topic Re: How to transform specification limits when doing a Box-Cox transform? in Discussions

How to transform specification limits when doing a Box-Cox transform?

moL — Thu, 08 Jun 2023 16:31:52 GMT

Hello all,

In this toy-example i want to do capability analysis on my data. I have a dataset which is a lognormal distribution. I can do a Box-Cox transformation of the data and it becomes a normal distribution. This is all well and good but when i try to transform the specification limits with the same transform used for the data, i do not get the same amount of datapoints within and outside the specification limits.

My question is: how do I correctly transform my specification limits using a Box-Cox transform?

Original data: (lognormal)

The Box-Cox transform (lambda almost 0 since the original data was a lognormal dist)

The transformed data with transformed (wrong) spec limits: (notice the different ratio of data inside and outside of the spec limits)

Best regards,

Mathias

Re: How to transform specification limits when doing a Box-Cox transform?

Mark_Bailey — Fri, 03 Mar 2023 16:42:37 GMT

How did you transform the specifications? Did you use the same formula as the Box-Cox transformation, or just take the logarithm?

Here is an example using Big Class. You can see the original weight, Log base 10 of weight, natural log of weight, and the Box-Cox transformed weight using lambda = 0.

Here is the formula saved for the Box-Cox transform of weight from Fit Least Squares:

Re: How to transform specification limits when doing a Box-Cox transform?

moL — Mon, 06 Mar 2023 07:41:12 GMT

Hi Mark,

Thank you for the answer, and sorry i could not get back to you before.

I wrongly believed that JMP simply uses the transform: y = (y^lambda-1)/(lambda) [1] for the data, and thus i used this for the spec limits as well. Using your tip i looked at the formula for the transformed column, and using this I get the correct transformed spec limits (with the same ratio of data inside and outside of the limits for the transformed and non-transformed data).

Just out of curiosity: why dosen't JMP use the simple transform [1] mentioned above? I seems like there is an additional factor which is multiplied.

Br Mathias

Re: How to transform specification limits when doing a Box-Cox transform?

Mark_Bailey — Mon, 06 Mar 2023 18:08:20 GMT

Computations set by JMP are simplified when possible for speed. The special case is faster than the general solution.

Also, function [1] above is only valid when lambda is not zero. Here is the transform function from the JMP Documentation for this transform:

Notice that the transformed response is the geometric mean of the response times the natural logarithm of the response when lambda is 0.