Contingency Tables
Prerequisites
Chi
Square Distribution,
OneWay Tables
Learning Objectives
 State the null hypothesis tested concerning contingency tables
 Compute expected cell frequencies
 Compute Chi Square and df
This section shows how to use Chi Square to test
the relationship between nominal variables for significance. For
example, Table 1 shows the data from the Mediterranean
Diet and Health case study.
The question is whether there is a significant
relationship between diet and outcome. The first step is to
compute the expected frequency for each cell based on the assumption
that there is no relationship. These expected frequencies are
computed from the totals as follows. We begin by computing the
expected frequency for the AHA Diet/Cancers combination. Note
that 22/605 subjects developed cancer. The proportion who developed
cancer is therefore 0.0364. If there were no relationship between
diet and outcome, then we would expect 0.0364 of those on the
AHA diet to develop cancer. Since 303 subjects were on the AHA
diet, we would expect (0.0364)(303) = 11.02 cancers on the AHA
diet. Similarly, we would expect (0.0364)(302) = 10.98 cancers
on the Mediterranean diet. In general, the expected frequency
for a cell in the ith row and the jth column is equal to
where Ei,j is the expected
frequency for cell i,j, Ti is the total
ith row, Tj is the total for the jth
column, and T is the total number of observations. For the AHA
Diet/Cancers cell, i = 1, j = 1, Ti =
303, Tj =
22, and T = 605. Table 2 shows the expected
frequencies (in parenthesis) for each cell in the experiment. Table
2 shows the expected frequencies (in parenthesis) for each cell
in the experiment.
The significance test is conducted by computing
Chi Square as follows.
The degrees of freedom is equal to (r1)(c1)
where r is the number of rows and c is the number of columns.
For this example, the degrees of freedom is (21)(41) = 3. The
Chi
Square calculator can be used to determine that the probability
value for a Chi Square of 16.55 with three degrees of freedom
is less 0.0009. Therefore, the null hypothesis of no relationship
between diet and outcome can be rejected.
A key assumption of the Chi Square test of independence
is that each subject contributes data to only one cell. Therefore
the sum of all cell frequencies in the table must be the same
as the number of subjects in the experiment. Consider an experiment
in which each of 16 subjects each attempted two anagram problems.
The data are shown in Table 3.
It would not be valid to use the Chi Square test
on these data since each subject contributed data to two cells:
one cell based on their performance on Anagram 1
and one cell based on their performance on Anagram 2. The total
of the cell frequencies in the table is 32 but the total number
of subjects is only 16.
The formula for Chi Square yields a statistic
that is only approximately a Chi Square distribution. In order
for the approximation to be adequate, the total number of
subjects should be at least 20. Some authors claim that the correction
for continuity should be used whenever an expected cell frequency
is below 5. Research in statistics has shown that this practice
is not advisable. For example, see:
Bradley, D. R., Bradley, T. D., McGrath, S. G., & Cutcomb,
S. D. (1979) Type I error rate of the chi square test of independence
in r x c tables that have small expected frequencies. Psychological
Bulletin, 86, 12001297.
The correction for continuity when applied to 2
x 2 contingency tables is called the Yates correction. The simulation 2
x 2 tables lets you explore the accuracy of the approximation
and the value of this correction.
