Lecture 3.11 © 2014 Michael Stuart Design and Analysis of Experiments Lecture 3.1 1.Review of Lecture 2.2 –2-level factors –Homework 2.2.1 2.A 2 3 experiment.

Lecture 3.11 © 2014 Michael Stuart Design and Analysis of Experiments Lecture 3.1 1.Review of Lecture 2.2 –2-level factors –Homework 2.2.1 2.A 2 3 experiment 3.2 4 in 16 runs with no replicates Certificate in Statistics Design and Analysis of Experiments

Lecture 3.12 © 2014 Michael Stuart 2 k Factorial Designs Designs with k factors each at 2 levels 2 k factor level combinations –2 2 = 4 –2 3 = 8 L H A H B L L H A H B L H L C Ref: Lecture 1.1 Multi-factor Designs Certificate in Statistics Design and Analysis of Experiments

Lecture 3.13 © 2014 Michael Stuart 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: 3x4x5 = 60 More reasons later! Certificate in Statistics Design and Analysis of Experiments

Lecture 3.14 © 2014 Michael Stuart 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. 2 k Factorial Designs Certificate in Statistics Design and Analysis of Experiments

Lecture 3.15 © 2014 Michael Stuart Design matrix Replicate 1 Replicate 2 Rows designate experimental conditions Design Point TemperatureCatalyst 1Low1 2High1 3Low2 4High2 5Low1 6High1 7Low2 8High2 Certificate in Statistics Design and Analysis of Experiments

Lecture 3.17 © 2014 Michael Stuart Effects as calculated by Minitab Estimated Effects and Coefficients for Yield Term Effect Coef SE Coef T P Constant 64.25 1.31 49.01 0.000 Temperature 23.0 11.50 1.31 8.77 0.001 Catalyst 1.5 0.75 1.31 0.57 0.598 Temperature*Catalyst 10.0 5.00 1.31 3.81 0.019 S = 3.70810 Effect = Coef x 2 SE(Effect) = SE(Coef) x 2 T effect: 23 = Mean Yield at HighT – Mean Yield at Low T Certificate in Statistics Design and Analysis of Experiments

Lecture 3.110 © 2014 Michael Stuart Classwork 2.2.2 Calculate a confidence interval for the Temperature effect. t 4,.05 = 2.78 CI:23  2.78 x 2.6 23  7.2 15.8 to 30.2 Certificate in Statistics Design and Analysis of Experiments

Lecture 3.111 © 2014 Michael Stuart Design matrix Rows designate experimental conditions Design Point TemperatureCatalyst 1Low1 2High1 3Low2 4High2 5Low1 6High1 7Low2 8High2 Certificate in Statistics Design and Analysis of Experiments

Lecture 3.112 © 2014 Michael Stuart Design matrix: generic notation + = "High";– = "Low" Design Point Temperature A Catalyst B 1–– 2+– 3–+ 4++ 5–– 6+– 7–+ 8++ Certificate in Statistics Design and Analysis of Experiments

Lecture 3.113 © 2014 Michael Stuart Design Matrix with Y’s Effect estimates: ¼ (Y 2 +Y 4 +Y 6 +Y 8 ) – ¼ (Y 1 +Y 3 +Y 5 +Y 7 ) ¼ (Y 3 +Y 4 +Y 7 +Y 8 ) – ¼ (Y 1 +Y 2 +Y 5 +Y 6 ) Design Point Temperature A Catalyst B Yield 1––Y1Y1 2+–Y2Y2 3–+Y3Y3 4++Y4Y4 5––Y5Y5 6+–Y6Y6 7–+Y7Y7 8++Y8Y8 Certificate in Statistics Design and Analysis of Experiments

Lecture 3.114 © 2014 Michael Stuart 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 ) Certificate in Statistics Design and Analysis of Experiments

Lecture 3.115 © 2014 Michael Stuart Design Matrix: applying the signs Effect estimates: Sum/4 Sum/4 Classwork 3.1.1: Check correspondence with Slide 13. Temperature A Catalyst B – Y 1 + Y 2 – Y 2 – Y 3 + Y 3 + Y 4 – Y 5 + Y 6 – Y 6 – Y 7 + Y 7 + Y 8 Certificate in Statistics Design and Analysis of Experiments

Lecture 3.116 © 2014 Michael Stuart Augmented Design Matrix with Y’s Interaction effect: (Y1 – Y2 – Y3 + Y4 + Y5 – Y6 – Y7 + Y8)/4 Classwork 3.1.2: Check correspondence with Slide 14 Certificate in Statistics Design and Analysis of Experiments

