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5. Process Improvement 5.5. Advanced topics 5.5.9. An EDA approach to experimental design 5.5.9.9. Cumulative residual standard deviation plot 5.5.9.9.6. Motivation: Why is the 1/2 in the Model?=-1> =-1> |
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| Presence of 1/2 term does not affect predictive quality of model |
The leading 1/2 is a multiplicative constant that we have chosen to
include in our expression of the linear model. Some authors and
software prefer to "simplify" the model by omitting this leading 1/2.
It is our preference to include the 1/2. This follows a hint
given on page 334 of
Box, Hunter, and
Hunter (1978) where they note that the coefficients that appear
in the equations are half the estimated effects.
The presence or absence of the arbitrary 1/2 term does not affect the predictive quality of the model after least squares fitting. Clearly, if we choose to exclude the 1/2, then the least squares fitting process will simply yield estimated values of the coefficients that are twice the size of the coefficients that would result if we included the 1/2. |
| Included so least squares coefficient estimate equal to estimated effect |
We recommend the inclusion of the 1/2 because of an additional property
that we would like to impose on the model; namely, we desire that:
For a given factor, say X2, the estimated least squares coefficient B2 and the estimated effect E2 are not in general identical in either value or concept. |
| Effect |
For factor X2, the effect E2 is defined as the change in the mean
response as we proceed from the "-" setting of the factor to the
"+" setting of the factor. Symbolically:
On the other hand, the estimated coefficient B2 in a model is defined as the value that results when we place the model into the least squares fitting process (regression). The value that returns for B2 depends, in general, on the form of the model, on what other terms are included in the model, and on the experimental design that was run. The least squares estimate for B2 is mildly complicated since it involves a behind-the-scenes design matrix multiplication and inversion. The coefficient values B that result are generally obscured by the mathematics to make the coefficients have the collective property that the fitted model as a whole yield a minimum sum of squared deviations ("least squares"). |
| Orthogonality |
Rather remarkably, these two concepts and values:
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| Why is 1/2 the appropriate multiplicative term in these orthogonal models? |
Given the computational simplicity of orthogonal designs, why then is
1/2 the appropriate multiplicative constant? Why not 1/3, 1/4, etc.?
To answer this, we revisit our specified desire that
Y in
the response Y as we proceed from the "-" setting
of X2 to the "+" setting of X2 (that is, we would like the estimated
coefficient B2 to be identical to the estimated effect E2 for factor
X2).
In general, the least squares estimate of a coefficient in a linear model will yield a coefficient that is essentially a slope:
= (change in response)/(change in factor levels)
Y), we
must account for and remove the change in X
( X).
What is the
-1*-1 = +1 -1*+1 = -1 +1*-1 = -1 +1*+1 = +1and hence the set of values that the 2-factor interactions (and all interactions) take on are in the closed set {-1,+1}. This -1 and +1 notation is superior in its consistency to the (1,2) notation of Taguchi in which the interaction, say X1*X2, would take on the values 1*1 = 1 1*2 = 2 2*1 = 2 2*2 = 4which yields the set {1,2,4}. To circumvent this, we would need to replace multiplication with modular multiplication (see page 440 of Ryan (2000)). Hence, with the -1,+1 values for the main factors, we also have -1,+1 values for all interactions which in turn yields (for all terms) a consistent X of
X =
(+1) - (-1) = +2
B = (and so to achieve our goal of having the final coefficients reflect Y only,
we simply gather up all of the 2's in the denominator and create a
leading multiplicative constant of 1 with denominator 2, that is, 1/2.
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| Example for k = 1 case |
For example, for the trivial k = 1 case, the obvious model
Y = c + ( )*X1
X)
* ( Y)*X1
Y)*X1
Y = c + (1/2)*(factor 1 effect)*X1 Y = c + (1/2)*(B*)*X1, with B* = 2B = E |