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SJS SDI_91 Design of Statistical Investigations Stephen Senn 9 Unbalanced Designs

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SJS SDI_92 Lack of Orthogonality So far we have been considering balanced designs –for example every treatment appears equally frequently in every block Sometimes we do not have such balance –by accident missing observations –by design

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SJS SDI_93 Consequences Some loss of efficiency –compared to some theoretical optimum CAUTION: this may not be obtainable in practice and may be why an unbalanced design has been chosen Complications in analysis –Sums of squares may depend on what other terms have been fitted so far only residual sum of squares has had this property

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SJS SDI_94 Exp_11 Senn 2002 Example 5.1 Cross-over trial in asthma Comparison of salbutamol, formoterol, placebo Trial run in six sequences Unequal numbers of patients per sequence

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SJS SDI_95 Exp_11 Sequences and Periods: Number of Observations I II III FSP SPF PFS FPS SFP PSF Patients by Sequence FSP SPF PFS FPS SFP PSF Note that although there are no missing data due to patients not having completed a sequence, the numbers of patients are unbalanced by sequence

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SJS SDI_97 Exp_11 Data 1 FSP FSP FSP FSP FSP SPF SPF SPF PFS PFS PFS PFS PFS PFS FPS FPS FPS FPS FPS FPS SFP SFP SFP SFP SFP PSF PSF PSF PSF PSF

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SJS SDI_98 Exp_11 Not Fitting Period > fit1 <- lm(fev1 ~ patient + treat) > summary(fit1, corr = F) Coefficients: Value Std. Error t value Pr(>|t|) …... treatS treatP Residual standard error: on 58 degrees of freedom Multiple R-Squared:

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SJS SDI_99 Exp_11 Fitting Period > fit2 <- update(fit1,. ~. + period) summary(fit2, corr = F) Call: lm(formula = fev1 ~ patient + treat + period) Coefficients: Value Std. Error t value Pr(>|t|) …... treatS treatP periodII periodIII Residual standard error: on 56 degrees of freedom Multiple R-Squared:

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SJS SDI_910 Exp_11 ANOVA > aov.1 <- aov(fev1 ~ patient + treat) > summary(aov.1) Df Sum of Sq Mean Sq F Value Pr(F) patient e-009 treat e+000 Residuals > aov.2 <- aov(fev1 ~ patient + period) > summary(aov.2) Df Sum of Sq Mean Sq F Value Pr(F) patient period Residuals >

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SJS SDI_911 Exp_11 ANOVA > aov.3 <- aov(fev1 ~ patient + period + treat) >summary(aov.3) Df Sum of Sq Mean Sq F Value Pr(F) patient period treat Residuals > aov.4 <- aov(fev1 ~ patient + treat + period) > summary(aov.4) Df Sum of Sq Mean Sq F Value Pr(F) patient treat period Residuals

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SJS SDI_912 Exp_11 ANOVA > ssType3(aov.3) Type III Sum of Squares Df Sum of Sq Mean Sq F Value Pr(F) patient period treat Residuals > ssType3(aov.4) Type III Sum of Squares Df Sum of Sq Mean Sq F Value Pr(F) patient treat period Residuals

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SJS SDI_913 Exp_11 Standard Errors Period effect not fitted Period effect fitted

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SJS SDI_914 Incomplete Blocks These designs arise when the number of treatments exceeds the number of units in a typical block Not possible to have every treatment in every block Each block receives a subset of the units These to be chosen in a sensible manner

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SJS SDI_915 Exp_12 Senn 2002 Example 7.2 Placebo (P) controlled cross-over design to compare two doses of formoterol –F12 : 12 mg in a single puff –F24: 24 mg in a single puff Patients could only be treated in two periods Incomplete blocks design 24 Patients to be allocated in equal numbers to each of six sequences

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SJS SDI_916 EXP_12 Sequences used P F12 F12 P P F24 F24 P F12 F24 F24 F12 The basic design is said to be that of balanced incomplete blocks. In this context balance has a special meaning: each pair of possible treatments appears equally often in every block Because this is a cross-over design and we are worried about period effects the design is also balanced by period (order) but that is another matter

