Lecture 4. Random Effects in Completely Randomized Design

Transcription

1 Lecture 4. Random Effects in Completely Randomized Design Montgomery: 3.9, 13.1 and Lecture 4 Page 1

2 Random Effects vs Fixed Effects Consider factor with numerous possible levels Want to draw inference on population of levels Not just concerned with levels in experiment Example of differences Fixed: Compare reading ability of 10 2nd grade classes in NY Select a = 10 specific classes of interest. Randomly choose n students from each classroom. Want to compareτ i (class-specific effects). Random: Compare variability among all 2nd grade classes in NY Randomly choose a = 10 classes from large number of classes. Randomly choose n students from each classroom. Want to assessσ 2 τ (class to class variability). Inference broader in random effects case: inference on population with randomly chosen levels 2 Lecture 4 Page 2

3 Random Effects Model (CRD) Same model as in fixed effects case y ij = µ+τ i +ǫ ij i = 1,2...a j = 1,2,...n i µ - grand mean τ i -ith treatment effect ǫ ij iid N(0,σ 2 ) But view treatment effects in different way Instead of τ i = 0, assume τ i iid N(0,σ 2 τ) {τ i } and{ǫ ij } independent Var(y ij ) =σ 2 τ +σ 2 3 Lecture 4 Page 3

4 The hypotheses are: Random Effects Model H 0 : στ 2 = 0 H 1 : στ 2 > 0 Partitioning of Total Sum of Squares identical E(MS E )=σ 2 E(MS Treatment )=σ 2 +nστ 2 UnderH 0,F 0 = MS Treatment /MS E F a 1,N a Same test as before: Direct comparison of variabilities (between vs within) Conclusions, however, pertain to entire population 4 Lecture 4 Page 4

5 Model Estimates Usually interested in estimating variances Use mean squares (known as ANOVA method) ˆσ 2 = MS E ˆσ τ 2 = (MS Treatment MS E )/n If unbalanced, replacenwith n 0 = (( n i ) 2 n 2 i )/((a 1) n i ) Estimate ofσ 2 τ can be negative - SupportsH 0? Use zero as estimate? - Is the model reasonable? - Bayesian approach (nonnegative prior) - Use estimation method that only gives nonnegative estimates 5 Lecture 4 Page 5

6 σ 2 : Same as fixed case σ 2 τ Confidence intervals (N a)ms E σ 2 (N a)ms E χ 2 α/2,n a σ 2 : Approximate Confidence Interval χ 2 N a ˆσ 2 τ = (MS Trt MS E )/n No exact calculation of CI available. (N a)ms E χ 2 1 α/2,n a Approximate CI based on Satterthwaite s Approximation: rˆσ 2 τ/ χ 2 α/2,r σ 2 τ rˆσ 2 τ/ χ 2 1 α/2,r r = (MS Trt MS E ) 2 MS 2 Trt /(a 1)+MS2 E /(N a) 6 Lecture 4 Page 6

7 Approximate Confidence interval CI for a variance component: σ 2 0 = E[MS MS ]/k MS = MS r + +MS s MS = MS u + +MS v No common mean squares terms shared byms andms. Notef i MS i /σi 2 = SS i/σi 2 ind χ 2 f i Point estimate ofσ0 2 : ˆσ2 0 = (MS MS )/k Satterthwaite s Approximation: rˆσ 2 0/σ 2 0 χ 2 r r = (ˆσ 2 0) 2 / i MS 2 i k 2 f i Approximate(1 α) 100% CI ofσ 2 0 rˆσ 2 0/ χ 2 α/2,r σ 2 0 rˆσ 2 0/ χ 2 1 α/2,r 7 Lecture 4 Page 7

8 Proportion ofσ 2 τ in Var(y ij ), i.e., Intraclass Correlation Coefficient (ICC) Common estimate if goal is to reduce variance Uses ratio of twoχ 2 distributions (i.e.,f dist) L L+1 σ2 τ σ 2 +σ 2 τ U U+1 L = 1 n U = 1 n ( ( ) MS Trt 1 MS E F α/2,a 1,N a MS Trt MS E F 1 α/2,a 1,N a 1 = 1 ( ) F 0 1 n F α/2,a 1,N a ) = 1 ( F 0 1 n F 1 α/2,a 1,N a ) Grand mean: µ Example: Average reading ability of 2nd grade class, y.. = (y 1. +y y a. )/a. y i. iid Normal. But what is the variance? CI forµ: y.. ±t α/2,a 1 MSTrt /(an) 8 Lecture 4 Page 8

