Basic Statistics

144 CATEGORICAL DATA: ANALYSIS OF TWO-WAY FREQUENCY TABLES

Table 11.3 Outcomes for Experimental and Standard Treatment: Two Samples

Treatment

Outcome Experimental Standard Total

Success 95 = a 80=b 175=a+b

Failure 10 = c 20=d 30=c+d

Total 105 =n1 100= n2 205 =n

Table 11.4 Exposure to Risk Factor for a Matched Case/Control Study

Control

Case Yes No Total

Yes 59 =a 23 = b 82

NO 8=c 10=d 18

Total 67 33 1oo=n

In Chapter 10 we presented statistical methods for examining differences in the
population proportions. Confidence intervals for nl - 7r2 were given as well as a two-
sided test of hypothesis that Ho : 7il = 7~2 and a similar one-sided test of hypothesis.
Another method of testing the two-sided hypothesis given in Section 10.6.2 will be
presented; it can also be used when there are more than two groups andor more than
two outcomes. The tests for equal population proportions are often called tests of
homogeneity.

11 .I .3 Tables Based on Matched or Paired Samples

In some biomedical studies, each subject is paired with another subject. For example,

in casekontrol studies a control subject is often paired with a case. Table 1 1.4 presents

an example from a hypothetical matched casekontrol study.

In Table 11.4 there are n pairs of matched cases and controls. In 59 instances,

both the case and the control were exposed to the risk factor, and in 10 instances

they both were not exposed. So for 69 pairs the results were a tie. For 23 pairs the

case was exposed and the control was not, and for 8 pairs the control was exposed

and the case was not. Thus, among the nontied pairs, the cases were exposed more

than the controls. Note that the same symbols have been used to depict the counts in

Table 11.1 as in Table 11.4, but what is counted is different. Here, ties and nonties

for matched pairs are given.

Note that the number of ties has no effect on the differences between pl, the

proportion of cases exposed to risk, and p2, the proportion of controls exposed to

risk. From Table 11.4, pl = (59 + 23)/100 = .82 and p2 = (59 + 8)/100 = .67,

so that the difference in proportions is .82 - .67 = .15. As expected, ignoring ties

23/100 - 8/100 = .15.