Fitting scheffe model in mixture + non mixture design

[#1]

Hi,

I have a custom design with 4 mixture variables and 13 non mixture variables. I am performing a screening experiment with 24 runs. When I use the "Fit Model" functionality with only main effects and choose "No intercept" to fit a model,

1) Do i get a scheffe model?

2) I am not including any interaction terms in the model as I am performing screening and so am interested in only the main effects. So, is there any benefit of developing a scheffe model in this case?

[#2 · Accepted solution]

Anytime you are building a model with mixture factors that add up to 1, you should be fitting a mixture model, such as the Scheffe mixture model. If you don't, you will get a singular matrix message. This is true even with just a main effects model. You really don't have a choice.

Remember that a Scheffe mixture model is not the same as a no-intercept model. When fitting a Scheffe mixture model in JMP, the Analysis of Variance report will state: Tested against reduced model Y=mean. If the model is a no-intercept model, the message would be: Tested against reduced model Y=0.

[#3]

Thanks Dan!

But when I click on "Fit model" under "Analyze" I do not get an Analysis of Variance report. I get the Least square fit parameter estimates. I observe that the mixture variabels are scaled (mean = 0 and range = 1) and their parameter estimates are also present. So, are these estimates the co-efficients of the Scheffe's model?

[#4]

With version 13 of JMP the Analysis of Variance report is closed by default. You need to open it by going to the red pop-up menu at the top of your report next to "Response Y" with Y being the name of your response variable. From there choose Regression Reports > Analysis of Variance. Remember that you sometimes will need to request the items that you need for your analysis from those red popup menus.

As for your parameter estimates, if JMP created the design, then it will automatically turn on pseudocomponent coding for your mixture factors as well as the "regular" coding for the non-mixture factors. This assumes that when you created the design you specified the natural units for your factors. This coding does not indicate if a mixture model was fit to the data or not.

[#5]

Thanks Dan!

I understand that for scheffe model the calculation of p-values for mixture variables is not straightforward as for a regular model because the t-test do not assess the statistical significance of the component effects. Is there any documentation which shows how the p-values are being calculated for mixture variables in case of a scheffe model? 

[#6]

As far as I know, the only place that the testing is covered is in our training course: Design and Analysis of Mixture Experiments.

The calculations are actually correct, it is the null hypothesis on the main effects that is not appropriate.

For a Scheffe model, the main effects should compare the parameter estimates against the overall mean of the data. The Parameter Estimates table compares to 0.

A quick solution: just form the confidence intervals for each parameter estimate. If the confidence interval contains the overall mean of the data, then that component has little effect. If the mean is outside of the confidence interval, that component has an effect.

Conclusions on any interaction effects would be just fine.

[#7]

Thanks Dan!

Regarding the quick solution, when you say compare the confidence interval of the parameter estimate with the overall mean of the data, which data you we referring to? Is it the data correponding to the dependent variable (Y) or the data corresponding to the mixture variables?

[#8]

I think I understand the quick solution now. Just writing it here to make sure that I understand correctly:

Let the estimates of my mixture variables X1,X2,X3,X4 be B1,B2,B3,B4 respectively and the std error of each of the estimates be SD1,SD2,SD3,SD4 respectively. Then the 95% CI of my estimates are (B1-2*SD1, B1+2*SD1), (B2-2*SD1, B2+2*SD2) and so on.Then I compute the average of the data as Y/n where n are the number of observations. Now if the inverval (B1-2*SD1,B1+2*SD1) contains the average of the data, that means that X1 is not significant and vice versa. Am I correct?

[#9]

You have the right idea. Of course the multiplier of 2 may or may not be correct depending on degrees of freedom and such. Therefore, it is easier to use JMP. Right-click on the parameter estimates table and choose Columns > Lower 95% and Upper 95%. That will give you the 95% confidence interval for each of the parameter estimates.