Single Factor Experiments Moving from comparing two samples (e.g. - - PowerPoint PPT Presentation

ST 516 Experimental Statistics for Engineers II

Single Factor Experiments

Moving from comparing two samples (e.g. two formulations of mortar) to more than two samples (e.g. 4 levels of RF power in etching a silicon wafer). Many ideas carry over: testing for real differences, etc. Some new complications: how to estimate differences when they are real.

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ST 516 Experimental Statistics for Engineers II

The data (etch-rate.txt): Power Obs1 Obs2 Obs3 Obs4 Obs5 160 575 542 530 539 570 180 565 593 590 579 610 200 600 651 610 637 629 220 725 700 715 685 710

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ST 516 Experimental Statistics for Engineers II

Boxplots

etchRate <- read.table("data/etch-rate.txt", header = TRUE) etchRateLong <- reshape(etchRate, varying = 2:6, idvar = "Code", v.names = "EtchRate", direction = "long", timevar = "Obs") boxplot(EtchRate ~ Power, etchRateLong, ylab = expression(paste("Rate (", ring(A), "/min)")), xlab = "Power (w)")

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ST 516 Experimental Statistics for Engineers II

160 180 200 220 550 600 650 700 Power (w) Rate (A °/min)

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ST 516 Experimental Statistics for Engineers II

Dot plots

plot(EtchRate ~ Power, etchRateLong, ylab = expression(paste("Rate (", ring(A), "/min)")), xlab = "Power (w)")

• 160

170 180 190 200 210 220 550 600 650 700 Power (w) Rate (A °/min)

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ST 516 Experimental Statistics for Engineers II

What do we see? Level of power has strong effect. No obvious outliers. Similar spreading at each level.

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ST 516 Experimental Statistics for Engineers II

Statistical Issues Test whether the observed differences could have been caused by chance (randomization of runs): Null hypothesis H0: no difference. If we reject H0 (and in this example, we expect to!), what differences exist? Point estimates, interval estimates.

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ST 516 Experimental Statistics for Engineers II

Statistical Models Notation: n replicates for each of a levels of the factor. yi,j = jth response for ith level of factor. The “means” model: yi,j = µi + ǫi,j, i = 1, 2, . . . , a, j = 1, 2, . . . , n. The “effects” model: yi,j = µ + τi + ǫi,j, i = 1, 2, . . . , a, j = 1, 2, . . . , n.

8 / 14 Single Factor Experiments Analysis of Variance

ST 516 Experimental Statistics for Engineers II

Broad Approach to Testing Null hypothesis H0: means model: µ1 = µ2 = · · · = µa effects model: τ1 = τ2 = · · · = τa = 0 Basic summaries: ¯ yi·, s2

i , i = 1, 2, . . . , a.

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ST 516 Experimental Statistics for Engineers II

Strategy: reject H0 if the ¯ yi·’s differ a lot. We could use max

i<i′ |¯

yi· − ¯ yi′·| . Instead, we effectively use

• i<i′

|¯ yi· − ¯ yi′·|2 . In either case, we need a denominator to compare with.

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ST 516 Experimental Statistics for Engineers II

Analysis of Variance (ANOVA)

Sums of squares and degrees of freedom (N = na):

a

• i=1

n

• j=1

(yi,j − ¯ y··)2 SSTotal df = N − 1 = n

a

• i=1

(¯ yi· − ¯ y··)2 SSTreatments df = a − 1 +

a

• i=1

n

• j=1

(yi,j − ¯ yi·)2 SSError df = N − a

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ST 516 Experimental Statistics for Engineers II

For each sum of squares, Mean Square = Sum of Squares degrees of freedom. MSTreatments is proportional to

• i<i′

|¯ yi· − ¯ yi′·|2. MSError is a pooled estimate of common variance σ2.

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ST 516 Experimental Statistics for Engineers II

E(MSError) = σ2, and E(MSTreatments)

• = σ2

under H0 > σ2

• therwise.

As a test statistic, we use F0 = MSTreatments MSError . Under H0, F0 is F-distributed with a − 1 and N − a degrees of freedom.

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ST 516 Experimental Statistics for Engineers II

In R

summary(aov(EtchRate ~ factor(Power), data = etchRateLong))

Output

Df Sum Sq Mean Sq F value Pr(>F) factor(Power) 3 66871 22290 66.797 2.883e-09 *** Residuals 16 5339 334

• Signif. codes:

0 *** 0.001 ** 0.01 * 0.05 . 0.1 1

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