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

492 CHAPTER 15 | Correlation
Substituting the totals in the formula gives
SP 5 oXY 2 oXoY
n
5 52 2 10s20d
4
= 52 – 50
= 2
Both formulas produce the same result, SP = 2.
■
The following example is an opportunity to test your understanding of the calculation
of SP (the sum of products of deviations).
EXAMPLE 15.2
Calculate the sum of products of deviations (SP) for the following set of scores. Use the
definitional formula and then the computational formula. You should obtain SP = 5 with
both formulas.
X
Y
0 1
3 3
2 3
5 2
0 1
■
■ Calculation of the Pearson Correlation
As noted earlier, the Pearson correlation consists of a ratio comparing the covariability
of X and Y (the numerator) with the variability of X and Y separately (the denominator).
In the formula for the Pearson r, we use SP to measure the covariability of X and Y. The
variability of X is measured by computing SS for the X scores and the variability of Y
is measured by SS for the Y scores. With these definitions, the formula for the Pearson
correlation becomes
r 5
SP
ÏSS X
SS Y
(15.3)
Note that you multiply SS for X by SS for Y in the denominator of the Pearson formula.
The following example demonstrates the use of this formula with a simple set of scores.
EXAMPLE 15.3
X
Y
0 2
10 6
4 2
8 4
8 6
The Pearson correlation is computed for the set of n = 5 pairs of scores shown in the
margin.
Before starting any calculations, it is useful to put the data in a scatter plot and make
a preliminary estimate of the correlation. These data have been graphed in Figure 15.4.
Looking at the scatter plot, it appears that there is a very good (but not perfect) positive
correlation. You should expect an approximate value of r = +.8 or +.9. To find the Pearson
correlation, we need SP, SS for X, and SS for Y. The calculations for each of these values,
using the definitional formulas, are presented in Table 15.1. (Note that the mean for the
X values is M X
= 6 and the mean for the Y scores is M Y
= 4.)