STAT22200 Chapter 14

Transcription

1 STAT00 Chapter 4 Yibi Huang Chapter 4 Incomplete Block Designs 4. Balanced Incomplete Block Designs (BIBD) Chapter 4 -

2 Incomplete Block Designs A Brief Introduction to a Class of Most Useful Designs in Practice Recall that for a randomized complete block design (RCBD) of g treatments, the size k of each block has to be g (or multiples of g). Under complete block design, each treatment occurs once and only once in each block, so that we can randomize and run all treatment combinations in each block. However, in practice, it often happens that the size of available blocks is smaller than the numbers of treatments (k < g). We cannot apply every treatment to every block. We then have Incomplete Block Design (IBD). IBD makes analysis harder, but sometimes cannot be avoided. The next best thing to a randomized complete block design (RCBD) is a Balanced Incomplete Block Design (BIBD). Chapter 4 -

3 An Example of Unavoidable Incomplete Blocks Eye irritation can be reduced with eyedrops. Three brands of eyedrops are to be compared for their ability to reduce eye irritation. As there is a strong individual effect, subjects should be used as blocks. If each subject can only be used during one treatment period, then the researchers must use one brand of drop in the left eye and another brand in the right eye. The design is restricted into blocks of size two. The study is force into incomplete blocks, with k = < 3 = g (block size) < (number of treatments) Chapter 4-3

4 Balanced Incomplete Block Designs (BIBD) BIBD not balanced in the general sense that all treatment-block combinations occur equally often. Rather they are balanced in the looser sense by the criteria described below. A balanced incomplete block design with g treatments, b blocks, k as the size of each block, r replications of each treatment, is a design satisfying the following: Incomplete: k < g. Balanced: Each treatment appears at most once per block. Each pair of treatments appear in a block the same number of times λ E.g., if treatments A and C appear in the same block 5 times, then treatments B and C must show up together in exactly 5 blocks. Chapter 4-4

5 The following design table is a BIBD. Block size k = 3, there are g = 5 treatments and b = 0 blocks. A B E A C D B E D C B D A C A E C B E D C A B D E A D C B E First Balancing Condition of BIBD If there are b blocks of size k each, then the total number of experimental units is N = bk. If there are g treatments, and each appears r times, then the total number of experimental units is N = rg. Therefore in a BIBD, we must have N = bk = rg. Chapter 4-5

6 Second Balancing Condition of BIBD Suppose each pair of treatments appears in the same block λ times. The second balancing condition r(k ) = λ(g ) comes from two different ways to count the total number of pairs including treatment i 0 in the experiment: g possible pairs, each appears in λ blocks, so λ(g ); treatment i 0 appears in r blocks. Within each block, there are k pairs including i 0, so r(k ) Remarks. In a BIBD, λ = r(k )/(g ) must be an integer. In the eyedrop example: g = 3, k =, r =. λ = r(k ) g = ( ) 3 = is a whole number, and we need b = rg/k = 3 blocks. Chapter 4-6

7 Example A Marketing Psychology Experiment Goal: to compare the effects of 5 advertisements, g = 5 Response: recall after being shown the ad. A subject is a block Block size: Subjects lose patience after being shown 3 ads, k = 3 What is the smallest feasible BIBD? b =? r =? r(k ) By N = rg = kb, and that is a whole number, we need g to make r = 3b 5 and r 4 both being whole numbers So r must be a multiple of and 3, and b be a multiple of and 5.Thus the smallest BIBD is with b = 0, and r = 6. The number of subjects (blocks) must be multiple of 0 (0, 0, 30,... ). Chapter 4-7

8 The following design table is a BIBD with g = 5, k = 3, b = 0, r = 6, λ = 3 (Oehlert Appendix C., p ) A B E A C D B E D C B D A C A E C B E D C A B D E A D C B E In the marketing psychology experiment, if we have b = 0 subjects (blocks), then we can do repetitions of the above design. One obvious randomization is to randomize subjects to columns, then randomize the order of treatments in each block based on the above design. Chapter 4-8

