20 20 8.06 Total 636.96 24

So contrast 1 (level 5 – level 4) and contrast 3 (level 1 – level 3) are significant.

Although the orthogonal contrast approach is widely used, the experimenter may not know in advance which levels to test or they may be interested in more than L-1 comparisons. A number of other methods are available for such testing.

These methods include:

Scheffe’s Method Least Significant Difference Method Duncan’s Multiple Range Test Newman-Keuls test

There is some disagreement about which is the best method, but it is best if all are applied only after there is significance in the overall F test.

Now let’s look at the random effects model.

Suppose there is a factor of interest with an extremely large number of levels. If the experimenter selects L of these levels at random, we have a random effects model or a components of variance model.

The linear statistical model is

as before, except that both andare random variables instead of simply .Because and are independent, the variance of any observation is

These two variances are called variance

components, hence the name of the model.

ijjij eYY

j ije

ije

j ije

)()()( ijij eVarVarYVar

The requirements of this model are that the are NID(0,σ2), as before, and that the are NID(0, ) and that and are independent. The normality assumption is not required in the random effects model.

As before, SSTotal = SStreatments + SSerror

And the E(MSerror) = σ2.

But now E(MStreatments) = σ2 + n

So the estimate of is

ijej

ije

2 j

2

n

MSMS errortreatments 2ˆ

2

The computations and the ANOVA table are the same as before, but the conclusions are quite different.

Let’s look at an example.

A textile company uses a large number of looms. The process engineer suspects that the looms are of different strength, and selects 4 looms at random to investigate this.

The results of the experiment are shown in the table below.

The ANOVA table is Source SS df MS p Looms 89.19 3 29.73 <0.001 Error 22.75 12 1.90 Total 111.94 15

Loom

1 98 97 99 96 97.5

2 91 90 93 92 91.5

3 96 95 97 95 95.75

4 95 96 99 98 97.0

95.44

jY

Y

In this case, the estimates of the variances are: =1.90

Thus most of the variability in the observations is due to variability in loom strength. If you can isolate the causes of this variability and eliminate them, you can reduce the variability of the output and increase its quality.

2e

96.64

90.173.29ˆ 2

86.896.690.1222 eij

When we studied the differences between two treatment means, we considered repeated measures on the same individual experimental unit.

With three or more treatments, we can still do this. The result is a repeated measures design.

Consider a repeated measures ANOVA partitioning the SSTotal.

This is the same as SStotal = SSbetween subjects + SSwithin subjects

The within-subjects SS may be further partitioned into SStreatment + SSerror .

2

1 1

2

1 1

2

1 1

)()()( i

n

i

J

jij

n

i

J

ji

n

i

J

jij YYYYYY

In this case, the first term on the RHS is the differences between treatment effects and the second term on the RHS is the random error.

2

1 1

2

1 1

2

1 1

)()()( YYYYYYYY ji

n

i

J

jij

n

i

J

jji

n

i

J

jij

Now the ANOVA table looks like this.

Source SS df MS p Between subjects n-1 Within Subjects n(J-1) Treatments J-1 Error (J-1)(n-1) Total Jn-12

1 1

2

1 1

2

1 1

2

1 1

2

1 1

)(

)(

)(

)(

)(

YY

YYYY

YY

YY

YY

n

i

J

jij

j

n

i

J

jiij

n

i

J

jj

n

i

J

jiij

n

i

J

ji

The test for treatment effects is the usual

but now it is done entirely within subjects.

This design is really a randomized complete block design with subjects considered to be the blocks.

error

treatment

MS

MS

Now what is a randomized complete blocks design?

Blocking is a way to eliminate the

effect of a nuisance factor on the comparisons of interest. Blocking can be used only if the nuisance factor is known and controllable.

Let’s use an illustration. Suppose we want to test the effect of four different tips on the readings from a hardness testing machine.

The tip is pressed into a metal test coupon, and from the depth of the depression, the hardness of the coupon can be measured.

The only factor is tip type and it has four levels. If 4 replications are desired for each tip, a completely randomized design would seem to be appropriate.

This would require assigning each of the 4x4 = 16 runs randomly to 16 different coupons.

The only problem is that the coupons need to be all of the same hardness, and if they are not, then the differences in coupon hardness will contribute to the variability observed.

Blocking is the way to deal with this problem.

In the block design, only 4 coupons are used and each tip is tested on each of the 4 coupons. So the blocking factor is the coupon, with 4 levels.

In this setup, the block forms a homogeneous unit on which to test the tips.

This strategy improves the accuracy of the tip comparison by eliminating variability due to coupons.

Because all 4 tips are tested on each coupon, the design is a complete block design. The data from this design are shown below.

Test coupon

Tip type 1 2 3 4

1 9.3 9.4 9.6 10.0

2 9.4 9.3 9.8 9.9

3 9.2 9.4 9.5 9.7

4 9.7 9.6 10.0 10.2

Now we analyze these data the same way we did for the repeated measures design. The model is

where βk is the effect of the kth block and the rest of the terms are those we already know.

jkkjjk eYY

Since the block effects are deviations from the grand mean,

just as

01

K

kk

01

J

jj

We can express the total SS as

which is equivalent to SStotal = SStreatments + SSblocks + SSerror

with df

N-1 = J-1 + K-1 + (J-1)(K-1)

J

j

K

kkjjk

J

j

K

kk

J

j

K

kj

kjjkk

J

j

K

kj

J

j

K

kjk

YYYYYYYY

YYYYYYYYYY

1 1

2

1 1

2

1 1

2

2

1 1

2

1 1

)()()(

)]()()[()(

The test for equality of treatment means is

and the ANOVA table is

Source SS df MS p Treatments SStreatments J-1 MStreatments

Blocks SSblocks K-1 MSblocks

Error SSerror (J-1)(K-1) MSerror

Total SStotal N-1

error

treatments

MS

MSF

For the hardness experiment, the ANOVA table is

Source SS df MS p Tip type 38.50 3 12.83 0.0009 Coupons 82.50 3 27.50 Error 8.00 9 .89 Total 129.00 15

As is obvious, this is the same analysis as the repeated measures design.

Now let’s consider the Latin Square design. We’ll introduce it with an example.

The object of study is 5 different formulations of a rocket propellant on the burning rate of aircraft escape systems. Each formulation comes from a batch of raw material large enough for only 5 formulations. Moreover, the formulations are prepared by 5 different operators, who differ in skill and experience.

The way to test in this situation is with a 5x5 Latin Square, which allows for double blocking and therefore the removal of two nuisance factors. The Latin Square for this example is

Batches of raw material

Operators

1 2 3 4 5

1 A B C D E

2 B C D E A

3 C D E A B

4 D E A B C

5 E A B C D

Note that each row and each column has all 5 letters, and each letter occurs exactly once in each row and column.

The statistical model for a Latin Square is

where Yjkl is the jth treatment observation in the kth row and the lth column.

jkllkjjkl eYY

Again we have SStotal=SSrows+SScolumns+SStreatments+SSerror

with df =

N = R-1 + C-1 + J-1 + (R-2)(C-1) The ANOVA table for propellant data is Source SS df MS p Formulations 330.00 4 82.50 0.0025 Material batches 68.00 4 17.00 Operators 150.00 4 37.50 0.04 Error 128.00 12 10.67 Total 676.00 24

So both the formulations and the operators were significantly different. The batches of raw material were not, but it still is a good idea to block on them because they often are different.

This design was not replicated, and Latin Squares often are not, but it is possible to put n replicates in each cell.

Now if you superimposed one Latin Square on another Latin Square of the same size, you would get a Graeco-Latin Square. In one Latin Square, the treatments are designated by roman letters. In the other Latin Square, the treatments are designated by Greek letters.

Hence the name Graeco-Latin Square.

A 5x5 Graeco-Latin Square is

Note that the five Greek treatments appear exactly once in each row and column, just as the Latin treatments did.

Batches of raw materi

al

Operators

1 2 3 4 5

1 Aα Bγ Cε Dβ Eδ

2 Bβ Cδ Dα Eγ Aε

3 Cγ Dε Eβ Aδ Bα

4 Dδ Eα Aγ Bε Cβ

5 Eε Aβ Bδ Cα Dγ

If Test Assemblies had been added as an additional factor to the original propellant experiment, the ANOVA table for propellant data would be

Source SS df MS p Formulations 330.00 4 82.50 0.0033 Material batches 68.00 4 17.00 Operators 150.00 4 37.50 0.0329 Test Assemblies 62.00 4 15.50 Error 66.00 8 8.25 Total 676.00 24

The test assemblies turned out to be nonsignificant.

