Experimental Design

Lecture Notes on MAT3021

Experimental Design

Janet Godolphin

Semester 2 2014-15

Contents

1 Comparisons of Samples 11.1 Principles of Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Comparison of Independent Samples . . . . . . . . . . . . . . . . . . . . . . . 2

1.2.1 Decomposition of the Total Sum of Squares . . . . . . . . . . . . . . . 31.2.2 The Analysis of Variance Table . . . . . . . . . . . . . . . . . . . . . 8

2 Incomplete Models 112.1 Incomplete Block Designs . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.2 Balanced Incomplete Blocks . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.2.1 Constructions of Balanced Incomplete Block Designs . . . . . . . . . . 192.2.2 Analysis of the Balanced Incomplete Block Design . . . . . . . . . . . 232.2.3 Choosing Balanced Incomplete Block Designs . . . . . . . . . . . . . 31

3 Analysis of Covariance 353.1 Separate regressions model . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363.2 Parallel regressions model . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383.3 Further simplifications of the model . . . . . . . . . . . . . . . . . . . . . . . 393.4 ANCOVA Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

4 Specialist Incomplete Models 434.1 Latin Square Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434.2 Youden Squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454.3 Cross-Over Designs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

4.3.1 Column-Complete Latin Square . . . . . . . . . . . . . . . . . . . . . 494.3.2 Analysis of Cross Over Experiments . . . . . . . . . . . . . . . . . . . 52

5 Factorial Experiments 555.1 Advantages of Factorial Experiments . . . . . . . . . . . . . . . . . . . . . . . 555.2 k Factors at Two Levels: Yates’Algorithm . . . . . . . . . . . . . . . . . . . . 56

5.2.1 The 22 Factorial Experiment . . . . . . . . . . . . . . . . . . . . . . . 575.2.2 The 23 Factorial Experiment . . . . . . . . . . . . . . . . . . . . . . . 585.2.3 The 2k Factorial Experiment . . . . . . . . . . . . . . . . . . . . . . . 595.2.4 The 2k Factorial with a Single Replicate . . . . . . . . . . . . . . . . . 60

5.3 Fractional Replication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 635.3.1 Half-Replicate Fractional Factorial . . . . . . . . . . . . . . . . . . . . 63

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5.3.2 Quarter-Replicate Fractional Factorial . . . . . . . . . . . . . . . . . . 675.3.3 The Resolution of a Design . . . . . . . . . . . . . . . . . . . . . . . . 69

5.4 Confounding in a 2k Factorial . . . . . . . . . . . . . . . . . . . . . . . . . . 705.5 Specialised Uses Of Fractional Factorial Experiments . . . . . . . . . . . . . . 745.6 Robust Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

5.6.1 Analysing the Effects of Design Factors on variability . . . . . . . . . 79

Chapter 1

Comparisons of Samples

1.1 Principles of Design

Statisticians do not usually perform experiments on their own because, in many aspects ofexperimental work, the statistician has no expert knowledge. However, many research organi-sations and commercial industries depend on the help of qualified statisticians for the planningof experiments and for the interpretation of experimental results.

For example, Pharmaceutical Companies are prevented by law from marketing new drugs unlessthey have undergone valid test trials which require appropriate statistical expertise.

Each member of a research team engaged on experimentation involving the collection and anal-ysis of data has a different role. The statistician needs to be involved at the beginning of theproblem to help define the experimental objectives. The statistician is also involved after exper-imentation in order to take charge of the interpretation of the experimental results.

Experiment Example: Daphnia Lifetimes in Cu ConcentrationsA study is undertaken to determine the effect of copper concentrations in water on the lifetimesof aquatic animals (daphnia).

Fifteen containers each contain one daphnia. Three copper concentration levels for the water inthe containers are chosen: no copper, 20 mg/l and 40 mg/l. The treatments (copper levels) areassigned at random to the experimental units (daphnia in containers) such that each treatment isused on five daphnia.

The daphnia are raised in the three concentrations and their lifetimes (in days) are recorded.

The Experimental Objectives(1) Recognition of the Problem requires a clear statement of the experimental problem, usuallyas a result of exchanges of views with experts in the field. The aim should be to make thestatement lucid and specific. The most common faults are vagueness and excessive ambition.Often it is helpful to classify the various requirements of the experiment as “major” and “minor”because compromises may have to be made.

(2) Choice of Experimental Units is required to ensure as far as possible that the stated objectivesof the experimental problem can be pursued and achieved satisfactorily. If possible the choice

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of units should be made to enable the treatment effects to show up most strikingly.

(3) The Number of Treatments needs to decided. The aim is to ensure that all relevant treatmentsare considered but that no effort and cost is wasted on examination of treatments which are oflimited interest. It is often necessary to understand the role that each treatment will play inpursuing the objectives of the experiment.

(4) The Choice of Experimental Design needs to be made with particular care. The term exper-imental design refers to the assignment of treatments, or treatment combination, to the exper-imental units so that analysis of the data is possible and is relevant to the problem. A majorconsideration is the precision of the experiment and this is often the principal concern whenchoosing between competing designs.

We start by looking at the simplest Complete Design, namely the One-Way Classification modelin detail.

1.2 Comparison of Independent Samples

In this Section we consider m independent samples, relating to m treatments;

y11, · · · , y1n1 , y21, · · · , y2n2 , · · · · · · , ym1, · · · , ymnm

of sizes n1, n2, · · · , nm respectively and let n1 + n2 + · · ·+ nm = N.Suppose that each sample comes from a normal population and that

E[yij] = µ+ τi wherem∑i=1

niτi = 0, and Var[yij] = σ2.

This model is known as the One-Way Classification model.

NotationThe following notation style is commonly used:

yi. represents the total of the observations under the ith treatment;

yi. represents the average of the observations under the ith treatment;

y.. represents the grand total of all the observations;

y.. represents the grand average of all the observations.

Objective of ExperimentThe objective is to compare the means of the m different samples, i.e. to compare the m treat-ment effects τ1, τ2, . . . , τm. This is the simplest experimental design although, strictly speaking,there is little attempt at design here. This is sometimes known as the Completely RandomisedDesign and is important as it arises quite frequently in practice.

RandomisationRandomisation is employed to avoid systematic bias, selection bias, accidental bias and evencheating by the experimenter (not necessarily badly intentioned).

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RemarkIf m = 2 there are two independent normal samples with a common variance but with possiblydifferent population means, and the comparison of the means is usually accomplished by thetwo-sample t−test. In this section we sketch out the mathematical details for the general casem ≥ 2.

The Null and Alternative HypothesesThe aim is to test the null hypothesis H0 : τ1 = τ2 = · · · = τm = 0 against the contraryhypothesis H1 : τi 6= 0 for at least one i.

1.2.1 Decomposition of the Total Sum of Squares

The mechanism employed to carry out the test involves analysis of variance. The term analysisof variance comes from a partitioning of total variability into its component parts.

The total sum of squares,

SSTotal =m∑i=1

ni∑j=1

(yij − y..)2,

is used as a measure of total variability. This is an intuitive measure since dividing SSTotal

by N − 1 gives the sample variance of the N observations. The total sum of squares can berewritten as:

m∑i=1

ni∑j=1

(yij − y..)2 =m∑i=1

ni∑j=1

[(yi. − y..) + (yij − yi.)]2

=m∑i=1

ni(yi. − y..)2 +m∑i=1

ni∑j=1

(yij − yi.)2 + 2m∑i=1

ni∑j=1

(yi. − y..)(yij − yi.)

=m∑i=1

ni(yi. − y..)2 +m∑i=1

ni∑j=1

(yij − yi.)2 (1.1)

Equation 1.1 demonstrates that SSTotal, the total variability in the data, can be partitioned intotwo parts, namely;

• a sum of squares of the differences between the treatment averages and the overall aver-age;

• a sum of squares of the differences of observations within treatments from the treatmentaverage.

We write Equation 1.1 symbolically as

SSTotal = SST + SSResid,

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where SST is called the Treatment sum of squares and SSResid is called the Residual or Errorsum of squares.

There are N observations; thus SSTotal consists of the sum of squares of N elements of theform yij − y... These elements are not all independent because

∑mi=1

∑ni

j=1(yij − y..) = 0. Infact exactly N − 1 of the elements are independent, implying that SSTotal has N − 1 degrees offreedom.

Similarly, SST consists of the sum of squares of m elements of the form yi. − y.. since there arem levels of treatment, but

∑mi=1 ni(yi. − y..) = 0, so SST has m− 1 degrees of freedom.

Finally, within any treatment there are ni replicates with∑ni

j=1(yij − yi.) = 0, providing ni − 1degrees of freedom with which to estimate the experimental error. In total, there are

∑mi=1(ni−

1) = N −m degrees of freedom for error. Note that:

df(SSTotal) = df(SST) + df(SSResid).

Considering the two terms on the right hand side of Equation 1.1, the residual sum of squarescan be represented as:

SSResid =m∑i=1

[ni∑j=1

(yij − yi.)2]. (1.2)

In this form it can be seen that the term within square brackets, is (ni − 1)s2i , where s2i is thesample variance of the observations involving the ith treatment. Thus, SSResid/(N −m) is anestimate of the common variance, σ2, obtained from comparisons between data within each ofthe m treatments. The quantity

SSResid

N −m

is called the residual mean square.

We now obtain the expectation of SSResid/(N − m) by a formal approach. The followingexpectations are useful:

4

E[y2ij]

= Var [yij] + {E [yij]}2

= σ2 + (µ+ τi)2 (1.3)

E[y2i.]

= Var [yi.] + {E [yi.]}2

=σ2

ni+ (µ+ τi)

2 (1.4)

E[y2..]

= Var [y..] + {E [y..]}2

=σ2

N+

1

N

m∑i=1

ni∑j=1

µ2

=σ2

N+ µ2 (1.5)

E

[SSResid

N −m

]=

1

N −mE

[m∑i=1

ni∑j=1

(yij − yi.)2]

=1

N −mE

[m∑i=1

ni∑j=1

(y2ij − 2yij yi. + y2i.)

]

=1

N −mE

[m∑i=1

ni∑j=1

y2ij − 2m∑i=1

niy2i. +

m∑i=1

niy2i.

]

=1

N −mE

[m∑i=1

ni∑j=1

y2ij −m∑i=1

niy2i.

]

=1

N −m

{m∑i=1

ni∑j=1

E[y2ij]−

m∑i=1

niE[y2i.]}

=1

N −m

{m∑i=1

ni(σ2 + (µ+ τi)

2)−

m∑i=1

ni

(σ2

ni+ (µ+ τi)

2

)}=

1

N −m{Nσ2 −mσ2

}= σ2 (1.6)

The quantitySST

m− 1

is called the treatment mean square. If there are no differences between them treatment effects,we can use

SST

m− 1=

∑mi=1 ni(yi. − y..)2

m− 1

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as an estimate of σ2.This statement is justified by consideration of the expectation of SST/(m− 1):

E

[SST

m− 1

]=

1

m− 1E

[m∑i=1

ni(yi. − y..)2]

=1

m− 1E

[m∑i=1

niy2i. −Ny2..

]

=1

m− 1

{m∑i=1

niE[y2i.]−NE

[y2..]}

=1

m− 1

{m∑i=1

ni

(σ2

ni+ (µ+ τi)

2

)−N

(σ2

N+ µ2

)}

=1

m− 1

{mσ2 +

m∑i=1

ni (µ+ τi)2 − σ2 −Nµ2

}

=1

m− 1

{(m− 1)σ2 +

m∑i=1

ni(µ2 + 2µτi + τ 2i

)−Nµ2

}

=1

m− 1

{(m− 1)σ2 +

m∑i=1

niτ2i

}

= σ2 +1

m− 1

m∑i=1

niτ2i (1.7)

Thus, SSResid/(N − m) is an unbiased estimator of σ2, and if there are no differences in thetreatment effects then SST/(m− 1) is also an unbiased estimator of σ2.

Note that if the treatment effects are not the same then the expectation of SST/(m−1) is greaterthan σ2.

A test of the hypothesis of no difference in treatment effects can be performed by comparingthe two mean squares SST/(m− 1) and SSResid/(N −m).

Statistical AnalysisWe now outline the theory behind a formal test of H0 : τ1 = τ2 = · · · = τm = 0 againstH1 : τi 6= 0 for at least one i.

Theorem 1.1. Let the n independent random variables X1, X2, · · · , Xn each have a normaldistribution with mean 0 and variance 1. Then

X21 +X2

2 + · · ·+X2n

has a chi-squared distribution with n degrees of freedom.

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Corollary 1.1. Let the n independent random variables X1, X2, · · · , Xn each have a normaldistribution with mean 0 and variance σ2. Then

1

σ2(X2

1 +X22 + · · ·+X2

n)

has a chi-squared distribution with n degrees of freedom.

Theorem 1.2. Let U be a χ2ν random variable and let

U = U1 + U2 + . . .+ Un,

where the Ui are χ2νi

random variables with

ν = ν1 + ν2 + . . .+ νn.

Then the random variables U1, U2, . . . , Un are independent.

Theorem 1.3. Let V , W be two independent χ2 random variables with v, w degrees of freedomrespectively. Then the ratio

V/v

W/w

has an F distribution with v and w degrees of freedom.

We have assumed that the εij are independent normal random variables with mean 0 and vari-ance σ2; it follows that the observations yij are independent normal random variables with meanµ+ τi and variance σ2.

It can be shown thatSSTotal/σ

2 ∼ χ2N−1.

Similarly, it can be shown thatSSResid/σ

2 ∼ χ2N−m

and that if the null hypothesis is true then

SST/σ2 ∼ χ2

m−1.

The degrees of freedom for SST/σ2 and SSResid/σ

2 add toN−1, the total number of degrees offreedom and so Theorem 1.2 implies that, if the null hypothesis is true, SST/σ

2 and SSResid/σ2

are independently distributed chi-square random variables.

Now Theorem 1.3 implies that under the null hypothesis

SST/(m− 1)

SSResid/(N −m)∼ Fm−1

N−m (1.8)

Note that if the null hypothesis is false, then the expected value of the numerator of the teststatistic (Equation 1.8) is greater than the expected value of the denominator, and we reject H0

on values of the test statistic that are larger than the 95th percentile of Fm−1N−m.

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1.2.2 The Analysis of Variance Table

The test procedure is summarized in an ANOVA table:

Source DF SS MS E[MS]

Treatments m− 1 SST SST/(m− 1) σ2 +∑m

i=1 niτ2i

m−1

Residual N −m SSResidSSResid

N−m σ2

Total N − 1 SSTotal

Practical ComputationsIf performing the calculations by hand, the sums of squares in the ANOVA table are most easilyobtained using the mean correction M.C. = Ny2... The Treatment sums of squares is

SST =m∑i=1

ni(yij − y..)2 =m∑i=1

y2i.ni−M.C.

and the Total sum of squares SSTotal =m∑i=1

ni∑j=1

(yij − y..)2 =m∑i=1

ni∑j=1

y2ij −M.C.

The Residual sum of squares, SSResid, is given by subtraction.

Example: Daphnia Lifetimes in Cu ConcentrationsThe lifetimes (in days) of the 15 daphnia are:

none 20 mg/` 40 mg/`60 58 4090 74 5874 50 2582 65 3077 68 42

Solution

M.C. = Ny2.. = 15× 59.532 = 53163.3

SST =m∑i=1

y2i.ni−M.C. =

3832

5+

3152

5+

1952

5− 53163.3 = 3624.5

SSTotal =m∑i=1

ni∑j=1

y2ij −M.C. = 58271− 53163.3 = 5107.7

SSResid = SSTotal − SST = 1483.2

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Thus the ANOVA table is:

Source DF SS MS F value

Treatments 2 3624.5 1812.3 14.7

Residual 12 1483.2 123.6

Total 14 5107.7

We compare the realisation of the test statistic, 14.7, with the 95th percentile of F 212 which is

3.885.

The test statistic is larger so we reject H0 and conclude that copper does affect the lifetimes ofdaphnia.

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10

Chapter 2

Incomplete Models

Remark on Incomplete DesignsAn Incomplete Design is one in which some combinations of the levels of all the factors are notobserved. Such an Incomplete Design could arise in practice if some of the observations in aComplete Design go missing (e.g. certain data may be spoiled or destroyed) but this is not thecontext which we consider here. Our primary concern is with designs which are deliberatelymade incomplete because of practical limitations on the experiment which are known in ad-vance. Choosing an Incomplete Design is part of the planning of the experiment. The principalaim is to create an incomplete design which is suitably balanced; how we define this balancewill be discussed in the context of the different incomplete designs which we will meet.

2.1 Incomplete Block Designs

Blocking is an extremely important design technique and is used extensively in industrial ex-perimentation. Blocking can be used to systematically eliminate the effect of a nuisance sourceof variability on the statistical comparisons among treatments. Examples of blocking factorsinclude; batches of raw material, factory, worker.The most common block design for t treatments arranged in b blocks is the randomised or com-plete block design. In this design, each treatment occurs once in each block, with the treatmentsbeing randomly assigned to the experimental units within each block. The observations yij areindependent and normally distributed with a common variance σ2 and with the expectations

E[yij] = µ+ τi + βj,

where τi is the ith treatment effect (i = 1, 2, . . . , t) and βj is the jth block effect (j =1, 2, . . . , b). This requires that the block-size is constant and equal to t (the number of dif-ferent treatments) and the number of replications of treatments is constant and equal to b (thenumber of blocks). Practical considerations may prevent these conditions from being realised.

ExampleIn an agricultural experiment there are six varieties of winter wheat and four different siteswhere wheat can be grown, and these four sites should be considered as blocks. Ideally, all six

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wheat varieties should be grown on all four sites but each site has space for only three or fourexperimental units, i.e. only three or four varieties of winter wheat can be compared. If the sixtreatments are designated A,B,C,D,E and F then one such design might be as follows:

D1 ≡

1 2 3 4A A A CB D B DC E E FD F

.

