topic Re: Confidence interval in Fit Model Actual by Predicted plot in Discussions

Confidence interval in Fit Model Actual by Predicted plot

vcl0124 — Tue, 16 Jan 2024 08:58:25 GMT

I would like to know how the confidence interval in Fit Model Actual by Predicted plot is calculated. I tried to manually plot Actual vs Predicted with Fit Y by X and the confidence interval is different with the one in Fit Model. I can manually calculate the confidence interval of Fit Y by X but cannot figure out how to calculate the one in Fit Model. Please advise. Thanks a lot!

Re: Confidence interval in Fit Model Actual by Predicted plot

MRB3855 — Tue, 16 Jan 2024 09:33:55 GMT

Hi @vcl0124 : It is a leverage plot.

https://community.jmp.com/t5/Discussions/how-to-read-actual-by-predicted-plot/td-p/390471

https://www.jmp.com/support/help/en/17.2/index.shtml#page/jmp/statistical-details-for-leverage-plots.shtml

Re: Confidence interval in Fit Model Actual by Predicted plot

vcl0124 — Thu, 18 Jan 2024 00:31:26 GMT

Hi MRB3855,

Thanks a lot for your advice! However, I found the z^2 in below equation should be (z - mean of predicted)^2. Do you agree?

Another question is about fit model with (1st plot below) and without (2nd plot below) interception. I believe the with interception one is the better model but the leverage plot is wider. How to understand this? Please advise. Thanks a lot!

Re: Confidence interval in Fit Model Actual by Predicted plot

MRB3855 — Thu, 18 Jan 2024 14:11:05 GMT

Hi @vcl0124 :

Q:I found the z^2 in below equation should be (z - mean of predicted)^2. Do you agree?

A: not according to Sall (1990). See JMP reference

https://www.jstor.org/stable/2684358

https://blogs.sas.com/content/iml/2022/07/11/confidence-bands-partial-leverage-plot.html

Q: I believe the interception one is the better model but the leverage plot is wider. How to understand this? Please advise.

A: This may be a bit involved for this forum; but , briefly, unless you have a really good reason to exclude the intercept, do not exclude it. The second plot you show, clearly shows the model is not correct so any inference drawn from that is not correct as well.

Here is a pretty good discussion(s).

https://stats.stackexchange.com/questions/7948/when-is-it-ok-to-remove-the-intercept-in-a-linear-regression-model

https://communities.sas.com/t5/SAS-Communities-Library/The-Why-How-and-Cautions-of-Regression-Without-an-Intercept/ta-p/894641

Re: Confidence interval in Fit Model Actual by Predicted plot

vcl0124 — Fri, 19 Jan 2024 00:59:45 GMT

Hi MRB3855,

Thanks a lot for your advice!

For z^2 or (z - mean of predicted)^2, I cannot read the paper. However, the example in the 2nd link seems to have mean of predicted = 0. Actually, if you look at the shape of the leverage plot of the with interception case, the confidence interval is the smallest at mean of predicted. Therefore, z^2 must be smallest at mean of predicted and that's why I suspected the z is shifted by mean of predicted (replace z by z - mean of predicted).

For the leverage plot of without interception, I figured out it is tested against y=0 instead of y=mean of y and that's why the confidence interval is narrower as y=0 is very unlikely. I actually think it is OK to test without interception model against y=mean which should give us more meaningful information about whether the model is valid, am I correct?

Re: Confidence interval in Fit Model Actual by Predicted plot

Newbie_1 — Tue, 14 Apr 2026 17:13:40 GMT

Here is the solution: It is a bit misleading to suggest that actual-vs.-predicted plots are a type of partial regression or leverage plot. That is technically correct but not very useful. Essentially, JMP is fitting the line y = a0 + a_1 y^hat + e and from this constructs the projected confidence intervals in y^hat. Doing this reproduces the band in JMP's actual-vs.-predicted plot. This is poorly documented and probably why this has been such a longstanding befuddlement in the JMP Community. On another note, if one were to superimpose the actual confidence intervals from the associated factor-space for each predicted point, one obtains discrete points that are potentially multi-valued in y^hat. One might presume that JMP's confidence interval is a least-squares solution in y^hat to the actual CI values, but that is not the case, though one could certainly do this and would derive slightly different results per the attached.

To make this clear, I've attached three files: a JMP file, and Excel File, and a PDF with a fuller description. I hope this is helpful.

Re: Confidence interval in Fit Model Actual by Predicted plot

Newbie_1 — Sat, 11 Apr 2026 16:02:55 GMT

Here is an updated write-up that corrects some typographical errors.