Latin Square Designs (LS) Clarice G. B. Demétrio

Transcription

1 Estatística Experimental: Session 4 1 Latin Square Designs (LS) based on Chris Brien s notes (University of South Australia, Adelaide), Clarice G. B. Demétrio Departamento de Ciências Exatas ESALQ/USP Piracicaba, SP, Brasil Clarice@esalq.usp.br April 2010

2 Estatística Experimental: Session 4 2 Course Outline V.A Design of Latin squares V.B Indicator-variable models and estimation for an LS VI.C Hypothesis testing using the ANOVA method for a Latin Square V.D Treatment differences IV.E Summary IV.F Exercise

3 Estatística Experimental: Session 4 3 V. Latin squares designs (LS) V.A Design of Latin squares Sometimes we need more than one type of blocks. In general call one sort of blocks rows and the other sort columns. Definition V.1: A Latin square design is one in which each treatment occurs once and only once in each row and each column so that the numbers of rows, columns and treatments are all equal. Clearly, the total number of observations is n = t 2. Suppose in a field trial moisture is varying across the field and the stoniness down the field. A Latin square can eliminate both sources of variability.

4 Estatística Experimental: Session 4 4 Example V.1 Sugarcane experiment Suppose there are five different varieties of sugarcane to be compared and suppose that moisture is varying across the field and that the stoniness down the field. A Latin square design for this would be as follows: 5 5 Latin Square Column Less stony of I A B C D E field II C D E A B Row III E A B C D IV B C D E A V D E A B C Stonier end of field Less More moisture moisture Varieties A, B, C, D, E

5 Estatística Experimental: Session 4 5 Even if one has not identified trends in two directions, a Latin square may be employed to guard against the problem of putting the blocks in the wrong direction. Latin squares may also be used when there are two different kinds of blocking variables, for example, animals and times. General principle is that one is interested in maximizing row and column differences so as to minimize the amount of uncontrolled variation affecting treatment comparisons. The major disadvantage with the Latin square is that you are restricted to having the number of replicates equal to the number of treatments.

6 Estatística Experimental: Session 4 6 Several fundamentally different Latin squares exist for a particular t for t = 4 there are three different squares. A collection of Latin squares for t = 3, 4,..., 9 is given in Appendix 8A of Box, Hunter and Hunter. To randomize these designs appropriately involves the following: 1. randomly select one of the designs for a value of t; 2. randomly permute the rows and then the columns; 3. randomly assign letters to treatments. Note: no nested.factors as Rows and Columns are to be randomized independently Hence they are not nested (they are crossed) Generally we will use R to obtain randomized layouts. General instructions given in Appendix B, Randomized layouts and sample size computations in R.

7 Estatística Experimental: Session 4 7 a) Obtaining a layout for an LS in R # Obtaining randomized layout for a LS library(dae) t <- 5 n <- t * t LSSugarCane.unit <- list(rows = t, Columns = t) Varieties <- factor(c("a", "B", "C", "D", "E", "C", "D", "E", "A", "B", "E", "A", "B", "C", "D", "B", "C", "D", "E", "A", "D", "E", "A", "B", "C"), labels =c("a", "B", "C", "D", "E")) LSSugarcane.lay <- fac.layout(unrandomized=lssugarcane.unit, randomized=varieties,seed=941) remove("varieties") LSSugarcane.lay Units Permutation Rows Columns Varieties A D C E B C A E B D B E D A C E

8 Estatística Experimental: Session C B D A D B A C E Thus the randomized layout is: 5 5 Latin Square Column Less stony of I A D C E B field II C A E B D Row III B E D A C IV E C B D A V D B A C E Stonier end of field Less More moisture moisture Varieties A, B, C, D, E

9 Estatística Experimental: Session 4 9 V.B Indicator-variable models and estimation for an LS Have to decide whether each of the factors Rows, Columns and Treatments are to be regarded as fixed or random. As for the RCBD, it happens that the analysis of the Latin square is essentially unaffected by which model is used. Generally, the Latin square involves t rows and columns so that there are n = t 2 observations in all.

