1. Review of Lecture level factors Homework A 2 3 experiment in 16 runs with no replicates

Transcription

1 Lecture Review of Lecture level factors Homework A 2 3 experiment in 16 runs with no replicates Lecture 3.1 1

2 2 k Factorial Designs Designs with k factors each at 2 levels 2 k factor level combinations 2 2 = 4 B H L L A H Ref: Lecture 1.1 Multi-factor Designs 2 3 = 8 H B L L A H L H C Lecture 3.1 2

3 Why use 2- level factorial designs? 3 factors each with 2 levels: 2 3 = 8 3 factors each with 3 levels: 3 3 = 27 3 factors each with 4 levels: 4 3 = 64 3 factors with levels 3, 4 and 5, respectively: More reasons later! 3x4x5 = 60 Lecture 3.1 3

4 2 k Factorial Designs A 2 2 experiment Project: optimisation of a chemical process yield Factors (with levels): operating temperature (Low, High) catalyst (C1, C2) Design: Process run at all four possible combinations of factor levels, in duplicate, in random order. Lecture 3.1 4

5 Design matrix Design Point Temperature Catalyst 1 Low 1 2 High 1 3 Low 2 Replicate 1 4 High 2 5 Low 1 6 High 1 7 Low 2 Replicate 2 8 High 2 Columns designate experimental factors Rows designate experimental conditions Lecture 3.1 5

6 Results (standard order) Standard Order Run Order Temperature Catalyst Yield 1 6 Low High Low High Low High Low High 2 80 Lecture 3.1 6

7 Effects as calculated by Minitab Estimated Effects and Coefficients for Yield Term Effect Coef SE Coef T P Constant Temperature Catalyst Temperature*Catalyst S = Effect = Coef x 2 SE(Effect) = SE(Coef) x 2 T effect: 23 = Mean Yield at HighT Mean Yield at Low T Lecture 3.1 7

8 Direct Calculation Temp Cat Rep Rep Variance = Degrees of Rep1 Rep2 1 2 ½(Rep1 Rep2)² Freedom Low High Low High Sum 55 4 Mean = s² s 3.7 YLow Y Y High Y 23 High Low SE(Y 2 High YLow ) 2s / 4 s / t Y High SE(Y High Y Low Y Low ) Lecture 3.1 8

9 Direct Calculation Temp Cat Rep Rep Variance = Degrees of Rep1 Rep2 1 2 ½(Rep1 Rep2)² Freedom Low High Low High Sum 55 4 Mean = s² s 3.7 YLow Y Y High Y 23 High Low SE(Y 2 High YLow ) 2s / 4 s / t Y High SE(Y High Y Low Y Low ) Lecture 3.1 9

10 Classwork Calculate a confidence interval for the Temperature effect. Y High Y Low 23 SE(Y High Y Low ) 2.6 t 4,.05 = 2.78 CI: x to 30.2 Lecture

11 Design matrix Design Point Temperature Catalyst 1 Low 1 2 High 1 3 Low 2 4 High 2 5 Low 1 6 High 1 7 Low 2 8 High 2 Lecture

12 Design matrix: generic notation Design Temperature Catalyst Point A B = "High"; = "Low" Lecture

13 Design Matrix with Y s Design Point Temperature A Catalyst B Yield 1 Y Y Y Y 4 5 Y Y Y Y 8 Main effect estimates: Â ¼ (Y 2 +Y 4 +Y 6 +Y 8 ) ¼ (Y 1 +Y 3 +Y 5 +Y 7 ) Bˆ ¼ (Y 3 +Y 4 +Y 7 +Y 8 ) ¼ (Y 1 +Y 2 +Y 5 +Y 6 ) Lecture

14 Design Matrix with Y s Design Point Temperature A Catalyst B Yield 1 Y Y Y Y 4 5 Y Y Y Y 8 Main effect estimates: Â ¼ (Y 2 +Y 4 +Y 6 +Y 8 ) ¼ (Y 1 +Y 3 +Y 5 +Y 7 ) Bˆ ¼ (Y 3 +Y 4 +Y 7 +Y 8 ) ¼ (Y 1 +Y 2 +Y 5 +Y 6 ) Lecture