Note that the ANOVA tables for the Latin Square and the Graeco-Latin Square designs are identical, except for the error term.

The SS(error) for the Latin Square design was decomposed to be both Test Assemblies and error in the Graeco-Latin Square. This is a good example of how the error term is really a residual. Whatever isn’t controlled falls into error.

Before we leave one-way designs, we should look at the regression approach to ANOVA. The model is

Using the method of least squares, we rewrite this as

ijjij eY

2

1 11 1

2 )( j

n

i

J

jij

n

i

J

jij

nj

YeE

Now to find the LS estimates of μ and τ j,

When we do this differentiation with respect to μ and τ j, and equate to 0, we obtain

for all j

0

0

E

E

0)ˆˆ(2

0)ˆˆ(2

1

1 1

j

J

jij

j

n

i

J

Jij

Y

Yj

After simplification, these reduce to

In these equations,

JJ

J

Ynn

Ynn

Ynn

YnnnN

.

2.2

1.1

21

ˆ............................ˆ

.

.

.

.................ˆ............ˆ

.............................ˆˆ

..ˆ...ˆˆˆ

jj YnY

YNY

.

..

These j + 1 equations are called the least squares normal equations.

If we add the constraint

we get a unique solution to these normal equations.

0ˆ1

J

jj

YY

Y

jj

ˆ

ˆ

It is important to see that ANOVA designs are simply regression models. If we have a one-way design with 3 levels, the regression model is

where Xi1 = 1 if from level 1 = 0 otherwise and Xi2 = 1 if from level 2 = 0 otherwise

Although the treatment levels may be qualitative, they are treated as “dummy” variables.

ijiiij eXXY 22110

Since Xi1 = 1 and Xi2 = 0,

so Similarly, if the observations are from

level 2,

so

ij

iji

e

eY

10

2101 )0()1(

110 Y

ij

iji

e

eY

20

2102 )1()0(

220 Y

Finally, consider observations from level 3, for which Xi1 = Xi2 = 0. Then the regression model becomes

so

Thus in the regression model formulation of this one-way ANOVA with 3 levels, the regression coefficients describe comparisons of the first two level means with the third.

ij

iji

e

eY

0

2103 )0()0(

30 Y

So

Thus, testing β1= β2 = 0 provides a test of the equality of the three means.

In general, for J levels, the regression model will have J-1 variables

and

322

311

30

YY

YY

Y

ijJiJiiij eXXXY 1,122110 ...

Jjj

J

YY

Y

0

Now what if you have two factors under test? Or three? Or four? Or more?

Here the answer is the factorial design. A factorial design crosses all factors. Let’s take a two-way design. If there are J levels of factor A and K levels of factor B, then all JK treatment combinations appear in the experiment.

Most commonly, J = K = 2.

In a two-way design, with two levels of each factor, we have, where -1 and +1 are codes for low and high levels, respectively

We can have as many replicates as we want in this

design. With n replicates, there are n observations in each cell of the design.

Factor A Factor B Response

-1 (low level) -1 (low level) 20+1 (high

level)-1 (low level) 40

-1 (low level) +1 (high level)

30+1 (high

level)+1 (high

level)52

SStotal = SSA + SSB + SSAB + SSerror

This decomposition should be familiar by now except for SSAB. What is this term? Its official name is interaction.

This is the magic of factorial designs. We find out about not only the effect of factor A and the effect of factor B, but the effect of the two factors in combination.

How do we compute main effects? The main effect of factor A is the difference between the average response at A high and the average response at A low,

Similarly, the B effect is the difference between the average response at B high and the average response at B low

2125462

3020

2

5240

1130412

4020

2

5230

So the main effect of factor A is 21 and the main effect of factor B is 11.

That is, changing the level of factor A from the low level to the high level brings a response increase of 21 units.

And changing the level of factor B from the low level to the high level increases the response by 11 units.

The plots below show the main effects of factors A and B.

Main Effect of Factor A

25

30

35

40

45

50

1 2

Factor A Level

Response

Main Effect of Factor B

25

30

35

40

45

50

1 2

Factor B Level

Response

Both A and B are significant, which you can see by the fact that the slope is not 0.

A 0 slope in the effect line that connects the response at the high level with the response at the low level indicates that it doesn’t matter to the response whether the factor is set at its high value or its low value, so the effect of such a factor is not significant.

Of course, the p value from the F test gives the significance of the factors precisely, but it is usually evident from the effects plots.

Now how do you compute the interaction effect? Interaction occurs when the difference in response between the levels of one factor are different at the different levels of the other factor. That is,

The first term here is the difference between the two levels of factor A at the low level of factor B. That is, 40 -20 = 20.

But the difference between the two levels of factor A at the high level of factor B is

52-30 = 22.

)()( 21221112 BABABABA

Then the size of the interaction effect is (22 -20) / 2 = 1 and the interaction is not significant. The interact

ion plot below shows almost parallel lines, which indicates no interaction.

AxB Interaction Effect

0

10

20

30

40

50

60

1 1.5 2

Level of Factor A

resp

onse

Now suppose the two factors are quantitative, like temperature, pressure, time, etc. Then you could write a regression model version of the design.

As before, X1 represents factor A and X2

represents factor B. X1X2 is the interaction term, and e is the error term.

The parameter estimates for this model turn out to be related to the effect estimates.

211222110 XXXXY

The parameter estimates are:

So the model is

5.02

1)(

2

1

5.52

11)(

2

1

5.102

21)(

2

1

5.354

52304020

12

2

1

0

ABeffect

Beffect

Aeffect

Y

2121 5.05.55.105.35ˆ XXXXY

With this equation, you can find all the effects of the design. For example, if you want to know the mean when both A and B are at the high (+1) level, the equation is

Now if you want the mean when A is at the high level and B is at the low level, the equation is

All you have to do is fill in the values of X1 and X2 with the appropriate codes, +1 or -1.

52)1)(1(5.0)1(5.5)1(5.105.35ˆ Y

40)1)(1(5.0)1(5.5)1(5.105.35ˆ Y

Now suppose the data in this experiment are:

Now let’s look at the main and interaction effects.

Factor A Factor B Response

-1 (low level)

-1 (low level)

20+1 (high

level)-1 (low

level)50

-1 (low level)

+1 (high level)

40+1 (high

level)+1 (high

level)12

The main effect of factor A is

and the main effect of factor B is

The interaction effect is

(50- 20) – (12-40)= 30+28 = 58 which is very high and is significant.

130312

4020

2

1250

935262

5020

2

1240

)()( 21221112 BABABABA

Now let’s look at the main effects of the factors graphically.

Main Effect of Factor B

20

25

30

35

40

1 2

Factor B level

Resp

onse

Main Effect of Factor A

20

25

30

35

40

1 2

Factor A level

Resp

onse

Clearly, factor A is not significant, which you can see by the approximately 0 slope.

Factor B is probably significant because the slope is not close to 0. The p value from the F test gives the actual significance.

Now let’s look at the interaction effect. This is the effect of factors A and B in combination, and is often the most important effect.

Interaction of factors A and B

0

10

20

30

40

50

60

1 1.5 2

Factor A

Response

Factor B - - lowFactor B - - high

Now these two lines are definitely not parallel, so there is an interaction. It probably is very significant because the two lines cross.

Only the p value associated with the F test can give the actual significance, but you can see with the naked eye that there is no question about significance here.

Interaction of factors is the key to the East, as we say in the West.

Suppose you wanted the factor levels that give the lowest possible response. If you picked by main effects, you would pick A low and B high.

But look at the interaction plot and it will tell you to pick A high and B high.

This is why, if the interaction term is significant, you never, never, never interpret the corresponding main effects. They are meaningless in the presence of interaction.

And it is because factorial designs provide the ability to test for interactions that they are so popular and so successful.

You can get response surface plots for these regression equations. If there is no interaction, the response surface is a plane in the 3rd dimension above the X1,X2 Cartesian space. The plane may be tilted, but it is still a plane.

If there is interaction, the response surface is a twisted plane representing the curvature in the model.

The simplest factorials are two-factor experiments.

As an example, a battery must be designed to be robust to extreme variations in temperature. The engineer has three possible choices for the plate material. He decides to test all three plate materials at three temperatures,

-15˚F, 70˚F, 125˚F. He tests four batteries at each combination of material type and temperature. The response variable is battery life.

Here are the data he got.

