Blocked Designs Recall the paired comparison design: two treatments - - PowerPoint PPT Presentation

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Blocked Designs Recall the paired comparison design: two treatments - - PowerPoint PPT Presentation

Feb 08, 2024 351 likes •529 views

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

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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.

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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 yi,j = µi + βj + ǫi,j, i = 1, 2; j = 1, 2, . . . , n.

  • r, in effects form,

yi,j = µ + τi + βj + ǫi,j, i = 1, 2; j = 1, 2, . . . , n.

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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).

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ST 516 Experimental Statistics for Engineers II

Statistical Analysis Same statistical model as for paired comparisons: yi,j = µ + τi + βj + ǫi,j, i = 1, 2, . . . , a; j = 1, 2, . . . , b. Here: yi,j = response for ith treatment in the jth block µ = overall mean τi = ith treatment effect βj = jth block effect ǫi,j = random error.

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ST 516 Experimental Statistics for Engineers II

Same general questions as for the completely random design (CRD): Are there any differences among treatments? H0 : τ1 = τ2 = · · · = τa = 0. If so, how to describe them: regression modeling for a quantitative factor; pairwise comparisons and other contrasts for a qualitative factor.

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ST 516 Experimental Statistics for Engineers II

In addition: Was blocking necessary? H0 : β1 = β2 = · · · = βb = 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.

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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(βj) = σ2

β, the null hypothesis becomes H0 : σ2 β = 0.

H0 implies not only that β1 = β2 = · · · = βb = 0, but in addition that all future β’s will also be zero.

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ST 516 Experimental Statistics for Engineers II

Estimating the parameters Under the natural constraints τi = βj = 0: E (¯ y··) = µ E (¯ yi·) = µ + τi E (¯ y·j) = µ + βj So the obvious estimates are: ˆ µ = ¯ y·· ˆ τi = ¯ yi· − ¯ y·· ˆ βj = ¯ y·j − ¯ y··

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ST 516 Experimental Statistics for Engineers II

Analysis of Variance Sums of squares: SSTotal =

a

  • i=1

b

  • j=1

(yi,j − ˆ µ)2 SSTreatments = b

a

  • i=1

ˆ τ 2

i

SSBlocks = a

b

  • j=1

ˆ β2

j

SSError =

a

  • i=1

b

  • j=1
  • yi,j − ˆ

µ − ˆ τi − ˆ βj 2

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ST 516 Experimental Statistics for Engineers II

The ANOVA identity: SSTotal df = N − 1 = SSTreatments a − 1 + SSBlocks b − 1 + SSError N − a − b + 1 = (a − 1)(b − 1)

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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)

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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

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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 ***

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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

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ST 516 Experimental Statistics for Engineers II

Random Effects If Block (Batch) is a random effect with variance σ2

β, then the

expected mean square is E (MSBlocks) = σ2 + aσ2

β.

So we estimate σ2

β by

ˆ σ2

β = MSBlocks − MSResiduals

a = 38.450 − 7.326 4 = 7.781.

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