11.3.3 - The Analysis of Mixture Designs

Example 11.3: Elongation of Yarn Section

close up of yarn threads

Ex11.5.MTW from the text.

This example has to do with the elongation of yarn based on its component fabrics. There are three components in this mixture and each component is a synthetic material. The mixture design was one that we had looked at previously. It is a simple lattice design of degree 2. This means that it has mixtures of 0, 1/2, 100%. The components of this design are made up of these three possibilities.

  StdOrder RunOrder PtType Blocks A B C Y
1 1 1 1 1 1.0 0.0 0.0 11.0
2 2 2 2 1 0.5 0.5 0.0 15.0
3 3 3 2 1 0.5 0.0 0.5 17.7
4 4 4 1 1 0.0 1.0 0.0 8.8
5 5 5 2 1 0.0 0.5 0.5 10.0
6 6 6 1 1 0.0 0.0 1.0 16.8
7 7 7 1 1 1.0 0.0 0.0 12.4
8 8 8 2 1 0.5 0.5 0.0 14.8
9 9 9 2 1 0.5 0.0 0.5 16.4
10 10 10 1 1 0.0 1.0 0.0 10.0
11 11 11 2 1 0.0 0.5 0.5 9.7
12 12 12 1 1 0.0 0.0 1.0 16.0
13 13 13 1 1 1.0 0.0 0.0 *
14 14 14 2 1 0.5 0.5 0.0 16.1
15 15 15 2 1 0.5 0.0 0.5 16.6
16 16 16 1 1 0.0 1.0 0.0 *
17 17 17 2 1 0.0 0.5 0.5 11.8
18 18 18 1 1 0.0 0.0 1.0 *

In the Minitab program, the first 6 runs show you the pure components, and in addition, you have the 5 mixed components. All of this was replicated 3 times so that we have 15 runs. There were three that had missing data.

You can also specify in more detail which type of points that you want to include in the mixture design using the dialog boxes in Minitab if your experiment requires this.

Analysis

In the analysis we fit the quadratic model ( the linear + the interaction terms). Remember we only have 6 points in this design, the vertex, the half-lengths, so we are fitting a response surface to these 6 points. Let's take a look at the analysis:

Regression for Mixtures: Y versus A, B, C

Estimated Regression Coefficients for Y (component proportions)
Term Coef SE Coef T P VIF
A 11.70 0.6037 * * 1.750
B 9.400 0.6037 * * 1.750
C 16.400 0.637 * * 1.750
A*B 19.000 2.6082 7.28 0.000 1.750
A*C 11.400 2.6082 4.37 0.002 1.750
B*C -9.600 2.6082 -3.68 0.005 1.750
S = 0.853750 PRESS = 18.295  
R-Sq = 95.14% R- Sq(pred) = 86.43% R-Sq(adj) = 92.43%

Analysis of variance for Y (component proportions)

Source DF Seq SS Adj SS Adj MS F P
Regression 5 128.296 128.2960 25.6592 35.20 0.000
Linear 2 57.629 50.9200 25.4600 34.93 0.000
Quadratic 3 70.667 70.669 23.5556 32.32 0.000
Residual Error 9 6.560 6.5600 0.7289    
Total 14 134.856        

Here we get 2 df linear, 3 df quaratic, these are the five regression parameters. If you look at the individual coefficients, six of them because they are is no intercept, three linear and three cross-product terms... The 9 df for error are from the triple replicates and the double replicates. This is pure error and there is no additional df for lack of fit in this full model.

If we look at the coutour service plot we get:

plot

We have the optimum somewhere between a mixture of A and C, with B essentially not contributing very much at all. So, roughly 2/3rds C and 1/3 A is what we would like in our mixture. Let's look at the optimizer to find the optimum values.

optimizer

It looks like A = about .3 and B = about .7, with B not contributing nothing to the mixture.

Unless I see the plot how can I use the analysis output? How else can I determine the appropriate levels?

