STAT 430 (Fall 2017): Tutorial 8

Transcription

1 STAT 430 (Fall 2017): Tutorial 8 Balanced Incomplete Block Design Luyao Lin November 7th/9th, 2017 Department Statistics and Actuarial Science, Simon Fraser University

2 Block Design Complete Random Complete Block Design Incomplete Block Design Balanced Incomplete Block Design (BIBD) Latin Square Design 1

3 Incomplete Block Design Why we need it? Advantage of it? Disadvantage of it? How to do it? 2

4 Why Incomplete Block Design? (Motivation) Block sizes Complete Block Design, every block can hold every treatment level at least once. Incomplete Block Design, the block size is less than the treatment level. Suppose we have 4 different treatment levels, but each block can only have 2 experiment units. Balanced Incomplete Block Design (BIBD) 3

5 Balanced Incomplete Block Design (BIBD) v treatment level r: each treatment appears in r blocks b blocks; b > r k each block size (# of units each block can have) λ: each pair of the treatment appears together in λ blocks (why we care about this?) total # of units: bk or vr 4

6 Balanced Incomplete Block Design (BIBD) v treatment level r each treatment appears in r blocks b blocks; b > r k each block size (# of units each block can have) λ: each pair of the treatment appears together in λ blocks total # of units: bk or vr 5

7 Properties of BIBD Requirements: (i) Each treatment must appear the same number of times in the design: vr = bk (ii) each pair of treatments appears together in λ blocks r(k 1) = λ(v 1) Advantage: all treatment contrasts estimable all pairwise comparisons are estimated with the same variance tends to give the shortest CIs for contrasts Disadvantage: for certain value of b, k, v, r, BIBD might not exist. If b = v = 8, r = k = 3, λ =? 6

8 r(k 1) = λ(v 1) for treatment say A, it appears in r blocks within those r blocks, there are in total r(k 1) treatments other than A itself on the other hand, we have v 1 other treatment levels A is supposed to appear with each of them in λ blocks so that λ(v 1) = r(k 1) v treatment level r each treatment appears in r blocks b blocks; b > r k each block size (# of units each block can have) λ: each pair of the treatment appears together in λ blocks total # of units: bk or vr 7

9 Revisit the example 8

10 How to analyze BIBD? Y hi = µ + θ h + τ i + ɛ ij ɛ i.i.d. N(0, σ 2 ) h = 1,..., b, i = 1,..., v (h, i) in the design τ i = 0 θ h = 0 Assumptions for Y hi N(µ + θ h + τ i, σ 2 ) normality equal variance independence no interaction effect (hard to verify) 9

11 Parameter Estimation: least square unadjusted estimate: ˆτ i = Ȳ.i E(Ȳ.1 Ȳ.2 ) = τ 1 τ ( 7 θ h 7 τ 1 τ 2 adjusted estimate (section ) (Intra-block equations) r(k 1)ˆτ i λ p i h=4 ˆτ p = kq i Q i = T i 1 n hi B h k T i is the total of response on treatment i B h is the total of response in block h n hi is 1 if i is in block h or 0 otherwise h 11 h=8 θ h ) 10

12 adjusted estimate (section ) r(k 1)ˆτ i λ p i ˆτ p = kq i along with i ˆτ i = 0 r(k 1)ˆτ i + λˆτ i = kq i kq i ˆτ i = r(k 1) + λ = kq i λv = kt i h n hib h λv because λ(v 1) = r(k 1) ˆτ i = kq i λv 11

13 Other Parameters Estimate Y hi = µ + θ h + τ i + ɛ ij ˆτ i = kq i λv ˆµ = G bk ; G is the grand total ˆσ 2 = mse = SSE bk b v+1 ˆθ h = B h k G v bk i=1 n hi ˆτ i k 12

14 ANOVA table bk b v 1 SSE = = bk 1 SStot = b 1 SSθ = 1 k v 1 SST adj = b h=1 i=1 b h=1 i=1 b h=1 i=1 v i=1 v n hi êhi 2 = v yhi 2 1 k b h=1 v Q i ˆτ 2 i F 0 = MsT adj MSE b h=1 i=1 b Bh 2 h=1 y 2 hi 1 bk G 2 B 2 h 1 bk G 2 v (y hi ˆµ ˆθ h ˆτ i ) 2 v i=1 Q i ˆτ 2 i 13

15 CI for the contrasts ˆτ i = k λv Q i Var(Q i ) = σ 2 Var( c i ˆτ i ) = c 2 i CI for c i ˆτ i : k λv σ2 ( k ) ci Q i ± w c 2 k i λv λv ˆσ2 w B = t tk b v+1,α/2m, w T = q v,bk b v+1,α / 2 Section has a good example for BIBD page

16 The Latin Square Design 15

17 Big picture The Latin Square Design 1. incomplete block design 2. one treatment 3. two blocking variables 4. only one single treatment is applied within each combination of blocking variables 5. the level of treatment = the level of two blockings operators Material A=24 B=20 C=19 D=24 E=24 2 B=17 C=24 D=30 E=27 A=36 3 C=18 D=38 E=26 A=27 B=21 4 D=26 E=31 A=26 B=23 C=22 5 E=22 A=30 B=20 C=29 D=31 16

18 why use a Latin Square? 1. impossible to use each treatment level for the same combination of blocking 2. for example, consider an experiment with four diets, each to be given to four cows in succession 3. each cow can only be given a single diet during a single time period. 4. one block factor is the cow, the other block factor is the time or the order time cow C A D B 2 A B C D 3 D C B A 4 B D A C 5. Latin square is not unique; as long as each letter (treatment) appears in each row and column exactly once. 17

19 R function latin = function (n, nrand = 20) { x = matrix ( LETTERS [1: n], n, n) x = t(x) for (i in 2:n) x[i, ] = x[i, c(i:n, 1:( i - 1))] if ( nrand > 0) { for (i in 1: nrand ) { x = x[ sample (n), ] x = x[, sample (n)] } } x } latin (5) [,1] [,2] [,3] [,4] [,5] [1,] "E" "D" "C" "B" "A" [2,] "D" "C" "B" "A" "E" [3,] "C" "B" "A" "E" "D" [4,] "A" "E" "D" "C" "B" [5,] "B" "A" "E" "D" "C" latin (4) [,1] [,2] [,3] [,4] [1,] "A" "C" "B" "D" [2,] "B" "D" "C" "A" [3,] "D" "B" "A" "C" [4,] "C" "A" "D" "B" 18

20 Model y ijk = µ + α i + τ j + β k + ɛ ijk y ijk : ith row, kth col, jth treatment; i, j, k = 1, 2,..., p µ overall mean, α i ith row effect; τ j jth treatment effect; β k kth column effect αi = τ j = β k = 0 ɛ ijk iid N(0, σ 2 ) N = p 2 SS T =SS rows + SS col + SS treatment + SS E p 2 1 =p 1 + p 1 + p 1 + (p 2)(p 1) 19

21 ANOVA SS Rows = 1 p SS T = i y : total for all = i y i : total for row i y k : total for row k i SS Treatment = 1 p y 2 i y 2 N y j k : total for treatment j yijk 2 y 2 N j k j k y ijk j y 2 j y 2 N SS Cols = 1 p k y 2 k y 2 N 20

22 Model Continued 1. Hull hypothesis: τ 1 = τ 2 =... = τ p = 0 2. Alternative hypothsis: at least one of them is not zero 3. Test statistics: 4. residuals are given as F = MS treatment MS E = SS treatmet/(p 1) SS E /(p 2)(p 1) e ijk = y ijk ŷ ijk = y ijk y i y j y k + 2y 21