BI 255 · Bethune-Cookman University · Fall 2026
All the tests covered so far have dealt with continuous measurements. But biological data are often categorical counts: how many offspring showed a given phenotype, how many patients fell into each disease category, how many sequences matched a particular motif.
The chi-square test is the workhorse for categorical count data. It comes in two forms:
By the end of this lab you will be able to:
chisq.test() against expected Mendelian ratiosGregor Mendel predicted that crossing two heterozygous plants (Aa × Aa) produces offspring in a 3:1 phenotypic ratio (dominant:recessive). A chi-square goodness of fit test asks: are our observed counts consistent with this predicted ratio?
# Observed offspring counts from a monohybrid cross (300 total offspring)
# Tall (dominant) vs short (recessive)
observed <- c(Tall = 233, Short = 67)
# Expected ratio 3:1 means 75% and 25% of total
total <- sum(observed)
expected_prop <- c(Tall = 0.75, Short = 0.25)
cat("Observed counts:", observed, "\n")Observed counts: 233 67 cat("Expected counts:", total * expected_prop, "\n")Expected counts: 225 75 # Chi-square goodness of fit test
# H₀: the observed frequencies match the 3:1 Mendelian ratio
# H₁: the observed frequencies deviate from the 3:1 ratio
chisq.test(observed, p = expected_prop)
Chi-squared test for given probabilities
data: observed
X-squared = 1.1378, df = 1, p-value = 0.2861| Component | Meaning |
|---|---|
X-squared |
The test statistic: Σ (O − E)² / E, summed over all categories |
df |
Degrees of freedom = number of categories − 1 |
p-value |
Probability of observing this X² or larger if H₀ is true |
A non-significant result (p > 0.05) means we fail to reject the Mendelian ratio — the data are consistent with 3:1. This does not prove the ratio is correct; it means the deviation we observed could plausibly be due to chance.
# Visualise observed vs expected
par(mar = c(5, 5, 4, 2))
counts_matrix <- rbind(Observed = observed,
Expected = total * expected_prop)
barplot(counts_matrix,
beside = TRUE,
col = c("#2C5F8A", "#A8C4D9"),
ylab = "Number of Offspring",
main = "Monohybrid Cross: Observed vs Expected (3:1)",
legend = TRUE,
args.legend = list(bty = "n"))
# Dihybrid cross (AaBb × AaBb) predicts 9:3:3:1 for two independent loci
# Seed shape (Round/Wrinkled) × Seed colour (Yellow/Green)
observed_dihybrid <- c("Round-Yellow" = 315,
"Round-Green" = 108,
"Wrinkled-Yellow" = 101,
"Wrinkled-Green" = 32)
# Expected proportions under 9:3:3:1
expected_dihybrid <- c(9/16, 3/16, 3/16, 1/16)
chisq.test(observed_dihybrid, p = expected_dihybrid)
Chi-squared test for given probabilities
data: observed_dihybrid
X-squared = 0.47002, df = 3, p-value = 0.9254The test of independence asks whether two categorical variables are associated. Unlike goodness of fit, the expected frequencies are calculated from the marginal totals rather than specified in advance.
Example: Is disease severity independent of treatment type?
# 200 patients classified by: Treatment (A, B, Placebo) × Outcome (Improved, No Change, Worse)
disease_table <- matrix(
c(52, 18, 10, # Treatment A
44, 22, 14, # Treatment B
20, 12, 8), # Placebo
nrow = 3,
byrow = TRUE,
dimnames = list(
Treatment = c("Drug A", "Drug B", "Placebo"),
Outcome = c("Improved", "No Change", "Worse")
)
)
addmargins(disease_table) Outcome
Treatment Improved No Change Worse Sum
Drug A 52 18 10 80
Drug B 44 22 14 80
Placebo 20 12 8 40
Sum 116 52 32 200# Chi-square test of independence
# H₀: treatment type and outcome are independent
# H₁: there is an association between treatment and outcome
chisq_result <- chisq.test(disease_table)
chisq_result
Pearson's Chi-squared test
data: disease_table
X-squared = 3.0587, df = 4, p-value = 0.5481# Inspect expected counts (required to be ≥ 5 in each cell)
round(chisq_result$expected, 1) Outcome
Treatment Improved No Change Worse
Drug A 46.4 20.8 12.8
Drug B 46.4 20.8 12.8
Placebo 23.2 10.4 6.4# Standardised residuals — which cells deviate most from expectation?
round(chisq_result$stdres, 2) Outcome
Treatment Improved No Change Worse
Drug A 1.64 -0.92 -1.10
Drug B -0.70 0.39 0.47
Placebo -1.15 0.64 0.77Standardised residuals larger than |2| indicate cells that contribute disproportionately to the chi-square statistic — these are the cells where observed counts deviate most from what would be expected under independence.
These residuals help you interpret where the association is coming from, not just whether one exists.
# Mosaic plot: area of each rectangle is proportional to cell frequency
mosaicplot(disease_table,
col = c("#2C5F8A", "#A8C4D9", "#C8102E"),
main = "Treatment vs Outcome: Mosaic Plot",
shade = TRUE,
las = 1)
The chi-square approximation is only reliable when expected cell counts are large enough. The standard rule is that all expected frequencies should be ≥ 5 (some texts say ≥ 1 with no more than 20% below 5).
