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STT 511-STT411: DESIGN OF EXPERIMENTS AND ANALYSIS OF VARIANCE Dr. Cuixian Chen

STT 511-STT411: DESIGN OF EXPERIMENTS AND ANALYSIS OF VARIANCE Dr. Cuixian Chen. Chapter 13: Experiments with Random Factors. Design of Engineering Experiments - Experiments with Random Factors. Text reference, Chapter 13 Previous chapters have focused primarily on fixed factors

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STT 511-STT411: DESIGN OF EXPERIMENTS AND ANALYSIS OF VARIANCE Dr. Cuixian Chen

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  1. Design & Analysis of Experiments 8E 2012 Montgomery STT 511-STT411:DESIGN OF EXPERIMENTS AND ANALYSIS OF VARIANCEDr. Cuixian Chen Chapter 13: Experiments with Random Factors

  2. Design & Analysis of Experiments 8E 2012 Montgomery

  3. Design of Engineering Experiments - Experiments with Random Factors • Text reference, Chapter 13 • Previous chapters have focused primarily on fixed factors • A specific set of factor levels is chosen for the experiment • Inference confined to those levels • Often quantitative factors are fixed • When factor levels are chosen at random from a larger population of potential levels, the factor is random • Inference is about the entire population of levels • Industrial applications include measurement system studies • It has been said that failure to identify a random factor as random and not treat it properly in the analysis is one of the biggest errors committed in DOE • The random effect model was introduced in Chapter 3 Design & Analysis of Experiments 8E 2012 Montgomery

  4. Chap 3.9: Random Effects Model in One way ANOVA

  5. Review: fundamental one-way ANOVA identity with fixed effect • The total variability in the data, as measured by the total corrected sum of squares, can be partitioned into a sum of squares of the differences between the treatment averages and the grand average, plus a sum of squares of the differences of observations within treatments from the treatment average. • Now, the difference between the observed treatment averages and the grand average is a measure of the differences between treatment means, whereas the differences of observations within a treatment from the treatment average can be due only to random error. Design & Analysis of Experiments 8E 2012 Montgomery Chapter 3

  6. 3.9 Random Effects Model in One way ANOVA • Text reference, page 116. • There are a large number of possible levels for the factor (theoretically an infinite number). • The experimenter chooses a of these levels at random. • Inference will be to the entire population of levels. • Example: What are the effects of repeated exposure to an advertising message (digital camera)? The answer may depend on length of the ad. • Consider different lengths of the ad. Eg: 1 second, 2 seconds, …, 10 seconds, …, 100 seconds, and so on. (Infinite possibilities) • Now suppose we randomly choose 2 levels of lengths: 15 seconds and 32 seconds. (a=2) Design & Analysis of Experiments 8E 2012 Montgomery

  7. Variance components Testing hypotheses about individual treatment effects is not very meaningful because they were selected randomly, we are more interested in the population of treatments. p-value=1-pf(F0,a-1, N-a) Design & Analysis of Experiments 8E 2012 Montgomery

  8. Covariance structure: Unlike the fixed effects case in which all of the observations yij are independent, in the random model the observations yij are only independent if they come from different factor levels. Design & Analysis of Experiments 8E 2012 Montgomery

  9. Observations (a = 3 and n = 2): Design & Analysis of Experiments 8E 2012 Montgomery

  10. > Design & Analysis of Experiments 8E 2012 Montgomery

  11. Note: ANOVA F-test is identical to the fixed-effects case. Design & Analysis of Experiments 8E 2012 Montgomery

  12. Estimating the variance components using the ANOVA method: Design & Analysis of Experiments 8E 2012 Montgomery

  13. Example 3.11: Random Effect Model • A textile company weaves a fabric on a large number of looms. It would like looms to be homogeneous so that it obtains a fabric of uniform strength. The process engineer suspects that, in addition to usual variation in strength within samples of fabric from the same loom, there may also be significant variations in strength between looms. To investigate this, she selects four looms at random and makes four strength determinations on fabric manufactured on each loom. This experiment is run in random order, and data obtained are shown in Table 3.17. Design & Analysis of Experiments 8E 2012 Montgomery

