cancel
Showing results for 
Show  only  | Search instead for 
Did you mean: 
Try the Materials Informatics Toolkit, which is designed to easily handle SMILES data. This and other helpful add-ins are available in the JMP® Marketplace
Choose Language Hide Translation Bar

Whole Model Leverage Details

Well, I must say, I am completely baffled as to how JMP is constructing the confidence band in the Actual by Predicted Plot. I have read all the statistical details. They claim that the Actual by Predicted Plot is actually a whole-model leverage plot, but I can construct nothing like it. When constructing the plot and confidence band, what exactly is L in the matrix equation

PatriotOGrady_0-1716491224342.png

?

Normally, a leverage plot shows the residuals of all predictors but one against the residuals of the one parameter alone.  But what could this possibly be when using the whole model against itself? The model without the mean vs. the mean? If so, the plot is not going to resemble an Actual by Predicted plot. Nor would the statistical details that follow for the confidence band make any sense.  I also wonder if one can construct upper and lower prediction limits in a similar way.

 

I am using JMP 17.2.0. Thanks for any help.

2 REPLIES 2
statman
Super User

Re: Whole Model Leverage Details

see here:

 

https://www.jmp.com/support/help/en/18.0/?os=mac&source=application#page/jmp/statistical-details-for...

 

"All models are wrong, some are useful" G.E.P. Box
MRB3855
Super User

Re: Whole Model Leverage Details

Hi @PatriotOGrady . L can be, for example, a 1x (p+1) row vector, and B is a (p+1) x 1 column vector. In a multiple regression setting, B would be a column vector with elements B0, B1, B2, …, Bp. This is the case for example where Y= B0 + B1*x1 + B2*x2 + …+ Bp*xp. And, for example, if you are testing H0: B2=0, then L is a row vector <0, 0, 1, 0, …, 0>. This results in LB= B2.

 

For the whole model test, L is a p x (p+1) matrix (each row similar to the row vector described above with all zeros except a “1” in the appropriate position) resulting in, effectively, H0: B1=B2=…=Bp=0.

 

You may want to review the Gauss-Markov model and the associated matrix/vector representations and operations.