158 CATEGORICAL DATA: ANALYSIS OF TWO-WAY FREQUENCY TABLESFor matched samples, the same correction factor can be used. Here, the formula
can be simplified so that the corrected xz is
(Ib-cl-1)2
b+c
x: =where /b - c/ denotes the positive difference.
11.4 CHI-SQUARE TESTS FOR LARGER TABLESSo far in this chapter, discussion of the use of measures of association and chi-square
tests has been restricted to frequency tables with two rows and two columns. To
illustrate the analysis of larger tables, we now present a hypothetical example of a
two-way table with four rows and four columns. In general, frequency tables can be
described as having r rows and c columns.
11.4.1 Chi-square for Larger Tables: Single Sample
Our example is from a hypothetical health survey in which a single sample of respon-
dents was taken from a population of adults in a county. The health survey included
questions on health status, access to health care, and whether or not the respondents
followed recommended preventive measures. The results from two of the questions
are given in Table 11.8. The first question was: In general, would you say your
health is excellent, good, fail; or poor? These four choices were coded 1, 2, 3, and
4, respectively. The second question was: Are you able to afford the kind of medical
care you should have? The possible answers were almost nevel; not often, often, or
always and again were coded 1,2,3, and 4. The data were analyzed using a statistical
program. Here, the results are given in Table 11.8. Both observed frequencies and
expected frequencies (in parentheses) are displayed. Note that these data are ordinal
data. Ordinal data are sometimes analyzed using the chi-square test when the authors
want to display the data in a two-way table.
The expected frequencies have been computed using the same method used for
tables with two rows and columns. In each cell, the row and column totals that the cell
falls in are multiplied together and the product is divided by the total frequency, 600.
For example, for respondents who reported that their health was fair and that they could
afford medical care often, the expected frequency is computed as 105(103)/600 =
18.025, which has been rounded off to 18.02, as reported in Table 11.8.
To test the null hypothesis that the answers to the health status question were
independent of the answers to the access to health care question for adults in the
county, we compute chi-square using the same formula as the earlier examples with
two rows and columns; now, however, there are 16 cells rather than 4.
all cells (observed - expected)2
x2= c expected