Z-Scores vs just dividing by SD I am an MSc student conducting a study on men’s pupillary responses to various sexual stimuli. Because pupils vary in size and degree of dilation, I’ve needed to standardize dilation measures within participants.
My approach is as follows:
(x - Baseline)/SD(x)
I did this individually for each participant’s data.
However, I realized that this is more of a pseudo Z-score, as the baseline mean is from a different distribution.
Would it make more sense to do:
If x - baseline = y,
(y - mean(y))/SD(y)
Because of the equipment available to me, it would be quite onerous to go back and reprocess my dataset.
In case it is worth knowing, I am only planning on running simple repeated measures analyses of variance.
Any advice would be appreciated. Thanks in advance.
 A: Since you transform each participant's measurements separately, the mathematical relationship between the two proposed standardizations is simple and easy to derive by rewriting the second formula:
$$
\begin{aligned}
\frac{y-\operatorname{E}(y)}{\operatorname{SD}(y)} 
= \frac{(x-x_0) - \operatorname{E}(x-x_0)}{\operatorname{SD}(x-x_0)}
= \frac{x - x_0 - \operatorname{E}(x) + x_0}{\operatorname{SD}(x)}
= \frac{x - \operatorname{E}(x)}{\operatorname{SD}(x)}
\end{aligned}
$$
Notice that the baseline cancels out in the numerator and adding a constant doesn't change the standard deviation. So the baseline is "ignored" in the second standardization.
The difference between the two formulas is equal to a shift of $\left(x_0 - \operatorname{E}(x)\right) / \operatorname{SD}(x)$.

Here is the R code to reproduce the figure:
set.seed(1234)

x <- rnorm(100)

baseline <- 0.111

y <- x - baseline

par(pty = "s")
plot(
  (x - baseline) / sd(x),
  (y - mean(y)) / sd(y)
)
abline(a = 0, b = 1, col = "red")

