Homework 3 - Solution

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1 STAT Spring 2011 Homework 3 - Solution Olga Vitek Each part of the problems 5 points 1. KNNL (Note: you can choose either the restricted or the unrestricted version of the model. Please state clearly which model you use.) Here is the unrestricted model below, Y ijk = µ + α i + β j + (αβ) ij + ε ijk where α i is the fixed coats effect. i α i = 0, i = 1, 2, 3 β j is the random batch effect, β j N (0, σβ 2 ), j = 1, 2, 3, 4 (αβ) ij is the interaction effect, (αβ) ij N(0, σαβ 2 ) ε ijk N (0, σ 2 ), k = 1, 2, 3, 4 a = 3, b = 4,, n = 4 (a) Test interaction effect : Analysis of Variance Table Response: value Df Sum Sq Mean Sq F value Pr(>F) coatf e-05 *** batchf e-05 *** coatf:batchf Residuals Hypothesis: H 0 : σαβ 2 = 0 vs H a : σαβ 2 > 0, i = 1, 2, 3, j = 1, 2, 3, 4 - Decision rule: Reject H 0 if F > F 0.95,6,36 = Conclusion: Since F = MSAB/MSE = 0.064, we do not reject H 0 and conclude that there are no interaction effects with p = X <- read.table("ch25pr17.txt", sep="", as.is=true, header=false) colnames(x)<-c("value","coat","batch","replicate") # Factor X$coatF<-factor(X$coat) X$batchF<-factor(X$batch) #ANOVA table fit1<-aov(value~coatf*batchf, data=x) anova(fit1) qf(0.095,6,36) # F(0.95,6,36) 1-pf(0.064,6,36) # p-value 1

2 (b) Test main effects: i. Factor A (the number of coats) main effect - Hypothesis: H 0 : α i = 0 vs H a : at least one α i is not 0, i = 1, 2, 3 - Decision rule: Reject H 0 if F = MSA/MSAB > F 0.95,2,6 = Conclusion: Since F = /0.309 = , we reject H 0 and conclude that there is Factor A main effect for the number of coats with p 0 ii. Factor B (batch) main effect - Hypothesis: H 0 : σ 2 β = 0 vs H a : σ 2 β > 0 - Decision rule: Reject H 0 if F = MSB/MSAB > F 0.95,3,6 = Conclusion: Since F = /0.309 = , we reject H 0 and conclude that there is main effect for Gender with p 0 (c) Comparisons i. CI for ˆD 1 = Ȳ2 Ȳ1 : ˆD 1 = Ȳ2 Ȳ1 = = , s 2 { ˆD} = MSAB/bn c 2 i = 0.309/(4 4) (1 + 1) = , s{ ˆD 1 } = = t(1 0.05/2, 2 3) = t(0.975, 6) = Therefore, 95% Confidence interval for ˆD 1 is ± ( ) = (3.2066, ), which does not include zero. ii. CI for ˆD 2 = Ȳ3 Ȳ2 : ˆD 2 = Ȳ3 Ȳ2 = = , s 2 { ˆD} = MSAB/bn c 2 i = 0.309/(4 4) (1 + 1) = , s{ ˆD 1 } = = t(1 0.05/2, 2 3) = t(0.975, 6) = Therefore, 95% Confidence interval for ˆD 2 is ± ( ) = ( , ), which includes zero. iii. Conclusion : at 90% joint family confidence, there is a significant difference between the number of coats=6 (Factor A=1) and the number of coats=8 (Factor A=2). But, there is no significant difference between the number of coats=8 (Factor A=2) and the number of coats=10 (Factor A=3). (d) 95% confidence interval for µ 2 using Satterthwaite procedure : Y ijk = µ + α i + β j + (αβ) ij + ε ijk µ ˆ 2 = µ + α 2 = = and based on unrestricted mixed model, the standard error for the estimate for µ 2 is ˆµ 2 = j=1 k=1 y 2jk = µ + α β j j=1 (αβ) 2j j=1 j=1 k=1 ε ijk and therefore ˆµ 2 N(µ + α 2, 1 4 σ2 β σ2 αβ σ2 ) 2

