Contents TAMS38 - Lecture 8 2 k p fractional factorial design Lecturer: Zhenxia Liu Department of Mathematics - Mathematical Statistics Example 0 2 k factorial design with blocking Example 1 2 k p fractional factorial design 2 4 1 fractional factorial design Example 2 The coming tasks 5 December, 2017 Example 0 - Glazing ceramic 3 Example 0 - continued 4 Model for 2 3 factorial design When glazing ceramic people use three color pigments A, B, C. To find a good combination, people deign a 2 3 experiment where each pigment occur in low or high dose. The glazing result are then being scored. Hence, the collected data, y, is the score for the different level combinations and only one observation for each combination. Y ijk = µ + τ i + β j + γ k + (τβ) ij + (τγ) ik + +(βγ) jk + (τβγ) ijk + ε ijk, where ε ijk N(0, σ) and independent for i = 1, 1, j = 1, 1, k = 1, 1 with the corresponding constraint. Note: there are 2 3 = 8 level combinations, which are (1) a b ab c bc abc Therefore we need 8 observations. The planning is complicated by the oven, where the glazing will take place. The oven only manages four objects at a time. Thus we have to divide the 8 level combinations into two batches. How?
Example 0 - continued 5 We can divide the 8 level combinations into two batches according to different levels of a level combination. The batches is a new factor to this model, which is a block factor. That is, we add a new block factor (batches) to this model, the new block factor is denoted by D. New model y ijkl = µ + τ i + β j + (τβ) ij + γ k + (τγ) ik + (βγ) jk + (τβγ) ijk + q l + ε ijkl with original setting and l = 1, 1 and the new constraint l q l = 0. Two batches = two blocks =? 2 k factorial design with blocking 7 2 k factorial design with blocking 2 k factorial design with blocking is that we add block factor batches to 2 k factorial design. Why do we study this? Restrictions caused by equipments or instruments. Batches of raw material. Time. Fees.... Usually we get batches in the following way For 2 2 factorial design, We let block factor C = AB. For 2 3 factorial design, We let block factor D = ABC. For 2 4 factorial design, We let block factor E = ABCD.... Example 0 - continued 6 We choose one level combination, and get the corresponding two batches or two blocks in the following way. For example,... We choose level combination a, i.e. we let D = A. D = 1 D = 1 First batch: a ab ac abc Block 1 l = 1 Second batch: (1) b c bc Block 2 l = 1 We choose level combination abc, i.e. we let D = ABC. D = 1 D = 1 First batch: a b c abc Block 1 l = 1 Second batch: (1) ab ac bc Block 2 l = 1 2 k factorial design with blocking 8 Now we use 2 3 factorial design to explain why it is better to choose block factor as D = ABC. Before we add block factor, we have the following point estimates equation. ȳ (1). ȳ a. ȳ b. ȳ ab. ȳ c. ȳ ac. ȳ bc. ȳ abc. } {{ } =ȳ I A B AB C AC BC ABC ˆµ 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ˆτ 1 1 1 1 1 1 1 1 1 ˆβ = 1 1 1 1 1 1 1 1 (τβ) 11 1 1 1 1 1 1 1 1 ˆγ 1 1 1 1 1 1 1 1 (τγ) 11 1 1 1 1 1 1 1 1 (βγ) 11 1 1 1 1 1 1 1 1 (τβγ) 111 That is, ȳ = ˆµI + ˆτ 1 A + ˆβ 1 B +... + (τβγ) 111 ABC =F = ˆξ
