STAT 506: Randomized complete block designs

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STAT 506: Randomized complete block designs Timothy Hanson Department of Statistics, University of South Carolina STAT 506: Introduction to Experimental Design 1 / 10

Randomized complete block designs Subjects placed into homogeneous groups, called blocks. All treatment combinations assigned randomly to subjects within blocks. Example: executives exposed to one of three methods (treatment, i = 1 utility method, i = 2 worry method, i = 3 comparison method) of quantifying maximum risk premium they would be willing to pay to avoid uncertainty in a business decision. Response is degree of confidence in the method on a scale from 0 (no confidence) to 20 (complete confidence). It is thought that confidence is related to age, so the subjects are blocked according to age (j = 1, 2, 3, 4, 5 from oldest to youngest). N = 15 subjects are recruited, with three subjects in each of the 5 age categories. Within each age category, the three subjects are randomly given one of the three treatments. 2 / 10

RCB designs, comments With thoughtful blocking, can provide more precise results than completely randomized design. There is only one replication for each pairing of treatment and block; need to assume no interaction between treatments and blocks to obtain estimate of σ 2. The blocking variable is observational, not experimental. Cannot infer causal relationship. Not a problem though...usually only care about treatments. Blocking works by making σ 2 smaller; block effects are variability in the data that normally go into σ 2 when they are ignored. 3 / 10

Model and inference One observation per block/treatment combination gives N = ab. Need to fit additive model to get SSE > 0 Y ij = µ + α i + β j + ɛ ij. Estimates obtained via LS as usual, Q(α, β) = a b (Y ij [µ + α i + β j ]) 2 i=1 j=1 minimized subject to α a = β b = 0 (SAS and R s default) or a i=1 α i = b j=1 β j = 0 (cfcdae s and Minitab s default). Your book uses the notation g treatments and r blocks; I m keeping a and b as before, because even though blocks are not treatments, the model is analyzed as an additive twoway ANOVA model. 4 / 10

ANOVA table Source SS df MS F p-value A SSA = b a i=1 (Ȳ i Ȳ ) 2 a 1 SSA MSA p a 1 MSE 1 B SSB = a b j=1 (Ȳ j Ȳ ) 2 b 1 SSB MSB p b 1 MSE 2 Error SSE = a bj=1 i=1 (Y ij Ȳ i Ȳ j + Ȳ ) 2 (a 1)(b 1) SSE (a 1)(b 1) Total SSE = a bj=1 i=1 (Y ij Ȳ )2 ab 1 Here, p 1 = P{F (a 1, (a 1)(b 1)) > MSA MSE } tests H 0 : α 1 = = α a = 0 (no blocking effect) and p 2 = P{F (b 1, (a 1)(b 1)) > MSB MSE } tests H 0 : β 1 = = β b = 0 (no treatment effect). These appear in SAS as Type III tests. If reject H 0 : β j = 0, then obtain inferences in treatment effects as usual, e.g. pairwise(f,treatment). 5 / 10

Diagnostics 1 Interaction (also called spaghetti or profile) plots of the y ij vs. treatment j, connected by block i are useful. Should be somewhat parallel if additive model is okay, but there is a lot of sampling variability here as ˆµ ij = y ij. 2 Standard R diagnostic panel: e ij vs. ŷ ij, normal probability plot of the {e ij }, etc. Can also look at e ij vs. either i or j, should show constant variance within blocks and treatments. 3 Tukey s test for additivity. 6 / 10

Tukey s test for additivity Reduced model is additive Y ij = µ + α i + β j + ɛ ij. Full model is Y ij = µ + α i + β j + Dα i β j + ɛ ij. This is more restrictive than using a general interaction (αβ) ij, leaves df to estimate error. a b i=1 j=1 ˆD = (Ȳ i Ȳ )(Ȳ j Ȳ ) a i=1 (Ȳ i Ȳ ) 2 b j=1 (Ȳ j Ȳ ). 2 SSAB = a i=1 j=1 b ˆD 2 (Ȳ i Ȳ ) 2 (Ȳ j Ȳ ) 2, and SSTO=SSA+SSB+SSAB*+SSE*. SSAB F = SSE F (1, ab a b), /(ab a b) if H 0 : D = 0 is true. I placed a script on the STAT 506 webpage to compute the p-value and D for you. 7 / 10

Confidence ratings, pp. 895-896 conf=c(1,2,7,6,12,5,8,9,13,14,8,14,16,18,17) method=factor(c(rep("utility",5),rep("worry",5),rep("comparison",5))) age=factor(rep(1:5,3)) d=data.frame(conf,method,age) par(mfrow=c(1,1)) with(d,interactplot(age,method,conf)) # parallel? with(d,interactplot(method,age,conf)) # parallel? source("http://people.stat.sc.edu/hansont/stat506/tukey.r") tukeys.add.test(d$conf,d$age,d$method) # accept additive model ok f=lm(conf~method+age,data=d) Anova(f,type=3) pairwise(f,method) lines(pairwise(f,method)) par(mfrow=c(2,2)) plot(f) 8 / 10

Sample size and power Russ Lenth s power applet uses a sd(a) = 1 a 1 i=1 α 2 i, sd(b) = 1 b 1 assuming a i=1 α i = b j=1 β j = 0. Same as sd(a) = 1 a 1 a ( µ i µ ) 2, sd(b) = i=1 b j=1 a βj 2, j=1 ( µ j µ ) 2, First one is SD[Block] and second SD[treatment] in the dialogue box. Also need an estimate of σ 2. The GUI will give you estimates of power to reject H 0 : α i = 0 (which we don t care about) and H 0 : β j = 0 (which we care about). 9 / 10

Example Say you are designing a RCBD where it σ 2, the b = 3 treatment effects are thought to be µ 1 = 15, µ 2 = 15, and µ 3 = 18. Then µ = 1 3 (15 + 15 + 18) = 16 and 1 sd(b) = 2 (1 + 1 + 4) = 3 1.73. Variability in blocking effects is thought to vary over a range of 6 units. Then, crudely sd(a) = range 4 = 6 4 = 1.5. Russ Lenth s dialogue box then shows us a = 10 blocks gives a power of about 90%. 10 / 10

Factorial treatments in a RCBD: dental pain Anesthesiologist studied effects of acupuncture and codeine on dental pain in N = 32 male subjects. Pain relief scores (higher = less pain) recorded; two factors A (placebo or codeine) and B (inactive vs. active acupuncture site). Subjects blocked on initial pain tolerance...why? library(cfcdae); library(car); library(lsmeans) pain=c(0.0,0.6,0.5,1.2,0.3,0.7,0.6,1.3,0.4,0.8,0.8,1.6,0.4,0.9,0.7,1.5, 0.6,1.5,1.0,1.9,0.9,1.6,1.4,2.3,1.0,1.7,1.8,2.1,1.2,1.6,1.7,2.4) tol=factor(rep(1:8,each=4)) # pain tolerance drug=factor(c(1,1,2,2,1,1,2,2,1,1,2,2,1,1,2,2,1,1,2,2,1,1,2,2,1,1,2,2,1,1,2,2)) acupuncture=factor(rep(1:2,16)) d=data.frame(pain,tol,drug,acupuncture) levels(d$drug)=c("placebo","codeine") levels(d$acupuncture)=c("inactive","active") with(d,interactplot(acupuncture:drug,tol,pain)) source("http://people.stat.sc.edu/hansont/stat506/tukey.r") tukeys.add.test(d$pain,d$tol,d$drug:d$acupuncture) f=lm(pain~tol+drug*acupuncture,data=d) Let s keep going... 11 / 10