Chi-Square Test Exercises in R: 20 Practice Problems with Solutions
These 20 chi-square R exercises cover goodness-of-fit, tests of independence, effect size with Cramer's V, standardized residuals, Yates' correction, Fisher's exact, and Monte Carlo p-values. Every exercise sets an expected output, asks you to save a named object, and hides a fully runnable solution with a short explanation.
Section 1. Goodness-of-fit (4 problems)
Exercise 1.1: Test a six-sided die for fairness from raw roll counts
Task: A casino floor manager has rolled a six-sided die 120 times to verify it's fair and observed counts c(15, 22, 18, 24, 19, 22) for faces 1 through 6. Run a goodness-of-fit chi-square test against equal proportions (1/6 each) and save the full chisq.test() result object to ex_1_1 so the p-value can be inspected.
Expected result:
#> Chi-squared test for given probabilities
#>
#> data: rolls
#> X-squared = 2.7, df = 5, p-value = 0.7461Difficulty: Beginner
A fairness check weighs each face's observed count against what perfect equal chance would predict across all six faces.
Call chisq.test() on the count vector and supply the p = argument with six equal probabilities.
Click to reveal solution
Explanation: chisq.test() switches to goodness-of-fit mode when you pass a single count vector plus a p = argument. The expected count for each face is n * p = 120 / 6 = 20. With p = 0.75 the data are very consistent with a fair die. If you omit p, the function defaults to equal proportions, so chisq.test(rolls) would produce the same result here.
Exercise 1.2: Test Mendel's 9:3:3:1 dihybrid ratio on 556 peas
Task: A genetics class is checking Mendel's classic dihybrid ratio of 9:3:3:1 (round-yellow, wrinkled-yellow, round-green, wrinkled-green) from observed counts c(315, 108, 101, 32) in a sample of 556 peas. Run a goodness-of-fit test against those expected proportions and save the test result to ex_1_2.
Expected result:
#> Chi-squared test for given probabilities
#>
#> data: peas
#> X-squared = 0.47, df = 3, p-value = 0.9254Difficulty: Intermediate
The expected proportions here are not equal, so the test needs the specific ratio you want to compare the counts against.
Hand chisq.test() the counts plus a p = argument built by dividing the integer ratio c(9, 3, 3, 1) by its total so it sums to 1.
Click to reveal solution
Explanation: The p vector must sum to 1, so divide the integer ratio by its total (16). A p-value of 0.93 is so high that some statisticians have famously suggested Mendel's published numbers are "too clean" to be real raw data. Goodness-of-fit assumes independent observations and expected counts of at least 5, both of which hold here.
Exercise 1.3: Extract chi-square statistic, df, and p-value as a named vector
Task: A junior analyst needs the chi-square statistic, degrees of freedom, and p-value from ex_1_2 packed into a single named numeric vector for a results table. Build a length-3 vector with names "chisq", "df", and "p" by pulling $statistic, $parameter, and $p.value and save it to ex_1_3.
Expected result:
#> chisq df p
#> 0.47000000 3.00000000 0.92540000Difficulty: Beginner
A fitted test object stores its statistic, degrees of freedom, and p-value as separate components you can pull out one at a time.
Build the result with c(chisq = ..., df = ..., p = ...), reading ex_1_2$statistic, $parameter, and $p.value, and strip stray names with unname().
Click to reveal solution
Explanation: $statistic and $parameter come back as named numeric vectors of length 1 (named "X-squared" and "df" respectively), so wrapping them in unname() keeps your vector's names clean. This pattern is the building block for any pipeline that loops chi-square tests and collects results into a tibble or data frame for reporting.
Exercise 1.4: Compare store satisfaction mix against the national baseline
Task: A retailer surveyed 400 customers and got satisfaction counts c(Very = 180, Some = 140, Not = 80) but the national mix is 0.40, 0.40, 0.20 according to headquarters. Run a goodness-of-fit chi-square test of the store's counts against the national proportions and save the result to ex_1_4 so the team can decide whether the store deviates.
