Chi-squared Tests
2026-02-04
Last updated: 2026-02-04
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| Rmd | dddb710 | Dave Tang | 2025-11-14 | Chi-squared tests |
Introduction
The chi-squared (\(\chi^2\)) test is one of the most widely used statistical tests for categorical data. There are two main types:
- Chi-squared test of independence: Tests whether two categorical variables are related
- Chi-squared goodness of fit test: Tests whether observed frequencies match expected frequencies
Common applications include:
- Is smoking related to lung cancer?
- Is gender related to voting preference?
- Does a die produce fair outcomes?
- Is a treatment associated with recovery?
The Chi-Squared Distribution
The chi-squared distribution arises when summing squared standard normal random variables. It has one parameter: degrees of freedom (df), which determines its shape.
x <- seq(0, 20, length = 200)
plot(x, dchisq(x, df = 1), type = "l", lwd = 2, col = 1,
ylim = c(0, 0.5), ylab = "Density", xlab = "Chi-squared value",
main = "Chi-squared distributions")
lines(x, dchisq(x, df = 2), lwd = 2, col = 2)
lines(x, dchisq(x, df = 5), lwd = 2, col = 3)
lines(x, dchisq(x, df = 10), lwd = 2, col = 4)
legend("topright", legend = paste("df =", c(1, 2, 5, 10)),
col = 1:4, lwd = 2)
As degrees of freedom increase, the distribution becomes more symmetric and shifts to the right.
Chi-Squared Test of Independence
This test determines whether there is a statistically significant association between two categorical variables.
Hypotheses
- Null hypothesis (\(H_0\)): The two variables are independent (no association)
- Alternative hypothesis (\(H_1\)): The two variables are not independent (there is an association)
The Test Statistic
The chi-squared statistic measures the discrepancy between observed and expected frequencies:
\[\chi^2 = \sum \frac{(O - E)^2}{E}\]
where:
- \(O\) = observed frequency in each cell
- \(E\) = expected frequency if variables were independent
The expected frequency for each cell is calculated as:
\[E = \frac{\text{row total} \times \text{column total}}{\text{grand total}}\]
Example 1: Smoking and Lung Cancer
Let’s test whether smoking is associated with lung cancer using a 2x2 contingency table.
data <- matrix(
c(60, 40, 30, 70),
nrow = 2,
byrow = TRUE
)
colnames(data) <- c("Cancer", "NoCancer")
rownames(data) <- c("Smoker", "NonSmoker")
data Cancer NoCancer
Smoker 60 40
NonSmoker 30 70Step 1: Calculate Expected Values
If smoking and cancer were independent, we would expect each cell to have:
\[E = \frac{\text{row total} \times \text{column total}}{\text{grand total}}\]
# Row and column totals
row_totals <- rowSums(data)
col_totals <- colSums(data)
grand_total <- sum(data)
row_totals Smoker NonSmoker
100 100 col_totals Cancer NoCancer
90 110 grand_total[1] 200# Calculate expected values
expected <- outer(row_totals, col_totals) / grand_total
expected Cancer NoCancer
Smoker 45 55
NonSmoker 45 55For example, the expected count for Smoker + Cancer is: \((100 \times 90) / 200 = 45\)
Step 2: Calculate the Chi-Squared Statistic
# Chi-squared = sum of (observed - expected)^2 / expected
chi_sq <- sum((data - expected)^2 / expected)
chi_sq[1] 18.18182Step 3: Run the Test
result <- chisq.test(data)
result
Pearson's Chi-squared test with Yates' continuity correction
data: data
X-squared = 16.99, df = 1, p-value = 3.758e-05Interpreting the Output
- X-squared = 16.99: The test statistic measuring how different observed values are from expected values
- df = 1: Degrees of freedom = (rows - 1) x (columns - 1) = 1 for a 2x2 table
- p-value = 3.76e-05: Probability of seeing this extreme a result if the null hypothesis were true
Since the p-value is very small (< 0.05), we reject the null hypothesis and conclude that smoking and lung cancer are associated.
