Statistical Methods for Psychology

two measures is .215, which has an associated probability under the null of .016. This

correlation is significant, and we can reject the null hypothesis of independence.

Some people may be concerned about the use of Pearson’s rin this situation because

“number of traumatic events” is such a discrete variable. In fact that is not a problem for

Pearson’s rand no less an authority than Agresti (2002) recommends that approach. Perhaps

you are unhappy with the idea of specifying a particular metric for Trauma, although you

do agree that it is an ordered variable. If so, you could calculate Kendall’s tau instead of

Pearson’s r. Tau would be the same for any set of values you assign to the levels of Trauma,

assuming that they increased across the levels of that variable. For our data tau would be

.169, with a probability of .04. So the relationship would still be significant even if we are

only confident about the order of the independent variable(s). (The appeal to Kendall’s tau

as a possible replacement for Pearson’s ris the reason why I included this material here

rather than in Chapter 9. Agresti, however, has pointed out that if the cell frequencies are

very different, there are negative consequences to using either Kendall’s tau or Spearman’s rs.

I recommend strongly that you simply use r.)

Agresti (2002, p. 87) presents the approach that we have just adopted and shows that

we can compute a chi-square statistic from the correlation. He gives

M^25 (N 2 1)r^2

where M^2 is a chi-square statistic on 1 degree of freedom, ris the Pearson correlation be-

tween Dropout and Trauma, and Nis the sample size. For our example this becomes

which has an associated probability under the null hypothesis of .016.

The probability value was already given by the test on the correlation, so that is noth-

ing new. But we can go one step further. We know that the overall Pearson chi-square on

4 dfis 9.459. We also know that we have just calculated a chi-square of 5.757 on 1 dfthat

is associated with the linearrelationship between the two variables. That linear relation-

ship is part of the total chi-square, and if we subtract the linear component from the overall

chi-square we obtain

df Chi-square

Pearson 4 9.459

Linear 1 5.757

Deviation from linear 3 3.702

The departure from linearity is itself a chi-square equal to 3.702 on 3 df, which has a

probability under the null of .295. Thus we do not have any evidence that there is anything

other than a linear trend underlying these data. The relationship between Trauma and

Dropout is basically linear, as can be seen in Figure 10.2.

Agresti (1996, 2002) has an excellent discussion of the approach taken here, and he

makes the interesting point that for small to medium sample sizes, the standard Pearson

chi-square is more sensitive to the negative effects of small sample size than is the ordinal

chi-square that we calculated. In other words, although some of the cells in the contingency

table are small, I am more confident of the ordinal (linear) chi-square value of 5.757 than I

can be of the Pearson chi-square of 9.459.

You can calculate the chi-square for linearity using SPSS. If you request the chi-square

statistic from the statistics dialog box, your output will include the Pearson chi-square, the

Likelihood Ratio chi square, and Linear-by-Linear Association. The SPSS printout of the

x^2 (1)=125(0.215^2 )=5.757

M^2 =x^2 (1)=(N 2 1)r^2

308 Chapter 10 Alternative Correlational Techniques