Here, practical considerations compete with the natural desire to achieve a complete block de-sign and it is necessary to use an incomplete block design where the number of treatments isgreater than the block-sizes. But what are good and bad properties of an incomplete blockdesign? To answer this question it is useful to note some common definitions.

Common ParametersA block design is equi-replicate if each treatment occurs the same number of times, say r. Itfollows that every treatment appears in r blocks (since no treatment can occur twice in a block).

A block design has a common block-size if each block contains the same number of experimentalunits, say k. In this case the block design is complete if k = t and incomplete if k < t.

A block design is binary if no treatment occurs more than once in a single block.

The concurrence λAB of two treatments A and B is the number of blocks in which A and Boccur together. A block design has

(t2

)concurrences.

An equi-replicate block design with a common block-size whose concurrences are equal is saidto be balanced.

ExamplesDesign D1 has t = 6 and b = 4. It is not equi-replicate and does not have a common block-size.The concurrences involving treatment A are λAB = 2, λAC = 1, λAD = 2, λAE = 2 andλAF = 1. Design D1 is not balanced.

Consider design D2, which is given by

D2 ≡

1 2 3 4 5 6 7A B A B E E FB D C C F F GC A D D G H H

.

D2 has t = 8 and b = 7. It is not equi-replicate but it does have a common block-size k = 3.The concurrences involving treatment A are λAB = λAC = λAD = 2 and λAE = λAF = λAG =λAH = 0. Design D2 is not balanced.

Design D3,, given by

D3 ≡1 2 1 3 5 2 3 66 4 5 7 9 9 4 810 8 7 12 11 10 11 12

,

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has t = 12 and b = 8. This design is equi-replicate with r = 2 and it has a common block-sizek = 3. The concurrences involving treatment 1 are λ1j = 0 except λ15 = λ16 = λ17 = λ1,10 =1. Design D3 is not balanced.

ConnectednessAn incomplete block design is said to be connected if it is possible to obtain a linear unbiasedestimator of every contrast involving two treatment effects, i.e. for each i, j = 1, 2, . . . , t thereis a vector a such that E[a′Y ] = τi − τj.

Remark on Connected DesignsConnectedness is related to concurrence, i.e. a design is disconnected if a subset of the treat-ments occur wholly in blocks which do not contain any of the remaining treatments. Dis-connectedness is a bad property of a design because it is impossible to test the hypothesisH0 : τ1 = τ2 = . . . = τt = 0 in this case.

Design D2 is disconnected since blocks 1 to 4 contain treatments A,B,C,D and blocks 5 to 7contain treatments E,F,G,H . In this design treatment A can be compared with B,C or D butnot with any of E,F,G or H .

Check for a Connected DesignConsider Block 1: put the treatments and β1 in S1.

Consider Block 2: if any treatments are in S1 put the remaining treatments and β2 in S1; other-wise put the treatments and β2 in S2.

Consider Block 3: if treatments are in both S1 and S2 let S1 = S1 ∪ S2, delete S2 and put theremaining treatments and β3 in S1; if treatments are in just one of S1 or S2, put the remainingtreatments and β3 in this set; if no treatments are in S1 or S2, put the treatments and β3 in S3.

Continue in this fashion for Block p: if several sets already contain one or more treatmentswhich are in block p then merge those sets and put the remaining treatments from block pand βp in that set; if any block p treatments are in just one set then put the remaining treat-ments and βp in that set; if no block p treatments are in any sets then form a new set with thesetreatments and βp.

This process algorithm terminates in two ways:

Completion occurs: when all t treatments are placed in S1 at any stage; we can conclude thatthe design is connected.

Completion occurs: when all b blocks are considered and the sets S1, · · · , Sd are formed. Ifd = 1 the design is connected. If d ≥ 2 the design is disconnected.

Illustration of the Checking ProcessThe checking process is applied to design D3. Sets are formed after considering each block,reading from left to right, as follows:

Block 1: S1 = {τ1, τ6, τ10, β1}.

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Block 2: S1 as before and S2 = {τ2, τ4, τ8, β2}.

Block 3: S1 = {τ1, τ5, τ6, τ7, τ10, β1, β3} and S2 as before.

Block 4: S1 = {τ1, τ3, τ5, τ6, τ7, τ10, τ12, β1, β3, β4} and S2 as before.

Block 5: S1 = {τ1, τ3, τ5, τ6, τ7, τ9, τ10, τ11, τ12, β1, β3, β4, β5} and S2 as before.

Block 6: S1 = {τ1, · · · , τ12, β1, · · · , β6} and S2 deleted.

Completion of the algorithm occurs after considering block 6 since all twelve treatment param-eters are in S1. Thus the design is connected.

Incidence MatrixThe structure of an incomplete block design can be seen from its incidence matrix N = {nij}which is a t× b matrix with nij = 1 if treatment i occurs in block j and nij = 0 otherwise.

Note that a complete block design has a t× b matrix of 1’s.

The incidence matrices can be quite revealing. Some properties of the design can be obtainedfrom the incidence matrix. For example, the incidence matrices of D1 and D2 are given by

1 1 1 01 0 1 01 0 0 11 1 0 10 1 1 00 1 0 1

and

1 1 1 0 0 0 01 1 0 1 0 0 01 0 1 1 0 0 00 1 1 1 0 0 00 0 0 0 1 1 00 0 0 0 1 1 10 0 0 0 1 0 10 0 0 0 0 1 1

.

ExamplesDesign D1:The number of 1’s indicates that there are 14 observations. The number of 1’s in each rowindicates that A,B,C,D,E and F are replicated 3, 2, 2, 3, 2 and 2 times respectively. Simi-larly, the number of 1’s in each column shows that blocks 1, 2, 3 and 4 have sizes 4, 4, 3 and 3respectively.

Count when two 1’s occur together in the first row and each other row to see that λAB =2, λAC = 1, λAD = 2, λAE = 2, and λAF = 1, and similarly for the other concurrences.

Design D2:Similarly, D2 has 21 experimental units, A,B,C,D and F are replicated 3 times, E,G and Hare replicated twice, D2 has a common block-size of k = 3 and the concurrences can be found.The incidence matrix also shows that D2 is disconnected since no treatment in the first fourblocks shares a positive concurrence with a treatment in the last three blocks.

Dual of an Incomplete Block DesignFor a given incomplete block design D, the dual of D is the incomplete block design in whichthe roles of treatments and blocks are interchanged.

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Thus a design with t treatments and b blocks has a dual with b treatments and t blocks and anincidence matrix which is the transpose of that of D. For example, the dual of D1 is given by

D4 ≡

1 2 3 4 5 6A A A A B BB C D B C DC D

.

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2.2 Balanced Incomplete BlocksIn this Section, we compare t treatments in b blocks when practical limitations on the experimentimply that only k treatments can be included in each block because the block-size k < t. Thissituation often arises in practice, so we consider how to arrange an experimental design to copewith this problem.

Definition of Balance in an Incomplete BlockAn arrangement of blocks and treatments is called a Balanced Incomplete Block if the followingconditions are satisfied:

(1) every block contains k experimental units;(2) there are more treatments t than the block-size k, i.e. k < t;(3) every treatment appears in r blocks;(4) each pair of treatments occurs together in a block λ times.

Balanced Incomplete Block Designs have very good optimality properties. Thus, if practicallimitations mean that a Complete Block Design is not feasible, then whenever possible, a Bal-anced Incomplete Block Design should be used.

Example of Balanced Incomplete Block DesignSuppose that we have a block-size of k = 3. The design B1 arranges 7 treatments in sevenblocks.

B1 ≡

1 2 3 4 5 6 7A B C D E F GB C D E F G AD E F G A B C

.

This design satisfies all four conditions:(1) k = 3;(2) k < t = 7;(3) r = 3;(4) every pair of treatments occurs together in a block once, i.e. λ = 1.

So B1 is a balanced incomplete block design.The incidence matrix is:

1 0 0 0 1 0 11 1 0 0 0 1 00 1 1 0 0 0 11 0 1 1 0 0 00 1 0 1 1 0 00 0 1 0 1 1 00 0 0 1 0 1 1

.

Conditions for a Balanced Incomplete Block DesignWe consider necessary conditions for a balanced incomplete block design. In this case eachrow of the incidence matrix has r elements unity due to the r blocks occupied by that treatment;each column has k elements unity because k treatments occupy that block. Hence

16

∑bj=1 nij = r and

∑ti=1 nij = k. Also

∑bj=1 nijn`j = λ.

This implies that N1b = r1t and 1′tN = k1′b.

Theorem 2.1. Consider a balanced incomplete block design for t treatments replicated r timesarranged in b blocks with block-size k such that λ is the number of times each pair of treatmentsoccurs together in a block. Then there are two identities:

(1) tr = bk and (2) λ(t− 1) = r(k − 1).

Furthermore, there is the inequality: (3) t ≤ b (Fisher’s Inequality).

Remark on Theorem 2.1Balanced incomplete block designs exist only for particular values of t, r, b, k and λ andTheorem 2.1 shows that, given any three of these parameters, we can determine the other two.This makes it easier to decide whether it may be possible to find a balanced design for a givenset of parameters. For example, it may be possible to construct a balanced incomplete blockdesign with t = 7, b = 7, k = 3 but it is not possible with t = 7, b = 6, k = 3 nor witht = 7, b = 8, k = 3.

Theorem 2.2. If N is the incidence matrix of a balanced incomplete block design with theparameters t, r, b, k and λ then

(i) NN ′ = (r − λ)It + λKt and (ii) KtN = k1t1′b.

where Kt is the t× t matrix of 1’s.

Conversely, if N is a matrix of zeros and ones satisfying (i) and (ii), then

t =r(k − 1)

λ+ 1 and b =

tr

k,

and, provided that k < t, then N is the incidence matrix of a balanced incomplete block designwith parameters t, r, b, k and λ.

It follows from Theorem 2.2 that NN ′ has the particularly simple form

NN ′ =

r λ λ · · · λλ r λ · · · λλ λ r · · · λ...

...... . . . ...

λ λ λ · · · r

=⇒ det(NN ′) = (r − λ)t−1rk.

Note that r 6= λ since otherwise k = t which contradicts k < t.

Existence of Balanced Incomplete BlocksThe identities specified in Theorem 2.1 are necessary for the existence of a balanced incompleteblock but they are not sufficient. For example, no balanced incomplete block exists for t =15, r = 7, b = 21, k = 5 and λ = 2 even though both identities and Fisher’s inequality inTheorem 2.1 are satisfied.

17

We now look at sufficient conditions for a balanced incomplete block design.

Unreduced Balanced Incomplete BlocksFor given integers t, k, where k < t, put in blocks all combinations of the t treatments taken kat a time. Thus we have

b =

(tk

), r =

(t− 1k − 1

)and λ =

(t− 2k − 2

).

Obviously, a design with these parameters always exists. It is called an Unreduced balancedincomplete block design.

ExampleThe following design B2 is an unreduced balanced incomplete block with parameters t = 6,k = 2 so that b = 15, r = 5 and λ = 1 :

B2 ≡1 2 3 4 5 6 7 8 9 10 11 12 13 14 15A A A A A B B B B C C C D D EB C D E F C D E F D E F E F F

Unreduced designs usually require rather a lot of blocks and replicates and the resulting designsmay be too large for practical purposes. For example, the unreduced design with t = 7 andk = 3 requires b = (73) = 35 blocks, but this is considerably more than the seven blocks whichwere required for B1.

Complementary Balanced Incomplete BlocksSuppose that a balanced incomplete block design B exists for parameters t, k (where k < t)and λ. A balanced incomplete block B has the same treatments and the same number of blocks(of possibly different block-size), where the blocks ofB contain all treatments omitted from thecorresponding blocks ofB. B is the Complementary design of the original balanced incompleteblock B.Note: the complementary design of B and the dual design of B are not the same!

Theorem 2.3. The complementary design of B always exists whenever b− 2r + λ > 0.The parameters of the complementary design B are given by

t = t, b = b, k = t− k, r = b− r and λ = b− 2r + λ,

where t, b, k, r and λ are the parameters of the original design B.

Examples of Complementary Balanced Incomplete BlocksComplementary designs of B1 and B2 are

B1 ≡

1 2 3 4 5 6 7C D E F G A BE F G A B C DF G A B C D EG A B C D E F

≡ Remainder of a 7× 7 Latin Square

18

B2 ≡

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15C B B B B A A A A A A A A A AD D C C C D C C C B B B B B BE E E D D E E D D E D D C C CF F F F E F F F E F F E F E D

2.2.1 Constructions of Balanced Incomplete Block Designs

Cyclic Constructions

Theorem 2.4. A cyclic design with numerical treatment labels, 0, 1, . . . , t, is balanced if andonly if the first block has the property that all non-zero treatments occur equally often amongits differences.

A first block which gives rise to equal frequencies for the differences is called a perfect differ-ence set.

Perfect Difference Sets

A perfect difference set (modulo ν) is a set of distinct numbers

{b1, b2, . . . , bk}

such that the non-zero differences

b1 − b2, b1 − b3, . . . , bk − bk−1

include each non-zero number (modulo ν) equally often.

Example The set {0, 1, 4, 6} is a perfect difference set (modulo 13), but the set {0, 1, 2, 3} isnot a perfect difference set (modulo 13).

If ν is a prime number of form 4n+ 3 then a perfect difference set (modulo ν) can be obtainedby writing down the squares 12, 22, 32, . . . and reducing the answers (modulo ν).

Example The non-zero squares (modulo 7) are

12 = 1; 22 = 4; 32 = 9 ≡ 2; 52 = 25 ≡ 4; 62 = 36 ≡ 1.

Thus, the non-zero squares are 1,2 and 4 and the non-squares are 3, 5 and 6 (modulo 7). Theset {1, 2, 4} is a perfect difference set (modulo 7).

The following four constructions use cyclic operations on sets of squares and non-squares.

Construction 1Let p be a prime number of the form 4n+ 3.

To obtain a symmetric balanced design with parameters

t = b = p, r = k = (p− 1)/2 and λ = (p− 3)/4,

19

take as first block the set of non-zero squares (modulo p); construct the other blocks using thecyclic construction.

Example of Construction 1Consider p = 11. The non-zero squares (modulo p) are 1, 3, 4, 5 and 9. Using these as the firstblock and employing a cyclic construction we obtain the balanced design

Block1 2 3 4 5 6 7 8 9 10 111 2 3 4 5 6 7 8 9 10 03 4 5 6 7 8 9 10 0 1 24 5 6 7 8 9 10 0 1 2 35 6 7 8 9 10 0 1 2 3 49 10 0 1 2 3 4 5 6 7 8

Construction 2

Let p be a prime number of the form 4n+ 1.


t = p, b = 2p, r = p− 1, k = (p− 1)/2 and λ = (p− 3)/2,

take the set of non-zero squares (modulo p), and the set of non-squares (modulo p) and use thecyclic construction to obtain p blocks from each set.

Example of Construction 2

Consider p = 5. The non-zero squares (modulo 5) are 1 and 4, the non-squares (modulo 5)are 2 and 3. Using these sets as two blocks and employing a cyclic construction we obtain thebalanced design

Blocks from Blocks fromnon-zero-squares non-squares1 2 3 4 0 2 3 4 0 14 0 1 2 3 3 4 0 1 2

Construction 3



t = p+ 1, b = 2p, r = p, k = (p+ 1)/2 and λ = (p− 1)/2,

take the set of squares (modulo p), including 0, and the set of non-zero squares (modulo p),together with an extra treatment z, and use the cyclic construction to obtain p blocks from eachset.


20

Consider p = 5. The squares (modulo 5) are 0,1 and 4, the non-zero squares (modulo 5) are 1and 4. The following design is obtained

Blocks from Blocks fromsquares non-squares and z

0 1 2 3 4 1 2 3 4 01 2 3 4 0 4 0 1 2 34 0 1 2 3 z z z z z

Construction 4



t = p+ 1, b = 2p, r = p, k = (p+ 1)/2 and λ = (p− 1)/2,

take the set of non-zero squares (modulo p),together with an extra treatment z, and the set ofnon-squares (modulo p), together with 0, and use the cyclic construction to obtain p blocks fromeach set.


Consider p = 7. The non-zero-squares (modulo 7) are 1, 2 and 4, the non-squares (modulo7) are 3, 5 and 6. Using sets 1, 2, 4, z and 3, 5, 6, 0 as two blocks and employing a cyclicconstruction we obtain the balanced design

Blocks from Blocks fromnon-zero-squares and z non-squares and 01 2 3 4 5 6 0 0 1 2 3 4 5 62 3 4 5 6 0 1 3 4 5 6 0 1 24 5 6 0 1 2 3 5 6 0 1 2 3 4z z z z z z z 6 0 1 2 3 4 5

Construction Using a Complete Set of Orthogonal Latin Squares

Theorem 2.5. The total number of mutually orthogonal Latin squares of order r is at mostr − 1. However, for some values of r the total number of mutually orthogonal Latin squares isless than r − 1.

Corollary 2.1. If p is a prime number or the power of a prime number then there exists acomplete set of p− 1 mutually orthogonal Latin squares of order p.

From Corollary 2.1 there is a complete set of mutually orthogonal Latin squares of order pwhenever p is a prime number or the power of a prime number, in particular whenever p =2, 3, 4, 5, 7, 8 or 9. This complete set is used to construct two series of balanced incompleteblocks.

SERIES (i): t = p2, k = p, r = p+ 1, b = p(p+ 1) and λ = 1.

21

Write out the p2 treatments in a p × p array and also provide a complete set of p − 1 mutuallyorthogonal latin squares of order p.

