Chi-Square Test Exercises in R: 10 Independence & Goodness-of-Fit Problems, Solved Step-by-Step

These 10 chi-square test exercises in R cover both goodness-of-fit and independence problems with runnable solutions, so you practise picking the right variant, reading the output, inspecting residuals, computing Cramér's V, and falling back to Fisher's exact when expected counts are small.

Which chi-square test matches your data layout?

One chi-square function in R, two very different questions. chisq.test() runs a goodness-of-fit test when you hand it a single count vector, and a test of independence when you hand it a two-way table. The decision hinges on whether you have one categorical variable or two. Here is the same function used both ways on two tiny datasets so you can see both calls before drilling into the 10 exercises.

ROne function, two chi-square questions

# Question 1 (goodness-of-fit): are these coin flips fair? coin_flips <- c(heads = 45, tails = 55) gof_res <- chisq.test(coin_flips, p = c(0.5, 0.5)) gof_res$p.value #> [1] 0.3173 # Question 2 (independence): is treatment related to outcome? toy_tab <- matrix(c(20, 10, 15, 25), nrow = 2, dimnames = list(drug = c("A", "B"), outcome = c("ok", "fail"))) ind_res <- chisq.test(toy_tab) ind_res$p.value #> [1] 0.01428

Both calls use the exact same function, and both return an object with the same fields ($statistic, $parameter, $p.value, $expected, $residuals, $stdres). The coin-flip p-value of 0.32 fails to reject the fair-coin hypothesis, while the 2×2 treatment table p-value of 0.014 flags a real association between drug and outcome. One function, one workflow, one result shape. The shape of the input is what switches between the two tests.

Here is the one-line decision rule:

You have...	Question	R call
One categorical variable + an expected distribution	Goodness of fit	`chisq.test(counts, p = expected)`
Two categorical variables in a two-way table	Independence / homogeneity	`chisq.test(two_way_table)`

Key Insight

A chi-square test of independence on a 2×2 table gives the same p-value as a two-proportion z-test squared. They are two faces of the same comparison. The chi-square statistic equals $z^2$, and chisq.test() without Yates' correction matches prop.test(..., correct = FALSE) exactly. Pick whichever framing communicates the result best for your audience.

Try it: A bag of M&Ms is supposed to contain Brown/Yellow/Red/Blue/Orange/Green in equal proportions. You open a bag and count the colours. Do you want a goodness-of-fit test or a test of independence? Set ex_choice to "gof" or "independence".

RYour turn: pick the test

# Try it: which test fits the M&M scenario? ex_choice <- "___" # replace with "gof" or "independence" ex_choice #> Expected: "gof"

Click to reveal solution

RM&M scenario solution

ex_choice <- "gof" ex_choice #> [1] "gof"

Explanation: You have one categorical variable (colour) with 6 levels, and an expected distribution (equal proportions). That's textbook goodness-of-fit. If you instead compared colour counts across two different bags or two factories, that would be independence/homogeneity.

How do you read chi-square output in R?

chisq.test() returns a list with everything you need: the test statistic, degrees of freedom, p-value, observed and expected counts, Pearson residuals, and standardised residuals. Most of the interesting reporting lives inside those fields, not in the printed block. Pulling them out by name gives one-line access for write-ups, assumption checks, and residual plots.

RExtract every useful field from a chi-square test

# Independence test on HairEyeColor collapsed over Sex he_tab <- margin.table(HairEyeColor, c(1, 2)) he_res <- chisq.test(he_tab) # The five fields you'll actually report he_res$statistic #> X-squared #> 138.2898 he_res$parameter #> df #> 9 he_res$p.value #> [1] 2.325027e-25 round(he_res$expected, 1)[1:2, ] #> Eye #> Hair Brown Blue Hazel Green #> Black 40.1 39.2 16.9 10.8 #> Brown 106.3 103.9 44.7 28.2 round(he_res$stdres, 2)[1:2, ] #> Eye #> Hair Brown Blue Hazel Green #> Black 6.21 -4.54 -0.57 -2.25 #> Brown 2.99 -4.77 2.28 -0.55

A statistic of 138.3 on 9 df (rows−1 × cols−1 = 3 × 3) gives a p-value of about $2.3 \times 10^{-25}$: hair and eye colour are not independent. The expected-count matrix shows what the null hypothesis would predict. If rows and columns were independent, Black/Brown-eyed would sit near 40 instead of the observed 68. The standardised residuals pin the effect to specific cells: Black/Brown is +6.2 standard deviations above expectation, Blond/Blue (not shown in this slice) is the other driver. You will reproduce that in Exercise 8.

