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During my calculations I need to use square roots but z-scores can be negative.

Is there a trick to transform them into positive value without missing the usefulness of z-scores?

What if I have not a single distribution but more than one to compare each other?

Thanks

Edit for details

I have two ranks calculated using z-scores (they rank student performance using different metrics) and I want to make a geometric average between the two rank, but as z-scores has negative values I cannot perform square root on negative results.

Eg: if student Foo has a score of 2 on the first rank and a score of -2 on the other i cannot make the square root of -4.

If there is a method to convert the scores into positive values I can perform geometric average.

For example when I use percentage I add one to the value divided by 100 (eg -80% and +50% become 0,8 and 1,5). Is there a trick like this that works on z-scores?

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    $\begingroup$ Welcome to our site. You need to explain how you are using these Z scores and what the calculations are intended to do, for otherwise there is no way to provide an objective answer. $\endgroup$ – whuber Oct 3 '18 at 18:30
  • $\begingroup$ I edited the question with more detsils $\endgroup$ – TheItalianJobless Oct 4 '18 at 5:01
  • $\begingroup$ It might help to explain why you want the geometric average $\endgroup$ – Juho Kokkala Oct 4 '18 at 6:08
  • $\begingroup$ The first reason is that it values higher students that are equally good on both metrics and not students with extremes scores. The second one is that I need to compute historical analysis also and it allows to rank higher students with regular good performance $\endgroup$ – TheItalianJobless Oct 4 '18 at 7:02
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Welcome to the site. Of course, there are lots of ways to transform z scores into positive values: You could take absolute value, you could square them, you could add 20 to all values, you could rank them and probably some other methods. Which method should you use?

In one comment (which should be added to your question), you say:

The first reason is that it values higher students that are equally good on both metrics and not students with extremes scores. The second one is that I need to compute historical analysis also and it allows to rank higher students with regular good performance

You could certainly rank them. You would then no longer need the geometric average. But, since ranks are ordinal, you might want the median rather than the mean. That is, by ranking, you have lost some information.

For the first problem, adding a large constant could work, but I'm not at all sure geometric mean is what you want. The geometric mean is most useful when different items have different ranges - for instance, if you were trying to take the mean of GPA (range 0 to 4) and SAT scores (range 0 to 1600). z scores already do that.

If you want to value "regular good performance" then you probably need at least two numbers.

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  • $\begingroup$ I'm fairly confident I need geometric mean for both computation (more for the second purspose) and yes I have more than two numbers to compare. There are disadvantages on adding a constant to z-scores to make them positive? In this case I have to add the absolute value of the smaller z-score between all the ranges (eg GPA and SAT in your example) or there are some best practices? $\endgroup$ – TheItalianJobless Oct 4 '18 at 12:08
  • $\begingroup$ PS: just for clarification, if I have to compare, on the line of your example, GPA, SAT and another test 'X', and I want to compute the average performance for a student on the 3 tests, using a geometric mean on raw scores has the same result as using arithmetic mean on z-scores of the 3 tests? I don't mean "absolute result" but same result on the final rank and relation and distance on the points of the rank. $\endgroup$ – TheItalianJobless Oct 4 '18 at 12:12
  • $\begingroup$ I'm not at all sure you do need geometric mean. You haven't given reasons for using it that really apply. The bad point of adding a constant is that it is arbitrary. $\endgroup$ – Peter Flom Oct 4 '18 at 12:27
  • $\begingroup$ @PeterFlom My understanding is that TheItalianJobless wants to use geometric mean because when the sum of a series is fixed, the geometric mean of the series takes its maximum value when the underlying elements are all equal; so a student who scored 1 in both tests will be favoured over a student who scored a 0.5 in one test and a 1.5 in the other, despite the fact that the arithmetic mean of the scores from the two students are the same. $\endgroup$ – Xiubo Zhang Oct 4 '18 at 12:46
  • $\begingroup$ @XiuboZhang you get the point. The decision to use geometric mean between the different ranks is not set in stone and I can reconsider it. However I think I must use it for the historical average. $\endgroup$ – TheItalianJobless Oct 4 '18 at 14:41

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