Lecture 12
Contingency Analysis
ABD 3e Chapter 9
Learning Objectives
- Construct and interpret contingency tables.
- Compute and interpret risk, relative risk (RR), odds, and odds ratio (OR).
- Select RR vs. OR based on study design (cohort vs. case-control).
- State and test hypotheses of independence using the chi-square test.
- Calculate expected counts and the chi-square statistic.
- Make decisions using \(P\)-values or critical values.
- Identify when alternative tests (e.g., Fisher’s exact test) are appropriate.
Contingency Analysis: Relationships Between Categorical Variables
- We now move from one categorical variable to two categorical variables
Examples:
- Treatment (Aspirin vs Placebo) × Cancer (Yes/No)
- Sex (Men/Women) × Survival (Yes/No)
- Smoking (Yes/No) × Lung Disease (Yes/No)
Core Question:
Does the distribution of one variable depend on the other?
If yes → association
If no → independence
Example: Does knowing a person’s sex tell you anything about the chance they survived the Titanic?
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Two Complementary Goals: Estimation and Hypothesis Testing Answer Difference Questions
1️⃣ Estimation
How strong is the relationship?
- Relative Risk (RR)
- Odds Ratio (OR)
These quantify the magnitude of association.
Primarily defined for 2×2 tables (binary exposure × binary outcome)
2️⃣ Hypothesis Testing
Is the relationship statistically detectable?
- Chi-square contingency test
This evaluates evidence for association.
Applicable to any r × c contingency table (two categorical variables with two or more categories each)
What is a Contingency Table?
- A contingency table summarizes the joint distribution of two categorical variables
- Rows = categories of one variable
- Columns = categories of the second variable
- Each cell = a count (frequency)
- Margins show row and column totals
- Allows us to:
- Compare conditional proportions
- Assess association vs independence
- Compute relative risk and odds ratios
Example Study: Aspirin and Cancer Risk
A large randomized study investigated whether regular aspirin use reduces cancer risk.
Two categorical variables:
- Treatment: (Aspirin vs. Placebo)
- Cancer Outcome: Cancer, No Cancer
Each participant falls into exactly one cell of a 2 × 2 contingency table.
Our goal:
Does cancer risk differ between the aspirin and placebo groups?
Conditional Probability (Risk)
A risk is a conditional probability:
\[
\text{Risk} = \operatorname{Pr}(\text{Cancer} \mid \text{Group})
\]
We compute risk as a proportion by dividing:
\[ p=\frac{\text{number with cancer in group}}{\text{total in group}} \]
Example: Risk of Cancer for each Group
Risk of cancer in the aspirin group:
\[
\operatorname{Pr}(\text{Cancer} \mid \text{Aspirin})
\]
\[
\hat{p}_1=\frac{1438}{1438+18496}=0.0721
\]
Risk of cancer in the placebo group:
\[
\operatorname{Pr}(\text{Cancer} \mid \text{Placebo})
\]
\[
\hat{p}_2=\frac{1427}{1427+18515}=0.0716
\]
Relative Risk (RR)
Now we compare risks across groups.
Relative risk of aspirin:
\[ RR = \frac{ \operatorname{Pr}(\text{Cancer} \mid \text{Aspirin}) }{ \operatorname{Pr}(\text{Cancer} \mid \text{Placebo}) } \]
\[ \hat{RR}=\frac{\hat{p}_1}{\hat{p}_2} \]
\[ =\frac{0.0721}{0.0716}=1.007 \]
Interpretation:
- \(RR = 1\) → same risk in both groups
- \(RR > 1\) → higher risk in aspirin group
- \(RR < 1\) → lower risk in aspirin group
Relative risk measures the magnitude of association between treatment and outcome.
Odds: A Different Way to Express Probability
Risk compares outcome to total.
Odds compare outcome to non-outcome.
Odds are not bounded between 0 and 1.
