Contents Factorial design TAMS38 - Lecture 6 Factorial design, Latin Square Design Lecturer: Zhenxia Liu Department of Mathematics - Mathematical Statistics 28 November, 2017 Complete three factor design Point estimates SS Hypothesis Additive three factor design Pairwise comparisons Example 1 Randomization and Block The coming tasks Complete randomized design Randomized block design Latin square design Factorial design 3 The factorial design is a design involving two or more factors in an experiment. Complete / Full factorial design = contains all possible level combinations of factors. E.g. Complete two-way ANOVA = Complete two factor design with analysis of variance. Additive factorial design = does not contain all possible level combinations of factors. E.g. Additive two-way ANOVA = Additive two factor design with analysis of variance. Factorial designs are efficient and provide extra information - the interactions between the factors, which can not be obtained when using single factor designs. Complete three factor design 4 An experiment is affected by three factors A, B, C with levels A 1,..., A a, B 1,..., B b and C 1,..., C c, respectively. Model There are n measurements / observations for each level combination A i B j C k. Let y ijkl be the l th observation for the level combination A i B j C k. There are abcn observations in total. Complete three factor design or Complete three-way ANOVA Y ijkl = µ + τ i + β j + γ k + (τβ) ij + (τγ) ik + (βγ) jk + (τβγ) ijk + ε ijkl = µ ijk + ε ijkl where ε ijkl N(0, σ) and independent for i = 1,..., a, j = 1,..., b, k = 1,..., c and l = 1,..., n.
Complete three factor design 5 Constraint a τ i = 0, 1 b β j = 0, 1 a (τβ) ij = 0, j, i=1 a (τγ) ik = 0 k, i=1 b (βγ) jk = 0 k, j=1 c γ k = 0, 1 b (τβ) ij = 0 i, j=1 c (τγ) ik = 0 i, k=1 c (βγ) jk = 0 j, k=1 a (τβγ) ijk = 0 j, k, i=1 b (τβγ) ijk = 0 i, k, j=1 c (τβγ) ijk = 0 i, j. k=1 Complete three factor design - SS 7 We can divide the total sum of squares (SS T ) into the following: SS A = nbc a ˆτ 2 i=1 i, df = a 1 SS B = nac b ˆβ j 2, df = b 1 j=1 SS C = nab c k=1 ˆγ2 k, df = c 1 SS AB = nc a b (τβ) 2 ij, df = (a 1)(b 1). SS ABC = n i a b SS E = i=1 j c j=1 i=1 j=1 k=1 l=1 k (τβγ) 2 ijk, df = (a 1)(b 1)(c 1) n (y ijkl ȳ ijk ) 2, dfe = abc(n 1) Complete three factor design - Point estimates 6 Point estimates ˆµ = ȳ ˆτ i = ȳ i ȳ ˆβj = ȳ j ȳ ˆγ k = ȳ k ȳ (τβ) ij = ȳ ij ˆτ i ˆβ j ˆµ = ȳ ij ȳ i ȳ j + ȳ (τγ) ik = ȳ i k ȳ i ȳ k + ȳ (βγ) jk = ȳ jk ȳ j ȳ k + ȳ (τβγ) ijk = ȳ ijk (τβ) ij (τγ) ik (βγ) jk ˆτ i ˆβ j ˆγ k ˆµ ˆσ =? = ȳ ijk }{{} =ˆµ ijk ȳ ij ȳ i k ȳ jk + ȳ i + ȳ j + ȳ k ȳ Note: The corresponding point estimator of the above point estimates are unbiased. Complete three factor design - SS 8 Remark of SS Under constraint and certain conditions just as before, it also can show that the various sums of squares are independent from each other, and sum of squares divided by σ 2 are χ 2 -distributed with the specified degrees of freedom. For example, (i) SS A, SS B..., and SS E (ii) SS E σ 2 χ 2 (abc(n 1)). are independent. (iii) If τ 1 =... = τ a = 0, then SS A σ 2 χ 2 (a 1). (iv)... Then we have ˆσ 2 = s 2 = SS E abc(n 1) = SS E dfe.
