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May 27, 2015 10:21 AM
(1991 views)

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May 28, 2015 2:17 PM
(3518 views)

Solution

I believe this is because your first response will be undefined in the logit transform, log(y/[1-y]). You will end up with a denominator = 0.

Did you see my note about adjusting for this? (Change the result to something close to 1; e.g. 0.995; something supported by your sense of no practical difference and measurement precision.)

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May 28, 2015 6:40 AM
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May 28, 2015 6:41 AM
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May 28, 2015 8:15 AM
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Michael,

Could you send the sample file back with the corrected transformation? I'm having trouble with it.

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May 28, 2015 8:49 AM
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Here you go.

Notice that Y=0 and Y=100 will be undefined (log(0) and denominator=0). I adjust these by ±0.01. You could use a smaller or larger adjustment, possibly relating it to the sensitivity and variability of your measurements. The smaller the adjustment, the more extreme the transformed value.

Also, as was pointed out in another reply, ln (log base e) is log in JMP.

Michael

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May 28, 2015 9:14 AM
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I deleted that post about ln versus log because, although JMP does not list ln in the list of functions (lnz though is listed), it does recognize ln if you type it into the formula box manually.

An alternate but similar approach is to generate a new column called, say, p, defined by formula y/100. Then the least squares model dialog lets you choose p for the Y box and if you select p in the Y box you can choose the logit transformation. This has the advantage that when you fit the model you don't have to manually unwind the logit to get to p . For example you can directly save the formula for p, and the profiler shows directly how p changes when the mixture changes. You still have to figure out what to do for p=0 or p=1. If you select Automatic Recalc in the model output you can change the value selected to replace 0 or 100 and watch the model change as a result.

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May 28, 2015 12:58 PM
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May 28, 2015 2:17 PM
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I believe this is because your first response will be undefined in the logit transform, log(y/[1-y]). You will end up with a denominator = 0.

Did you see my note about adjusting for this? (Change the result to something close to 1; e.g. 0.995; something supported by your sense of no practical difference and measurement precision.)

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May 29, 2015 6:29 AM
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Yes. It all makes sense to me now. Thank you Michael. I now have a more practical analysis.

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May 28, 2015 9:14 AM
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Also, to convert the predicted transformed responses (Y’) back to native units (Y):

Y = 100*Exp(Y’) / (1 + Exp(Y’))

I notice when I ran your model, the two largest responses (runs 1 & 12) had very large residuals. This is likely due to the logarithmic nature of the transformation. You can reduce this somewhat with the adjustment to the 100% response. If your reporting as integers is indicative of precision, increase the adjustment to 0.5%, suggesting the precision makes the difference between 99.5 and 99.99 of no practical significance.