Plate material type

Temperature (˚F)

-15 70 125

1 130 34 20

74 40 70

155 80 82

180 75 58

2 150 136 25

159 122 70

188 106 58

126 115 45

3 138 174 96

110 120 104

168 150 82

160 139 60

The model here is

Both factors are fixed so we have the same constraints as before

and

In addition,

ijkkjkjijk YY

J

ijj 0

K

ikk 0

J

ij

K

kkjkj 0

1

The experiment has n = 4 replicates, so there are nJK total observations.

n

YY

nJ

Y

Y

nK

YY

nJK

Y

Y

n

iijk

jk

n

i

J

jijk

k

n

i

K

kijk

j

n

i

J

j

K

kijk

1

1 1

1 1

1 1 1

The total sum of squares can be partitioned into four components:

SStotal = SSA + SSB + SSAB +SSe

2

1 1 1

2

1 1

2

1

2

1

2

1 1 1

)()()()()( jk

n

i

J

j

K

kijkkj

J

j

K

kjk

K

kk

J

jj

n

i

J

j

K

kijk YYYYYYnYYnJYYnKYY

and the ANOVA table is Source SS df MS p A SSA J-1

B SSB K-1

AB SSAB (J-1)(K-1)

Error SSe JK(n-1)

Total SStotal JKn -1

For the battery life experiment,

Material type

Temperature (˚F)

-15 70 125

1 134.75 57.25 57.50 83.17

2 155.75 119.75 49.50 108.33

3 144.00 145.75 85.50 125.08

144.83 107.58 64.17kY

jY

53.105Y

The ANOVA table is

Source SS df MS p Material 10,683.72 2 5,341.86 0.0020 Temperature 39,118.72 2 19,558.36 0.0001 Interaction 9,613.78 4 2,403.44 0.0186 Error 18,230.75 27 675.21 Total 77,646.97 35

Because the interaction is significant, the only plot of interest is the interaction plot.

Interaction Plot for Material Type vs Temperature

Type 1

Type 1 Type 1

Type 2

Type 2

Type 2

Type 3 Type 3

Type 3

30

50

70

90

110

130

150

170

- 20 30 80 130

Temperature

Batt

ery

Lif

e i

n h

ours

Although it is not the best at the lowest temperature, Type 3 is much better than the other two at normal and high temperatures. Its life at the lowest temperature is just an average of 12 hours less than the life with Type 2.

Type 3 would probably provide the design most robust to temperature differences.

Suppose you have a factorial design with more than two factors. Take, for example, a three-way factorial design, where the factors are A, B, and C.

All the theory is the same, except that now you have three 2-way interactions, AB, AC, BC, and one 3-way interaction, ABC.

Consider the problem of soft-drink bottling. The idea is to get each bottle filled to a uniform height, but there is variation around this height. Not every bottle is filled to the same height.

The process engineer can control three variables during the filling process: percent carbonation (A), operating pressure (B), and number of bottles produced per minute or line speed (C).

The engineer chooses three levels of carbonation (factor A), two levels of pressure (factor B), and two levels for line speed (factor C). This is a fixed effects design. He also decides to run two replicates.

The response variable is the average deviation from the target fill height in a production run of bottles at each set of conditions. Positive deviations are above the target and negative deviations are below the target.

The data are

Operating pressure (B)

Percent carbonation (A)

25 psi 30 psi

line speed (C)

line speed (C)

200 250 200 250

10 -3 -1 -1 1

-1 0 0 1

12 0 2 2 6

1 1 3 5

14 5 7 7 10

4 6 9 11

The 3–way means are

Operating pressure (B)

Percent carbonation (A)

25 psi 30 psi

line speed (C)

line speed (C)

200 250 200 250

10 -2 -.5 -.5 1

12 .5 1.5 2.5 5.5

14 4.5 6.5 8 10.5

The 2-way means are

B (low) B (high)

A 25 psi 30 psi

10 -1.25 0.25

12 1.00 4.00

14 5.50 9.25

C (low) C (high)

A 200 250

10 -1.25 0.25

12 1.50 3.50

14 6.25 8.50

C (low)

C (high)

B 200 250

25 psi 1.00 2.50

30 psi 3.33 5.67

The main effect means are

Factor A Mean

10 % -0.500

12 % 2.50014 % 7.375

Factor B Mean

25 psi 1.75

30 psi 4.50

Factor C Mean

200 2.167

250 4.083

The ANOVA table is Source SS df MS p A 252.750 2 126.375 <0.0001 B 45.375 1 45.375 <0.0001 C 22.042 1 22.042 0.0001 AB 5.250 2 2.625 0.0557 AC 0.583 2 0.292 0.6713 BC 1.042 1 1.042 0.2485 ABC 1.083 2 0.542 0.4867 Error 8.500 12 0.708 Total 336.625 23

So the only significant effects are those for A, B, C, AB. The AB interaction is barely significant, so interpretation must be tempered by what we see in the A and B main effects. The plots are shown next.

The plots are Factor A

- 1

0

1

2

3

4

5

6

7

8

10 12 14

Level of A

Response

Factor B

-1

0

1

2

3

4

5

6

7

8

20 25 30 35

Level of B

Response

Factor C

- 1

0

1

2

3

4

5

6

7

8

150 200 250 300

Level of C

Resp

onse

AB Interaction

B1: 25 psi

B1: 25 psi

B1: 25 psi

B2: 30 psi

B2: 30 psi

B2: 30 psi

-2

0

2

4

6

8

10

10 12 14

Level of A

Resp

onse

Our goal is to minimize the response. Given the ANOVA table and these plots, we would choose the low level of factor A, 10% carbonation, and the low level of factor B, 25 psi. This is true whether we look at the two main effects plots or the interaction plot. This is because the interaction is barely significant.

We would also choose the slower line speed, 200 bottles per minute.

Now suppose you do an experiment where you suspect nonlinearity and want to test for both linear and quadratic effects.

Consider a tool life experiment, where the life of a tool is thought to be a function of cutting speed and tool angle. Three levels of each factor are used. So this is a 2-way factorial fixed effects design.

The three levels of cutting speed are 125, 150, 175. The three levels of tool angle are 15˚, 20˚, 25˚. Two replicates are used and the data are shown below.

Tool Angle(degrees)

Cutting Speed (in/min)

125 150 175

15 -2 -3 2

-1 0 3

20 0 1 4

2 3 6

25 -1 5 0

0 6 -1

The ANOVA table for this experiment is

Source SS df MS p Tool Angle 24.33 2 12.17 0.0086 Cut Speed 25.33 2 12.67 0.0076 TC 61.34 4 15.34 0.0018 Error 13.00 9 1.44 Total 124.00 17

The table of cell and marginal means isFacto

r TFactor C

125 150 175

15˚ -1.5 -1.5 2.5 -0.167

20˚ 1.0 2.0 5.0 2.667

25˚ -0.5 5.5 -0.5 1.500

-0.33 2.0 2.33kY

jY

Tool Angle Factor

- 0.5

0

0.5

1

1.5

2

2.5

3

15 20 25

Level of T

Response

Cutting Speed Factor

- 0.5

0

0.5

1

1.5

2

2.5

3

125 150 175

Level of C

Response

Clearly there is reason to suspect quadratic effects here. So we can break down each factor’s df into linear and quadratic components.

We do this by using orthogonal contrasts. The contrast for linear is

-1, 0, =1 and the contrast for quadratic is +1, -2, +1.

We need a table of factor totals to proceed. For factor T,

Now applying the linear and quadratic contrasts to these sums,

Factor T Sum of Obs

15 -1

20 16

25 9

Factor T

Sum of Obs

Linear Quadratic

15 -1 -1 +1

20 16 0 -2

25 9 +1 +1

Contrast 10 -24

Now to find the SS due to these two new contrasts,

33.8

)2)(3)(2(

10 2

3

1

2

23

1

jj

jjj

lin

cnJ

Yc

SS

16

)6)(3)(2(

24 2

3

1

2

23

1

jj

jjj

quad

cnJ

Yc

SS

Now we can do the same thing for factor C. The table of sums with the contrasts included is

Now for the SS due to each contrast,

Factor C Sum of Obs

Linear Quadratic

125 -2 -1 +1

150 12 0 -2

175 14 +1 +1

Contrast 16 -12

33.21

)2)(3)(2(

16 2

3

1

2

23

1

kk

kkk

lin

cnK

Yc

SS

0.4)6)(3)(2(

12 2

3

1

2

23

1

kk

kkk

quad

cnK

Yc

SS

Now we can write the new ANOVA table

Source SS df MS p Tool angle 24.33 2 12.17 0.0086 Linear 8.33 1 8.33 0.0396 Quad 16.00 1 16.00 0.0088 Cut Speed 25.33 2 12.67 0.0076 Linear 21.33 1 21.33 0.0039 Quad 4.00 1 4.00 0.1304 TC 61.34 4 15.34 0.0018 Error 13.00 9 1.44 Total 124.00 17

Now see how the df for each of the factors has been split into its two components, linear and quadratic. It turns out that everything except the quadratic for Cutting Speed is significant.