3 s 2 (ˆµ 2 ) = ˆσ2 β b + ˆσ2 αβ + ˆσ2 b bn = anˆσ2 β + anˆσ2 αβ + aˆσ2 = ((a 1)nσ2 β + (a 1)σ2 ) + (σ 2 + anσβ 2 + nσ2 αβ ) abn abn = a 1 abn MSAB + 1 abn MSB 3 1 = (3 4 4) (0.309) + 1 (3 4 4) (50.951) = s(ˆµ 2 ) = and the degrees of freedom are t(1-0.05/2, 3)= , Therefore, 95% confidence interval for µ 2 is df = ( 2 48 MSAB MSB)2 ( 2 48 MSAB)2 6 + ( 1 48 MSB)2 3 = ± ( ) = ( , ) The average market value for a pearl with 8 coats is between and with 95% confidence interval. (e) 90% confidence interval for σβ 2 using MLS procedure : MLS procedure designed to estimate a linear combination of two expected mean squares for balanced studies of the form : Using the unrestricted model, L = c 1 E{MS1} + c 2 E{MS 2 }, c 1 > 0, c 2 < 0 ˆL = ˆσ β 2 = c 1 MS 1 + c 2 MS 2 = MSB MSAB an = MSB an MSAB an = = So, c 1 = 1 12, c 2 = 1 12, MS 1 = MSB, MS 2 = MSAB Using below (with formula in p. 1045), 90% confidence interval for σβ ) by R result, σβ 2 is not much different from σ2. is ( # 25.17(e) : MLS procedure alpha=0.1 dfab=6 dfb=3 c1=1/(3*4) c2=-1/(3*4) MSAB=0.309 MSA=

4 MSB= F1<-qf(1-alpha/2, dfb, Inf) F2<-qf(1-alpha/2, dfab, Inf) F3<-qf(1-alpha/2, Inf, dfb) F4<-qf(1-alpha/2, Inf, dfab) F5<-qf(1-alpha/2, dfb, dfab) F6<-qf(1-alpha/2, dfab, dfb) G1<-1.0-1/F1 G2<-1.0-1/F2 G3<-((F5-1)^2-(G1*F5)^2-(F4-1)^2)/F5 G4<-F6*(((F6-1)/F6)^2-((F3-1)/F6)^2-G2^2) HL<-sqrt((G1*c1*MSB)^2+((F4-1)*c2*MSAB)^2-(G3*c1*c2*MSB*MSAB)) HU<-sqrt(((F3-1)*c1*MSB)^2+(G2*c2*MSAB)^2-(G4*c1*c2*MSB*MSAB)) L=c1*MSB+c2*MSAB c(l-hl, L+HU) 2. KNNL (a) ML estimates of parameters : Here is the model fit by R below, Y ijk = µ 1 + α i + β j + (αβ) ij + ε ijk where α i is the fixed coats effect (the mean difference from the first level), α 1 = 0, i = 2, 3 β j is the random batch effect, β j N (0, σβ 2 ), j = 1, 2, 3, 4 (αβ) ij is the interaction effect, (αβ) ij N(0, σαβ 2 ) ε ijk N (0, σ 2 ), k = 1, 2, 3, 4 a = 3, b = 4,, n = 4 The estimates of the parameters of the fixed effects model fit in R are ˆµ 1 = , ˆα 1 = 0, ˆα 2 = , ˆα 3 = The corresponding parameters of the unrestricted model above are ˆµ = = , ˆα 1 = = , ˆα 2 = = , ˆα 3 = = (and therefore α i = 0) The estimates of variances are the same in both models ˆσ 2 β = , ˆσ2 αβ = 0, ˆσ2 = ˆσ αβ 2 = 0. If an estimated variance component is zero, this provides evidence that the true value of the parameter is on the boundary of the parameter space, and the likelihood ratio test should not be used. Linear mixed model fit by maximum likelihood Formula: value ~ 1 + coatf + (1 batchf) + (1 coatf:batchf) Data: X AIC BIC loglik deviance REMLdev Random effects: Groups Name Variance Std.Dev. 4