9 Example A supplier delivers several hundred batches of raw material to a company each year. The company is interested in a high yield from each batch of raw material (percent usable). Therefore, to investigate the consistency of this supplier, an experiment is done where five batches were selected at random and three yield determinations were made on each batch. Batch Lecture 4 Page 9

10 Source of Sum of Degrees of Mean Variation Squares Freedom Square F 0 Between Within Total Highly significant result (F.05,4,10 = 3.48) ˆσ τ 2 = ( )/3 = % (=11.71/( )) is attributable to batch differences Time to improve consistency of the batches 10 Lecture 4 Page 10

11 95% CI forσ 2 ( SS E, χ 2.025,10 SS E χ 2.975,10 ) = (18.00/20.48, 18.00/3.25) = (0.879, 5.538) 95% CI forσ 2 τ r = ( ) / /10 = 3.62 χ ,3.62 = (1.62)χ 2.025,3 +.62χ ,4 = = χ ,3.62 = (1.62)χ 2.975,3 +.62χ ,4 = =.3812 ( / , /.3812) = (4.0527, ) SAS allows noninteger degrees of freedom forχ 2 with :CINV(p,df) 11 Lecture 4 Page 11

12 95% CI for Intraclass Correlation L = 1 ( ) U = 1 3 ( / = = L L+1 = ) = = U U +1 = using property that 95% CI: (0.545, 0.984) F 1 α/2,v1,v 2 = 1/F α/2,v2,v 1 12 Lecture 4 Page 12

13 Using SAS data example; input batch percent cards; ; proc glm data=example; class batch; model percent=batch; random batch; output out=diag r=res p=pred; proc gplot data=diag; plot res*pred; proc mixed data=example cl; class batch; model percent = ; random batch / vcorr; run; 13 Lecture 4 Page 13

14 Dependent Variable: PERCENT Sum of Mean Source DF Squares Square F Value Pr > F Model Error Corrected Total Source DF Type I SS Mean Square F Value Pr > F BATCH Source BATCH Type III Expected Mean Square Var(Error) + 3 Var(BATCH) GLM : Uses least squares for estimation Not designed for random effects...must compute variance estimates / CIs by hand MIXED: Uses restricted maximum likelihood (REML) Often preferred to ML because it produces unbiased estimates of covariance parameters by taking into account the loss of degrees of freedom in estimating fixed effects Usually residual variance profiled out of the likelihood 14 Lecture 4 Page 14

15 The MIXED Procedure Estimated V Correlation Matrix for batch 1 Row Col1 Col2 Col Covariance Parameter Estimates Cov Parm Estimate Alpha Lower Upper BATCH Residual Fit Statistics -2 Res Log Likelihood 62.8 AIC (smaller is better) 66.8 AICC (smaller is better) 67.8 BIC (smaller is better) Lecture 4 Page 15

16 Negativeσ 2 τ Estimate Example data new; input school subj cards; ; proc glm data=new; class school; model score = school; random school; proc mixed data=new cl; class school; model score = ; random school; run; 16 Lecture 4 Page 16

17 The GLM Procedure Sum of Source DF Squares Mean Square F Value Pr > F Model Error Corrected Total R-Square Coeff Var Root MSE score Mean Source DF Type I SS Mean Square F Value Pr > F school Source DF Type III SS Mean Square F Value Pr > F school Source school Type III Expected Mean Square Var(Error) + 10 Var(school) Parameter Estimates using ANOVA method ˆσ 2 = 8.42 ˆσ 2 τ = = /10 = Lecture 4 Page 17

18 The MIXED Procedure Covariance Parameter Estimates Cov Parm Estimate Alpha Lower Upper school 0... Residual Fit Statistics -2 Res Log Likelihood AIC (smaller is better) AICC (smaller is better) BIC (smaller is better) Lecture 4 Page 18

19 In log file for PROC MIXED analysis, Word of Caution NOTE: Convergence criteria met. NOTE: Estimated G matrix is not positive definite. Caused by variance estimated to be zero Suggests possibly removing this term from the model Decision often an issue in more complicated models Can remove positivity constraint (NOBOUND) proc mixed cl nobound; class school; model score = ; random school ; run; Includes null model LRT to determine if it is necessary to model the covariance structure 19 Lecture 4 Page 19