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Practical Interpretation With a quantitative factor, like power in - - PowerPoint PPT Presentation

Nov 08, 2023 403 likes •644 views

ST 516 Experimental Statistics for Engineers II Practical Interpretation With a quantitative factor, like power in the etch-rate example, typically use regression modeling: y = 0 + 1 x + or perhaps y = 0 + 1 x + 2 x 2 + . With

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ST 516 Experimental Statistics for Engineers II

Practical Interpretation

With a quantitative factor, like power in the etch-rate example, typically use regression modeling: y = β0 + β1x + ǫ

  • r perhaps

y = β0 + β1x + β2x2 + ǫ. With a qualitative factor, like “method” in the peak discharge rate example, we typically focus on comparisons among means.

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ST 516 Experimental Statistics for Engineers II

Comparing Means A contrast is a linear combination of treatment means:

a

  • i=1

ciµi, with

a

  • i=1

ci = 0. Examples are: µ1 − µ2, where c1 = 1, c2 = −1, c3 = ... = ca = 0; µ1 − 1

2(µ2 + µ3), where c1 = 1, c2 = c3 = − 1 2, c4 = ... = ca = 0.

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ST 516 Experimental Statistics for Engineers II

Testing a Contrast Many hypotheses about treatment means can be written in terms of a contrast: H0 :

a

  • i=1

ciµi = 0 for appropriate contrast constants c1, c2, . . . , ca. We estimate a

i=1 ciµi by a

  • i=1

ci ˆ µi =

a

  • i=1

ci ¯ yi· = C.

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ST 516 Experimental Statistics for Engineers II

C is an unbiased estimator, with V (C) = σ2 n

a

  • i=1

c2

i ,

which we estimate by ˆ V (C) = MSE n

a

  • i=1

c2

i .

The test statistic is therefore t0 = C

  • ˆ

V (C) =

a

  • i=1

ci ¯ yi·

  • MSE

n

a

i=1 c2 i

.

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ST 516 Experimental Statistics for Engineers II

Compare t0 with the t-distribution with df = N − a. Equivalently, compare F0 = t2

0 with the F-distribution with

df = 1, N − a. Confidence interval for C:

a

  • i=1

ci ¯ yi· ± tα/2,N−a

  • MSE

n

a

  • i=1

c2

i .

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ST 516 Experimental Statistics for Engineers II

Multiple Contrasts Sometimes we test several contrasts in the same experiment, or equivalently set up CIs for those contrasts; error rate becomes an issue. Control experiment-wise error rate for all possible contrasts using Scheff´ e’s method: replace tα/2,N−a with

  • (a − 1)Fα,a−1,N−a.

Confidence interval becomes

a

  • i=1

ci ¯ yi· ±

  • (a − 1)Fα,a−1,N−a
  • MSE

n

a

  • i=1

c2

i .

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ST 516 Experimental Statistics for Engineers II

Pairwise Comparisons Often the only contrasts we consider are pairwise comparisons µi − µj. To control experiment-wise error rate for all pairwise comparisons, Tukey’s method gives shorter intervals than Scheff´ e’s: CI is ¯ yi· − ¯ yj· ± qα(a, f )

  • MSE

n , where qα(a, f ) is a percent point of the studentized range statistic, and f = N − a is the degrees of freedom in MSE.

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Fisher’s Least Significant Difference The basic t-statistic for comparing µi with µj is t0 = ¯ yi· − ¯ yj·

  • MSE
  • 1

ni + 1 nj

. So we declare µi and µj to be significantly different if |¯ yi· − ¯ yj·| > tα/2,N−a

  • MSE

1 ni + 1 nj

  • .

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ST 516 Experimental Statistics for Engineers II

That is, tα/2,N−a

  • MSE

1 ni + 1 nj

  • is the least significant difference, or LSD.

Notes In a balanced design, ni = nj = n, so the LSD is the same for every pair µi and µj. The LSD method has comparison-wise error rate α; it does not control experimentwise error rate.