Complete three factor design - Hypothesis 9 Complete three factor design - Hypothesis 10 Hypothesis on main effects A, B, C, and interaction effects. For example 1. Test on factor A H 0A : τ 1 =... = τ a = 0 i.e. No effect of factor A Test statistic v A = SS A/(a 1) SS E /dfe, if v A > F α (a 1, dfe), reject H 0A. 2. Test on interaction between A and B H 0AB : (τβ) ij = 0 i, j i.e. no interaction between A and B Test statistic v AB = SS AB/[(a 1)(b 1)] SS E /dfe. If v AB > F α ((a 1)(b 1), dfe), then reject H 0AB. 3. Test on interaction between A, B and C H 0ABC : (τβγ) ijk = 0 i, j, k 4.... i.e. no interaction between A, B and C Test statistic v ABC = SS ABC/[(a 1)(b 1)(c 1)] SS E /dfe. If v ABC > F α ((a 1)(b 1)(c 1), dfe), then reject H 0ABC. Additive three factor design 11 Model =? It depends!!! Additive three factor design or Additive three-way ANOVA Compare to full/complete design Y ijkl = µ + τ i + β j + γ k + (τβ) ij + (τγ) ik + (βγ) jk + (τβγ) ijk + ε ijkl How many additive models can we get? The answer is ( 4 1) + ( 4 2) + ( 4 3) + ( 4 4) = 15. Remark: The formulas for SS in additive design is the same as in complete design except SS E the sum of squares of errors of additive design. Additive three factor design - SS 12 If we have additive design Y ijkl = µ + τ i + β j + γ k + (τγ) ik + (βγ) jk + (τβγ) ijk + ε ijkl Then we have SS E = SS AB + SS E and degrees of freedom for SS E is dfe = (a 1)(b 1) + abc(n 1). If we have additive design Y ijkl = µ + τ i + β j + γ k + (τγ) ik + (βγ) jk + ε ijkl Then we have SS E = SS AB + SS ABC + SS E and degrees of freedom for SS E is dfe = (a 1)(b 1) + (a 1)(b 1)(c 1) + abc(n 1)....
Three factor design - Pairwise comparisons 13 Example 1 14 Pairwise comparisons for Three factor design We also can apply, e.g., Tukey s method or t-interval method to do Pairwise comparisons for main effects A, B, C and interaction effects. For example, For main effect A, I τi τ j = ȳ i... ȳ j... ± q α (a, dfe) s n. For interaction effects between B and C, let m ij = µ + β i + γ j + (βγ) ij, I mij m kl = ȳ.ij. ȳ.kl. ± q α (bc, dfe) s n. One wants to examine three factors that influence the production of a certain kind of units. A: skill levels of staff. Levels: A1, A2, A3 (decreasing) B: the complexity of the method. Levels: B1 (high), B2 (low) C: tiredness. Levels C1 (am), C2 (pm) A full factorial design with two observations per cell was conducted and a measurement y of quality and quantity were determined. Note: n =? Example 1 - continued 15 Example 1 - continued 16 MTB > print Y Data Display Y 92 88 88 84 78 82 78 77 70 74 69 71 69 66 63 67 62 59 60 58 64 60 62 59 MTB > set c2 DATA> (1 2 3)8 DATA> end MTB > set c3 DATA> 3(1 1 1 1 2 2 2 2) DATA> end MTB > set c4 DATA> 6(1 1 2 2) DATA> end MTB > anova Y=A B C. ANOVA: Y versus A, B, C Factor Type Levels Values A fixed 3 1, 2, 3 B fixed 2 1, 2 C fixed 2 1, 2 Analysis of Variance for Y Source DF SS MS F P A 2 2151.58 1075.79 195.60 0.000 B 1 104.17 104.17 18.94 0.001 C 1 32.67 32.67 5.94 0.031 A*B 2 116.58 58.29 10.60 0.002 A*C 2 3.08 1.54 0.28 0.760 B*C 1 0.17 0.17 0.03 0.865 A*B*C 2 1.08 0.54 0.10 0.907 Error 12 66.00 5.50 Total 23 2475.33 S = 2.34521 R-Sq = 97.33% R-Sq(adj) = 94.89%
Example 1 - continued 17 Example 1 - additive design 18 MTB > ANOVA Y = A B C ; SUBC> Means A B C. ANOVA: Y versus A, B, C Factor Type Levels Values A fixed 3 1, 2, 3 B fixed 2 1, 2 C fixed 2 1, 2 Analysis of Variance for Y Source DF SS MS F P A 2 2151.58 1075.79 260.03 0.000 B 1 104.17 104.17 25.18 0.000 A*B 2 116.58 58.29 14.09 0.000 C 1 32.67 32.67 7.90 0.012 Error 17 70.33 4.14 Total 23 2475.33 S = 2.03402 R-Sq = 97.16% R-Sq(adj) = 96.16% Example 1 - compare full / additive design 19 Output for Full design v.s. Additive design Analysis of Variance for Y=A B C Example 1 - interaction 20 Source DF SS MS F P A 2 2151.58 1075.79 195.60 0.000 B 1 104.17 104.17 18.94 0.001 C 1 32.67 32.67 5.94 0.031 A*B 2 116.58 58.29 10.60 0.002 A*C 2 3.08 1.54 0.28 0.760 B*C 1 0.17 0.17 0.03 0.865 A*B*C 2 1.08 0.54 0.10 0.907 Error 12 66.00 5.50 Total 23 2475.33 Analysis of Variance for Y=A B C Source DF SS MS F P A 2 2151.58 1075.79 260.03 0.000 B 1 104.17 104.17 25.18 0.000 A*B 2 116.58 58.29 14.09 0.000 C 1 32.67 32.67 7.90 0.012 Error 17 70.33 4.14 Total 23 2475.33 The skilled workers (A1) seem to perform worse when they change from B1 to B2. The workers A3 obtain approximately the same results for B1 and B2.
