Statistics for the Behavioral Sciences by Frederick J. Gravetter, Larry B. Wallnau ISBN 10: 1305504917 ISBN 13: 9781305504912

144 CHAPTER 5 | z-Scores: Location of Scores and Standardized Distributions
Note that the individual with a score of X = 3 is located exactly at the mean in the
X distribution and this individual is transformed into z = 0, exactly at the mean in
the z-distribution.
3. After the transformation, the standard deviation becomes σ = 1. For these z-scores,
∑z = 0 and
∑z 2 = (−1.50) 2 + (1.50) 2 + (1.00) 2 + (−0.50) 2 + (0) 2 + (−0.50) 2
= 2.25 + 2.25 + 1.00 + 0.25 + 0 + 0.25
= 6.00
Using the computational formula for SS, substituting z in place of X, we obtain
SS 5Sz 2 2 sSzd2
N
5 6 2 s0d2
6 5 6.00
For these z-scores, the variance is s 2 5 SS N 5 6 5 1.00 and the standard deviation is
6
s5Ï1.00 5 1.00
Note that the individual with X = 5 is located above the mean by 2 points, which is
exactly one standard deviation in the X distribution. After transformation, this individual
has a z-score that is located above the mean by 1 point, which is exactly one standard
deviation in the z-score distribution.
■
Be sure to use the μ
and s values for the
distribution to which X
belongs.
Note that Dave’s z-score
for biology is 12.0,
which means that his
test score is 2 standard
deviations above the
class mean. On the
other hand, his z-score
is 11.0 for psychology,
or 1 standard deviation
above the mean. In
terms of relative class
standing, Dave is doing
much better in the
biology class.
■ Using z-Scores for Making Comparisons
One advantage of standardizing distributions is that it makes it possible to compare
different scores or different individuals even though they come from completely different
distributions. Normally, if two scores come from different distributions, it is impossible to
make any direct comparison between them. Suppose, for example, Dave received a score
of X = 60 on a psychology exam and a score of X = 56 on a biology test. For which course
should Dave expect the better grade?
Because the scores come from two different distributions, you cannot make any direct
comparison. Without additional information, it is even impossible to determine whether
Dave is above or below the mean in either distribution. Before you can begin to make comparisons,
you must know the values for the mean and standard deviation for each distribution.
Suppose the biology scores had μ = 48 and σ = 4, and the psychology scores had
μ = 50 and σ = 10. With this new information, you could sketch the two distributions,
locate Dave’s score in each distribution, and compare the two locations.
Instead of drawing the two distributions to determine where Dave’s two scores are
located, we simply can compute the two z-scores to find the two locations. For psychology,
Dave’s z-score is
z 5 X 2m
s
For biology, Dave’s z-score is
z 5
5
56 2 48
4
60 2 50
10
5 10
10 511.0
5 8 4 512.0
Notice that we cannot compare Dave’s two exam scores (X = 60 and X = 56) because
the scores come from different distributions with different means and standard deviations.
However, we can compare the two z-scores because all distributions of z-scores have the
same mean (μ = 0) and the same standard deviation (σ = 1).