structure balance and anova tables for experiments which ...rab/daebeam.pdf · an experiment on an...

Structure balance and ANOVA tables forexperiments which are randomized in stages

R. A. Bailey

[email protected]

joint work with C. J. Brien, University of South Australia

DAE, MemphisNovember 2007

An experiment on an industrial process (Gelman, 2005) 4 Variations

20 Machines6 Times in M

-4 treatments 120 runs

runs tiersource dfMean 1Machines 19Times[Machines] 100

treatments tiersource dfMean 1Variations 3

runs tier

treatments tier

source df

source df

Mean 1

Mean 1

Machines 19

Variations 3Residual 16

Times[Machines] 100






runs tier treatments tiersource df source dfMean 1

Mean 1

Machines 19


Times[Machines] 100






runs tier treatments tiersource df source dfMean 1 Mean 1Machines 19


Times[Machines] 100






runs tier treatments tiersource df source dfMean 1 Mean 1Machines 19 Variations 3

Residual 16

Times[Machines] 100







Residual 16Times[Machines] 100

An experiment on an industrial process: skeleton anova



The skeleton analysis of variance shows

I which “block” term the Variations term is confounded with,hence the likely magnitude of the variance of the estimator of thecontrast between any two variations

I the relevant residual term, and its degrees of freedom,hence the likely precision of the estimator of that variance,and information about the power of a hypothesis test.

An experiment on an industrial process: skeleton anova



The skeleton analysis of variance showsI which “block” term the Variations term is confounded with,

hence the likely magnitude of the variance of the estimator of thecontrast between any two variations

I the relevant residual term, and its degrees of freedom,hence the likely precision of the estimator of that variance,and information about the power of a hypothesis test.

An agricultural experiment (Gelman, 2005)

5 Varieties

2 Fertilizers

5 Rows5 Columns2 Subplots in R, C

- d⊥XXX

-

10 treatments 50 subplots

subplots tier

treatments tier

source df

source df

Mean 1

Mean 1

Rows 4Columns 4Rows#Columns 16

Varieties 4Residual 12

Subplots[R,C] 25

Fertilizers 1Varieties#Fertilizers 4Residual 20


5 Varieties

2 Fertilizers


- d⊥XXX

-


subplots tier treatments tiersource df source dfMean 1

Mean 1

Rows 4Columns 4Rows#Columns 16


Subplots[R,C] 25



5 Varieties

2 Fertilizers


- d⊥XXX

-


subplots tier treatments tiersource df source dfMean 1 Mean 1Rows 4Columns 4Rows#Columns 16


Subplots[R,C] 25



5 Varieties

2 Fertilizers


- d⊥XXX

-


subplots tier treatments tiersource df source dfMean 1 Mean 1Rows 4Columns 4Rows#Columns 16 Varieties 4

Residual 12

Subplots[R,C] 25



5 Varieties

2 Fertilizers


- d⊥XXX

-



Residual 12

Subplots[R,C] 25 Fertilizers 1Varieties#Fertilizers 4

Residual 20


5 Varieties

2 Fertilizers


- d⊥XXX

-



Residual 12Subplots[R,C] 25 Fertilizers 1

Varieties#Fertilizers 4

Residual 20


5 Varieties

2 Fertilizers


- d⊥XXX

-



Residual 12Subplots[R,C] 25 Fertilizers 1

Varieties#Fertilizers 4Residual 20

Basic idea for a single randomization

set Γ of objects‘treatments’

set Ω of objects

‘plots’

-allocate by randomization

vector space VΓ = RΓ vector space VΩ = RΩVΓ→≤ VΩ

Decomposition P of VΩ

into orthogonal subspacesDecomposition Q of VΓ

into orthogonal subspaces

Q further decomposes P

I What happens if the design (allocation) is not orthogonal?I What happens if there are 2 or more stages of (random)

allocation?



set Ω of objects

‘plots’

-

allocate by randomization







allocation?



set Ω of objects

‘plots’

-


vector space VΓ = RΓ vector space VΩ = RΩ

VΓ→≤ VΩ






allocation?



set Ω of objects

‘plots’

-








allocation?



set Ω of objects

‘plots’

-







I What happens if the design (allocation) is not orthogonal?