Here are the five fields you will reach for most often:

Field	What it is	When to use
`$statistic`	The chi-square statistic $X^2$	Report alongside df and p-value
`$parameter`	Degrees of freedom	Needed to compute or verify p manually
`$p.value`	Two-sided tail probability	The decision number
`$expected`	Expected counts under H₀	Assumption check (all ≥ 5)
`$stdres`	Standardised residuals	Find cells that drive the effect,	z	> 2

Tip

broom::tidy() turns any R test into a one-row data frame. broom::tidy(he_res) gives a neat statistic / p.value / parameter / method row that slots straight into reports, markdown tables, or a bigger grid of models. Great when you are running the same test across many subsets.

Warning

If chisq.test() prints "Chi-squared approximation may be incorrect", stop and check $expected. That warning appears when any expected cell is below 5 and the large-sample approximation is unreliable. Switch to simulate.p.value = TRUE (any size) or fisher.test() (2×2) instead of reporting the printed number.

Try it: Extract just the p-value from a goodness-of-fit test on mtcars$cyl against the uniform probabilities c(1/3, 1/3, 1/3). Save it to ex_p.

RYour turn: extract the cyl GOF p-value

# Try it: GOF on mtcars$cyl against uniform, keep only the p-value ex_p <- chisq.test(table(mtcars$cyl), p = ___)$p.value # fill in the p vector ex_p #> Expected: a p-value near 0.32 (you will fail to reject uniform)

Click to reveal solution

Rcyl GOF solution

ex_p <- chisq.test(table(mtcars$cyl), p = c(1/3, 1/3, 1/3))$p.value ex_p #> [1] 0.3149

Explanation: table(mtcars$cyl) gives counts 11, 7, 14 for 4/6/8 cylinders across 32 cars. Expected uniform = 10.67 each. The chi-square statistic is about 2.31 on 2 df, and $p = 0.31$, so we fail to reject uniform. The 32-car sample is simply too small to rule out uniformity even though the observed counts look uneven.

Practice Exercises

Ten capstone problems, ordered roughly easier → harder. Every exercise uses a distinct ex<N>_ prefix so solutions do not clobber earlier tutorial state.

Exercise 1: Goodness-of-fit with equal expected counts

Test whether the 4/6/8 cylinder distribution in mtcars is uniform. Save the test object to ex1_res and print it. State the decision against α = 0.05.

RExercise 1 starter: mtcars$cyl uniform GOF

# Exercise 1: GOF on mtcars$cyl vs uniform # Hint: build the table with table(), pass p = rep(1/3, 3) ex1_tab <- table(mtcars$cyl) ex1_tab #> 4 6 8 #> 11 7 14 # Write your code below:

Click to reveal solution

RExercise 1 solution

ex1_tab <- table(mtcars$cyl) ex1_res <- chisq.test(ex1_tab, p = rep(1/3, 3)) ex1_res #> #> Chi-squared test for given probabilities #> #> data: ex1_tab #> X-squared = 2.3125, df = 2, p-value = 0.3148

Explanation: Expected counts are 10.67 each. The observed 11/7/14 are within about ±3 of expected, so the statistic is modest (2.31) and p = 0.31. We fail to reject uniformity. The lesson: an uneven-looking sample is not automatically evidence against uniformity, small samples demand more disagreement before rejecting.

Exercise 2: Goodness-of-fit with unequal expected (Mendelian 9:3:3:1)

Classic pea-plant data: in a dihybrid cross, Mendelian theory predicts the four phenotypes in ratio 9:3:3:1. Observed counts are c(315, 108, 101, 32). Test whether the observed counts are consistent with Mendel's ratio.

