Z-scores, standardized tests and population means I have been using z-scores by subtracting the mean (and then dividing by the SD) of the sample, whereas I recently read they actually need to be the population mean and SD. Assuming you are transforming raw scores to z-scores in order to compare within-sample variables (various test scores of the same sample of subjects), does it really invalidate the comparison if you have been using sample rather than population statistics?
Also, is it ever realistic that one would know the population statistics? I always assumed this is more of a theoretical concept than an actual statistic. For instance, in order for the population mean of the IQ test to be set to 100, surely this didn't imply testing every single human on the planet and adjusting the raw-to-standardized tables to reflect the population average!
Many thanks.
 A: If you are using the sample mean and standard deviation to standardise, then strictly speaking you have a $t$-statistic.
The difference between this and a $z$-statistic using the population mean and standard deviation matters more when you have a smaller sample.
A: If your purpose really is as simple as making comparisons within your dataset, then your dataset is your population of interest, not merely a sample from it. This also answers your second question.
If you're interested in generalizing the within-subjects differences across various tests that you find in your dataset to the larger population of people just like your subjects (the population from which you sampled your subjects), then you need to work with sample statistics. Technically, standardizing scores with sample means and $SD$s produces $t$-statistics, not $z$-scores, as I've mentioned in my recent response to Triangular distribution, and as @Henry said just now :)
I wouldn't say the difference invalidates your comparisons altogether, but since $t$-statistics have different distributions depending on your degrees of freedom ($\nu$), inferences you base on the probability of differences among $t$-stats would also depend on $\nu$, whereas this isn't true of $z$-scores. Here's Wikipedia's pdf of Student's $t$ (only the PDF for $\nu=\infty$ is the same as that of $z$):
 by Skbkekas.
