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Design of Statistical Investigations

Stephen Senn

7. Orthogonal Designs

Two (plus) Blocking Factors

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Exp_5 (Again)

• This experiments was run in two sequences– Formoterol followed by salbutamol– Salbutamol followed by formoterol

• Suppose that values in second period tend to be higher or lower than those in first

• Differences formoterol -salbutamol will be affected one way or other depending on sequence

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Exp_5A Further Factor

• For each PEF reading we have accounted for– the patient it was measured under– the treatment the patient was on

• We have not accounted for the period

• We now look at an analysis that does

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Exp_5Fitting period

#ANOVA fitting treat and patient# Code factor for period period<-factor(c(rep(1:2,n)))#Perform ANOVAfit3<-aov(pef~patient+period+treat)summary(fit3)multicomp(fit3,focus="treat",error.type="cwe",method="lsd")

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Exp_5Comparison of 3 Models

> summary(fit1) Df Sum of Sq Mean Sq F Value Pr(F) treat 1 13388.5 13388.46 2.56853 0.1220902Residuals 24 125100.0 5212.50 > summary(fit2) Df Sum of Sq Mean Sq F Value Pr(F) patient 12 115213.5 9601.12 11.65357 0.000079348 treat 1 13388.5 13388.46 16.25053 0.001665618Residuals 12 9886.5 823.88 > summary(fit3) Df Sum of Sq Mean Sq F Value Pr(F) patient 12 115213.5 9601.12 12.79457 0.0000890 period 1 984.6 984.62 1.31211 0.2763229 treat 1 14035.9 14035.92 18.70444 0.0012048

Residuals 11 8254.5 750.41

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Exp_5: Three Fits95 % non-simultaneous confidence intervals for specified linear combinations, by the Fisher LSD method

critical point: 2.0639 intervals excluding 0 are flagged by '****' Estimate Std.Error Lower Bound Upper Bound salbutamol-formoterol -45.4 28.3 -104 13.1

critical point: 2.1788 Estimate Std.Error Lower Bound Upper Bound salbutamol-formoterol -45.4 11.3 -69.9 -20.9 ****

critical point: 2.201

Estimate Std.Error Lower Bound Upper Bound salbutamol-formoterol -46.6 10.8 -70.3 -22.9 ****

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Why the Different Estimate?

• Mean difference Salbutamol-Formoterol for 7 patients in seq 1 is -30.71

• Mean difference Formoterol-Salbutamol for 6 patients in seq 1 is -62.50

• The weighted average of these is -45.4

• The un-weighted average is -46.6

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Weighted

• The weighted average weights the mean difference in a sequence by the number of patients

• Thus the difference from each patient is weighted equally

• This makes sense if there is no period effect.• Why make a distinction between sequences if this

is the case?

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Un-weighted

• The un-weighted average weights means equally• Since there are more patients in the first sequence

their individual influence is down-weighted.• This makes no sense unless we regard the results

as not exchangeable by sequence• However, if there is a period effect then they are

not exchangeable

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How the Un-weighted Approach Adjust for Bias

• Suppose there is a difference between period one and two and this difference is additive and equal to .

• Every patient will have their treatment difference affected by

• Those in one sequence will have added.

• Those in other sequence will have subtracted.

• Averaged over the sequences this cancels out

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How the Variances are Affected

1 2

1 2

2 2

1 2

1 2

2 2 221 2

1 2 1 2

22 1 2

1 2 1 21 2

var , var

ˆ ,2

1ˆvar

4 4

1, ,

2 4 4

d dseq seq

seq seq

d d dd

d

d dn n

d dn n n

n n

n n n n n

n n n niff n n n n

n n n

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The Effect of Modelling In general the variance of a treatment contrast may be expressed as the product of two factors: q and 2.

For example, for the common two-sample t case

q = (1/n1 + 1/n2) and 2 is the variance of the original observations within treatment groups.

If further terms are added to the model the value of q will at best remain the same but in general will increase.

If these terms are explanatory, however, they will reduce the value of 2 .

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Variances in the Linear Model 12

1,1 1,2 1, 1

2,1 2,2, ,

1,1 1, 1

2,

ˆvar( ) ,

,

ˆvar( ) ,

k

i j j i

k k k

h h h

a a a

a aa a

a a

q q a

A A X X

A

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Efficient Experimentation and Modelling

• Two go hand in hand

• Choose explanatory factors for model– Will reduce variance, 2.

• Design experiment taking account of model– Minimise adverse effect on q.

• Randomise– Subject to constraints above

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Exp_5 and Efficiency

• In this case there were 14 patients initially• Split 7 and 7 by sequence• But one (patient 8) dropped out• Hence the design is unbalanced• Note that balance is not the be all and end all• 7 and 6 is better than 6 and 6, although 6 and 6 is

balanced

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Efficiency in General

• Balance in some sense produces efficiency

• Equal numbers per treatment etc– Provided all contrasts are of equal interest

• Treatments orthogonal to blocks

• Construct treatment plan if possible so that this happens

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Latin Squares

• Suppose that we have two blocking factors each at r levels.

• We also have r treatments and we wish to allocate these efficiently.

• How should we do this?

• One solution is to use a so-called Latin Square

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More Than one Blocking Factor- Examples

• Agricultural field trials– Rows and columns of a field

• Cross-over trials– Patients and periods

• Lab experiments– Technicians and days

• Fuel efficiency– Drivers and cars

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Latin Squares Examples

2 23 3

4 4

5 5

A B C DA B C

A B B D A CB C A

B A C A D BC A B

D C B A

A B C D E

B C E A D

C A D E B

D E B C A

E D A B C

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Latin Square5 x 5

Latin Square: 5 levels. B A C D E A E B C D C B D E A D C E A B E D A B C

Produced by SYSTAT

Each treatment given once to each row and once to each column

Completely orthogonal

No adverse effect on q

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Design Matrices and Orthogonality

• Such orthogonality in design is reflected in the “design matrices” used for coding for the linear model.

• This is illustrated on the next few slides for the case of a 4 x 4 Latin square.

• Four subjects are treated in four periods with four treatment.

• The coding of the design matrix in Mathcad is illustrated.

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Exp_7

• A four period cross-over in four subjects

• The four sequences chosen form a Latin square.

Treatment sequences for designSeq

1

4

3

2

2

3

1

4

3

2

4

1

4

1

2

3

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