WIXLIB

yield90yield90TWO-FACTOR FACTORIAL WITH BLOCKSTWO-FACTOR FACTORIAL WITH BLOCKSThe GLM ProcedureThe GLM ProcedureDistribution of yield88Distribution of yield889494928692869090yieldyieldTWO-FACTOR FACTORIAL WITH BLOCKS8888The GLM Procedure8684250868484260 of eld90yieldMeanStd Dev2506 88.51666672.098014932606 88.30000003.44325427Level of270 6pressN89.5666667Level oftempNyield1.74661578yieldL6 85.78333331.11250468MeanStd Dev2506 88.51666672.09801493H6 The91.53333331.74661578GLM Procedure2606 88.30000003.44325427L6 85.78333332706 89.56666672.8380744694MDistributionof yield6 89.06666671.07455417TWO-FACTOR FACTORIAL WITH BLOCKSThe GLM ProcedureThe GLM ProcedureDistribution of yield949292yield90yield9088H 2608488yield9086H 2501.11250468TWO-FACTOR FACTORIAL WITH BLOCKS9284Std Dev6 91.5333333M6 89.0666667 1.07455417Level ofTWO-FACTORFACTORIAL WITH BLOCKStempNMeanStd Dev2.838074469486MeanHDistribution of yield88MMtempyieldLevel ofpressNLtempH 270H 260H 250H 270L 250L 250L 26086LL 260270M25088L 270M 270M 260M 25084M 2601M 270temp*pressyieldLevel of Level oftemppressNHyieldMeanStd DevH2502 90.40000001.83847763H2602 91.70000002.12132034H270Level of Levelof 2 92.5000000L2502 86.2000000temppressNL2602 84.6000000L250MM2702 86.55000002502 88.95000002602 88.6000000Level ofdayNyield86MeanStd Dev19 87.94444442.3016902029 28141.06066017Std Dev2 388291.70000001.83847763HM260H2702 92.50000001.69705627L2502 86.20000000.14142136L2602 84.60000000.84852814L2702 86.55000001.06066017M2502 88.95000000.63639610M2602 88.60000001.83847763M2702 12dayyieldLevel ofdayNMeanStd Dev19 87.94444442.3016902029 89.64444442.99337231

TWO-FACTOR FACTORIAL WITH BLOCKSThe GLM ProcedureFit Diagnostics for 20-1-2-1.0848688909284Predicted yield0.0Cook's D0.5Residual0.7Predicted .092.505Predicted ualTWO-FACTOR FACTORIALWITH BLOCKS30ANALYSIS OF MODEL WITHTWO-FACTORBLOCK INTERACTIONS4220010Observations18Parameters10Error DF8MSE0.5313R-Square0.9666Adj R-Square 0.9289The GLM Procedure-2Dependent Variable: yield-40-1.6 -0.800.81.6SourceResidualDFModel0.0 Sum0.4 0.8of 0.0 0.4 0.8SquaresProportionMeanLess Square F Value Pr F13 126.3861111Error4Corrected Total9.72200850.683333356.910.00070.170833317 127.0694444R-Square Coeff Var Root MSE yield Mean0.994622Source0.4654790.41332088.79444DF Type III SS Mean Square F Value Pr Ftemp2 99.85444444press2temp*press449.92722222292.26 3055566.520.0484FACTORIALWITH BLOCKSdayTWO-FACTOR1 13.0050000013.0050000076.13 0.0010ANALYSIS day*tempOF MODEL WITHTWO-FACTORBLOCKINTERACTIONS2 2.543333331.271666677.44 0.0448day*press21.023333330.51166667TheGLM Procedure3.00SourceType III Expected Mean SquaretempVar(Error) 3 Var(day*temp) Q(temp,temp*press)pressVar(Error) 3 Var(day*press) Q(press,temp*press)0.1603temp*press Var(Error) Q(temp*press)dayVar(Error) 3 Var(day*press) 3 Var(day*temp) 9 Var(day)day*tempVar(Error) 3 Var(day*temp)day*pressVar(Error) 3 Var(day*press)171