Expected result:
#> Chi-squared test for given probabilities
#>
#> data: sat
#> X-squared = 5, df = 2, p-value = 0.08208Difficulty: Intermediate
Judge the store's three-category mix against the externally supplied national baseline rather than against equal shares.
Run chisq.test() on the sat vector with p = c(0.40, 0.40, 0.20).
Click to reveal solution
Explanation: Expected counts are 400 * p = c(160, 160, 80). The "Very" cell contributes most of the chi-square value because the store has 20 more "Very" responses than expected. A p-value of 0.08 lands in borderline territory: not significant at 0.05 but worth flagging. With a larger sample size the same proportional gap would push the p-value lower.
Section 2. Tests of independence (4 problems)
Exercise 2.1: Test independence of drug and outcome in a 2x2 table
Task: A pharmaceutical analyst wants to know whether drug A and drug B produce different success rates in a small trial of 100 patients. Using the 2x2 matrix in the chunk below (rows = drug, columns = outcome), run chisq.test() without Yates' correction and save the result object to ex_2_1 so success rates can be compared.
Expected result:
#> Pearson's Chi-squared test
#>
#> data: drug_tab
#> X-squared = 5.7692, df = 1, p-value = 0.01632Difficulty: Beginner
For a two-by-two table R nudges the statistic by default, but here you want the plain, unadjusted version.
Call chisq.test() on drug_tab and set correct = FALSE.
Click to reveal solution
Explanation: correct = FALSE turns off Yates' continuity correction, which R applies by default to 2x2 tables. Without correction, the chi-square statistic equals z^2 from a two-proportion z-test, so chisq.test(..., correct = FALSE) matches prop.test(..., correct = FALSE). Drug A's success rate (60%) versus B's (36%) gives a p-value of 0.016, comfortably below the conventional 0.05 cutoff.
Exercise 2.2: Test whether Titanic passenger class is independent of survival
Task: A historian asks whether passenger class on the Titanic was related to survival. Collapse the built-in Titanic 4D array over Sex and Age using margin.table() to get a Class by Survived 4x2 table, then run chisq.test() and save the full result object to ex_2_2 for further analysis later in the hub.
Expected result:
#> Pearson's Chi-squared test
#>
#> data: titanic_tab
#> X-squared = 190.4, df = 3, p-value < 2.2e-16Difficulty: Intermediate
Once the four-way array has been collapsed to a single two-way table, an independence test reads it directly.
Pass the collapsed titanic_tab straight to chisq.test() with no extra arguments.
Click to reveal solution
Explanation: margin.table(Titanic, c(1, 4)) sums the 4-way array over dimensions 2 (Sex) and 3 (Age), leaving the Class by Survived 2-way table. Yates' correction is not applied because the table is larger than 2x2. With df = (4-1) * (2-1) = 3 and a chi-square statistic of 190, the p-value is effectively zero: class and survival are strongly associated, exactly what the lifeboat allocation pattern would predict.
Exercise 2.3: Test wool and tension on the high-break face of warpbreaks
Task: A production engineer wants to know whether wool type and tension setting are independent when looms produce a high break rate. Bin warpbreaks$breaks into "low" (under 25) and "high" (25 or more), build the 3D table wool by tension by bin, slice the bin == "high" face, run chisq.test() on that 2x3 table, and save the result to ex_2_3.
Expected result:
#> Pearson's Chi-squared test
#>
#> data: high_face
#> X-squared = 5.5043, df = 2, p-value = 0.0638
#> Warning: Chi-squared approximation may be incorrectDifficulty: Advanced
Small expected counts on this sliced face will trigger a noisy approximation message you can quietly silence.
Wrap the chisq.test(high_face) call inside suppressWarnings().
Click to reveal solution
Explanation: A 3D table indexed by three variables can be sliced like a matrix: tab[, , "high"] grabs the wool by tension face for high-break looms only. Some cells will have small expected counts on this sliced face, which is why chisq.test() warns about the approximation. Exercise 4.2 shows how to switch to fisher.test() when those warnings show up.