5 coatf:batchf (Intercept) e e-07 batchf (Intercept) e e+00 Residual e e+00 Number of obs: 46, groups: coatf:batchf, 12; batchf, 4 Fixed effects: Estimate Std. Error t value (Intercept) coatf coatf X<-X[-c(3,38),] fit2.25<- lmer(value ~ 1 + coatf + (1 batchf) + (1 coatf:batchf), data=x, REML=FALSE) summary(fit2.25) (b) additive model: no chang in the estimates after dropping the interaction term. Linear mixed model fit by maximum likelihood Formula: value ~ 1 + coatf + (1 batchf) Data: X AIC BIC loglik deviance REMLdev Random effects: Groups Name Variance Std.Dev. batchf (Intercept) Residual Number of obs: 46, groups: batchf, 4 Fixed effects: Estimate Std. Error t value (Intercept) coatf coatf fit2.25b<- lmer(value ~ 1 + coatf + (1 batchf), data=x, REML=FALSE) summary(fit2.25b) (c) Test factor B (batch) main effect by likelihood ratio test - Hypothesis: H 0 : σ 2 β = 0 vs H a : σ 2 β > 0 - Decision rule: Reject if G 2 > χ 2 (1 α,p q) = χ2 (0.95,1) = Conclusion: G 2 = , therefore reject H 0 and conclude that there is significant Factor B (batch) effect with p 0. fit2.25c.ha<- lmer(value ~ 1 + coatf + (1 batchf), data=x, REML=FALSE) fit2.25c.h0<- lm(value ~ 1 + coatf, data=x) 5

6 ## likelihood ratio test teststat <- as.numeric(2*( loglik(fit2.25c.ha) - loglik(fit2.25c.h0))) teststat pchisq(teststat, 1, lower=false) qchisq(1-0.05,1) (d) Test factor A (The number of coats) main effect by likelihood ratio test - Hypothesis: H 0 : α i = 0 vs H a : at least one α i is not 0, i = 1, 2, 3 - Decision rule: Reject if G 2 > χ 2 (1 α,p q) = χ2 (0.95,2) = Conclusion: G 2 = , therefore reject H 0 and conclude that there is significant Factor A (the number of coats) effect with p 0. fit2.25d.ha<- lmer(value ~ 1 + coatf + (1 batchf), data=x, REML=FALSE) fit2.25d.h0<- lmer(value ~ 1 + (1 batchf), data=x, REML=FALSE) ## likelihood ratio test teststat <- as.numeric(2*( loglik(fit2.25d.ha) - loglik(fit2.25d.h0))) teststat pchisq(teststat, 2, lower=false) qchisq(1-0.05,2) (e) Since distributional form of ML estimates are unknown, we used bootstrap. I got 95% confidence interval for σβ 2 is ( , ) getcoef.boot <- function(x) { select.id <- sample(1:nrow(x), nrow(x), replace=true) x.boot <- x[sort(select.id),] lmer.boot <-lmer(value ~ 1 + coatf + (1 batchf) + (1 coatf:batchf), data=x.boot, REML=FALSE) VarCorr(lmer.boot)$batch[1,1] } n <- 100 sigma2b.boot <- rep(na, n) for (i in 1:n) { cat("bootstrap iteration:", i, "\n") sigma2b.boot[i] <- getcoef.boot(x) } left <- VarCorr(fit2.25)$batch[1,1] - abs(quantile(sigma2b.boot, /2) - VarCorr(fit2.25)$batch[1,1] right <- VarCorr(fit2.25)$batch[1,1] + abs(quantile(sigma2b.boot,0.05/2) - VarCorr(fit2.25)$batch[1,1] ) c(left, right) 3. [Methods Qualifying Exam, January 2010: use paper and pencil.] A 3 x 4 factorial study was designed to investigate the effects of 3 fertilization methods (factor A) and 4 seeding rates (factor B) on the yield of sugar beets. To this end they conducted an experiment with the combinations of all levels of both factors. Each treatment combination was replicated in three plots. The two tables below show the mean yield values ȳ ij for each treatment combination, as well as some summaries. 6