Accessing Test Components
The test object contains useful information:
# Observed values
result$observed Cancer NoCancer
Smoker 60 40
NonSmoker 30 70# Expected values (same as our manual calculation)
result$expected Cancer NoCancer
Smoker 45 55
NonSmoker 45 55# Residuals: (observed - expected) / sqrt(expected)
result$residuals Cancer NoCancer
Smoker 2.236068 -2.0226
NonSmoker -2.236068 2.0226Residuals show which cells contribute most to the chi-squared statistic. Large positive residuals indicate more observations than expected; large negative residuals indicate fewer.
Visualising the Result
x2 <- result$statistic
df <- result$parameter
x <- seq(0, x2 + 3, length = 500)
plot(x, dchisq(x, df), type = "l", lwd = 2,
ylab = "Density", xlab = "Chi-squared value",
main = paste0("Chi-squared distribution (df = ", df, ")"))
abline(v = x2, col = "red", lwd = 2)
# Shade the rejection region
x_shade <- seq(x2, x2 + 3, length = 100)
polygon(c(x_shade, rev(x_shade)),
c(dchisq(x_shade, df), rep(0, length(x_shade))),
col = rgb(1, 0, 0, 0.3), border = NA)
text(x2, 0.1, paste("p =", format.pval(result$p.value, digits = 2)),
pos = 4, col = "red")
| Version | Author | Date |
|---|---|---|
| 503ceec | Dave Tang | 2025-11-14 |
The red line shows our test statistic. The shaded area represents the p-value: the probability of getting a value this extreme or more extreme under the null hypothesis.
Example 2: Larger Contingency Table
Chi-squared tests work for tables larger than 2x2. Let’s examine the relationship between education level and job satisfaction:
# Create a 3x3 table
job_data <- matrix(
c(20, 25, 15, # High school
30, 40, 20, # Bachelor's
15, 30, 25), # Graduate
nrow = 3,
byrow = TRUE
)
colnames(job_data) <- c("Low", "Medium", "High")
rownames(job_data) <- c("HighSchool", "Bachelors", "Graduate")
job_data Low Medium High
HighSchool 20 25 15
Bachelors 30 40 20
Graduate 15 30 25result2 <- chisq.test(job_data)
result2
Pearson's Chi-squared test
data: job_data
X-squared = 5.1406, df = 4, p-value = 0.2732For a 3x3 table, df = (3-1) x (3-1) = 4.
# View expected values
round(result2$expected, 1) Low Medium High
HighSchool 17.7 25.9 16.4
Bachelors 26.6 38.9 24.5
Graduate 20.7 30.2 19.1The p-value of 0.273 suggests a marginally significant association between education and job satisfaction.
Example 3: No Association
To understand what “no association” looks like, let’s create data where the variables are independent:
set.seed(1984)
# Generate independent data
n <- 200
gender <- sample(c("Male", "Female"), n, replace = TRUE)
preference <- sample(c("Tea", "Coffee"), n, replace = TRUE)
indep_table <- table(gender, preference)
indep_table preference
gender Coffee Tea
Female 53 50
Male 51 46chisq.test(indep_table)
Pearson's Chi-squared test with Yates' continuity correction
data: indep_table
X-squared = 0.00028872, df = 1, p-value = 0.9864With a large p-value, we fail to reject the null hypothesis. There is no evidence of an association between gender and beverage preference in this simulated data.
Chi-Squared Goodness of Fit Test
This test compares observed frequencies to expected frequencies based on a theoretical distribution.
Example: Is a Die Fair?
Suppose we roll a die 120 times and observe the following counts:
observed <- c(18, 23, 16, 21, 24, 18)
names(observed) <- 1:6
# Expected: equal probability for each face
expected_prob <- rep(1/6, 6)
# Goodness of fit test
gof_result <- chisq.test(observed, p = expected_prob)
gof_result
Chi-squared test for given probabilities
data: observed
X-squared = 2.5, df = 5, p-value = 0.7765The large p-value suggests we cannot reject the hypothesis that the die is fair.
barplot(
rbind(observed, expected_prob * sum(observed)),
beside = TRUE,
col = c("steelblue", "coral"),
names.arg = 1:6,
xlab = "Die Face",
ylab = "Frequency",
main = "Observed vs Expected Frequencies",
legend.text = c("Observed", "Expected"),
args.legend = list(x = "topright")
)
Example: Testing a Genetic Ratio
In genetics, Mendel’s law predicts a 9:3:3:1 ratio for certain crosses. Let’s test whether observed data follows this ratio:
observed_plants <- c(315, 108, 101, 32)
names(observed_plants) <- c("Round-Yellow", "Round-Green",
"Wrinkled-Yellow", "Wrinkled-Green")
# Expected ratio 9:3:3:1
expected_ratio <- c(9, 3, 3, 1) / 16
mendel_test <- chisq.test(observed_plants, p = expected_ratio)
mendel_test
Chi-squared test for given probabilities
data: observed_plants
X-squared = 0.47002, df = 3, p-value = 0.9254The p-value of 0.925 indicates the observed data is consistent with the expected 9:3:3:1 ratio.