When to use: 2 variables, each with 2 categories
Risk (probability):
\[
\operatorname{Pr}(\text{Cancer})
\]
Odds:
\[
\frac{\operatorname{Pr}(\text{Cancer})}
{\operatorname{Pr}(\text{No Cancer})}
\]
\[
=\frac{\operatorname{Pr}(\text{Cancer})} {1 - \operatorname{Pr}(\text{Cancer})}
\]
Odds of developing cancer while taking aspirin
Success
\[
\hat{p}_1=\frac{1438}{1438+18496}=0.0721
\]
Failure
\[
1-\hat{p}_1=1-0.0721=0.9279
\]
Odds of success
\[
\hat{O}_1=\frac{\hat{p}_1}{1-\hat{p}_1}=\frac{0.0721}{0.9279}=0.0777
\]
Odds of developing cancer while taking placebo
Success
\[
\hat{p}_2=\frac{1427}{1427+18515}=0.0716
\]
Failure
\[
1-\hat{p}_2=1-0.0716=0.9274
\]
Odds of success
\[
\hat{O}_2=\frac{\hat{p}_2}{1-\hat{p}_2}=\frac{0.0721}{0.9284}=0.0771
\]
Odds Ratio (OR)
The odds ratio compares two odds:
\[ \hat{OR} = \frac{\text{Odds}(\text{Cancer} \mid \text{Aspirin})} {\text{Odds}(\text{Cancer} \mid \text{Placebo})} \]
Interpretation:
- \(OR = 1\) → same odds in both groups
- \(OR > 1\) → higher odds in numerator group
- \(OR < 1\) → lower odds in numerator group
For the aspirin study:
\[
\hat{OR}
=
\frac{0.0777}{0.0771}
=
1.008
\]
The odds of cancer are essentially the same in the aspirin and placebo groups.
Odds ratio shortcut
The following is a shortcut formula where 𝑎, 𝑏, 𝑐, and 𝑑 refer to the observed frequencies in the cells of the contingency table:
\[
\hat{OR}=\frac{a/c}{b/d}=\frac{ad}{bc}
\]
Standard error for odds ratio
- The sampling distribution for the odds ratio is highly skewed, so we convert the odds ratio to its natural log, \(\operatorname{ln}(\hat{OR})\)
\[
\operatorname{SE}[\operatorname{ln}(\hat{OR})]=\sqrt{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}
\]
- This is called the standard error of the log-odds ratio
For the aspirin example:
\[
\operatorname{SE}[\operatorname{ln}(\hat{OR})]=\sqrt{\frac{1}{1438}+\frac{1}{1427}+\frac{1}{18496}+\frac{1}{18515}}=0.03878
\]
Confidence interval for odds ratio
- We can approximate a confidence interval for the log-odds ratio:
\[
\operatorname{ln}(\hat{OR}) \pm Z \times \operatorname{SE}[\operatorname{ln}(\hat{OR})]
\]
where \(Z=1.96\) for a 95% CI and \(Z=2.58\) for a 99% CI
To get the CI for the odds ratio itself, you have to take the antilog of the upper and lower bounds:
\[
e^x < \operatorname{OR} < e^y
\]
Odds ratio interpretation
If OR=1 there is no association between exposure and outcome
If 95% CI includes 1, results are not statistically significant
Relative Risk vs. Odds Ratio
Relative Risk (RR)
- Requires estimating risk (probability)
- Risk requires a meaningful denominator:
- number with outcome
- total number at risk
Best used when
- Cohort studies
- Randomized experiments
- The total population at risk is known
Odds Ratio (OR)
- Does not require estimating population risk
- Can be computed even when totals at risk are unknown
Best used when
- Case-control studies
- Logistic regression
If the outcome is rare, \(OR \approx RR\)
Case-Control Study
Investigates associations between an exposure and an outcome.
Start with individuals who already have the outcome (cases)
Select a comparison group without the outcome (controls)
Look backward to determine prior exposure status
The total population at risk is unknown, so risk cannot be estimated directly. Use odds ratio instead.