7 Level of A Level of B ȳ i Source Mean Squares Fertilizer (A) Seeding rate (B) Interaction (AB) Error (a) i. State the ANOVA model, and the corresponding assumptions, that can be used to answer research questions in this experiments. The 2-way fixed effects ANOVA model is Y ijk = µ + α i + β j + (αβ) ij + ε ijk where µ is the overall mean α i is the deviation of fertilizer i from the overall mean, i α i = 0 β i is the deviation of seedling rate j from the overall mean, j β j = 0 (αβ) ij is the non-additive effect (i.e. the interaction) i (αβ) ij = j (αβ) ij = 0 ε ijk N (0, σ 2 ) is the random error ii. Test whether the change in yield due to the use of different fertilizers depends on the seedling rate. State the null and the alternative hypotheses, and the conclusions. Use the significance level of 5%. H 0 : (αβ) ij = 0 for all i and j. H a : at least one (αβ) ij 0 F = We fail to reject H 0. MS(AB) MS(Error) = = < F (1 0.05, 6, 24) = iii. Test whether their is a difference in yield between the fertilizers. State the null and the alternative hypotheses, and the conclusions. Use the significance level of 5%. H 0 : α i = 0 for all i. H a : at least one α i 0 We reject H 0. F = MS(A) MS(Error) = = < F (1 0.05, 2, 24) = (b) Test whether the use of fertilizers 1 and 2 results in a same mean yield. State the null and the alternative hypotheses, and the conclusions. Use the significance level of 5%. We test H 0 : µ 2 µ 1 = 0 against H a : µ 2 µ 1 0 The contrast estimated by ˆµ 2 ˆµ 1 = ȳ 2 ȳ 1 = =

8 and The test statistic s 2 {ˆµ 2 ˆµ 1 } = 2 MSE 12 = = Therefore we reject H = > Student 24 (1 0.05) = 1.71 (c) Suppose now that instead of focusing on 4 levels of seeding rate, the researchers were interested in variations in yield that can result from varying the seeding rate over the entire operational range. Therefore, the four seeding rates above are not the fixed levels of interest, but a random sample of all possible levels. i. Modify the model above to reflect the different experimental setting, and state the assumptions. The 2-way mixed effects ANOVA model is Y ijk = µ + α i + β j + (αβ) ij + ε ijk where µ is the overall mean α i is the deviation of fertilizer i from the overall mean, i α i = 0 β i is the deviation of seedling rate j from the overall mean, β i N (0, σ 2 β ) (αβ) ij is the joint non-additive effect (i.e. the interaction) (αβ) ij N (0, σ 2 αβ ) ε ijk N (0, σ 2 ) is the random error ii. Using the modified model, test whether the change in yield due to the use of different fertilizers depends on the seedling rate. State the null and the alternative hypotheses, and the conclusions. Use the significance level of 5%. H 0 : σ(αβ) 2 = 0. H a : σ(αβ) 2 0. Same test statistic and conclusion as in (a,ii). iii. Using the modified model, test whether their is a difference in yield between the fertilizers. State the null and the alternative hypotheses, and the conclusions. Use the significance level of 5%. Same H 0 and H a as in (a,iii). Test statistic We reject H 0. F = MS(A) MS(AB) = = > F (1 0.05, 2, 6) =

