Combining z-scores in clinical practice

Question

I am a clinical neuropsychologist working with children with neurodevelopmental disorders. I have a quite naive question but we are unable to solve it with my colleagues (we are not very good in statistics).

Here is our problem: In our structure, all children complete several tests assessing the same latent variable. For example, children systematically perform 4 different tests assessing inhibition. Sometimes, we would like to group results of these tests in one overall interpretable variable. Although each test gives an z score, we understand that we cannot take the average of the 4 z-scores obtained by each child and interpret it as an average z score.

First, I have found several papers providing solutions but they systematically require unknown values (for example, Evans 1996 requires to know the correlations between the tests and, unfortunately, we don't have access to precise correlations between the tests at each children's age group).

Second, I have read on this forum that we could obtain an overall z score even if the correlations between the tests are unknown : Combining Z Scores by Weighted Average. Sanity Check Please? In essence, this second solution proposes to divide the averaged z score by its standard deviation. To compute the standard deviation, this solution requires to know only the weights attributed to each test.

I would like to know whether this second solution could be applied to our problem. For example, if a child obtains 4 z-scores (respectively, -1, -1.4, -0.8, -1.9; for which we attribute the same weight), based upon rpatel's formula, is it accurate to say that the overall z score (Zw) of this child is: Zw = (mean (-1, -1.4, -0.8, -1.9))/(SQRT((0.25)²+(0.25)²+(0.25)²+(0.25)²))=-2.55 (i.e., the child performs worse than 99% of his peers?)

I am not able to understand the maths behind the two approaches, so I prefer asking your confirmation rather than making a mistake.

Thank you very much for your help! BF

Answer 1

First, it is unlikely that several different tests of inhibition would be totally uncorrelated across the population of your subjects. Because we're dealing with Z-scores, I suppose that means they're not completely independent tests. So if Evans 1996 says you'd need to know correlations to get a meaningful composite Z-score, that is correct.

Second, as far as I can see, the link is assuming that the four z-scores are completely independent. Suppose we use that independence assumption to get a combined score that weights each of the tests equally. Then we have four independent random variables $Z_1, Z_2, Z_3, Z_4$ each with $E(Z_i) = 0,$ and $Var(Z_i) = SD(Z_i) = 1.$ Let $A = \frac 14\sum_i Z_i.$

Then $$E(A) = E\left(\frac 14 \sum Z_i\right) = \frac 14 E\left(\sum Z_i\right)\\ = \frac 14 \sum E(Z_i) = \frac 14(0+0+0+0) = 0.$$ And $$V(A) = V\left(\frac 14 \sum Z_i\right) = \frac{1}{16} V\left(\sum Z_i\right)\\ = \frac{1}{16}\sum V(Z_i) = \frac{1}{16}(1+1+1+1) = \frac{4}{16} = \frac 14.$$

Addendum: Here are data simulated in R for four positively correlated tests administered to 50 subjects.

set.seed(2019); n = 50
v1 = rnorm(n,50,3); v2 = rnorm(n,60,4)
v3 = rnorm(n,40,2); v4 = rnorm(n,50,2)
w = rnorm(n,0,3)
x1 = v1+w; x2 = v2+w; x3 = v3+w; x4 = v4+w
MAT = cbind(x1,x2,x3,x4)
cor(MAT)

          x1        x2        x3        x4
x1 1.0000000 0.5622012 0.6479422 0.6513025
x2 0.5622012 1.0000000 0.5262790 0.6410636
x3 0.6479422 0.5262790 1.0000000 0.6738916
x4 0.6513025 0.6410636 0.6738916 1.0000000

Means and standard deviations of the scores for the 50 subjects are found below, and from them, the z-scores for each subject relative to the rest of the group of 50.

a = rowMeans(MAT); s = apply(MAT,1,sd)
z = (min(a)-mean(a))/sd(a)

Subject #27 had the lowest such z-score (-2.06), which (not surprisingly) puts that subject at about the 2nd percentile. Also, #27's scores on the four tests are shown below, followed by the corresponding individual z-scores relative to the group of 50, and percentages in a normal population below these z-scores.

z.27 = (min(a)-mean(a))/sd(a);  z.27;  pnorm(z.27)
[1] -2.06086
[1] 0.01965821

MAT[27,]
      x1       x2       x3       x4 
39.49311 51.58849 33.49084 44.05014 
(MAT[27,]-mean(a))/sd(a)
            x1         x2         x3         x4 
    -2.8288067  0.6598278 -4.5600233 -1.5144369 
round(pnorm((MAT[27,]-mean(a))/sd(a)),4)
    x1     x2     x3     x4 
0.0023 0.7453 0.0000 0.0650 

Thus, in effect, one way to derive the z-score $-2.06$ as a 'combination' of z-scores $-2.83, 0.65 -4.56,$ and $-1.51$ is to use this subject's individual exam scores in the context of the other 49 subjects.