Cohort Study
- Follows individuals forward in time to assess whether an exposure is associated with an outcome.
- Begin with individuals classified by exposure status (Exposed, Unexposed)
- Follow both groups over time
- Record who develops the outcome
- The total number at risk in each group is known, so risk can be estimated directly.
- We can compute relative risk.
- Types: Prospective or Retrospective (see diagram)
The \(\chi^2\) contingency test
The \(\chi^2\) contingency test is the most commonly used test of association between two categorical
It tests the goodness of fit to the data of the null model of independence of variables.
RR and OR allow us to estimate magnitude of association, but do not test whether an association may be caused by chance alone.
Example: Consider the life cycle of the trematode E. californensis
- Trematode worms Euhaplorchis californensis use three hosts during their life cycle
- Mature worms in birds lay eggs that pass out in bird’s feces
- Horn snails Cerithidea californica eat the eggs
- Eggs hatch and grow to another life stage in the snail, sterilizing it
- Californa killifish Fundulus parvipinnis eat the snail
- Parasite develops to the next life stage and encysts in the brain
- Birds eat infected killifish, worm matures in the bird
Research on fish behavior
Researchers have observed that infected fish spend excessive time near the water surface
They may be more vulnerable to bird predation, which would benefit the worm
Lafferty and Morris (1996) tested whether bird predation varies with severity of infection
Fish placed into outdoor pens open to bird predation, with fish of varying infection intensity:
highly infected
lightly infected
not infected
Observed frequencies of fish eaten or not eaten by birds according to trematode infection level
Table 1. Observed Frequencies.
| Eaten by birds |
1 |
10 |
37 |
| Not eaten by birds |
49 |
35 |
9 |
Question:
Is being eaten by birds (outcome) independent from infection level?
Steps to hypothesis testing
- State hypotheses
- Calculate test statistic ( \(\chi_2\) )
- Calculate row, column, and grand totals
- Calculate expected proportions assuming independence
- Calculate expected frequencies assuming independence
- Calculate difference between observed and expected frequencies ( \(\chi_2\) )
- Calculate \(P\)-value
- Interpret hypotheses (2 ways):
- \(P<\alpha\)
- \(\chi^2_{df}>\chi^2_{crit}\)
STEP 1: State the hypotheses
\(H_0\) : Parasite infection level and being eaten are independent
\(H_A\) : Parasite infection level and being eaten are not independent
STEP 2: Calculate the test statistic ( \(\chi^2\) )
Goal: calculate the \(\chi^2\) from the data to see how different the observed frequencies are from the expected frequencies
Start with: contingency table of observed frequencies
Next step: 2a. Calculate row, column, and grand totals
Table 1. Observed Frequencies.
STEP 2A: Calculate row, column, and grand totals
Sum the values in each row to get row totals
Sum the values in each column to get column totals
Sum the row or column totals to get the grand total
Table 1. Observed Frequencies.
STEP 2B: Calculate expected proportions assuming independence
- Goal: calculate the values in table 2
- First, calculate the marginal values assuming independence
- For each column or row, divide frequency by grand total to get expected proportion
Table 1. Observed Frequencies.
Table 2. Expected Proportions.
Example 1: Probability of being not infected
\[
\hat{\operatorname{Pr}}[\text{Not infected}]=
\]
\[
\frac{50}{141}=
\]
\[
0.3546
\]
Table 1. Observed Frequencies.
Table 2. Expected Proportions.
Example 2: Probability of not being eaten by birds
\[
\hat{\operatorname{Pr}}[\text{Eaten by birds}]=
\]
\[
\frac{48}{141}=
\]
\[
0.3404
\]
Table 1. Observed Frequencies.
Table 2. Expected Proportions.
Repeat for all columns and rows
Table 1. Observed Frequencies.
Table 2. Expected Proportions.
