Computing Z-scores using SPSS
Analyze  Descriptives  Descriptives
Move ‘age’ into the variable(s) box and select the “Save standardized values as
variables” option, shown above.
This will give us descriptive statistics on the age variable, while at the same time saving
the standardized value of the variable ‘age.’
QUERY: What does this procedure involve and why would we want to do this?
To get the descriptives on both age and the zscore of age, go back to Analyze 
Descriptives  Descriptives and put age and zage in the variable(s) box. Unclick “Save
standardized values as variables” otherwise it will do the same thing all over again.
Descriptive Statistics
AGE OF RESPONDENT
Zs core: AGE OF
RESPONDENT
Valid N (lis twise)
N
1385
Minimum
18
Maximum
89
Mean
44.94
Std. Deviation
17.080
1385
-1.57725
2.57978
.0000000
1.00000000
1385
QUERY: Why is the mean equal to zero and the standard deviation equal to 1?
To inspect the scores of zage, use the Case Summaries procedure. Analyze  Reports 
Case Summaries
Move age and zage into the variable(s) window
Let’s look at only 10 cases, so make sure the box to the left of the option is checked and
type 10 in the box to the right. Click ‘OK’
Case Summariesa
1
2
3
4
5
6
7
8
9
10
Total
N
AGE OF
RESPO
NDENT
60
27
21
35
43
29
39
45
29
41
10
Zs core:
AGE OF
RESPO
NDENT
.88184
-1.05030
-1.40160
-.58190
-.11351
-.93320
-.34770
.00359
-.93320
-.23061
10
a. Limited to first 10 cas es.
QUERY:
 Can you tell me the mean age using this chart?1
 What do the positive and negative signs on the zscores indicate? 2
 How many standard deviations below the mean is age = 27? How can we verify
this? 3
 Can you calculate the standard deviation of age from this chart?4
1
Yes, the mean has been recalculated to equal 0, so by finding 0 on the chart, we note that the mean age is
approximately 45 years.
2
Ans. The positive signs indicate that the observed value falls above the mean. The negative sign means
that it falls higher than the mean. If we look at the value corresponding to -.23061, it is age 41, which
indeed falls below the mean. Verify that a zscore = .88184 falls above the mean.
3
We see that age = 27 falls about in standard deviation below the mean. We can verify this if we knew the
standard deviation. From the output, we know that the standard deviation is 17.080. Therefore 1 standard
deviation below the mean is ~ 45 – (1)17.080 ~ 27.
4
Yes, 1 standard deviation from the mean is about 45 – 27 ~ 18.