10 Estatística Experimental: Session 4 10 Maximal model The maximal model when all are fixed, is Ψ R+C+T = E(Y) = X R β + X C γ + X T τ and var(y) = V Y = σ 2 I n where Y is the t 2 -vector of random variables for the response variable observations, β is the t-vector of parameters specifying a different mean response for each row, X R is the t 2 t matrix indicating the row from which an observation came, γ is the t-vector of parameters specifying a different mean response for each column, X C is the t 2 t matrix indicating the column from which an observation came, τ is the t-vector of parameters specifying a different mean response for each treatment, X T is the t 2 t matrix indicating the observations that received each of the treatments. Our model also involves assuming Y N(Ψ R+C+T, V Y ).

11 Estatística Experimental: Session 4 11 Example V.2 A 3 3 Latin square For example, suppose that a 3 3 Latin square with the following arrangement of treatments was being considered: 3 3 Latin Square Column I A B C Row II C A B III B C A Suppose the random vector Y is ordered in standard order for Rows then Columns.

12 Estatística Experimental: Session 4 12 Then, for this example, Y = Y 11A Y 12B Y 13C Y 21C Y 22A Y 23B Y 31B Y 32C Y 33A X R = X C = X T = Note that X R = I t 1 t and X C = 1 t I t, but that X T cannot be written as a direct product.

13 Estatística Experimental: Session 4 13 The model for the expectation is still of the form E(Y) = Xθ, with X = [X R X C X T ] and θ = [β γ τ ] = [β 1, β 2,, β t, γ 1, γ 2,, γ t, τ 1, τ 2,, τ t ]. From the theory of linear models (see chapter XI, Brien s notes) the estimators for Ψ T = E(Y) = X B β + X C γ + X T τ is given by ˆΨ R+C+T = R + C + T 2Ḡ where R = X R (X R X R) 1 X R Y = M RY C = X C (X C X C) 1 X C Y = M CY T = X T (X T X T ) 1 X T Y = M T Y Ḡ = X G (X G X G) 1 X G Y = M GY are the t 2 -vectors of row, column, treatment and grand means, respectively. That is, M R, M C, M T and M G are the row, column, treatment and grand mean operators, respectively. So once again the fitted values are functions of means.

14 Estatística Experimental: Session 4 14 Note that, if the data in the vector Y has been arranged in standard order for Rows then Columns the mean operators can be written as M G = n 1 J t J t = t 2 J n M R = t 1 I t J t M C = t 1 J t I t where is called the direct product operator and if A r and B r are square matrices of order r and c, respectively, a 11 B... a 1r B A r B c = a r1 B... a rr B In this case it is not possible to write M T as a direct product of I and J matrices as the treatments will not be in a systematic order expressible in this form.

15 Estatística Experimental: Session 4 15 X RX R = X CX C = X T X T = t t t, M R = 1 t J t 0 t t... 0 t t 0 t t 1 t J t... 0 t t t t 0 t t... 1 t J t, M C = 1 t I 1 t t I 1 t... t I t 1 t I 1 t t I 1 t... t I t t I t 1 t I 1 t... t I t

16 Estatística Experimental: Session 4 16 Example V.3 A 3 3 Latin square (continued) For this example, R = R 1 R 1 R 1 R 2 R 2 R 2 R 3 R 3, C = C 1 C 2 C3 C 1 C 2 C 3 C 1 C 2, T = T A T B T C T C T A T B T B T C, Ḡ = Ḡ Ḡ Ḡ Ḡ Ḡ Ḡ Ḡ Ḡ R 3 C 3 t where R k=1 j. = Y jk, C.k = t t t j=1 k=1 Ḡ = Y jk t 2. T A t j=1 Y jk, t T l = t l=1 Y jk (l) t Ḡ and

