Lecture 2.11 © 2016 Michael Stuart Design and Analysis of Experiments Lecture 2.1 1.Review –Minute tests 1.2 –Homework –Experimental factors with several.

Lecture 2.11 © 2016 Michael Stuart Design and Analysis of Experiments Lecture 2.1 1.Review –Minute tests 1.2 –Homework –Experimental factors with several levels 2.Analysis of Variance 3.Randomised blocks design: illustrations 4.Randomised blocks design and analysis: case study Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.14 © 2016 Michael Stuart Exercise 1.2.1 Process Development Study Process:pellet making Requirement:specification limits for pellet size Problem:proportion meeting specification too low Proposal:change machine speed from A to B Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.15 © 2016 Michael Stuart Postgraduate Certificate in Statistics Design and Analysis of Experiments Replication Naive experiment: run process once at speed A, run process once at speed B, calculate response difference Q:is response difference due to change or chance?

Lecture 2.16 © 2016 Michael Stuart Postgraduate Certificate in Statistics Design and Analysis of Experiments Replication Improved experiment: run process twice at speed A, run process twice at speed B, calculate mean response at each speed, difference in mean responses measures change effect calculate response difference at each speed, mean of response differences measures chance effect

Lecture 2.18 © 2016 Michael Stuart Exercise 1.2.1 Formal test: Numerator measures change effect, Denominator measures chance effect. Carry out the test using the results from the first two runs at each speed. Compare with test using complete data Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.19 © 2016 Michael Stuart Process Development Study Variable N Mean StDev Speed B 2 78.75 3.61 Speed A 2 75.25 2.47 Speed BSpeed A 76.273.5 81.377.0 Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.110 © 2016 Michael Stuart Replication Two measurements per sample provides a valid test –but not a powerful test More measurements per sample provides –more precision in estimating within-sample variation, i.e., estimating  and, therefore, –more power in testing between-sample variation. Recall the discussion in Base Module Chapter 4: –11 replications needed to detect a 5% improvement Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.112 © 2016 Michael Stuart A multi-level experimental factor Filter membrane improvement project Four membrane types: A:current standard B:newly developed alternative C:OEM 1 D:OEM 2 Criterion:failure pressure level (kPa) Objectives:(i)is Type B better than Type A? (ii)are OEM membranes better? Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.113 © 2016 Michael Stuart Filter membrane improvement project Procedure:from each of 10 production batches of each membrane type, sample 5 membranes, for each sample of 5, run the filtering process using each membrane, increasing pressure until membrane failure, calculate sample mean failure pressure reading. Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.114 © 2016 Michael Stuart Filter membrane improvement project the response the experimental factor the factor levels the treatments an experimental unit an observational unit unit structure treatment assignment replication burst strength membrane type A, B, C, D 5 process test runs process test run simple no information 10 Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.115 © 2016 Michael Stuart Results Mean burst strengths (failure pressure level, kPa) of 10 samples from each of 4 filter membrane types Classwork 1.2.3 Make dotplots of the breaking strengths Ref: Membrane strength.xls Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.116 © 2016 Michael Stuart Initial data analysis Burst strength (kPa) of 10 samples of each of four filter membrane types Variable Membrane N Mean StDev Minimum Maximum Range Strength A 10 93 4.8 85 103 19 B 10 96 3.4 91 101 11 C 10 85 4.3 77 93 16 D 10 90 2.8 86 96 9 Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.118 © 2016 Michael Stuart One-way ANOVA: Strength versus Membrane Source DF SS MS F P Membrane 3 709.2 236.4 15.54 0.000 Error 36 547.8 15.2 Total 39 1257.0 S = 3.901 F 3,36;0.05 ≈ 2.85 Conclusion: Differences between means are highly statistically significant: process variation has standard deviation of almost 4. Comparing several means Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.119 © 2016 Michael Stuart One-way ANOVA: Strength versus Membrane Source DF SS MS F P Membrane 3 709.2 236.4 15.54 0.000 Error 36 547.8 15.2 Total 39 1257.0 S = 3.901 F 3,36;0.05 ≈ 2.85 Conclusion: Differences between means are highly statistically significant: process variation has standard deviation of almost 4. Comparing several means Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.120 © 2016 Michael Stuart Analysis of Variance Explained Decomposing Total Variation Expected Mean Squares Ref: Base Module Chapter 5, §5.1 Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.121 © 2016 Michael Stuart Decomposing Total Variation Elementary decomposition: Analysis of Variance decomposition: SSTO = SSM + SSE DFTO=DFM + DFE total deviation membrane deviation error deviation =+ Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.122 © 2016 Michael Stuart Decomposing Total Variation Classwork 1.2.4 Confirm the degrees of freedom (DF) and sum of squares (SS) decompostion and confirm the calculation of the mean squares and the F-ratio in the membrane analysis of variance table. One-way ANOVA: Strength versus Membrane Source DF SS MS F P Membrane 3 709.2 236.4 15.54 0.000 Error 36 547.8 15.2 Total 39 1257.0 Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.123 © 2016 Michael Stuart Expected Mean Squares All  i =  ↔ E(MSM) = E(MSE) All  i ≠  ↔ E(MSM) > E(MSE) Hence, MSM ≈ MSE suggests all  i ≈ , and MSM >> MSE suggests all  i ≠ . F = measures by how much MSM exceeds MSE Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.125 © 2016 Michael Stuart Multiple comparisons Confidence interval for difference between means: If 0 is not within the interval, then 0 is more than 2SE from so is more than 2SE from 0, that is means are statististically significantly different Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.126 © 2016 Michael Stuart Interpreting multiple comparisons Membrane B mean is significantly stronger than Membranes C and D means and close to significantly stronger than Membrane A mean. Membrane C mean is significantly less strong than the other three means. Membranes A and D means are not significantly different. Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.127 © 2016 Michael Stuart Multiple comparisons explained Simultaneous confidence intervals slightly wider than individual confidence intervals. –level of confidence in several intervals simultaneously versus –level of confidence in a single interval; –more opportunities for being wrong Widening intervals increases confidence. –extent of widening chosen to compensate for reduction in confidence involved. Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.129 © 2016 Michael Stuart Report Postgraduate Certificate in Statistics Design and Analysis of Experiments Reminder of objectives of the experiment: (i)is the company's newly developed membrane Type B better than the standard Type A? (ii)is there any advantage in introducing other companies' membranes? Answer (ii):NO Answer (i):"Some evidence" that B is better than A, but other factors may be more important. Possibly make further comparisons.