I What happens if there are 2 or more stages of (random)allocation?



set Ω of objects

‘plots’

-








allocation?

A pair of orthogonal decompositions

A decomposition P of VΩ into n pairwise orthogonal subspaces≡ a set of real Ω×Ω matrices P1, . . . , Pn which

I are symmetric (Pi = P>i )I are idempotent (P2

i = Pi)I are mutually orthogonal (PiPj = 0 if i 6= j)I sum to I.

A decomposition Q of VΓ into m pairwise orthogonal subspaces≡ a set of real Γ×Γ matrices Q1, . . . , Qm which . . .

Given an allocation of Γ to Ω, we can regard subspaces of VΓ assubspaces of VΩ and hence regard Q1, . . . , Qm as Ω×Ω matrices.



I are symmetric (Pi = P>i )

I are idempotent (P2i = Pi)

I are mutually orthogonal (PiPj = 0 if i 6= j)I sum to I.






i = Pi)

I are mutually orthogonal (PiPj = 0 if i 6= j)I sum to I.






i = Pi)I are mutually orthogonal (PiPj = 0 if i 6= j)

I sum to I.






i = Pi)I are mutually orthogonal (PiPj = 0 if i 6= j)I sum to I.



A pair of orthogonal decompositions: continued

P1 +P2 + · · ·+Pn = I (identity for VΩ)

Q1 +Q2 + · · ·+Qm = IQ (identity for VΓ in VΩ)

The design is orthogonal ifeach subspace in Q is contained in a subspace in P;that is, for each Qi there is some j such that

I QiPj = PjQi = Qi

I QiPk = PkQi = 0 if k 6= j.

The designs in the preceding examples are both orthogonal.




The design is orthogonal ifeach subspace in Q is contained in a subspace in P;

that is, for each Qi there is some j such thatI QiPj = PjQi = Qi






The design is orthogonal ifeach subspace in Q is contained in a subspace in P;that is, for each Qi there is some j such that

I QiPj = PjQi = Qi



Structure balance (following Nelder’s ‘general balance’)

P P1 +P2 + · · ·+Pn = I (identity for VΩ)

Q Q1 +Q2 + · · ·+Qm = IQ (identity for VΓ in VΩ)

DefinitionA structure Q is structure-balanced in relation to a structure Pif there are scalars λPQ for P in P and Q in Q such that

(i) QPQ = λPQQ for all P in P and all Q in Q, and

(ii) Q1PQ2 = 0 for all P in P and all Q1 6= Q2 in Q.

The structure Q is orthogonal in relation to P if (i) and (ii) hold witheach λPQ equal to either 1 or 0.The λPQ are called efficiency factors and are summarized in theP×Q efficiency matrix ΛPQ.







The structure Q is orthogonal in relation to P if (i) and (ii) hold witheach λPQ equal to either 1 or 0.

The λPQ are called efficiency factors and are summarized in theP×Q efficiency matrix ΛPQ.







The structure Q is orthogonal in relation to P if (i) and (ii) hold witheach λPQ equal to either 1 or 0.The λPQ are called efficiency factors and are summarized in theP×Q efficiency matrix ΛPQ.

Structure balance (using James and Wilkinson)

Fix P and Q.

Then QPQ = λPQQ, soI every vector in Im(Q) makes angle cos−1(λPQ) with Im(P);I λ

−1PQPQP is the matrix of orthogonal projection onto P(ImQ);

I write PBQ for λ−1PQPQP (the part of P explained by Q).

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Im(P)

Im(Q)

Im(PBQ)

Fix Q, let P vary.I ∑P = I implies that ∑P λPQ = 1.