17 Estatística Experimental: Session 4 17 b) Alternative expectation models There are 8 possible different models for the expectation that we consider: Ψ G = E(Y) = X G µ (no treatment, row or column differences) Ψ R = E(Y) = X R β (row differences only) Ψ C = E(Y) = X C γ (column differences only) Ψ R+C = E(Y) = X R β + X C γ (additive row and column differences) Ψ T = E(Y) = X T τ (treatment differences only) Ψ R+T = E(Y) = X R β + X T τ (additive row and treatment differences) Ψ C+T = E(Y) = X C γ + X T τ (additive column and treatment differences) Ψ R+C+T = E(Y) = X R β + X C γ + X T τ (additive row, column and treatment differences)

18 Estatística Experimental: Session 4 18 The estimators of the fitted values under the different models are: ˆΨ G = Ḡ ˆΨ R = R ˆΨ C = C ˆΨ R+C = R + C Ḡ ˆΨ T = T ˆΨ R+T = R + T Ḡ ˆΨ C+T = C + T Ḡ ˆΨ R+C+T = R + T + C 2Ḡ That is, they are all functions of the four mean vectors for this design.

19 Estatística Experimental: Session 4 19 We note the following marginality relations between the models: Ψ G Ψ R, Ψ C, Ψ R+C, Ψ T, Ψ R+T, Ψ C+T, Ψ R+C+T Ψ R Ψ R+C, Ψ R+T, Ψ R+C+T Ψ C Ψ R+C, Ψ C+T, Ψ R+C+T Ψ T Ψ R+T, Ψ C+T, Ψ R+C+T Ψ R+C, Ψ R+T, Ψ C+T Ψ R+C+T

20 Estatística Experimental: Session 4 20 V.C Hypothesis testing using the ANOVA method for an LS An analysis of variance will be used to choose between the eight alternative expectation models for an LS. a) Analysis of an example Example V.1 Sugarcane data: The following data are from an 5 5 LS experiment with 5 varieties (A: Co 290, B: Co 421, C: Co 419, D: POJ 2878, E: CO 36-13) of sugarcane. The response variable was yield (kg/plot) of sugarcane. 5 5 Latin Square Column Row Totals 432 D 518 A 458 B 583 C 331 E C 478 E 524 A 550 B 400 D E 384 B 556 C 297 D 420 A B 500 D 313 E 486 A 501 C A 660 C 438 D 394 E 318 B 2325 Column Totals T A = 2463, T B = 2204, T C = 3024, T D = 2067, T E = 2005

21 Estatística Experimental: Session 4 21 Yields Yields Yields A B C D E Rows Columns Figure 1: Sugarcane LS data. Boxplots Treatments It would appear that there are yield mean differences between the rows, columns and varieties.

22 Estatística Experimental: Session 4 22 Example V.1 Sugarcane LS experiment (continued): Mean Variance StandardError A B C D E The differences between the treatment variances are not too large. s 2 max s 2 min = = 4.5

23 Estatística Experimental: Session 4 23 The hypothesis test for this example is: Step 1: Set up hypotheses a) H 0 : τ 1 = τ 2 =... = τ 5 (or X T τ not required in the model) H 1 : not all population Treatment means are equal b) H 0 : β 1 = β 2 =... = β 5 (or X R β not required in the model) H 1 : not all population Row means are equal b) H 0 : γ 1 = γ 2 =... = γ 5 (or X C γ not required in the model) H 1 : not all population Column means are equal Set α = 0.05.

24 Estatística Experimental: Session 4 24 Step 2: ANOVA table using R Source df SSq MSq F Prob Rows Columns * Rows:Columns Varieties *** Residual Total Step 3: Decide between hypotheses It would appear that there are significant differences between the varieties and columns but not between the rows.

25 Estatística Experimental: Session 4 25 b) Sums of squares for the analysis of variance The estimators of the sum of squares for the LS ANOVA are the sums of squares of the following vectors: Total or Units Deviation: D G = Y Ḡ = (M U M G )Y = Q U Y with Q U = M U M G Rows Effect: R e = R Ḡ = (M R M G )Y = Q R Y with Q R = M R M G Columns Effect: C e = C Ḡ = (M C M G )Y = Q C Y with Q C = M C M G Rows:Columns Deviation: D R+C = Y R C + Ḡ = (M U M R M C + M G )Y = Q RC Y with Q RC = M U M R M C + M G Treatments Effect: T e = T Ḡ = (M T M G )Y = Q T Y with Q T = M T M G Residual Deviation: D R+C+T = Y R C T+2Ḡ = Y R e C e T e Ḡ = (M U M R M C M T +2M G )Y = Q RCRes Y with Q RCRes = M U M R M C +2M G where D s are n-vectors of deviations from Y and R e, C e and T e are n-vectors of Rows, Columns and Treatment effects.