Example 1 - sample means 21 Example 1 - pairwise comparisons 22 Means A N Y C N Y 1 8 83.375 1 12 72.000 2 8 68.625 2 12 69.667 3 8 60.500 The simultaneous confidence level is at least 90%. B N Y 1 12 72.917 2 12 68.750 A B N Y 1 1 4 88.000 1 2 4 78.750 2 1 4 71.000 2 2 4 66.250 3 1 4 59.750 3 2 4 61.250 Note: Values for factor C suggest that there are better results in the morning than in the afternoon. Randomization and Block 23 Think about which factors influence the results. How can eliminate or at least reduce the influence of possible disturbing factors? Randomization: Block: We randomly assign the levels of the factors to the experimental units or we randomly choose experimental units to each level of the factors. It can reduce the influence of the disturbing factors such that it at least not systematically influence. We consider block factors which are the factors may affect the measurements, but not a factor of interest, such as the times, the operators and the temperatures. It can reduce or eliminate the errors. Example 2 24 One wants to compare the performance of four different varieties of wheat V 1, V 2, V 3, V 4 and have at his disposal a field divided into boxes: Forest (sv. skog) Meadow (sv. äng) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Lake (sv. sjö) Arable (sv. åker)
Complete randomized design 25 Randomization: We randomly choose four boxes for V 1, four for V 2, four for V 3 and four for V 4. For example, V 1 : 04, 08, 15, 02 V 2 : 07, 01, 14, 05 V 3 : 13, 16, 11, 06 V 4 : 03, 09, 10, 12 Model 1: Completely randomized design - One-Way ANOVA One factor: Wheat. where i τ i = 0 and ε ij N(0, σ). Y ij = µ + τ i + ε ij is assumed to be independent and Randomized block design 26 If the rows on the field suspected to be different, then we consider the rows as block factor. We randomly choose the boxes within the row for each wheat type. Randomization in rows = Randomization in one block factor V 2 V 1 V 3 V 4 V 3 V 1 V 2 V 4 V 3 V 4 V 1 V 2 V 4 V 2 V 1 V 3 Note: τ i describe effects of wheat. Randomized block design 27 Latin square design 28 Model 2 Randomized block design - additive two-way ANOVA Two factors: Wheat and block factor rows. Y ij = µ + τ i + β j + ε ij where i τ i = 0, j β j = 0 and ε ij is assumed to be independent and N(0, σ). Note: τ i β j describe effects of wheat. describe effect of block factor rows. If there are differences both between rows and columns on the field, then we consider both rows and columns as block factors. We randomly choose the boxes within the row for each wheat type and within the column for each wheat type, respectively, such that every wheat type occurs exactly once in each row and exactly once in each column. Then we make a design according to a Latin square: Randomization in both rows and columns = Randomization in two block factors V 2 V 4 V 3 V 1 V 4 V 3 V 1 V 2 V 3 V 1 V 2 V 4 V 1 V 2 V 4 V 3
Latin square design 29 Model 3: Latin square design Three factors: Wheat, block factor rows and block factor columns. Y ijk = µ + τ i + β j + γ k + ε ijk, i τ i = 0, j β j = 0, k γ k = 0 and ε ijk is assumed to be independent and N(0, σ). Note: τ i β j γ k describe effects of wheat. describe effect of block factor rows. describe effect of the other block factor columns. Latin square design 30 Assume we have three factors A, B, C, with levels A 1,..., A p, B 1,..., B p and C 1,..., C p, respectively. We are interested in factor A, thereby factors B and C are block factors. Model: Latin square design Y ijk = µ + τ i + β j + γ k + ε ijk, i τ i = 0, j β j = 0, k γ k = 0 and ε ijk N(0, σ) and are independent for i = 1,..., p, j = 1,..., p and k = 1,..., p. Note: τ i describe effects of factor A. β j describe block effect of factor B. γ k describe block effect of factor C. Latin square design 31 Latin square design - Point estimates 32 Point estimates B C A 1 A p... A 2 A 2 A 4... A 3............ A 3 A 1... A p ˆµ = ȳ, ˆτ i = ȳ i ȳ, Note: Every level of A occurs exactly once in each row and exactly once in each column. The data is a p p-square. ˆβ j = ȳ j ȳ, ˆγ k = ȳ k ȳ. ˆσ =?