9 Still a Balanced design The following table shows that, in the marketing psychology example, in the 0 blocks, each treatment appears 6 times. Block Treatment A B C D E If we have b = 0 subjects (blocks), we can do repetitions of the above design, and each treatment appears r = 6 = times. Chapter 4-9

10 The following table shows that, in the 0 blocks, each treatment pair appear λ = 3 times. Block Treatment-pair AB AC AD AE BC BD BE CD CE DE If we have b = 0 subjects (blocks), we can do repetitions of the above design, and each treatment pair appears λ = 3 = 6 times. Chapter 4-0

11 Models for BIBD The model for BIBD looks familiar: y ij = µ + α i + β j + ε ij (treatment) (block) (i.i.d. N(0, σ )) for i =,..., g, and j =,..., b with g i= α i = b j= β j = 0. additive model (no trt-block interaction) Not all y ij exist because of incompleteness Nonorthogonality between treatments and blocks, just like unbalanced factorial designs. Thus, Type I Sum of squares will change with the order of terms in the model. Chapter 4 -

12 Parameter Estimates for BIBD Let I ij = {, if treatment i appears in block j, 0, otherwise. and define Q i = y i I ij y j, k j Q j = y j r I ij y i i the least square estimates for µ, α i, β j are µ = y N, α i = kq i λg, βj = kq j λb Remark: Can verify that i Q i = 0 i α i = 0 Chapter 4 -

13 ANOVA for BIBD (Type I Sum of Squares!) Source d.f. SS MS F -value Block b SS block MS block (MS block /MSE) Treatment g SS Trt MS Trt MS trt /MSE Error N g b+ SSE MSE Total N SS total {, if treatment i appears in block j, Let I ij = 0, otherwise. Then SS total = g i= i b j= I ij(y ij y ) SS block = k b j y ) j= (unadjusted) SS Trt = k Qi = λg α i λg k (adjusted for block) SSE = SS total SS block SS trt i Chapter 4-3

14 Pairwise Comparisons Estimate of α i α i is SE( α i α i ) = α i α i = k λg (Q i Q i ) MSE ( ) k λg t-statistic = α i α i SE with df = df of MSE Chapter 4-4

15 Example of BIBD Problem 4.3 on. p The State Board of Education has adopted basic skills tests for high school graduation. One of these is a writing test. The student writing samples are graded by professional graders, and the board is taking some care to be sure that the graders are grading to the same standard. We examine grader differences with the following experiment. There are 5 graders available. We select 30 writing samples at random, each writing sample will be graded by 5 graders. Thus each grader will grade 30 5/5 = 6 samples. Data file: Questions of Interest Do the 5 graders grade consistently? If graders don t grade consistently, how should we adjust the grades? If graders don t grade consistently, can we identify which graders fail to follow the grading standard? Chapter 4-5

16 38 Incomplete Block Designs Exam Grader Score Exam Grader Score Problem 4.4 Analyze these data to determine if graders differ, and if so, how. Be sure to describe the design. Here a exam is a writing sample. Which factor is the treatment factor? Graders or writing samples? Thirty consumers are asked to rate the softness of clothes washed by ten different detergents, but each consumer rates only four different detergents. The design and responses are given below: Which factor is the block factor? Graders or writing samples? Is this a BIBD? Trts Softness Trts Softness A B C D A B C D A B E F A B E F A C G H A C G H A D I J A D I J Chapter 4-6

17 Y ij = µ + α i + β j + ε ij (grader) (exam) As writing samples differ in levels, we expect β j not all equal. If the graders grade consistently, they should give the same score to a writing sample, i.e., α = α = = α 5 > pr4.3 = read.table( " h=t) > EXAM = as.factor(pr4.3$exam) > GRADER = as.factor(pr4.3$grader) > anova(lm(score ~ GRADER + EXAM, data=pr4.3)) Df Sum Sq Mean Sq F value Pr(>F) GRADER <.e-6 *** EXAM <.e-6 *** Residuals > anova(lm(score ~ EXAM + GRADER, data=pr4.3)) Df Sum Sq Mean Sq F value Pr(>F) EXAM <.e-6 *** GRADER e-08 *** Residuals Why are the two ANOVA tables different? Which one should we look at? Chapter 4-7