We obtain b = p2 + p blocks of size k = p as follows:

2p blocks from the p× p array of treatments:

A set of p blocks of size p is obtained as the p rows of the array; a second set of p blocks isobtained as the p columns of the array;

p blocks from each of the p− 1 Latin squares:

Superimpose one Latin square on the p × p array of treatments; a set of p blocks of size p isobtained if all treatments common to a particular label in the square are placed in a block.

Example Using Latin Squares of Order 3The 3× 3 array and two mutually orthogonal Latin squares of order 3 are:

Array Orthogonal Latin Squares

A B CD E FG H I

S1 ≡0 1 21 2 02 0 1

and S2 ≡0 1 22 0 11 2 0

so we get four sets of three blocks of size k = 3 for the following design B3:

Blocks1 2 3 4 5 6 7 8 9 10 11 12A D G A B C A B C A B CB E H D E F F D E E F DC F I G H I H I G I G H

It can be checked that this design is a balanced incomplete block design with the parameterst = 9, b = 12, k = 3, r = 4 and λ = 1.

Remark on Construction using MOLSThis technique gives a rather limited set of balanced incomplete block designs but it is useful insuggesting designs that would not be found by other means. The same applies to a continuationof the method to give further balanced incomplete block designs, which we consider next.

SERIES (ii): t = p2 + p+ 1, k = p+ 1, r = p+ 1, b = p2 + p+ 1 and λ = 1.

p2 + p blocks from the method of Series (i):

Let p(p + 1) blocks of a design for t = p2 treatments be arranged in p + 1 sets of p blocks ineach set, as described in Series (i). Now let a new treatment be added to all of the blocks in aparticular set in such a way that the treatment added is different for each set.

1 further block:

Another block is obtained by putting all of the p+ 1 new treatments together in a block.

Example Using Latin Squares of Order 3 (continuation)

22

Adding the new treatments J,K, L,M to the previous example we get thirteen blocks of sizek = 4 for the following design B4:

Blocks1 2 3 4 5 6 7 8 9 10 11 12 13A D G A B C A B C A B C JB E H D E F F D E E F D KC F I G H I H I G I G H LJ J J K K K L L L M M M M

It can be checked that this design is a balanced incomplete block design with the parameterst = b = 13, k = r = 4 and λ = 1. Note that this is a useful design; by contrast, the unreducedbalanced incomplete block has

(134

)= 715 blocks instead of the 13 blocks specified here.

2.2.2 Analysis of the Balanced Incomplete Block Design

Let yij denote the observation corresponding to the ith treatment in the jth block (i = 1, · · · , t; j =1, · · · , b). These data are independent and normally distributed with expected values E[yij] =µ+ τi + βj and variance σ2 such that∑t

i=1 τi =∑b

j=1 βj = 0.

Theorem 2.6. The maximum likelihood estimator of τi is given by

τi = 1λtT ∗i = 1

λt

{kTi −

∑bj=1 nijBj

}(i = 1, · · · , t),

where Ti = the ith treatment total and Bj = the jth block total.

Theorem 2.7. The treatment sum of squares, adjusted for blocks, is given by

SSttments(adj) =λt

k

t∑i=1

τ 2i =1

kλt

t∑i=1

(T ∗i )2.

Further ComputatationsThe sum of squares due to blocks, unadjusted for treatments, and the Total sum of squaresrequire the mean correction M.C., obtained in the usual way.Then

SSTotal =t∑i=1

b∑j=1

y2(ij) −M.C.,

and

SSblocks(unadj) =1

k

b∑j=1

B2j −M.C.

23


The Analysis of Variance table for the balanced incomplete block design is:

Source DF SS MS E[MS]Treatments t− 1 SSttments(adj) SSttments(adj)/(t− 1) σ2 + λt

k(t−1)∑t

1 τ2i

Blocks b− 1 SSblocks(unadj)

Residual tr − t− b+ 1 SSResid SSResid/(tr − t− b+ 1) σ2

Total tr − 1 SSTotal

Example: BIB Design for Catalyst ExperimentFour catalysts A, B, C and D are being investigated. The purpose of the investigation is todetermine if the percentage yield for a chemical process is a function of the type of catalystused.

Each batch of raw material is only large enough to permit trials using three of the four catalyststo be run.

Variations in the batches of raw material may affect the performance of the catalysts so a Bal-anced Incomplete Block Design is used with batches as blocks.

The percentage yields data are:

Block(batch of raw material)1 2 3 4

A 73 A 74 B 67 A 71C 73 B 75 C 68 B 72D 75 C 75 D 72 D 75221 224 207 218

For this design λ = 2, t = 4, k = 3.

M.C.= 63075. The treatment totals, adjusted for blocks, are

Ti ΣnijBj T ∗i τiA 218 663 -9 -1.25B 214 649 -7 -0.875C 216 652 -4 -0.5D 222 646 20 2.5

The total sum of squares is

SSTotal = 63156−M.C. = 81;

The unadjusted block sum of squares is

SSblocks(unadj) = 63130−M.C. = 55;

24

The treatment sum of squares, adjusted for blocks, is given by

SSttments(adj) =1

kλt

t∑i=1

(T ∗i )2 = 22.75.

Then the ANOVA table is

Source DF SS MS F PTreatments (adjusted 3 22.75 7.58 11.66 0.0107

for blocks)Blocks (unadjusted) 3 55

Residual 5 3.25 0.65Total 11 81

The conclusion is that the catalyst used does affect the percentage yield.

To obtain more information about the differences between the catalysts we can decomposeSSttments(adj) into 3 separate component sums of squares, such that each component has onedegree of freedom and each represents a comparison between treatment levels which may be ofinterest.

This is done using the method of orthogonal contrasts.

Orthogonal ContrastsA contrast in the treatment effects τ1, τ2, . . . , τt is a linear combination

∑ti=1 ciτi such that∑t

i=1 ci = 0.

Two contrasts∑t

i=1 c1iτi and∑t

i=1 c2iτi are orthogonal iff∑t

i=1 c1ic2i = 0.

For a Balanced Incomplete Block Design involving t treatments, SSttments(adj) can be split intot− 1 components, each with one degree of freedom and each pertaining to one of a set of t− 1mutually orthogonal contrasts.

We distinguish between cases where the Treatment levels are qualitative and where they arequantitative.

Qualitative Factor LevelsA Qualitative factor does not have its levels ordered on a numerical scale; its levels are justlabelled (1), (2), (3), ... or A, B, C, ... etc.

Examples are:

• Operators 1,2,3

• Machines X,Y,Z

• Methods ‘New’, ‘Old’, ‘Control’

In practice, a number of particular linear comparisons of qualitative factor levels will usuallysuggest themselves.

25

Example: Catalyst Experiment ContinuedIn this case the treatment factor has four levels. Catalysts A and B are the standard ones andhave a chemical element in common. C and D are new catalysts which have a different elementin common.

Thus it is sensible to compare:

A versus B; C versus D; A and B versus C and D.

There are exactly 3 independent linear comparisons, each with one degree of freedom.

The comparisons can be represented by the linear contrast vectors

• c1 = (1, −1, 0, 0)T

• c2 = (0, 0, 1, −1)T

• c3 = (1, 1, −1, −1)T

Note that these vectors are orthogonal.

Theorem 2.8.∑t

i=1 ciτi is an unbiased estimator of the linear contrast∑t

i=1 ciτi, where

t∑i=1

ciτi =1

λt

t∑i=1

ciT∗i , and Var

[t∑i=1

ciτi

]=kσ2

λt

t∑i=1

c2i .

Theorem 2.9. If∑t

i=1 ci1τi, · · · ,∑t

i=1 ci,t−1τi are t− 1 mutually orthogonal contrasts,

SSttments(adj) = q1 + · · ·+ qt−1,

where

qj =λt

k

(∑t

i=1 cij τi)2∑t

i=1 c2ij

(1 ≤ j ≤ t− 1).

An alternative way of expressing qj in terms of T ∗i is:

qj =1

kλt

(∑t

i=1 cijT∗i )2∑t

i=1 c2ij

.

Now suppose that the residual mean square σ2 has r degrees of freedom and is independent ofSSttments(adj). We have the following result:

Theorem 2.10. The statistics q1, . . . , qt−1 are mutually independent random variables and areindependent of σ2. Furthermore, if E[

∑ti=1 cijτi ] = 0, i.e. the expected value of the jth

treatment comparison is zero, then qj/σ2 has the F-distribution with 1,r degrees of freedom.

26

Example: Catalyst Experiment ContinuedHere the treatment factor ‘catalyst’ with levels A, B, C and D is clearly a qualitative factor.

We can use Theorems 2.8 and 2.9 to quantify the three comparisons outlined and to split up theSSttments(adj) into three parts, one pertaining to each comparison.

The significance of these parts can then be assessed using Theorem 2.10.

Comparison 1: A versus B

c1 = (1, −1, 0, 0)T

Thus

q1 =1

24

(T ∗1 − T ∗2 )2

2= 0.0833.

q1, can be used to test Ho : τA = τB against the alternative H1 : τA 6= τB.

The test statistic is q1s2

, where s2 is the estimate of σ2. From the ANOVA table produced earlier,we get s2 = 0.65, on 5 degrees of freedom.

Under H0, the test statistic is a random observation from F (1, 5). The value of the test statisticis 0.128.

To obtain the significance probability use the command ‘1-pf(0.128,1,5)’ in R. This gives theprobability 0.735 which is much larger than 0.05.

Thus we accept H0 and conclude that there is no evidence of difference between catalysts Aand B.

Comparison 2: C versus D

c2 = (0, 0, 1, −1)T

q2 =1

24

(T ∗3 − T ∗4 )2

2= 12.

q2, can be used to test Ho : τC = τD against the alternative H1 : τC 6= τD.


= 18.462.UsingR the significance probability is ‘1-pf(18.462,1,5)’. Thisgives a significance probability of 0.00774 which is much less than 0.05.

Thus we reject Ho in favour of H1 and conclude that catalysts C and D give rise to significantlydifferent percentage yields times.

Comparison 3: A and B versus C and D

c3 = (1, 1, −1, −1)T

q3 =1

24

(T ∗1 + T ∗2 − T ∗3 − T ∗4 )2

4= 10.6667.

q3, can be used to test Ho : τA + τB = τC + τD against the alternative H1 : τA + τB 6= τC + τD.

27


= 16.410. The significance probability is ‘1-pf(16.41,1,5)’. This gives asignificance probability of 0.00982 which is much less than 0.05.

Thus we reject Ho in favour of H1 and conclude that catalysts A, B give rise to significantlydifferent reaction times compared to C, D.

Quantitative Factor LevelsA Quantitative factor has its t levels specified in numerical terms. Examples are:

• ‘Temperature’ in degrees centigrade with observations taken at 30oC, 40oC, 50oC

• ‘Time’ specified at three-hourly intervals

• ‘pH value’ specified at different levels of acidity

There is a clearly defined way of ordering the levels of a quantitative factor. It will be assumedfor the sake of simplicity that the t levels are ordered in arithmetic progression, i.e. the intervalsbetween these levels are the same.

For a quantitative factor at t levels the treatment sums of squares can be separated into up tot− 1 components representing linear, quadratic, cubic, . . . etc trend. This is used to identify anappropriate polynomial model.

The usual approach to determine the contrast and hence sum of squares for each trend com-ponent is to employ coefficients of orthogonal polynomials. These are mutually orthogonalcontrasts which pick out polynomial trends that could be present in the observations.

The contrast vectors for linear, quadratic, cubic, ... polynomials are denoted, as previously, byc1, c2, c3, ... etc.

Note, however, that the contrasts are different for different values of t.

ExampleSuppose the treatment factor has t = 5 levels and that these five levels occur in arithmeticprogression. Without loss of generality, we take these levels to be

{x1, x2, x3, x4, x5} = { − 2, − 1, 0, 1, 2}.

The polynomials p1(x) = a + bx, p2(x) = a + bx + cx2, p3(x) = a + bx + cx2 + dx3 andp4(x) = a+ bx+ cx2 + dx3 + ex4, which take values at these five levels, are represented by thefour contrast vectors

c1 =

a− 2ba− ba

a+ ba+ 2b

, c2 =

a− 2b+ 4ca− b+ c

aa+ b+ ca+ 2b+ 4c

, c3 =

a− 2b+ 4c− 8da− b+ c− d

aa+ b+ c+ d

a+ 2b+ 4c+ 8d

and

28

c4 =

a− 2b+ 4c− 8d+ 16ea− b+ c− d+ e

aa+ b+ c+ d+ e

a+ 2b+ 4c+ 8d+ 16e

.

To find c1: The coefficients of c1 must sum to zero, i.e. 1T5 c1 = 0 ⇒ 5a = 0 but b isarbitrary.Take b = 1 giving:

c1 = (−2 − 1 0 1 2)T .

To find c2: 1T5 c2 = 0 ⇒ 5a + 10c = 0. Also c2 is orthogonal to c1 so that cT1 c2 = 0 ⇒10b = 0.Choose c = 1⇒ a = −2 giving:

c2 = (2 − 1 − 2 − 1 2)T .

To find c3: 1T5 c3 = 0⇒ 5a+ 10c = 0. Furthermore:

to be orthogonal to c1 then cT1 c3 = 0⇒ 10b+ 34d = 0;

to be orthogonal to c2 then cT2 c3 = 0⇒ 14c = 0⇒ a = 0.

Put a = c = 0, b = −175d and choose d = 5

6giving:

c3 = (−1 2 0 − 2 1)T .

To find c4: A similar argument gives:

c4 = (1 − 4 6 − 4 1)T .

The following table gives the vectors c1, . . . , ct−1 of mutually orthogonal contrasts, which arecoefficients of orthogonal polynomials, for the t treatment levels which range in values fromt = 3 to t = 6. Theorems 2.9 and 2.10 apply as before.

(3) (4) (5) (6)-5 5 -5 1 -1

-2 2 -1 1 -3 -1 7 -3 5-3 1 -1 -1 -1 2 -4 -1 -4 4 2 -10

-1 1 -1 -1 3 0 -2 0 6 1 -4 -4 2 100 -2 1 -1 -3 1 -1 -2 -4 3 -1 -7 -3 -51 1 3 1 1 2 2 1 1 5 5 5 1 12 6 20 4 20 10 14 10 70 70 84 180 28 252

Example: BIB Design for Temperature Effect on Cleaning Process

29

A cleaning process uses enzymes. The purpose of an investigation is to determine if the timefor completion of the process is a function of temperature.

Four industrial machines are available in which the cleaning process is carried out. Four temper-atures are to be used for the experiment; 5◦C, 10◦C, 15◦C and 20◦C. It is desirable to completethe experiment in one day. There is only time to run the process three times in each machine.

Therefore a balanced incomplete block design is used with four blocks, each of size three.

The design and times for completion (in minutes) are as follows:

Block(Industrial Machine)1 2 3 4

Temp Time Temp Time Temp Time Temp Time5 57 5 112 5 95 10 62

10 62 10 37 15 87 15 3715 61 20 137 20 112 20 137

180 286 294 236

For this design λ = 2, t = 4, k = 3.

M.C.= 82668. The treatment totals, adjusted for blocks, are calculated as for the catalyst exper-iment to give:

T ∗1 = 32, T ∗2 = −219, T ∗3 = −155 and T ∗4 = 342.

The total sum of squares is

SSTotal = 96616−M.C. = 13948;

The unadjusted block sum of squares is

SSblocks(unadj) = 85443−M.C. = 2775;


SSttments(adj) =1

kλt

t∑i=1

(T ∗i )2 = 7916.


Source DF SS MS F PTreatments (adjusted 3 7916 2639 4.05 0.0831

for blocks)Blocks (unadjusted) 3 2775

Residual 5 3258 652Total 11 13948

30

From this analysis we cannot reject the null hypothesis that temperature has no effect on thereaction time. However, we can use the method of orthogonal contrasts to examine the data inmore detail.

The experiment involves a treatment factor with 4 levels in arithmetic progression. We considerthe 3 comparisons given by the linear, quadratic and cubic contributions represented by theorthogonal polynomials for t = 5 levels which are given in the table.

By using Theorem 2.9 as before, the four sums of squares associated with these comparisonsare q1 = 2058.4, q2 = 5828.2 and q3 = 29.

Linear Comparisonq1, can be used to test the null hypothesisHo : −3τ5−τ10+τ15+3τ20 = 0 against the alternativeH1 : −3τ5 − τ10 + τ15 + 3τ20 6= 0.


= 2058.4652

= 3.16. Using Theorem 2.10, under H0, the statistic q1s2

is arandom observation from an F distribution with 1 and 5 degrees of freedom. The significanceprobability is ‘1-pf(3.16,1,5)’.

This gives a significance probability of 0.1356 which is not significant at the 5% level. Thus weaccept Ho.

Quadratic Comparisonq2, can be used to test the null hypothesis Ho : τ5 − τ10 − τ15 + τ20 = 0 against the alternativeH1 : τ5 − τ10 − τ15 + τ20 6= 0.


= 8.94. The significance probability is 0.0304. Thus we reject Ho andconclude that there is a significant quadratic component.

Cubic Comparisonq3, can be used to test the null hypothesisHo : −τ5+3τ10−3τ15+τ20 = 0 against the alternativeH1 : −τ5 + 3τ10 − 3τ15 + τ20 6= 0.


= 0.04. The significance probability is 0.8493. Thus we accept Ho andconclude that there is no cubic component.

It therefore appears that for the Catalyst experiment the temperature effect can be modelled bya quadratic.

2.2.3 Choosing Balanced Incomplete Block Designs

When choosing a suitable balanced incomplete block design there are two additional pointswhich should be considered.

Point (1) Looking for a Resolvable Design:

In practice it is sometimes possible to perform experiments one complete replicate at a time. Inthis case the experiment has a built-in safety feature, i.e. if it becomes necessary to discontinuetesting then all treatments will have occurred equally often. So we need to look for a resolvablebalanced incomplete block design, although this only applies when the number of treatments isa multiple of the block size, i.e. t = mk, m an integer⇒ b = mr.

