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Fitting a model for a two-factor experimental design

chabotb

I'm trying to fit a model for a two-factor experimental design (2 factors). This is Problem 5.4 in D.C Montgomery book (Design and analysis of experiments 8th ed. Chap 5.The yield of a chemical process is being studied. The two most important variables are thought to be the pressure and the temperature. Three levels of each factor are selected, and a factorial experiment with two replicates is performed. The yield data are as follows:

The solution of this problem is met when Temperature and Pressure are set at "Nominal". I was expecting to set them at "Continous" but if I do that the ANOVA is wrong.

1) Why do I need to set factors to "Nominal" instead of "Continous" to get a proper ANOVA and fitting estimates?

2) If I want to find a regression model from the data, I need to set Temperature and Pressure at "Continous". Why?

Accepted answers

P_Bartell

I'm not familiar with the exact and full problem context as stated in the book so am only guessing here...in JMP variables can have different data types that align more or less with the analytic modeling methods for those types of data. In JMP, some (but certainly not all...which is what makes JMP such a useful application!) of those data types are labeled 'Nominal' and 'Continuous'. Perhaps what you are reading in the problem in the book is 'nominal' refers to the preferred or current settings for manufacturing process under study and NOT the data type label used in JMP?

These types of experimental factors as laid out in the problem are typically considered 'Continuous' in a JMP data type sense, which in turn opens up a variety of modeling methods, one of which is called Standard Least Squares in JMP...which in turn leads to a good old fashioned linear regression model.

All comments

P_Bartell

statman

There is more than one way to analyze data sets. Not necessarily "wrong". In the ANOVA approach, what you are estimating is the effect of each term in the model. For the two factor experiment you are referring to, each factor has 2 degrees of freedom (or 8 for the model effects), so the model is:

Y = T + P + TP + Error

If you use regression for analysis, you will be able to assign those 2 degrees of freedom for each factor to the linear and the quadratic effects:

Y = T + P + TP + T*T + P*P + E

Georg

I think the answer to both of your questions is: due to the workflow in JMP.

If you want to do ANOVA (analysis of variance, difference among group means in a sample):

Different groups in X are only existing (or unambiguous), when X is nominal.

So when using nominal X in "fit x by y" you end up in ANOVA.

Regression needs, that you can calculate with your x values. This only is the case for continuous x.

Simply by means of the data type of x and y you tell JMP, what analysis you want to perform. Have a look at the picture in the lower left, when starting the platform "fit y by x". I think, this is a nice and effective interface, letting JMP offer the methods that fit to your modeling types. However it may take some time to get used to it.