Statistical Methods for Psychology

276 Chapter 9 Correlation and Regression

Males Females

r .50 .40

r .549 .424

N 53 53

Since is less than , we fail to reject and conclude, that

with a two-tailed test at a5.05, we have no reason to doubt that the correlation between

smoking and life expectancy is the same for males as it is for females.

I should point out that it is surprisingly difficult to find a significant difference between

two independent rs for any meaningful comparison unless the sample size is quite large.

Certainly I can find two correlations that are significantly different, but if I restrict myself

to testing relationships that might be of theoretical or practical interest, it is usually diffi-

cult to obtain a statistically significant difference.

Testing the Hypothesis That rEquals Any Specified Value

Now that we have discussed the concept of we are in a position to test the null hypothe-

sis that ris equal to any value, not just to zero. You probably can’t think of many situations

in which you would like to do that, and neither can I. But the ability to do so allows us to

establish confidence limits on r, a more interesting procedure.

As we have seen, for any value of r, the sampling distribution of is approximately

normally distributed around (the transformed value of r) with a standard error of.

From this it follows that

is a standard normal deviate. Thus, if we want to test the null hypothesis that a sample rof

.30 (with N 5 103) came from a population where r5.50, we proceed as follows

Since 5 22.39 is more extreme than 56 1.96, we reject at a5.05 (two-

tailed) and conclude that our sample did not come from a population where r5.50.

Confidence Limits on r

We can move from the preceding discussion to easily establish confidence limits on rby

solving that equation for rinstead of z. To do this, we first solve for confidence limits on

, and then convert to r.

z=

r¿2r¿

B

1

N 23

r¿ r¿

zobt z.025 H 0

z=

.310 2 .549

0.10

=-0.239>0.10=-2.39

N= 103 sr¿= 1 > 1 N 23 =0.10

r=.50 r¿=.549

r=.30 r¿=.310

z=

r¿2r¿

B

1

N 23

1

r¿ 1 N 23

r¿

r¿,

zobt=0.625 z.025= 6 1.96 H 0

z=

.549 2 .424

B

1

5323

1

1

5323

=

.125

B

2

50

=

.125

1

5

=0.625

¿