TWO-FACTOR FACTORIAL WITH BLOCKSANALYSIS OF MODEL WITH TWO-FACTOR BLOCK INTERACTIONSThe GLM ProcedureTests of Hypotheses for Mixed Model Analysis of VarianceDependent Variable: yieldSourceDF Type III SS Mean Square* tempError: MS(day*temp)299.85444449.92722222.5433331.271667F ValuePr F39.260.0248F ValuePr F5.380.1567* This test assumes one or more other fixed effects are zero.SourceDF Type III SS Mean Square* pressError: MS(day*press)25.5077782.75388921.0233330.511667* This test assumes one or more other fixed effects are zero.SourceDF Type III SS Mean Square F Value Pr .000.1603Error: MS(Error)40.6833330.170833SourcedayErrorDF Type III SS Mean Square F Error: MS(day*temp) MS(day*press) - MS(Error)SAS CodeDM ’LOG; CLEAR; OUT; CLEAR;’;ODS GRAPHICS ON;ODS PRINTER PDF file ’C:\COURSES\ST541\TWOBLOCK.PDF’;OPTIONS NODATE *** TWO-FACTOR FACTORIAL WITH BLOCKS ***;****************************************;DATA in;DO day 1 TO 2;DO temp ’L’ , ’M’ , ’H’;DO press 250 TO 270 BY 10;INPUT yield @@; OUTPUT;END; END; END;CARDS;86.3 84.0 85.8 88.5 87.3 89.0 89.1 90.2 91.386.1 85.2 87.3 89.4 89.9 90.3 91.7 93.2 93.7PROC GLM DATA in PLOTS (ALL);CLASS day temp press;MODEL yield temp press day / SS3;MEANS press temp day;172Pr F0.0728

RANDOM day / TEST;OUTPUT OUT diag P pred R resid;TITLE ’TWO-FACTOR FACTORIAL WITH BLOCKS’;PROC UNIVARIATE DATA diag NORMAL ;VAR resid;*** MODEL INCLUDING BLOCK INTERACTIONS ***;PROC GLM DATA in;CLASS day temp press;MODEL yield temp press day@2 / SS3;RANDOM day day*temp day*press / TEST;TITLE2 ’ANALYSIS OF MODEL WITH TWO-FACTOR BLOCK INTERACTIONS’;RUN;4.17Expected Means Squares (EMS) Algorithm for Balanced Designs (Supplemental) Classify each effect as either a fixed effect or a random effect. A fixed effect is represented by the EMS componentPaPτ2(fixed effect)2For example: i 1 id.f. for the fixed effecta 1 A random effect is represented by a variance component with its associated subscripts. For example:στ2 . An interaction containing at least one random effect is considered random and is represented by avariance component. For example:– If factor A is random and factor B is random, then the A*B interaction is random. Theassociated variance component is στ2β .– If factor A is fixed and factor B is random, then the A*B interaction is random. The associatedvariance component is στ2β .– If factor A is fixed with a levels and factor B is fixed with b levels, then the A*B interaction isPa Pb2i 1 τ βiji 1fixed. The associated EMS component is.(a 1)(b 1)STEP 1: Prepare the EMS Table1. Set up a row for each model effect that has one or more subscripts. Write the error term ij.m as a‘nested’ effect m(ij.) . We say that m(ij.) is the mth replicate for treatment combination ij . . . Wewill use parentheses () to indicate which effects are nested. Nested effects will be discussed in detaillater in the course.2. Set up columns for each subscript. Above each subscript, write the number of levels associated withthe factor and whether the factor is fixed (F) or random (R).3. For each row effect, subscripts are divided into three classes: live, dead, or absent. A subscript is live if it is present and is not in parentheses. A subscript is dead if it is present and is in parentheses. A subscript is absent if it is not present.Example of Step 1: Consider the two-factor factorial design. Factor A is fixed with a levels and factor Bis random with b levels. n replicates were taken for each of the ab combinations of the levels of A and B.We will be using the mixed model yijk µ αi βj αβij ijk .173