Now guess what! There are 4 df for the interaction term and why not split them into linear and quadratic components as well. It turns out that you can get TlinClin, TlinCquad, TquadClin, and TquadCquad.

These 4 components use up the 4 df for the interaction term.

There is reason to believe the quadratic component in the interaction, as shown below, but we’ll pass on this for now.

Interaction of Tool Angle and Cutting Speed

Speed 125

Speed 125

Speed 125

Speed 150

Speed 150

Speed 150

Speed 175

Speed 175

Speed 175

- 2

- 1

0

1

2

3

4

5

6

15 20 25

Tool Angle

Response

Interaction of Tool Angle and Cutting Speed

Angle 15 Angle 15

Angle 15

Angle 20

Angle 20

Angle 20

Angle25

Angle25

Angle25

- 2

- 1

0

1

2

3

4

5

6

125 150 175

Cutting Speed

Response

Now let’s talk about blocking in a factorial design. The concept is identical to blocking in a 1-way design. There is either a nuisance factor or it is not possible to completely randomize all the runs in the design.

For example, there simply may not be enough time to run the entire experiment in one day, so perhaps the experimenter could run one complete replicate on one day and another complete replicate on the second day, etc. In this case, days would be a blocking factor.

Let’s look at an example. An engineer is studying ways to improve detecting targets on a radar scope. The two factors of importance are background clutter and the type of filter placed over the screen.

Three levels of background clutter and two filter types are selected to be tested. This is a fixed effects 2 x 3 factorial design.

To get the response, a signal is introduced into the scope and its intensity is increased until an operator sees it. Intensity at detection is the response variable.

Because of operator availability, an operator must sit at the scope until all necessary runs have been made. But operator differ in skill and ability to use the scope, so it makes sense to use operators as a blocking variable.

Four operators are selected for use in the experiment. So each operator receives the 2 x 3 = 6 treatment combinations in random order, and the design is a completely randomized block design. The data are:

Operators

1 2 3 4

Filter type

1 2 1 2 1 2 1 2

Ground clutte

r

Low 90 86 96 84 100

92 92 81

Medium 102 87 106

90 105

97 96 80

High 114 93 112

91 108

95 98 83

Since each operator (block) represents the complete experiment, all effects are within operators. The ANOVA table is

Source SS df MS p

Within blocks 1479.33 5 295.87 <0.000001 Ground clutter 335.58 2 167.79 <0.0003 Filter type 1066.67 1 1066.67 <0.0001 GF interaction 77.08 2 38.54 0.0573 Between blocks 402.17 3 134.06 <0.0002 Error 166.33 15 11.09 Total 2047.83 23

The effects of both the background clutter and the filter type are highly significant. Their interaction is marginally significant.

As suspected, the operators are significantly different in their ability to detect the signal, so it is good that they were used as blocks.

Now let’s look at the 2k factorial design. This notation means that there are k factor

s, each at 2 levels, usually a high and a low level. These factors may be qualitative or quantitative.

This is a very important class of designs and is widely used in screening experiments. Because there are only 2 levels, it is assumed that the response is linear over the range of values chosen.

Let’s look at an example of the simplest of these designs, the 22 factorial design.

Consider the effect of reactant concentration (factor A) and amount of catalyst (factor B) on the yield in a chemical process. The 2 levels of factor A are: 15% and 25%. The 2 levels of factor B are: 1 pound and 2 pounds. The experiment is replicated three times.

Here are the data.

This design can be pictured as rectangle. 20 30 +1

factor B

-1 26.67 33.33 -1 factor A +1

Factor A Factor B Replicate 1 Replicate 2 Replicate 3

-1 (low) -1 (low) 28 25 27

+1 (high)

-1 (low) 36 32 32

-1 (low) +1 (high)

18 19 23

+1 (high)

+1 (high)

31 30 29

The interaction effect can also be derived from this table.

Multiplying the A and B factor level codes gets the AB interaction. This is always the way interactions are derived. Now averaging according to the AB coefficients gives the interaction effect.

Factor A Factor B AB interaction

-1 (low) -1 (low) (-1)(-1)= +1

+1 (high) -1 (low) (+1)(-1)= -1

-1 (low) +1 (high) (-1)(+1)= -1

+1 (high) +1 (high) (+1)(+1)= +1

Now we can find the effects easily from the table below. A B AB Replicate

average

-1 -1 +1 26.67

+1 -1 -1 33.33

-1 +1 -1 20

+1 +1 +1 30

67.12

34.3

2

)30)(1()20)(1()33.33)(1()67.26)(1(

52

10

2

)30)(1()20)(1()33.33)(1()67.26)(1(

33.82

67.16

2

)30)(1()20)(1()33.33)(1()67.26)(1(

ABeffect

Beffect

Aeffect

Because there are only first-order effects, the response surface is a plane. Yield increases with increasing reactant concentration (factor A) and decreases with increasing catalyst amount (factor B).

The ANOVA table is

Source SS df MS p A 208.33 1 208.33 <0.0001 B 75.00 1 75.00 <0.0024 AB 8.33 1 8.33 0.1826 Error 31.34 8 3.92 Total 323.00 11

It is clear that both main effects are significant and that there is no AB interaction.

The regression model is

where the β coefficients are ½ the effects, as before. 27.5 is the grand average of all 12 observations.

21 2

00.5

2

33.85.27ˆ XXY

Now let’s look at the 23 factorial design. In this case, there are three factors, each at 2 levels. The design is

Run A B C AB AC BC ABC

1 -1 -1 -1 (-1)(-1)= +1 (-1)(-1)= +1 (-1)(-1)= +1 (-1)(-1)(-1)= -1

2 +1 -1 -1 (+1)(-1)= -1 (+1)(-1)= -1 (-1)(-1)= +1 (+1)(-1)(-1)= +1

3 -1 +1 -1 (-1)(+1)= -1 (-1)(-1)= +1 (+1)(-1)= -1 (-1)(+1)(-1)= +1

4 +1 +1 -1 (+1)(+1)= +1

(+1)(-1)= -1 (+1)(-1)= -1 (+1)(+1)(-1)= -1

5 -1 -1 +1 (-1)(-1)= +1 (-1)(+1)= -1 (-1)(+1)= -1 (-1)(-1)(+1)= +1

6 +1 -1 +1 (+1)(-1)= -1 (+1)(+1)= +1

(-1)(+1)= -1 (+1)(-1)(+1)= -1

7 -1 +1 +1 (-1)(+1)= -1 (-1)(+1)= -1 (+1)(+1)= +1

(-1)(+1)(+1)= -1

8 +1 +1 +1 (+1)(+1)= +1

(+1)(+1)= +1

(+1)(+1)= +1

(+1)(+1)(+1)= +1

Remember the beverage filling study we talked about earlier? Now assume that each of the 3 factors has only two levels.

So we have factor A (% carbonation) at levels 10% and 12%.

Factor B (operating pressure) is at levels 25 psi and 30 psi.

Factor C (line speed) is at levels 200 and 250.

Now our experimental matrix becomes

Run

A: Percent carbonati

on

B: Operatin

g pressure

C: Line spee

d

Replicate 1

Replicate 2

Total of obs

1 10 25 200 -3 -1 -4

2 12 25 200 0 1 1

3 10 30 200 -1 0 -1

4 12 30 200 2 3 5

5 10 25 250 -1 0 -1

6 12 25 250 2 1 3

7 10 30 250 1 1 2

8 12 30 250 6 5 11

And our design matrix is

From this matrix, we can determine all our effects by applying the linear codes and dividing by 8, the number of terms being averaged.

