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Is it possible to use z-score as a number to compare two different data sets (none of them with normal distribution) and identify which one has an average z-score higher than another? Thanks!

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    $\begingroup$ Z-score is (x-mean)/sd, so every variable will have average z-score = 0... $\endgroup$ – Tim Jan 5 '16 at 14:15
  • $\begingroup$ That's right, every variable will have average = 0. I think I failed to express myself, I'll reformulate:The average z-score that I mentioned is related to the mean of all z-scores of a data series. For example, the z-scores to my three observations are: 2, 2, 7, so that series' average z-score is 3,66. Is it ok to compare thar number to the average z-scores from another series: 5, 3, 9 (avg z-score: 5,66)? $\endgroup$ – mrncst Jan 5 '16 at 16:57
  • $\begingroup$ It is unclear what you mean. Average z-score is zero, if you have other average z-score than you are doing something wrong. $\endgroup$ – Tim Jan 5 '16 at 17:09
  • $\begingroup$ Got it. I'm trying to find an unique value that allows me to compare two different data sets regarding dispersion. What if I evaluate (the maxium z-score of a data series) - (the minimum z-score from the same data series). Is it possible to use this number for comparision? $\endgroup$ – mrncst Jan 7 '16 at 16:16
  • $\begingroup$ But you want to compare what exactly? Comparing z-scores of two points from two different variables can tell you only that one of them is more standard deviations away from mean of its sample comparing to the other. $\endgroup$ – Tim Jan 8 '16 at 21:32
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$z$-scores are used for standardization of variables, so that they have mean equal to zero and standard deviation equal to one, which is achieved by subtracting mean and dividing by standard deviation. Since after converting variables to $z$-scores they have the same mean and standard deviation, there is no point in comparing the means. However, your intuition that converting to $z$-score leads both variables to be similarly scaled so we can somehow compare them is partially correct since a similar kind of manipulation is used in Student's $t$-test

$$ t = \frac{ \bar X - \bar Y }{ \sqrt{s_X^2 + s_Y^2} \times \sqrt{\frac{1}{n}} } $$

where $\bar X, \bar Y$ are empirical means, $s_X^2, s_Y^2$ are empirical (squared) standard deviations and $n$ is sample size (for equal sized samples). As for normality of the samples, check the following threads:

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    $\begingroup$ Helpful, but it is worth emphasising that the term normalization, which is unfortunately overloaded throughout several areas of mathematics, has absolutely nothing to do with the normal distribution. For that reason alone, standardization is arguably a better term, although it too needs to be kept close to the equation that explains what it is precisely. $\endgroup$ – Nick Cox Jan 5 '16 at 14:35
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Your questions and comments seems unclear. But I'll try to answer, what I could understand from "I'm trying to find an unique value that allows me to compare two different data sets regarding dispersion".

To compare the dispersion of differently scaled data sets (many be different units like age in years and weight in pounds), one should use relative measures of dispersion.

One such unit-less (dimension-less) measure is coefficient of variation (CV), which is given by SD/Mean.

It is also known as relative standard deviation. It has some limitations too, like:

  1. If mean is close to zero, CV will approach to infinity and so it is sensitive to small changes in the mean.
  2. Unlike the standard deviation, it cannot be used directly to construct confidence intervals for the mean.
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  • $\begingroup$ It helped me a lot! Thank, this is the conclusion I was trying to find. $\endgroup$ – mrncst Jan 13 '16 at 13:11
  • $\begingroup$ That's good. Try to ask straight forward query in simple words. Your question should be rephrased as "how to compare variation of two different data sets?" $\endgroup$ – Dr Nisha Arora Jan 14 '16 at 11:15

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