RExercise 2 starter: Mendel 9:3:3:1 GOF

# Exercise 2: GOF against 9:3:3:1 # Hint: convert 9:3:3:1 to probabilities by dividing by 16 ex2_obs <- c(315, 108, 101, 32) ex2_exp <- c(9, 3, 3, 1) / 16 # Write your code below:

Click to reveal solution

RExercise 2 solution

ex2_obs <- c(315, 108, 101, 32) ex2_exp <- c(9, 3, 3, 1) / 16 ex2_res <- chisq.test(ex2_obs, p = ex2_exp) ex2_res #> #> Chi-squared test for given probabilities #> #> data: ex2_obs #> X-squared = 0.470, df = 3, p-value = 0.9254

Explanation: Total n = 556. Expected counts under 9:3:3:1 are 312.75, 104.25, 104.25, 34.75, within fractions of the observed values. The p-value of 0.93 says the data are fully consistent with Mendel's ratio. This dataset is famous because Fisher later argued the fit was too good, but for our purposes it's a textbook pass.

Exercise 3: Goodness-of-fit with a Monte Carlo p-value

You have 6 weekday-preference counts from a small survey: c(3, 4, 2, 5, 3, 2, 1) for Monday through Sunday. With such small expected counts (average ≈ 2.9), the chi-square approximation is unreliable. Use a Monte Carlo p-value via simulate.p.value = TRUE and B = 10000.

RExercise 3 starter: Monte Carlo GOF

# Exercise 3: Monte Carlo p-value for a sparse GOF ex3_obs <- c(3, 4, 2, 5, 3, 2, 1) # Write your code below:

Click to reveal solution

RExercise 3 solution

set.seed(503) ex3_obs <- c(3, 4, 2, 5, 3, 2, 1) ex3_res <- chisq.test(ex3_obs, p = rep(1/7, 7), simulate.p.value = TRUE, B = 10000) ex3_res #> #> Chi-squared test for given probabilities with simulated p-value #> (based on 10000 replicates) #> #> data: ex3_obs #> X-squared = 3.8, df = NA, p-value = 0.7359

Explanation: Under uniform preference each expected count is 2.86, below the usual ≥5 threshold. The simulated p-value (~0.74) says observed counts at least this uneven would happen about 74% of the time under uniform, no evidence for a weekday preference.

Note

Monte Carlo p-values are approximate and depend on B. With B = 10000 the Monte Carlo error is roughly $1/\sqrt{10000} = 0.01$; raise B for tighter estimates. df prints as NA because the simulated test does not use the asymptotic chi-square distribution.

Exercise 4: 2×2 independence with Yates' correction (default)

From HairEyeColor, build a 2×2 table of Hair (Black vs Blond) × Sex (Male vs Female). Run chisq.test() with default settings (Yates' continuity correction is applied automatically for 2×2). Report the statistic, df, and p-value.

RExercise 4 starter: 2x2 independence with Yates

# Exercise 4: 2x2 Hair(Black,Blond) x Sex # Hint: margin.table over Eye, then slice to rows Black + Blond ex4_tab <- margin.table(HairEyeColor, c(1, 3))[c("Black", "Blond"), ] # Write your code below:

Click to reveal solution

RExercise 4 solution

ex4_tab <- margin.table(HairEyeColor, c(1, 3))[c("Black", "Blond"), ] ex4_tab #> Sex #> Hair Male Female #> Black 56 52 #> Blond 46 81 ex4_res <- chisq.test(ex4_tab) ex4_res #> #> Pearson's Chi-squared test with Yates' continuity correction #> #> data: ex4_tab #> X-squared = 5.396, df = 1, p-value = 0.02019

Explanation: Among Black-haired people the sample is slightly male-leaning (56/108), while Blonds skew female (81/127). The Yates-corrected p-value of 0.020 rejects independence at α = 0.05, there is a Hair × Sex association in this classic dataset.

Exercise 5: Same 2×2 without Yates' correction

Re-run the test from Exercise 4 with correct = FALSE. Compare the statistic and p-value, which direction does the correction move them, and by how much?

RExercise 5 starter: 2x2 without Yates

# Exercise 5: same ex4_tab, correct = FALSE # Hint: reuse ex4_tab from Exercise 4 # Write your code below:

Click to reveal solution

RExercise 5 solution

ex5_res <- chisq.test(ex4_tab, correct = FALSE) ex5_res #> #> Pearson's Chi-squared test #> #> data: ex4_tab #> X-squared = 5.989, df = 1, p-value = 0.01441 c(yates_on = 0.02019, yates_off = ex5_res$p.value) #> yates_on yates_off #> 0.02019 0.01441

Explanation: Yates' correction shrinks each observed-minus-expected difference by 0.5 before squaring, which lowers the statistic (from 5.99 to 5.40) and raises the p-value (from 0.014 to 0.020). The correction makes the 2×2 test more conservative, easier to miss real effects, harder to raise false alarms. For small 2×2 tables the correction is standard; for large ones it barely matters.