Exercise 2.4: Test independence of iris Species and binned Sepal.Length
Task: A botany student wants to know whether iris Species is related to flower size. Bin iris$Sepal.Length into "short" (under 5.5), "medium" (5.5 to under 6.5), and "long" (6.5 and above), build the Species by size table, run chisq.test() on the 3x3 table, and save the full result to ex_2_4.
Expected result:
#> Pearson's Chi-squared test
#>
#> data: iris_tab
#> X-squared = 90.81, df = 4, p-value < 2.2e-16Difficulty: Intermediate
With the species-by-size table already built, the independence test needs nothing beyond the table itself.
Hand iris_tab to chisq.test() directly.
Click to reveal solution
Explanation: Degrees of freedom equal (3 - 1) * (3 - 1) = 4. The p-value is essentially zero because setosa is overwhelmingly short while virginica is overwhelmingly medium or long, with versicolor straddling between. The cells driving the chi-square statistic are exactly where Species and binned size disagree most, which exercise 3.4 quantifies through per-cell contributions.
Section 3. Effect size and residuals (4 problems)
Exercise 3.1: Compute Cramer's V for the Titanic class by survived table
Task: A data scientist reporting on ex_2_2 needs Cramer's V as the effect size measure alongside the p-value. Compute V = sqrt(chi_square / (n * (min(rows, cols) - 1))) using ex_2_2$statistic for the chi-square value and sum(ex_2_2$observed) for n, with rows = 4 and cols = 2, then save the single numeric value to ex_3_1.
Expected result:
#> [1] 0.294Difficulty: Intermediate
Effect size rescales the raw statistic by the sample size and the smaller table dimension so it lands between 0 and 1.
Plug ex_2_2$statistic and sum(ex_2_2$observed) for n into sqrt(chisq / (n * (min(4, 2) - 1))), then round() to 3 places.
Click to reveal solution
Explanation: Cramer's V rescales chi-square onto the 0-to-1 interval so it can be compared across tables of different sizes and sample sizes. By Cohen's rule of thumb for a 1-df table, V around 0.1 is small, 0.3 medium, 0.5 large. The Titanic table lands at V = 0.29, a medium-sized effect: class explains a real but not overwhelming share of the survival variation. The vcd::assocstats() function will print V along with phi and the contingency coefficient if you prefer not to compute it by hand.
Exercise 3.2: Extract standardized residuals for the Titanic table
Task: A reviewer asks which cells of the Titanic Class by Survived table drive the chi-square result. Pull the matrix of standardized residuals from ex_2_2$stdres, round to two decimals, and save the rounded 4x2 matrix to ex_3_2 so cells with |z| greater than 2 can be flagged as deviating from independence.
Expected result:
#> Survived
#> No Yes
#> 1st -9.46 9.46
#> 2nd -1.45 1.45
#> 3rd 5.94 -5.94
#> Crew 4.41 -4.41Difficulty: Intermediate
The fitted test already holds a matrix of per-cell deviations scaled to behave like z-scores under the null.
Read ex_2_2$stdres and wrap it in round(..., 2).
Click to reveal solution
Explanation: $stdres returns Pearson residuals adjusted to have asymptotic variance 1, so each cell behaves like a z-score under the null of independence. Values outside +/-2 are unusual: 1st class survived far more (z = 9.5) and 3rd class died far more (z = 5.9) than independence would predict. Use $residuals instead for unadjusted Pearson residuals if you want the raw (O-E)/sqrt(E).
Exercise 3.3: Inspect expected counts before trusting a small contingency test
Task: A statistician wants the matrix of expected counts before running chi-square on the mtcars cyl by gear contingency table to decide if the approximation is trustworthy. Run chisq.test() (wrap with suppressWarnings()), pull $expected, round to two decimals, and save the matrix to ex_3_3 so cells with expected counts below 5 can be located.