Lecture 2.130 © 2016 Michael Stuart Brief management report Membrane Type C can be eliminated from our inquiries. Membrane Type D shows no sign of being an improvement on the existing Membrane Type A and so need not be considered further. Membrane Type B shows some improvement on Membrane Type A but not enough to recommend a change. It may be worth while carrying out further comparisons between Membranes Types A and B. Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.131 © 2016 Michael Stuart Fisher on Analysis of Variance Table "a convenient method of arranging the arithmetic" (so don't show it in management reports!) Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.133 © 2016 Michael Stuart Part 3 Randomized complete blocks design Example 1:treating crops with fertiliser to improve yield. Four fertilisers being tested: divide a single field into four plots (experimental units) to form one block, assign treatments at random to the four plots, repeat with several other fields to form several blocks, choose blocks in varying locations, for generalising. Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.135 © 2016 Michael Stuart Randomized blocks design Example 2:treating long spools of rubber to improve abrasion resistance. Four treatments being tested: cut a single piece into four experimental units to form one block, assign treatments at random to the four units, repeat with several other pieces to form several blocks. Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.139 © 2016 Michael Stuart Randomized blocks design Block 1Block 2Block 3Block 4etc. Blocking accounts for anticipated variation patterns along the length of the spool of rubber Randomization allows for unanticipated sources of variation within blocks, e.g., side to side, diagonal, any other BACAADDB DCBDCBCA Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.140 © 2016 Michael Stuart Randomized blocks design Example 3:assessing process changes. Five versions of the process being assessed: assess the five versions on five successive days in a working week, Randomize the time order in which the versions are used, repeat over several weeks to form several blocks. NB:Pairing = Blocking with two units per block Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.141 © 2016 Michael Stuart Randomized block design Where replication entails increased variation, replicate the full experiment in several blocks so that non-experimental variation within blocks is as small as possible, –comparison of experimental effects subject to minimal chance variation, variation between blocks may be substantial, –comparison of experimental effects not affected Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.142 © 2016 Michael Stuart Illustrations of blocking variables Agriculture: fertility levels in a field or farm, moisture levels in a field or farm, genetic similarity in animals, litters as blocks, etc. Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.143 © 2016 Michael Stuart Illustrations of blocking variables Clinical trials (stratification) age, sex, height, weight, social class, medical history etc. Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.144 © 2016 Michael Stuart Illustrations of blocking variables Clinical trials body parts as blocks, hands, feet, eyes, ears, different treatments applied to the same individual at different times, cross-over, carry-over, correlation, etc. Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.145 © 2016 Michael Stuart Illustrations of blocking variables Industrial trials multiple machines, multiple test laboratories, time based blocks, time of day, day of week, shift etc. Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.147 © 2016 Michael Stuart Case Study Reducing yield loss in a chemical process Process: chemicals blended, filtered and dried Problem:yield loss at filtration stage Proposal:adjust initial blend to reduce yield loss Plan: –prepare five different blends –use each blend in successive process runs, in random order –repeat at later times (blocks) Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.148 © 2016 Michael Stuart Classwork 1.2.5: What were the response: experimental factor(s): factor levels: treatments: experimental units: observational units: unit structure: treatment allocation: replication: Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.149 © 2016 Michael Stuart Classwork 1.2.5: What were the response: experimental factor(s): factor levels: treatments: experimental units: observational units: unit structure: treatment allocation: replication: yield loss Blend A, B, C, D, E process runs 3 blocks of 5 units random order of blends within blocks 3 Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.150 © 2016 Michael Stuart Unit Structure Block 1Block 2Block 3 Postgraduate Certificate in Statistics Design and Analysis of Experiments Unit 1 Unit 2 Unit 3 Unit 4 Unit 5 Unit 1 Unit 2 Unit 3 Unit 4 Unit 5 Unit 1 Unit 2 Unit 3 Unit 4 Unit 5