Exercise 6: R×C independence on mtcars (cyl × gear)

Build the table of cylinders × gears from mtcars using table(), and test independence. Expect a small-count warning, this exercise sets up Exercise 7.

RExercise 6 starter: mtcars cyl x gear

# Exercise 6: independence on mtcars cylinders vs gears ex6_tab <- table(cyl = mtcars$cyl, gear = mtcars$gear) # Write your code below:

Click to reveal solution

RExercise 6 solution

ex6_tab #> gear #> cyl 3 4 5 #> 4 1 8 2 #> 6 2 4 1 #> 8 12 0 2 ex6_res <- suppressWarnings(chisq.test(ex6_tab)) ex6_res #> #> Pearson's Chi-squared test #> #> data: ex6_tab #> X-squared = 18.04, df = 4, p-value = 0.001214

Explanation: The 3×3 table has several small cells (the 8 cyl / 4 gear cell is 0), so R warns "Chi-squared approximation may be incorrect". The point estimate (p = 0.0012) still suggests strong dependence, eight-cylinder engines pair with 3-speed gearboxes, four-cylinders with 4-speeds, but treat the exact p-value with caution. A Monte Carlo variant would confirm whether the effect is robust to the small-cell issue.

Exercise 7: Extract observed, expected, residuals into a tidy data frame

Take the result object from Exercise 6 (ex6_res) and pull observed, expected, residuals, and stdres into a single tidy data frame with row and column labels. The goal is a report-ready per-cell breakdown.

RExercise 7 starter: tidy per-cell components

# Exercise 7: per-cell tidy summary for ex6_res # Hint: as.data.frame() on each matrix, cbind or tibble::tibble # Write your code below:

Click to reveal solution

RExercise 7 solution

ex7_df <- data.frame( cyl = rep(rownames(ex6_res$observed), times = ncol(ex6_res$observed)), gear = rep(colnames(ex6_res$observed), each = nrow(ex6_res$observed)), observed = as.vector(ex6_res$observed), expected = round(as.vector(ex6_res$expected), 2), residual = round(as.vector(ex6_res$residuals), 2), stdres = round(as.vector(ex6_res$stdres), 2) ) head(ex7_df, 4) #> cyl gear observed expected residual stdres #> 1 4 3 1 5.16 -1.83 -3.12 #> 2 6 3 2 3.28 -0.71 -1.18 #> 3 8 3 12 6.56 2.12 3.82 #> 4 4 4 8 4.12 1.91 3.15

Explanation: Flattening the four matrices into one data frame gives you every diagnostic per cell at a glance. The largest positive stdres (+3.82 for 8 cyl / 3 gear) and the largest negative (-3.12 for 4 cyl / 3 gear) are the cells that produce the significant overall p-value. That's the information you cannot recover from print(ex6_res) alone.

Exercise 8: Find the residual hotspots in HairEyeColor

Build the 4×4 Hair × Eye table from HairEyeColor collapsed over Sex, run an independence test, and return the hair-eye combinations where the standardised residual exceeds 2 in absolute value. Save the hotspots to ex8_hot.

RExercise 8 starter: residual hotspots

# Exercise 8: which hair-eye combinations drive the p-value? # Hint: which(abs(stdres) > 2, arr.ind = TRUE) ex8_tab <- margin.table(HairEyeColor, c(1, 2)) # Write your code below:

Click to reveal solution

RExercise 8 solution

ex8_tab <- margin.table(HairEyeColor, c(1, 2)) ex8_res <- chisq.test(ex8_tab) ex8_idx <- which(abs(ex8_res$stdres) > 2, arr.ind = TRUE) ex8_hot <- data.frame( hair = rownames(ex8_tab)[ex8_idx[, "row"]], eye = colnames(ex8_tab)[ex8_idx[, "col"]], stdres = round(ex8_res$stdres[ex8_idx], 2) ) ex8_hot[order(-abs(ex8_hot$stdres)), ] #> hair eye stdres #> 7 Blond Blue 11.16 #> 6 Blond Brown -9.42 #> 1 Black Brown 6.21 #> 2 Brown Brown 2.99 #> 3 Black Blue -4.54 #> 5 Red Green 2.69 #> 4 Black Green -2.25 #> 8 Blond Hazel -2.55

Explanation: The biggest hotspot, Blond+Blue with stdres = +11.2, says that combination appears about 11 standard deviations more often than independence would predict. Reporting the three or four biggest residuals alongside the test statistic is good practice, it turns a number into an interpretable finding.