  14. Example 3.11: Random Effect Model #install.packages("GAD") exp3.11<-read.table("http://people.uncw.edu/chenc/STT411/dataset%20backup/Strength-Data.txt",header = TRUE); library(GAD); ## General ANOVA Design ## ## For random effect with any combination of orthogonal/nested and fixed/random ## factors y<- exp3.11$Strength; ## “as.random” works the same way as as.factor, but assigns an additional class informing that it is a random factor. gad(lm(y~ as.random(exp3.11$Loom) )); anova(lm(y~factor(exp3.11$Loom))) ## For regular ANOVA with fixed effect ## Design & Analysis of Experiments 8E 2012 Montgomery

  15. Example 3.11: Random Effect Model #install.packages("GAD") exp3.11<-read.table("http://people.uncw.edu/chenc/STT411/dataset%20backup/Strength-Data.txt",header = TRUE); library(GAD); ## General ANOVA Design ## ## For random effect with any combination of orthogonal/nested and fixed/random factors ## y<- exp3.11$Strength; ## “as.random” works the same way as as.factor, but assigns an additional class informing that it is a random factor. gad(lm(y~ as.random(exp3.11$Looms))); anova(lm(y~as.factor(exp3.11$Loom))) ## For regular ANOVA with fixed effect ## Design & Analysis of Experiments 8E 2012 Montgomery

  16. Example 3.11: Random Effect Model From the ANOVA, we conclude that the looms in the plant differ significantly. Q: How to estimate Design & Analysis of Experiments 8E 2012 Montgomery

  17. Estimators • The ANOVA variance component estimators are momentestimators. • Normality not required to estimate variance component. • They are unbiased estimators. • Finding confidence intervals on the variance components is “clumsy”. • Negative estimates can occur – this is “embarrassing”, as variances are always non-negative. Design & Analysis of Experiments 8E 2012 Montgomery

  18. Confidence interval for sigma^2 Design & Analysis of Experiments 8E 2012 Montgomery

  19. Confidence interval for the interclass correlation coefficient interclass correlation coefficient gives proportion of variance of an observation that is the results of difference b/w treatments. Design & Analysis of Experiments 8E 2012 Montgomery

  20. Design & Analysis of Experiments 8E 2012 Montgomery

  21. Is there significant variation in the calcium content from batch to batch? • Estimate the components of variance, by POINT estimate method. • F • Find a 95% confidence interval for mu. • Find a 95% confidence interval for sigma.

  22. Is there significant variation in the Uniformity at different wafer positions? • Estimate the components of variance. • F Design & Analysis of Experiments 8E 2012 Montgomery

  23. Chap 4.1: Random Blocks and Fixed Treatments (page 151)

  24. Review: ANOVA for RCBD with fixed treat/block • Suppose single factor, a treatments (factor levels) and b blocks (treat/block are both fixed) • A statistical model (effects model) for the RCBD is • The relevant (fixed effects) hypotheses are • Model Assumption: Constraints: Design & Analysis of Experiments 8E 2012 Montgomery

  25. Review: ANOVA for RCBD with fixed treat/block The degrees of freedom for the sums of squares in are as follows: Therefore, ratios of sums of squares to their degrees of freedom result in mean squares and the ratio of the mean square for treatments to the error mean square is an F statistic that can be used to test the hypothesis of equal treatment means Design & Analysis of Experiments 8E 2012 Montgomery

  26. Random Blocks and Fixed Treatments • There are many situations where either treatments or blocks (or both) are random factors. • It is very common to find that the blocks are random. • For conclusions from the experiment to be valid across the population of blocks that the ones selected for the experiments were sampled from. Design & Analysis of Experiments 8E 2012 Montgomery