Fix P and Q. Then QPQ = λPQQ, so

I every vector in Im(Q) makes angle cos−1(λPQ) with Im(P);I λ



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Im(P)

Im(Q)

Im(PBQ)



Fix P and Q. Then QPQ = λPQQ, soI every vector in Im(Q) makes angle cos−1(λPQ) with Im(P);

I λ−1PQPQP is the matrix of orthogonal projection onto P(ImQ);


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

Im(P)

Im(Q)

Im(PBQ)



Fix P and Q. Then QPQ = λPQQ, soI every vector in Im(Q) makes angle cos−1(λPQ) with Im(P);I λ



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.......................................................................................................................................................................................................................... ........... ............. ............... ................. ................. ................. ................. ................. ................. ............... ............. .............................. Im(P)

Im(Q)

Im(PBQ)


Structure balance

Fix P, let Q vary.I Q1PQ2 = 0 implies that

P(Im(Q1)) ⊥ P(Im(Q2)), so

PBQ1 and PBQ2 correspond to orthogonal subspaces of Im(P).

I WriteP`Q = P−∑

QPBQ,

so that P`Q corresponds to Im(P)∩V⊥Γ

(residual in P).

So PBQ1, PBQ2, . . . , PBQm, P`Q decompose P orthogonally.

P BQ = PBQ : P ∈P, Q ∈Q, λPQ 6= 0∪P`Q : P ∈P .

Structure balance

Fix P, let Q vary.I Q1PQ2 = 0 implies that

P(Im(Q1)) ⊥ P(Im(Q2)), so

PBQ1 and PBQ2 correspond to orthogonal subspaces of Im(P).I Write

P`Q = P−∑Q

PBQ,

so that P`Q corresponds to Im(P)∩V⊥Γ

(residual in P).

So PBQ1, PBQ2, . . . , PBQm, P`Q decompose P orthogonally.

P BQ = PBQ : P ∈P, Q ∈Q, λPQ 6= 0∪P`Q : P ∈P .

Toy example from micorarrays

slides1 2 3 4 5 6 7 8

red 0 1 0 2 0 3 0 4green 1 0 2 0 3 0 4 0

observations

treatments

source df

eff source df

Mean 1

1 Mean 1

Colours 1Slides 7

12 Within-rest 3

Residual 4

Colours # Slides 7

1 Control-v-rest 112 Within-rest 3

Residual 3


slides1 2 3 4 5 6 7 8

red 0 1 0 2 0 3 0 4green 1 0 2 0 3 0 4 0

observations treatmentssource df eff source dfMean 1 1 Mean 1Colours 1Slides 7

12 Within-rest 3

Residual 4

Colours # Slides 7

1 Control-v-rest 112 Within-rest 3

Residual 3


slides1 2 3 4 5 6 7 8

red 0 1 0 2 0 3 0 4green 1 0 2 0 3 0 4 0

observations treatmentssource df eff source dfMean 1 1 Mean 1Colours 1Slides 7

12 Within-rest 3

Residual 4

Colours # Slides 7 1 Control-v-rest 1

12 Within-rest 3

Residual 3


slides1 2 3 4 5 6 7 8

red 0 1 0 2 0 3 0 4green 1 0 2 0 3 0 4 0

observations treatmentssource df eff source dfMean 1 1 Mean 1Colours 1Slides 7 1

2 Within-rest 3

Residual 4

Colours # Slides 7 1 Control-v-rest 112 Within-rest 3

Residual 3


slides1 2 3 4 5 6 7 8

red 0 1 0 2 0 3 0 4green 1 0 2 0 3 0 4 0

observations treatmentssource df eff source dfMean 1 1 Mean 1Colours 1Slides 7 1

2 Within-rest 3Residual 4

Colours # Slides 7 1 Control-v-rest 112 Within-rest 3

Residual 3

Experiments with two randomizations

Laboratory animals (White, 1975) 2 Drugs 10 Days

60 Animals- -

Meat-loaves (T. B. Bailey)

2 Rosemary3 Irradiation

3 Blocks6 Meatloaves in B

3 Replicates12 Panellists in R6 Time-orders in R

t -QQ

- d⊥XXX

-

6 treatments 18 meatloaves 216 tastings

Structure balance when one arrow follows another

R −→Q −→P

TheoremIf Q is structure-balanced in relation to P(with efficiency matrix ΛPQ)and R is structure-balanced in relation to Q(with efficiency matrix ΛQR)then

I R is structure-balanced in relation to P and ΛPR = ΛPQΛQR ;I R is structure-balanced in relation to P BQ (and Λ . . . );I Q BR is structure-balanced in relation to P (and Λ . . . );I (P BQ)BR = P B (Q BR).