18 Model diagnostic for y ij = µ + α i + β j + ε ij λ log Likelihood 95% Fitted value Residual Exam Factor Residual Grader Factor Residual Chapter 4-8

19 As the 5 graders grade inconsistently, how to adjust the scores? Based on the model y ij = µ + α i + β j + ε ij, the score of the ith writing sample should be µ + β j, and is estimated by µ + β j. How to get β j in R? Recall that by default, R reports estimates of parameters using the baseline constraints α = β = 0, not the sum-to-zero constraints g i= α i = b j= β j = 0. One can use constrasts() and contr.sum() to force R using the the sum-to-zero constraints. > contrasts(exam) = contr.sum(30) > contrasts(grader) = contr.sum(5) > lm = lm(score ~ EXAM + GRADER, data=pr4.3); lm$coef (Intercept) EXAM EXAM... (omitted) EXAM8 EXAM9 GRADER... (omotted) GRADER Why there are neither estimates for exam30, nor for grader 5? Chapter 4-9

20 > muhat = lm$coef[] > betahat = vector("numeric",length=30) > betahat[:9] = lm$coef[:30] > betahat[30] = -sum(betahat[:9]) > adjustedscore = muhat + betahat; adjustedscore [] [9] [7] [5] > adjustedscore = array(adjustedscore,dim=c(,30)) > adjustedscore = data.frame(adjustedscore) > colnames(adjustedscore) = :30 > adjustedscore Chapter 4-0

21 How to identify inconsistent graders? We can do pairwise comparisons for the grader effects α i α i using the t-statistic = α i α i where SE ( ) ( ) k 5 SE = MSE = λg 5 with df = df of MSE = 96. The t-statistic needs to be at least t 0.05, to be significant at 5% level. That is, α i α i needs to be at least t 0.05,96 SE to be significant at 5% level. Chapter 4 -

22 How to identify inconsistent graders? We don t need to use the formula for estimates of α i s because we can obtain them in R α i α i = k λg (Q i Q i ) > lm$coef[3:54] GRADER GRADER GRADER3 GRADER4 GRADER5 GRADER6 GRADER7 GRADER GRADER9 GRADER0 GRADER GRADER GRADER3 GRADER4 GRADER5 GRADER GRADER7 GRADER8 GRADER9 GRADER0 GRADER GRADER GRADER3 GRADER > sort(lm$coef[3:54]) GRADER3 GRADER5 GRADER6 GRADER6 GRADER5 GRADER4 GRADER8 GRADER GRADER9 GRADER GRADER9 GRADER3 GRADER4 GRADER8 GRADER0 GRADER GRADER7 GRADER GRADER GRADER7 GRADER0 GRADER GRADER GRADER We can see that Grader, 3, 4, and 5 have their α i s differ from most of other α i s by more than They appeared to be inconsistent with other graders. Chapter 4 -

23 exam score Grader 3 always gave the lowest score among the 5 graders grading the same exam Grader 4 always gave the highest score, substantially higher than the scores given by the other graders grading the same exam. Grader tend to give higher scores, Grader 5 tended to give lower score, but not as much as Grader 3 and 4. Chapter 4-3

24 Problem 4. Incomplete Block Design but Not BIBD Milk can be strained through filter disks to remove dirt and debris. Filters are made by surface-bonding fiber webs to both sides of a disk. Goal of the experiment: to know how the construction of the filter affects the speed of milk flow through the filter Treatments: We have a 4 factorial structure for the filters. The 4 factors are fiber weight (normal or heavy), loft (thickness of the filter, normal or low), bonding solution on bottom surface (A or B), and bonding solution on top surface (A or B). Treatments through 6 are the factor-level combinations in standard order. Chapter 4-4