Run

A B C AB AC BC ABC Replicate 1

Replicate 2 Total of obs

1 -1 -1 -1 +1 +1 +1 -1 -3 -1 -4

2 +1 -1 -1 -1 -1 +1 +1 0 1 1

3 -1 +1 -1 -1 +1 -1 +1 -1 0 -1

4 +1 +1 -1 +1 -1 -1 -1 2 3 5

5 -1 -1 +1 +1 -1 -1 +1 -1 0 -1

6 +1 -1 +1 -1 +1 -1 -1 2 1 3

7 -1 +1 +1 -1 -1 +1 -1 1 1 2

8 +1 +1 +1 +1 +1 +1 +1 6 5 11

The effects are

5.08

4

8

)11)(1()2)(1()3)(1()1)(1()5)(1()1)(1()1)(1()4)(1(

5.08

4

8

)11)(1()2)(1()3)(1()1)(1()5)(1()1)(1()1)(1()4)(1(

25.08

2

8

)11)(1()2)(1()3)(1()1)(1()5)(1()1)(1()1)(1()4)(1(

75.08

6

8

)11)(1()2)(1()3)(1()1)(1()5)(1()1)(1()1)(1()4)(1(

75.18

14

8

)11)(1()2)(1()3)(1()1)(1()5)(1()1)(1()1)(1()4)(1(

25.28

18

8

)11)(1()2)(1()3)(1()1)(1()5)(1()1)(1()1)(1()4)(1(

38

24

8

)11)(1()2)(1()3)(1()1)(1()5)(1()1)(1()1)(1()4)(1(

ABCeffect

BCeffect

ACeffect

ABeffect

Ceffect

Beffect

Aeffect

The ANOVA table is

Source SS df MS pA: Percent carb 36.00 1 36.00 <0.0001B: Op Pressure 20.25 1 20.25 <0.0005C: Line speed 12.25 1 12.25 0.0022AB 2.25 1 2.25 0.0943AC 0.25 1 0.25 0.5447BC 1.00 1 1.00 0.2415ABC 1.00 1 1.00 0.2415Error 5.00 8 0.625Total 78.00 15

There are only 3 significant effects, factors A, B, and C. None of the interactions is significant.

The regression model for soft-drink fill height deviation is

Because the interactions are not significant, they are not included in the regression model. So the response surface here is a plane at each level of line speed.

321

3322110

2

75.1

2

25.2

2

300.1

ˆ

XXX

XXXY

All along we have had at least 2 replicates for each design so we can get an error term. Without the error term, how do we create the F-ratio to test for significance?

But think about it. A 24 design has 16 runs. With 2 replicates, that doubles to 32 runs. The resources need for so many runs are often not available, so some large designs are run with only 1 replicate.

Now what do we do for an error term to test for effects?

The idea is to pool some high-level interactions under the assumption that they are not significant anyway and use them as an error term. If indeed they are not significant, this is OK. But what if you pool them as error and they are significant? This is not OK.

So it would be nice to know before we pool, which terms are actually poolable. Thanks to Cuthbert Daniel, we can do this. Daniel’s idea is to do a normal probability plot of the effects.

All negligible effects will fall along a line and those that do not fall along the line are significant. So we may pool all effects that are on the line. The reasoning is that the negligible effects, like error, are normally distributed with mean 0 and variance σ2 and so will fall along the line.

Let’s look at an example of a chemical product. The purpose of this experiment is to maximize the filtration rate of this product, and it is thought to be influenced by 4 factors: temperature (A), pressure (B), concentration of formaldehyde (C), and stirring rate (D).

The design matrix and response are:Run A B C D AB AC BC AD BD CD ABC AB

DACD BCD ABCD Filt

rate

1 -1 -1 -1 -1 +1 +1 +1 +1 +1 +1 -1 -1 -1 -1 +1 45

2 +1 -1 -1 -1 -1 -1 +1 -1 +1 +1 +1 +1 +1 -1 -1 71

3 -1 +1 -1 -1 -1 +1 -1 +1 -1 +1 +1 +1 -1 +1 -1 48

4 +1 +1 -1 -1 +1 -1 -1 -1 -1 +1 -1 -1 +1 +1 +1 65

5 -1 -1 +1 -1 +1 -1 -1 +1 +1 -1 +1 -1 +1 +1 -1 68

6 +1 -1 +1 -1 -1 +1 -1 -1 +1 -1 -1 +1 -1 +1 +1 60

7 -1 +1 +1 -1 -1 -1 +1 +1 -1 -1 -1 +1 +1 -1 +1 80

8 +1 +1 +1 -1 +1 +1 +1 -1 -1 -1 +1 -1 -1 -1 -1 65

9 -1 -1 -1 +1 +1 +1 +1 -1 -1 -1 -1 +1 +1 +1 -1 43

10 +1 -1 -1 +1 -1 -1 +1 +1 -1 -1 +1 -1 -1 +1 +1 100

11 -1 +1 -1 +1 -1 +1 -1 -1 +1 -1 +1 -1 +1 -1 +1 45

12 +1 +1 -1 +1 +1 -1 -1 +1 +1 -1 -1 +1 -1 -1 -1 104

13 -1 -1 +1 +1 +1 -1 -1 -1 -1 +1 +1 +1 -1 -1 +1 75

14 +1 -1 +1 +1 -1 +1 -1 +1 -1 +1 -1 -1 +1 -1 -1 86

15 -1 +1 +1 +1 -1 -1 +1 -1 +1 +1 -1 -1 -1 +1 -1 70

16 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 96

From this matrix, we can estimate all the effects and then do a normal probability plot of them. The effects are:

A= 21.625 AB= 0.125 ABC= 1.875 B= 3.125 AC=-18.125 ABD= 4.125 C= 9.875 AD= 16.625 ACD=-1.625 D= 14.625 BC= 2.375 BCD=-2.625

BD= -0.375 ABCD=1.375

CD= -1.125

The best stab at a normal probability plot is

normal probability plot of effects

0

10

20

30

40

50

60

70

80

90

100

- 30 - 20 - 10 0 10 20 30

effects

cum

ula

tive

pro

babili

ty

There are only 5 effects that are off the line. These are, in the upper right corner: C, D, AD, A, and in the lower left corner, AC. All of the points on the line are negligible, behaving like residuals.

Because we drop factor B and all its interactions, we now get an ANOVA table with the extra observations as error.

Source SS df MS p A 1870.56 1 1870.56 <0.0001 C 390.06 1 390.06 <0.0001 D 855.56 1 855.56 <0.0001 AC 1314.06 1 1314.06 <0.0001 AD 1105.56 1 1105.56 <0.0001 CD 5.06 1 5.06 ACD 10.56 1 10.56 Error 179.52 8 22.44 Total 5730.94 15

Essentially, we have changed the design from a 24 design with only 1 replicate to a 23 design with two replicates.

This is called projecting a higher-level design into a lower-level design. If you start with an unreplicated 2k design, then drop h of the factors, you can continue with a 2k-h design with 2h replicates.

In this case, we started with a 24 design, dropped h=1 factor, and ended up with a 24-1 design with 21 replicates.

The main effects plots areFactor A

55

60

65

70

75

80

85

- 1 0 1

Level of A

response

Factor C

55

60

65

70

75

80

- 1 0 1

Level of C

response

Factor D

55

60

65

70

75

80

- 1 0 1

Level of D

response

The two significant interaction plots are

AC Interaction

C low

C low

C highC high

45

50

55

60

65

70

75

80

85

90

- 1 0 1

Level of A

response

AD Interaction

D low

D low

D high

D high

45

50

55

60

65

70

75

80

85

90

95

100

- 1 0 1

Level of A

response

Now we are going to talk about the addition of center points to a 2k design. In this case, we are looking for quadratic curvature, so we must have quantitative factors.

The center points are run at 0 for each of the k factors in the design. So now the coefficients are -1, 0, +1. We have the same n replicates at the center points as at the -1 and +1 points.

Now let’s go back to the box we used earlier to describe a 22 design.

+1 factor B -1 -1 +1 factor A At each corner, we have a point of the desi

gn, for example, (A-,B-), (A-,B+), (A+,B-), and (A+,B+).

Now we can add center points to this design to see if there is quadratic curvature.

+1

factor B 0 o

-1 -1 0 +1

factor A Now in addition to the 4 points we had earlier, we

have n observations at the center point: (A0, B0).

Now if we average the 4 factorial points to get , and then average the n center points to get , we can tell if there is a quadratic effect by the size of

.

If this difference is small, then the center points lie on the plane established by the factorial points. If, this difference is large, then there is quadratic curvature present.

factorialY

centerY

centerfactorial YY

A single df SS for pure quadratic curvature is given by

where nF is the number of design points in the 2k design and nC is the number of replicates at the center point.

This SSquad = MSquad can be tested for significance by MSerr

or.

Let’s look at an example.

cF

centerfactorialcFquad nn

YYnnSS

2)(

A chemical engineer is studying the yield of a process. The two factors of interest are reaction time (A) and reaction temperature (B).

The engineer decides to run a 22 unreplicated factorial and include 5 center points to test for quadratic curvature.

The design then has reaction time at 30, 35, 40 minutes and reaction temp at 150˚C, 155˚C, 160˚C. So the design points and the yield data are

40 41.5 +1

factor B 0

-1 39.3 40.9 -1 0 +1 factor A

40.3 40.5 40.7 40.2 40.6

The ANOVA table for this experiment is

Source SS df MS p A (time) 2.4025 1 2.4025 0.0017 B (temp) 0.4225 1 0.4225 0.0350 AB 0.0025 1 0.0025 0.8185 Pure quad 0.0027 1 0.0027 0.8185 Error 0.1720 4 0.0430 Total 3.0022 8

In this design, = 40.425 and = 40.46. Since the difference is very small, there

is no quadratic effect, so the center points may be used to get an error term to test each of the effects.

