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SJS SDI_61 Design of Statistical Investigations Stephen Senn 6. Orthogonal Designs Randomised Blocks II

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SJS SDI_62 Exp_5 Alternative Analyses We will now show three alternative analyses of example Exp_5 First two of these are equivalent to the analysis already done –First is only equivalent because there are only two treatments –Not equivalent for three or more treatments

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SJS SDI_63 Matched Pairs t-test Reduce data to a single difference d –per patient –between treatments Analyse these differences using a t-test for a single sample

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SJS SDI_64 Matched Pairs t-test (cont) Under H 0, the population mean difference d is zero Hence a significance test may be based on the statistic

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SJS SDI_65 Exp_5 Matched Pairs t-test

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SJS SDI_66 Exp_5 Calculations

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SJS SDI_67 Exp_5 Matched Pairs Using S-Plus #First split data into two columns #Matched by patient i<-sort.list(patient) data2<-cbind(patient=patient[i],pef=pef[i],treat=treat[i]) pefTREAT<-split(data2[,2],data2[,3]) #Perform matched pairs t-test t.test(pefTREAT$2", pefTREAT$1", alternative="two.sided", mu=0, paired=T, conf.level=.95)

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SJS SDI_68 Exp_5 S-Plus Output Paired t-Test data: pefTREAT$"2" and pefTREAT$"1" t = , df = 12, p-value = alternative hypothesis: true mean of differences is not equal to 0 95 percent confidence interval: sample estimates: mean of x - y

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SJS SDI_69 Exp_5 using a Linear Model Approach > fit3 <- lm(pef ~ patient + treat) > summary(fit3, corr = F) Call: lm(formula = pef ~ patient + treat) Residuals: Min 1Q Median 3Q Max e Coefficients: Value Std. Error t value Pr(>|t|) (Intercept) patient patient etc. treat

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SJS SDI_610

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SJS SDI_611 Exp_5 A Non-Parametric Approach Wilcoxon signed ranks test Calculate difference Ignore sign Rank Re-assign sign Calculate sum of negative (or positive) ranks

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SJS SDI_612 Exp_5 Signed Rank Calculations * = tie arbitrarily broken

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SJS SDI_613 Exp_5 Hypothesis Test Suppose H 0 true P = 1/2 any difference is positive or negative 2 13 = 8192 different possible patterns of - and + How many produce sum of negative ranks as low as that seen?

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SJS SDI_614 Possible Assignments of Negative Ranks With Equal or Lower Score No negative ranks: 1 case 1,2,3,4,5,6 only: 6 cases 1+(2,3,4,5): 4 cases 2+(3,4): 2 cases 1+2+3: 1 case Total = = 14 cases 14/2 13 = Two sided P-value = 2 x =0.0034

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SJS SDI_615 Exp_5 SPlus Output > wilcox.test(pefTREAT$"1", pefTREAT$"2", alternative = "two.sided", mu = 0, paired = T, exact = T, correct = T ) Wilcoxon signed-rank test cannot compute exact p-value with ties in: wil.sign.rank(dff, alternative, exact, correct) data: pefTREAT$"1" and pefTREAT$"2" signed-rank normal statistic with correction Z = , p-value = alternative hypothesis: true mu is not equal to 0

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SJS SDI_616 Why the Discrepancy? P = by hand, SPlus SPlus is using asymptotic approximation StatXact gives an accurate calculation StatXact output Exact Inference: One-sided p-value: Pr { Test Statistic.GE. Observed } = Pr { Test Statistic.EQ. Observed } = Two-sided p-value: Pr { | Test Statistic - Mean |.GE. | Observed - Mean | = Two-sided p-value: 2*One-Sided =

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SJS SDI_617 Orthogonal (See Marriott A Dictionary of Statistical Terms) Mathematical meaning is perpendicular –as in co-ordinate axes Statistical variates are orthogonal if independent Experimental design is orthogonal if certain variates or linear combinations are independent –rectangular arrays are orthogonal

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SJS SDI_618 Randomised Blocks and Orthogonality Randomised blocks are examples of orthogonal designs –Rectangular arrays –Balanced Consequences –Treatment sum of squares does not change as blocks are fitted –Design is efficient

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SJS SDI_619 Orthogonality and Regression

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SJS SDI_620 Orthogonality and Regression 2

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SJS SDI_621 Orthogonality and Regression 3 Here we have

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SJS SDI_622 Orthogonality Addition of patients has not increased variance multiplier for treatments Variance is as it would be had patients not been included patient and treat are orthogonal The factors are balanced

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SJS SDI_623 Exp_6 Another Example of a Two- Way Layout Classic attempt (Cushny & Peebles, 1905) to investigate optical isomers Subjects: eleven patients insane asylum Kalamazoo Outcome: hours of sleep gained Data subsequently analysed by Student (1908) in famous t-test paper.

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SJS SDI_624 The Cushny and Peebles Data B=L-Hyosciamine HBr C=L-Hyoscine HBr D=R-Hyoscine HBr

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SJS SDI_625 Features Treatments provide one dimension Patients provide the others Patients are the blocks Control is no treatment Main interest is difference between optical isomers Other treatment is positive control

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SJS SDI_626 Cushny and Peebles Data

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SJS SDI_627 Plotting Points Important to show three dimensions of data –Outcome –Treatment –Block (patients) This has been done here by using –Patient as a pseudo-dimension X –Outcome as the Y dimension –Treatment by colour and symbols

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SJS SDI_628 Points Clear difference between treatments and control Some suggestion of difference to active control Little suggestion of difference between isomers

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SJS SDI_629

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SJS SDI_630 Exp_6 SPlus Analysis #Analysis of first 10 patients #Cushny and peebles data patient<-factor(rep(c("1","2","3","4","5","6", "7","8","9","10"),4)) treat<-factor(c(rep("A",10),rep("B",10),rep("C",10), rep("D",10))) sleep<-c(0.6,3.0,4.7,5.5,6.2,3.2,2.5,2.8,1.1,2.9, 1.3,1.4,4.5,4.3,6.1,6.6,6.2,3.6,1.1,4.9, 2.5,3.8,5.8,5.6,6.1,7.6,8.0,4.4,5.7,6.3, 2.1,4.4,4.7,4.8,6.7,8.3,8.2,4.3,5.8,6.4) fit1<-aov(sleep~patient+treat) summary(fit1)

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SJS SDI_631 Exp_6 SPlus Output summary(fit1) Df Sum of Sq Mean Sq F Value Pr(F) patient treat Residuals

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SJS SDI_632

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SJS SDI_633 Calculation of Standard Errors Here we have Note that this applies as a consequence of orthogonality The multiplier 2/r is the same whether or not we fit patient in addition to treat

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SJS SDI_634 > multicomp(fit1, focus = "treat", error.type = "cwe", method = "lsd") 95 % non-simultaneous confidence intervals for specified linear combinations, by the Fisher LSD method critical point: response variable: sleep intervals excluding 0 are flagged by '****' Estimate Std.Error Lower Bound Upper Bound A-B A-C **** A-D **** B-C **** B-D **** C-D

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SJS SDI_635 Questions Perform a matched pairs analysis comparing B and C and ignoring data from A and D. Compare it to the pair-wise contrast for B & C obtained above? Why are the results not the same? What are the advantages and disadvantages of these approaches?

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