# Rare allele study: small expected counts
rare_table <- matrix(
c(3, 2,
15, 20),
nrow = 2,
byrow = TRUE,
dimnames = list(
c("Rare allele", "Common allele"),
c("Case", "Control")
)
)
# Chi-square will warn about small expected counts
chisq.test(rare_table)
Pearson's Chi-squared test with Yates' continuity correction
data: rare_table
X-squared = 0.05772, df = 1, p-value = 0.8101# Fisher's exact test — no minimum cell size requirement
# Use this whenever expected counts are < 5 in any cell
fisher.test(rare_table)
Fisher's Exact Test for Count Data
data: rare_table
p-value = 0.6419
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
0.1986056 26.3241941
sample estimates:
odds ratio
1.965285 | Situation | Use |
|---|---|
| 2×2 table, all expected counts ≥ 5 | Chi-square (or Fisher’s — both fine) |
| 2×2 table, any expected count < 5 | Fisher’s exact test |
| Larger tables (r × c), all expected counts ≥ 5 | Chi-square |
| Larger tables with small cells | Consider collapsing categories or using simulation |
R’s chisq.test() will warn you when expected counts are small, but it will still give you a (potentially inaccurate) result. Always check chisq_result$expected and switch to fisher.test() if needed.
library(ggplot2)
library(dplyr)
# Convert to data frame for ggplot2
disease_df <- as.data.frame(as.table(disease_table))
names(disease_df) <- c("Treatment", "Outcome", "Count")
# Calculate proportions within each treatment
disease_df <- disease_df |>
group_by(Treatment) |>
mutate(Proportion = Count / sum(Count)) |>
ungroup()
disease_df$Outcome <- factor(disease_df$Outcome,
levels = c("Improved", "No Change", "Worse"))
ggplot(disease_df, aes(x = Treatment, y = Proportion, fill = Outcome)) +
geom_col(position = "fill", colour = "white", linewidth = 0.3) +
scale_fill_manual(values = c("Improved" = "#2C5F8A",
"No Change" = "#A8C4D9",
"Worse" = "#C8102E")) +
scale_y_continuous(labels = scales::percent) +
labs(title = "Outcome Proportions by Treatment Group",
x = "Treatment",
y = "Proportion of Patients") +
theme_classic(base_size = 13)
Close your notes. Answer these on your own — you have 3 minutes. We’ll go through the answers together after.
1. You cross two dihybrid plants and observe 280 Round-Yellow, 130 Round-Green, 85 Wrinkled-Yellow, and 35 Wrinkled-Green offspring. Which test would you use to determine if this follows a 9:3:3:1 Mendelian ratio?
a) t-test b) Chi-square test of independence c) Chi-square goodness of fit d) Pearson correlation2. In a 3×2 chi-square test of independence with df = 2, you obtain X² = 7.8. At α = 0.05, is this significant? (Critical value for χ²(2) = 5.99.)
3. You run chisq.test() on a 2×2 table and R returns a warning: “Chi-squared approximation may be incorrect.” What should you do?
4. A chi-square test of independence on a treatment × outcome table gives X² = 14.3, df = 4, p = 0.006. Standardised residuals show that the “Drug A / Improved” cell has a residual of +3.2. What does this mean?
5. True or False: For a chi-square goodness of fit test with 4 categories, the degrees of freedom equals 4.
1. c) Chi-square goodness of fit — You have a specific expected distribution (9:3:3:1) from Mendelian theory, and you want to test whether your observed counts match it. Goodness of fit compares observed frequencies to specified expected proportions. A test of independence would be used if you had two categorical variables and no specific expected ratio.
2. Yes, X² = 7.8 is significant because 7.8 > 5.99 (the critical value for χ²(2) at α = 0.05). The p-value is less than 0.05, so we reject H₀ and conclude there is a significant association between the two categorical variables.
3. Check the expected frequencies using chisq_result$expected. If any expected cell count is less than 5 (especially in a 2×2 table), switch to fisher.test() which makes no assumptions about minimum cell sizes and gives an exact p-value. Fisher’s exact test is always valid; the chi-square approximation is not reliable with small expected counts.
4. A standardised residual of +3.2 for the Drug A / Improved cell means that patients on Drug A improved substantially more than would be expected if treatment and outcome were independent. The magnitude (|3.2| > 2) confirms this cell is a major driver of the overall significant result. Drug A appears to be the most effective treatment.
5. False — df = number of categories − 1 = 4 − 1 = 3. In a goodness of fit test, one degree of freedom is lost because the expected frequencies must sum to the total observed count (a constraint on the data).
Before you leave, make sure you can:
A genetics study crosses AaBbCc × AaBbCc (three independent loci). The expected phenotypic ratios are 27:9:9:9:3:3:3:1 (there are 8 phenotypic classes from three loci with independent assortment). Simulate 500 offspring under these ratios using sample() or rmultinom(), then add small random deviations and test whether your observed counts are consistent with the expected 27:9:9:9:3:3:3:1 ratio. What is the df?