Example 11.4: Gasoline Production Section

container of gasoline

Pr11-31.MTW from text

This example focuses on the production of an efficient gasoline mixture. The response variable is miles per gallon (mpg) as a function of the 3 components in the mixture. The data set contains these 14 points - which has duplicates at the centroid, labeled (1/3, 1/3, 1/3), and the three vertices, labeled (1,0,0), (0,1,0), and (0,0,1).

  StdOrder RunOrder PtType Blocks A B C Y-mpg
1 1 1 1 1 1.00000 0.00000 0.00000 24.5
2 2 2 2 1 0.50000 0.50000 0.00000 25.1
3 3 3 2 1 0.50000 0.00000 0.50000 24.3
4 4 4 1 1 0.00000 1.00000 0.00000 24.8
5 5 5 2 1 0.00000 0.50000 0.50000 23.5
6 6 6 1 1 0.00000 0.00000 1.00000 22.7
7 7 7 0 1 0.33333 0.33333 0.33333 24.8
8 8 8 -1 1 0.66667 0.16667 0.16667 24.2
9 9 9 -1 1 0.16667 0.66667 0.16667 23.7
10 10 10 -1 1 0.16667 0.16667 0.66667 23.7
11 11 11 1 1 1.00000 0.00000 0.00000 25.1
12 12 12 1 1 0.00000 1.00000 0.00000 23.9
13 13 13 1 1 0.00000 0.00000 1.00000 23.6
14 14 14 0 1 0.33333 0.33333 0.33333 24.1

This is a degree 2 design that has points at the vertices, middle of the edges, the center, and axial points, which are interior points, (2/3, 1/6, 1/6), (1/6, 2/3, 1/6) and (1/6, 1/6, 2/3). Also the design includes replication at the vertices and the centroid.

If you analyze this dataset without having first generated the design in Minitab, you need to tell Minitab some things about the data since you're importing it.

Analysis of variance for Y-mpg (component proportions)

Source DF Seq SS Adj SS Adj MS F P
Regression 5 4.2224 4.2224 0.84449 3.90 0.043
Linear 2 3.9247 2.7487 1.37433 6.35 0.022
Quadratic 3 0.2978 0.2978 0.09925 0.46 0.719
Residual Error 8 1.7319 1.7319 0.21648    
Lack-of-Fit 4 0.4969 0.4969 0.12421 0.40 0.800
Pure Error 4 1.2350 1.2350 0.30875    
Total 13 5.9543        

The model shows a linear term significant, the quadratic terms not significant, and the lack of fit, ( a total of 10 points and we are fitting a model sex parameters - 4 df), it shows that there is no lack of fit from the model. It is not likely that it would make any difference.

If we look at the contour plot for this data:

plot

We can see that the optimum looks to be about 1/3, 2/3 between components A and B. Component C does not play hardly any role at all. Next, let's look at the optimizer for this data where we want to maximize a target of about 24.9.

Minitab output

And, again, we can see that component A at the optimal level is about 2/3rds and component B is at about 1/3rd. Component C plays no part, as a matter of fact if we were to add it to the gasoline mixture it would probably lower our miles per gallon average.

Let's go back to the model and take out the factors related to component C and see what happens. When this occurs we get the following contour plot...

plot

... and the following analysis:

Analysis of variance for Y-mpg (component proportions)

Source DF Seq SS Adj SS Adj MS F P
Regression 3 4.0812 4.0812 1.3604 7.26 0.007
Linear 2 3.9247 3.1548 1.5774 8.42 0.007
Quadratic 1 0.1566 0.1566 0.1566 0.84 0.382
Residual Error 10 18731 18731 0.1873    
Lack-of-Fit 6 0.6381 0.6381 0.1063 0.34 0.882
Pure Error 4 1.2350 1.2350 0.3088    
Total 13 5.9543        

Our linear terms are still significant, our lack of fit is still not significant. the analysis is saying that linear is adequate for this situation and this set of data.

One says 1 ingredient and the other says a blend - which one should we use?

I would like look at the variance. ...

24.9 is the predicted value.

By having a smaller, more parimonious model you decrease the variance. This is what you would expect with a model with fewer parameters. The standard error of the fit is a function of the design, and for this reason, the fewer the parameters the smaller the variance. But is also a function of residual error which gets smaller as we throw out terms that were not significant.