So now this unreplicated design has an error term from

the replicated center points that lie on the same plane as the factorial points.

factorialYcenterY

0430.015

)46.40(

1

)(5

1

25

1

2

center

c

c

centercc

error

errorerror

Y

n

YY

df

SSMS

In this experiment, a first-order model is appropriate because there is no quadratic effect and no interaction. But suppose we have a situation where quadratic terms will be required and we have the following second-order model

2222

2111211222110 XXXXXXY

But this gives 6 values of β to estimate and the 22 design with center points has only 5 independent runs. So we cannot estimate the 6 parameters unless we change the design.

So we augment the design with 4 axial runs and create a central composite design to fit the second-order model.

The central composite design for a 22 factorial looks like this in our box format x2,

*(0,α)

(-,+) (+,+)

(-α,0)* *(α,0) X1

(-,-) (+,-)

*(0,-α)

o (0,0)

We’ll talk about central composite designs later, when we cover response surface methodology.

Now we’ll move on to fractional 2k factorials. A fractional factorial is a ½ fraction or a ¼ fraction or an 1/8 fraction of a complete factorial. Fractional factorials are used when a large number of factors need to be tested and higher-order interactions are considered unlikely.

Fractional factorials are widely used as screening experiments, where we try to identify which factors have a major effect and which factors are not relevant.

They are often used in the early stages of a project when the major features of the project are little understood.

They are often followed by sequential studies to explore the project further.

A ½ fraction is obtained as follows. Suppose you have three factors of interest and need a 23 design (8 runs) but for whatever reason, you cannot make 8 runs. You can however make 4 runs.

So instead of a 23 design, you use a 23-1 design or 4 runs. This 23-1 design is called a ½ fraction of the 23 design.

To create the 23-1 design, set up a 22 design and put the third factor into the AB interaction column.

Now factor C is confounded with AB. You

cannot separate the effect of the AB interaction from the effect of the C factor. In other words, C is aliased with AB.

Run Factor A Factor B Factor C = AB

1 -1 -1 +1

2 +1 -1 -1

3 -1 +1 -1

4 +1 +1 +1

What may be the consequences of this confounding?

1. The AB effect and the C effect are both large and significant but are in the opposite direction so they cancel each other out.

2. The AB effect and the C effect are both small but in the same direction, so the effect looks significant, but neither AB nor C separately is significant.

3. One effect is significant and the other is not, but you cannot tell which one is significant.

But this isn’t all. Now where are the AC and the BC interactions? Well, multiplying the codes we get

So a fractional factorial doesn’t just confound the AB effect with the C effect, it also confounds all main effects with 2-way interactions. When effects are confounded, they are called aliased. Now since A is aliased with BC, the first column actually estimates A+BC. Similarly, the second column estimates B+AC because B and AC are aliases of one another. The third column estimates the sum of the two aliases C and AB.

Run Factor A + BC

Factor B + AC

Factor C + AB

1 -1 -1 +1

2 +1 -1 -1

3 -1 +1 -1

4 +1 +1 +1

Now there are some better fractional designs, but you have to look at the generator to see them.

C=AB is called the generator of the design. Since C = AB, multiply both sides by C to get I= ABC, so I= ABC is the defining relation for the design. The defining relation is the set of columns equal to I.

ABC is also called a word. The length of the generator word tells you the resolution of the design. The generator ABC is a 3-letter word so this design is of resolution III.

What does design resolution mean? Design resolution tells you the degree of confounding in the design. There are three levels of resolution.

Resolution III: Main effects are aliased with 2-factor interactions and 2-factor interactions may be aliased with one another.

Resolution IV: Main effects are not aliased with 2-factor interactions, but 2-factor interactions are aliased with one another.

Resolution V: main effects are not aliased with 2-factor interactions and 2-factor interactions are not aliased with one another. But main effects and 2-way interactions may be aliased with higher-way interactions.

Of course, we would like to have the highest resolution design possible under the circumstances.

You can also use the defining relation to get the aliasing. In this example, where I = ABC, we can get the aliases by multiplying any column by the generator.

Alias of A: A*ABC = IBC=BC so A is aliased with BC

Alias of B: B*ABC = AIC= AC so B is

aliased with AC.

Let’s look at 24-1 factorial, a resolution IV design. First we create a 23 design.

Then we alias the 4th factor with the highest level interaction.

Run

A B C AB AC BC D=ABC

1 -1 -1 -1 +1 +1 +1 -1

2 +1 -1 -1 -1 -1 +1 +1

3 -1 +1 -1 -1 +1 -1 +1

4 +1 +1 -1 +1 -1 -1 -1

5 -1 -1 +1 +1 -1 -1 +1

6 +1 -1 +1 -1 +1 -1 -1

7 -1 +1 +1 -1 -1 +1 -1

8 +1 +1 +1 +1 +1 +1 +1

The generator for this design is D=ABC. To get the defining relation, multiply both sides by D to get I = ABCD. Since the defining relation word here is length 4, this is a resolution IV design.

Now let’s look at the aliases for this design.

Alias for A: A*ABCD = BCD Alias for B: B*ABCD = ACD Alias for C: C*ABCD = ABD Alias for D: D*ABCD = ABC Alias for AB: AB*ABCD = CD Alias for AC: AC*ABCD = BD Alias for BC: BC*ABCD = AD

After all the aliasing, the design is

Note that the main effects are aliased with 3-factor interactions and the 2-factor interactions are aliased with one another, so this is a resolution IV design.

Run A+BCD B+ACD C+ABD AB+CD AC+BD BC+AD D+ABC

1 -1 -1 -1 +1 +1 +1 -1

2 +1 -1 -1 -1 -1 +1 +1

3 -1 +1 -1 -1 +1 -1 +1

4 +1 +1 -1 +1 -1 -1 -1

5 -1 -1 +1 +1 -1 -1 +1

6 +1 -1 +1 -1 +1 -1 -1

7 -1 +1 +1 -1 -1 +1 -1

8 +1 +1 +1 +1 +1 +1 +1

Now let’s look at a 25-1 factorial, a resolution V design. First we create the 24 design, and then place the 5th factor in the highest-level interaction column.

Run A B C D AB AC AD BC BD CD ABC ABD ACD BCD E=ABCD

1 -1 -1 -1 -1 +1 +1 +1 +1 +1 +1 -1 -1 -1 -1 +1

2 +1 -1 -1 -1 -1 -1 -1 +1 +1 +1 +1 +1 +1 -1 -1

3 -1 +1 -1 -1 -1 +1 +1 -1 -1 +1 +1 +1 -1 +1 -1

4 +1 +1 -1 -1 +1 -1 -1 -1 -1 +1 -1 -1 +1 +1 +1

5 -1 -1 +1 -1 +1 -1 +1 -1 +1 -1 +1 -1 +1 +1 -1

6 +1 -1 +1 -1 -1 +1 -1 -1 +1 -1 -1 +1 -1 +1 +1

7 -1 +1 +1 -1 -1 -1 +1 +1 -1 -1 -1 +1 +1 -1 +1

8 +1 +1 +1 -1 +1 +1 -1 +1 -1 -1 +1 -1 -1 -1 -1

9 -1 -1 -1 +1 +1 +1 -1 +1 -1 -1 -1 +1 +1 +1 -1

10 +1 -1 -1 +1 -1 -1 +1 +1 -1 -1 +1 -1 -1 +1 +1

11 -1 +1 -1 +1 -1 +1 -1 -1 +1 -1 +1 -1 +1 -1 +1

12 +1 +1 -1 +1 +1 -1 +1 -1 +1 -1 -1 +1 -1 -1 -1

13 -1 -1 +1 +1 +1 -1 -1 -1 -1 +1 +1 +1 -1 -1 +1

14 +1 -1 +1 +1 -1 +1 +1 -1 -1 +1 -1 -1 +1 -1 -1

15 -1 +1 +1 +1 -1 -1 -1 +1 +1 +1 -1 -1 -1 +1 -1

16 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1

This is a resolution V design because the generator is E=ABCD, which makes the defining relation I = ABCDE. Now let’s check the aliases.