9 iv. Test whether the use of fertilizers 1 and 2 results in a same mean yield, given this new model specification. State the null and the alternative hypotheses, and the conclusions. Use the significance level of 5%. Compare with the results of (b). -Hypothesis: H 0 : L = µ 1 µ 2 = 0, H a : L 0 ˆL = Ȳ1 Ȳ2 = s(ˆl) = nσ2 αβ +σ2 nb We reject H 0 c 2 i = MSAB 12 2 = = ˆL s{ˆl} = > t (0.975,6) = [Methods Qualifying Exam, January 2005: use paper and pencil.] The school superintendent is concerned about the development of technology skills in middle school. Since there are 3 middle schools in his district, all of which go about this instruction differently, he decided to assess if there were any differences across schools. He first compiled a long list of tech skills and randomly selected 5 to be used in his study. He then randomly selected 20 students from each school and assigned each to one of the five tasks so that there were 4 students per task per school. Each student then performed the skill and a score between 0 and 100 was assigned. (a) If a two-way ANOVA is to be used for the analysis, should it be treated as a fixed effects, random effects, or mixed effects model? Explain. School should be treated as fixed effect since all must be chosen. Skill effect should be treated as random effect since skills were selected from a long list. (b) Complete the following ANOVA table and determine which effects are significant at the α =.05 level. State your conclusions, making sure to estimate all variances and describing any additional mean comparisons youd like to perform. In order to calculate the distribution of the F statistic, we consider the unrestricted model Y ijk = µ + α i + β j + (αβ) ij + ε ijk where µ is the overall mean α i is the deviation of school i from the overall mean, i = 1, 2, 3, i α i = 0 β i is the deviation of skill j from the overall mean, j = 1,, 5, β i N (0, σβ 2) (αβ) ij is the non-additive effect (i.e. the interaction), (αβ) ij N (0, σαβ 2 ) ε ijk N (0, σ 2 ) is the random error, k = 1, 2, 3, 4 a = 3, b = 5, n = 4 Source DF SS MS EMS F SSA School(fixed) a-1 = df = 220 A 2 = 110 σ 2 + bn P α 2 i /(a 1) + nσ2 MSA αβ MSAB = = 5 SSB Skill(random) b-1 = df = 96 B 4 = 24 σ2 + anσ 2 β + nσ2 MSB αβ MSAB = = 1.09 SSAB School Skill (a-1)(b-1)= df = 176 AB 8 = 22 σ 2 + nσ 2 MSAB αβ MSE = = 2.2 SSE Error ab(n-1)= dfe = = 10 σ2 9

10 i. school effect : - Hypothesis : H 0 : α 1 = α 2 = α 3 = 0 - Decision rule : Reject if F = MSA/MSAB > F (0.95,2,8) = Conclusion : F = MSA/MSAB = 110/22 = 5 > , Reject H 0 and conclude that there is school effect. ii. skill effect : - Hypothesis : H 0 : σ β = 0 - Decision rule : Reject if F = MSB/MSAB > F (0.95,4,8) = Conclusion : F = MSB/MSE = 24/22 = 1.09 < , Do not reject H 0 and conclude that there is no significant school effect. iii. interaction effect : - Hypothesis : H 0 : σ αβ = 0 - Decision rule : Reject if F = MSAB/MSE > F (0.95,8,45) = Conclusion : F = MSB/MSE = 22/10 = 2.2 > , Reject H 0 and conclude that there is significant school and skill interaction effect. To estimate those parameters we have ˆσ 2 = MSE = 10, ˆσ 2 αβ = MSAB MSE n ˆσ 2 β = MSB MSAB an = = = = 3, and (c) If the grand skill level of the middle schools (average over the three schools) is of interest, describe how one would construct a 95% confidence interval. Let µ be the overall mean. Then we can construct the 95% confidence interval based on t 45 distribution. The overall average is ˆµ = 1 60 indicating that 3 5 i=1 j=1 k=1 y ijk = µ j=1 β j i=1 j=1 5 (αβ) ij i=1 j=1 k=1 ˆµ N(µ, 1 5 σ2 β σ2 αβ σ2 ) s 2 {ˆµ} = = 0.6 Then, we can estimated accordingly and so a 95% confidence interval for µ is derived based on 4 degree of freedom. ε ijk 5. [Methods Qualifying Exam, January 2009: use paper and pencil.] The following dataset concerns an experiment where six pullets were placed into each of 12 pens. Four blocks were formed from groups of three pens based on location. Three treatments were applied. The number of eggs produced was recorded. treat block eggs 1 O O O O E E E E F F F F

