cancel
Showing results for 
Show  only  | Search instead for 
Did you mean: 
  • Instantly extract effect sizes, F-ratios, and FDR-adjusted p-values from your models with the Calculate Effects Sizes extension, available now in the JMP Marketplace!
  • New to JMP? Join us Sept. 23-24 for the Early User Edition of Discovery Summit, tailor-made for new users. Register now for free!

Discussions

Solve problems, and share tips and tricks with other JMP users.
Choose Language Hide Translation Bar
scjphysicist
Level I

JMP bug or alternate calculation in Life Distribution scale confidence limits?

Dear JMP,

I found a discrepancy between a JMP result and the result in the textbook you cite as a reference.  Is this a bug or does JMP use a different calculation than the textbook?

JMP Pro 17.2.0 > Reliability and Survival > Life Distribution platform, fit to a Weibull, Wald confidence intervals > in the Parametric Estimate – Weibull section, all the numbers agree except the scale and Weibull ß confidence intervals (screenshot below).

Reference book is Meeker, Escobar, and Pascual, second edition, 2022.  (In JMP documentation, an earlier version of this book is cited as the source for these calculations.)  Section 8.3.2 Wald Confidence Intervals for Model Parameters, Example 8.13 Wald Confidence Interval for the Shock Absorber Weibull Shape Parameter, data set attached in JMP format.  Book gets for the scale factor [0.201, 0.498] (compare [0.173, 0.460] below) and for the Weibull beta parameter [2.01, 4.97] (compare 2.175, 5.780] below).

All the other parameters agree to at least 3 significant figures, including the location and Weibull alpha confidence intervals, all the standard errors, and the covariance matrix; it is only the scale and Weibull beta that disagrees.  The book says (equation 8.7) that the scale LCL, for example, is calculated as  (scale best estimate) / w,  where  w = exp[ (z score of 95% confidence) * (standard error) / (scale best estimate) ].  All those numbers are given by both the book and JMP and all agree, and the calculation gives the result in the book, not the JMP result.  So, either JMP has a bug, or it is using a different calculation, in which case, what is JMP’s calculation?

Thank you,

Scott

scjphysicist_0-1766101605906.png

 

1 ACCEPTED SOLUTION

Accepted Solutions

Re: JMP bug or alternate calculation in Life Distribution scale confidence limits?

For the LCLs on the scale and the Weibull beta specifically JMP is building Wald CIs on the extreme‑value parameters on the so called "identity scale," and then transforming endpoints for them. Whereas, Meeker–Escobar–Pascual (2022) seem to  construct Wald CIs for strictly positive parameters on the log scale directly and then back‑transform, which yields a different multiplicative form (albeit arguably more intuitive based on the information/ data table shared).

So, JMP fits these: 
μ^=location=10.230, σ=scale=0.316, SE(μ)=0.1099, SE(σ)=0.0732

Then directly computes:
Lower and Upper CI on location:
CI 0.95​(µ)=µ^ ± zSE(µ^) = 10.230 ± 1.96(0.1099) = (10.015,10.445)

Lower and Upper CI on scale:
CI 0.95​(σ)=σ^ ± zSE(σ^) = 0.316 ± 1.96(0.0732) = (0.173, 0.460) 

from the reported Estimates and Std Errors that it has fitted as shown above and in the attached summary table appended to the data shared:

PatrickGiuliano_0-1771291149354.png

The Lower and Upper CI on Weibull α are computed this way which gives the same result as in Meeker and Escobar:
CI 0.95(α) = exp(LCL 0.95​(µ)), exp(UCL 0.95​(µ)) = (exp(10.015), exp(10.445)) = (22347.770, 34380.492)


But the Lower and Upper CI on Weibull β, computed by 'inversion' of the sigma confidence limit estimates, differs from what you showed per Meeker and Escobar:
CI 0.95(β)=(1/UCL 0.95​(σ^), ((1/LCL 0.95​(σ^)) = (1/0.460, 1/0.173)=(2.175, 5.780).

The attached modified version of this data with the summary table from Life Distribution appended to it shows the calculations in column formulas next to JMP's reported output, and next to the results calculated using the formulas shared here: 

PatrickGiuliano_1-1771291254817.png

 

View solution in original post

4 REPLIES 4

Re: JMP bug or alternate calculation in Life Distribution scale confidence limits?

Hi @scjphysicist Thanks for your question! I'm looking into this now and will provide an update for you soon.   

-JMP Technical Support

Re: JMP bug or alternate calculation in Life Distribution scale confidence limits?

For the LCLs on the scale and the Weibull beta specifically JMP is building Wald CIs on the extreme‑value parameters on the so called "identity scale," and then transforming endpoints for them. Whereas, Meeker–Escobar–Pascual (2022) seem to  construct Wald CIs for strictly positive parameters on the log scale directly and then back‑transform, which yields a different multiplicative form (albeit arguably more intuitive based on the information/ data table shared).

So, JMP fits these: 
μ^=location=10.230, σ=scale=0.316, SE(μ)=0.1099, SE(σ)=0.0732

Then directly computes:
Lower and Upper CI on location:
CI 0.95​(µ)=µ^ ± zSE(µ^) = 10.230 ± 1.96(0.1099) = (10.015,10.445)

Lower and Upper CI on scale:
CI 0.95​(σ)=σ^ ± zSE(σ^) = 0.316 ± 1.96(0.0732) = (0.173, 0.460) 

from the reported Estimates and Std Errors that it has fitted as shown above and in the attached summary table appended to the data shared:

PatrickGiuliano_0-1771291149354.png

The Lower and Upper CI on Weibull α are computed this way which gives the same result as in Meeker and Escobar:
CI 0.95(α) = exp(LCL 0.95​(µ)), exp(UCL 0.95​(µ)) = (exp(10.015), exp(10.445)) = (22347.770, 34380.492)


But the Lower and Upper CI on Weibull β, computed by 'inversion' of the sigma confidence limit estimates, differs from what you showed per Meeker and Escobar:
CI 0.95(β)=(1/UCL 0.95​(σ^), ((1/LCL 0.95​(σ^)) = (1/0.460, 1/0.173)=(2.175, 5.780).

The attached modified version of this data with the summary table from Life Distribution appended to it shows the calculations in column formulas next to JMP's reported output, and next to the results calculated using the formulas shared here: 

PatrickGiuliano_1-1771291254817.png

 

scjphysicist
Level I

Re: JMP bug or alternate calculation in Life Distribution scale confidence limits?

Thank you, Patrick!  Now that you show me what JMP is doing, it is somewhat obvious, and I feel like I should have seen it myself.

That does mean that JMP does something a little bit different from Meeker and Escobar.  These are two expert opinions, and it is hard for me to judge which to use.  After some thought, my philosophy is this:  My company is paying for JMP for several reasons, and one of those reasons is your (JMP's) judgment on the correct statistical calculation to use in each instance.  (There are plenty of other examples of this judgment in JMP, like which goodness-of-fit test to use for each distribution type.)  So, I think we should trust your (JMP's) judgment and go with the JMP version of the calculation.

So, thanks again for the detailed and thoughtful technical response!  (Note: I want to give you a Kudo, but it doesn't seem to accept my input.)

-Scott

Craige_Hales
Super User

Re: JMP bug or alternate calculation in Life Distribution scale confidence limits?

And a kudo to you as well. The kudo issue is reported.

Craige

Recommended Articles