Assumptions and Conditions
The chi-squared test has several assumptions:
- Independence: Observations must be independent
- Random sampling: Data should come from a random sample
- Expected frequencies: All expected cell counts should be at least 5
Checking Expected Frequencies
# Check expected values
result$expected Cancer NoCancer
Smoker 45 55
NonSmoker 45 55# All expected values >= 5, so assumption is met
all(result$expected >= 5)[1] TRUEWhen Assumptions Are Violated
If expected frequencies are too small (< 5), consider:
- Fisher’s exact test: For 2x2 tables with small samples
- Combining categories: Merge cells to increase expected counts
- Simulation-based p-value: Use
chisq.test(..., simulate.p.value = TRUE)
# Small sample example
small_data <- matrix(c(3, 2, 1, 4), nrow = 2)
rownames(small_data) <- c("Treatment", "Control")
colnames(small_data) <- c("Success", "Failure")
small_data Success Failure
Treatment 3 1
Control 2 4# Chi-squared test warns about small expected values
chisq.test(small_data)Warning in chisq.test(small_data): Chi-squared approximation may be incorrect
Pearson's Chi-squared test with Yates' continuity correction
data: small_data
X-squared = 0.41667, df = 1, p-value = 0.5186# Use Fisher's exact test instead
fisher.test(small_data)
Fisher's Exact Test for Count Data
data: small_data
p-value = 0.5238
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
0.218046 390.562917
sample estimates:
odds ratio
4.918388 Effect Size: Cramer’s V
A significant p-value tells us an association exists, but not how strong it is. Cramer’s V measures effect size for chi-squared tests:
\[V = \sqrt{\frac{\chi^2}{n \times (k - 1)}}\]
where \(n\) is the sample size and \(k\) is the smaller of the number of rows or columns.
# Calculate Cramer's V for the smoking example
chi_sq <- result$statistic
n <- sum(data)
k <- min(nrow(data), ncol(data))
cramers_v <- sqrt(chi_sq / (n * (k - 1)))
cramers_vX-squared
0.291461 Interpretation guidelines for Cramer’s V:
| V | Interpretation |
|---|---|
| 0.1 | Small effect |
| 0.3 | Medium effect |
| 0.5 | Large effect |
A Cramer’s V of 0.29 indicates a small to medium effect size. # Yates’ Continuity Correction
For 2x2 tables, R applies Yates’ continuity correction by default. This makes the test more conservative (larger p-values) but can be overly conservative with moderate sample sizes.
# With Yates' correction (default)
chisq.test(data)
Pearson's Chi-squared test with Yates' continuity correction
data: data
X-squared = 16.99, df = 1, p-value = 3.758e-05# Without Yates' correction
chisq.test(data, correct = FALSE)
Pearson's Chi-squared test
data: data
X-squared = 18.182, df = 1, p-value = 2.008e-05The uncorrected version gives a slightly smaller p-value. For large samples, the difference is negligible.
Summary
| Test | Purpose | R Function |
|---|---|---|
| Test of independence | Association between two categorical variables | chisq.test(table) |
| Goodness of fit | Compare observed to expected frequencies | chisq.test(x, p = prob) |
| Fisher’s exact test | Small samples (expected < 5) | fisher.test(table) |
Key points:
- Chi-squared tests assess relationships between categorical variables
- The test statistic measures how much observed differs from expected
- Larger chi-squared values indicate stronger evidence against independence
- Check that expected frequencies are at least 5
- Use Cramer’s V to measure effect size
- For small samples, use Fisher’s exact test instead
sessionInfo()R version 4.5.0 (2025-04-11)
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Running under: Ubuntu 24.04.3 LTS
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time zone: Etc/UTC
tzcode source: system (glibc)
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