---
title: "Lab 14: Contingency Tables, Mosaic Plots, Odds Ratios & Chi-square Independence"
subtitle: "BI 255 · Bethune-Cookman University · Fall 2026"
author: "Dr. Rosie Stanbrook-Buyer"
date: "September 25, 2026"
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toc-title: "In This Lab"
toc-depth: 3
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---
```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)
```
------------------------------------------------------------------------
## Associations Between Categorical Variables
Many important biological questions involve two categorical variables: Does a genetic variant increase disease risk? Are a drug's side effects more common in one sex? Is smoking associated with a particular cancer type? These questions are answered by analysing **contingency tables** — cross-tabulations of two categorical variables.
This lab covers the construction and visualisation of contingency tables, the chi-square test for independence, Fisher's exact test for small samples, and the odds ratio as a measure of association strength.
::: callout-note
## Learning Objectives
By the end of this lab you will be able to:
1. Build and interpret a two-way contingency table
2. Visualise associations using `mosaicplot()`
3. Test for independence using `chisq.test()`
4. Interpret chi-square output: test statistic, degrees of freedom, p-value
5. Apply Fisher's exact test when sample sizes are small
6. Calculate and interpret an odds ratio
:::
------------------------------------------------------------------------
## Part 1: Contingency Tables in Genetics
A classic application is testing whether a genetic variant is associated with disease. Suppose a case-control study genotypes 500 individuals for a single nucleotide polymorphism (SNP). Each individual is classified by genotype (carriers of the minor allele vs homozygous reference) and disease status.
```{r}
# Genotype data: 500 individuals
# Rows: Disease status (Case / Control)
# Columns: Genotype (Carrier = at least one minor allele / Non-carrier)
snp_table <- matrix(
c(120, 80, # Cases: carrier, non-carrier
90, 210), # Controls: carrier, non-carrier
nrow = 2,
byrow = TRUE,
dimnames = list(
Status = c("Case", "Control"),
Genotype = c("Carrier", "Non-carrier")
)
)
snp_table
```
```{r}
# Margins: row totals and column totals
addmargins(snp_table)
```
```{r}
# Row proportions: within cases, what fraction are carriers?
round(prop.table(snp_table, margin = 1), 3)
```
::: callout-note
## Reading a Contingency Table
The table above is called a **2×2 contingency table**. Each cell contains the count of individuals with that combination of row and column categories.
The key question: **is genotype independent of disease status?**
If independent, the proportion of carriers among cases should be the same as among controls. Looking at the proportions: carriers make up 60% of cases but only 30% of controls. This suggests an association — but is it statistically significant?
:::
------------------------------------------------------------------------
## Part 2: Mosaic Plots
A mosaic plot visualises a contingency table as a set of rectangles, where the area of each rectangle is proportional to the cell count.
```{r}
mosaicplot(snp_table,
col = c("#2C5F8A", "#A8C4D9"),
main = "SNP Genotype by Disease Status",
xlab = "Disease Status",
ylab = "Genotype",
border = "white")
```
::: callout-tip
## Reading a Mosaic Plot
- The **width** of each column is proportional to the total in that row category
- The **height** of each segment within a column is proportional to the proportion in that cell
- If there were **no association**, the segment heights would be the same in every column
- Segments of unequal height signal a potential association
:::
------------------------------------------------------------------------
## Part 3: The Chi-Square Test for Independence
The chi-square test asks: **are the observed cell counts consistent with what we would expect if the two variables were independent?**
```{r}
# Chi-square test of independence
chi_result <- chisq.test(snp_table)
chi_result
```
```{r}
# Expected counts under independence (what we would see if there were no association)
chi_result$expected
```
```{r}
# Observed vs expected
cat("Observed:\n")
print(snp_table)
cat("\nExpected under independence:\n")
round(chi_result$expected, 1)
```
::: callout-important
## Interpreting the Chi-Square Test
The chi-square statistic measures how far the observed counts deviate from the expected counts under independence:
X² = sum( (Observed − Expected)² / Expected )
A large X² → large deviation from independence → small p-value → evidence against independence.
