Assume that the correlation between quantitative and verbal SAT
scores in a given population is 0.60. In other words,
r=0.60
r
0.60
. If 12
students were sampled randomly, the sample correlation,
rr, would
not be exactly equal to 0.60. Naturally different samples of 12
students would yield different values of
rr. The distribution of
values of rr after repeated samples
of 12 students is the sampling distribution of
rr.
The shape of the sampling distribution of
rr for the above example is shown
in Figure 1. You can see that the
sampling distribution is not symmetric: It is
negatively skewed. The reason for the skew
is that rr cannot take on values
greater than 1.0 and therefore the distribution cannot extend as
far in the positive direction as it can in the negative
direction. The greater the value of
rr, the more pronounced the skew.
Figure 2 shows the sampling
distribution for
r=0.90
r
0.90
. This distribution has a very short positive tail and
a long negative tail.
Referring back to the SAT example, suppose you wanted to know
the probability that in a sample of 19 students, the sample
value of rr would be 0.75 or
higher. You might think that all you would
need to know to compute this probability is the mean and
standard error of the sampling distribution of
rr. However, since
the sampling distribution is not normal, you would still not be
able to solve the problem. Fortunately, the statistician Fisher
developed a way to transform rr to
a variable that is normally distributed with a known standard
error. The variable is called
z′
z′
and the formula for the transformation is given below.
z′=0.5ln1+r1-r
z′
0.5
1
r
1
r
The details of the formula are not important here since normally
you will use either a table or a
computer program to do the transformation. What is important is
that
z′
z′
is normally distributed and has a standard error of
1N-3
1
N
3
where NN is the number of pairs of scores.
Let's return to the question of determining the probability of
getting a sample correlation of 0.75 or above in a sample of 12
from a population with a correlation of 0.60. The first step is
to convert both 0.60 and 0.75 to
z′
z′
s. From a table, the values
are 0.6931 and 0.9730 respectively. The standard error of
z′
z′
for
N=12
N
12
is 0.333. Therefore the question is reduced to the
following: given a normal distributing with a mean of 0.6931 and
a standard deviation of 0.333, what is the probability of
obtaining a value of 0.9730 or higher? The answer can be found
directly from the applet Calculate Area
for a given X to be 0.20. Alternatively, you could use
the formula:
Z=X-ms=0.9730-0.69310.333=0.8405
Z
X
m
s
0.9730
0.6931
0.333
0.8405
and use a table to find that the area above 0.8405 is 0.20.
Introduction to Sampling Distributions
