Can Spearman's correlation be run on z scores? I understand the logic of standardising raw data that is based on different scales into z scores so that they can compared, for example comparing a score of 75 out of 100 in one test versus 65 out of 120 in another test.


*

*Can I run the Spearman's correlation test on the z scores then?  (i.e. forget about the raw data). 

*How do I then relate the result back to the raw data?

 A: Spearman's correlation is just Pearson's correlation using ranks (see the Wikipedia page), so any transformation of the data that preserves their ordering (and so gives the same ranks) will give precisely the same value for Spearman's correlation.
Pearson's correlation doen't have that property, but it is the case that it is unaffected by any linear transformation of either variable (or both), say $x^*=ax+b$ and $y^*=cy+d$.  This includes the z-score transformation (which I call standardizing).
A: Spearman's correlation on z-scores is the same as it is on raw scores. Here's a little R code to demonstrate the idea:
> # Create two correlated random variables with means and standard deviations
> # that are clearly not z-scores (i.e., not mean = 0, sd = 1):
> set.seed(4444)
> x <- rnorm(100, mean = 100, sd =3)
> y <- x + rnorm(100, mean =50, sd = 2)
> 
> # Create z-score versions of the variables:
> zx <- scale(x)
> zy <- scale(y)
> 
> # Calculate Spearman's correlation on both raw and z-score versions of the
> # variables: 
> # Note that they are the same value.
> cor(x, y, method="spearman")
[1] 0.7756736
> cor(zx, zy, method="spearman")
          [,1]
[1,] 0.7756736
> 
> 
> # Note that this also holds for Pearson's correlation:
> cor(x, y, method="pearson")
[1] 0.8393452
> cor(zx, zy, method="pearson")
          [,1]
[1,] 0.8393452

... 
> # Another way of thinking about it is that Pearson's correlation is 
> # equivalent to the standardised beta in a linear regression 
> # involving one variable predicting the other (i.e., a regression
> # coefficient as if the two predictors were z-scores):
> coef(lm(zy~zx))[2]
       zx 
0.8393452 
> coef(lm(zx~zy))[2]
       zy 
0.8393452 
> 
> # In the context of Spearman's correlation, you can think of the 
> # correlation as the 
> # standardised regression coefficient for the variables after converting  
> # each variable to ranks:
> rzx <- rank(zx)
> rzy <- rank(zy)
> 
> coef(lm(rank(rzy)~rzx))[2]
      rzx 
0.7756736 
> coef(lm(rzx~rzy))[2]
      rzy 
0.7756736 
> 