Lecture 3.118 © 2014 Michael Stuart 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 Certificate in Statistics Design and Analysis of Experiments

Lecture 3.119 © 2014 Michael Stuart Lecture 3.1 1.Review of Lecture 2.2 –2-level factors –Homework 2.2.1 2.A 2 3 experiment 3.2 4 in 16 runs with no replicates Certificate in Statistics Design and Analysis of Experiments

Lecture 3.120 © 2014 Michael Stuart Homework 2.2.1 Test the statistical significance of and calculate confidence intervals for the Catalyst effect and the Temperature by Catalyst interaction effect. Certificate in Statistics Design and Analysis of Experiments

Lecture 3.121 © 2014 Michael Stuart Class work 3.1.4: t-tests of effects B: AB: BABYield –+60 ––72 +–52 ++83 –+54 ––68 +–45 ++80 Certificate in Statistics Design and Analysis of Experiments From Slide 9,

Lecture 3.123 © 2014 Michael Stuart Lecture 3.1 1.Review of Lecture 2.2 –2-level factors –Homework 2.2.1 2.A 2 3 experiment 3.2 4 in 16 runs with no replicates Certificate in Statistics Design and Analysis of Experiments

Lecture 3.124 © 2014 Michael Stuart Part 2A 2 3 experiment 3 factors each at 2 levels Project: 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. Certificate in Statistics Design and Analysis of Experiments

Lecture 3.131 © 2014 Michael Stuart Calculating interaction effects, the extended design matrix Classwork 3.1.6: Complete the missing columns. Calculate the TK and TCK interactions Certificate in Statistics Design and Analysis of Experiments

Lecture 3.134 © 2014 Michael Stuart Classwork 3.1.7 Calculate the t-ratio for the T effect and the 3-factor interaction. What conclusions do you draw? Certificate in Statistics Design and Analysis of Experiments

Lecture 3.135 © 2014 Michael Stuart Minitab analysis Estimated Effects for Yield Term Effect SE T P T 23.0 1.414 16.26 0.000 C -5.0 1.414 -3.54 0.008 K 1.5 1.414 1.06 0.320 T*C 1.5 1.414 1.06 0.320 T*K 10.0 1.414 7.07 0.000 C*K 0.0 1.414 0.00 1.000 T*C*K 0.5 1.414 0.35 0.733 S = 2.82843 Certificate in Statistics Design and Analysis of Experiments

Lecture 3.138 © 2014 Michael Stuart Exercise 3.1.1 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. Certificate in Statistics Design and Analysis of Experiments

Lecture 3.140 © 2014 Michael Stuart Lecture 3.1 1.Review of Lecture 2.2 2.A 2 3 experiment 3.2 4 in 16 runs with no replicates –Normal plot, Pareto chart, Lenth's method –Reduced model method –Design projection method Certificate in Statistics Design and Analysis of Experiments

Lecture 3.141 © 2014 Michael Stuart Project:Improving the filtration rate of a chemical manufacturing process Key factor:Formaldehyde concentration. Problem:Reducing the level of formaldehyde concentration reduces the filtration rate to an unacceptably low level. Proposal:Raise levels of –Temperature, –Pressure and –Stirring rate. Design:2 4 unreplicated, random run order Part 32 4 in 16 runs, no replicates Certificate in Statistics Design and Analysis of Experiments

Lecture 3.145 © 2014 Michael Stuart Minitab / DOE / Factorial / Analyse Factorial Design Estimated Effects Term Effect T 21.625 P 3.125 F 9.875 S 14.625 T*P 0.125 T*F -18.125 T*S 16.625 P*F 2.375 P*S -0.375 F*S -1.125 T*P*F 1.875 T*P*S 4.125 T*F*S -1.625 P*F*S -2.625 T*P*F*S 1.375 Certificate in Statistics Design and Analysis of Experiments

Lecture 3.146 © 2014 Michael Stuart 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! Certificate in Statistics Design and Analysis of Experiments

Lecture 3.150 © 2014 Michael Stuart 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 ) Certificate in Statistics Design and Analysis of Experiments

Lecture 3.151 © 2014 Michael Stuart Calculating PSE Term Effect T 21.625 P 3.125 F 9.875 S 14.625 T*P 0.125 T*F -18.125 T*S 16.625 P*F 2.375 P*S -0.375 F*S -1.125 T*P*F 1.875 T*P*S 4.125 T*F*S -1.625 P*F*S -2.625 T*P*F*S 1.375 Certificate in Statistics Design and Analysis of Experiments

