Contingency Analysis (Part 2)

Our data in this chapter consists of two categorical variables.

We are interested in:

• Estimating association in 2 x 2 tables (i.e. special case of two categoricals with only 2 levels each)
• Testing if there is an association (or dependence) between two categorical variables [\( \chi^2 \! \) contingency test]

Before moving on to hypothesis testing, know that there is another measure of association: relative risk. More specific context…

Definition: Relative risk is the probability of an undesired outcome in the treatment group divided by the probability of the same outcome in a control group.

\[ \hat{RR} = \frac{\mathrm{Pr[undesired \ | \ treatment]}}{\mathrm{Pr[undesired \ | \ control]}} = \frac{\frac{a}{a+c}}{\frac{b}{b+d}} \]

dataTable <- matrix(c(19, 122, 2473, 7768), nrow = 2, byrow = TRUE, dimnames = list(c("Cancer", "No cancer"), c("Coffee", "No coffee")))
(dTmargins <- addmargins(dataTable, margin=1))

          Coffee No coffee
Cancer        19       122
No cancer   2473      7768
Sum         2492      7890

(RR <- (dTmargins[1,1]/dTmargins[3,1]) / (dTmargins[1,2]/dTmargins[3,2]))

[1] 0.4930861

Now using epitools package. Strangely enough, we need the table to be both transposed and levels reversed for the riskratio function to work.

(RR_result <- riskratio(t(dataTable), rev = "both", method = "wald"))

RR_result$measure[-1,]

 estimate     lower     upper 
0.4930861 0.3047198 0.7978932 

Remember, the odds ratio is 0.4891915. Hmmm, so close!!!

The relative risk and odds ratio will be very close when the undesired outcome is rare in the population.

Which one do we use to measure association?

Definition: A case-control study is a method of observational study in which a sample of individuals having a disease or other focal condition (“cases”) is compared to a second sample of individuals who do not have the condition (“controls”).

Total number of cases and controls are chosen by researcher and not sampled randomly from the population. This means we can't estimate probability of undesired result in control and treatment. Thus, we cannot estimate relative risk.

Thus begins our venture into parasitology…

Toxoplasma gondii is a protozoan parasite that can infect the brains of many birds and mammals, including humans… In humans, toxoplasmosis may be associated with some mental illnesses, and it may be associated with risky behavior. Yereli et al. (2006) compared the prevalence of Toxoplasma gondii in a sample of 185 drivers between 21 and 40 years old who had been involved in a driving accident (cases) with a sample of 185 drivers of similar age and sex who had not had accidents (control). The researchers were interested in whether Toxoplasma infection may cause a change in the probability of an accident.

             infected uninfected
accidents          61        124
no accidents       16        169

Question: Why can’t we estimate relative risk?

Answer: Because we can’t estimate the necessary probabilities!!!

\[ \mathrm{Pr[accidents \ | \ infected]} = \frac{\mathrm{Pr[accidents \ AND \ infected]}}{\mathrm{Pr[infected]}} \]

There exists a scaling factor \( x \) that scales the “no accidents” counts that will make the proportions representative (i.e. with no bias) of the population. The odds ratio thus becomes

\[ OR = \frac{61\times 169x}{124\times 16x} = \frac{61\times 169}{124\times 16} = 5.1960685. \]

Ah hah! So, \( x \) cancels out, and the odds ratio can be computed without worrying about under- or over-representation of the levels of the “cases” in the population.

Discuss: Turn the result \( OR = 5.1960685 \) into a statement.

Answer: The odds of having an accident when infected with toxoplasma is about 5 times larger than having an accident when not infected.

Let's check if this has statistical significance or not by computing a confidence interval on the odds ratio:

oddsratio(toxTable, method="wald")$measure[-1,]

estimate    lower    upper 
5.196069 2.859352 9.442394 

estimate    lower    upper 
5.196069 2.859352 9.442394 

Discuss: Given this 95% confidence interval, what do you conclude?

Conclusion: Since the 95% confidence interval does not contain 1, an odds ratio of 1 is not a plausible parameter, and thus there is evidence that there is increased odds of having an accident when infected with toxoplasma.

Estimation of association for 2 x 2 contingency tables: odds ratios and relative risks

What about hypothesis testing?

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

Remember: Hypothesis tests will give you a yes/no answer, but will NOT give you the magnitude of the effect if there is one (hence, always use confidence intervals as well, if possible)

\( \chi^2 \) contingency test

Example 9.4

Many parasites have more than one species of host, so the individual parasite must get from one host to another to complete its life cycle. Trematodes of the species Euhaplorchis californiensis use three hosts during their life cycle.

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 as follows:

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

If we call our two categorical variable “fate” and “infection status”, then what we want to know is are these two variables 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. The number of degrees of freedom is given by

\[ \begin{align} df & = rc - 1 - (r-1) - (c - 1) \\ & = rc - r - c + 1 \\ & = (r-1)\times(c-1) \end{align} \]

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 larger than 2 x 2, you can combine row and/or columns.
• If table is 2 x 2, you can use Fisher's exact test.
• You can use a permutation test (see Chapter 13).

Expected frequency table

Assumptions seem to be 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 p. 251)

    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.

Fisher’s exact test: (2 x 2 tables only) Examines the independence of two categorical variables, even with small expected values

\( G \)-test: (any table) Derived from principles of likelihood.

     Pro: Great with complicated experimental designs with multiple explanatory variables.
     Con: Can be less accurate for small sample sizes.

Situation: You've found a statistically significant association between two categorical variables using the \( \chi^2 \) contingency test.

Question: Where is the association? In other words, in which levels of the categories is the association present and how large is the association?

We need to estimate the magnitude of the association, which the \( P \)-value does not give us!

We can estimate odds ratios or relative risks for 2 \( \times \) 2 sub-tables within the contingency table by either subsetting or collapsing.