31

Definition of Resolution Class and Resolvable DesignA Resolution Class in a balanced incomplete block design is a set of blocks which togethercontain each treatment precisely once.

If all blocks of a balanced incomplete design can be partitioned into mutually disjoint resolutionclasses then the design is said to be Resolvable.

Example of Resolvable Balanced Incomplete BlocksThe earliest recorded example of a balanced incomplete block was probably that of the ‘school-girls problem’ posed by the Reverend T. P. Kirkman in 1850:

A schoolmistress has fifteen girl pupils and she wishes to take them on a daily walk. The girlsare to walk in five rows of three girls each, but no two girls should walk together (i.e. in thesame row as each other) more than once a week. Can this be done?

In other words, is there a resolvable balanced incomplete block with t = 15, k = 3 and r = 7?The answer is yes. One solution, which divides the 35 blocks into seven resolution classes offive blocks per class is as follows:

Replicate1 2 3 4

A B C A D E A F G A H ID H L B H J B I K B L OE J N C M O C L N C E FF K M F I N D J O D K NG I O G K L E H M G J M

Replicate5 6 7

A J K A L M A N OB M N B D F B E GC D G C I J C H KE I L E K O D I MF H O G H N F J L

Point (2) Modifying the Design Parameters:

Usually t and k are considered fixed but if these parameters can be amended then a more suitabledesign may emerge.

For example, if t = 8 and k = 5 the unreduced design has 56 blocks⇒ bk = 280 experimentalunits.

But:

if t→ 6 and k = 5, ∃ a design with b = 6 blocks⇒ bk = 30;

if t→ 9 and k = 5, ∃ a design with b = 18 blocks⇒ bk = 90;

32

if k → 4 and t = 8, ∃ a design with b = 14 blocks⇒ bk = 56.

For reference, the following table gives a list of balanced incomplete designs using not morethan 150 (= tr = bk) experimental units.

This table contains 42 balanced incomplete block designs with block sizes ranging from k = 2to k = 11 and with numbers of treatments ranging from t = 3 to t = 25.

Special designs are picked out by brief comments; these comments are:

U ≡ the balanced incomplete block design is unreduced;

R ≡ the balanced incomplete block design is Resolvable;

Y ≡ a Youden Square also exists.

k t b r bk Comment k t b r bk Comment2 3 3 2 6 U; Y 4 16 20 5 80 R2 4 6 3 12 U 5 6 6 5 30 U; Y2 5 10 4 20 U 5 9 18 10 902 6 15 5 30 U; R 5 10 18 9 902 t (t2) t− 1 t2 − t U 5 11 11 5 55 Y3 4 4 3 12 U; Y 5 21 21 5 105 Y3 5 10 6 30 U 5 25 30 6 150 R3 6 10 5 30 6 7 7 6 42 U; Y3 6 20 10 60 U; R 6 9 12 8 723 7 7 3 21 Y 6 10 15 9 903 9 12 4 36 R 6 11 11 6 66 Y3 10 30 9 90 6 16 16 6 96 Y3 13 26 6 78 6 16 24 9 1443 15 35 7 105 R 7 8 8 7 56 U; Y4 5 5 4 20 U; Y 7 15 15 7 105 Y4 6 15 10 60 U 8 9 9 8 72 U; Y4 7 7 4 28 Y 8 15 15 8 120 Y4 8 14 7 56 R 9 10 10 9 90 U; Y4 9 18 8 72 9 13 13 9 117 Y4 10 15 6 60 10 11 11 10 110 U; Y4 13 13 4 52 Y 11 12 12 11 132 U; Y

33

34

Chapter 3

Analysis of Covariance

IntroductionAnalysis of covariance is a technique that can be useful in improving the precision of an ex-periment.

A covariate is a quantitative variable whose value is thought to influence the response under theassigned treatment.

For example, if we have an experiment comparing the effect of different diets on pigs, theresponse of interest could be the amount of weight gained by a pig when fed on one of t possiblediets for a fixed period of time. However, variables such as the initial weight of the pig or theage of the pig, could also affect the response. Here ‘initial weight of pig’ and ‘age of the pig’are possible covariates.

For simplicity we shall consider only models with one treatment factor and one covariate.

Considering the weight gain in pigs example; with ‘weight gain’ as the response and t diets ofinterest, the one-way model is: yij = µ + τi + εij , for i = 1, 2, . . . , t and j = 1, 2, . . . , r. Notethat, for simplicity, we have assumed r observations for each diet.

If we now include the covariate ‘initial weight of pig’, then, assuming a linear relationshipbetween the response and the covariate, the model is: yij = µ+αi+βixij+εij for i = 1, 2, . . . , t; and j = 1, 2, . . . , r.

If the covariate ‘initial weight of pig’ does have an affect on the response and we fail to includeit in the model, then the covariate could cause the residual mean square to be inflated whichwould make true differences in the response due to the treatments (diets in this case) harder todetect. That is, we experience a loss of precision.

If we plot the data there are a number of possibilities. For convenience we shall illustrate withtwo levels of the treatment factor. (see Figure 3.1)

35

◦ ◦ ◦◦ ◦ ◦

◦ ◦?? ?

??? ?

?

separateregressions

◦ ◦ ◦◦ ◦ ◦

◦ ◦? ? ?? ?

? ? ?parallelregressions

◦ ◦ ◦◦ ◦◦ ◦ ◦

? ? ?? ? ?

? ?

coincidentregressions ◦ ◦ ◦

◦ ◦ ◦◦ ◦

?? ? ? ? ? ? ?

concurrentregressionsat x = 0

Figure 3.1: Illustration of possible regression lines

3.1 Separate regressions model

The full model, which fits a separate regression line to each of the t levels of the treatmentfactor, is given by

yij = µ+ αi + βixij + εij

for i = 1, . . . , t and j = 1, . . . , r. We let n = rt. Note that we are assuming an equireplicatedesign. We make the usual assumptions about the errors. The different models are illustratedin Figure 3.1. Parallel regressions correspond to β1 = · · · = βt. Coincident regressions haveα1 = · · · = αt and β1 = · · · = βt. Regressions concurrent at x = 0 have α1 = · · · = αt.For the full model we can estimate the parameters using least squares, which gives the max-imum likelihood estimates under normality. The method of least squares involves estimatingparameters so that the following sum of squares is minimised when the unknown parametersare replaced by the estimates:

S =t∑i=1

r∑j=1

(yij − µ− αi − βixij)2.

There are 2t + 1 parameters. The normal equations, obtained by differentiating S with respectto µ, αi and βi are: ∑

i

∑j

(yij − µ− αi − βixij) = 0 (3.1)∑j

(yij − µ− αi − βixij) = 0 (3.2)∑j

xij(yij − µ− αi − βixij) = 0 (3.3)

36

Now∑t

i=1(3.2) = (3.1) so there are only 2t independent equations. If we choose µ = 0 thenwe obtain the following 2t normal equations∑

j

(yij − αi − βixij) = 0∑j

xij(yij − αi − βixij) = 0

but for each value of i these are just the normal equations for a simple linear regression model.Thus the solutions are

αi = yi· − βixi· and βi =[Sxy]i[Sxx]i

.

Note that S is minimised by minimising separately within each level of treatment factor, that is:

Smin =t∑i=1

[Smin]i.

For the simple linear regression model it is known that E[[Smin]i] = (r − 2)σ2. Hence

E[Smin] = t(r − 2)σ2.

ThusSmin

t(r − 2)

is an unbiased estimate of σ2. The fitted values are given by

µij = yi· + βi(xij − xi·)

and the residuals byeij = yij − yi· − βi(xij − xi·).

To check the adequacy of the full model we can plot eij versus

• µij for homogeneity of variance or general lack of fit;

• xij for linearity in x;

• omitted covariates to see if they should be included;

• in a normal plot.

We should also do these checks separately for each level. If there is inadequacy in one groupthen this casts doubt on the full model.We can also plot eij versus i to check for homogeneity of variance between levels. A moreformal test of this can be done by finding s2i = [Smin]i/(r − 2) and testing the hypotheses thatthe variances for the different levels are the same by Bartlett’s test.

37

3.2 Parallel regressions model

We now consider a model which says that the β’s are all the same. Suppose the full model isadequate, then we want to test the hypothesis H0 : β1 = · · · = βt versus an alternative that atleast two are different. Thus we are asking if the model

yij = µ+ αi + βxij + εij

is adequate. We can find the normal equations as before by differentiating

S =∑∑

(yij − µ− αi − βxij)2

with respect to µ, αi and β. We obtain∑i

∑j

(yij − µ− αi − βxij) = 0 (3.4)∑j

(yij − µ− αi − βxij) = 0 (3.5)∑i

∑j

xij(yij − µ− αi − βxij) = 0 (3.6)

As before∑

(3.5) = (3.4) so there are t+ 1 independent equations in t+ 2 unknowns. Wetake µ = 0. We write the estimate of β as βw to distinguish it from the estimate of β in thecoincident regressions model. Solving the normal equations (3.5) and (3.6) we have∑

j

yij = rαi + βw∑j

xij (3.7)∑i

∑j

xijyij =∑i

αi∑j

xij + βw∑i

∑j

x2ij. (3.8)

Thus from (3.7)αi = yi· − βwxi·

and substituting for αi in (3.8) we have

βw

(∑i

∑j

x2ij − r∑i

x2i·

)=∑i

∑j

xijyij − r∑i

xi·yi·.

Hence

βw =

∑i[Sxy]i∑i[Sxx]i

.

Recall that for the full model

βi =[Sxy]i[Sxx]i

and therefore we can write

βw =

∑i βi[Sxx]i∑i[Sxx]i

.

38

The overall estimate of β is a weighted average of the separate estimates βi with weights [Sxx]iwhich are proportional to (var[βi])

−1.The fitted values are given by

µij = yi· + βw(xij − xi·)

and the residuals byyij − yi· − βw(xij − xi·).

Thus the residual sum of squares is

Smin =∑i

∑j

{yij − yi· − βw(xij − xi·)

}2

.

The associated degrees of freedom are the number of observations (n) minus the number ofindependent parameters fitted, i.e. rt− (t+ 1) = t(r− 1)− 1. We can rewrite the residual sumof squares as

Smin =∑i

∑j

(yij − yi·)2 + β2w

∑i

∑j

(xij − xi·)2 − 2βw∑i

[∑j

(yij − yi·)(xij − xi·)

]=

∑i

∑j

(yij − yi·)2 + β2w

∑i

[Sxx]i − 2βw∑i

[Sxy]i

=∑i

[Syy]i − β2w

∑i

[Sxx]i (3.9)

The residual sum of squares under the full model can be written as∑i

[Smin]i =∑i

{[Syy]i − β2i [Sxx]i}

on t(r − 2) degrees of freedom. The difference in these residual sums of squares is the extrasum of squares due to lack of parallelism which is given by∑

i

β2i [Sxx]i − β2

w

∑i

[Sxx]i.

The necessary F statistic is therefore{∑i

β2i [Sxx]i − β2

w

∑i

[Sxx]i

}/(t− 1)s2

which under H0 has an F distribution with t− 1 and t(r − 2) degrees of freedom.

3.3 Further simplifications of the modelSuppose the parallel regressions model is accepted. There are two possibilities for further sim-plification of the model. The first is to consider whether the covariate has any effect at all, i.e.

39

test the hypothesis that β = 0. The second is to consider whether there are any differences inthe treatments, i.e. test the hypothesis that α1 = · · · = αt.In considering whether the covariate has any effect, we are asking in effect whether the model

yij = µ+ αi + εij

is adequate. This is just a one-way model. Its residual sum of squares is given by

RSS(A) =∑i

∑j

(yij − yi·)2 =∑i

[Syy]i

with t(r−1) degrees of freedom. The residual sum of squares for the parallel regressions modelRSS(A+X) is given by (3.9) and hence the extra sum of squares due to including the covariatein the model, written SS(X|A), is

β2w

∑i

∑j

(xij − xi·)2

on one degree of freedom. It follows that the F statistic for testing the hypothesis that β = 0 is{β2w

∑i

∑j

(xij − xi·)2}/s2 (3.10)

which has an F distribution with 1 and t(r − 2) degrees of freedom under the hypothesis. Notethat this test is conditional on the treatments being different. If we accept that the covariate hasno effect then we may go on to test if the treatments have no effect by comparing SS(A) withs2.We now consider whether the treatments are different, conditional on the covariate having aneffect. We are asking if the model

yij = α + βxij + εij

is adequate, i.e. whether one overall regression equation is an adequate representation of thedata. This is the model of coincident regressions in Figure 3.1. The residual sum of squares is

RSS(X) = Syy − β2Sxx

where Syy =∑

i

∑j(yij − y··)2, Sxx =

∑i

∑j(xij − x..)2 and β = Syy/Sxx.

Note thatSyy =

∑i

[Syy]i + r∑i

(yi· − y··)2.

It follows that the sum of squares due to the hypothesis that the α’s are all the same is

SS(A|X) = SS(A+X)− SS(X)

= RSS(X)−RSS(A+X)

= r∑i

(yi· − y··)2 − β2Sxx + β2w

∑i

[Sxx]i

40

on t− 1 degrees of freedom and the F statistic for testing the hypothesis is{r∑i

(yi· − y··)2 − β2Sxx + β2w

∑i

[Sxx]i

}/(t− 1)s2

which has an F distribution with t − 1 and t(r − 2) degrees of freedom under the hypothesis.Again if we do accept that the treatments are the same then we can now test the hypothesis thatthe covariate has no effect by comparing SS(X) with s2.In general it does matter in which order we do the two tests outlined above, but there are somecircumstances in which SS(A|X) = SS(A) and also SS(X) = SS(X|A). In this case we sayA and X are orthogonal.

3.4 ANCOVA TableWe can summarise the results of 3.1, 3.2 and 3.3 in an analysis of covariance table as illustratedbelow:

Source DF SS

A (between groups) t− 1 r∑

i(yi. − y..)2

X after A (X|A) 1 β2w

∑i[Sxx]i

X 1 β2Sxx

A after X (A|X) t− 1 r∑

i(yi· − y··)2 − β2Sxx + β2w

∑i[Sxx]i

A+X t β2w

∑i[Sxx]i + r

∑i(yi. − y..)2

Parallelism (A.X|A+X) t− 1∑

i β2i [Sxx]i − β2

w

∑i[Sxx]i

Full model (A+X + A.X) 2t− 1∑

i β2i [Sxx]i + r

∑i(yi. − y..)2

Residual t(r − 2)∑

i[Syy]i −∑

i β2i [Sxx]i

Total rt− 1 Syy =∑

i[Syy]i + r∑

i(yi. − y..)2

The mean squares and F statistics can be calculated in the usual way.

41

42

Chapter 4

Specialist Incomplete Models

4.1 Latin Square Design

The Experimental Definition of a Latin SquareThe Latin Square Design is an incomplete three-way classification with special features:

1. the model will involve three factors, each at t levels;

2. each level of each factor will be observed t times;

3. the t2 observations are arranged in a t × t square, with two of the factors representing‘rows’ and ‘columns’, such that each level of the third factor appears once in each rowand once in each column;

4. the model is additive, i.e. no interactions exist between the factors.

Remarks on the Experimental Definition of a Latin SquareNote that if condition (1) is satisfied in a complete three-way classification then we would expecta total of t3 observations. But in a Latin Square only t2 different factor level combinations areused − a very substantial saving of resources. This is the main advantage of the Latin square:we reduce the experimental resources by a multiple of 1

twithout losing the ability to evaluate

separately the effects of any potential changes in level of any of the three factors.

The second advantage, which is derived from conditions (2) and (3), is that the t2 observationsform a genuine square, i.e. they lie in a plane. This reflects the historical fact that Latin squaresfirst found extensive practical use in the 1930s in agricultural experiments concerned with theplanting of crops. Here two of the factors simply defined position in a two-dimensional coordi-nate system representing location on a field. The third factor would be ‘treatments’ at t levels.The levels of this third factor were indicated by letters (Latin letters in fact, hence giving thename ‘Latin Square’).

It is usual for the two factors ‘rows’ and ‘columns’ in the square to be two kinds of blocks.However, in some more recent industrial applications it does happen that two or even three

43

factors are of interest in themselves and not introduced merely to model effects due to a two-dimensional system of blocking. Care is needed in such cases, however. If real interactions arepresent the effect is to swell the size of σ2, which means a loss of sensitivity.

The special feature of the Latin Square design, which is implied by condition (2), is that allthree factors (‘rows’, ‘columns’, ‘treatments’) must appear at the same number t of levels.This inflexibility may create practical difficulties or may otherwise limit the possibilities of theexperimental objectives.

Example: Experiment Requiring a Latin Square DesignConsider a biological experiment where five different types of drug (Factor ‘treatments’ at fivelevels), sayA,B,C,D andE are to be applied to specimen cultures in a laboratory. Twenty-fivespecimens are available but, for practical reasons, only five specimens can be treated on any oneday.

It is suspected, however, that there may be systematic variation between observations obtainedon different days. To eliminate the day-to-day effect we consider each group of 5 specimenstested on the same day as one block. We therefore assign the specimens at random to each day,and, on each day, we subject one specimen to each treatment A,B,C,D and E.

It is possible however that the time of day also affects the results. To remove the time-of-dayeffect we need to arrange the testing programme so that each treatment occurs not only onceevery day but also once at a particular time of day. Denoting the five parts of a working day byt1, t2, t3, t4 and t5 we need to organise the test programme so that each treatment occurs onceand only once in each row and column, as shown in the following arrangement:

Time Mon Tue Wed Thu Frit1 A B C D Et2 B C D E At3 C D E A Bt4 D E A B Ct5 E A B C D

This is just one of many 5×5 Latin Square arrangements that we could choose. Each design hasthe property that each row and each column is a complete replicate set of the five treatment lev-els. The experiment is worthwhile if differences between rows and between columns representmajor sources of systematic variation.