**Degrees of freedom** for a 2×2 table = (rows − 1) × (cols − 1) = 1.
The **assumption** of the chi-square test: expected counts in every cell should be ≥ 5. If not, use Fisher's exact test instead.
:::
### Examining the Residuals
```{r}
# Standardised residuals: large absolute values indicate which cells drive the association
round(chi_result$stdres, 2)
```
::: callout-note
## Standardised Residuals
Standardised residuals above |2| indicate cells that contribute most to the chi-square statistic. Here, the excess of carriers among cases (and deficit among controls) are the main drivers of the association.
:::
------------------------------------------------------------------------
## Part 4: Fisher's Exact Test
When expected cell counts fall below 5 (common in rare disease studies or when sample sizes are small), the chi-square approximation is not reliable. Fisher's exact test calculates the exact p-value without any approximation.
```{r}
# Suppose we have a much smaller sample from a rare disease study
rare_table <- matrix(
c(8, 2,
3, 12),
nrow = 2,
dimnames = list(Status = c("Case", "Control"),
Exposure = c("Exposed", "Unexposed"))
)
rare_table
addmargins(rare_table)
```
```{r}
# Chi-square would give unreliable results here (some cells < 5)
# Always use Fisher's exact test for small samples
fisher.test(rare_table)
```
::: callout-tip
## When to Use Fisher's Exact Test
Use Fisher's exact test when:
- Any **expected** cell count is < 5
- Total sample size is small (< 20–30)
- You want an exact p-value rather than an asymptotic approximation
Fisher's test is always valid; chi-square is an approximation that breaks down in small samples.
:::
------------------------------------------------------------------------
## Part 5: Odds Ratios
The **odds ratio (OR)** quantifies the strength of association in a 2×2 table. It compares the odds of the outcome in one group to the odds in another.
For our SNP table:
```{r}
# Odds ratio calculation
# Odds of disease for carriers = cases with carrier / controls with carrier
# Odds of disease for non-carriers = cases with non-carrier / controls with non-carrier
a <- snp_table["Case", "Carrier"]
b <- snp_table["Control", "Carrier"]
c <- snp_table["Case", "Non-carrier"]
d <- snp_table["Control", "Non-carrier"]
OR <- (a / b) / (c / d)
cat("Odds Ratio:", round(OR, 3), "\n")
```
```{r}
# Fisher's test also reports the OR and its confidence interval
fisher_result <- fisher.test(snp_table)
cat("OR from Fisher's test:", round(fisher_result$estimate, 3), "\n")
cat("95% CI:", round(fisher_result$conf.int, 3), "\n")
```
::: callout-important
## Interpreting the Odds Ratio
| OR value | Interpretation |
|----------|---------------|
| OR = 1 | No association — equal odds in both groups |
| OR > 1 | Increased odds of outcome in the exposed/carrier group |
| OR < 1 | Decreased odds (protective association) |
For our SNP: OR ≈ 3.5 means carriers have approximately 3.5 times the odds of disease compared to non-carriers.
The 95% CI for the OR is critical: if it does not cross 1.0, the association is statistically significant at α = 0.05.
