14. Linear Mixed-Effects Models for Data from Split-Plot Experiments

Transcription

1 14. Linear Mixed-Effects Models for Data from Split-Plot Experiments opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics 51 1 / 3

2 Start with a Field Field opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics 51 2 / 3

3 Partition the Field into locks Field lock 1 lock 2 lock 3 lock 4 opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics 51 3 / 3

4 Partition Each lock into Plots Field lock 1 lock 2 lock 3 lock 4 opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics 51 4 / 3

5 Randomly ssign Genotypes to Plots within locks Field lock 1 Genotype Genotype Genotype lock 2 Genotype Genotype Genotype lock 3 Genotype Genotype Genotype lock 4 Genotype Genotype Genotype opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics 51 5 / 3

6 Partition Each Whole Plot into Split Plots Field lock 1 Genotype Genotype Genotype lock 2 Genotype Genotype Genotype lock 3 Genotype Genotype Genotype lock 4 Genotype Genotype Genotype opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics 51 6 / 3

7 Randomly ssign Fertilizer mounts within Split Plots Field lock 1 lock 2 lock 3 lock 4 Genotype Genotype Genotype Genotype Genotype Genotype Genotype Genotype Genotype Genotype Genotype Genotype opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics 51 7 / 3

8 n Example Split-Plot Experiment Field Whole Plot or Main Plot lock 1 lock 2 lock 3 lock 4 Genotype Genotype Genotype Genotype Genotype Genotype Genotype Genotype Genotype Genotype Genotype Genotype Split Plot or Sub Plot opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics 51 8 / 3

9 This experiment has two factors: genotype and fertilizer amount. Genotype has levels,, and. Fertilizer has levels, 5, 1, 15 lbs. N / acre. Genotype is called the whole-plot (or main-plot) factor because its levels are randomly assigned to whole plots (main plots). Fertilizer is called the split-plot factor because its levels are randomly assigned to split plots within each whole plot. opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics 51 9 / 3

10 Experimental Units in Split-Plot Designs Whole plots are the whole-plot experimental units because the levels of the whole-plot factor (genotype) are randomly assigned to whole plots. The split-plots are the split-plot experimental units because the levels of the split-plot factor (amount of fertilizer) are randomly assigned to split plots within each whole plot. Thus, we have two different sizes of experimental units in split-plot experimental designs. opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics 51 1 / 3

11 Same Treatment Structure in an RD Field lock lock lock lock opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics / 3

12 Same Treatment Structure in an RD Field opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics / 3

13 Why Use a Split-Plot Design? Split-plot designs usually arise because logistical constraints make a RD or RD impractical. For example, it may be easier to change from one fertilizer level to another as a tractor drives through a field, while it may be more difficult to change from planting one genotype to planting another. In the engineering literature, split-plot designs are sometimes called designs with hard-to-change factors. opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics / 3

14 Recognizing Designs with Split-Plot Structures Many variations on split-plot designs are used for practical reasons. Examples include split-split-plot designs and split-block designs, but the names of these designs are not so important. Pay close attention to the experimental unit to which the levels of each factor are randomly assigned to recognize split-plot-like design structures. opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics / 3

15 Split-plot designs may not involve plots of land. Suppose eight pairs of mice from eight litters are housed in eight cages so that each cage holds two mice from the same litter. Suppose diets 1 and 2 are randomly assigned to the litters with four litters per diet. Within each cage, suppose drugs 1 and 2 are randomly assigned to the mice with one mouse per drug. opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics / 3

16 Split-Plot Experimental Design Diet 1 Drug 2 Drug 1 Diet 2 Drug 2 Drug 1 Diet 1 Drug 1 Drug 2 Diet 1 Drug 1 Drug 2 Diet 2 Drug 1 Drug 2 Diet 2 Drug 2 Drug 1 Diet 2 Drug 2 Drug 1 Diet 1 Drug 2 Drug 1 opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics / 3

17 Diet is the whole-plot treatment factor. Litters are the whole-plot experiment units. Drug is the split-plot treatment factor. Mice are the split-plot experiment units. opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics / 3

