How do I compute z-scores in SPSS?
This is done automatically using the DESCRIPTIVES procedure wither by clicking on Analyze>Descriptive Statistics>Descriptives and clicking the box called save standardized values as variables or running syntax with the desired variable(s) replacing age iq in the below macro call. The standardised variables for age and iq are saved respectively as variables called Zage and Ziq.
DEFINE !ZSCORE ( X !CMDEND). DESCRIPTIVES VARIABLES= !X /SAVE /STATISTICS=MEAN STDDEV MIN MAX . !ENDDEFINE. !ZSCORE X=AGE IQ.
To provide a more stable measure composites can be formed with unit-weighted z scores of related measures (See Ackerman and Cianciolo (2000) p.264). This composite adjusts for differences in raw score variance related to different scales of raw scores which could lead to scores with larger variance diluting the influence of scores with smaller variance in the sum of raw scores.
A one sample t-test on z-scores versus t-tests on group means for comparing a patient group with a control
On a related note there are two ways we could compare a patient group mean to that fo a control and it is not immediately apparent how they differ or which we should use. We could obtain z-scores for each patient using control mean and standard deviation and see if this quantity differs from zero via a one-sample t-test. We could also do an unpaired t-test between the patient and control group means. In the below we show that the z-score approach, unlike the unpaired t-test, ignores variation in the control group and for this reason suggest using the unpaired t-test when we are interested in comparing overall patient and control group means.
To see the z-score approach does not use the control group variance we can express the one-sample t-test on the z-scores as the below, using SD to denote the standard deviation, SE to signify the standard error and $$\bar{x}_text{P}$$ and $$\bar{x}_text{C}$$ to represent the patient and control group means respectively.
$$\frac{\mbox{Mean z-score}}{\mbox{Standard error of mean z-score}}$$ =
$$\frac{\frac{\bar{x}_text{P} - \bar{x}_text{C}}{SD_text{c}}}{\frac{1}{SD_text{C}}SE(\bar{x}_text{P})}$$ =
$$\frac{\bar{x}_text{P}-\bar{x}_{C}}{SE(\bar{x}_{P})}$$
which does not include SD($$x_text{C}$$).
Reference
Ackerman PL & Cianciolo AT (2000) Cognitive, perceptual speed, and psychomotor determinants of individual differences during skill acquisition. Journal of Experimental Psychology: Applied 6 (4), 259-290.