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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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