R −→Q −→P


I R is structure-balanced in relation to P and ΛPR = ΛPQΛQR ;

I R is structure-balanced in relation to P BQ (and Λ . . . );I Q BR is structure-balanced in relation to P (and Λ . . . );I (P BQ)BR = P B (Q BR).


R −→Q −→P


I R is structure-balanced in relation to P and ΛPR = ΛPQΛQR ;I R is structure-balanced in relation to P BQ (and Λ . . . );

I Q BR is structure-balanced in relation to P (and Λ . . . );I (P BQ)BR = P B (Q BR).


R −→Q −→P


I R is structure-balanced in relation to P and ΛPR = ΛPQΛQR ;I R is structure-balanced in relation to P BQ (and Λ . . . );I Q BR is structure-balanced in relation to P (and Λ . . . );

I (P BQ)BR = P B (Q BR).


R −→Q −→P


I R is structure-balanced in relation to P and ΛPR = ΛPQΛQR ;I R is structure-balanced in relation to P BQ (and Λ . . . );I Q BR is structure-balanced in relation to P (and Λ . . . );I (P BQ)BR = P B (Q BR).

Laboratory animals: skeleton anova

2 Drugs 10 Days

60 Animals- -

animals tier

days tier treatments tier

source df

source df source df

Mean 1

Mean 1 Mean 1

Animals 59

Days 9 Drugs 1Residual 8

Residual 50

I Differences between Drugs are confounded with differencesbetween Days and differences between Animals.

I The 8-df Residual gives the variability between Days plus thevariability between Animals.


2 Drugs 10 Days

60 Animals- -

animals tier days tier

treatments tier

source df source df

source df

Mean 1 Mean 1

Mean 1

Animals 59 Days 9

Drugs 1Residual 8

Residual 50




2 Drugs 10 Days

60 Animals- -

animals tier days tier treatments tiersource df source df source dfMean 1 Mean 1 Mean 1Animals 59 Days 9 Drugs 1

Residual 8

Residual 50




2 Drugs 10 Days

60 Animals- -

animals tier days tier treatments tiersource df source df source dfMean 1 Mean 1 Mean 1Animals 59 Days 9 Drugs 1

Residual 8Residual 50



Meatloaves: skeleton anova

2 Rosemary3 Irradiation

3 Blocks6 Meatloaves in B

3 Replicates12 Panellists in R6 Time-orders in R

t -QQ

- d⊥XXX

-

6 treatments 18 meatloaves 216 tastings

tastings tier meatloaves tier treatments tiersource df source df source dfMean 1 Mean 1 Mean 1Replicates 2 Blocks 2Panellists[Reps] 33Time-orders[Reps] 15P#T[Reps] 165 Meatloaves[B] 15 Rosemary 1

Irradiation 2R# I 2Residual 10

Residual 150

Structure balance when two arrows have the same end (1)

Q −→P ←−R

There are three possibilities.

Unrandomized-inclusive randomizationsThe outcome of the randomization Q −→P is known;the design for R −→P and method of randomizing R −→P bothuse knowledge of P BQ.Assume that Q is structure-balanced in relation to P ,and that R is structure-balanced in relation to P BQ.Use the decompostion (P BQ)BR.


Q −→P ←−R


Unrandomized-inclusive randomizationsThe outcome of the randomization Q −→P is known;the design for R −→P and method of randomizing R −→P bothuse knowledge of P BQ.

Assume that Q is structure-balanced in relation to P ,and that R is structure-balanced in relation to P BQ.Use the decompostion (P BQ)BR.


Q −→P ←−R


Unrandomized-inclusive randomizationsThe outcome of the randomization Q −→P is known;the design for R −→P and method of randomizing R −→P bothuse knowledge of P BQ.Assume that Q is structure-balanced in relation to P ,and that R is structure-balanced in relation to P BQ.Use the decompostion (P BQ)BR.