26 Estatística Experimental: Session 4 26 Theorem V.1: For a latin squares design, the sums of squares in the analysis of variance for Units, Rows, Columns, Rows:Columns, Treatments and Residual are given by the quadratic forms Y Q U Y = D G D G, Y Q R Y = R er e, Y Q C Y = C ec e, Y Q RC Y = D RC D RC, Y Q T Y = T et e and Y Q RCRes Y = D R+C+T D R+C+T, respectively, where Q U = M U M G, Q R = M R M G, Q C = M C M G, Q RC = M RC M R M C + M G, Q T = M T M G, Q RCRes = M U M R M C M T + 2M G, Y = I n Y = M U Y = (I t I t )Y, Ḡ = 1 t J 2 n Y = M G Y = (t 2 J t J t )Y, R = M R Y = (t 1 I t J t )Y, C = M C Y = (t 1 J t I t )Y, T = M T Y. All the M s and Q s are symmetric (M = M, Q = Q ) and idempotent (MM = M, QQ = Q).

27 Estatística Experimental: Session 4 27 So the analysis of variance table is constructed as follows: Source df SSq MSq F Prob Rows t 1 Y Q R Y s 2 R Columns t 1 Y Q C Y s 2 C Rows:Columns (t 1) 2 Y Q RC Y Treatments t 1 Y Q T Y s 2 T Residual (t 1)(t 2) Y Q RCRes Y s 2 RC Res Total t 2 1 Y Q U Y s 2 R s 2 RC Res s 2 C s 2 RC Res s 2 T s 2 RC Res p R p C p T where s 2 R = Y Q R Y t 1, s2 C = Y Q C Y t 1 s 2 RC Res = Y Q RCRes Y (t 1)(t 2)., s2 T = Y Q T Y t 1 and

28 Estatística Experimental: Session 4 28 Y Q U Y = n i=1 Y i 2 correction, where correction = ( n i=1 Y i) 2 n Y Q R Y = 1 t Y Q C Y = 1 t Y Q T Y = 1 t t Rj 2 correction, where R j is total for row j j=1 t Ck 2 correction, where C k is total for column k k=1 t T(l) 2 correction, where T (l) is total for treatment l l=1 Y Q RCRes Y = Y Q U Y Y Q R Y Y Q C Y Y Q T Y

29 Estatística Experimental: Session 4 29 c) Expected mean squares The expected means squares when both Rows and Columns effects are fixed are given in the following table: Source df MSq E(MSq) F Rows t 1 Y Q R Y t 1 σ 2 + q R (Ψ) Columns t 1 Y Q C Y t 1 σ 2 + q C (Ψ) Rows:Columns (t 1) 2 Y Q T Y Treatments t 1 t 1 σ 2 + q T (Ψ) Residual (t 1)(t 2) Y Q RCRes Y (t 1)(t 2) σ2 Total t 2 1 s 2 R s 2 RC Res s 2 C s 2 RC Res s 2 T s 2 RC Res where q R (Ψ) = Ψ Q R Ψ t 1 q C (Ψ) = Ψ Q C Ψ t 1 = t(β j β.) 2 t 1 = t(γ k γ.) 2 t 1,, q T (Ψ) = Ψ Q T Ψ t 1 = t(τ l τ.) 2 t 1.

