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Mixture Design with Response in %

bwanders

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Sep 20, 2013

I have attached an example mixture design with response data in percentages.  When you use standard least squares, it is possible to have predictions outside of 0 and 100%.  Isn't there way to constrain responses to 0 and 1 or 0 and 100?.  The generalized linear model can't take mixture effects. 

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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.)

9 REPLIES
michael_bresnic

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Jun 30, 2014

The logit transformation of the response may be suitable: y' =  ln ( [y - lower limit]  / [upper limit - y] ); for results between 0 and 1 y' - ln ( y / [1-y] ).  This will not accommodate responses = 0 or 1, but does ensure y' stays within the boundaries.

michael_bresnic

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Jun 30, 2014

Oops, for results between 0 and 1 y' = ln ( y / [1-y] ). 

bwanders

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Sep 20, 2013

Michael,

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

michael_bresnic

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Jun 30, 2014

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

mpb

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Jun 23, 2011

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.

bwanders

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Sep 20, 2013

Originally, I started with a probability column (y/100).  Look at the JMP alert I get when I try to run the model after I have chosen the logit transformation.

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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.)

bwanders

Community Trekker

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Sep 20, 2013

Yes.  It all makes sense to me now.  Thank you Michael.  I now have a more practical analysis.

michael_bresnic

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Jun 30, 2014

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.