# Transformations of Negatives

Six Sigma – iSixSigma Forums Old Forums General Transformations of Negatives

Viewing 14 posts - 1 through 14 (of 14 total)
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• #35178

Statman
Member

I have a column of continuous data that includes both positive and negative values and is not normal.  Is there a transformation for this distribution?  Can I simply square the data points?

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#98223

Mikel
Member

Is this a test or something you need to know?

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#98224

Statman
Member

It is a real problem, with real data

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#98225

Ken Feldman
Participant

What do you want to do with the data?  If you are just adding them up why transform?  What is the significance of non normality?  Are you trying to just do some descriptive statistics?  BTW, thanks for some of the kind comments in your previous posts.

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#98227

Mikel
Member

Since most transformations are some form of taking the existing data to some power (can be 1), negatives don’t work real well unless you like those imaginary numbers.
I’ve always taught this and it takes a few hours, but the Cliff Note version –
1) Look at the data in a histogram and see that it is a smooth continuous distribution. If it is not, it will not transform no matter how many tricks you think of.
2) If you have negative numbers only, multiply all numbers by -1.
3) If you have negatve and positive numbers, add enough to every value to make every number > 0.
4) Use a transform program like the Box Cox in Minitab.
5) If the transformed data does not look normal, look again at the histogram. You may need to adjust how much you added.
6) The same thing can be true with a small distribution of very large numbers – you may need to subtract before transforming.
Most people are taught these days if not normal, throw into Box Cox. That is terrible advice. There is tremendous information contained in “non normal” data sets. Most progress is in understanding why it is not normal. The least progress is made in trying to force a transformation.

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#98274

Gabriel
Participant

I don’t think that squaring the data is a good idea. A -2 and a 2 will give you the same transformed output, so you are loosing info.
I donhave much experience on this, but I think a thing that can work is to use any transformation with an offset, as Darth said. If the trnsformation function is Xtransformed=f(Xoriginal), then use Xtransformed=f(Xoriginal-Xo). You can try several values of Xo and find the best one (for example, the one that gives you the highest p value in a goodnes-of-fit (normality) test or the one that maximizes the f(L(Xo)) function (MLE method).
I’ve seen this method used (with the MLE method) for cases where zero was the physical limit (such as ovality), the lognormal transformation was being used, and some individuals had the value zero (where the log is not defined). In fact, even in cases where no individual was zero, pretty better fit was obteined using an MLE offset.

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#98276

Ken Feldman
Participant

Old Darth just wanted to know what you planned to do with the data before suggesting how best to accomplish it.  Gabriel and Stan provided good options but without the intent it is just an exercise.

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#98277

Statman
Member

Thanks Gabriel, this is a big help

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#98279

Gabriel
Participant

Tell me what you get.

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#98282

Anonymous
Guest

http://www.stat.umn.edu/arc/yjpower.pdf

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#98284

Statman
Member

You ask a good question however, and I think that Stan has given you very sound advice.  Adding a constant to make all values positive before you transform is usually the safest method.
Cheers,
Statman (the original)

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#98309

1918wtny
Participant

Thanks to all, I have changed my screen name from statman since it was already taken.

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#99098

Graeme
Participant

I realise that this has question has been closed out… but what I’d like to ask is

what kind of data have you measured? i.e. Temprature, Cycle Time (naturally it’s not this), etc
how many data points?

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#99112

HSigma
Participant

Granted, I have not read all the posted threads, but if you have negative numbers, so what?  If you can get negatives, then negatives are a viable data form…transforming to get away from having negatives is meaningless.  And, as at least one other observed, you lose information in doing ANYTHING to make the numbers all positive.
Remember, positie simply refers to the right of zero; negative is to the left of zero.  Kind of like the Z and t distributions; these are centered on zero.  Ergo, they possess positive AND negative values…and they serve a great purpose!
Personally, I feel that all Likert Scales should have the neutral response be zero, with disagrees being negatives and agrees being positive.  Thus, get an average respondant score of, say, -2.8…Pretty darned negative, huh?  1.1, positive but you have room for improvement; 0.2, practically neutral on the average.
Which segues us to your normality question…of course, not knowing what you are trying to do, assuming that you NEED normal data, take samples and then find averages, and use the averages as your decision variables; Control Charts 101 (a la Central Limit Theorem).

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