ST 516 Experimental Statistics for Engineers II Blocked Designs Recall the paired comparison design: two treatments applied to the same experimental material. E.g. hardness testing Treatment: two types of pointed tip pressed into a sample of metal with known force; Pairs: both types tested on the same samples. 1 / 15 Blocked Designs Randomized Complete Block Design
ST 516 Experimental Statistics for Engineers II E.g. shear strength of girders Treatment: two methods; Pairs: both methods tested on the same girders. Statistical model is y i , j = µ i + β j + ǫ i , j , i = 1 , 2; j = 1 , 2 , . . . , n . or, in effects form, y i , j = µ + τ i + β j + ǫ i , j , i = 1 , 2; j = 1 , 2 , . . . , n . 2 / 15 Blocked Designs Randomized Complete Block Design
ST 516 Experimental Statistics for Engineers II More Than Two Treatments If feasible, apply all treatments to each unit. Treatments are assigned randomly: time order of runs; subsamples of metal coupon. This is the randomized complete block design (RCBD). 3 / 15 Blocked Designs Randomized Complete Block Design
ST 516 Experimental Statistics for Engineers II Statistical Analysis Same statistical model as for paired comparisons: y i , j = µ + τ i + β j + ǫ i , j , i = 1 , 2 , . . . , a ; j = 1 , 2 , . . . , b . Here: y i , j = response for i th treatment in the j th block µ = overall mean τ i = i th treatment effect β j = j th block effect ǫ i , j = random error. 4 / 15 Blocked Designs Randomized Complete Block Design
ST 516 Experimental Statistics for Engineers II Same general questions as for the completely random design (CRD): Are there any differences among treatments? H 0 : τ 1 = τ 2 = · · · = τ a = 0. If so, how to describe them: regression modeling for a quantitative factor; pairwise comparisons and other contrasts for a qualitative factor. 5 / 15 Blocked Designs Randomized Complete Block Design
ST 516 Experimental Statistics for Engineers II In addition: Was blocking necessary? H 0 : β 1 = β 2 = · · · = β b = 0. But note: because treatments are randomized only within blocks, testing this hypothesis has problems: F -statistic may not be F -distributed under the null hypothesis; Corresponding P -value may not be valid. 6 / 15 Blocked Designs Randomized Complete Block Design
ST 516 Experimental Statistics for Engineers II Also note: blocks are often viewed as a random sample from a population of possible blocks. That would mean that the β ’s are random variables , not constants. If Var( β j ) = σ 2 β , the null hypothesis becomes H 0 : σ 2 β = 0. H 0 implies not only that β 1 = β 2 = · · · = β b = 0, but in addition that all future β ’s will also be zero. 7 / 15 Blocked Designs Randomized Complete Block Design
ST 516 Experimental Statistics for Engineers II Estimating the parameters Under the natural constraints � τ i = � β j = 0: E (¯ y ·· ) = µ E (¯ y i · ) = µ + τ i E (¯ y · j ) = µ + β j So the obvious estimates are: µ = ¯ ˆ y ·· ˆ τ i = ¯ y i · − ¯ y ·· ˆ β j = ¯ y · j − ¯ y ·· 8 / 15 Blocked Designs Randomized Complete Block Design
ST 516 Experimental Statistics for Engineers II Analysis of Variance Sums of squares: a b � � µ ) 2 SS Total = ( y i , j − ˆ i =1 j =1 a � τ 2 SS Treatments = b ˆ i i =1 b � ˆ β 2 SS Blocks = a j j =1 a b � 2 � � � τ i − ˆ SS Error = y i , j − ˆ µ − ˆ β j i =1 j =1 9 / 15 Blocked Designs Randomized Complete Block Design
ST 516 Experimental Statistics for Engineers II The ANOVA identity: SS Total = SS Treatments + SS Blocks + SS Error df = N − 1 a − 1 b − 1 N − a − b + 1 = ( a − 1)( b − 1) 10 / 15 Blocked Designs Randomized Complete Block Design
ST 516 Experimental Statistics for Engineers II Example: yield (%) of a manufacturing process; factors are Extrusion Pressure and Batch (of resin, the raw material in the process). Data file: graft.txt R commands graft <-read.table("data/graft.txt", header = TRUE); graftLong <- reshape(graft, varying = c("Batch1", "Batch2", "Batch3", "Batch4", "Batch5", "Batch6"), v.names = "Yield", timevar = "Batch", direction = "long") # boxplots: par(mfcol = c(1, 2)) plot(Yield ~ factor(Batch) + factor(Pressure), data = graftLong) 11 / 15 Blocked Designs Randomized Complete Block Design
ST 516 Experimental Statistics for Engineers II R command for ANOVA summary(aov(Yield ~ factor(Batch) + factor(Pressure), data = graftLong)) Output Df Sum Sq Mean Sq F value Pr(>F) factor(Batch) 5 192.252 38.450 5.2487 0.005532 ** factor(Pressure) 3 178.171 59.390 8.1071 0.001916 ** Residuals 15 109.886 7.326 --- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 12 / 15 Blocked Designs Randomized Complete Block Design
ST 516 Experimental Statistics for Engineers II R command for effects summary(lm(Yield ~ factor(Batch) + factor(Pressure), data = graftLong)) Call: lm(formula = Yield ~ factor(Batch) + factor(Pressure), data = graftLong) Residuals: Min 1Q Median 3Q Max -3.5708 -1.3333 -0.3167 1.1417 4.1792 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 90.721 1.657 54.735 < 2e-16 *** factor(Batch)2 2.050 1.914 1.071 0.301043 factor(Batch)3 3.300 1.914 1.724 0.105201 factor(Batch)4 2.850 1.914 1.489 0.157175 factor(Batch)5 -2.375 1.914 -1.241 0.233684 factor(Batch)6 6.750 1.914 3.527 0.003050 ** factor(Pressure)8700 -1.133 1.563 -0.725 0.479457 factor(Pressure)8900 -3.900 1.563 -2.496 0.024713 * factor(Pressure)9100 -7.050 1.563 -4.512 0.000414 *** 13 / 15 Blocked Designs Randomized Complete Block Design
ST 516 Experimental Statistics for Engineers II R command for (Tukey) pairwise comparisons TukeyHSD(aov(Yield ~ factor(Batch) + factor(Pressure), data = graftLong), "factor(Pressure)") Output Tukey multiple comparisons of means 95% family-wise confidence level Fit: aov(formula = Yield ~ factor(Batch) + factor(Pressure), data = graftLong) $‘factor(Pressure)‘ diff lwr upr p adj 8700-8500 -1.133333 -5.637161 3.370495 0.8854831 8900-8500 -3.900000 -8.403828 0.603828 0.1013084 9100-8500 -7.050000 -11.553828 -2.546172 0.0020883 8900-8700 -2.766667 -7.270495 1.737161 0.3245644 9100-8700 -5.916667 -10.420495 -1.412839 0.0086667 9100-8900 -3.150000 -7.653828 1.353828 0.2257674 14 / 15 Blocked Designs Randomized Complete Block Design
ST 516 Experimental Statistics for Engineers II Random Effects If Block (Batch) is a random effect with variance σ 2 β , then the expected mean square is E (MS Blocks ) = σ 2 + a σ 2 β . So we estimate σ 2 β by β = MS Blocks − MS Residuals = 38 . 450 − 7 . 326 σ 2 ˆ = 7 . 781 . 4 a 15 / 15 Blocked Designs Randomized Complete Block Design
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