Expected result:
#> gear
#> cyl 3 4 5
#> 4 5.16 4.13 1.72
#> 6 3.28 2.62 1.09
#> 8 6.56 5.25 2.19Difficulty: Intermediate
Before trusting a test on a sparse table, look at the count each cell would hold if the variables were independent.
Run suppressWarnings(chisq.test(cyl_gear)), pull its $expected component, and round() to 2 decimals.
Click to reveal solution
Explanation: Five of the nine expected counts sit below 5, so the chi-square approximation is unreliable: R warns about this whenever you fit a small table. The cure is either to collapse sparse rows or columns into broader categories, or to switch to Monte Carlo simulation (exercise 4.3) or Fisher's exact (exercise 4.2). Use $expected to make this decision explicitly rather than guessing from row and column totals.
Exercise 3.4: Rank cells by their contribution to the chi-square statistic
Task: A teaching assistant wants the per-cell contribution to the chi-square statistic for the Titanic Class by Survived table to highlight which cells dominate. Compute (O - E)^2 / E from ex_2_2$observed and ex_2_2$expected, round to one decimal, and save the 4x2 matrix to ex_3_4 so the largest contributors can be ranked.
Expected result:
#> Survived
#> No Yes
#> 1st 47.6 74.7
#> 2nd 1.0 1.5
#> 3rd 16.8 26.4
#> Crew 8.7 13.7Difficulty: Advanced
Each cell's share of the total statistic is its squared gap between observed and expected, scaled by the expected value.
Compute (O - E)^2 / E from ex_2_2$observed and ex_2_2$expected, then round() to 1 decimal.
Click to reveal solution
Explanation: Each cell of (O-E)^2/E is its share of the total chi-square statistic: the matrix sums to the reported X-squared = 190.4. The "1st class survived" cell contributes 74.7, the single largest piece, confirming the qualitative story that first-class passengers survived at far higher rates than independence would predict. Combine this with $stdres from exercise 3.2 to get both the size and direction of each deviation.
Section 4. Small samples and alternatives (4 problems)
Exercise 4.1: Compare the drug 2x2 result with and without Yates' correction
Task: A textbook author wants to show how Yates' continuity correction shifts a borderline 2x2 result. Using the same drug_tab matrix from exercise 2.1, run chisq.test() with correct = TRUE and with correct = FALSE, extract both p-values, and save them as a length-2 named numeric vector with names "yates" and "no_yates" to ex_4_1.
Expected result:
#> yates no_yates
#> 0.02637923 0.01631752Difficulty: Intermediate
Run the same two-by-two test twice, once with the continuity adjustment on and once off, and collect both tail probabilities.
Build a named vector calling chisq.test() with correct = TRUE and correct = FALSE, taking $p.value from each.
Click to reveal solution
Explanation: Yates' continuity correction shrinks |O - E| by 0.5 before squaring, which pulls the chi-square statistic down and the p-value up. For a 2x2 table near the significance threshold the difference can flip the conclusion, as here: both p-values are below 0.05 but the corrected one is noticeably less significant. Most modern guides recommend correct = FALSE for tables with all expected counts at least 5; Yates was designed for the small-sample era before computers made fisher.test() cheap.
Exercise 4.2: Switch to Fisher's exact test on a sparse 2x2 table
Task: A pilot trial has tiny cell counts so chisq.test() warns the approximation may be incorrect. Run fisher.test() on the inline 2x2 matrix given below (rows = treatment, columns = outcome), save the full result object to ex_4_2, and verify the exact p-value falls well below 0.05 even at this sample size.
Expected result:
#> Fisher's Exact Test for Count Data
#>
#> data: pilot_tab
#> p-value = 0.04762
#> alternative hypothesis: true odds ratio is not equal to 1
#> 95 percent confidence interval:
#> 1.024092 Inf
#> sample estimates:
#> odds ratio
#> InfDifficulty: Intermediate
When cells are too sparse for the chi-square approximation, switch to a test that computes the probability exactly.
