Comparing z-scores from different data sets 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!
 A: $z$-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:


*

*Is normality testing 'essentially useless'?  

*How to choose between t-test or non-parametric test e.g. Wilcoxon in small samples  

*T-test for non normal when N>50?
A: 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:


*

*If mean is close to zero, CV will approach to infinity and so it is
sensitive to small changes in the mean.

*Unlike the standard deviation, it cannot be used directly to
construct confidence intervals for the mean.

