[R] cubic complete Scheffe mixture models
Cleber N.Borges
klebyn at yahoo.com.br
Thu Mar 15 23:36:03 CET 2018
Hello everyone
I'm trying to use Scheffe's complete cubic model (mixture design).
In the bibliographies, they indicate that the term is of the type: A * B
* (A-B).
But I see that trying to adjust the three cubic terms results in
singularities.
I know this implies not having the inverse matrix:
solve (t (X)% *% X) does not exist.
The bibliographies show all three cubic terms.
So my question:
How to circumvent the singularity and estimate the coefficients for
these terms in Scheffe's complete cubic model in the modeling of
Chemical Blends?
Thanks in advance for any help.
Thank you so much
Cleber Borges
#### MY_DATA:
H2O TOL MET Response
1 1.0000000 0.0000000 0.0000000 0.39056923
2 0.0000000 1.0000000 0.0000000 -1.49611856
3 0.0000000 0.0000000 1.0000000 -0.33922649
4 0.5000000 0.5000000 0.0000000 -0.65517683
5 0.5000000 0.0000000 0.5000000 -1.23281158
6 0.0000000 0.5000000 0.5000000 0.51523665
7 0.3333333 0.3333333 0.3333333 -0.93516022
8 0.3333333 0.3333333 0.3333333 -0.84429633
9 0.6666667 0.1666667 0.1666667 0.55931646
10 0.6666667 0.1666667 0.1666667 0.47999083
11 0.1666667 0.6666667 0.1666667 0.07182446
12 0.1666667 0.6666667 0.1666667 -0.01813801
13 0.1666667 0.1666667 0.6666667 1.79895957
14 0.1666667 0.1666667 0.6666667 1.70503083
# my try:
reg <- lm( Response ~ -1 + H2O*TOL*MET + I(H2O*TOL*(H2O-TOL)) +
I(H2O*MET*(H2O-MET)) + I(TOL*MET*(TOL-MET)), data=MY_DATA )
summary( reg )
##
Coefficients: (1 not defined because of singularities)
Estimate Std. Error t value Pr(>|t|)
H2O 0.6279 1.1070 0.567 0.595
TOL -1.2588 1.1070 -1.137 0.307
MET -0.1019 1.1070 -0.092 0.930
I(H2O * TOL * (H2O - TOL)) 11.8720 13.8902 0.855 0.432
I(H2O * MET * (H2O - MET)) -15.7052 13.8902 -1.131 0.309
I(TOL * MET * (TOL - MET)) NA NA NA NA
H2O:TOL 0.5398 5.4316 0.099 0.925
H2O:MET -4.0846 5.4316 -0.752 0.486
TOL:MET 6.6810 5.4316 1.230 0.273
H2O:TOL:MET 2.0005 31.3024 0.064 0.952
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