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ST 516 Experimental Statistics for Engineers II

R command

TukeyHSD(aov(EtchRate ~ factor(Power), etchRateLong))

Output

Tukey multiple comparisons of means 95% family-wise confidence level Fit: aov(formula = EtchRate ~ factor(Power), data = etchRateLong) $‘factor(Power)‘ diff lwr upr p adj 180-160 36.2 3.145624 69.25438 0.0294279 200-160 74.2 41.145624 107.25438 0.0000455 220-160 155.8 122.745624 188.85438 0.0000000 200-180 38.0 4.945624 71.05438 0.0215995 220-180 119.6 86.545624 152.65438 0.0000001 220-200 81.6 48.545624 114.65438 0.0000146

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ST 516 Experimental Statistics for Engineers II

Graph

plot(TukeyHSD(aov(EtchRate ~ factor(Power), etchRateLong)))

50 100 150 220−200 220−180 200−180 220−160 200−160 180−160

95% family−wise confidence level

Differences in mean levels of factor(Power)

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Sample Size

Using Operating Characteristic curves: β is the probability of type II error: β = 1 − P{Reject H0|H0 is false} = 1 − P{F0 > Fα,a−1,N−a|H0 is false}. Charts plot β against Φ, where Φ2 = n a

i=1 τ 2 i

aσ2 . So if we know σ2 and a set of treatment effects τ1, τ2, . . . , τa for which we want the type II error to be β, we can find the smallest acceptable n.

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Note: often the resulting n is too large for the experiment to be

  • feasible. The experimenter must accept higher β or larger τ’s.

Alternative to using OC charts: using length of confidence interval; for comparing two means, confidence interval is ¯ yi· − ¯ yj· ± tα/2,N−a

  • MSE × 2

n, If we know σ2, can choose n to give desired width. Still often gives infeasible (too large) n.

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ST 516 Experimental Statistics for Engineers II

The Regression Approach

Recall the “effects” model: yi,j = µ + τi + ǫi,j, i = 1, 2, . . . , a, j = 1, 2, . . . , ni. Note that we now allow for unbalanced data (unequal ni).

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Suppose we put all the y’s in a response vector: Y =                          y1,1 y1,2 . . . y1,n1 y2,1 y2,2 . . . y2,n2 . . . ya,1 ya,2 . . . ya,na                         

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To keep track of the group that an observation belongs to, we also create a design matrix X. The row for an observation in group i consists of all zeroes, except for a single 1 at position i. For convenience, we add a first column with all 1’s.

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X =                          1 1 . . . 1 1 . . . . . . . . . . . . . . . . . . 1 1 . . . 1 1 . . . 1 1 . . . . . . . . . . . . . . . . . . 1 1 . . . . . . . . . . . . . . . . . . 1 . . . 1 1 . . . 1 . . . . . . . . . . . . . . . 1 . . . 1                         

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We also put all the parameters µ, τ1, τ2, . . . , τa into a parameter vector: β =        µ τ1 τ2 . . . τa        With an error vector ǫ constructed like Y, we have Y = Xβ + ǫ.

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This is a multiple regression equation. The least squares estimates ˆ µ, ˆ τ1, ˆ τ2, . . . , ˆ τa satisfy the least squares normal equations X′Xˆ β = X′Y. If X′X were non-singular, we could solve these equations: ˆ β = (X′X)−1X′Y. But it isn’t, so we can’t...

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The first column of X is the sum of the other columns, which makes

  • ne column redundant.

More formally, if b =        +1 −1 −1 . . . −1        , then (X′X)b = X′(Xb) = 0. So X′X is singular.

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If we leave out one column of X, the remaining columns have no redundancy, and the reduced X′X matrix is non-singular. Leaving out a column means leaving the corresponding parameter out

  • f the model.

Leave out the first column ⇔ leave out µ ⇔ the means model. Leave out the second column ⇔ leave out τ1 ⇔ the effects model with the first level as the baseline. Leave out the last column ⇔ leave out τa ⇔ the effects model with the last level as the baseline.

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The R function lm() is a general-purpose multiple regression function, and uses this approach to estimation. Because the formula EtchRate ~ factor(Power) has a factor on the right hand side, lm() internally creates a design matrix X with one column for each level of the factor except the first (thus imposing the constraint τ1 = 0), and solves the normal equations. This looks complicated, but it does not require balanced data, and generalizes easily to models with more than one factor.

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