Alias of A: A*ABCDE = BCDE Alias of B: B*ABCDE = ACDE Alias of C: C*ABCDE = ABDE Alias of D: D*ABCDE = ABCE Alias of E: E*ABCDE = ABCD

Alias of AB: AB*ABCDE = CDE Alias of AC: AC*ABCDE = BDE Alias of AD: AD*ABCDE = BCE Alias of BC: BC*ABCDE = ADE Alias of BD: BD*ABCDE = ACE Alias of CD: CD*ABCDE = ABE Alias of AE: AE*ABCDE = BCD Alias of BE: BE*ABCDE = ACD Alias of CE: CE*ABCDE = ABD Alias of DE: DE*ABCDE = ABC

Now the design isRun A +

BCDEB +ACDE

C +ABDE

D +ABCE

AB +CDE

AC + BDE

AD +BCE

BC +ADE

BD +ACE

CD +ABE

ABC+DE

ABD+CE

ACD+BE

BCD+AE

E +ABCD

1 -1 -1 -1 -1 +1 +1 +1 +1 +1 +1 -1 -1 -1 -1 +1

2 +1 -1 -1 -1 -1 -1 -1 +1 +1 +1 +1 +1 +1 -1 -1

3 -1 +1 -1 -1 -1 +1 +1 -1 -1 +1 +1 +1 -1 +1 -1

4 +1 +1 -1 -1 +1 -1 -1 -1 -1 +1 -1 -1 +1 +1 +1

5 -1 -1 +1 -1 +1 -1 +1 -1 +1 -1 +1 -1 +1 +1 -1

6 +1 -1 +1 -1 -1 +1 -1 -1 +1 -1 -1 +1 -1 +1 +1

7 -1 +1 +1 -1 -1 -1 +1 +1 -1 -1 -1 +1 +1 -1 +1

8 +1 +1 +1 -1 +1 +1 -1 +1 -1 -1 +1 -1 -1 -1 -1

9 -1 -1 -1 +1 +1 +1 -1 +1 -1 -1 -1 +1 +1 +1 -1

10 +1 -1 -1 +1 -1 -1 +1 +1 -1 -1 +1 -1 -1 +1 +1

11 -1 +1 -1 +1 -1 +1 -1 -1 +1 -1 +1 -1 +1 -1 +1

12 +1 +1 -1 +1 +1 -1 +1 -1 +1 -1 -1 +1 -1 -1 -1

13 -1 -1 +1 +1 +1 -1 -1 -1 -1 +1 +1 +1 -1 -1 +1

14 +1 -1 +1 +1 -1 +1 +1 -1 -1 +1 -1 -1 +1 -1 -1

15 -1 +1 +1 +1 -1 -1 -1 +1 +1 +1 -1 -1 -1 +1 -1

16 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1

So the main effects are aliased with 4-factor interactions and the 2-factor interactions are aliased with the 3-factor interactions.

This all seems better than a resolution III design, on the surface, but remember that these effects are still confounded and all the consequences of confounding are still there.

In all of the cases we have been talking about, we have had ½ fractions. If necessary to clear up any ambiguities, we can always run the other ½ fraction.

If the original ½ fraction has defining relatio

n I=ABCD, the complementary ½ fraction has defining relation I = -ABCD.

Remember the filtration rate experiment we talked about earlier which was an unreplicated 24 factorial design. We used the normal probability plot of effects to determine that temperature (A), concentration of formaldehyde (C), and stirring rate (D) were significant, along with AC and AD interactions.

Now what would have happened if we had run a ½ fraction instead of the full factorial?

Since the original design was 24, we use instead a 24-1 ½ fraction. The design now is

and the generator is D = ABC.

Run A+BCD

B+ACD C+ABD AB+CD

AC+BD

BC+AD D+ABC Filt rate

1 -1 -1 -1 +1 +1 +1 -1 45

2 +1 -1 -1 -1 -1 +1 +1 100

3 -1 +1 -1 -1 +1 -1 +1 45

4 +1 +1 -1 +1 -1 -1 -1 65

5 -1 -1 +1 +1 -1 -1 +1 75

6 +1 -1 +1 -1 +1 -1 -1 60

7 -1 +1 +1 -1 -1 +1 -1 80

8 +1 +1 +1 +1 +1 +1 +1 96

The effect of the first column is A+BCD=(-1)45+(+1)100+(-1)45+(+1)65

+(-1)75+(+1)60+(-1)80 +(+1)96 = 76/4 = 19 The other effects are found in the same wa

y. They are:

B+ACD = 1.5 AB+CD = -1.0 C+ABD = 14.0 AC+BD = 18.5 D+ABC = 16.5 AD+BC = 19.0

But these effects are all confounded. The engineer suspects that because the B column effect is small and the A, C, and D column effects are large that the A, C, and D effects are significant.

He also thinks that the significant interactions are AC and AD, not BD or BC because the B effect is so small.

He may suspect, but is he right?

Let’s do the complementary ½ fraction and find out. For the complementary ½ fraction, the generator is D = -ABC. So the effect confounding is

Run A-BCD B-ACD C-ABD AB-CD AC-BD BC-AD D-ABC Filt rate

1 -1 -1 -1 +1 +1 +1 +1 43

2 +1 -1 -1 -1 -1 +1 -1 71

3 -1 +1 -1 -1 +1 -1 -1 48

4 +1 +1 -1 +1 -1 -1 +1 104

5 -1 -1 +1 +1 -1 -1 -1 68

6 +1 -1 +1 -1 +1 -1 +1 86

7 -1 +1 +1 -1 -1 +1 +1 70

8 +1 +1 +1 +1 +1 +1 -1 65

Now we can find the effects from this design the same way as from the original one.

A-BCD = 24.25 AB-CD = 1.25 B-ACD = 4.75 AC-BD = -17.75 C-ABD = 5.75 AD-BC = 14.25 D-ABC = 12.75

Now we can resolve the ambiguities by combining the effects from the original ½ fraction and those from the complementary ½ fraction.

Since the original ½ fraction estimates A+BCD and the complementary ½ fraction estimates A-BCD, we can isolate A by averaging the two estimates. This gives

[(A+BCD) + (A-BCD)]/2 = 2A/2 = A

Similarly we can isolate the BCD effect by [(A+BCD) – (A-BCD)]/2 = 2BCD/2 = BCD.

The unconfounded estimates are

So we can unconfound the effects by doing the co

mplementary ½ fraction. This should not be surprising because the complete fractional has no confounding.

Column

Original

estimate

Complementary

estimate

1/2(Orig + Comp)

1/2(Orig – Comp)

A 19 24.25 21.63 → A -2.63 → BCD

B 1.5 4.75 3.13 → B -1.63 → ACD

C 14 5.75 9.88 → C 4.13 → ABD

D 16.5 12.75 14.63 → D 1.88 → ABC

AB -1 1.25 0.13 → AB -1.13 → CD

AC 18.5 -17.75 -18.13 → AC -0.38 → BD

AD 19 14.25 16.63 → AD 2.38 → BC

Now let’s look at the ¼ fraction design. The designation for a ¼ fraction is 2k-2 fract

ional design.

To make a ¼ fraction design, say a 26-2, we first create a 24 design and associate t

he extra two variables with the highest-level interactions. This means that a ¼ fraction will have two generators.

In the 26-2 example, we may associate factor E with ABC and factor F with BCD. The two generators are E=ABC and F=BCD.

Therefore the two defining relations are I=ABCE and I=BCDF. To get the complete defining relation, we use all columns = I, so the complete defining relation is the above two and their interaction: I=ABCE=BCDF=ADEF.

Because the smallest word here is length 4, this is a resolution IV design.

To find the aliases for each effect, multiply each word in the complete defining

relation by that effect. For example,

Aliases of A: A*ABCE= BCE A*BCDF= ABCDF A+ADEF= DEF So A= BCE=ABCDF=DEF.

In ¼ fraction designs, each effect has a number of aliases. The complete alias structure for the 26

-2 design with I=ABCE=BCDF=ADEF is

A=BCE=DEF=ABCDF AB=CE=ACDF=BDEF B=ACE=CDF=ABDEF AC=BE=ABDF=CDEF C=ABE=BDF=ACDEF AD=EF=BCDE=ABCF D=BCF=AEF=ABCDE AE=BC=DF=ABCDEF E=ABC=ADF=BCDEF AF=DE=BCEF=ABCD F=BCD=ADE=ABCEF BD=CF=ACDE=ABEF ABD=CDE=ACF=BEF BF=CD=ACEF=ABDE ACD=BDE=ABF=CEF

There are three complementary fractions for the I=ABCE=BCDF=ADEF design. They have defining relations:

I = ABCE =-BCDF =-ADEF I =-ABCE = BCDF =-ADEF I =-ABCE =-BCDF = ADEF

In the first and third complementary fractions, the expression –BCDF means that F is placed in the BCD column and all the signs in the BCD column are reversed.