  27. Ch 4.1: Random Blocks and Fixed Treatments It is exactly same test statistic we used when blocks were fixed. Design & Analysis of Experiments 8E 2012 Montgomery

  28. Example 1 Q: Find the components of variance for the following example. Design & Analysis of Experiments 8E 2012 Montgomery

  29. Example 1 Design & Analysis of Experiments 8E 2012 Montgomery

  30. Example 2 Q1: List all the Null and Alternative hypotheses. Q2: Input all data into R and conduct the ANOVA analysis. Q3: Based on the outputs, draw conclusions to hypothsis. Q4: Find the components of variance. Design & Analysis of Experiments 8E 2012 Montgomery

  31. Example 2 Q1: List all the Null and Alternative hypotheses. Q2: Draw conclusions? Q3: Find the components of variance. Block=c(rep(1, 4), rep(2, 4), rep(3, 4), rep(4, 4), rep(5,4)); Treat=rep(c("A", "B", "C", "D"), 5) y=c(89, 88, 97, 94, 84, 77, 92, 79, 81, 87, 87, 85, 87, 92, 89, 84, 79, 81, 80, 88); cbind(y, Treat, Block) library(GAD); Block = as.random(Block); Treat = as.fixed(Treat); gad(lm(y~Treat+Block)); ##anova(lm(y~factor(Treat)+factor(Block))); Design & Analysis of Experiments 8E 2012 Montgomery

  32. Example 3 Q1: List all the Null and Alternative hypotheses. Q2: Draw conclusions? Q3: Find the components of variance. Treat=c(rep(1, 4), rep(2, 4), rep(3, 4), rep(4, 4)); Block=rep(c("A", "B", "C", "D"), 4) y=c(-2, -1, 1, 5, -1, -2, 3, 4, -3, -1, 0, 2, 2, 1, 5, 7); cbind(y, Treat, Block) library(GAD); Block = as.random(Block); Treat = as.fixed(Treat); gad(lm(y~Treat+Block)); ##anova(lm(y~factor(Treat)+factor(Block))); Design & Analysis of Experiments 8E 2012 Montgomery

  33. Chap 13: Experiments with Random Factors

  34. Review: Chapter 5 -- Two-factor Factorial Design Statistical (effects) model: Other models (means model, regression models) can be useful. is overall mean effect, is effect of ith level of row factor A, is effect of jth level of column factor B, is the effect of interaction between and , and is a random error component. Design & Analysis of Experiments 8E 2012 Montgomery

  35. Review: Chapter 5 -- Two-factor Factorial Design Design & Analysis of Experiments 8E 2012 Montgomery

  36. Review: ANOVA Table for Two-factor Factorial Design – Case1: Both are with Fixed Effects • The reference distribution for F0 is the F((numerator df), (denominator df)) distribution • Reject the null hypothesis (equal treatment means) if p-value<0.05, • Where p-value=1-pf(F0, (numerator df), (denominator df)) Design & Analysis of Experiments 8E 2012 Montgomery

  37. Design of Engineering Experiments - Experiments with Random Factors • Text reference, Chapter 13 • Previous chapters have focused primarily on fixed factors • A specific set of factor levels is chosen for the experiment • Inference confined to those levels • Often quantitative factors are fixed (why?) • When factor levels are chosen at random from a larger population of potential levels, the factor is random • Inference is about the entire population of levels • Industrial applications include measurement system studies • It has been said that failure to identify a random factor as random and not treat it properly in the analysis is one of the biggest errors committed in DOX • The random effect model was introduced in Chapter 3 Design & Analysis of Experiments 8E 2012 Montgomery

  38. Case 2: Experiments with Two Random Effects • Two factors, factorial experiment, both factors random (Section 13.2, pg. 574) • The model parameters are NID random variables. • Random effects model. Design & Analysis of Experiments 8E 2012 Montgomery