Effectαiβjαβij k(ij)FaiComponentP 2αi /(a 1)σβ22σαβσ2RbjRnkEMSSTEP 2: Filling in the rows of the EMS Table:1. Write 1 in each column containing dead subscripts.Effectαiβjαβij k(ij)ComponentP 2αi /(a 1)σβ22σαβσ2FaiRbj11RnkEMS2. If any row subscript corresponds to a random factor (R), then write 1 in all columns with a matchingsubscript. Otherwise, write 0 in all columns with a matching subscript.Effectαiβjαβij k(ij)ComponentP 2αi /(a 1)σβ22σαβσ2Fai011RbjRnk1111EMS3. For the remaining missing values, enter the number of factor levels for that column.Effectαiβjαβij k(ij)ComponentP 2αi /(a 1)σβ22σαβσ2Fai0a11Rbjb111Rnknnn1EMSSTEP 3: Obtaining the EMS1. Consider all rows containing all of the subscripts in the row effect.2. For each row, delete all columns containing live subscripts. Take the product of the remaining valuesand the component.3. EMS the sum of these products.In the example, for αi , there is only one subscript i. In the model αi , αβij , and k(ij) all contain subscripti. We delete the i column and generate the three productsPbn αi22(a)from the αi row(b) nσαβfrom the αβij row(c) σ 2 from the k(ij) row.a 1174

EffectαiComponentP 2αi /(a 1)βjαβij k(ij)σβ22σαβσ2FaiRbjRnk0bna11111nn1EMS2σ 2nσαβPbn αi2 a 1In the example, for βj , there is only one subscript j. In the model βj , αβij , and k(ij) all contain subscriptj. We delete the j column and generate the three products(a) anσβ2 from the βj row,Effect2(b) nσαβfrom the αβij row,αiComponentP 2αi /(a 1)βjαβij k(ij)σβ22σαβσ2(c) σ 2 from the k(ij) row.FaiRbjRnkEMS0bn2σ 2 nσαβ a11111nn12σ 2 nσαβPbn αi2a 1 anσβ2In the example, for αβij , there are two subscripts i and j. In the model αβij and k(ij) all contain subscriptsi and j. We delete the i and j columns and generate the two products2(a) nσαβfrom the αβij rowEffectandαiComponentP 2αi /(a 1)βjαβij k(ij)σβ22σαβσ2(b) σ 2 from the k(ij) row.FaiRbjRnkEMS20bnσ a11111nn1σ2 σ2 2nσαβ2nσαβ2nσαβPbn αi2 a 1 anσβ2Finally, for k(ij) , there are three subscripts i, j, and k. In the model only k(ij) contains all three subscripts.We delete the i, j, and k columns. This leaves only σ 2 .EffectαiComponentP 2αi /(a 1)βjαβij k(ij)σβ22σαβσ2FaiRbjRnkEMS20bnσ a11111nn1σ2 σ2 σ21752nσαβ2nσαβ2nσαβPbn αi2 a 1 anσβ2

Next consider the three-factor factorial design for which all three factors are random. The number of factorlevels are a, b, and c with n replicates taken for each three-factor treatment combination. The full threefactor-factorial model isyijkl µ τi βj γk τ βij τ γik βγjk τ βγijk ijklFrom Step 1 and Step 2, we haveEffectτiβjγkτ βijτ γikβγjkτ βγijk l(ijk)Effectτiβjγkτ βijτ γikβγjkτ βγijk l(ijk)Effectτiβjγkτ βijτ γikβγjkτ βγijk �2στ 2βστ2γ2σβγ2στ 1c11111Rnlnnnnnnn1EMSEMSσ 2 nστ2βγσ 2 nστ2βγσ 2 nστ2βγσ 2 nστ2βγσ 2 nστ2βγσ 2 nστ2βγσ 2 nστ2βγσ2176 bnστ2γ cnστ2β bcnστ22 anσβγ cnστ2β acnσβ22 anσβγ bnστ2γ abnσγ22 cnστ β bnστ2γ2 anσβγ

