Basic Statistics

CHI-SQUARE TESTS FOR FREQUENCY TABLES: TWO-BY-TWO TABLES 155

For two samples, if we compute the expected frequency for A, we take

175(105) - (a + b)(a + c)

• 175
  205 205 n

A = -(105) =

Numerically, the expected frequency is computed in the same way although the null
hypothesis and the method of sampling are different. From a numerical standpoint,
the chi-square test for two samples is identical to that for the single-sample test. (This
statement holds true even when there are more than two rows and/or columns.) Again
we compute
all cells (observed - expected)

x2= c expected

in the same way as before. For the frequencies in Table 1 1.7, we have

(95 - 89.63)’ (80 - 85.37)’ (10 - 15.37)’ (20 - 14.63)’

= 89.63 + 85.37 + 15.37 + 14.63

or

X’ = 4.50

With Q = .05 using Table A.4, we reject the null hypothesis of equal proportions in
the two populations since the computed chi-square is larger than the tabled chi-square
with 1 d.f. (4.50 > 3.84).
By comparing the observed and expected values in Table 11.7, we see a higher
observed frequency of successes than expected in the experimental group and a lower
observed frequency of successes than expected in the standard treatment group. This
indicates that the experimental treatment may be better than the standard treatment.

11.3.3 Chi-square Test for Matched Samples: Two-by-Two Tables

In Table 11.4 there are 100 pairs of individually matched cases and controls presented
in a table with two rows and two columns. The appropriate null hypothesis is that
there is no association between getting the disease (being a case or a control) and
being exposed or not exposed to the risk factor. In other words, in the populations
from which the cases and controls were sampled, the number of pairs where the case is
exposed and the control is not (23) is equal to the number of pairs where the control is
exposed but the case is not (8). As described in Section 11.1.3, on tables for matched
samples, the counts entered into Table 1 1.4 are quite different from those in the other
tables, so a different x2 test is used.
To test this null hypothesis, we compute

x^2 =--- (b-

bit

or

• = 7.26

(23 -8)’ - 225

23+8 31

x2 = -