The 2 k Factorial Design A design with k factors each with just 2 - - PowerPoint PPT Presentation

ST 435/535 Statistical Methods for Quality and Productivity Improvement / Statistical Process Control

The 2k Factorial Design

A design with k factors each with just 2 levels has 2k treatments, and is known as a 2k design. 2k designs have the fewest treatments among all k-factor designs. They are widely used, often supplemented by extra design points to provide more information about the effects of the factors.

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ST 435/535 Statistical Methods for Quality and Productivity Improvement / Statistical Process Control

Example: The router experiment Problem: too much variability in dimensions of notches in PC boards. The experiment: vibration during cutting was used as an indicator of notch variability. Factors: Bit size (A), with levels

1 16” and 1 8”;

Speed (B), with levels 40 and 80 rpm. Replicated: n = 4 replications of each of the 22 = 4 treatments.

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ST 435/535 Statistical Methods for Quality and Productivity Improvement / Statistical Process Control

In R:

router <- read.csv("Data/Table-13-08.csv") summary(aov(Vibration ~ A * B, router))

Both factors have significant effects, and the interaction is also significant. The interaction plot:

with(router, interaction.plot(A, B, Vibration))

Vibration is low when A (bit size) is at its low level, for either level of B (speed).

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ST 435/535 Statistical Methods for Quality and Productivity Improvement / Statistical Process Control

Regression equation To predict what the vibration would be for other settings, we need a regression equation: E(Y ) = β0 + β1x1 + β2x2 + β1,2x1x2. It is convenient to use coded variables: x1 =

• +1

if A is at its high level −1 if A is at its low level x2 =

• +1

if B is at its high level −1 if B is at its low level

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ST 435/535 Statistical Methods for Quality and Productivity Improvement / Statistical Process Control

router$x1 <- ifelse(router$A == "+", 1, -1) router$x2 <- ifelse(router$B == "+", 1, -1) summary(lm(Vibration ~ x1 * x2, router))

The fitted equation is ˆ y = 23.83 + 8.32x1 + 3.77x2 + 4.36x1x2 Note that the P-value associated with each coefficient is the same as the corresponding P-value in the analysis of variance table.

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ST 435/535 Statistical Methods for Quality and Productivity Improvement / Statistical Process Control

The part of the data set involving x1 and x2 is:

x1 x2

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+1

• 1

+1

• 1

+1 +1 +1 +1 +1 +1 +1 +1 +1

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ST 435/535 Statistical Methods for Quality and Productivity Improvement / Statistical Process Control

Add a column of 1’s and column for x1x2:

1 x1 x2 x1x2 1

• 1
• 1

+1 1

• 1
• 1

+1 1

• 1
• 1

+1 1

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+1 1 +1

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1 +1 +1 +1 1 +1 +1 +1 1 +1 +1 +1 1 +1 +1 +1

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ST 435/535 Statistical Methods for Quality and Productivity Improvement / Statistical Process Control

These four columns (divided by 16) are the coefficients needed to compute ˆ β0, ˆ β1, ˆ β2, and ˆ β1,2, respectively. The estimates could be calculated by hand using this table. We’ll use tables like this later to study the effects of blocking and to construct fractionally replicated designs.

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ST 435/535 Statistical Methods for Quality and Productivity Improvement / Statistical Process Control

Physical variables The equation may be rewritten in terms of the physical variables x∗

1 = bit size, in inches

x∗

2 = speed, in rpm

router$bitsize <- ifelse(router$A == "+", 1/8, 1/16) router$speed <- ifelse(router$B == "+", 80, 40) summary(lm(Vibration ~ bitsize * speed, router))

The fitted equation is ˆ y = 26.78 − 152x∗

1 − 0.465x∗ 2 + 6.97x∗ 1x∗ 2.

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ST 435/535 Statistical Methods for Quality and Productivity Improvement / Statistical Process Control

The equation in physical variables is more useful for predicting the response under new conditions of bit size and speed. However, its statistical interpretation is not as clear as when coded variables are used. Note that x∗

2 = 60 + 20x2,

and that x∗

1 is similarly a linear function of x1. These relationships

can be used to transform the equation in x1 and x2 into the equation in x∗

1 and x∗ 2, without actually refitting the model.

10 / 23 Factorial Experiments for Process Design The 2k Factorial Design

ST 435/535 Statistical Methods for Quality and Productivity Improvement / Statistical Process Control

More general equations This regression equation is a special case of the general second-order model E(Y ) = β0 + β1x1 + β2x2 + β1,2x1x2 + β1,1x2

1 + β2,2x2 2

with β1,1 = β2,2 = 0. This equation cannot be fitted to these data. Extra runs are needed, such as: Center points: x1 = x2 = 0; Axial points: x1 = 0, x2 = ±α and x2 = 0, x1 = ±α, where α could be 1, but is often larger.

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ST 435/535 Statistical Methods for Quality and Productivity Improvement / Statistical Process Control

More than 2 factors The design and analysis of experiments with k > 2 two-level factors builds on the 22 case. The basic statistical model includes interaction terms for more than two factors at a time, which are difficult to interpret. In practice, it is often found that only a few factors are “active”, and that no high-order interactions are important.