Call fisher.test() on pilot_tab.
Click to reveal solution
Explanation: Fisher's exact computes the p-value from the hypergeometric distribution rather than relying on the chi-square approximation, so it works regardless of expected-cell counts. The odds ratio reported here is Inf because the table has a zero cell, making the conditional MLE infinite. Use fisher.test() whenever any expected count is below 5 or whenever a table has a zero margin and you still need a defensible p-value.
Exercise 4.3: Compute a Monte Carlo p-value for the sparse cyl by gear table
Task: A simulation enthusiast wants a Monte Carlo p-value for the mtcars$cyl by mtcars$gear table because so many expected cells sit below 5. Set the seed to 42, run chisq.test() with simulate.p.value = TRUE and B = 10000, and save the full result object to ex_4_3 so the simulated p-value can be compared with the asymptotic one.
Expected result:
#> Pearson's Chi-squared test with simulated p-value (based on 10000
#> replicates)
#>
#> data: cyl_gear
#> X-squared = 18.036, df = NA, p-value = 0.000999Difficulty: Advanced
Instead of leaning on the reference distribution, resample many tables with the same margins to build the p-value empirically.
Call chisq.test() on cyl_gear with simulate.p.value = TRUE and B = 10000.
Click to reveal solution
Explanation: simulate.p.value = TRUE resamples B random tables with the same row and column margins and counts how often the simulated chi-square statistic equals or exceeds the observed one. With B = 10000, the smallest p-value reportable is 1 / (B + 1) = 0.0001. The simulated approach avoids the chi-square approximation entirely, so the result is trustworthy even with the small expected counts found in exercise 3.3. The reported df = NA is intentional: a Monte Carlo test does not use the chi-square reference distribution.
Exercise 4.4: Flag cells whose expected count violates the rule of 5
Task: An audit team wants a logical matrix flagging which cells of a 3x3 contingency table have expected counts under 5, the standard chi-square approximation threshold. Run chisq.test() on the inline matrix below (wrap with suppressWarnings()), compare $expected < 5 element-by-element, and save the resulting 3x3 logical matrix to ex_4_4.
Expected result:
#> [,1] [,2] [,3]
#> [1,] TRUE FALSE TRUE
#> [2,] TRUE FALSE FALSE
#> [3,] TRUE FALSE TRUEDifficulty: Intermediate
Fit the test, then ask of every expected-count cell whether it clears the minimum-count threshold of 5.
Run suppressWarnings(chisq.test(audit_tab)) and compare its $expected component with < 5.
Click to reveal solution
Explanation: Five of the nine cells fall below the 5-count threshold (column 2 holds the only cells safely above), so the asymptotic p-value will be unreliable. A common heuristic is that the chi-square approximation needs at least 80% of expected counts of 5 or more; this table fails it. Pivot to Fisher's exact or Monte Carlo simulation, or collapse columns 1 and 3 if they're semantically combinable.
Section 5. Real-world workflows (4 problems)
Exercise 5.1: Build a contingency table from raw row-level survey data
Task: A marketing analyst has 30 customer records stored as a tibble with region (one of "N", "S", "E", "W") and purchase ("yes" or "no") columns and needs a two-way frequency table before running chi-square. Use table() on the two columns of the survey tibble built below and save the resulting 4x2 table to ex_5_1 exactly as table() produces it.
Expected result:
#> purchase
#> region no yes
#> E 3 4
#> N 5 3
#> S 2 6
#> W 4 3Difficulty: Beginner
Long-format rows must be cross-tabulated into a two-way frequency layout before any test can run.
Use table() on the survey$region and survey$purchase columns.
Click to reveal solution
Explanation: table() is the standard bridge between long-format data and the wide contingency-table layout that chisq.test() expects. Passing dnn = c(...) names the rows and columns of the resulting table so downstream printing and residual inspection stay self-documenting. If you prefer named arguments, table(region = survey$region, purchase = survey$purchase) works the same way. Always inspect the table before testing to catch typos in factor levels or unexpected NA rows.
