Randomized Complete Block Design (RCBD)RCBD: examples and modelEstimates, ANOVA table and f-testsChecking assumptionsRCBD with subsampling: ModelLatin square designDesign and modelANOVA tableMultiple Latin squaresOutline1Randomized Complete Block Design (RCBD)RCBD: examples and modelEstimates, ANOVA table and f-testsChecking assumptionsRCBD with subsampling: Model2Latin square designDesign and modelANOVA tableMultiple Latin squaresRandomized Complete Block Design (RCBD)Suppose a slope difference in the field is anticipated. We blockthe field by elevation into 4 rows and assign irrigation treatmentrandomly within each block (row). Ex:> sample(c("A","B","C","D"))[1] "D" "A" "B" "C"B A C DD A B CC B D AA C D BRCBD modelresponse ∼ treatment + block + errorHere block=row, and error=variation at theplotlevel.no treatment:block interaction.Treatments and blocks are crossed factors.RCBD modelModel: response ∼ treatment + block + errorYi= µ + αj[i]+ βk[i]+ eiwith ei∼ iid N (0, σ2e)µ = population mean across treatments,αj= deviation of irrigation method j from the mean,constrained toPaj=1αj= 0. Fixed treatment effects.βk= fixed block effect (categorical), k = 1, . . . , bconstrained toPbk=1βk= 0. or random effect withβk∼ iid N (0, σ2β).Soil moisture: a = 4, b = 4. Total of ab = 16 observations.Seedling emergence exampleCompare 5 seed disinfectant treatments using RCBD with 4blocks. In each plot, 100 seeds were planted.Response: # plants that emerged in each plot.BlockTreatment 1 2 3 4 Mean (¯yj·)Control 86 90 88 87 87.75Arasan 98 94 93 89 93.50Spergon 96 90 91 92 92.25Semesan 97 95 91 9293.75Fermate 91 93 95 95 93.50Mean (¯y·k) 93.6 92.4 91.6 91.0¯y··= 92.15Model:Yi= µ + αj[i]+ βk[i]+ eiwith ei∼ iid N (0, σ2e)αj: seed treatment effect, βk: block effect.Seedling emergence examplePopulation mean for trt j and block k: µjk= µ + αj+ βkPredicted means, or fitted values: ˆµjk= ˆµ + ˆαj+ˆβk. How?BlockTrt 1 2 · · · b ¯µj·1 µ + α1+ β1µ + α1+ β2µ + α1+ βbµ + α12 µ + α2+ β1µ + α2+ β2µ + α2+ βbµ + α2· · ·· · · · · ·a µ + αa+ β1µ + αa+ β2µ + αa+ βbµ + αa¯µ·kµ + β1µ + β2µ + βbµEstimated coefficients (balance: 1 obs/trt/block):ˆµ =¯y··ˆαj=¯yj·−¯y··ˆβk=¯y·k−¯y··if fixed block effectsANOVA table with RCBDSource df SS MS IE(MS)Block b − 1 SSBlk MSBlk σ2e+ aPbk=1β2kb−1(fixed)σ2e+ aσ2β(random)f testTrt a − 1 SSTrt MSTrt σ2e+ bPaj=1α2ja−1f testError (b − 1)(a − 1) SSErr MSErr σ2eTotal ab − 1 SSTotSSBlk: involves (¯y.k− y..)2over all blocks kSSTrt: involves (¯yj.− y..)2over all treatments jSSErr: involves (yij− ˆµij)2from all residualsSSTot: involves (yij−¯y..)2Why not include an interaction Block:Treatment in the model?It would take(b − 1)(a − 1)df and there would remain0df forMSErr.Debate: fixed vs. random block effectsEx: does it make sense to view the 4 specific rows blockedby elevation as randomly selected from a largerpopulation?Ex: 4 dosages of a new drug are randomly assigned to 4mice in each of the 20 litters: RCBD with a = 4 dosagetreatments and b = 20 litters, for a total of ab = 80observations. Here, blocks (litters) can be considered asrandom samples from the population of all litters that couldbe used for the study.In RCBD, the choice fixed vs. random blocks does notaffect the testing of the trt effect. In more complicateddesigns, it could.If we can use the simpler analysis with fixed effects, it isokay to use it!F test for block variabilityEstimation, if random block effects: ˆσ2β=MSBlk − MSErraANOVA tableTest for the block effects (uncommon):F =MSBlkMSErron df = b − 1, (b − 1)(a − 1)but even if there appears to be non-significant differencesbetween blocks, we would keep blocks into the model, to reflectthe randomization procedure.Other commonly used blocking factors: observers, time, farm,stall arrangement etc. The general guideline to choose blocksis scientific knowledge.F-tests for treatment effectsTo test H0: αj= 0 for all j (i.e., no treatment effect), use the factthat under H0,F =MSTrtMSErr∼ Fa−1, (b−1)(a−1)ANOVA tableSource df SS MS F p-valueTreatments 4 102.30 25.58 3.598 0.038Blocks 3 18.95 6.32 0.889 0.47Error 12 85.30 7.11Total 19 206.55ANOVA in R with RCBD> emerge = read.table("seedEmergence.txt", header=T)> str(emerge)’data.frame’: 20 obs. of 3 variables:$ treatment: Factor w/ 5 levels "Arasan","Control",..: 2 1 5 4 3 2 1 5 4 3 ...$ block : int 1 1 1 1 1 2 2 2 2 2 ...$ emergence: int 86 98 96 97 91 90 94 90 95 93 ...> emerge$block = factor(emerge$block)Make sure blocks are treated as categorical! They should beassociated with b − 1 = 3 df in the ANOVA table or LRT.ANOVA in R with RCBD> fit.lm = lm( emergence ˜ treatment + block, data=emerge)> anova(fit.lm)Df Sum Sq Mean Sq F value Pr(>F)treatment 4 102.300 25.575 3.5979 0.03775*block 3 18.950 6.317 0.8886 0.47480Residuals 12 85.300 7.108> fit.lm = lm( emergence ˜ block + treatment, data=emerge)> anova(fit.lm)Df Sum Sq Mean Sq F value Pr(>F)block 3 18.95 6.3167 0.8886 0.47480treatment 4 102.30 25.5750 3.5979 0.03775*Residuals 12 85.30 7.1083> drop1(fit.lm)Single term deletionsDf Sum of Sq RSS AIC F value Pr(F)<none> 85.30 45.009block 3 18.95 104.25 43.021 0.8886 0.47480treatment 4 102.30 187.60 52.772 3.5979 0.03775*ANOVA in R with RCBDHere, the output of anova() does not depend on the orderin which treatment and block are given.Here, type I sums of squares (sequential, anova) and typeIII sums of squares (drop1) are equal.Because the design is balanced.Significant effect of treatmentsNon-significant differences between blocks, but still keepblocks in the model.Note: aov() could have been used in place of lm().Model assumptionsThe model assumes:1Errors eiare independent, have homogeneous variance,and a normal distribution.2Additivity: means are µ + αj+ βk, i.e. the trt differencesare the same for every block and the block differences arethe same for every trt. No interaction.Extra assumption for the ANOVA table and f-test: balance.In particular, they assumecompleteness: each trt appears atleast once in each block. That is n ≥ 1 per trt and block.Example of an incomplete block design for b = 4, a = 4:B A CD A BC B DA C DModel diagnosticsCheck that residuals (ri= yi−ˆyi):approximately have a normal distribution,no pattern (trend, unequal variance) across blocks.no pattern (trend, unequal variance) across treatments.plot(fit.lm)88 90 92