Calculate cell proportions by multiplying marginal proportions
Use multiplication rule:
If two events are independent (null hypothesis), probability of both occurring is probability of one times probability of the other
\[
\operatorname{Pr}[\text{not infected and eaten}]\\
=\operatorname{Pr}[\text{not infected}]\times\operatorname{Pr}[\text{eaten}]\\
=0.3546 \times 0.3404
=0.1207
\]
Table 1. Observed Frequencies.
Table 2. Expected Proportions.
Repeat for each cell proportion
Use multiplication rule:
If two events are independent (null hypothesis), probability of both occurring is probability of one times probability of the other
Table 1. Observed Frequencies.
Table 2. Expected Proportions.
STEP 2C: Calculate expected frequencies assuming independence
\[
\operatorname{Expected}[\text{not infected and eaten}]=
\\\operatorname{Pr}[\text{not infected and eaten}]\times G=
\\0.1207 \times 141 =
\\17.0
\]
Repeat for each cell
Table 3. Expected Frequencies
Table 2. Expected Proportions.
STEP 2D: Calculate \(\chi^2\) test statistic
Use the tables for Observed and Expected frequencies to calculate the test statistic
\[
\chi^2 = \sum_{i=1}^{r} \sum_{j=1}^{c} \frac{(O_{ij} - E_{ij})^2}{E_{ij}}
\]
Where:
- \(r\) is the number of rows
- \(c\) is the number of columns
- \(O_{ij}\) is the observed frequency in row \(i\) and column \(j\)
- \(E_{ij}\) is the expected frequency in row \(i\) and column \(j\)
Table 3. Expected Frequencies
Table 1. Observed Frequencies.
STEP 2D: Calculate \(\chi^2\) test statistic
Use the tables for Observed and Expected frequencies to calculate the test statistic
\[ \chi^2 = \sum_{i=1}^{r} \sum_{j=1}^{c} \frac{(O_{ij} - E_{ij})^2}{E_{ij}} \]
\[
= \frac{(1-17)^2}{17} + \frac{(49-33)^2}{33} + \\
\frac{(10-15.3)^2}{15.3} + \frac{(35-29.7)^2}{29.7} + \\
\frac{(37-15.7)^2}{15.7} + \frac{(9-30.3)^2}{30.3}
\]
\[
= 69.5
\]
Table 3. Expected Frequencies
Table 1. Observed Frequencies.
STEPS 3-4: \(P\)-value and Interpretation
- Decide a priori (beforehand) on a significance level, e.g. \(\alpha=0.05\), then…
Method 1 - exact P-value
- Calculate exact \(P\)-value using computer
- If \(P<\alpha\) then reject \(H_0\)
Method 2 - statistical table
Calculate degrees of freedom
\(\operatorname{df}=(r-1)(c-1)\)
Look up critical value in a table
If \(\chi^2_{df}>\operatorname{critical value}\) then reject \(H_0\)
How to look up a critical value in a chi-square table
- See any chi-square distribution table
- Know your 𝑑𝑓 and 𝛼
- For example, at 𝑑𝑓=2 𝛼=0.05, The critical value is Χ^2=5.991
Decision rule: \(\chi^2_{df}>\operatorname{critical value}\)
\[
\chi^2 = 69.5
\]
\[
\operatorname{critical value} = 5.994
\]
Therefore, reject \(H_0\), conclude parasite infection level and being eaten are not independent
Wrapping Up: Contingency Analysis
- Chi-square test of independence compares observed vs. expected frequencies
- Expected counts computed from row and column totals
- When assumptions are strained
- Yates correction: continuity adjustment sometimes recommended for 2 × 2 tables
- Fisher’s exact test: preferred when expected counts are small (any expected count < 5)
- Alternative framework
- G-test (likelihood ratio test)
- Based on likelihood ratios rather than squared deviations
- Extends more naturally to complex models and multiple explanatory variables
- Still sensitive to small expected counts
- Choose the method based on sample size, expected counts, and study design
- Always interpret results in the biological context of the association