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ST 435/535 Statistical Methods for Quality and Productivity Improvement / Statistical Process Control

Example: Surface finish Response is the surface finish of a metal part. Factors, each with two levels: A: feed rate; B: depth of cut; C: tool angle. Two replications of each treatment. In R:

finish <- read.csv("Data/Table-13-12.csv") summary(aov(Finish ~ A * B * C, finish))

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ST 435/535 Statistical Methods for Quality and Productivity Improvement / Statistical Process Control

Note that every line that involves C has a P-value greater than 0.18, so it appears that C has no effect, either overall or on the effects of A and B. If C is ignored, the design collapses (or projects) onto a 2 × 2 design with 4 replicates.

with(finish, interaction.plot(A, B, Finish))

The interaction plot shows that the finish is best (lowest) when A (feed rate) is at its low level, and then the level of B (depth of cut) is unimportant.

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ST 435/535 Statistical Methods for Quality and Productivity Improvement / Statistical Process Control

Single replicate In the model with all interactions, the degrees of freedom for error are a × b × ... × (n − 1), where n is the number of replications of each treatment. When n = 1, we have no estimate of σ2, so we cannot estimate the standard error of any parameter, and we cannot test hypotheses about the parameters or set up confidence intervals. Various procedures have been proposed to distinguish important effects from the rest. The simplest and most widely used is the half-normal plot of the estimated effects.

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ST 435/535 Statistical Methods for Quality and Productivity Improvement / Statistical Process Control

Example: Plasma etching Four factors: Gap (A): spacing between anode and cathode; Pressure (B): in the etching vessel; Flow (C): flow rate of hexafluoroethane; Power (D): applied to the cathode. A single completely randomized run of the 24 design

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ST 435/535 Statistical Methods for Quality and Productivity Improvement / Statistical Process Control

In R:

plasma <- read.csv("Data/Table-13-15.csv") summary(aov(Rate ~ A * B * C * D, plasma)) library(gplots) qqnorm(aov(Rate ~ A * B * C * D, plasma), label = TRUE)

No F-ratios or P-values are given. In the half-normal plot, A, D, and their interaction are the most prominent points.

with(plasma, interaction.plot(A, D, Rate))

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ST 435/535 Statistical Methods for Quality and Productivity Improvement / Statistical Process Control

Reduced model Because no third- or fourth-order interaction is large, we could fit a reduced model with only main effects and two-factor interactions:

summary(aov(Rate ~ (A + B + C + D)^2, plasma))

Note that this model leaves 5 degrees of freedom for residuals, so now we get hypothesis tests, which show that only A, D, and the AD interaction are significant. However, those degrees of freedom come from the omitted interactions, which were omitted because they were small, so the residual mean square might be biased downwards.

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ST 435/535 Statistical Methods for Quality and Productivity Improvement / Statistical Process Control

Center points Because the AD interaction seems important, we might want to fit a complete second order model. As noted earlier, that’s not possible with the two-level design. The plasma experiment actually included 4 center points in the design:

plasmaC <- read.csv("Data/Table-13-18.csv") summary(aov(Rate ~ (A + B + C + D)^2 + I(A^2) + I(B^2) + I(C^2) + I(D^2), plasmaC))

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ST 435/535 Statistical Methods for Quality and Productivity Improvement / Statistical Process Control

Note that A2, B2, C 2, and D2 all have the same values at all design points (+1 at the factorial points and 0 at the center points), so their effects cannot be distinguished; only the first is shown, but it represents the sum of all the squared terms. The sum of squares on the I(A^2) line is called the pure quadratic

• r curvature sum of squares.

The corresponding F-statistic tests the null hypothesis H0 : β1,1 + β2,2 + β3,3 + β4,4 = 0, where βi,i is the coefficient of x2

i .

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ST 435/535 Statistical Methods for Quality and Productivity Improvement / Statistical Process Control

Pure error The 4 center-point runs provide 1 degree of freedom for testing for curvature, and 3 degrees of freedom for estimating σ2. Because they reflect variability among the 4 replicates, they are unaffected by the choice of a model: pure error. By contrast, the other 5 degrees of freedom for residuals come from assuming that all high-order interactions are zero: lack of fit error. Some software breaks out these two types of error; in R, you can use the highest-order interaction to capture the lack-of-fit error:

summary(aov(Rate ~ (A + B + C + D)^2 + I(A^2) + factor(A):factor(B):factor(C):factor(D), plasmaC))

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Blocking and confounding Factorial designs may require many runs (individual experiments), and complete randomization is not always possible. For example, runs may be spread over several days, with one complete replicate carried out on each day. To allow for variation from day to day, days are treated as a blocking factor, and extra parameters are included in the model to estimate its impact.

22 / 23 Factorial Experiments for Process Design The 2k Factorial Design

ST 435/535 Statistical Methods for Quality and Productivity Improvement / Statistical Process Control

Example: Surface finish For example, suppose that the two replications of the paint finish experiment were carried out on separate days:

finish$Day <- rep(c(1, 2), 8) summary(aov(Finish ~ Day + A * B * C, finish))

We usually assume, as here, that the blocking factor does not interact with the experimental factors, so including just removes 1 degree of freedom from the residuals. If blocks are too small to run a complete replication, a fraction of the experiment is run in each block.

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