2 k factorial design with blocking 9 2 k factorial design with blocking 10 If we get two batches by setting block factor D = A, then the right hand side of point estimates equation will have one more term: ˆq 1 1 1 1 1 1 1 1 1 = ˆq 1 D = ˆq 1 A level combination: Then we get new point estimates equation on next page. (1) l = 1 a l = 1 b l = 1 ab l = 1 c l = 1 ac l = 1 bc l = 1 abc l = 1 ȳ (1). ȳ a. ȳ b. ȳ ab. ȳ c. ȳ ac. ȳ bc. ȳ abc. } {{ } =ȳ I A B AB C AC BC ABC ˆµ 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ˆτ 1 + ˆq 1 1 1 1 1 1 1 1 1 ˆβ = 1 1 1 1 1 1 1 1 (τβ) 11 1 1 1 1 1 1 1 1 ˆγ 1 1 1 1 1 1 1 1 (τγ) 11 1 1 1 1 1 1 1 1 (βγ) 11 1 1 1 1 1 1 1 1 (τβγ) 111 Which means that we can only estimate the sum (τ 1 + q 1 ) (not the parameters separately). This is of course not good ( why?) and we must do in some other way. =F = ˆξ 2 k factorial design with blocking 11 Now we set block factor D = ABC, then we get the point estimates equation ȳ (1). ȳ a. ȳ b. ȳ ab. ȳ c. ȳ ac. ȳ bc. ȳ abc. } {{ } =ȳ I A B AB C AC BC ABC ˆµ 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ˆτ 1 1 1 1 1 1 1 1 1 ˆβ = 1 1 1 1 1 1 1 1 (τβ) 11 1 1 1 1 1 1 1 1 ˆγ 1 1 1 1 1 1 1 1 (τγ) 11 1 1 1 1 1 1 1 1 (βγ) 11 1 1 1 1 1 1 1 1 (τβγ) 111 + ˆq 1 =F Remark ) The point estimates ( (τβγ)111 + ˆq 1 can not be separated. The block effect D and the ABC-effect can not be separated. = ˆξ Example 1 - Chemical process 12 For a chemical process one have measured the level of contamination in the finished product, then each of the four factors A, B, C and D has been varied on two levels. The experiment was conducted in two blocks, where the division between the blocks were done following the rule Resultat: H = ABCD. (1) 1.43 c 1.53 d 1.54 cd 1.84 a 1.35 ac 1.61 ad 1.67 acd 1.70 b 1.22 bc 1.35 bd 1.48 bcd 1.48 ab 1.35 abc 1.27 abd 1.45 abcd 1.59
Example 1 - continued 13 Example 1- continued 14 (a) Which observations are included in each block? Block H = 1 : Block H = 1 : a, b, c, abc, d, abd, acd, bcd (1), ab, ac, bc, ad, bd, cd, abcd (b) What effects seem to be most significant? Justify your answer. Is there a difference between the blocks? MTB > let c17=c17/100 MTB > copy c17 m2 MTB > copy c1-c16 m1 MTB > trans m1 m3 MTB > mult m3 m2 m4 MTB > copy m4 c18 MTB > let c19=c18/16 MTB > set c20 DATA> 1:16 DATA> end MTB > sort c19 c20 c21 c22 MTB > print c21-c22 Data Display Row C21 C22 1-0.09250 3 2-0.03125 7 3-0.01125 6 4-0.00500 14 5-0.00125 11 6 0.00125 10 7 0.00250 8 8 0.00250 12 9 0.00375 13 10 0.00750 15 11 0.00750 2 12 0.00875 4 13 0.04875 16 14 0.05500 5 15 0.10250 9 16 1.49125 1 Example 1 - continued 15 Example 1 - continued 16 MTB > copy c21 c23; SUBC> omit 16. MTB > nscores c23 c24 MTB > plot c24*c23 MTB > ANOVA Y = B C D H; SUBC> Means B C D H. ANOVA: Y versus B, C, D, H Analysis of Variance for Y Source DF SS MS F P B 1 0.136900 0.136900 231.05 0.000 C 1 0.048400 0.048400 81.69 0.000 B*C 1 0.015625 0.015625 26.37 0.000 D 1 0.168100 0.168100 283.71 0.000 H 1 0.038025 0.038025 64.18 0.000 Error 10 0.005925 0.000593 Total 15 0.412975 S = 0.0243413 R-Sq = 98.57% R-Sq(adj) = 97.85% b) The most significant effects are number 3, 7, 9, 5, 16, i.e., B, BC, D, C and (ABCD + H). And there is significant difference between the blocks.