Remark on the ExampleIn the same way it follows that a t × t Latin Square is found by “moving each row along oneplace to the left”. However, it is required practice to choose the square and allocate treatmentsby the process of randomisation.

Analysis of the Latin Square DesignLet y(ijk) denote the observation corresponding to the ith row, jth column and kth letter. Obvi-ously i, j and k each take values from 1 to t, but the subscripts are placed in brackets becauseonly two are needed to identify the observation.

44

These data are assumed independent and normally distributed with a common variance σ2 andwith the expected values

E[y(ijk)] = µ+ ρi + γj + τk,

satisfying the constraints∑t

i=1 ρi =∑t

j=1 γj =∑t

k=1 τk = 0.

The maximum likelihood estimators are µ = y..., the overall mean, and

µ+ ρi = yi.., µ+ γj = y.j. and µ+ τk = y..k,

the row, column and treatment means. The ANOVA table is:

Source DF SS MS E[MS]Rows t− 1 SSR SSR/(t− 1) σ2 + t

t−1∑t

i=1 ρ2i

Columns t− 1 SSC SSC/(t− 1) σ2 + tt−1∑t

j=1 γ2j

Treatments t− 1 SST SST/(t− 1) σ2 + tt−1∑t

k=1 τ2k

Residual (t− 1)(t− 2) SSResidSSResid

(t−1)(t−2)Total t2 − 1 SSTotal

Practical ComputationsThese sums of squares are computed using the mean correction M.C. = t2y2.... The Row,Column and Treatment sums of squares are

SSR =1

t

t∑i=1

y2i.. −M.C., SSC =1

t

t∑j=1

y2.j. −M.C., SST =1

t

t∑j=1

y2..k −M.C.

and the Total sum of squares SSTotal =t∑i=1

t∑j=1

t∑k=1

y2(ijk) −M.C..


4.2 Youden Squares

A Youden Square can be regarded as a balanced incomplete block with the additional require-ment that the design is balanced with respect to a second form of blocking. Thus if we consider“columns” as blocks, then we need to add one further condition for a Youden Square:

(1) every column contains k experimental units;

(2) there are more treatments t than the column-size k, i.e. k < t;

(3) every treatment appears in r columns;

(4) each pair of treatments occurs together in a column λ times.

(5) each of the k rows contains each of the t treatments precisely once.

45

It follows that a Youden Square is a particular kind of incomplete Latin square which isformed by deleting some of the rows of the Latin square. Thus we require b = t and r = k ⇒λ = k − 1, i.e. a Youden Square is formed from a balanced incomplete block and is specifiedby just two parameters t and k.

Examples of Youden SquaresIf ‘row’ was considered as a second blocking factor then designs B1 and Bc

1 in Section 2.2would be Youden Squares.

Existence of Youden SquaresAn examination of the table of balanced incomplete blocks in Section 2.2 shows that YoudenSquares divide into two kinds: (1) unreduced designs consisting of a Latin square with one rowremoved, and (2) certain Latin squares with several rows removed. For type (1) designs we havethat

Theorem 4.1. Every latin square with one row removed gives a Youden square which is anunreduced balanced incomplete block with b = t, k = r = t− 1 and λ = t− 2.

Remark on Theorem 4.1Note that b− 2r + λ = 0 so from Theorem 2.3 none of the designs of type (1) have a comple-mentary balanced incomplete block.

The practical value of Youden squares comes when t is large. For example, if t = 13 it wouldbe unlikely that a latin square of order 13 would be contemplated because it would requiret2 = 169 experimental units. But a Youden square of type (1) would still require t(t− 1) = 156experimental units! In these cases, Youden squares of type (2) are more useful. (In fact designB4 has t = 13 and k = 4 so altogether there are 52 experimental units.)

Analysis of Youden Square DesignThe model for the Youden Square design is the same as that for a latin square in Section 4.1,except that the parameters γj range from 1 to k instead of 1 to t. The analysis of varianceis the same as for a balanced incomplete block plus a term SSrows which is calculated in astraightforward way. To illustrate:

46

Example: Treatments on car tyresAn experiment has been performed to compare the effects of five different rubber treatments oncross-ply motor tyres. Five cars were available for the experiment and it was felt that positionsof the car wheels could affect the results. A suitable measure of tyre-wear was made, and thedata were

cars 1 2 3 4 5 Totals Sumsqsfront n/s C 31.1 A 11.7 B 39.6 E 20.5 D 16.2 119.1 3354.95front o/s A 43.4 D 13.3 E 17.5 C 72.1 B 58.8 205.1 11022.55rear n/s B 36.2 E 16.2 A 22.4 D 14.5 C 61.4 150.7 6054.85rear o/s E 33.5 C 26.1 D 25.3 B 78.2 A 42.5 205.6 10365.04Totals 144.2 67.3 104.8 185.3 178.9 680.5 30797.39

The treatment totals, adjusted for blocks, and the sums of squares are

Ti ΣnijBj T ∗i τiA 120.0 495.2 -15.2 -1.01 SS Total = 30797.39−M.C. = 7643.4;B 212.8 613.2 238.0 15.87C 190.7 575.7 187.1 12.47 SScars(unadj) = 2507.8 (usual way);D 69.3 536.3 -259.1 -17.27E 87.7 501.6 -150.8 -10.05 SSwheels = 3029.2 (usual way).

Sum 680.5 2722.0 0.0 0.0


SSttments(adj) =1

kλt

t∑i=1

(T ∗i )2 = 3029.2.


Source DF SS MS F PTreatments 4 3029.2 757.3 5.98 0.016

cars 4 2507.8wheels 3 1092.5 364.2 2.87 0.103

Residual 8 1013.8 126.7Total 19 7643.4

Comparison between the τ ’s will suggest the differences between treatments, depending on thefour orthogonal contrasts which are deemed to be of special interest in this example.

47

4.3 Cross-Over Designs

Reasons for Cross-Over Designs

Up to this point we have arranged that each experimental unit has resulted in a single obser-vation. Such an arrangement usually makes good sense in order to assure that the observationsare independent random variables and therefore maximize the information in the experimentfor the subsequent analysis. However there are circumstances where it is sensible to adopt adifferent arrangement with the following two features:

1. the experiment is performed over several time periods and an observation is made at theend of each period;

2. each experimental unit, hereafter called a subject, receives all treatments such that a dif-ferent treatment is applied for each period.

In general, therefore, there are s subjects receiving t treatments over t periods of time, resultingin st observations altogether. This arrangement makes sense if it is suspected that the differencesbetween subjects are likely to be greater than the differences between treatments, i.e. subjectsneed to be considered as blocks for the sake of precision. It is also customary for the periods tobe considered as blocks, i.e. we have a two-way blocking arrangement as occurs with a Latinor Youden square.

Example: Milk Yield for Cows [Williams, 1949]The classic example of a cross-over design is an experiment to determine diet treatments whichaffect the milk yield of cows. There are t = 6 diet treatments, that are labelled 0, 1, 2, 3, 4 and5, and s = 6 subjects (cows) in the design which is arranged over 6 periods as the followingLatin square:

SUBJECTS1 2 3 4 5 6

1 0 1 2 3 4 52 1 2 3 4 5 0

PERIODS 3 5 0 1 2 3 44 2 3 4 5 0 15 4 5 0 1 2 36 3 4 5 0 1 2

There are 36 observations from the experiment, each of which is the milk yield for one ofthe cows in one of the periods. Milk yield differs markedly from cow to cow, i.e. cows arenot homogeneous material, so the Williams experiment should give more precise estimates oftreatment differences than a similar (Latin square) experiment using 36 different cows. This isthe principal reason for choosing this cross-over design.However, the Williams experiment has the useful practical feature that only six cows are neededin the experiment, thus freeing up other cows for normal farming purposes and therefore makinghighly efficient use of the available resources. Of course, the practical disadvantage is that the

48

experiment must take a substantial amount of time to complete. In this case the diets wereapplied for a week and the cows were rested for a week so the experiment took eleven weeks intotal.

Carry Over EffectsThe other feature of the cross over design is the lack of dependence within each subject fromone period to the next. This dependence is included in the model by assuming that the effect ofeach treatment persists into the next period to create a carry over effect. Thus for each subjectthere are two treatments which are applied in each period except the first, i.e. in periods 2 to t,which are:

1. the direct treatment effect which measures the usual treatment applied for that subject inthat period;

2. the carry over treatment effect which measures the treatment carried over for that subjectfrom the previous period.

Only the direct treatment effect applies in the first period.

Thus the number of parameters in the model are: t direct effects, t carry over effects (i.e. 2ttreatment effects), s subject effects, t period effects (i.e. s + t block effects). With the mean µthis gives a total of s+ 3t+ 1 parameters in the model. However, it is necessary to impose twofurther restrictions before we obtain a cross over design suitable for experimentation.

Uniformly Balanced DesignsDefinition A cross over design is said to be uniform if each treatment occurs equally often ina period.

Remark The construction of a cross over design ensures that each treatment occurs once andonce only for every subject. A uniform design ensures that we get the same symmetry for everyperiod. It is clear that a cross over design is uniform if and only if s is an integer multiple of t.

Example The Williams design is uniform because s = t and it can be seen by inspection thateach treatment occurs exactly once in each period.

Definition A cross over design is said to be balanced for carry over effects if, within thesubjects, each treatment precedes every other treatment the same number of times.

Example The Williams design is balanced because it can be seen by inspection that eachtreatment precedes every other treatment within the subjects the same number of times, i.e.exactly once.

Definition A cross over design is said to be uniformly balanced if the design is both uniformand balanced.

4.3.1 Column-Complete Latin SquareIn this section we explore in more detail the properties of a uniformly balanced design and de-scribe a method for obtaining such a design in a practical situation. We begin with the definitionof a column-complete Latin square.

49

Definition A Latin square is said to be column-complete if every ordered pair of distincttreatments appears in adjacent positions precisely once in the columns of the square.

Remark By using a counting argument it can be seen that this is the same as the definition ofbalance for carry over effects in the particular case where s = t. So if the number of subjectsis equal to the number of treatments (which is equal to the number of periods) then we get auniformly balanced design if we can find a column-complete Latin square of order t.

Theorem 4.2. A column-complete Latin square of order 2m can be found for every integerm ≥ 1.

Construction of Column-Complete Latin Squares

Let there be t = 2m treatments labelled 0, 1, . . . , t − 1 = 2m − 1 and let the first column ofthe square have treatments arranged in the order

0 1 2m− 1 2 2m− 2 3 . . . m− 1 m+ 1 m.

Form the second column by adding unity to each treatment (mod 2m), form the third columnby adding unity to each treatment in the second column (mod 2m) and continue in this way byadding unity to each column until t columns are formed. It is clear that this gives a Latin square.It can also be shown that every ordered pair of distinct treatments appears in adjacent positionsprecisely once in the columns of the square.

The Williams Latin square of order 6 is column-complete and was constructed by this method.There is another column- complete Latin square of order 6 and this is given below:

SUBJECTS1 2 3 4 5 6

1 0 1 2 3 4 52 2 3 4 5 0 1

PERIODS 3 1 2 3 4 5 04 4 5 0 1 2 35 5 0 1 2 3 46 3 4 5 0 1 2

Remark The above construction shows that a column-complete Latin square of order t canalways be formed if t is an even integer. However, the position is rather more complicated if tis an odd integer.

Theorem 4.3. No column-complete Latin square of order 3, 5 or 7 exists.

Remark However, a column-complete Latin square of order 9 does exist and one example ofsuch a column-complete Latin square is given below:

50

SUBJECTS1 2 3 4 5 6 7 8 9

1 0 1 2 3 4 5 6 7 82 8 6 7 5 3 4 0 1 23 6 7 8 1 2 0 3 4 54 5 3 4 2 0 1 8 6 7

PERIODS 5 3 4 5 8 6 7 1 2 06 7 8 6 0 1 2 5 3 47 4 5 3 7 8 6 2 0 18 2 0 1 6 7 8 4 5 39 1 2 0 4 5 3 7 8 6

Remark To conclude this discussion it is necessary to consider the existence of a uniformlybalanced design for those cases where no column-complete Latin squares exist, i.e. cases wheret is odd, in particular t = 3, 5 or 7.

Theorem 4.4. A uniformly balanced design based on two Latin squares of order 2m + 1 canbe found for every integer m ≥ 1.

Construction of a Uniformly Balanced Design for 2m+ 1 TreatmentsLet there be t = 2m + 1 treatments labelled 0, 1, . . . , t − 1 = 2m and let the first column ofthe square have treatments arranged in the order

0 1 2m 2 2m− 1 3 . . . m− 1 m+ 2 m m+ 1.

Again form the second and subsequent columns by adding unity to each treatment in the previ-ous column (mod 2m + 1) until t columns are formed. It is clear that this gives a Latin square.It can be shown that

(2m+1

2

)ordered pairs of distinct treatments appears in adjacent positions

precisely twice in the columns of the square. Then just reverse the order of the columns and theremaining ordered pairs of distinct treatments appears in adjacent positions precisely twice inthe columns of the square. This gives a uniformly balanced design with s = 2t.

The following pair of Latin squares of order 5, which together give a uniformly balanced design,is constructed by this method:

SUBJECTS1 2 3 4 5 6 7 8 9 10

1 0 1 2 3 4 0 1 2 3 4PERIODS 2 1 2 3 4 0 4 0 1 2 3

3 4 0 1 2 3 1 2 3 4 04 2 3 4 0 1 3 4 0 1 25 3 4 0 1 2 2 3 4 0 1

Note that, in this example, columns 6, 7, 8, 9 and 10 are the reverse order of the columns3, 4, 5, 1 and 2 respectively.

51

4.3.2 Analysis of Cross Over ExperimentsCross over experiments are used extensively in different areas of experimental research, in-cluding agriculture, biology, psychology and medicine. The subjects used in these experimentsusually tend to be living entities, such as human beings involved in clinical trials or in psy-chological research, or different types of animal involved in agricultural or medical research oreven agricultural entities such as planted fields. The basic idea is that the subjects in a crossover experiment receive all t treatments in sequence over t different periods of time.One feature of experiments based on cross over designs is that there are two sets of treatmentsto consider. Normally the set of direct treatment effects are of primary interest, but the Analysisof Variance table should include both sets of treatments (direct and carry over) and the two setsof blocks (periods and subjects). The computation of the sums of squares is difficult to do byhand but it is a straightforward task for most modern computer packages.

Example: Milk Yield for CowsA design based on a column-complete Latin square of order 6 is used for an experiment toinvestigate the effect of six treatments on the milk yield of cows. The data are superimposed onthe Latin square design of order 6 from section 4.3.1.

SUBJECTS1 2 3 4 5 6

1 0 56.7 1 57.3 2 53.7 3 55.7 4 58.5 5 58.12 2 56.7 3 57.7 4 57.1 5 60.7 0 60.2 1 55.7

PERIODS 3 1 54.4 2 55.2 3 59.2 4 56.7 5 61.3 0 58.94 4 54.4 5 58.1 0 58.9 1 59.9 2 54.4 3 56.65 5 58.9 0 60.2 1 58.9 2 56.6 3 59.1 4 59.66 3 54.5 4 60.2 5 59.6 0 59.6 1 59.8 2 57.5

In the ANOVA table for this experiment no entry can be calculated for ‘Periods’. The usualestimate of this term involves the carry over effects, which are present in periods 2 to 6 but not inperiod 1, therefore no unbiased estimate of ‘periods’ exists. (We say: “periods are confoundedwith carry over effects”.) The other three terms can be estimated and the completed table is asfollows:

Anova Table for Milk Yield of Cowsdegrees

Source of Variation of sums of squares mean squares F-valuefreedom

subjects 5 39.8684 7.97 13.20periods (unadj.) 5 21.7114 – –direct effects 5 68.6581 13.73 22.74carry over effects 5 23.4363 4.69 7.76residual 15 9.0593 0.60total 35 165.0620

The sums of squares entries in the Anova table do not sum to the Total sums of squares, whichresults from the fact that the doubling up of types of treatments in periods 2 to 6 gives a non-orthogonal model. The entries in the Anova table suggest that direct treatment effects are highly

52

significant, i.e. there is a clear difference between the effects of the six treatments and it ispossible to investigate further why such differences exist. Also there are significant differencesbetween subjects, suggesting that it is sensible to treat separate cows as blocks.

(ii)∑t

i=1 ciτi is an unbiased estimator of the linear contrast∑t

i=1 ciτi, where

t∑i=1

ciτi =1

λt

t∑i=1

ciT∗i , and Var

[t∑i=1

ciτi

]=kσ2

λt

t∑i=1

c2i .

Remark on Theorem 2.6 (ii)This result implies that SSttments(adj), the treatment sum of squares adjusted for blocks, can bedivided into t − 1 components, each with a single degree of freedom, by using a set of t − 1mutually orthogonal contrasts.

53

54

Chapter 5

Factorial Experiments

5.1 Advantages of Factorial Experiments

Remarks on the Factorial Concept

Factorial experiments with k factors are formally the same as complete k−way classifications.In complete k−way classifications observations are taken at all treatment combinations of allfactors.

k = 2 : Examples of two−way classifications from three different experimental environmentsare as follows:

Medical Experiment: study the effects of a programmes of measured exercise (Factor Aat a levels) and b different types of varied diet (Factor B at b levels) on blood sugar levels indiabetics;

Chemical Experiment: study the effects of a different levels of local pressure (Factor A ata levels) and b different levels of temperature (Factor B at b levels) on viscosity of an adhesive;

Engineering Experiment: study the effects of a different levels of engine speed (Factor Aat a levels) and b separate categories of oil type (Factor B at b levels) on the life span of pistonrings.