Lecture 3.152 © 2014 Michael Stuart Calculating PSE, ignore  signs Term Effect T 21.625 P 3.125 F 9.875 S 14.625 T*P 0.125 T*F 18.125 T*S 16.625 P*F 2.375 P*S 0.375 F*S 1.125 T*P*F 1.875 T*P*S 4.125 T*F*S 1.625 P*F*S 2.625 T*P*F*S 1.375 Certificate in Statistics Design and Analysis of Experiments

Lecture 3.153 © 2014 Michael Stuart No. TermEffect 1 T*P0.125 2 P*S0.375 3 F*S1.125 4 T*P*F*S 1.375 5 T*F*S1.625 6 T*P*F1.875 7 P*F2.375 8 P*F*S2.625 9 P3.125 10 T*P*S4.125 11 F9.875 12 S14.625 13 T*S16.625 14 T*F18.125 15 T21.625 Calculating PSE, order the effects Certificate in Statistics Design and Analysis of Experiments

Lecture 3.154 © 2014 Michael Stuart No. TermEffect 1 T*P0.125 2 P*S0.375 3 F*S1.125 4 T*P*F*S 1.375 5 T*F*S1.625 6 T*P*F1.875 7 P*F2.375 8 P*F*S2.625 9 P3.125 10 T*P*S4.125 11 F9.875 12 S14.625 13 T*S16.625 14 T*F18.125 15 T21.625 x 2.5 = 9.84x 1.5 = 3.94 x 1.5 = 2.625 avge = 1.75 Calculating PSE, order the effects s0s0 PSE Certificate in Statistics Design and Analysis of Experiments

Lecture 3.155 © 2014 Michael Stuart From Excel, find median(Absolute Values) = 2.625, so initial SE is s 0 = 1.5 × 2.625 = 3.9375. 2.5 × s 0 = 9.84375, 5 values exceed. The median of the remaining 10 is mean of 1.625 and 1.875 = 1.75 Hence, PSE = 1.5 × 1.75 = 2.625. Check Slide 47 Calculating PSE Certificate in Statistics Design and Analysis of Experiments

Lecture 3.156 © 2014 Michael Stuart Assessing statistical significance Critical value for effect is t.05,df × PSE df ≈ (number of effects)/3 t.05,5 = 2.57 PSE = 2.625 Critical value = 6.75 Check Slide 49 Certificate in Statistics Design and Analysis of Experiments

Lecture 3.157 © 2014 Michael Stuart Estimating  PSE = 2.625 is the (pseudo) standard error of an estimated effect. SE(effect) =  (s 2 /8 + s 2 /8) = s/2. s ≈ 2 × 2.625 = 5.25 Certificate in Statistics Design and Analysis of Experiments

Lecture 3.158 © 2014 Michael Stuart Lecture 3.1 1.Review of Lecture 2.2 2.A 2 3 experiment 3.2 4 in 16 runs with no replicates –Normal plot, Pareto chart, Lenth's method –Reduced model method –Design projection method Certificate in Statistics Design and Analysis of Experiments

Lecture 3.159 © 2014 Michael Stuart Reduced Model method Select identified terms for a fitted model –omitted terms provide basis for estimating  Estimate effects –ANOVA used to calculate s Check diagnostics Certificate in Statistics Design and Analysis of Experiments

Lecture 3.160 © 2014 Michael Stuart Reduced Model method Y equalsoverall mean plus T effect plus F effect plus S effect plus TF interaction effect plus TS interaction effect plus chance variation Certificate in Statistics Design and Analysis of Experiments

Lecture 3.161 © 2014 Michael Stuart Estimated effects (Minitab) Estimated Effects and Coefficients for R (coded units) Term Effect Coef SE Coef T P Constant 70.063 1.104 63.44 0.000 T 21.625 10.812 1.104 9.79 0.000 F 9.875 4.938 1.104 4.47 0.001 S 14.625 7.312 1.104 6.62 0.000 T*F -18.125 -9.062 1.104 -8.21 0.000 T*S 16.625 8.313 1.104 7.53 0.000 S = 4.41730 Certificate in Statistics Design and Analysis of Experiments

Lecture 3.162 © 2014 Michael Stuart Analysis of Variance (basis for s) Source DF Adj SS Adj MS F-Value P-Value Model 5 5535.8 1107.16 56.74 0.000 Linear 3 3116.2 1038.73 53.23 0.000 T 1 1870.6 1870.56 95.86 0.000 F 1 390.1 390.06 19.99 0.001 S 1 855.6 855.56 43.85 0.000 2-Way Int'ns 2 2419.6 1209.81 62.00 0.000 T*F 1 1314.1 1314.06 67.34 0.000 T*S 1 1105.6 1105.56 56.66 0.000 Error 10 195.1 19.51 Total 15 5730.9 Certificate in Statistics Design and Ana'iments