Example 1- continued 17 (c) Estimate the expected value of the level combination that looks to be the best. Example 1 - continued 18 c) Better value is lower value. Means B N Y -1 8 1.5838 1 8 1.3987 C N Y -1 8 1.4363 1 8 1.5463 D N Y -1 8 1.3887 1 8 1.5938 H N Y -1 8 1.4425 1 8 1.5400 B C N Y -1-1 4 1.4975-1 1 4 1.6700 1-1 4 1.3750 1 1 4 1.4225 2 k p fractional factorial design 19 B-level 1, C-level 1, D-level 1, block 1 give µ 1 1 1 1 = 1.3750 0.10250 0.04875 1.224 2 4 1 fractional factorial design 20 2 k p fractional factorial design 2 k p fractional factorial design is a two level factorial design which consists of a carefully chosen subset (fraction) of a full two level factorial design. That is, 2 k p fractional factorial design is 2 p = 1 2 p of a full 2 k factorial design. k = the number of factors investigated. p describes the size of the fraction of the full factorial design used. p is also the number of generators (generator =???). Remark: The fractional factorial design is useful when we can t afford enough time and money to replicate the full factorial design. 2 4 1 fractional factorial design Suppose we want to do a 2 4 factorial design with 4 factors A, B, C, D which needs 16 observations. But we only can take 8 observations. How can we choose 8 level combinations from 16 level combinations? How can we estimate 16 unknown parameters from 8 observations? How can we estimate 16 effects from 8 observations? 2 4 1 fractional factorial design is generated from a full 2 3 factorial design by choosing an alias structure. (alias structure=???). Remark: A fractional factorial design is generated from a full factorial design by choosing an alias structure.
2 4 1 fractional factorial design 21 We apply the rule D = ABC to get the levels for factor D. For 2 4 1 fractional design, we have the following 16 level combinations. (1), a, b, ab, c, ac, bc, abc d ad bd abd cd acd bcd abcd We choose 8 level combinations by applying the rule D = ABC on the first 8 level combinations which are level combinations for the full 2 3 design. We get the 8 level combinations: 2 4 1 design: (1), ad, bd, ab, cd, ac, bc, abcd Now we can take one observation from each level combination, then we get 8 observations: y (1) y ad y bd y ab y cd y ac y bc y abcd Note: Of course, you also can choose other rules. 2 4 1 fractional factorial design 23 2 4 1 fractional factorial design 22 In the full 2 3 design for the factors A, B, C, we have the following point estimates equation. ȳ (1). ȳ a. ȳ b. ȳ ab. ȳ c. ȳ ac. ȳ bc. ȳ abc. } {{ } =ȳ I A B AB C AC BC ABC ˆµ 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ˆτ 1 1 1 1 1 1 1 1 1 ˆβ = 1 1 1 1 1 1 1 1 (τβ) 11 1 1 1 1 1 1 1 1 ˆγ 1 1 1 1 1 1 1 1 (τγ) 11 1 1 1 1 1 1 1 1 (βγ) 11 1 1 1 1 1 1 1 1 (τβγ) 111 Where ABC is the column vector in F for (τβγ) 111. =F The rule D = ABC gives the point estimate for the sum (τβγ) 111 + δ 1. 2 4 1 fractional factorial design 24 = ˆξ Aliases The rule D = ABC gives the point estimate for the sum (τβγ) 111 + δ 1. We call that the main effect D and the interaction effect ABC are aliases, or the main effect D is aliased with the interaction effect ABC. There are more aliases according to the rule D = ABC. Aliases means that two or more parameters can not be separated as they get exactly the same coefficient vector in the point estimates equation. Let I, A, B, AB,... be the column vectors in the design matrix F. We know that AB = A B if the multiplication is elementwise. Further, BC = B C, A 2 = I, B 2 = I etc. Hence, D = ABC D 2 = DABC I = ABCD Generator I = ABCD is called the generator for the design. Note: For 2 4 1 design, the number of generator is 1, i.e., p = 1. The generator gives more aliases, thereby we get the alias structure.