These examples of a two−way classification model at a and b levels may be described in thiscontext as a factorial experiment with two factors, one at a levels and the other at b levels. Moreconcisely, each of the three examples could be described as an a× b factorial experiment.

Furthermore, if a = b in one of these examples then the experiment would be simply called ana2 factorial.

k = 3 : Another more general Chemical Experiment studies the effects of a levels of localpressure (Factor A at a levels), b levels of temperature (Factor B at b levels) and c levels ofconcentrations of an adhesive reagent (Factor C at c levels) on adhesive viscosity,

This is a three−way classification or, equivalently, a factorial experiment with three factors ata, b and c levels. More concisely, it is an a × b × c factorial experiment. If a = b = c this

55

experiment is an a3 factorial.

Agricultural Example of 23 Factorial:

A Fertiliser Trial has three fertiliser factors A, B and C denoting nitrogeneous fertiliser, phos-phate and potash which were applied to plots of corn. A simple experiment has each of thethree factors with only 2 levels, i.e. either absent or present, so that we are considering a 23

factorial experiment. There are 23 = 8 treatment combinations, represented by the followingtable:

TreatmentA B C Combination

absent absent absent (1)present absent absent aabsent present absent bpresent present absent ababsent absent present cpresent absent present acabsent present present bcpresent present present abc

All eight fertiliser combinations were replicated three times to produce a 3−replicate 23 facto-rial experiment, giving 24 observations ≡ 24 yields of corn.

If there is no interaction between A and the other factors it is possible to estimate the increasein yield due to factor A by using all 24 observations; it will be estimated by the difference ofthe mean of 12 observations with A absent from the mean of 12 observations with A present.We get the same precision as if all 24 observations had been used to test for A alone. Similarremarks apply to estimates of the effects of B and C. This is a big advantage of a factorialexperiment over the alternative approach which considers just one factor at a time.

The advantages are even more marked when interaction is actually present. The factorial ex-periment enables us to estimate all main effects and any two factor or three factor interactionby using a suitable linear contrast. If instead, separate experiments are carried out for factorsA, B and C then estimates for the main effects can be quite misleading and no estimates ofinteraction are obtained.

5.2 k Factors at Two Levels: Yates’Algorithm

We confine ourselves to factorial experiments with k factors, each at just two levels. It followsthat there are 2k treatment combinations altogether (which we can think of as 2k treatments) anda total of r2k observations, where r is the number of replications of each treatment combination.We begin by looking at some particular values of k.

56

5.2.1 The 22 Factorial Experiment

Consider two factors A, B, each at 2 levels high and low. The treatment combinations are

TreatmentA B Combination

low low (1)high low alow high bhigh high ab

We define the three effects to be

Effect of Factor A = (ab− b) + (a− (1)) = (a− 1)(b+ 1);Effect of Factor B = (ab− a) + (b− (1)) = (a+ 1)(b− 1);

Interaction AB = (ab− b)− (a− (1)) = (a− 1)(b− 1).

These three effects are mutually orthogonal contrasts with weights given by:

Factor A Factor B Interaction AB(1) -1 -1 +1a +1 -1 -1b -1 +1 -1ab +1 +1 +1

The contrast representing Interaction AB is the ordinary arithmetic product of contrasts repre-senting Factor A and Factor B, i.e. AB = A×B. Similarly A = B × AB and B = A× AB.

Example: An Agricultural 22 factorial experiment on Sugar Beet.

The two factors for this experiment are:

A ≡ Sulphate of Ammonia (high level = 3cwt/acre) (low level = none);B ≡ Depth of ploughing (high level = 11 in. deep) (low level = 7 in deep).

Five replicates of a 22 factorial are arranged in five blocks. Ploughing took place in late January,Sulphate of Ammonia was applied in late April and seed was sown early in May. The harvestedyields of sugar beet (cwt/acre) are:

blocks 1 2 3 4 5 Totals Sumsqs(1) 41.1 37.5 39.6 40.5 36.2 194.9 7614.31a 48.4 43.3 46.5 49.1 42.8 230.1 10622.35b 43.2 40.2 42.4 44.5 41.5 211.8 812982.54ab 48.1 48.1 52.3 54.2 48.5 251.2 12652.40

Totals 180.8 169.1 180.8 188.3 169.0 888.0 39871.60

Main Effect for Factor A = 251.2-211.8+230.1-194.9 = 74.6;Main Effect for Factor B = 251.2+211.8-230.1-194.9 = 38.0;Effect for Interaction AB = 251.2-211.8-230.1+194.9 = 4.2.

57

These three effects are orthogonal contrasts so these contributions will add up to SSTreatments. Itis seen that

74.62

20+ 382

20+ 4.22

20= 278.258 + 72.2 + 0.882 = 351.34 = SSTreatments;

and SSBlocks and SSTotal are 70.295 and 444.4 respectively. The ANOVA Table is

Source DF SS MS FA 1 278.258 278.258 146.67B 1 72.200 72.200 38.06AB 1 0.882 0.882 0.46

blocks 4 70.295 17.574 7 9.26Residual 12 22.765 1.897=s2

Total 19 444.400

Remark on the Example

It is evident that there is a marked response to the fertiliser and a smaller but still highly signif-icant response to the depth of ploughing. The interaction is not significant. This means that theresponse to Sulphate of Ammonia is the same whether ploughing is shallow or deep and that thedifference between the effects of deep or shallow ploughing is the same whether the fertiliser ispresent or not.

5.2.2 The 23 Factorial Experiment

In a 23 factorial experiment with factorsA, B andC, each at two levels, there are eight treatmentcombinations which we write here in Standard Order as

(1) a b ab c ac bc abc.

There are seven factorial effects which are defined as follows:

Effect Treatment CombinationA abc− bc+ ab− b+ ac− c+ a− (1) = (a− 1)(b+ 1)(c+ 1)B abc− ac+ ab− a+ bc− c+ b− (1) = (a+ 1)(b− 1)(c+ 1)AB abc− bc+ ab− b− ac+ c− a+ (1) = (a− 1)(b− 1)(c+ 1)C abc− ab+ ac− a+ bc− b+ c− (1) = (a+ 1)(b+ 1)(c− 1)AC abc− bc+ ac− c− ab+ b− a+ (1) = (a− 1)(b+ 1)(c− 1)BC abc− ac+ bc− c− ab+ a− b+ (1) = (a+ 1)(b− 1)(c− 1)ABC abc− ac− bc+ c− ab+ a+ b− (1) = (a− 1)(b− 1)(c− 1)

These seven effects are mutually orthogonal contrasts with weights given by:

58

A B AB C AC BC ABC(1) −1 −1 +1 −1 +1 +1 −1a +1 −1 −1 −1 −1 +1 +1b −1 +1 −1 −1 +1 −1 +1ab +1 +1 +1 −1 −1 −1 −1c −1 −1 +1 +1 −1 −1 +1ac +1 −1 −1 +1 +1 −1 −1bc −1 +1 −1 +1 −1 +1 −1abc +1 +1 +1 +1 +1 +1 +1

Note that any contrast representing an effect can be obtained by ordinary arithmetic multiplica-tion of two contrasts representing other effects modulo 2. Thus the product of any two columnsyields another column included in the table. We have, for example

AC = A× C ABC = AB × C A = B × AB = ABC ×BC and so on.

5.2.3 The 2k Factorial Experiment

For k factors at 2 levels there are 2k treatment combinations, giving r2k observations where ris the number of replicates of each treatment combination.

There are 2k − 1 mutually orthogonal contrasts which represent 2k − 1 factorial effects.

To associate these contrasts with the factorial effects we need to consider a combinatorial iden-tity. The combinatorial identity is given by

2k =k∑i=0

(ki

)which gives

(k1

)+

(k2

)+

(k3

)+ · · ·+

(kk

)= 2k − 1.

We associate each binomial coefficient with a class of factorial effects, i.e.

there are

(k1

)= k main effects;(

k2

)=

1

2k(k − 1) two− factor interactions;(

k3

)=

1

6k(k − 1)(k − 2) three− factor interactions;

· · ·(kk

)= 1 k−factor interaction.

(5.1)

Theorem 5.1. (Yates’ Algorithm)

59

The data corresponding to the 2k treatment combinations in standard order are entered in themargin as Column 0.

For a 2k factorial it is necessary to compute k columns (Columns 1,2, · · · k) by the followingmethod:

• Column i is obtained from Column (i − 1) by dividing the data of Column (i − 1) into2k−1 pairs;

• The top half of Column i is obtained by adding the pairs of the previous column;

• the bottom half of Column i is obtained by subtracting the first term in each pair from thesecond term.

The entries in Column k contain the grand total plus the 2k − 1 factorial effects. The grandtotal occupies the first position in Column k and each of the factorial effects is situated in thesame position in Column k as the corresponding treatment combination in Column 0.

N.B.: It is essential that the 2k treatment combinations are taken in standard order.vspace0.1in

Illustration of Yates’ Algorithm for 23 Factorial

Eight treatment totals corresponding to a 23 factorial experiment are written below in Column0. The application of Yates’ Algorithm then proceeds:

Column 0 Column 1 Column 2 Column 3(1) 425 851 3172 9331 Totala 426 2321 6159 333 Ab 1118 2679 86 2271 Bab 1203 3480 247 105 ABc 1283 1 1470 2987 Cac 1396 85 801 161 ACbc 1673 113 84 −669 BCabc 1807 134 21 −63 ABC

5.2.4 The 2k Factorial with a Single Replicate

If k is reasonably large then the number of treatment combinations in a 2k factorial experimentmay stretch the resources or the budgets of many experiments. There should be r2k observa-tions, where r is the number of replicates, however when it comes to deciding on the value of rit may be felt that it is inadvisable to allow r > 1. Then it is necessary to consider a 2k factorialexperiment with a single replicate., i.e. there are just 2k observations.

But if there are 2k observations then there are only 2k − 1 degrees of freedom. This raisesthe problem that the number of degrees of freedom is insufficient to estimate both the full set

60

of 2k − 1 factorial effects and the residual variance σ2. This problem is usually resolved bydeciding in advance what factorial effects can be considered negligible and can be ignored,thereby releasing this number of degrees of freedom for the residual variance.

Definition of Suppressed Factorial Effect

A factorial effect which is assumed to be zero will be said to be suppressed.

We consider two examples of suppressed factorial effects:

(1) It may be known that one factor is unlikely to interact with any of the others; then allsecond and higher order interactions involving this factor can be suppressed. In this case wehave

Number of factorial effects = k +∑k

i=2

(k − 1i

)= 2k−1;

(5.2)

Number of suppressed effects =∑k

i=1

(k − 1i

)= 2k−1 − 1.

(5.3)

(2) In many experiments it is reasonable to assume that all third and higher order interactionscan be suppressed. In this case we have

Number of factorial effects = k +

(k2

)=

(k + 1

2

);

Number of suppressed effects =k∑i=3

(ki

)= 2k −

(k + 1

2

)− 1.

(5.4)

Example: Analysis of a one-replicate 25 Factorial

A 25 factorial experiment was carried out to investigate the purification of a product by a steamdistillation process. The five factors, each at 2 levels, were concentration of material (A), rateof distillation (B), volume of solution (C), stirring rate (D) and solvent-to-water ratio (E). Bdoes not interact with A,C,D,E.

The residual acidity of material from using one replicate of each of the 32 treatment combina-tions was determined and given below in coded form.

61

A0 A1

D0 D1 D0 D1

E0 E1 E0 E1 E0 E1 E0 E1

C0 9 3 11 8 10 9 13 7B0

C1 3 5 7 7 5 6 10 7C0 8 4 9 8 6 6 16 6

B1

C1 6 4 7 5 10 10 13 6

Yates’ algorithm gave the following factorial effects:

Col 0 Col 1 Col 2 Col 3 Col 4 Col 5(1) 9 19 33 57 143 244 Totala 10 14 24 86 101 36 Ab 8 8 49 47 23 4 Bab 6 16 37 54 13 8 AB†

c 3 24 22 5 7 −22 Cac 5 25 25 18 −3 10 ACbc 6 17 29 15 7 18 BC†

· · · · · · · · · · · · · · · · · · · · · · · ·

† ≡ entries included in SSResidual. The ANOVA table is

Source DF SS MS F PA 1 40.500 40.500 11.57∗∗ < 0.01B 1 0.500 0.500 < 1C 1 15.125 15.125 4.32 < 0.1AC 1 3.125 3.125 < 1D 1 40.500 40.500 11.57∗∗ < 0.01AD 1 0.500 0.500 < 1CD 1 3.125 3.125 < 1ACD 1 0.125 0.125 < 1E 1 55.125 55.125 15.75∗∗ < 0.01AE 1 3.125 3.125 < 1CE 1 12.500 12.500 3.57 < 0.1ACE 1 0.500 0.500 < 1DE 1 15.125 15.125 4.32 < 0.1ADE 1 28.125 28.125 8.04∗ < 0.05CDE 1 0.500 0.500 < 1ACDE 1 4.500 4.500 1.29Residual 15 52.500 3.500

Total 31 275.500

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5.3 Fractional Replication

Consider again factorial experiments with k factors, each at only two levels. If k is large wemay not wish to consider even a single replicate of a 2k factorial if it is possible to suppressmany factorial effects. For example, a single replicate of a 28 factorial requires 256 treatmentcombinations. But only eight degrees of freedom correspond to main effects and only 28 to two-factor interactions. The remaining 219 degrees of freedom correspond to interactions involvingthree or more factors and, if higher order interactions are negligible and can be suppressed, then219 degrees of freedom are available for estimating σ2. This suggests that we consider using afraction of the replicate to estimate the 36 ‘nonsuppressed’ factorial effects.In general Fractional Factorial designs use only one-half, one-quarter or even smaller fractionsof the 2k treatment combinations. They are used for one or more of several reasons including

• the number of treatment combinations exceeds resources

• information is required only on main effects and low-order interactions

• screening studies are needed to identify a small number of important factors amongst alarge number of factors

• an assumption is made that only a few factors are important

5.3.1 Half-Replicate Fractional Factorial

Half-Replicate of a 23 Factorial

We begin by looking at a half-replicate of a 23 factorial. Of course, it is unlikely that a half-replicate of a 23 factorial would ever be needed in practice, nevertheless the discussion of thiscase is useful to introduce the main points.

Eight treatment combinations (1), a, b, ab, c, ac, bc, abc would be used for a complete repli-cate but only four of these are required in the half-replicate of a 23 factorial. Which four treat-ment combinations should be chosen from among the 70 possibilities? One suggestion is tochoose the first four treatment combinations from the standard order, i.e. (1), a, b, ab; but thissuggestion is very unwise since these are the four treatment combinations at the lower level offactor C so it would become impossible to estimate the main effect C.

The approach we adopt is based on the assumption that the three-factor interaction ABC canbe suppressed. The linear contrast for this effect is

ABC ≡ (a− 1)(b− 1)(c− 1) = abc− ac− bc+ c− ab+ a+ b− (1).

Choose four treatment combinations with the same sign in this linear contrast; say the fourcombinations with positive signs, which are a b c abc. Then the seven factorial effects areestimated by linear combinations of these four treatment combinations, with weights given by:

63

A B AB C AC BC ABCa +1 −1 −1 −1 −1 +1 +1b −1 +1 −1 −1 +1 −1 +1c −1 −1 +1 +1 −1 −1 +1abc +1 +1 +1 +1 +1 +1 +1

There are three points which emerge from these estimates.

• It is not possible to estimate the interaction ABC because the weights in the linear com-bination are all +1, i.e. we do not have a contrast.

• The estimates of A, B and C are still mutually orthogonal contrasts.

• However, these same mutually orthogonal contrasts also estimate the two-factor interac-tions, i.e. the estimates of pairs of factorial effects are identical:

A = BC B = AC and C = AB.

To summarise: the price to be paid for considering a 12−replicate rather than a full replicate 23

factorial is that each main effect has the same estimate as a two-factor interaction and that all ofthe information on the single three-factor interaction is lost.

Remark on the Half-Replicate of a 23 FactorialIf we had chosen the four treatment combinations with the same negative sign in the ABCcontrast, i.e {(1) ab ac bc}, then we would have arrived at the same three conclusions asabove.

Half-Replicate of a 2k FactorialThe procedure just described can be applied to a general 2k factorial. Select a particular factorialeffect and choose 2k−1 treatment combinations in this linear contrast which have the same sign.These 2k−1 treatment combinations form the 1

2−replicate of the 2k factorial, and all information

on this factorial effect is lost. The remaining 2k − 2 factorial effects divide into 2k−1 − 1 pairsof mutually orthogonal contrasts, with each pair being represented by the same linear contrast.

Definition of Defining ContrastThe factorial effect which divides the full set of 2k treatment combinations into a 1

2−replicate,

for which all information is lost, is called a defining contrast.

Definition of AliasTwo factorial effects are said to be aliases of each other if they are represented by the samelinear contrast of the available treatment combinations.

Definition of Generalised InteractionThe generalised interaction of two factorial effects is the combination of all letters that appearin the two effects, cancelling all letters that appear twice.

64

Theorem 5.2. In a 12−replicate of a 2k factorial, the alias of a given factorial effect is the

generalised interaction of this factorial effect with the defining contrast.

Example: Half-Replicate of a 23 FactorialThe approach we described above to obtain the 1

2−replicate of a 23 factorial is to choose the

factorial effect ABC as defining contrast. Thus all information on ABC is lost. From Theorem4.31 the alias of A is the generalised interaction of A with ABC, i.e. the alias of A is A ×ABC = A2BC = BC. Similarly, the alias of B is B ×ABC = AB2C = AC and the alias ofC is C × ABC = ABC2 = AB.

Example 1: Poor Half-Replicate of a 23 FactorialIf we had (unwisely) chosen the four treatment combinations {(1) a b ab} as the 1

2−replicate

then this is equivalent to choosing C as defining contrast so that all information on C is lost.From Theorem 5.2 , the alias of A is AC, the alias of B is BC and the alias of AB is ABC.