18 Diet i = 1, 2, Drug j = 1, 2, Litter k = 1, 2, 3, 4 (within each Diet i) y ijk = µ + α i + β j + γ ij + l ik + e ijk (i = 1, 2; j = 1, 2; k = 1,..., 4) µ + α i + β j + γ ij = mean for Diet i and Drug j l ik = random litter effect = whole-plot exp. unit random effect e ijk = random error effect = split-plot exp. unit random effect opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics / 3

19 y = y 111 y 121 y 112 y 122 y 113 y 123 y 114 y 124 y 211 y 221 y 212 y 222 y 213 y 223 y 214 y 224 β = µ α 1 α 2 β 1 β 2 γ 11 γ 12 γ 21 γ 22 u = l 11 l 12 l 13 l 14 l 21 l 22 l 23 l 24 e = e 111 e 121 e 112 e 122 e 113 e 123 e 114 e 124 e 211 e 221 e 212 e 222 e 213 e 223 e 214 e 224 opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics / 3

20 [ ] X = 1, I 1, 1 I, I 1 I Z = I y = Xβ + Zu + e opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics 51 2 / 3

21 [ ] ([ ] u N, e [ ] [ ]) σl 2I G = σei 2 R Var(Zu) = ZGZ = σlzz 2 [ ] [ = σl 2 I 1 I [ ] = σl 2 I ] = lock Diagonal with blocks [ σ 2 l σ 2 l σ 2 l σ 2 l ] opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics / 3

22 Var(y) = ZGZ + R = σl 2 I 11 + σei = lock Diagonal with blocks [ σ 2 l + σ2 e σ 2 l σ 2 l σ 2 l + σ2 e ] opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics / 3

23 Thus, the covariance between two observations from the same litter is σ 2 l and the correlation is σ l 2. σl 2+σ2 e These computations can also be done using the non-matrix expression of the model. i, j, Var(y ijk ) = Var(µ + α i + β j + γ ij + l ik + e ijk ) = Var(l ik + e ijk ) = σ 2 l + σ 2 e. opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics / 3

24 ov(y i1k, y i2k ) = ov(µ + α i + β 1 + γ i1 + l ik + e i1k, µ + α i + β 2 + γ i2 + l ik + e i2k ) = ov(l ik + e i1k, l ik + e i2k ) = ov(l ik, l ik ) + ov(l ik, e i2k ) + ov(e i1k, l ik ) + ov(e i1k, e i2k ) = ov(l ik, l ik ) = Var(l ik ) = σl. 2 opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics / 3

25 ack to the Traditional Split-Plot Experimental Design Field lock 1 lock 2 lock 3 lock 4 Genotype Genotype Genotype Genotype Genotype Genotype Genotype Genotype Genotype Genotype Genotype Genotype opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics / 3

26 Model for Data from the Traditional Split-Plot Experiment Genotype i = 1, 2, 3, Fertilizer j = 1, 2, 3, 4, lock k = 1, 2, 3, 4 y ijk = µ ij + b k + w ik + e ijk µ ij = mean for Genotype i, Fertilizer j b k = random block effect w ik = random whole-plot exp. unit effect e ijk = random error = random split-plot exp. unit effect opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics / 3

27 To express the model precisely in vector and matrix form as y = Xβ + Zu + e, we will sort the data first by lock, then Genotype, and then Fertilizer: y = [y 111, y 121, y 131, y 141, y 211, y 221, y 231, y 241,..., y 314, y 324, y 334, y 344 ] e = [e 111, e 121, e 131, e 141, e 211, e 221, e 231, e 241,..., e 314, e 324, e 334, e 344 ] opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics / 3

28 X = I 12 12, β = µ 11 µ 12 µ 13 µ 14 µ 21 µ 22 µ 23 µ 24 µ 31 µ 32 µ 33 µ 34 opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics / 3

29 [ ] Z = I 1, I u = [ b w ] = b 1. b 4 w 11 w 21. ([ N ], [ σ 2 b I σ 2 w I ]) w 34 opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics / 3

30 b σb 2 I w N, σw 2 I e [ u e ] ([ N σe 2 I ] [ ]) G, R opyright c 219 Dan Nettleton (Iowa State University) 14. Statistics 51 3 / 3