Contingency Analysis

Our data in this chapter consists of two categorical variables.

We may be interested in:

• Estimating the magnitude of association between two categorical variables:
  • odds ratio and relative risk (2 x 2 tables)
  • NOT covered in class
• Testing if there is an association (or dependence) between two categorical variables:
  • \( \chi^2 \! \) contingency test

\( \chi^2 \) contingency test

• \( H_{0} \): There is no association between two categorical variables
• \( H_{A} \): There is an association between two categorical variables

Example 9.4

Lafferty and Morris (1996) tested the hypothesis that infection influences risk of predation by birds. A large outdoor tank was stocked with three kinds of killfish: unparasitized, lightly infected, and heavily infected. This tank was left open to foraging by birds. The numbers of fish eaten according to their levels of parasitism is given below:

          Uninfected Lightly Highly Sum
Eaten              1      10     37  48
Not eaten         49      35      9  93
Sum               50      45     46 141

Calling our two categorical variables “fate” and “infection status”, what we want to know is whether these two variables are independent.

How do we compute the frequency table we would expect if the variables were independent? If two variables are independent, we have:

\[ \mathrm{Pr[A \ and \ B]} = \mathrm{Pr[A]}\times\mathrm{Pr[B]} \]

The observed table (in relative frequencies) is given by:

Assuming independence, the expected (relative frequencies) table should have:

Expected frequency table

Observed frequency table

The \( \chi^2 \) contingency test is just a special case of the \( \chi^2 \) goodness-of-fit test, so the test statistic is the same.

\[ \chi^2 = \sum_{i=1}^{r}\sum_{j=1}^{c} \frac{(Observed_{ij}-Expected_{ij})^2}{Expected_{ij}}, \]

where \( r \) and \( c \) are number of rows and columns, respectively.

\[ \chi^2 = \frac{(1-17.0)^2}{17.0}+\frac{(10-15.3)^2}{15.3}+...+\frac{(9-30.3)^2}{30.3} \]

The number of degrees of freedom is given by

\[ df = (r-1)\times(c-1) = (2-1)\times(3-1) = 2 \]

Assumptions of the \( \chi^2 \) contingency test are the same as the \( \chi^2 \) goodness-of-fit test.

What do you do if the assumptions are violated?

• If table is larger than 2 x 2, you can sometimes combine row and/or columns.
• If table is 2 x 2, you can use Fisher's exact test.
• You can use a permutation test (not discussed in class, but you may see Chapter 13).

Expected frequency table

The assumptions are met, so let's use R.

First, let's see how the table was created in R:

(parTable <- matrix(c(1, 10, 37, 49, 35, 9), 
                    nrow = 2, 
                    byrow = TRUE, 
                    dimnames = list(c("Eaten", "Not eaten"), 
                                    c("Uninfected", "Lightly", "Highly"))))

          Uninfected Lightly Highly
Eaten              1      10     37
Not eaten         49      35      9

chisq.test(parTable, correct = FALSE)

Note: correct = FALSE means no Yates correction (see W&S p. 251: Correction for continuity)

    Pearson's Chi-squared test

data:  parTable
X-squared = 69.756, df = 2, p-value = 7.124e-16

Conclusion?

Conclusion: Since \( p \)-value is less than 0.05 (actually less than 0.001), then we can reject the null hypothesis of independence.