30 Estatística Experimental: Session 4 30 The expected means squares when both Rows and Columns effects are random are given in the following table: Source df MSq E(MSq) F Rows t 1 Y Q R Y t 1 σ 2 + tσr 2 Rows t 1 Y Q C Y t 1 σ 2 + tσc 2 Rows:Columns (t 1) 2 Treatments t 1 Y Q T Y t 1 σ 2 + q T (Ψ) Residual (t 1)(t 2) Y Q RCRes Y (t 1)(t 2) σ2 Total t 2 1 s 2 R s 2 RC Res s 2 C s 2 RC Res s 2 T s 2 RC Res

31 Estatística Experimental: Session 4 31 Also is possible to show that (Chapter XII of Chris Brien s notes) - Y Q RCRes Y σ 2 χ 2 (t 1)(t 2) - Y Q T Y σ 2 χ 2 t 1, under H 0 - Y Q RCRes Y σ 2 - F = and Y Q T Y σ 2 Treatments MSq Residual MSq - Y Q R Y σ 2 χ 2 t 1, under H 0 - Y Q RCRes Y - F = σ 2 and Y Q R Y are independent F t 1,(t 1)(t 2) σ 2 are independent Rows MSq Residual MSq F t 1,(t 1)(t 2) - Y Q RCRes Y - F = σ 2 and Y Q C Y σ 2 are independent Columns MSq Residual MSq F t 1,(t 1)(t 2)

32 Estatística Experimental: Session 4 32 d) Summary of the hypothesis test Step 1: Set up hypotheses a) H 0 : τ 1 = τ 2 =... = τ t (or X T τ not required in the model) H 1 : not all population Treatment means are equal b) H 0 : β 1 = β 2 =... = β t (or X R β not required in the model) H 1 : not all population Row means are equal b) H 0 : γ 1 = γ 2 =... = γ t (or X C γ not required in the model) H 1 : not all population Column means are equal Set α = 0.05 (or 0.01).

33 Estatística Experimental: Session 4 33 Step 2: Calculate test statistic Source df SSq MSq E(MSq) F Prob Rows t 1 Y Q R Y s 2 R σ 2 + q R (Ψ) Columns t 1 Y Q C Y s 2 C σ 2 + q C (Ψ) v Rows:Columns (t 1) 2 Y Q RC Y s 2 R s 2 RC Res s 2 C s 2 RC Res p R p C Treatments t 1 Y Q T Y s 2 T σ 2 + q T (Ψ) Residual (t 1)(t 2) Y Q RCRes Y s 2 RC Res σ 2 Total t 2 1 Y Q U Y s 2 T s 2 RC Res p T where s 2 R = Y Q R Y t 1 s 2 RC Res = Y Q RCRes Y (t 1)(t 2)., s2 C = Y Q C Y t 1, s2 T = Y Q T Y t 1 and Step 3: Decide between hypotheses If P(F F 0 ) = p α then the evidence suggests that the null hypothesis H 0 should be rejected.

34 Estatística Experimental: Session 4 34 e) Comparison with traditional two-way ANOVA The above analysis of variance table and the traditional Latin-square ANOVA table are essentially the same at any rate the values of the F -statistics are exactly the same. Source df Source in LS ANOVA Rows t 1 Between Rows Columns t 1 Between Columns Rows:Columns (t 1) 2 Treatments t 1 Between Treatments Residual (t 1)(t 2) Error Total t 2 1 The two tables have in common four sources that are labelled differently; they differ in that our table includes the line Rows:Columns.

35 Estatística Experimental: Session 4 35 Rows:Columns reflects the variation of column differences from row to row or the differences in row-column combinations after overall row and overall column differences have been removed - it is uncontrolled variation in the units. The Rows:Columns sum of squares is partitioned into Treatments and Residual sums of squares. The advantage of the table we have presented is that it exhibits the confounding in the experiment. The indenting of Treatments under Rows:Columns signifies that treatment differences are confounded or mixed-up with row-column differences, adjusted for overall row and overall column differences, as a result of the randomization of treatments to row-column combinations. This is not obvious from the traditional table.

36 Estatística Experimental: Session 4 36 IV.D Diagnostic checking We have assumed a model on which the analysis outlined above is based, that Y N(Ψ, σ 2 I) where, for the maximal model, Ψ R+C+T = E(Y) = X R β + X C γ + X T τ. For this model to be appropriate requires a set of behaviours as: a) the response is operating additively, that is, that a treatment has about the same additive effect on each unit; b) that the variability of the units within the row (column) are the same for each row (column); c) each observation displays the covariance implied by the model (independence for Rows and Column fixed and equal correlation within rows (columns) for Rows (Columns) random); and d) that the response of the units is normally distributed.