15 Design Matrix with Y s Design Temperature Catalyst Yield Point A B 1 Y Y Y Y 4 5 Y Y Y Y 8 Interaction effect estimates: Â at High B ½ (Y 4 +Y 8 ) ½ (Y 3 +Y 7 ) Â at Low B ½ (Y 2 +Y 6 ) ½ (Y 1 +Y 5 ) Lecture

16 The interaction effect A effect at High B: ½ (Y 4 + Y 8 ) ½ (Y 3 + Y 7 ) A effect at Low B: ½ (Y 2 + Y 6 ) ½ (Y 1 + Y 5 ) ½ difference: ¼ (Y 4 + Y 8 ) ¼ (Y 3 + Y 7 ) ¼ (Y 2 + Y 6 ) + ¼ (Y 1 + Y 5 ) = ¼ (Y 4 + Y 8 + Y 1 + Y 5 ) ¼ (Y 3 + Y 7 + Y 2 + Y 6 ) Lecture

17 Design Matrix: applying the signs Temperature A Catalyst B Y 1 Y 1 + Y 2 Y 2 Y 3 + Y 3 + Y 4 + Y 4 Y 5 Y 5 + Y 6 Y 6 Y 7 + Y 7 + Y 8 + Y 8 Effect estimates: Â Sum/4 Bˆ Sum/4 Classwork 3.1.1: Check correspondence with Slide 14. Lecture

18 Augmented Design Matrix with Y s Interaction effect: Design Point A B AB Yield 1 + Y Y Y Y Y Y Y Y 8 (Y1 Y2 Y3 + Y4 + Y5 Y6 Y7 + Y8)/4 Classwork 3.1.2: Check correspondence with Slide 16 Lecture

19 Augmented Design Matrix with Data Design Point A B AB Yield Classwork 3.1.3: Calculate the estimate of the AB interaction. Lecture

20 Dual role of the design matrix Prior to the experiment, the rows designate the design points, the sets of conditions under which the process is to be run. After the experiment, the columns designate the contrasts, the combinations of design point means which measure the main effects of the factors. The extended design matrix facilitates the calculation of interaction effects Lecture

21 Lecture Review of Lecture level factors Homework A 2 3 experiment in 16 runs with no replicates Lecture

22 Homework Test the statistical significance of and calculate confidence intervals for the Catalyst effect and the Temperature by Catalyst interaction effect. Lecture

23 B AB Yield t Y Y SE(Y Y ) Lecture

24 Lecture Review of Lecture level factors Homework A 2 3 experiment in 16 runs with no replicates Lecture

25 Project: Part 2 A 2 3 experiment 3 factors each at 2 levels optimisation of a chemical process yield Factors (with levels): operating Temperature T ( C) (160, 180) raw material Concentration C (%) (20, 40) catalyst K (A, B) Design: Process run at all eight possible combinations of factor levels, in duplicate, in random order. Lecture

26 A 2 3 experiment Design matrix (standard order) Factor Design Point A = T B = C C = K Run order for design points (in duplicate) Run T C K Lecture

27 A 2 3 experiment Results, in standard order T C K Yield Mean SD Ref: PilotPlant.xls Lecture

28 Initial analysis Calculating effects Calculating s Minitab analysis A 2 3 experiment Lecture

29 Initial analysis Lecture

30 Initial analysis Lecture

31 Calculating main effects T C K Mean Classwork 3.1.4: Calculate K main effect Lecture

32 Calculating interaction effects, the extended design matrix Design Point T C K TC TK CK TCK Mean Classwork 3.1.5: Complete the missing columns. Calculate the TK and TCK interactions Lecture

33 Calculating s T C K Yield Variance =½(diff) 2 Lecture

34 Calculating s T C K Yield Variance =½(diff) Total 64 s 2 8 s Lecture

35 Classwork Calculate the t-ratio for the Temperature effect and the 3-factor interaction. What conclusions do you draw? Lecture

