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

SECTION 5.6 | Computing z-Scores for Samples 149
s, in place of the population parameters μ and s. The following example demonstrates the
transformation of Xs and z-scores for a sample.
EXAMPLE 5.10
In a sample with a mean of M = 40 and a standard deviation of s = 10, what is the z-score
corresponding to X = 35 and what is the X value corresponding to z = +2.00?
The score, X = 35, is located below the mean by 5 points, which is exactly half of
the standard deviation. Therefore, the corresponding z-score is z = −0.50. The z-score,
z = +2.00, corresponds to a location above the mean by 2 standard deviations. With a
standard deviation of s = 10, this is a distance of 20 points. The score that is located
20 points above the mean is X = 60. Note that it is possible to find these answers using
either the z-score definition or one of the equations (5.3 or 5.4).
■
■ Standardizing a Sample Distribution
If all the scores in a sample are transformed into z-scores, the result is a sample of z-scores.
The transformed distribution of z-scores will have the same properties that exist when a
population of X values is transformed into z-scores. Specifically,
1. The sample of z-scores will have the same shape as the original sample of scores.
2. The sample of z-scores will have a mean of M z
= 0.
3. The sample of z-scores will have a standard deviation of s z
= 1.
Note that the set of z-scores is still considered to be a sample (just like the set of
X values) and the sample formulas must be used to compute variance and standard deviation.
The following example demonstrates the process of transforming the scores from a
sample into z-scores.
EXAMPLE 5.11
We begin with a sample of n = 5 scores: 0, 2, 4, 4, 5. With a few simple calculations, you
should be able to verify that the sample mean is M = 3, the sample variance is s 2 = 4,
and the sample standard deviation is s = 2. Using the sample mean and sample standard
deviation, we can convert each X value into a z-score. For example, X = 5 is located
above the mean by 2 points. Thus, X = 5 is above the mean by exactly 1 standard deviation
and has a z-score of z = +1.00. The z-scores for the entire sample are shown in the
following table.
X
z
0 −1.50
2 −0.50
4 +0.50
4 +0.50
5 +1.00
Again, a few simple calculations demonstrate that the sum of the z-score values is
∑z = 0, so the mean is M z
= 0.
Because the mean is zero, each z-score value is its own deviation from the mean. Therefore,
the sum of the squared deviations is simply the sum of the squared z-scores. For this
sample of z-scores,
SS = ∑z 2 = (−1.50) 2 + (−0.50) 2 + (+0.50) 2 + (0.50) 2 + (+1.00) 2
= 2.25 + 0.25 + 0.25 + 0.25 + 1.00
= 4.00