37 Estatística Experimental: Session 4 37 constant variance plot e i against fitted values ˆψ i and look at pattern of spread often easier to see any pattern by using absolute values e i tabulate sample variance of e i for any obvious subsets, e.g. by any classifying factors. Normality Plot the ordered residuals e (i) against the quantiles of a standard normal distribution, Φ 1 (i/(n + 1). A straight line = normality. Normal-plot with simulated envelope. Nonadditivity Use Tukey s one-degree-of-freedom-for-nonadditivity.

38 Estatística Experimental: Session 4 38 Example V.1 Sugarcane data (continued): Source df SSq MSq F Prob Rows Columns * Rows:Columns Varieties *** Residual Nonadditivity Deviation Total The hypotheses for the one-degree-of-freedom is: H 0 : Rows, Columns and Varieties are additive H 1 : Rows, Columns and Varieties are nonadditive The null hypothesis cannot be rejected and there is no evidence of transformable nonadditivity.

39 Estatística Experimental: Session 4 39 Normal Q Q Plot res fit Theoretical Quantiles Studentized Residuals(LSSugarcane.lm) Sample Quantiles t Quantiles Figure 2: Sugarcane LS. Residual plots

40 Estatística Experimental: Session 4 40 IV.E Treatment differences For the purposes of the scientist the effect of rows and columns are not of primary interest. Rather, attention is likely to be focused on variety differences which can be investigated using the variety means. Example V.1 Sugarcane data (continued) The variety means (kg/plot) are: Varieties A B C D E As the variety levels are qualitative a multiple comparison procedure will be used to examine the differences between the varieties if they were significant.

41 Estatística Experimental: Session 4 41 Tukey s HSD procedure Hypotheses: H 0 : µ k = µ k versus H 1 : µ k µ k w(5%) = q t,ν,α 1 r Residual MSq = = Varieties E D B A Means D B A C * 191.4* 164.0* 112.2*

42 Estatística Experimental: Session % family wise confidence level E D E C D C E B D B C B E A D A C A B A Differences in mean levels of Treats

43 Estatística Experimental: Session 4 43 Conclusion: Evidence at the 5% level of significance that µ C µ A, µ B, µ D, µ E There is one variety (Co 419) more productive that differs from the others. E D B A C b b b b a or

44 Estatística Experimental: Session 4 44 R program for LS Sugarcane data library(dae) library(car) LSSugarcane.dat <- data.frame(rows=factor(rep(c(1,2,3,4,5), times=c(5,5,5,5,5))), Columns = factor(rep(c(1,2,3,4,5), times=5))) LSSugarcane.dat$Treats <- factor(c("d","a","b","c","e","c","e","a","b","d", "E","B","C","D","A","B","D","E","A","C","A","C","D","E","B")) LSSugarcane.dat$Yields <- c(432,518,458,583,331,724,478,524,550,400, 489,384,556,297,420,494,500,313,486,501,515,660,438,394,318) LSSugarcane.dat attach(lssugarcane.dat) # boxplots par(mfrow=c(1,3)) boxplot(split(yields, Rows), style.bxp = "old", xlab = "Rows", ylab = "Yields", medchar = T, medpch = 8) boxplot(split(yields, Columns), style.bxp = "old", xlab = "Columns", ylab = "Yields", medchar = T, medpch = 8) boxplot(split(yields, Treats), style.bxp = "old", xlab = "Treatments", ylab = "Yields", medchar = T, medpch = 8) par(mfrow=c(1,1)) # Variety means, variances and standard erros v <- tapply(yields, Treats,var) data.frame(mean= tapply(yields, Treats,mean), Variance=v,StandardError= sqrt(v)) rm(v) # ANOVA using aov without error term LSSugarcane.NoError.aov <- aov(yields ~ Rows + Columns + Treats, LSSugarcane.dat) summary(lssugarcane.noerror.aov) # ANOVA using aov with error term

