Basic Statistics

154 CATEGORICAL DATA: ANALYSIS OF TWO-WAY FREQUENCY TABLES

Table 11.7
Frequencies"

Outcomes for Experimental and Standard Treatment: Expected

Treatment

Outcome Experimental Standard Total

Success 95(89.63) SO(85.37) 175

Failure lO(15.37) 20(14.63) 30

Total 105= ni loo= 122 205 =n

"The numbers in parentheses are the expected values under the null hypothesis.

11.3.2 Chi-square Test for Two Samples: Two-by-Two Tables

In Table 1 1.3 we presented data from a clinical trial where one group of patients was
randomly assigned to receive the experimental treatment and the other the standard
treatment. The outcome was classified as a success or a failure. Hence, the table has
only two rows and two columns. There were n1 = 105 patients in the experimental
treatment group and n2 = 100 patients in the standard treatment group. The question
being asked is whether or not the treatments differ in their proportion of successes.
The null hypothesis to be tested is that the proportion of successes is the same in both
populations.
The total number of observed successes was 175 and the total of the two sample
sizes was 205. If the two treatments were equally effective (Ho is true), the expected
proportion of successes in each group would be 175/205 = ,8537. Of the 105
patients given the experimental treatment, we would expect .8537(105) = 89.63 to
be successes if the null hypothesis is true. Similarly, of the patients given the standard
treatment, we would expect .8537(100) = 85.37 to be classified as successes. The
proportion of failures overall is 1 - .8537 = .1463 (30/205 in the total column). So
in the second row we would expect .1463(105) = 15.37 failures in the experimental
group and 14.63 failures in the standard treatment group. The results are summarized
in Table 1 1.7.
Just as in the smoking and low vital capacity case where we had a single sample, in
this clinical trial example with two samples we have to compute the expected value for
only one cell and the others can be obtained by subtraction from the row and column
totals. For example, for the experimental group, if we compute the expected number
of successes (89.63), the expected number of failures can be obtained by subtracting
89.63 from 105. This implies that we will have 1 d.f. for testing.
In the single-sample case, we computed the expected frequency by multiplying
the row total by the column total, and then divided by the total sample size for any
cell in the table (see Table 11.6). For example, we computed

A=- 30(21) - - (a + b)(a + c)

120 n