Lecture 22 Mixed Effects Models III Nested designs

Transcription

1 Lecture 22 Mixed Effects Models III Nested designs 94

2 Introduction: Crossed Designs The two-factor designs considered so far involve every level of the first factor occurring with every level of the second factor. Example: examine the effect of applying nitrogen at 6 different levels on the yield of three varieties of wheat there are 18 possible combinations of treatments. Experiments of this type are called crossed designs. 95

3 N 1 N 2 N 3 N 4 N 5 N 6 V 1 X X X X X X V 2 X X X X X X V 3 X X X X X X 96

4 Nested Designs Suppose however that it was known in advance that the three varieties react quite differently to nitrogen. variety 1 needed very low levels of nitrogen (say N = 1 or 2 units) variety 2 needs an intermediate level of nitrogen (say N = 3 or 4 units) variety 3 requires high levels say (N = 5 or 6 units). 97

5 Six levels of nitrogen in the experiment but only two levels of nitrogen are being applied to each variety of wheat There are (only) six combinations of treatments being considered instead of the possible 18 treatment combinations in a crossed design. We then say that the levels of the factor nitrogen are nested within the levels of the factor, variety and we call this type of design a nested design. 98

6 N 1 N 2 N 3 N 4 N 5 N 6 V 1 X X V 2 X X V 3 X X Notice a characteristic of this type of design is that although the same number of levels of factor B may be applied to each level of factor A, the levels of B that are applied differ between levels of A. 99

7 In this example both factors might be considered as fixed.why? 100

8 Example Examine the characteristics of the offspring of bulls of a particular breed of cattle. Each bull would be mated with a number of dams and the required characteristic measured on the resulting offspring. Here dams would be nested within bulls since different dams would be mated with each bull. Here we have an example in which both factors could be considered random. Why? 101

9 Model Suppose factor A has a levels and factor B has b levels and there are n replicates of each treatment combination. Assume that the levels of factor B are nested within levels of A which we will denote by B(A). Then Y ijk = µ + α i + β j(i) + ǫ ijk (10) for i = 1, 2,...,a, j = 1, 2,...,b and k = 1, 2,...,n. 102

10 The notation j(i) indicates the j th level of B is nested within the i th level of A. If A and B are both random then α i N(0,σA), 2 β j(i) N(0,σB) 2 and ǫ ijk N(0,σ 2 ). 103

11 AOV Table Source df Factor A a 1 Factor B within A a(b 1) Error ab(n 1) 104

12 The expected mean squares for this model are given below under the assumption that A and B are both random. E(MS) df Both Random E(MSA) a 1 nbσ 2 A + nσ 2 B(A) + σ2 E(MSB(A)) a(b 1) nσ 2 B(A) + σ2 Error ab(n 1) σ 2 105

13 Example Seven specimens were sent to 6 laboratories in 3 separate batches and each analysed for an (unnamed) Analyte. Each analysis was duplicated. The data below are for specimen 1. The purpose of the study was to assess components of variation in co-operative trials. Hence laboratories and batches were regarded as random. 106

14 Laboratory Batch

15 Variance components The variation in concentration of the analyte is a combination of the three variances due to: measurements, batches and laboratories. The three components of variance are due to the distribution of: lab means around the overall mean batch means around the i th lab mean individual measurements around the j th batch mean. 108

16 R Output A and B both random ######## R code ######## # Read Data analyte <- read.table( analyte.txt,header=t) # Declare factors analyte$lab <- factor(analyte$lab) analyte$batch <- factor(analyte$batch) # Fit nested model analyte.lm <- lm(conc~lab/batch,data=analyte) print(anova(analyte.lm)) ######## Output ######## Analysis of Variance Table (F-values removed) Response: conc Df Sum Sq Mean Sq lab lab:batch Residuals

17 Exercises Confirm the following variance components (a = 6, b = 3, n = 2): ˆσ lab 2 = ˆσ 2 batch(lab) = ˆσ 2 =

18 Confirm that the variance components obtained from lme (edited output below) are the same as those from the aov output. 111

19 Partial Solution We have that ˆσ 2 = MSE = Now we find ˆσ 2 batch(lab) MSbatch(lab) ˆσ2 = n = 2 = = from lme 112

20 ######## R code ######## library(nlme) analyte.lme<-lme(conc~1,random=~1 lab/batch, data=analyte) print(summary(analyte.lme)) ######## Output ######## Random effects: Formula: ~1 lab (Intercept) StdDev: Formula: ~1 batch %in% lab (Intercept) Residual StdDev:

21 Comments Is the nested term (batch within lab) significant? 114

22 # remove nested term analyte2.lme<-lme(conc~1,random=~1 lab, data=analyte) # compare the two models print(anova(analyte2.lme,analyte.lme)) ################################# # Edited Output Model df Test L.Ratio p-value analyte2 1 3 analyte vs The model with batches may be marginally better (P=0.056) than the simpler model. 115

23 The P value from the aov is

24 > analyte2.lm<-lm(conc~lab,data=analyte) > anova(analyte2.lm) Analysis of Variance Table Response: conc Df Sum Sq Mean Sq F value Pr(>F) lab e-12 Residuals > > anova(analyte2.lm,analyte.lm) Analysis of Variance Table Model 1: conc ~ lab Model 2: conc ~ lab/batch Res.Df RSS Df Sum of Sq F Pr(>F)

25 Furthermore, the confidence intervals for the variance component for batch (0.038, 0.142) suggest that batches should be included in the model. 118

26 # Approx 95% CI for variance components print(intervals(analyte.lme)) ############################# Approximate 95% confidence intervals Random Effects: Level: lab lower est. upper sd((intercept)) Level: batch lower est. upper sd((intercept)) Within-group standard error: lower est. upper

27 Note, however, that there is more variability amongst labs (ˆσ 2 lab = 0.06) than among batches nested in labs (ˆσ 2 batch(lab) = 0.005). 120

28 Conclusion When dealing with random effects, we are interested in estimating the variance components. The use of lme readily provides these variance components (more easily than if we used the aov results.) Furthermore, in the case of unbalanced data, the variance components cannot be estimated using aov, only using lme. 121

29 The F-ratios reported in the aov output are based on the assumption that all variables are fixed and so can not be used to test hypotheses for all terms in a mixed model. The term which is required in the denominator for main effects in a mixed effects model is not MSE, as it is for a fixed effects model. 122