45 Estatística Experimental: Session 4 45 LSSugarcane.aov <- aov(yields ~ Rows + Columns + Treats + Error(Rows * Columns), LSSugarcane.dat) summary(lssugarcane.aov) #Compute Drivers and Cars Fs and p-values Rows.F <- 7620/2843 Rows.p <- 1 - pf(rows.f, 4, 12) Columns.F < /2843 Columns.p <- 1 - pf(columns.f, 4, 12) data.frame(rows.f, Rows.p, Columns.F, Columns.p) # ANOVA using lm LSSugarcane.lm <- lm(yields ~ Rows + Columns + Treats, LSSugarcane.dat) anova(lssugarcane.lm) # Diagnostic checking fit<-fitted(lssugarcane.lm) res<-residuals(lssugarcane.lm) data.frame(rows, Columns, Treats, Yields, res, fit) par(mfrow=c(1,2)) plot(fit, res, pch = 16) qqnorm(res, pch = 16) qqline(res) par(mfrow=c(1,1)) # normal-plot with simulation envelope library(car) qq.plot(lssugarcane.lm,simulate=true,reps=999) #click on the point, use ESC in R to esc # Test of nonadditivity - Tukey # directly from the package diagnostic.checking.r tukey.1df(lssugarcane.aov, LSSugarcane.dat, error.term = "Rows:Columns") # or using lm LSSugarcane.lm <- lm(yields ~ Rows + Columns + Treats) anova(lssugarcane.lm, test="f") LP2 <- (predict(lssugarcane.lm ))^2 LSSugarcane.lm2 <-update(lssugarcane.lm,.~. +LP2) anova(lssugarcane.lm,lssugarcane.lm2, test="f") # Multiple comparisons - Tukey s HSD procedure

46 Estatística Experimental: Session 4 46 model.tables(lssugarcane.aov, type="means") q <- qtukey(0.95, 5, 12) q w <- qtukey(0.95, 5, 12)*summary(LSSugarcane.lm)$sigma/sqrt(5) w summary(lssugarcane.aov <- aov(yields ~ Rows + Columns + Treats)) TukeyHSD(LSSugarcane.aov, "Treats", ordered = TRUE) plot(tukeyhsd(lssugarcane.aov, "Treats")) # Tukeys test using laercio library require(laercio) # same as library(laercio) LTukey(LSSugarcane.aov,"Treats",conf.level=0.95) TUKEY TEST TO COMPARE MEANS Confidence level: 0.95 Dependent variable: Yields Variation Coefficient: % Independent variable: Factors Means C a A b B b D b E 401 b Treats detach(lssugarcane.dat) search()

47 Estatística Experimental: Session 4 47 IV.H Summary In this chapter we have: described how to design an experiment using a Latin square design; formulated a linear model using indicator variables to describe the results from a Latin square design; given the estimators of the parameters in the linear model, the fitted values and the residuals as functions of M or mean operator matrices; outlined an ANOVA hypothesis test for choosing between expectation models; the partition of the total sums of squares was given with the sums of squares expressed as the sums of squares of the elements of vectors and as quadratic forms where the matrices of the quadratic forms, Q matrices, are symmetric idempotents;

48 Estatística Experimental: Session 4 48 discussed the differences between the analyses where Rows and Columns are fixed and where they are random; the expected mean squares under the alternative expectation models are used to justify the choice of F test statistic; shown how to obtain a layout and the analysis of variance in R; discussed procedures for checking the adequacy of the proposed models; subsequent to the hypothesis test, examined treatment differences in detail;

49 Estatística Experimental: Session 4 49 Exercise Pig experiment The following data are from a 4 4 LS experiment at Animal Science Department, ESALQ/USP with 4 treatments (A Castration at 56 old, B without castration, C Castration at 7 days old, D Castration at 21 days old). The response variable was weight gain (kg), after 252 days. The rows were litters and the columns were initial weight intervals (inside the litters) Columns Litters A B C D B 96.5 D 77.9 A C C 94.9 A D 96.0 B D C B 97.6 A