Definition of an Estimable Factorial EffectA factorial effect is estimable if and only if the given effect has an alias which is a factorialeffect that can be suppressed.

Example 2: Half-Replicate of a 24 FactorialThe factors are A, B, C, D. We choose the highest-order interaction ABCD as definingcontrast. The eight treatment combinations that have the same +1 weights in the linear contrastABCD ≡ (a− 1)(b− 1)(c− 1)(d− 1) are

(1) ab ac ad bc bd cd abcd.

Each main effect has a three-factor interaction as alias, i.e.

A = A× ABCD = BCD B = ACD C = ABD and D = ABC.

It follows that if all three-factor interactions are suppressed then all main effects are estimable.But this is not true of other non-negligible factorial effects.

Each two-factor interaction has another two-factor interaction as alias, i.e

AB = CD AC = BD and AD = BC,

so, in general, the two-factor interactions are not estimable.

If all interactions can be suppressed then 7 degrees of freedom associated with the 12−replicate

are divided into four for estimable main effects and three for the estimate of σ2. The sumsof squares due to these different components would be derived from an adaptation of Yates’algorithm. To use the algorithm it is essential that the treatment combinations are placed in thecorrect order.

Yates’ algorithm for Half-Replicate of 24 Factorial

To use the algorithm it is necessary to arrange the 12−replicate of the 24 factorial as a full-

replicate of a 23 factorial by “ignoring” one of the factors. Thus, if we “ignore” d then the

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following treatment combinations are written down in standard order.

(1) ad bd ab cd ac bc abcd.

We call this arrangement pseudo-standard order. The data corresponding to these eight treat-ment combinations in pseudo-standard order are entered in the margin as Column 0.

For a 12−replicate of a 24 factorial it is necessary to compute three columns Columns 1,2 and 3

by the usual method of adding and subtracting pairs from Column (i− 1) to give Column i.

The entries in Column 3 contain seven factorial effects paired with their aliases plus the grandtotal. The grand total occupies the first position in Column 3. Each factorial effect, togetherwith its alias, is situated in the same position in Column 3 as the corresponding treatment com-bination, ignoring d, in Column 0.

Illustration of Yates’ Algorithm

Eight treatment totals in pseudo-standard order are written in Column 0, then Columns 1,2 and3 are formed in the usual way:

Column 0 Column 1 Column 2 Column 3 Estimate(1) · · · · · · · · · Total −−ad · · · usual process · · · A = BCD Main Abd · · · of adding · · · B = ACD Main Bab · · · and subtracting · · · AB = CD Residualcd · · · in pairs · · · C = ABD Main Cac · · · · · · · · · AC = BD Residualbc · · · · · · · · · BC = AD Residualabcd · · · · · · · · · ABC = D Main D

Example 3: Half-Replicate of a 25 Factorial

The factors are A, B, C, D, E. Construct a 12−replicate using ABCDE as defining contrast.

Each main effect has a four-factor interaction as alias and each two-factor interaction has athree-factor interaction as alias; for example

A× ABCDE = BCDE etc. CD × ABCDE = ABE etc.

If we suppress three-factor and higher-order interactions then five main effects and ten two-factor interactions are estimable, i.e. all non-negligible effects are estimable. Unfortunately,this leaves no degrees of freedom for estimating σ2.

Example 4: Half-Replicate of a 26 Factorial

The factors are A, B, C, D, E, F . A 12−replicate uses ABCDEF as defining contrast. Each

main effect has a five-factor interaction as alias and each two-factor interaction has a four-factorinteraction as alias. The (63) = 20 three-factor interactions split into ten alias pairs. If wesuppress all interactions consisting of three or more factors then all non-negligible effects areestimable and ten degrees of freedom are available for estimating σ2.

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The 32 treatment combinations are all 32 even combinations, given by

(1) ab ac ad ae af bc bd be bf cd ce cf de df ef abcd abce abcf

abde abdf abef acde acdf acef adef bcde bcdf bcef bdef cdef abcdef

Alternatively, all the odd treatment combinations could form the 12−replicate.

5.3.2 Quarter-Replicate Fractional Factorial

Example 5: Quarter-Replicate of a 26 Factorial

The method for choosing a 14−replicate of a 26 factorial is to choose two 1

2−replicates and

take treatment combinations common to both. For example, if we choose defining contrastsABCDEF and ABCDE then the 1

4−replicate is

(1) ab ac ad ae bc bd be cd ce de abcd abce abde acde bcde.

But ‘f ’ is missing from all 16 treatment combinations, i.e. we have derived unintentionallya 1

4−replicate where we cannot estimate F . Note that F is the generalised interaction of the

chosen defining contrasts ABCDEF and ABCDE.

Theorem 5.3. Any two factorial effects may be used as defining contrasts to divide a 2k factorialinto a 1

4−replicate. Their generalised interaction also acts as a defining contrast and is not

estimable. Any factorial effect which is not a defining contrast has three aliases which are thegeneralised interactions of this factorial effect with the three defining contrasts.

Definition of an Estimable Factorial Effect

A factorial effect is estimable if and only if the given effect has three aliases which are factorialeffects that can all be suppressed.

To obtain a 14−replicate for a 26 factorial choose two four-factor interactions as defining con-

trasts whose generalised interaction is also a four-factor interaction.

Example 6: Quarter-Replicate of a 26 Factorial

Use ABCE and ABDF , which have interaction CDEF , as defining contrasts. With thischoice of defining contrasts, all main effects have aliases which are three-factor or five-factorinteractions:

A = BCE = BDF = ACDEF D = ABCDE = ABF = CEFB = ACE = ADF = BCDEF E = ABC = ABDEF = CDFC = ABE = ABCDF = DEF F = ABCEF = ABD = CDE.

Some two-factor interactions are aliases of each other, e.g.

AB = CE = DF = ABCDEF BC = AE = ACDF = BDEF.

If we can suppress all interactions involving three or more factors then all main effects areestimable but some two-factor interactions are not. On the other hand, if we can suppress all

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interactions then we have six estimable main effects and nine degrees of freedom for estimatingσ2.

Selecting the Quarter-Replicate

A simple method for obtaining the 14−replicate which contains the treatment combination (1)

is to derive all even combinations of a, b, c, d, e, f simultaneously in two defining contrasts;thus:

a b c d e fABCE 1 1 1 0 1 0ABDF 1 1 0 1 0 1

The sixteen treatment combinations in the 14−replicate containing (1) are

(1) ab ce df acd bcd aef bef ade bde acf bcf abce abdf cdef abcdef.

Selection of these elements is helped by the following property:

Lemma 5.1. The treatment combinations in a fractional replicate containing (1) form a sub-group of a multiplicative Abelian group, i.e. elements are closed with respect to multiplicationmodulo 2.

The analysis is carried out using Yates’ algorithm by arranging the sixteen treatment combina-tions in pseudo-standard order after “ignoring” two factors:

(1) aef bef ab ce acf bcf abce df ade bde abdf cdef acd bcd abcdef

which occupy Column 0. Columns 1,2,3 and 4 are obtained as usual. Entries in Column 4 arethe grand total and 15 factorial effects with their aliases.

Example 7: Quarter-Replicate of 27 Factorial

For a 27 factorial the factors are A, B, C, D, E, F, G. A useful choice for the definingcontrasts is the pair ABCDE and ABCFG, with interaction DEFG, since we arrive at thefollowing conclusions.

All main effects have second and higher order interactions as aliases.

Fifteen first order interactions have second and higher order interactions as aliases, but six firstorder interactions are aliased with each other:

DE = ABC = ABCDEFG = FG,DF = ABCEF = ABCDG = EG,DG = ABCEG = ABCDF = EF.

Therefore, if we can suppress only second and higher order interactions then it follows thatnot all non-negligible effects are estimable. However, if we can suppress all second and higherorder interactions and all first order interactions involving one of factors D, E, F or G then allnon-negligible effects are estimable in this particular case.

Example 8: Quarter-Replicate of 28 Factorial

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Factors for a 28 factorial are A, B, C, D, E, F, G, H . Choose defining contrasts ABCDEand ABCFGH , with generalised interaction DEFGH . Each main effect has third or higherorder interactions as aliases, each two-factor interaction has second or higher order interactionas aliases. If we can suppress second and higher order interactions then all non-negligiblefactorial effects are estimable. The degrees of freedom are

main effects 8two-factor interactions 28

Residual 27Total 63

5.3.3 The Resolution of a Design

Suppose we have a 2k−p design, that is a 1/2p fraction of a 2k factorial design. The designrequires p defining contrasts which need to be independent in the sense that no defining contrastcan be obtained as an interaction between two or more other defining contrasts. The Resolu-tion of the design is the smallest number of ‘letters’ involved in a defining contrast or in aninteraction between two or more defining contrasts.

Example Fraction Defining contrasts ‘Smallest’ defining contrast Resolutionor interaction

1 1/2 C C 12 1/2 ABCD ABCD 43 1/2 ABCDE ABCDE 54 1/2 ABCDEF ABCDEF 65 1/4 ABCDEF, ABCDE F 16 1/4 ABCE, ABDF ABCE 47 1/4 ABCDE, ABCFG DEFG 48 1/4 ABCDE, ABCFGH ABCDE 5

Designs of resolution III, IV and V are particularly important.

• Resolution III designs These designs have no main effect aliased with any other maineffect, but main effects are alised with two factor interactions and two factor interactionsmay be aliased with each other.

• Resolution IV designs These designs have no main effect aliased with any other maineffect or with any two factor interactions, but two factor interactions are aliased with eachother.

• Resolution V designs These designs have no main effect or two factor interactions aliasedwith any other main effect or two factor interactions, but two factor interactions are aliasedwith three factor interactions.

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5.4 Confounding in a 2k Factorial

Confounding is a useful device for arranging a factorial experiment, whether fractionally repli-cated or not, into blocks when the prevailing circumstances are such that the block-size issmaller than the number of treatment combinations. The purpose of confounding is to en-able certain non-negligible factorial effects to be estimated with higher precision than wouldotherwise be possible with no arrangement of the experiment into blocks. We shall see that thisis achieved by sacrificing information on a number of selected contrasts.

Example: Confounding a 23 Factorial into Pairs of Blocks

Consider three factors A,B,C, each at two levels, in r complete replicates of a 23 factorialinvolving treatment combinations (1), a, b, ab, c, ac, bc, abc. Altogether there are 8r obser-vations, where r is the number of replications of each treatment combination; if blocks of eightreasonably homogeneous experimental units can be formed then the use of randomised blocks,or possibly an 8× 8 latin square if r = 8, is probably the most sensible approach.

Suppose, however, that it is only possible to form 2r homogeneous blocks of size four. Then itis natural to arrange these blocks in pairs such that, for each pair of blocks, one half-replicateof the 23 factorial is contained in one block and the other half-replicate is contained in theother block. But how should the eight treatment combinations be separated into the two half-replicates? A poor suggestion is to take the first four combinations from standard order as halfreplicate so that r blocks contain (1), a, b, ab, and the other r blocks contain c, ac , bc , abc.This suggestion is poor because the four treatment combinations at the low level of C are in rblocks and the four at the high level of C are in the other r blocks; so the contrast estimatingthe main effect C would also be measuring differences between these blocks. We say that C isconfounded with blocks.

We should avoid confounding the main effects or the two-factor interactions if at all possible.A much better suggestion is to confound interaction ABC with blocks, particularly as ABCis likely to be negligible. So if we suppose that r = 3 the experiment involves six blocks asfollows:

Block 1 Block 2 Block 3 Block 4 Block 5 Block 6(1) a (1) a (1) aab b ab b ab bac c ac c ac cbc abc bc abc bc abc

The whole experiment consists of 24 observations derived from eight treatment combinationsallocated at random to these three pairs of blocks. Blocks 1, 3, 5 have the property that the fourelements are closed with respect to multiplication modulo 2 (c.f. 5.4). These blocks contain (1)and are known as the principal block. Blocks 2, 4 and 6 contain the other half-replicate; theelements are found by multiplying modulo 2 the principal block by any one of a, b, c or abc.

This situation holds generally for a 2k factorial experiment confounded in 2t blocks of size 2k−t

when t < k and is described by the following theorem.

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Theorem 5.4. In a complete replicate of a 2k factorial experiment confounded in 2t blocks ofsize 2k−t, the treatment combinations in the principal block include (1) and form a subgroup ofthe group of all 2k treatment combinations.

The remaining 2t − 1 blocks are cosets of the principal block, i.e. they can be formed bymultiplying modulo 2 the elements of the principal block by other treatment combinations notcontained in the principal block.

Example: Confounding a 24 Factorial in Two Blocks of size 8

If there are four factors A,B,C,D, each at two levels, in a 24 factorial it is usual to confoundinteraction ABCD with blocks. Then the principal block consists of the eight even treatmentcombinations.

The two blocks are:

principal block: (1) ab ac ad bc bd cd abcdsecond block: a b c d abc abd acd bcd

The second block consists of the remaining set of treatment combinations once the principalblock is removed. Note, however, that the second block can be obtained directly by multiplyingmodulo 2 the eight treatment combinations in the principal block by any remaining treatmentcombination not contained in the principal block. For example, multiplication by a gives thesecond block in the order specified above.

Example: Analysis of two replicates of a 24 Factorial in four blocks

Suppose that a 24 factorial experiment is performed in circumstances where four blocks areavailable but the block-size is restricted to eight cells. It is decided to choose two replicatesof the blocking arrangement specified above, i.e. two replicates of a 24 factorial experimentarranged in four blocks with interaction ABCD confounded with blocks.

The data for this experiment are given by:

block 1 ab bc ac (1) abcd ad bd cd Total2.1 3.7 4.0 1.5 6.3 3.3 2.8 5.2 28.9

block 2 b abc acd abd c a bcd d Total2.7 7.3 8.8 5.9 5.3 6.4 6.8 5.5 48.7

block 3 ad ac (1) ab cd abcd bd bc Total5.2 6.1 3.2 3.7 7.4 9.1 4.2 5.8 44.7

block 4 acd abc c d bcd abd a b Total9.2 8.0 8.2 6.7 6.9 6.5 7.1 3.1 55.7

Yates’ algorithm gave the following factorial effects:

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Col 0 Col 1 Col 2 Col 3 Col 4(1) 4.7 18.2 29.8 78.2 178.0 Totala 13.5 11.6 48.4 99.8 20.0 Ab 5.8 23.6 40.1 11.2 −8.2 Bab 5.8 24.8 59.7 8.8 5.8 ABc 13.5 20.7 8.8 −5.4 38.2 Cac 10.1 19.4 2.4 −2.8 −1.0 ACbc 9.5 30.6 1.7 0.4 7.6 BCabc 15.3 29.1 7.1 5.4 5.2 ABC†

d 12.2 8.8 −6.6 18.6 21.6 Dad 8.5 0.0 1.2 19.6 −2.4 ADbd 7.0 −3.4 −1.3 −6.4 2.6 BDabd 12.4 5.8 −1.5 5.4 5.0 ABD†

cd 12.6 −3.7 −8.8 7.8 1.0 CDacd 18.0 5.4 9.2 −0.2 11.8 ACD†

bcd 13.7 5.4 9.1 18.0 −8.0 BCD†

Blocks/abcd 15.4 1.7 −3.7 −12.8 −30.8 ABCD††

† ≡ entries in SSResidual. †† ≡ entry in SSBlocks.

The sum of squares for each factorial effect is found from Theorem 1.62.

For example, SSA = 20.02/(2× 16) = 12.50.

SSBlocks is obtained in the usual way.

The ANOVA table for this experiment is given by

Source DF SS MS F PA 1 12.50 12.50 19.85∗∗ < 0.001B 1 2.10 2.10 3.34 < 0.1C 1 45.60 45.60 72.43∗∗ < 0.001D 1 14.58 14.58 23.16∗∗ < 0.001AB 1 1.05 1.05 1.67 > 0.1AC 1 0.03 0.03 < 1AD 1 0.18 0.18 < 1BC 1 1.81 1.81 2.87 > 0.1BD 1 0.21 0.21 < 1CD 1 0.03 0.03 < 1

Blocks 3 48.31 16.10 25.58∗∗

Residual 15 11.33 0.63Total 31 137.74

It is clear that all main effects and two-factor interactions are estimated with higher precisionbecause of this arrangement in blocks.

Lemma can be applied to find the principal block. When arranging a 2k factorial in 2t blocks ofsize 2k−t when t < k, the remaining blocks can be obtained using Theorem .

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Example: Confounding a 25 Factorial in Four Blocks of size 8

To arrange five factors A,B,C,D,E, each at two levels, into four blocks of size eight it isnecessary to choose two factorial effects to confound with blocks and then their generalisedinteraction will also be confounded with blocks. The four blocks will each contain a quarterreplicate. For instance, if ABC and CDE are confounded with blocks then ABDE is alsoconfounded with blocks. To get the principal block, find all treatment combinations whichhave an even number of letters in common with ABC and CDE. (It is sufficient to find threeindependent members of the group since the remaining four are found by taking products.)

a b c d eABC 1 1 1 0 0CDE 0 0 1 1 1

Blocks 2, 3 and 4 are then found by multiplying modulo 2 the principal block by other treatmentcombinations not in the principal block. We get:

principal block (1) ab de acd abde bcd ace bcesecond block a b ade cd bde abcd ce abcethird block c abc cde ad abcde bd ae be

fourth block d abd e ac abe bc acde bcde

If a factorial experiment is conducted using a single replicate of a 25 factorial arranged in thesefour blocks of size eight there will be 31 degrees of freedom overall, which are distributed asfollows:

5 for all main effects;10 for all two-factor interactions;8 for three-factor interactions except ABC, CDE;4 for four-factor interactions except ABDE;1 for interaction ABCDE;3 for ABC, CDE, ABDE which are confounded with Blocks.