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SJS SDI_917 EXP_12 The sad reality Two incorrect packs were picked up. –One was for correct sequence –One was not Numbers of Observations Period Sequence 1 2 F12F F12P 5 5 F24F F24P 4 4 PF PF F12 F24 has one fewer patient F12 P has one more

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SJS SDI_918 EXP_12 The Data 6 F12F F12F F12F F12P F12P F12P F12P F12P F24F F24F F24F F24F F24P F24P F24P F24P PF PF PF PF PF PF PF PF

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SJS SDI_919

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SJS SDI_920 Exp_12 Analysis 1 > fit1 <- lm(FEV1 ~ patient + period + treat) > summary(fit1, corr = F) Call: lm(formula = FEV1 ~ patient + period + treat)... Coefficients: Value Std. Error t value Pr(>|t|) (Intercept) patient patient period treatF treatP

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SJS SDI_921 Exp_12 Analysis 2 > aov1 <- aov(FEV1 ~ patient + period + treat) > summary(aov1) Df Sum of Sq Mean Sq F Value Pr(F) patient period treat Residuals > ssType3(aov1) Type III Sum of Squares Df Sum of Sq Mean Sq F Value Pr(F) patient period treat Residuals

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SJS SDI_922 Exp_12 Analysis 3 > aov2 <- aov(FEV1 ~ patient + treat + period) > summary(aov2) Df Sum of Sq Mean Sq F Value Pr(F) patient treat period Residuals > ssType3(aov2) Type III Sum of Squares Df Sum of Sq Mean Sq F Value Pr(F) patient treat period Residuals

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SJS SDI_923 Standard Errors Consider the standard error of the contrast F24 versus F12 This is given as How could this be calculated? There are two sequences in which these drugs could be compared –F12F24 with 3 patients –F24F12 with 4 patients

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SJS SDI_924 However Thus the standard error we have from fitting the regression model is actually lower than that produced by a naïve argument.

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SJS SDI_925 Questions Exp_12 Why is the SE produced by the regression analysis lower than that produced by using the pooled MSE and the direct comparison of the means? What would the treatment estimate be if this naïve approach was used? How does it compare to that produced? What further information is the regression approach taking into account?

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SJS SDI_926 Block Size and Comparisons Suppose that the block size is k (there are k units per block) and that there are b blocks in total and bk units in total Suppose that we have v treatments and r replicates. There must also be rv units in total Hence rv = bk = N. Each block permits k(k-1)/2 comparisons. There are bk(k-1)/2 in total. However, there are v(v-1)/2 possible pair-wise comparisons.

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SJS SDI_927 Block Size and Comparisons Let be the average number of repetitions of the pair-wise comparisons in the design. Hence Obviously unless this is an integer, it will not be possible to balance the blocks. If v-1 is a multiple of k-1 then it becomes particularly easy to balance the blocks

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SJS SDI_928 Exp_13 It was desired to compare three doses each of two formulations of formoterol to placeo –ISF 6, ISF12, ISF24 –MTA6, MTA12,MTA24 –Placebo There are thus seven treatments Maximum number of acceptable periods was deemed to be five

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SJS SDI_929 Exp_13 Possible solution Since 7-1 = 6 is twice 4-1 = 3 use design in 4 periods If seven sequences are used it will also be possible to make the treatments uniform on the periods There are (7 6)/2 = 21 possible pair-wise comparisons of treatments Each patient provides (4 3)/2 = 6 possible comparison There are 7 6 = 42 = 2 21 such comparisons per set of seven sequences

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SJS SDI_930 A Balanced Design Uniform on the Periods for 7 treatments in 4 periods

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SJS SDI_931 Questions Exp_13 Exp_13 was in fact run using five periods and 21 sequences Check that such a design can be balanced An alternative considered was to use five periods and seven sequences Show that such a design cannot be balanced Why might it be preferable to the design in four periods and seven sequences?

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