Lecture 3.165 © 2014 Michael Stuart Lecture 3.1 1.Review of Lecture 2.2 2.A 2 3 experiment 3.2 4 in 16 runs with no replicates –Normal plot, Pareto chart –Lenth's method –Reduced model method –Design projection method Certificate in Statistics Design and Analysis of Experiments

Lecture 3.166 © 2014 Michael Stuart Design Projection Method Pressure not statistically significant, exclude leave 2 3, duplicated TFSFR –––45 –––48 +––71 +––65 –+–68 –+–80 ++–60 ++–65 ––+43 ––+45 +–+100 +–+104 –++75 –++70 +++86 +++96 Certificate in Statistics Design and Analysis of Experiments

Lecture 3.169 © 2014 Michael Stuart Design Projection Method Minitab analyis Estimated Effects and Coefficients for R (coded units) Term Effect Coef SE Coef T P Constant 70.063 1.184 59.16 0.000 T 21.625 10.812 1.184 9.13 0.000 F 9.875 4.938 1.184 4.17 0.003 S 14.625 7.312 1.184 6.18 0.000 T*F -18.125 -9.062 1.184 -7.65 0.000 T*S 16.625 8.313 1.184 7.02 0.000 F*S -1.125 -0.562 1.184 -0.48 0.647 T*F*S -1.625 -0.813 1.184 -0.69 0.512 S = 4.73682 Certificate in Statistics Design and Analysis of Experiments

Lecture 3.172 © 2014 Michael Stuart Identify optimal operating conditions T*F*S Mean SE Mean – – – 46.50 3.349 + – – 68.00 3.349 – + – 74.00 3.349 + + – 62.50 3.349 – – + 44.00 3.349 + – + 102.00 3.349 – + + 72.50 3.349 + + + 91.00 3.349 Certificate in Statistics Design and Analysis of Experiments

Lecture 3.173 © 2014 Michael Stuart Identify optimal operating conditions Calculate confidence interval for optimum yield. CI = 102.0  2.2 × 3.35 = ( 94.63, 109.37 ) Classwork 3.1.8 Calculate a confidence interval for the 'next best' yield. Certificate in Statistics Design and Analysis of Experiments

Lecture 3.174 © 2014 Michael Stuart Identify optimal operating conditions T*F*S Mean SE Mean – – – 46.50 3.349 + – – 68.00 3.349 – + – 74.00 3.349 + + – 62.50 3.349 – – + 44.00 3.349 + – + 102.00 3.349 – + + 72.50 3.349 + + + 91.00 3.349 Certificate in Statistics Design and Analysis of Experiments

Lecture 3.177 © 2014 Michael Stuart Laboratory 1, Thursday March 12, 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. Certificate in Statistics Design and Analysis of Experiments

Lecture 3.178 © 2014 Michael Stuart Laboratory 1, Thursday March 12, 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. Certificate in Statistics Design and Analysis of Experiments

Lecture 3.179 © 2014 Michael Stuart Laboratory 1, Thursday March 12, 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 –review the Learning Objectives and –check whether they have been achieved. Certificate in Statistics Design and Analysis of Experiments

Lecture 3.180 © 2014 Michael Stuart Laboratory 1, Thursday March 12, 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. Certificate in Statistics Design and Analysis of Experiments

Lecture 3.181 © 2014 Michael Stuart 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 Analysis Certificate in Statistics Design and Analysis of Experiments

Lecture 3.11 © 2014 Michael Stuart Design and Analysis of Experiments Lecture 3.1 1.Review of Lecture 2.2 –2-level factors –Homework 2.2.1 2.A 2 3 experiment.

Similar presentations

Presentation on theme: "Lecture 3.11 © 2014 Michael Stuart Design and Analysis of Experiments Lecture 3.1 1.Review of Lecture 2.2 –2-level factors –Homework 2.2.1 2.A 2 3 experiment."— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Lecture 3.11 © 2014 Michael Stuart Design and Analysis of Experiments Lecture 3.1 1.Review of Lecture 2.2 –2-level factors –Homework 2.2.1 2.A 2 3 experiment.

Similar presentations

Presentation on theme: "Lecture 3.11 © 2014 Michael Stuart Design and Analysis of Experiments Lecture 3.1 1.Review of Lecture 2.2 –2-level factors –Homework 2.2.1 2.A 2 3 experiment."— Presentation transcript:

Similar presentations

About project

Feedback