Similarly, in the second and third complementary fractions, the expression –ABCE means that E is placed in the ABC column and all the signs in the ABC column are reversed.

The alias structure for the effects will now change. For example, in the first complementary fraction, the aliases of A =BCE=-DEF=-ABCDF.

Whole tables of these fractional factorials exist, where you can find 1/8 fractions, 1/16 fractions, etc. So if you have a large number of factors to study and wish a small design and a huge headache, consult these tables.

In fact, there are 3k designs, which can be fractionalized. In these designs, there are three levels of each factor and k factors.

These designs work pretty much the same way as the 2k fractionals, except that there are complex alias relationships in 3k-1 designs that require the assumption of no interaction to be useful.

In addition, the 3-level designs are quite large even for a modest number of factors, so they tend to be used only occasionally. Mostly they are used for testing quadratic relationships, but they are not the best way to do so.

On the other hand, the 2k designs and their fractionals are used quite extensively in industrial experimentation, despite the confounding in the fractionals.

The most vigorous proponent of fractional factorials is a Japanese gentleman named Genichi Taguchi.

Taguchi has been rightly credited with popularizing experimental design in manufacturing. He has incorrectly credited himself with creating what he calls orthogonal arrays, which really are fractional factorials. Taguchi calls them L8, L16, etc. designs, depending on the number of runs.

Taguchi is a proponent of quality engineering and manufacturing. His design philosophy is that all products and processes should be robust to various forms of noise during their use.

For example, airplanes should fly as well in thunderstorms as they do in clear skies and cars should drive as well in the rain and snow as they do in good weather.

In addition to promoting robust design, Taguchi emphasizes the reduction of variability in manufacturing and emphasizes the importance of minimizing cost. For all of this, he deserves great credit.

Taguchi has designed his experiments to cover both controllable factors and uncontrollable noise.

He puts the controllable factors in an inner array and the noise factors in the outer array. So his design looks like

Y

Taguchi sees each system as

Variation

Signal Response

Noise factors

System

Inner Array (L8)

Outer Array (L4)

D 1 1 2 2 S-NE 1 2 1 2

Run A B C

1 1 1 1 resp

resp

resp

resp

2 2 1 1 resp

resp

resp

resp

3 1 2 1 resp

resp

resp

resp

4 2 2 1 resp

resp

resp

resp

5 1 1 2 resp

resp

resp

resp

6 2 1 2 resp

resp

resp

resp

7 1 2 2 resp

resp

resp

resp

8 2 2 2 resp

resp

resp

resp

Y

Taguchi proposes to use both the mean and a measure of variation he calls the SN ratio.

In each case, the combination of factors represented by the mean is the average over all noise combinations. This is what makes it robust.

Taguchi chooses the best combination of conditions by looking at plots of factor effects. He does not believe in significance tests.

Taguchi also proposes analyzing the signal-to-noise ratio for each combination of conditions. His S-N ratio for larger-the-better is

If smaller is better, the smaller-the-better SN is

n

i iL Yn

SN1

2

11log10

n

iiS Y

nSN

1

21log10

Taguchi believes that the SN ratios separate location from variability. When he analyzes them as response variables in an ANOVA, he thinks he is both optimizing the response and reducing the variability around it.

This has been shown to be completely incorrect. But Taguchi adherents still plot SN for each effect just as they do for .

Y

Taguchi also does not believe in interactions, although they are sometimes present in the experiments he has designed. He claims that if the engineer is working at the “energy level of the system,” there are no interactions.

But since Taguchi eyeballs marginal means plots and SN plots to pick the winners, he clearly misses out on some of the best combinations if there are interactions.

Another criticism of the Taguchi approach is that his combined inner and outer array set-up produces very large designs.

A better strategy might be to use a single design that has both controllable and noise factors and look at their interactions.

Now let’s look further at random effects factorial experiments.

You already know that a random effects model has factors with very many levels, and that the levels used in the experiment are chosen at random from all those available.

Let’s take a two-factor factorial where both factors are random.

In this experiment, the model is

where j = 1,2, …, J k = 1,2, …, K i = 1,2, …, n replicates

and the model parameters, , , , and are all independent normally distrib

uted random variables with mean 0 and variances , , , and .

ijkkjkjijk YY

2 2

2 2

jk kj

ijk

The SS and MS for each factor are calculated exactly the same as for the fixed effects model. But for the F ratios, we must examine the expected MS for each of the variance components.

22

222

222

2

)(

)(

)(

)(

nMSE

JnnMSE

KnnMSE

MSE

AB

B

A

E

To test each of these effects, we form the following F-ratios

Interaction effect:

A effect:

B effect:

Note that the main effects tests are different from those in the fixed-effects model.

AB

A

MS

MSF

E

AB

MS

MSF

AB

B

MS

MSF

Now most people do random effects models more to estimate the variance components than to test for significance. These estimates are

Jn

MSMS

Kn

MSMS

n

MSMS

MS

ABB

ABA

EAB

E

2

2

2

2

Gage R&R studies are a common industrial application of random effects models to test a measurement system.

In a typical experiment of this sort, there are J parts to be measured by some gage and K operators to do the measurement with n repetitions of the measurement.

In this example, there are J=20 parts to be measured, K=3 operators, and n=2 repetitions.

The data arePar

tOperator 1 Operator 2 Operator 3

1 21 20 20 20 19 21

2 24 23 24 24 23 24

3 20 21 19 21 20 22

4 27 27 28 26 27 28

5 19 18 19 18 18 21

6 23 21 24 21 23 22

7 22 21 22 24 22 20

8 19 17 18 20 19 18

9 24 23 25 23 24 24

10 25 23 26 25 24 25

11 21 20 20 20 21 20

12 18 19 17 19 18 19

13 23 25 25 25 25 25

14 24 24 23 25 24 25

15 29 30 30 28 31 30

16 26 26 25 26 25 27

17 20 20 19 20 20 20

18 19 21 19 19 21 23

19 25 26 25 24 25 25

20 19 19 18 17 19 17

The total variability can be divide into that due to parts, to operators, and to the gage itself.

is the variance component for parts

is the variance component for operators

is the variance component for interaction of parts and operators

is the random experimental error variance

2

22222 Y

2

2

2

A gage R&R study is a repeatability and reproducibility study.

The repeatability part of this is given by because this reflects variation when the same

part is measured by the same operator.

The reproducibility part is given by because this reflects the additional variability in t

he system from different operators using the gage.

2

22

The ANOVA table for this study is Source SS df MS p A: Parts 1185.43 19 62.39 <0.00001 B: Operators 2.62 2 1.31 0.1730 AB 27.05 38 0.71 0.8614 Error 59.50 60 0.99 Total 1274.60 119

The estimates of the variance components are

Notice that one of the variance components is negative, which is impossible because variances are positive by definition.

015.0)2)(20(

71.031.1

28.10)2)(3(

71.039.62

14.02

99.071.0

99.0

2

2

2

2

Jn

MSMS

Kn

MSMS

n

MSMS

MS

ABB

ABA

EAB

E

What can you do about this?

Well, you could just call it 0 and leave the other components unchanged.

Or you could notice that the interaction is insignificant and redo the ANOVA for a reduced model excluding the interaction term.

The redone ANOVA table is Source SS df MS p A: Parts 1185.43 19 62.39 <0.00001 B: Operators 2.62 2 1.31 0.2324 Error 86.55 98 0.88 Total 1274.60 119

and the new variance components are

0108.0)2)(20(

88.031.1

25.10)2)(3(

88.039.62

88.0

2

2

2

Jn

MSMS

Kn

MSMS

MS

ABB

ABA

E

Then the gage variance is

and the total variance is

So most of the variance is due to variability in th

e product. Very little is due to operator variability or nonrepeatability from part to part.

8908.0

0108.088.0

22

1408.11

25.100108.088.0

222

Of course, it had to come to this. If factors can be fixed and they can be random, there are certainly going to be studies that are a combination of fixed and random factors. These are called mixed models.

Let’s look at a simple case where there is one fixed factor A and one random factor B.

The model is

where is fixed so

The other effects are random. However, summing the interaction component over the fixed effect = 0. That is,

ijkkjkjijk YY

j 01

J

jj

01

k

J

jj

This restriction implies that some of the interaction elements at different levels of the fixed factor are not independent.

This restriction makes the model a restricted model. The expected MS are

22

1

2

22

22

2

)(

1)(

)(

)(

JnMSE

J

Kn

nMSE

nMSE

MSE

B

J

jj

A

AB

E

This implies that the F-ratio for the fixed factor is

But the tests for the random factor B and the AB interaction are

and

AB

A

MS

MSF

E

B

MS

MSF

E

AB

MS

MSF