:::
------------------------------------------------------------------------
## Part 6: A Full Worked Example — Drug Side Effects
A clinical trial of a new antibiotic records whether patients experienced gastrointestinal side effects, stratified by sex:
```{r}
side_effects <- matrix(
c(45, 105, # Female: side effect yes, no
28, 122), # Male: side effect yes, no
nrow = 2,
byrow = TRUE,
dimnames = list(Sex = c("Female", "Male"),
SideEffect = c("Yes", "No"))
)
cat("Contingency table:\n")
addmargins(side_effects)
```
```{r}
# Row proportions
cat("\nRow proportions:\n")
round(prop.table(side_effects, margin = 1), 3)
```
```{r}
# Chi-square test
chi_se <- chisq.test(side_effects)
chi_se
```
```{r}
# Odds ratio via Fisher's test
fisher_se <- fisher.test(side_effects)
cat("Odds Ratio:", round(fisher_se$estimate, 3), "\n")
cat("95% CI:", round(fisher_se$conf.int, 3), "\n")
```
```{r}
# Visualise
mosaicplot(side_effects,
col = c("#C8102E60", "#2C5F8A60"),
main = "Gastrointestinal Side Effects by Sex",
xlab = "Sex",
ylab = "Side Effect",
border = "white")
```
------------------------------------------------------------------------
## 3-Minute Knowledge Check
*Close your notes. Answer these on your own — you have 3 minutes. We'll go through the answers together after.*
::: callout-caution
## Knowledge Check Questions
**1.** In a 2×2 contingency table, the degrees of freedom for a chi-square test equals:
a) 1 b) 2 c) n − 1 d) (rows × cols) − 1
**2.** You find that one expected cell count in your 2×2 table is 3.2. Which test should you use?
a) Chi-square test b) t-test c) Fisher's exact test d) ANOVA
**3.** An odds ratio of 0.4 with a 95% CI of (0.2, 0.8) indicates:
a) No association b) A significant protective association c) A significant risk factor d) An inconclusive result because OR < 1
**4.** In a mosaic plot, what does it mean if the segment heights are the same in every column?
**5.** True or False: The chi-square test of independence tests whether the mean of one variable differs across levels of another variable.
:::
::: {.callout-caution collapse="true"}
## Answers (Only reveal if you have answered the above questions)
**1.** a) 1 — For a 2×2 contingency table, df = (2−1) × (2−1) = 1. For larger tables, df = (r−1)(c−1).
**2.** c) Fisher's exact test — The chi-square approximation is unreliable when any **expected** count is < 5. Fisher's exact test is always valid regardless of sample size and should be used here.
**3.** b) A significant protective association — OR < 1 means the exposed group has lower odds of the outcome. The 95% CI does not include 1.0 (it runs from 0.2 to 0.8), so the association is statistically significant at α = 0.05. The factor under study appears to reduce risk.
**4.** If segment heights are equal across columns, the proportion of each category is the same in every group — this is exactly what **independence** looks like in a mosaic plot. No visual difference between columns = no association.
**5.** **False** — The chi-square test of independence tests whether **two categorical variables are statistically independent** — i.e., whether the distribution of one variable differs across levels of the other. It does not involve means. Tests involving means (continuous outcomes) use t-tests or ANOVA.
:::
------------------------------------------------------------------------
## Lab 14 Checklist
Before you leave, make sure you can:
- [ ] Build a contingency table with `matrix()` and label it with `dimnames`
- [ ] Compute marginal totals with `addmargins()` and proportions with `prop.table()`
- [ ] Visualise a contingency table with `mosaicplot()`
- [ ] Run a chi-square test with `chisq.test()` and interpret the output
- [ ] Identify when expected counts are too small and use `fisher.test()` instead
- [ ] Calculate an odds ratio from a 2×2 table manually and via `fisher.test()`
- [ ] Interpret an OR with a 95% CI: what does it mean if the CI crosses 1?
::: callout-tip
## Bonus Challenge
Find a published 2×2 contingency table from a genetics or epidemiology paper (Google Scholar: "SNP case control contingency table"). Re-create the table in R, run both the chi-square and Fisher's exact tests, calculate the odds ratio, and produce a mosaic plot. Write two sentences interpreting the association.
:::
------------------------------------------------------------------------
## Before Next Class (Week 7 — Monday)
- Monday (Lab 15): The Normal Distribution and Central Limit Theorem
- Read **R for Data Science Ch. 9–10**
- Make sure you are comfortable with `pnorm()` and `qnorm()` — we will use them extensively
------------------------------------------------------------------------