11 An analysis on this dataset using lme in nlme package provides the output (Note: lme is an older implementation of mixed effects models in R, and has a slightly different model syntax. If the syntax presents challenges, you can refit the model with lmer using the data above.) > remlme1 <- lme(eggs~treat,random=~1 block,data=eggprod,method=reml) > summary(remlme1) Linear mixed-effects model fit by REML Data: eggprod AIC BIC loglik Random effects: Formula: ~1 block (Intercept) Residual StdDev: Fixed effects: eggs ~ treat Value Std.Error DF t-value p-value (Intercept) treatf treato (a) Describe the model used in the above analysis (in terms of Y = Xβ + Zγ + ε, where X includes all covariates with fixed effects and Z includes all covariates with random effects), and report the estimates of all parameters describing the model. Y = Xβ + Zγ + ε, where γ N(0, σ 2 D = I 3 σblock 2 ) and ε N(0, σ2 I) Y O Y O Y O Y O Y E γ 1 β E Y = Y E2 Y E3, X = , ˆβ = β F = Y E β O Y F Y F Y F Y F ˆσ = ,ˆσ block = β E : reference treatment, treate effect β F : the difference between treatf and treate β O : the difference between treato and treate ˆγ : random block effects vector , Z = , ˆγ = γ 2 γ 3 γ 4, (b) Give reason to use REML instead of ML to fit the model. REML produces less biased estimates for the variance components associated with the factors, especially for factors with smaller number of levels and in balanced case it coincides with ANOVA estimates. 11

12 (c) A new pen is added to the first block and treated with treatment F. Please predict the number of eggs for this pen along with an estimate of the variability in this prediction. What if this new pen is placed in a new location (i.e., a fifth block)? > fixed.effects(remlme1) (Intercept) treatf treato > random.effects(remlme1) (Intercept) Ŷ block1 = = , V ar(ŷfirstblock) = ˆσ 2 = = Ŷ block5 = = , V ar(ŷfifthblock) = ˆσ 2 + ˆσ 2 block = = (d) We also tried other methods and other models on this dataset, and got the log- likelihood function values: > remlme2 <- lme(eggs~1,random=~1 block,data=eggprod,method=reml) > loglik(remlme2) log Lik > mlme1 <- lme(eggs~treat,random=~1 block,data=eggprod,method=ml) > loglik(mlme1) log Lik > mlme2 <- lme(eggs~1,random=~1 block,data=eggprod,method=ml) > loglik(mlme2) log Lik Please determine the significance of the treatment using a likelihood ratio test (assuming χ 2 distribution). Should you improve the accuracy of this p-value? If yes, please state your approach. G 2 = 2{logL(R) logl(f )} = 2 { } = > χ 2 2 = , Therefore, there is a significant treatment effect. (e) Please propose an approach to testing whether there is a significant difference between the blocks, and state how you would like to calculate your test statistics and the p-value. - Hypothesis : H 0 : σ 2 = 0, H a : σ 2 > 0 - LR test is not appropriate because the standard derivation of the asymptotic χ 2 distribution for the likelihood ratio statistic depends on the null hypothesis lying in the interior of the parameter space and this assumption is broken when we test if a variance is zero. The null distribution in these circumstances is unknown in general and we must resort to numerical methods if we wish for precise testing. If you do use the χ 2 distribution with the usual degrees of freedom, then the test will tend to be conservative, p-values will tend to be larger. An alternative is to use bootstrap to estimate the sampling distribution and p-value. 12