Lecture 2.151 © 2016 Michael Stuart Unit Structure Block 1Block 2Block 3 Postgraduate Certificate in Statistics Design and Analysis of Experiments Unit 1_1 Unit 1_2 Unit 1_3 Unit 1_4 Unit 1_5 Unit 2_1 Unit 2_2 Unit 2_3 Unit 2_4 Unit 2_5 Unit 3_1 Unit 3_2 Unit 3_3 Unit 3_4 Unit 3_5 Blocks Units Units nested in Blocks

Lecture 2.152 © 2016 Michael Stuart Randomization procedure 1.enter numbers 1 to 5 in Column A of a spreadsheet, headed Run, 2.enter letters A-E in Column B, headed Blend, 3.generate 5 random numbers into Column C, headed Random 4.sort Blend by Random, 5.allocate Treatments as sorted to Runs in Block I, 6.repeat Steps 3 - 5 for Blocks II and III. Go to Excel Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.153 © 2016 Michael Stuart Part 2 Randomised blocks analysis Exploratory analysis Analysis of Variance Block or not? Diagnostic analysis –deleted residuals Analysis of variance explained Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.155 © 2016 Michael Stuart Initial data analysis Little variation between blocks More variation between blends Disturbing interaction pattern; see later Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.156 © 2016 Michael Stuart Formal Analysis Analysis of Variance: Loss vs Block, Blend Source DF SS MS F P Block 2 1.648 0.824 0.94 Blend 4 11.556 2.889 3.31 0.071 Error 8 6.992 0.874 Total 14 20.196 Classwork 1.2.6: Confirm the calculation of Total DF, Total SS, MS(Block), MS(Blend), MS(Error) F(Block), F(Blend) Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.157 © 2016 Michael Stuart Formal Analysis Analysis of Variance: Loss vs Block, Blend Source DF SS MS F P Block 2 1.648 0.824 0.94 Blend 4 11.556 2.889 3.31 0.071 Error 8 6.992 0.874 Total 14 20.196 Classwork 1.2.6: Confirm the calculation of Total DF, Total SS, MS(Block), MS(Blend), MS(Error) F(Block), F(Blend) Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.158 © 2016 Michael Stuart Assessing variation between blends F(Blends)= 3.3 F 4,8;0.1 = 2.8 F 4,8;0.05 = 3.8 p = 0.07 F(Blends) is "almost statistically significant" Multiple comparisons: All intervals cover 0; Blends B and E difference "almost significant" Ref: Lecture Note 1.2, p. 20. Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.159 © 2016 Michael Stuart Assessing variation between blocks F(Blocks) = 0.94 < 1; MS(Blocks) < (MS(Error) differences between blocks consistent with chance variation; Source DF SS MS F P Block 2 1.648 0.824 0.94 0.429 Blend 4 11.556 2.889 3.31 0.071 Error 8 6.992 0.874 Total 14 20.196 Source DF SS MS F P Blend 4 11.556 2.889 3.34 0.055 Error 10 8.640 0.864 Total 14 20.196 Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.160 © 2016 Michael Stuart Was the blocking effective? Source DF SS MS F P Block 2 1.648 0.824 0.94 Blend 4 11.556 2.889 3.31 0.071 Error 8 6.992 0.874 Total 14 20.196 S = 0.9349 Source DF SS MS F P Blend 4 11.556 2.889 3.34 0.055 Error 10 8.640 0.864 Total 14 20.196 S = 0.9295 Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.161 © 2016 Michael Stuart Was the blocking effective? F(Blocks) < 1 Blocks MS smaller than Error MS When blocks deleted from analysis –Residual standard deviation almost unchanged and –F(Blends) almost unchanged Blocking NOT effective. Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.162 © 2016 Michael Stuart Block or not? Source DF SS MS F P Block 2 1.648 0.824 0.94 0.429 Blend 4 11.556 2.889 3.31 0.071 Error 8 6.992 0.874 Total 14 20.196 Source DF SS MS F P Blend 4 11.556 2.889 3.34 0.055 Error 10 8.640 0.864 Total 14 20.196 Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.163 © 2016 Michael Stuart Block or not? Source DF SS MS F P Block 2 1.648 0.824 0.94 0.429 Blend 4 11.556 2.889 3.31 0.071 Error 8 6.992 0.874 Total 14 20.196 Source DF SS MS F P Blend 4 11.556 2.889 3.34 0.055 Error 10 8.640 0.864 Total 14 20.196 Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.164 © 2016 Michael Stuart Block or not? Omitting blocks increases DF(Error), therefore increases precision of estimate of , and increases power of F(Blends) F 4,8:0.10 = 2.8; F 4,8:0.05 = 3.8 F 4,10:0.10 = 2.6F 4,10:0.05 = 3.5 Smaller critical value easier to exceed, more power. Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.165 © 2016 Michael Stuart Block or not? Statistical theory suggests no blocking. Practical knowledge may suggest otherwise. Quote from Davies et al (1956): "Although the apparent variation among the blocks is not confirmed (i.e. it might well be ascribed to experimental error), future experiments should still be carried out in the same way. There is no clear evidence of a trend in this set of trials, but it might well appear in another set, and no complication in experimental arrangement is involved". Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.166 © 2016 Michael Stuart Diagnostic plots The diagnostic plot, residuals vs fitted values –checking the homogeneity of chance variation The Normal residual plot, –checking the Normality of chance variation Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.167 © 2016 Michael Stuart Diagnostic analysis Postgraduate Certificate in Statistics Design and Analysis of Experiments One exceptional case –likely to be related to interaction pattern. see Slide 55 −resist deletion and refitting!