Exercise 5.2: Run a homogeneity test across four regions
Task: A growth team wants to know whether purchase rates differ across four regions, a homogeneity question (different groups, same outcome distribution) that uses the same chisq.test() call as independence. Given the 4x2 matrix below where rows are regions and columns are yes/no counts, run chisq.test() and save the full result to ex_5_2.
Expected result:
#> Pearson's Chi-squared test
#>
#> data: region_tab
#> X-squared = 18.18, df = 3, p-value = 0.0004022Difficulty: Intermediate
A homogeneity question uses the very same machinery as an independence test, so the table goes straight in.
Pass region_tab to chisq.test() with no extra arguments.
Click to reveal solution
Explanation: Independence and homogeneity differ in their sampling story: independence assumes one sample classified two ways, while homogeneity assumes one sample per group with the outcome distribution compared across groups. The arithmetic is identical, which is why R uses one function for both. With p = 0.0004 the regions do not share the same yes/no split; exercise 5.3 finds which pairs drive the result.
Exercise 5.3: Bonferroni-adjust pairwise 2x2 chi-square p-values
Task: A consultant wants pairwise 2x2 chi-square tests between every pair of the four regions in region_tab to localize where purchase rates differ, then Bonferroni-adjust the 6 raw p-values to control the family-wise error rate. Use combn(rownames(region_tab), 2) to enumerate pairs, run chisq.test() on each 2x2 sub-table, run p.adjust(..., method = "bonferroni"), and save the named vector to ex_5_3.
Expected result:
#> N_vs_S N_vs_E N_vs_W S_vs_E S_vs_W E_vs_W
#> 0.0027075050 1.0000000000 0.0941748210 0.0011533230 0.4923404964 0.6121013380Difficulty: Advanced
Localize the difference by testing every region pair, then shrink each result to account for running many tests at once.
Enumerate pairs with combn(), test each two-by-two slice with chisq.test(), and feed the raw p-values to p.adjust(method = "bonferroni").
Click to reveal solution
Explanation: With 6 pairwise tests and a nominal alpha of 0.05, Bonferroni multiplies each raw p-value by 6 and caps the result at 1, so significance requires raw p below 0.0083. Only the N-vs-S and S-vs-E pairs survive: S is the outlier region with the highest purchase rate. Bonferroni is conservative; if many comparisons matter, consider method = "BH" to control the false-discovery rate instead.
Exercise 5.4: Assemble a publication-ready one-line chi-square summary
Task: A journal submission needs a one-line summary of the Titanic Class by Survived test combining the chi-square statistic, degrees of freedom, p-value, and Cramer's V for ex_2_2. Build a single character string formatted as "X^2(df) = stat, p < .001, V = v" with two-decimal rounding using sprintf(), and save the string to ex_5_4.
Expected result:
#> [1] "X^2(3) = 190.40, p < .001, V = 0.29"Difficulty: Advanced
Assemble the statistic, its degrees of freedom, the p-value, and the effect size into one formatted reporting line.
Use sprintf("X^2(%d) = %.2f, p < .001, V = %.2f", df, stat, V), reading the values from ex_2_2.
Click to reveal solution
Explanation: sprintf() produces APA-style reporting with stable formatting that survives copy-paste into a manuscript. Hard-coding p < .001 is acceptable because R prints < 2.2e-16 for very small p-values: keep the literal threshold whenever the actual p drops below 0.001. Wrap this block in a function report_chisq() if you compute many such summaries; pair it with report_chisq_inline() that returns LaTeX-style $X^2$ for R Markdown reports.
What to do next
- Compare two means instead of categorical splits in T-Test Exercises in R.
- Test more than two group means with ANOVA Exercises in R.
- Measure association between numeric variables in Correlation Exercises in R.
- Drill the broader testing toolkit in Hypothesis Testing Exercises in R.
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