36 Minitab analysis Estimated Effects for Yield Term Effect SE T P T C K T*C T*K C*K T*C*K S = Lecture

37 Diagnostic analysis 3 Normal Probability Plot of the Residuals (response is Y) Deleted Residual Score 1 2 Lecture

38 Diagnostic analysis 3 Residuals Versus the Fitted Values (response is Y) Deleted Residual Fitted Value Lecture

39 Exercise An experiment was run to assess the effects of three factors on the life of a cutting tool A: Cutting speed B: Tool geometry C: Cutting angle. The full 2 3 design was replicated three times. The results are shown in the next slide and are available in Excel file Tool Life.xls. Carry out a full analysis and report. Lecture

40 Exercise Cutting Speed Tool Geometry Cutting Angle Tool Life Ref: Tool Life.xls Exercise 3.1.2: Web Exercises Lecture

41 Lecture Review of Lecture A 2 3 experiment in 16 runs with no replicates Normal plot, Pareto chart, Lenth's method Reduced model method Design projection method Lecture

42 Part in 16 runs, no replicates Project: Improving the filtration rate of a chemical manufacturing process Key factor: Formaldehyde concentration. Problem: Proposal: Design: Reducing the level of formaldehyde concentration reduces the filtration rate to an unacceptably low level. Raise levels of Temperature, Pressure and Stirring rate. 2 4 unreplicated, random run order Lecture

43 Data Temperature Pressure Formaldehyde Stirring Filtration Concentration Rate Rate Low Low Low Low 45 High Low Low Low 71 Low High Low Low 48 High High Low Low 65 Low Low High Low 68 High Low High Low 60 Low High High Low 80 High High High Low 65 Low Low Low High 43 High Low Low High 100 Low High Low High 45 High High Low High 104 Low Low High High 75 High Low High High 86 Low High High High 70 High High High High 96 Ref: Formaldehyde.xls Lecture

44 Initial analysis Lecture

45 Initial analysis Lecture

46 Minitab / DOE / Factorial / Analyse Factorial Design Estimated Effects Term Effect T P F S T*P T*F T*S P*F P*S F*S T*P*F T*P*S T*F*S P*F*S T*P*F*S Lecture

47 No replication: alternative analyses Normal plots of effects if no effects present, estimated effects reflect chance variation, follow Normal model a few real effects will appear as exceptions in a Normal plot Lenth method alternative estimate of s, given few real effects Best approach: combine both! Lecture

48 Normal Effects Plot Effect Normal Plot of the Effects A C D AD Effect Type Not Significant Significant Factor A B C D Name T P F S AC Score 1 2 Lenth's PSE = Lecture

49 Alternative view: Pareto Chart vital few versus trivial many (Juran) Lecture

50 Minitab Pareto Chart Lecture

51 Lenth's method Given several Normal values with mean 0 and given their absolute values (magnitudes, or values without signs), then it may be shown that SD(Normal values) 1.5 median(absolute values). Denote this by s 0 Given a small number of effects with mean 0, then SD(Normal values) is inflated. Refinement: PSE 1.5 median(abs values < 2.5 s 0 ) Lecture

52 Calculating PSE Term Effect T P F S T*P T*F T*S P*F P*S F*S T*P*F T*P*S T*F*S P*F*S T*P*F*S Lecture

53 Calculating PSE, ignore signs Term Effect T P F S T*P T*F T*S P*F P*S F*S T*P*F T*P*S T*F*S P*F*S T*P*F*S Lecture

54 Calculating PSE, order the effects No. Term Effect 1 T*P P*S F*S T*P*F*S T*F*S T*P*F P*F P*F*S P T*P*S F S T*S T*F T Lecture

55 Calculating PSE, order the effects No. Term Effect 1 T*P P*S F*S T*P*F*S T*F*S T*P*F P*F P*F*S P T*P*S F S T*S T*F T PSE avge = 1.75 x 1.5 = s 0 x 1.5 = 3.94 x 2.5 = 9.84 Lecture

56 Calculating PSE From Excel, find median(absolute Values) = 2.625, so initial SE is s 0 = = s 0 = , 5 values exceed. The median of the remaining 10 is mean of and = 1.75 Hence, PSE = = Check Slide 50 Lecture

