Statistical Methods for Psychology

subsequent sexual abuse. Both variables have been scored as 0, 1 variables—an individual

received instruction, or she did not, and she was either abused, or she was not.

The appropriate correlation coefficient is the coefficient, which is equivalent to

Pearson’s rcalculated on these data. Again, special formulae exist for those people who

can be bothered to remember them, but they will not be considered here.

From Table 10.2 we can see that the correlation between whether a student receives in-

struction on how to avoid sexual abuse in school, and whether he or she is subsequently

abused, is 2 .1094, with a^25 .012. The correlation is in the right direction, but it does

not look terribly impressive. But that may be misleading. (I chose to use these data pre-

cisely because what looks like a very small effect from one angle, looks like a much larger

effect from another angle.) We will come back to this issue shortly.

Significance of f

Having calculated , we are likely to want to test it for statistical significance. The appro-

priate test of against : 5 0 is a chi-square test, since is distributed as on 1 df.

For our data,

which, on one df, is clearly significant. We would therefore conclude that we have convinc-

ing evidence of a relationship between sexual abuse training and subsequent abuse.

The Relationship Between fand x^2

The data that form the basis of Table 10.2 could be recast in another form, as shown in

Table 10.3. The two tables (10.2 and 10.3) contain the same information; they merely dis-

play it differently. You will immediately recognize Table 10.3 as a contingency table. From

it, you could compute a value of to test the null hypothesis that the variables are inde-

pendent. In doing so, you would obtain a of 9.79—which, on 1 df, is significant. It is

also the same value for x^2 that we computed in the previous subsection.

x^2

x^2

x^2 =Nf^2 =818( 2 .1094^2 )=9.79

f H 0 r Nf^2 x^2

f

f

f

300 Chapter 10 Alternative Correlational Techniques

Table 10.2 Calculation of ffor Gibson’s data

X: 0 5 Instruction

1 5 No Instruction

Y: 0 5 Sexual Abuse

1 5 No Sexual Abuse

Partial data:

X: 00010100010010

Y: 00101001100100

Calculations (based on full data set):

5 0.3888 sX 5 0.4878 covXY 52 0.0169

5 0.8863 sY 5 0.3176 N 5818

f^2 =.012

f=r=

covXY

sXsY

=

- 0.0169

(.4878)(.3176)

=-.1094

Y

X