How does JMP do BoxCox transformations?

[#1]

I have recently been converting JMP formulas into Excel using the python code option. I have been unable to actually transfer any formula (MLR) involving a BoxCox transformation. It have manually done the calculations following the formula and have gone numbers in the multimillions instead of 100 thousands where it should be. If someone knows if the formula is already converted back or If I need to and what I need to do to apply the inverse BoxCox to get it back in normal terms. I don't know if I am missing something or if the Python Code through the publish formula option is missing something. I was hoping that someone could explain how JMP preforms the Box Cox transformation so that I can better understand it to potentially fix it. I tried to attach the code of the formula but it wouldn't allow me to post. Any help would be appreciated.

[#2]

@DWS  It looks ok, though I don’t know what the geometric mean is so I can’t verify it completely.

You can check it though. As I indicated a few times, from the box-cox transform analysis in JMP, save the prediction formula (red triangle at top of output). It will create a new column of predictions based on that back transformation. And the formula saved to that column should match what you have above. 

[#3]

@DWS  And FWIW, we often round lambda to the nearest 0.5. It’s easier to interpret, and you often see in the Box Cox plot that the difference is negligible. In your case, you may want to choose lambda = -1.  

[#4]

@DWS  Or in general, see pic ( and g is the geometric mean of Y).

[#5]

Hi @DWS : This may be helpful as well.

https://community.jmp.com/t5/Discussions/How-to-understand-Lambda1-and-Lambda2-for-Box-Cox-transform...