25 Response: filtration time when pouring a measured amount of milk experiment are given below (data from Connor 958): through the disk considerable variation from farm to farm, so block on farm. 6 farms were selected considerable variation from milking to milking, so only one milking per farm. Only 3 filters can be used at a single milking. So blocks size 3 Is this a BIBD? Actually, there is a second blocking factor order Sixteen farms were selected. At each farm there will be three strain at one milking, with the milk strained first with one filter, then a second, a third. Each treatment will be used three times in the design: once as a filter, once as second, and once as third. The treatments and responses for Treatments and Responses Filtration time Farm First Second Third What type of design is this? Analyze the data and report your findings on influence of the treatment factors on straining time. The Chapter State Board 4 - of 5Education has adopted basic skills tests for h

26 pr4. = read.table( " pr4.$wt = factor(pr4.$weight, labels=c("normal", "heavy")) pr4.$lt = factor(pr4.$loft, labels=c("normal","low")) pr4.$bn = factor(pr4.$bottom, labels=c("a","b")) pr4.$tp = factor(pr4.$top, labels=c("a","b")) pr4.$ord = as.factor(pr4.$order, labels=c("st","nd", "3rd")) pr4.$fm = as.factor(pr4.$farm) Chapter 4-6

27 Model diagnostic for log Likelihood % Y ijk = µ + α i + β j + γ k +ε ij (trt) (farm) (order) 0 λ Normal Q Q Plot Residuals Fitted Values Sample Quantiles Theoretical Quantiles Chapter 4-7

28 After removing observation # 39, log Likelihood After Removing Observation #39 95% 0 λ Sample Quantiles 0 Normal Q Q Plot 0 Theoretical Quantiles Residuals Fitted Values Chapter 4-8

29 > anova(lm) Analysis of Variance Table Df Sum Sq Mean Sq F value Pr(>F) fm e-05 *** ord wt e-07 *** lt bn tp *** wt:lt wt:bn *** lt:bn wt:tp lt:tp ** bn:tp wt:lt:bn wt:lt:tp wt:bn:tp lt:bn:tp wt:lt:bn:tp Residuals Chapter 4-9

30 > library(car) > Anova(lm, type=) Anova Table (Type II tests) Sum Sq Df F value Pr(>F) fm *** ord wt e-07 *** lt bn tp *** wt:lt wt:bn *** lt:bn wt:tp lt:tp ** bn:tp wt:lt:bn wt:lt:tp wt:bn:tp lt:bn:tp wt:lt:bn:tp Residuals Chapter 4-30

31 > lma = lm(time ~ fm + ord + wt*lt*bn*tp, data=pr4.[-39,]) > anova(lma) Df Sum Sq Mean Sq F value Pr(>F) fm e-07 *** ord * wt e-09 *** lt * bn ** tp e-06 *** wt:lt wt:bn e-07 *** lt:bn wt:tp lt:tp ** bn:tp wt:lt:bn ** wt:lt:tp ** wt:bn:tp lt:bn:tp wt:lt:bn:tp Residuals Chapter 4-3

32 > library(car) > Anova(lma, type=) Anova Table (Type II tests) Response: time Sum Sq Df F value Pr(>F) fm e-07 *** ord wt e-09 *** lt * bn ** tp e-06 *** wt:lt wt:bn e-06 *** lt:bn wt:tp lt:tp ** bn:tp wt:lt:bn ** wt:lt:tp ** wt:bn:tp lt:bn:tp wt:lt:bn:tp Residuals Chapter 4-3

33 mean of time wt heavy normal farm mean of time lt low normal farm Chapter 4-33

34 mean of time tp B A farm mean of time bn A B farm Chapter 4-34

35 > lm3a = lm(time ~ fm + ord + wt*lt*(bn+tp), data=pr4.[-39,]) > Anova(lm3a, type=) Anova Table (Type II tests) Sum Sq Df F value Pr(>F) fm e-08 *** ord wt e-0 *** lt * bn ** tp e-06 *** wt:lt wt:bn e-07 *** wt:tp lt:bn lt:tp * wt:lt:bn ** wt:lt:tp ** Residuals > anova(lm3a,lma) Analysis of Variance Table Model : time ~ fm + ord + wt * lt * (bn + tp) Model : time ~ fm + ord + wt * lt * bn * tp Res.Df RSS Df Sum of Sq F Pr(>F) Chapter 4-35