This is the simplest example when no ratio of EMS exists for defining F -statistics to test certain nullhypotheses. In all of our previous examples, when H0 is trueE(numerator MS) E(denominator MS). For this example, We have exact tests for2 0,(1) H0 : σβγ(2) H0 : στ2γ 0, and(3) H0 : στ2β 0if we use M Sτ βγ as the denominator of the F -statistic. We also have an exact test for(4) H0 : στ2βγ 0 if we use M SE as the denominator of the F -statistic. We do not have exact tests for(5) H0 : στ2 0,(6) H0 : σβ2 0, and(7) H0 : σγ2 0because when (5), (6), or (7) is true, there is no matching EMS. Thus, there is no exact F -test.4.18Approximate F-Tests If we want to make inferences about effects for which there is no exact F -test, we can use approximate F -tests, a procedure attributed to Sattherthwaite (1946). In this procedure, the goal is to find linear combinations of EMS that are equal when H0 is true.That is, findM S 0 M Sr · · · M S sandM S 00 M Su · · · M Svsuch the E(M S 0 ) E(M S 00 ) equals the effect of interest.M S0 The test statistic would be F which is approximately distributed as F (p, q) where the degreesM S 00of freedom p and q arep (M Sr · · · M Ss )2M Sr2 /fr · · · M Ss2 /fsandq (M Su · · · M Sv )2.M Su2 /fu · · · M Sv2 /fvfi is the degrees of freedom associated with mean square M Si . It is unlikely that p and q are integers. Let the A, B, and C be the factors in a three-factor factorial design, and A, B, and C are random.Consider the following:M S 0 M SA M SABCandThen, E(M S 0 ) E(M S 00 ) 177M S 00 M SAB M SAC

Therefore, to test H0 : στ2 0, we can use F p (M SA M SABC )22M SAa 1 M S0with d.f.M S 00and2M SABC(a 1)(b 1)(c 1)q (M SAB M SAC )22M SAB(a 1)(b 1) 2M SAC(a 1)(c 1). Note that when H0 : στ2 0 is true, E(M S 0 ) E(M S 00 ). Note also the choice of M S 0 and M S 00 is not unique. For example, suppose we defineM S 0 M SAM S 00 M SAB M SAC M SABCandThen,E(M S 0 ) E(M S 00 ) σ 2 nστ2βγ bnστ2γ cnστ2β bcnστ2 σ 2 nστ2βγ bnστ2γ cnστ2β bcnστ2 Therefore, to test H0 : στ2 0, we can use F p (M SA )22M SAa 1 a 1andM S0with d.f.M S 00q (M SAB M SAC M SABC )22M SAB(a 1)(b 1) 2M SAC(a 1)(c 1) 2M SABC(a 1)(b 1)(c 1). Again, note that when H0 : στ2 0 is true, E(M S 0 ) E(M S 00 ).4.18.1Determining Exact and Approximate F -tests using SAS Prior to running an experiment that contains random effects, I recommend that you determine whatexact tests and approximate F -tests can be performed. Once you have a design:1. Generate random responses using a random number generator.2. Analyze the random responses in SAS using the RANDOM statement with the /TEST opion.3. Examine the output. For each test of a model effect, see if the denominator of the F -statisticsis a single mean square (exact F -test) or is a linear combination of means squares (approximateF -test). Consider the three factor factorial example and factors A, B, and C are random with a 3, b 2,c 2, and n 3.– There are exact tests for the A B, A C, B C, and A B C effects.– There are only approximate tests for the A, B, and C effects.DATA in;DO A 1 to 3;DO B 1 to 2;DO C 1 to 2;DO rep 1 TO 3;MU A B C;Y MU RANNOR(542057); OUTPUT;END; END; END; END;PROC GLM DATA in;CLASS A B C;MODEL Y A B C / SS3;RANDOM A B C / TEST;TITLE ’THE RANDOM EFFECTS MODEL WITH A, B, AND C RANDOM’;RUN;178

