Johnson transformation: The PPU has been got even if USL not?

#1
Hi,

I got an unexpected situation today with this analysis:
a transformated distribution with the Johnson method gave me the PPU value even if the USL value cannot be obtained, due to a negative Ln.
How can Minitab calculate the PPU if no USL has been transformed?
As attachment the graphical result and the data collection (it's a circularity dimension, so geometric, with the boundary limit below zero).

30 sample values of first new ring production:

0,015 0,04 0,023 0,037 0,035 0,037 0,037 0,029 0,038 0,032 0,031 0,039 0,038 0,032 0,033 0,031 0,025 0,031 0,021 0,04 0,029 0,038 0,038 0,018 0,039 0,025 0,035 0,03 0,036 0,04
 

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bobdoering

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#2
Recalculate it with the target as 0, boundary 0, no LSL. See attached analysis. The variation is too far from 0 to be beta. The PPU is calculating the "distance from the spec", which is kind of a waste, since the point of Ppk is to see if the variation is centered within the specification, and you don't want to be centered here.
 

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Last edited:
#3
Sorry Bob, but I didn't understand what you mean.
Specially about "The variation is too far from 0 to be beta" and the distribution centering taht I could not want to be.
Could you explain it with easier concepts and with not so much implicit considerations, please?
I'm a little bit newby.......
Thank you.
 

bobdoering

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#4
This animation may help show how as the beta distribution approached the hard limit (0 in this case) it "smashes" into a skewed distribution. The further away it gets the more it acts like a normal distribution - in fact curve fitting may assign normal to it the worse it is. In this case 0 is the target or optimum value - not a central value as it is for a normally distributed variation within a bilateral tolerance. This is key to the whole point of a Ppk calculation - its is generally used to determine of the distribution is not only within specification (Pp), but also centered (Ppk).
.
 

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