Superimposed Experiment in a Row-Column Design

5 Viruses XXXz@

@@@ 10 Rootstocks *

3 Blocks

10 Trees in B

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

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5 treatments

10 rootstocks 30 trees

trees tier rootstocks tier treatments tiersource df source df eff source dfMean 1 Mean 1 Mean 1Blocks 2Trees[Blocks] 27 Rootstocks 9 1

6 Viruses 4Residual 5

Residual 18 56 Viruses 4

Residual 14


Q −→P ←−R

Independent randomizations

The two structurally balanced designs are chosen so that,for all P except the Mean,either every PQ is zero or every PR is zero.Thus Q and R do not interfere with each other in P .

I Q is structure-balanced in relation to P BR;I R is structure-balanced in relation to P BQ;I (P BR)BQ = (P BQ)BR.


Q −→P ←−R



I Q is structure-balanced in relation to P BR;

I R is structure-balanced in relation to P BQ;I (P BR)BQ = (P BQ)BR.


Q −→P ←−R



I Q is structure-balanced in relation to P BR;I R is structure-balanced in relation to P BQ;

I (P BR)BQ = (P BQ)BR.


Q −→P ←−R



I Q is structure-balanced in relation to P BR;I R is structure-balanced in relation to P BQ;I (P BR)BQ = (P BQ)BR.


Q −→P ←−R

Coincident randomizationsThe two structurally balanced designs are chosen so that,for all P, Q, R

I either PQ is zero (after P, ignore Q)I or PR is zero (after P, ignore R)I or PBQ = P (after P, do Q before R)I or PBR = P (after P, do R before Q).

Then the decompositions P BQ and P BR are compatible in thesense that if A ∈P BQ and B ∈P BR then AB = BA. Hence

AB : A ∈P BQ, B ∈P BR

gives an orthogonal decomposition of VΩ.


Q −→P ←−R



Then the decompositions P BQ and P BR are compatible in thesense that if A ∈P BQ and B ∈P BR then AB = BA.

Hence




Q −→P ←−R



Then the decompositions P BQ and P BR are compatible in thesense that if A ∈P BQ and B ∈P BR then AB = BA. Hence



Summary

I We can extend this to three or more randomizations,so long as each one is structure-balanced.

I We get formulae for expected mean squares:each is a sum of terms from different tiers,with coefficients which depend on the efficiency factors.

I The anova tables help to compare different potential designs . . .I . . . and so are useful for design as well as for analysis.I For example, in a two stage experiment with partial confounding

of some treatment contrasts in each stage, is it better to partiallyconfound

I the same treatment contrasts with block terms at both stages (thusobtaining very poor precision on those contrasts)

I or some treatment contrasts with Stage 1 blocks and differenttreatment contrasts with Stage 2 blocks (thus decreasing thenumber of residual degrees of freedom)?

I In many cases, the models can be justified by the randomizationemployed.

Summary



I The anova tables help to compare different potential designs . . .

I . . . and so are useful for design as well as for analysis.I For example, in a two stage experiment with partial confounding





Summary



I The anova tables help to compare different potential designs . . .I . . . and so are useful for design as well as for analysis.

I For example, in a two stage experiment with partial confoundingof some treatment contrasts in each stage, is it better to partiallyconfound




Summary



I The anova tables help to compare different potential designs . . .I . . . and so are useful for design as well as for analysis.I For example, in a two stage experiment with partial confounding





References

1. C. J. Brien and R. A. Bailey: Multiple randomizations (withdiscussion). Journal of the Royal Statistical Society, Series B 68(2006), 571–609.

2. C. J. Brien and R. A. Bailey: Decomposition tables formultitiered experiments. Submitted for publication.

3. A. Gelman: Analysis of variance—Why it is more importantthan ever (with discussion). Annals of Statistics 33 (2005), 1–53.

4. A. T. James and G. N. Wilkinson: Factorization of the residualoperator and canonical decomposition of nonorthogonal factorsin the analysis of variance. Biometrika 58 (1971), 279–294.

5. J. A. Nelder: The analysis of randomized experiments withorthogonal block structure. I. Block structure and the nullanalysis of variance. II. Treatment structure and the generalanalysis of variance. Proceedings of the Royal Society ofLondon, Series A 283 (1965), 147–178.

6. R. F. White: Randomization and the analysis of variance.Biometrics, 31 (1975), 552–572.

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