nav[aria-label="Primary Navigation"] { padding: 0; & ul { list-style: none; width: 100%; display: flex; flex-direction: row; justify-content: start; align-items: start; gap: 30px; padding: 0; & li { margin: 0; } & ul li { list-style: none; } } }

Shape Parameter for a Lognormal Distribution in Parametric Reliability Distribution

Jed_Campbell

I'm having a hard time grasping what "Shape Parameter" means in the DOE > Sample Size > Reliability > Parametric Reliability Distribution when using a Lognormal Distribution.

The help says:

Shape Parameter

Specifies the scale parameter for the failure time distribution.

which isn't helping me, and combing through all the sources I could find only made my confusion worse. Could someone shed light on this, please? There's a bit more to this than I've explained, but I'd like unbiased answer(s) before I give more background.

I have a sample story problem below, with everything except the shape parameter populated.

ACME Pharmaceuticals manufactures Prednisolone acetate eye drops. Researchers have found, through shelf life studies, that the life distribution for these eye drops is lognormal with a scale parameter of 0.7. Reliability at 950 hours is estimated at 98%. The minimum acceptable reliability is 95% with 90% confidence. You will determine the test time for a sample size of 250 for a reliability demonstration with one failure allowed.

Accepted answers

calking

Actually, in this instance the Shape parameter is 1/Scale parameter for the lognormal distribution. Might be good to open a ticket asking for clearer labeling (I can try and work on that for 19.1). If you put in 98% for the demonstration reliability, the test time matches what you get in the old Sample Size and Power platform (that's the Add-in you're showing) when you use 1/0.7 as the shape parameter.

All comments

Di_Michelson

Hey Jed - I think about it this way: if X ~ lognormal(mu, sigma), then Y = log(X) ~ normal(mu, sigma). The sigma parameter is the scale parameter. So it's the standard deviation of the related normal distribution. Does that help?

Jed_Campbell

So, does that mean that I should just enter the supplied 0.7 as the shape parameter?

Di_Michelson

I think so!

Jed_Campbell

Thanks @Di_Michelson!

I'm asking because a customer is using an Add-In that got a different answer. I wasn't sure if I was just missing something in how to populate the platform in JMP. Here's a screenshot of the add-in:

vs JMP 19:

In JMP, changing the Shape/Scale Parameter to 1.43 (which is 1/0.7) gives the same result as the add-in. I'm pretty sure the Add-in was wrong. Yikes.

calking

Jed_Campbell

Thanks for the clarification, @calking ! I'll document some of these "unclarities" and a few other ideas as an internal wish list item, and I'll reference this post.

Edit: Here's a link to the wish list item: Improvements to Reliability Explorers - JMP User Community