It follows that if all second- and higher-order interactions can be suppressed then there are13 degrees of freedom available for deriving the residual mean square which is required forestimating σ2.

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5.5 Specialised Uses Of Fractional Factorial Experiments

So far we have only considered how factors and interactions of factors affect the location of theresponse. We now demonstrate, by means of examples, how fractional factorial experiments canbe used to gain information on which factors affect both the mean and variability of a response.

Screening Experiments

In the initial stages of the design of a product or process there are often a very large numberof factors that may affect the response. Fractions of 2k designs are particularly useful in suchsituations because, if the experimenter assumes that most interactions are not likely to be signif-icant, they provide a large amount of information on the effects that the individual factors haveon the mean and variability of the response for a small number of trials.

Once important factors have been identified, further experiments can be designed involving onlythe important factors and these can be planned to accommodate interactions between factors.

Experiments in which the experimenter is trying to determine which of a large number of factorsaffect the response are termed ‘screening’ experiments.

Saturated designs are the most extreme designs used in screening. These are designs that useonly n = p + 1 treatment combinations to estimate the effects on the mean and variability onthe response of up to p factors independently. It will generally be necessary to assume thatmost interactions are negligible. In fact if information is wanted on p factors, it is necessary toassume that all interactions are negligible.

Below is a saturated design that can accommodate up to 15 factors, each at two levels, with16 observations. Each factor is allocated to a column of the array. If fewer than 15 factors areunder consideration then different allocations of the factors to columns may lead to differentaliasing schemes, and sometimes these aliasing schemes may not be equally good.

Note that the column headings identify which columns are products of which other columns.

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Columns1 2 12 3 13 23 123 4 14 24 124 34 134 234 1234

-1 -1 1 -1 1 1 -1 -1 1 1 -1 1 -1 -1 1-1 -1 1 -1 1 1 -1 1 -1 -1 1 -1 1 1 -1-1 -1 1 1 -1 -1 1 -1 1 1 -1 -1 1 1 -1-1 -1 1 1 -1 -1 1 1 -1 -1 1 1 -1 -1 1-1 1 -1 -1 1 -1 1 -1 1 -1 1 1 -1 1 -1-1 1 -1 -1 1 -1 1 1 -1 1 -1 -1 1 -1 1-1 1 -1 1 -1 1 -1 -1 1 -1 1 -1 1 -1 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 -11 -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 11 -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 11 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 -11 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Epitaxial Layer ExperimentAn experiment was conducted to try to reduce the average thickness and the variability of thethickness of the ‘epitaxial layer’ deposited onto silicon wafers during the manufacture of inte-grated circuit devices. Eight factors labelled A, B, . . . , H were identified that might affect theaverage and variability of the thickness of the layer. The following table shows the eight factors,their prior levels and the experimental levels.

Factors Prior Level Experimental levelsLow (0) High(1)

A (rotation method) oscillating continuous oscillatingB (wafer batch) 668G4 678D4

C (deposition temperature) 1215◦C 1210◦C 1220◦CD (deposition time) low high lowE (arsenic flow rate) 57% 55% 59%

F (acid etch temperature) 1200◦C 1180◦C 1215◦CG (acid flow rate) 12% 10% 14%H (nozzle position) 4 2 6

Observations on the thickness of the layer were taken at sixteen different treatment combina-tions. The treatment combinations were selected via the orthogonal array above. The assign-ment of factors to columns that was used is indicated at the foot of the copy of the array below.

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Columns1 2 12 3 13 23 123 4 14 24 124 34 134 234 1234

-1 -1 1 -1 1 1 -1 -1 1 1 -1 1 -1 -1 1-1 -1 1 -1 1 1 -1 1 -1 -1 1 -1 1 1 -1-1 -1 1 1 -1 -1 1 -1 1 1 -1 -1 1 1 -1-1 -1 1 1 -1 -1 1 1 -1 -1 1 1 -1 -1 1-1 1 -1 -1 1 -1 1 -1 1 -1 1 1 -1 1 -1-1 1 -1 -1 1 -1 1 1 -1 1 -1 -1 1 -1 1-1 1 -1 1 -1 1 -1 -1 1 -1 1 -1 1 -1 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 -11 -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 11 -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 11 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 -11 1 1 1 1 1 1 1 1 1 1 1 1 1 1A B C D E F G H

The experiment is a 28−4 experiment. Sets of four independent defining contrasts can be foundby identifying four independent relationships between the columns of the array. This is maderelatively easy by the column labellings.

For example,

(1)(2) ≡ (3)(123)

and so AB is aliased with CD which implies that ABCD is a defining contrast.

A set of defining contrasts is ABCD, ABEF , ACEG and BCEH . Note that, for example, Dis aliased with ABC. This is a resolution IV design and there is considerable aliasing betweenthe 2-factor interactions. For example, the contrast listed in column 12 of the array measuresthe 2-factor interactions AB, CD, EF and GH as well as some higher order interactions.

At each of the sixteen treatment combinations r observations of the thickness of the epitaxiallayer were taken. The jth observation on the ith treatment combination is denoted by yij . Fromthe observations at each of the treatment combinations, two different summary statistics werecalculated:

• The average of the r observations, yij;

• The log of the sample variance of the r observations, log(s2i ).

The results are:

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Treatment Average Log VarianceCombination Response Response

A B C D E F G H yij log(s2i )0 0 0 0 0 0 0 0 14.821 -0.44250 0 0 0 1 1 1 1 14.888 -1.19890 0 1 1 0 0 1 1 14.037 -1.43070 0 1 1 1 1 0 0 13.880 -0.65050 1 0 1 0 1 0 1 14.165 -1.42300 1 0 1 1 0 1 0 13.860 -0.49690 1 1 0 0 1 1 0 14.757 -0.32670 1 1 0 1 0 0 1 14.921 -0.62701 0 0 1 0 1 1 0 13.972 -0.34671 0 0 1 1 0 0 1 14.032 -0.85631 0 1 0 0 1 0 1 14.843 -0.43691 0 1 0 1 0 1 0 14.415 -0.31311 1 0 0 0 0 1 1 14.878 -0.61541 1 0 0 1 1 0 0 14.932 -0.22921 1 1 1 0 0 0 0 13.907 -0.11901 1 1 1 1 1 1 1 13.914 -0.8625

The experimenters first analysed the log variance response. The contrast estimate for the logsample variance response for factor A is obtained using:

1

8(−1,−1,−1,−1,−1,−1,−1,−1, 1, 1, 1, 1, 1, 1, 1, 1)ls2i = 0.352

where ls2i is the 16× 1 column vector containing the log (s2i ) values.A table giving all the contrast estimates for the log sample variance response is:

Contrast A B C D E F G HEstimate 0.352 0.122 0.105 -0.249 -0.012 -0.072 -0.101 -0.566

This was done first for the following reasons:• In most processes, minimising the variability of the response is desirable;• There will generally be a smaller number of factors that affect the variability of the re-

sponse than affect the average response.The idea is to minimise the variability by careful choice of the levels of the factors that have aneffect on variability and then to use the factors that have an effect on the average response butnot a marked effect on the variability to shift the location of the response into a desired range.The contrast estimates for factors A and H are considerably larger, in absolute value, than thosefor the other factors.Since the contrast estimate for A is positive and the log variance response is to be reduced, Ashould be set at its low level.Since the contrast estimate for H is negative H should be set at its high level.Analysing the average response in the same way, the following contrast estimates for the meanresponse variable were obtained :

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Contrast A B C D E F G HEstimate -0.055 0.056 -0.109 -0.836 -0.067 0.060 -0.098 0.142

The contrast estimate for factor D is much larger, in absolute value, than those for the otherfactors. Since the contrast estimate for D is negative and it is desired to reduce the averageresponse, D should be set at its high level.A follow up experiment was carried out and confirmed the results above.Note that to obtain the summary statistics for each treatment combination it was necessary tohave at least two observations at each combination.

Using Orthogonal Arrays For 4-level FactorsThe following orthogonal array could be used for an experiment for up to seven 2-level factors.

Columns1 2 3 12 13 23 123

-1 -1 -1 1 1 1 -1-1 -1 1 1 -1 -1 1-1 1 -1 -1 1 -1 1-1 1 1 -1 -1 1 -11 -1 -1 -1 -1 1 11 -1 1 -1 1 -1 -11 1 -1 1 -1 -1 -11 1 1 1 1 1 1

We demonstrate how the array can be used to accommodate a 4-level factor. A 4-level factorrequires three columns to represent three orthogonal contrasts, with one column being the in-teraction of the other two columns. So, for example, a 4-level factor could be represented bycolumns (1), (2) and (12).Considering a 22 × 4 experiment, with factors A and B at two levels and factor C at fourlevels. Using the array given will be a half factorial design since 8 of the 16 possible treatmentcombinations will be used.First Column Allocation: Allocate factor A to column (1), factor B to column (2) and factorC to columns (23) and (123). Call (23) C1 and (123) C2, then C1C2 the remaining contrast forthe 4-level factor is represented by column (1) and is thus aliased with A. The defining relationis:

I ≡ AC1C2.

Thus the design has Resolution 2.Second Column Allocation: Allocate factorA to column (1), factorB to column (2) and factorC to columns (12) and (23). Call (12) C1 and (23) C2, then C1C2 the remaining contrast for the4-level factor is represented by column (13). The defining relation is:

I ≡ ABC1.

Thus this design has resolution 3 and is to be preferred over the first column allocation.

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5.6 Robust Design

Statistically designed experiments are widely used in industry for product and process design,development and improvement. In the 1980’s an important application of designed experiments,namely robust design, was introduced by a Japanese engineer Dr. Genichi Taguchi. In robustdesign the focus is on one or more of the following:

• Designing products or processes that are insensitive to environmental factors that canaffect performance once the product or process is put into operation. An example isthe formulation of an exterior paint that should exhibit long life irregardless of weatherconditions. That is, the product should be robust against a wide range of temperature,humidity and precipitation factors that affect the wear and finish of the paint.

• Designing products so that they are insensitive to variability of the components of thesystem. An example is designing an electronic amplifier so that the output voltage isas close as possible to the desired target regardless of the variability in the resistors,transistors and power supplies that are the components of the device.

• Determining the operating conditions for a process so that critical product characteristicsare as close as possible to the desired target value and so that the variability around thistarget is minimized. For example, in semiconductor manufacturing we would like theoxide thickness on a wafer to be as close as posssible to the target mean thickness, andwe would also like the variability in thickness across the wafer to be as small as possible.

Factors included in the experimentation are classified as either design factors or noise factors.Design factors are factors that are easy and inexpensive to control in the manufacture of theproduct. Noise factors are those that may affect the performance of a product but that aredifficult or impossible to control during ordinary operating conditions.

Examples of noise factors include factors such as the climate in which a product is to be used,variation in the raw materials and the wear of the components and materials over the life of theproduct.

A combination of levels of the design factors is a potential product design and is said to berobust if the product functions well despite uncontrolled variation in the levels of the noisefactors. Thus, in robust design we pay considerable attention to how the variability of theresponse changes as the factor levels change.

5.6.1 Analysing the Effects of Design Factors on variability

It is assumed that noise factors can be controlled for the purposes of experimentation, for exam-ple under laboratory test conditions, although they cannot be controlled in ordinary operatingconditions. The experiment is designed so that the same noise factor combinations are observedfor each design factor combination. The list of design factor combinations in the experimentis called the design array. The list of noise factor combinations is called the noise array. One

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important objective is to determine which combination of design factors is least sensitive tochanges in the noise factors.

Suppose there are v design factor combinations and u noise factor combinations in the design.Denote a particular design factor combinationby w and a noise factor combination by x, and thecorresponding observation by ywx . For each design factor combination there are u observations,one for each noise factor combination.

Denote the sample mean and sample variance of these u observations by

yw . =

∑x ywxu

and s2w =

∑x (ywx − yw .)2

u − 1.

Observations ywx can be assumed to follow the model:

Ywx = µ+ τwx + εwx ,

εwx ∼ N (0, σ2),

where the εwx ’s are mutually independent and w = 1, . . . , v ; x = 1, . . . , u.

The sample means Yw . are independent, normal random variables each with variance σ2/u.Thus we can write Yw . as:

Yw . = µ+ αw + εw ,

εw ∼ N (0, σ2/u),

where the εw ’s are mutually independent and w = 1, . . . , v .

Now we can analyse the effects of the design factors averaged over the noise factors via theusual analysis of variance, but with yw , w = 1, . . . , v , as the v observations.

Analysing the sample variances is a more difficult problem. The noise factors have been sys-tematically, rather than randomly, varied in the experiment. Despite this, s2w can be used as ameasure of the variability of the design factor combination w over the levels of the noise factors.

It is not appropriate to carry out the usual analysis of variance using s2w as the response variablebecause it can be shown that Var(S2

w) is not constant. In fact Var(S2w) increases with E(S2

w).However, if the distribution of ln(S2

w) is approximately normal, we can use this as a responsevariable in an analysis of variance to analyse the effect of the design factors on the variabilityof the response as the levels of the noise factors change.

Since there is only one value of yw . and one value of ln(s2w) for each design factor combinationw, the effects of the design factors on yw . and ln(s2w) must be analysed like a single replicatedesign. This means that in order to obtain an analysis of variance for yw . or ln(s2w) we must beable to assume that some interactions between design factors are negligible.

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Torque Optimisation experimentIn an experiment carried out by ITT Automotive to maximise the efficiency of car seat tracks,one purpose was to stabilise frame torque of a car seat track with a target value of 14 ± 4inch-pounds. The experiment involved two design factors: ‘anvil type’ (factor A) with levels1, 2, 3, and ‘rivet diameter’ (factor B) with levels 1, 2, 3. These are design factors because thebest combination of their levels is desired to be known for future production. In addition, theexperiment involved two different machines. ‘Machine’ (factor M ) is regarded as a noise factorbecause it will be necessary to use both machimes at the same settings of the design factors inthe production process. For each of the 9 design factor combinations two measurement weretaken on each machine. Settings of the design factors are required that give a response withinthe required range which exhibits as little variability as possible across machines.

The data, yij.., s2ij and ln(s2ij) are:

AB combination yij11 yij12 yij21 yij22 yij.. s2ij ln(s2ij)

11 16 21 24 18 19.75 12.2500 2.505512 38 40 36 38 38.00 2.6667 0.980813 48 60 42 40 47.50 81.0000 4.394521 8 10 16 12 11.50 11.6667 2.456722 22 28 16 18 21.00 28.0000 3.332223 28 34 16 16 23.50 81.0000 4.394531 8 14 8 6 9.00 12.0000 2.484932 18 24 8 14 16.00 45.3333 3.814033 20 14 16 14 16.00 8.0000 2.0794

Note that factors A and B are both 3-level factors.

We start by using a three-way complete model to examine the raw data;

Yijkl = µ+ τijk + εijkl,

= µ+ αi + βj + γk + (αβ)ij + (αγ)ik + (βγ)jk + (αβγ)ijk + εijkl.

The model assumptions may well not be valid, however the list of p-values from the standardanalysis of variance table for the raw data enables us to gain an impression of the importanteffects. Interactions between design factors and the noise factor are of particular interest.

The ANOVA table is:

Source DF SS MS F value p-valueA 2 3012.7222 1506.3611 116.62 0.0001B 2 1572.0555 786.0277 60.85 0.0001AB 4 470.4444 117.6111 9.11 0.0003M 1 240.2500 240.2500 18.60 0.0004AM 2 5.1666 2.5833 0.20 0.8205BM 2 197.1666 98.5833 7.63 0.0040ABM 4 170.6666 42.6666 3.30 0.0339

Residual 18 232.5000 12.9166Total 35 5900.9722

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Note from the ANOVA table that the AM interaction appears to be negligible. However, theBM interaction effect does appear to be important and the ABM interaction should probablybe considered as well. The idea is that by appropriate choice of the levels of the design factorsA and B it may be possible to exploit these interactions to dampen the effect of the noise factoron response variability.

Next consider the effects of the design factors on response variability. Figure 5.1 gives an ABinteraction plot for these data, with ln(s2ij) plotted against the level of factor A, and the level offactor B as the label. The plot indicates a strong AB interaction effect for the response variableln(s2ij). This indication corresponds to an ABM interaction in the raw data.

1 -

2 -

3 -

4 -

ln(s2ij) Labels: j (Level of B)

1 2 3Level of A, (i)

1. . . . . . 1. . . . . . 1

2.

..

..

.2. . . . . . 2

3. . . . . . 3.

..

..

.3

Figure 5.1: Effects of A and B on ln(s2ij)

Similarly Figure 5.2 gives the interaction plot for the effects of A and B on mean response.This Figure suggests that there is an interaction between the two factors when averaged over thenoise factor levels, i.e. this gives a strong indication of an AB interaction averaged over M .

Note that Figure 5.2 strongly suggests interaction between the design factors A and B so wecannot assume that the interaction is negligible and consequently we cannot carry out the stan-dard analysis of variance using either yij or ln(s2ij) as response varaibles.Design factors A and B apparently affect both the mean response and the response variability.The experimenter needs to choose a design factor combination that is satisfactory with respectto both. Since the target response is between 10 and 18 inch-pounds, we see that acceptablelevels of mean response are obtained only for treatment combinations 21, 32 and 33. However,combination 33 yielded a much lower sample variance. No inferential statistical methods wereused and so the recommendation is that some additional observations should be taken to confirmthat combination 33 is a good choice.

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15 -

30 -

45 -

yij. Labels: j (Level of B)

1 2 3Level of A, (i)

1. . . . . . 1. . . . . . 1

2.

..

..

.2. . . . . . 2

3.

..

..

.3. . . . . . 3

Figure 5.2: Effects of A and B on yij.

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