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

208 CHAPTER 7 | Probability and Samples: The Distribution of Sample Means
Caution: When computing
z for a single score,
use the standard deviation,
σ. When computing
z for a sample mean, you
must use the standard
error, σ M
.
EXAMPLE 7.4
However, we are now finding a location within the distribution of sample means. Therefore,
we must use the notation and terminology appropriate for this distribution. First, we
are finding the location for a sample mean (M) rather than a score (X). Second, the standard
deviation for the distribution of sample means is the standard error, σ M
. With these changes,
the z-score formula for locating a sample mean is
z 5 M 2 m
s M
(7.3)
Just as every score (X) has a z-score that describes its position in the distribution of scores,
every sample mean (M) has a z-score that describes its position in the distribution of
sample means. Specifically, the sign of the z-scores tells whether the sample mean is
above (+) or below (–) μ and the numerical value of the z–score is the distance between
the sample mean and μ in terms of the number of standard errors. When the distribution of
sample means is normal, it is possible to use z-scores and the unit normal table to find the
probability associated with any specific sample mean (as in Example 7.3). The following
example is an opportunity for you to test your understanding of z-scores and probability
for sample means.
A sample of n = 4 scores is selected from a normal distribution with a mean of μ = 40 and
a standard deviation of σ = 16. The sample mean is M = 42. Find the z-score for this sample
mean and determine the probability of obtaining a sample mean larger than M = 42.
This is, find p(M > 42) for n = 4. You should obtain z = 0.25 and p = 0.4013. ■
The following example demonstrates that it also is possible to make quantitative predictions
about the kinds of samples that should be obtained from any population.
EXAMPLE 7.5
Once again, the distribution of SAT scores forms a normal distribution with a mean of
μ = 500 and a standard deviation of σ = 100. For this example, we are going to determine
what kind of sample mean is likely to be obtained as the average SAT score for a random
sample of n = 25 students. Specifically, we will determine the exact range of values that is
expected for the sample mean 80% of the time.
We begin with the distribution of sample means for n = 25. This distribution is normal
with an expected value of μ = 500 and, with n = 25, the standard error is
s M
5 s Ïn 5 100
Ï25 5 100
5 5 20
(Figure 7.6). Our goal is to find the range of values that make up the middle 80% of the
distribution. Because the distribution is normal we can use the unit normal table. First, the
80% in the middle is split in half, with 40% (0.4000) on each side of the mean. Looking
up 0.4000 in column D (the proportion between the mean and z), we find a corresponding
z-score of z = 1.28. Thus, the z-score boundaries for the middle 80% are z = +1.28 and
z = –1.28. By definition, a z-score of 1.28 represents a location that is 1.28 standard deviations
(or standard errors) from the mean. With a standard error of 20 points, the distance
from the mean is 1.28(20) = 25.6 points. The mean is μ = 500, so a distance of 25.6 in
both directions produces a range of values from 474.4–525.6.
Thus, 80% of all the possible sample means are contained in a range between 474.4 and
525.6. If we select a sample of n = 25 students, we can be 80% confident that the mean
SAT score for the sample will be in this range.
■
The point of Example 7.5 is that the distribution of sample means makes it possible to
predict the value that ought to be obtained for a sample mean. We know, for example, that a