57 Assessing statistical significance Critical value for effect is t.05,df PSE df (number of effects)/3 t.05,5 = 2.57 PSE = Critical value = 6.75 Check Slide 50 Lecture

58 Estimating s PSE = is the (pseudo) standard error of an estimated effect. SE(effect) = (s 2 /8 + s 2 /8) = s/2. s = 5.25 Lecture

59 Lecture Review of Lecture A 2 3 experiment in 16 runs with no replicates Normal plot, Pareto chart, Lenth's method Reduced model method Design projection method Lecture

60 Reduced Model method Select identified terms for a fitted model omitted terms provide basis for estimating s check degrees of freedom Estimate effects ANOVA used to calculate s Check diagnostics Lecture

61 Reduced Model method Y equals overall mean plus T effect plus F effect plus S effect plus TF interaction effect plus TS interaction effect plus chance variation Lecture

62 Estimated effects (Minitab) Estimated Effects and Coefficients for R (coded units) Term Effect Coef SE Coef T P Constant T F S T*F T*S S = Lecture

63 Analysis of Variance (basis for s) Source DF SS MS F-Value P-Value T F S T*F T*S Error Total Lecture

64 Diagnostics 2 Versus Fits (response is R) 1 Deleted Residual Fitted Value Lecture

65 Diagnostics 3 Normal Probability Plot (response is R) 2 Deleted Residual Score 1 2 Lecture

66 Lecture Review of Lecture A 2 3 experiment in 16 runs with no replicates Normal plot, Pareto chart Lenth's method Reduced model method Design projection method Lecture

67 Design Projection Method Pressure not statistically significant, exclude leave 2 3, duplicated T F S FR Lecture

68 Design Projection Method Initial analysis Lecture

69 Design Projection Method Initial analysis Lecture

70 Design Projection Method Minitab analyis Estimated Effects and Coefficients for R (coded units) Term Effect Coef SE Coef T P Constant T F S T*F T*S F*S T*F*S S = Lecture

71 Design Projection Method Minitab analyis Lecture

72 Design Projection Method Minitab analyis Lecture

73 Identify optimal operating conditions List means and SE's for 8 design points: T*F*S Mean SE Mean Lecture

74 Identify optimal operating conditions Calculate confidence interval for optimum yield. CI = = ( 94.63, ) Classwork Test the statistical significance of the difference between the optimal yield and the 'next best' yield Calculate a confidence interval for the 'next best' yield. Lecture

75 Identify optimal operating conditions T*F*S Mean SE Mean Lecture

76 Lecture Review of Lecture A 2 3 experiment in 16 runs with no replicates Lecture

77 Laboratory 1, Thursday February 4, 6-8 Lecture

78 Laboratory 1, Thursday February 4, 6-8 Students will work in pairs (or threes where necessary) but not singly, with a view to exploiting synergy in solving Laboratory problems, promoting collaborative learning. Laboratory tasks involve analysing, discussing and reporting on the results of designed experiments. Laboratory handouts give detailed guidance on the use of Mintab. Lecture

79 Laboratory 1, Thursday February 4, 6-8 The laboratory handouts include the following: Invitations to consider the results of Minitab analysis and their statistical and substantive interpretations are printed in italics. Take some time for this; consult your neighbour or tutor. Enter your responses in a Word document, as a log of your work and as draft contributions to a report on the experiments and their analyses. Lecture

80 Laboratory 1, Thursday February 4, 6-8 The laboratory handouts include Learning Objectives. Students should check these objectives as they proceed through the Laboratory. At the end of each Laboratory, students are invited to and review the Learning Objectives check whether they have been achieved. Lecture

81 Laboratory 1, Thursday February 4, 6-8 A Feedback document will be circulated at the end of the Laboratory, containing solutions to the Laboratory tasks asides, constituting Extra Notes on relevant topics. Laboratory handout will appear on the module web page tomorrow, Feedback document will appear a week later. There will be a Minute Test at the end. Lecture

82 Reading EM 5.3, 5.4, 5.6 DCM 6-2, 6-3 to p.221, 6.5 to p. 237 (BHH 5.14 (Lenth plots) and all of Ch. 5!) Extra Notes: Lenth s PSE Lecture