2 4 1 fractional factorial design 25 Alias structure The alias structure determines which effects are confounded with each other. We apply generator I = ABCD, we get the following aliases. more aliases I = ABCD A = A 2 BCD = BCD B = AB 2 CD = ACD AB = A 2 B 2 CD = CD C = ABD AC = BD BC = AD alias structure base effect + ABCD effect A effect + BCD effect B effect + ACD effect AB effect + CD effect C effect + ABD effect AC effect + BD effect BC effect + AD effect ABC = D ABC effect + D effect. 2 4 1 fractional factorial design 26 Hence, 8 observations corresponding to the following parameters in the point estimates equation. y (1) y ad y bd y ab y cd y ac y bc y abcd = F ˆµ + ( τβγδ) 1111 ˆτ 1 + ( βγδ) 111 ˆβ 1 + ( τγδ) 111 ( τβ) 11 + ( γδ) 11 ˆγ 1 + ( τβδ) 111 ( τγ) 11 + ( βδ) 11 ( βγ) 11 + ( τδ) 11 ( τβγ) 111 + ˆδ 1 Note: we get the corresponding 16 effects: base + ABCD effect, A + BCD effect, B + ACD effect, AB + CD effect, C + ABD effect, AC + BD effect, BC + AD effect, ABC + D effect. 2 4 1 fractional factorial design 27 We summarize how to do 2 4 1 fractional design. The 16 level combinations in the 2 4 1 fractional design are (1) a b ab c ac bc abc d ad bd abd cd acd bcd abcd Apply the rule D = ABC to (1) a b ab c ac bc abc We choose these 8 level combinations to take observation (1), ad, bd, ab, cd, ac, bc, abcd Apply the generator I = ABCD, we get the corresponding effects: base + ABCD effect, A + BCD effect, B + ACD effect AB + CD effect, C + ABD effect, AC + BD effect BC + AD effect, ABC + D effect. Note: The red effects from 2 3 design. 2 k p fractional factorial design 28 Resolution Resolution is amount of the factors in the generator that has the smallest number of factors. The design above has resolution IV. Remark: Resolution IV designs are the designs in which no main effects aliased with any other main effect or with any two-factor interaction, but two-factor interactions are aliased with each other.
Example 2 29 Example 2- continued 30 In 2 5 1 fractional factorial design with factors A, B, C, D and E. The factor E was applied according to the rule E = ABCD. Results: e 1.43 c 1.53 d 1.54 cde 1.84 a 1.35 ace 1.61 ade 1.67 acd 1.70 b 1.22 bce 1.35 bde 1.48 bcd 1.48 abe 1.35 abc 1.27 abd 1.45 abcde 1.59 What effects seem to be most significant? Justify your answer. See minitab. MTB > let c17=c17/100 MTB > copy c17 m2 MTB > copy c1-c16 m1 MTB > trans m1 m3 MTB > mult m3 m2 m4 MTB > copy m4 c18 MTB > let c19=c18/16 MTB > set c20 DATA> 1:16 DATA> end MTB > sort c19 c20 c21 c22 MTB > print c21-c22 Data Display Row C21 C22 1-0.09250 3 2-0.03125 7 3-0.01125 6 4-0.00500 14 5-0.00125 11 6 0.00125 10 7 0.00250 8 8 0.00250 12 9 0.00375 13 10 0.00750 15 11 0.00750 2 12 0.00875 4 13 0.04875 16 14 0.05500 5 15 0.10250 9 16 1.49125 1 Example 2 - continued 31 Example 2 - continued 32 MTB > ANOVA Y = B C D E; SUBC> Means B C D E. ANOVA: Y versus B, C, D, E Analysis of Variance for Y MTB > copy c21 c23; SUBC> omit 16. MTB > nscores c23 c24 MTB > plot c24*c23 Source DF SS MS F P B 1 0.136900 0.136900 231.05 0.000 C 1 0.048400 0.048400 81.69 0.000 B*C 1 0.015625 0.015625 26.37 0.000 D 1 0.168100 0.168100 283.71 0.000 E 1 0.038025 0.038025 64.18 0.000 Error 10 0.005925 0.000593 Total 15 0.412975 S = 0.0243413 R-Sq = 98.57% R-Sq(adj) = 97.85%
2 k p fractional factorial design 33 The difference between 2 k factorial design with blocking and 2 k p fractional factorial design. For example, 2 4 design with one block factor v.s. 2 4 1 fractional design. 2 4 factorial design with one block factor There are 4 factors A, B, C, D and 16 observations, then add one more block factor E - batches. There is no interaction with the block factor E. 2 4 1 fractional factorial design There are 4 factors A, B, C, D with 8 observations, then apply the rule, say D = ABC, to get the levels of factor D. There are interactions with factor D. The coming tasks 34 Lab 5 is on Thursday 8:00-10:00. 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. Home Assignment 3 must be submitted by 5 PM on Friday, Dec. 15th. Note: Try to avoid designs like this. Always verify your results with extra observations. Thank you 35 Thank you!