Latin square design - SS 33 Latin square design - Hypothesis 34 We can divide the total of sum of squares (SS T ) into the followings and SS E = i,j,k SS A = p p (ȳ i ȳ ) 2, df = p 1, i=1 SS B = p p j=1 (ȳ j ȳ ) 2, df = p 1, SS C = p p k=1 (ȳ k ȳ ) 2, df = p 1 (y ijk ˆµ ˆτ i ˆβ j ˆγ k ) 2 = i,j,k(y ijk ȳ i ȳ j ȳ k + 2ȳ ) 2, dfe = (p 2)(p 1). Hypothesis 1. 2. Test on factor A ˆσ 2 = s 2 = SS E (p 2)(p 1) = SS E dfe H 0A : τ 1 =... = τ p = 0 i.e. A levels are equaivalent Test statistic v A = SS A/(p 1) SS E /dfe if v A > F α (p 1, dfe), reject H 0A. Latin square design - Pairwise comparisons 35 Example 3 36 Pairwise comparisons We also can apply, e.g., Tukey s method or t-interval method to do Pairwise comparisons for factors A, B, C. For example, I τi τ j = ȳ i.. ȳ j.. ± q α (a, dfe) s n = ȳ i.. ȳ j.. ± q α (p, dfe) s p Remark: Graeco-Latin Square design The purpose of a study was to see if music during work effected the production in a factory. Four different music programs A, B, C and D were compared with no music at all, E. Each program was played for a day and one wanted five replicates for each program, i.e., the study lasted for five weeks. Since there can also be variations between the days and between the weeks the design for the study was chosen to be a latin square. Three factors:
Example 3 - continued 37 Example 3 - continued 38 Results: Week Monday Tuesday Wednesday Thursday Friday 1 A 133 B 139 C 140 D 140 E 145 2 B 136 C 141 D 143 E 146 A 139 3 C 140 A 138 E 142 B 139 D 139 4 D 129 E 132 A 137 C 136 B 140 5 E 132 D 144 B 143 A 142 C 142 Note that each music program is present one time in each row and in each column. a) According to which model were data analyzed? Does the music seem to have no effect on production? Perform an appropriate test level 5%. b) Is production worse on Mondays than on other days? Motivate your answer with use of confidence interval on the simultaneous confidence level at least 80%. You can assume that even before you saw the result you suspected that production would be the worst on Mondays. c) Does the normal distribution assumption seem to be reasonable? Explain your answer briefly. Data are analyzed using Minitab, see below. Example 3 - continued 39 Example 3 - continued 40 MTB > GLM Y = week day music; SUBC> Means week day music. General Linear Model: Y versus week, day, music Analysis of Variance for Y, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P week 4 123.44 123.44 30.86 3.07 0.059 day 4 177.84 177.84 44.46 4.42 0.020 music 4 11.84 11.84 2.96 0.29 0.876 Error 12 120.72 120.72 10.06 Total 24 433.84 S = 3.17175 R-Sq = 72.17% R-Sq(adj) = 44.35% Least Squares Means for Y week Mean SE Mean 1 139.4 1.418 2 141.0 1.418 3 139.6 1.418 4 134.8 1.418 5 140.6 1.418 day 1 134.0 1.418 2 138.8 1.418 3 141.0 1.418 4 140.6 1.418 5 141.0 1.418 music 1 137.8 1.418 2 139.4 1.418 3 139.8 1.418 4 139.0 1.418 5 139.4 1.418
Example 3 - continued 41 Example 3 - continued 42 a) Week i, day j, music k give observations y ijk where Y ijk = µ + τ i + β j + γ k + ε ijk b) We construct intervals I βj β 1 by applying t interval method. 1 ( 4 1) α 80%, then α = 5%. We have and ε ijk N(0, σ) and independent for i = 1,..., 5, j = 1,..., 5 and k = 1,..., 5. Note: p = 5. We test H 0M : γ 1 =... = γ 5 = 0. The effect of music seems to be negligible. Production seems to be lower on Mondays. Example 3 - continued 43 Example 3 - continued 44 c) Neither the normal probability plot or histogram is really good, but the normal distribution can probably be good enough approximation.
The coming tasks 45 Thank you 46 Lab 4 is on Friday Download and print the labs. Work in a group of two or three. Once you have done or tried all the tasks, you can queue. Then a teacher will check your solutions and may ask a few questions. If the teacher thinks your solutions and replies are OK, then you will sign a passing sheet for each lab. Thank you! Lesson 7 2.3.1, 2.3.2, 2.3.3 and 2.2.4. Home Assignment 2 must be submitted by 5 PM on Friday, Dec. 1st.