Can we compare statistically z-scores? Isn't their mean is always zero? Let's say I have two variables of any distribution for which the population average and variance is assumed/known/fixed for some topic (common in psychology). Using this I can calculate z-scores for both variables and easily compare them.
But is it possible to perform any inference on such transformed data? Their population mean is assumed to be zero, even if sample one may be non-zero (collected all big values or all small ones).
I noticed some people tested it with t-test, while others told them it's pointless, because even if they find some statistically significant difference, it's a priori known to have both zero mean and unit variance. We could only compare the shapes of their empirical cumulative distribution functions, maybe quantiles, but not means, is this correct?

Answer from comments:
If you have two groups, A and B, and you pool the values, and apply a z-transformation to the pooled values, then it makes potential sense to compare the resulting z-scores for the two groups, with e.g. a ttest.
A z-transformation is a linear transformation, so the distribution of the pooled z scores will be similar to the distribution of the pooled original values.
On the other hand, applying a z transformation to A and B separately will result in two groups of mean 0 and standard deviation 1, so it probably won't make sense to compare these.
I think it's helpful to play with these scenarios with toy data.