Lecture 2.168 © 2016 Michael Stuart Initial data analysis Little variation between blocks More variation between blends Disturbing interaction pattern; see later Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.170 © 2016 Michael Stuart Decomposing Total Variation Analysis of Variance: Loss vs Block, Blend Source DF SS MS F P Block 2 1.648 0.824 0.94 Blend 4 11.556 2.889 3.31 0.071 Error 8 6.992 0.874 Total 14 20.196 SS(TO) = SS(Block) + SS(Blend) + SS(Error) Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.171 © 2016 Michael Stuart Model for analysis Yield loss includes –a contribution from each blend plus –a contribution from each block plus –a contribution due to chance variation. Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.172 © 2016 Michael Stuart Model for analysis Y =  +  +  +  where  is the overall mean,  is the blend effect, above or below the mean, depending on which blend is used,  is the block effect, above or below the mean, depending on which block is involved  represents chance variation Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.176 © 2016 Michael Stuart Decomposing Total Variation statistical residual format mathematically simplified format SSTO = SS(Blocks) + SS(Blends) + SS(Error) Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.177 © 2016 Michael Stuart Expected Mean Squares F(Blends) = tests equality of blend means F(Blocks) = assesses effectiveness of blocking Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.178 © 2016 Michael Stuart Minute test –How much did you get out of today's class? –How did you find the pace of today's class? –What single point caused you the most difficulty? –What single change by the lecturer would have most improved this class? Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.179 © 2016 Michael Stuart Reading EM Ch. 4, §7.2 Base Module, §2.2, §2.4, §§4.1- 4.3, §5.1 MGM §2.1, §§3.1,3,2 DCM§2-4.1 to §2-4.3, §2.5, §§3.1 to 3.4, §3.5.7, §4.1 DV §3.5, §§4.2.1-4.2.3, §4.3.2, §4.4.1, §§4.4.4-4.4.6, §10.3, §10.4 (with back references) Postgraduate Certificate in Statistics Design and Analysis of Experiments

Lecture 2.11 © 2016 Michael Stuart Design and Analysis of Experiments Lecture 2.1 1.Review –Minute tests 1.2 –Homework –Experimental factors with several.

Similar presentations

Presentation on theme: "Lecture 2.11 © 2016 Michael Stuart Design and Analysis of Experiments Lecture 2.1 1.Review –Minute tests 1.2 –Homework –Experimental factors with several."— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Lecture 2.11 © 2016 Michael Stuart Design and Analysis of Experiments Lecture 2.1 1.Review –Minute tests 1.2 –Homework –Experimental factors with several.

Similar presentations

Presentation on theme: "Lecture 2.11 © 2016 Michael Stuart Design and Analysis of Experiments Lecture 2.1 1.Review –Minute tests 1.2 –Homework –Experimental factors with several."— Presentation transcript:

Similar presentations

About project

Feedback