THE RANDOM EFFECTS MODEL WITH A, B, AND C RANDOMThe GLM ProcedureSource Type III Expected Mean SquareAVar(Error) 3 Var(A*B*C) 6 Var(A*C) 6 Var(A*B) 12 Var(A)BVar(Error) 3 Var(A*B*C) 9 Var(B*C) 6 Var(A*B) 18 Var(B)A*BVar(Error) 3 Var(A*B*C) 6 Var(A*B)CVar(Error) 3 Var(A*B*C) 9 Var(B*C) 6 Var(A*C) 18 Var(C)A*CVar(Error) 3 Var(A*B*C) 6 Var(A*C)B*C RANDOMVar(Error) EFFECTS 3 Var(A*B*C) 9 Var(B*C)THEMODELWITH A, B, AND C RANDOMA*B*C Var(Error) 3 Var(A*B*C)The GLM ProcedureTests of Hypotheses for Random Model Analysis of VarianceDependent Variable: YSourceDF Type III SS Mean 78F ValuePr F15.290.6679Error: MS(A*B) MS(A*C) - MS(A*B*C) 22E-17*MS(Error)SourceDF Type III SS Mean F ValuePr F3.990.3872Error: MS(A*B) MS(B*C) - MS(A*B*C) 11E-17*MS(Error)SourceDF Type III SS Mean Square F Value Pr 80.510.6606B*C12.6260132.6260131.640.3287Error: MS(A*B*C)23.2020691.601035SourceDF Type III SS Mean 377F ValuePr F11.570.4048Error: MS(A*C) MS(B*C) - MS(A*B*C) 11E-17*MS(Error)SourceA*B*CError: MS(Error)DF Type III SS Mean Square F Value Pr F23.2020691.6010352416.3618930.6817461792.350.1171

Consider the three factor factorial example and factors A and B are fixed, and factor C is randomwith a 3, b 2, c 2, and n 3.– There are exact tests for the A, B, A B, A C, B C, and A B C effects.– There is an approximate test for the C effect. The ‘Q’ components in the EMS correspond to fixed effects.PROC GLM DATA in;CLASS A B C;MODEL Y A B C / SS3;RANDOM C A*C B*C A*B*C / TEST;TITLE ’THE RANDOM EFFECTS MODEL WITH A, B FIXED AND C RANDOM’;THE RANDOM EFFECTS MODEL WITH A, B FIXED AND C RANDOMThe GLM ProcedureSource Type III Expected Mean SquareAVar(Error) 3 Var(A*B*C) 6 Var(A*C) Q(A,A*B)BVar(Error) 3 Var(A*B*C) 9 Var(B*C) Q(B,A*B)A*BVar(Error) 3 Var(A*B*C) Q(A*B)CVar(Error) 3 Var(A*B*C) 9 Var(B*C) 6 Var(A*C) 18 Var(C)A*CVar(Error) 3 Var(A*B*C) 6 Var(A*C)B*CVar(Error) 3 Var(A*B*C)9 Var(B*C)THE RANDOMEFFECTSMODEL WITHA, B FIXED AND C RANDOMA*B*C Var(Error) 3 Var(A*B*C)The GLM ProcedureTests of Hypotheses for Mixed Model Analysis of VarianceDependent Variable: YSourceDF Type III SS Mean Square* AError: MS(A*C)219.2712079.63560321.6447970.822398F ValuePr F11.720.0786F ValuePr F3.700.3052* This test assumes one or more other fixed effects are zero.SourceDF Type III SS Mean Square* BError: MS(B*C)19.7164439.71644312.6260132.626013* This test assumes one or more other fixed effects are zero.SourceDF Type III SS Mean Square F Value Pr 80.510.6606B*C12.6260132.6260131.640.3287Error: MS(A*B*C)23.2020691.601035SourceDF Type III SS Mean 377F ValuePr F11.570.4048Error: MS(A*C) MS(B*C) - MS(A*B*C) 11E-17*MS(Error)SourceA*B*CError: MS(Error)DF Type III SS Mean Square F Value Pr F23.2020691.6010352416.3618930.6817461802.350.1171

The complete interaction model for a three-factor completely randomized design is: y ijkl (35) { is the baseline mean, { i, j, and k are the main factor e ects for A, B, and C, respectively. { ( ) ij, ( ) ik and ( ) jk are the two-factor interaction e ects for interactions AB, AC, and BC, respectiv