  39. Case 2: Testing Hypotheses - Random Effects Model • Standard ANOVA partition is appropriate. Relevant hypotheses: • Form of the test statistics depend on the expectedmeansquares: Test effects of A and B: different from two fixed effects Modes • The reference distribution for F0 is the F((numerator df), (denominator df)) distribution • Reject the null hypothesis (equal treatment means) if p-value<0.05, • Where p-value=1-pf(F0, (numerator df), (denominator df)) Design & Analysis of Experiments 8E 2012 Montgomery

  40. Case 2: Estimating the Variance Components – Two Factor Random model • As before, we can use the ANOVA method; equate expected mean squares to their observed values: • These are moment estimators. • Potential problems with these estimators. Design & Analysis of Experiments 8E 2012 Montgomery

  41. Example 13.1 A Measurement Systems Capability Study • Statistically designed experiments are frequently used to investigate the sources of variability that affect a system. • A common industrial application is to use a designed experiment to study the components of variability in a measurement system. • These studies are often called gauge capability studies or gauge repeatability and reproducibility (R&R) studies because these are the components of variability that are of interest (for more discussion of gauge R&R studies, see the supplemental material for this chapter). Design & Analysis of Experiments 8E 2012 Montgomery

  42. Example 13.1 A Measurement Systems Capability Study • Gauge capability (or R&R) is of interest • The gauge is used by an operator to measure a critical dimension on a part • Repeatability is a measure of the variability due only to the gauge • Reproducibility is a measure of the variability due to the operator • See experimental layout, Table 13.1. This is a two-factor factorial (completely randomized) with both factors (operators, parts) random – a random effects model Design & Analysis of Experiments 8E 2012 Montgomery

  43. Example 13.1: ANOVA with Two Random Factors Design & Analysis of Experiments 8E 2012 Montgomery

  44. Example 13.1: ANOVA with Two Random Factors Design & Analysis of Experiments 8E 2012 Montgomery

  45. Example 13.1: ANOVA with Two Random Factors Q1: List all Null and alternatives Q2: If the F values and p-values are missing, may we still find them? Q3: Find the components of variance. Q4: The reduced model? Q5:The model assumptions. Design & Analysis of Experiments 8E 2012 Montgomery

  46. Example 13.2 SummaryMinitab Solution – Balanced ANOVA When both factors are random: • There is a large effect of parts (not unexpected) • Small operator effect • No Part – Operator interaction • Negative estimate of the Part – Operator interaction variance component • Fit a reduced model with the Part – Operator interaction deleted Design & Analysis of Experiments 8E 2012 Montgomery

  47. Example 13.1: Reduced model: • ANOVA with Two Random Factors Design & Analysis of Experiments 8E 2012 Montgomery

  48. Example 13.1 Minitab Solution – Reduced Model Q1: Find the components of variance. Q2: Compare with the one we have to original model? Original Model Reduced Model Design & Analysis of Experiments 8E 2012 Montgomery

  49. Example 13.1 Minitab Solution – Reduced Model • Estimating gauge capability: • If interaction had been significant? The variability in the gauge appears small relative to the variability in the product. This is generally a desirable situation, implying that the gauge is capable of distinguishing among different grades of product. Design & Analysis of Experiments 8E 2012 Montgomery

  50. 2nd example on two random factors model Q1: The model assumptions. Q2: Verify the F-value Q3: Find the components of variance. Q4: List all Null and alternatives Q5: Need a reduced model? ## 13.2 Two factor factorial with random factor ## ## Input data into R by hand… library(GAD); y<- Battery$Life; mat<- as.random(Battery$Material); temp<- as.random(Battery$Temp); gad(lm(y~mat*temp)); ## anova(lm(y~factor(Battery$Temp)*factor(Battery$Material))); Design & Analysis of Experiments 8E 2012 Montgomery

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