Key Insight

The overall chi-square statistic tells you the whole table disagrees with independence, but only residuals tell you where. A p-value of $10^{-25}$ is just one number. The residuals matrix is the story: Blond+Blue over-represented, Blond+Brown under-represented, and Black+Brown over-represented. Without residuals you have significance but no narrative.

Exercise 9: Effect size, Cramér's V on HairEyeColor

The chi-square statistic grows with sample size, so "significant" ≠ "meaningful". Cramér's V rescales it to the range \[0, 1\], interpretable the way a correlation is. Compute V for the HairEyeColor Hair × Eye table using the formula below and interpret against Cohen's thresholds.

The formula is:

$$V = \sqrt{\dfrac{\chi^2}{n \cdot (k - 1)}}$$

Where:

$\chi^2$ is the test statistic from chisq.test()
$n$ is the total sample size, $\sum n_{ij}$
$k$ is $\min(\text{rows}, \text{cols})$

RExercise 9 starter: Cramér's V

# Exercise 9: compute Cramér's V from ex8_res # Hint: sqrt(statistic / (n * (min(dim(tab)) - 1))) # Write your code below:

Click to reveal solution

RExercise 9 solution

ex9_v <- sqrt(unname(ex8_res$statistic) / (sum(ex8_tab) * (min(dim(ex8_tab)) - 1))) round(ex9_v, 3) #> [1] 0.279

Explanation: With $\chi^2 = 138.3$, $n = 592$, and $k = 4$, the formula gives $V = \sqrt{138.3 / (592 \cdot 3)} = 0.279$. Reporting V alongside the p-value prevents the common trap of mistaking tiny, significant effects in huge samples for practically important ones.

Tip

Cramér's V reads like a correlation, with conventional thresholds. For $\text{df}^* = \min(r,c) - 1 = 3$, Cohen's rules of thumb are 0.06 (small), 0.17 (medium), 0.29 (large). V = 0.28 lands in the "large" band for this table size, the hair-eye effect is real, practically meaningful, and not just a consequence of the n = 592 sample.

Exercise 10: Raw data → `xtabs()` → chi-square

Until now every exercise started from a pre-built table. Real data usually arrives one row per observation. Build the table with xtabs(), then run the independence test on it.

RExercise 10 starter: xtabs pipeline

# Exercise 10: raw data frame -> xtabs -> chisq.test # Hint: xtabs(~ var1 + var2, data = df) ex10_df <- data.frame( smoker = rep(c("yes", "no"), each = 50), disease = c(rep("yes", 30), rep("no", 20), rep("yes", 12), rep("no", 38)) ) head(ex10_df) #> smoker disease #> 1 yes yes #> 2 yes yes #> 3 yes yes #> 4 yes yes #> 5 yes yes #> 6 yes yes # Write your code below:

Click to reveal solution

RExercise 10 solution

ex10_tab <- xtabs(~ smoker + disease, data = ex10_df) ex10_tab #> disease #> smoker no yes #> no 38 12 #> yes 20 30 ex10_res <- chisq.test(ex10_tab) ex10_res #> #> Pearson's Chi-squared test with Yates' continuity correction #> #> data: ex10_tab #> X-squared = 12.57, df = 1, p-value = 0.000392

Explanation: xtabs() takes a formula and a data frame and returns a contingency table in exactly the shape chisq.test() wants. This is the pipeline for any real-world categorical analysis: tidy long data → xtabs() → chisq.test(). The synthetic cohort here gives a Yates-corrected p-value below 0.001, strong evidence that smoking and disease are associated.

Complete Example: Hair × Eye Colour Survey

Let's stitch everything together: build the table, check assumptions, run the test, quantify the effect size, and localise the drivers.

REnd-to-end independence analysis on HairEyeColor

# Step 1 - collapse over Sex to a 2-way Hair x Eye table he_full <- margin.table(HairEyeColor, c(1, 2)) he_full #> Eye #> Hair Brown Blue Hazel Green #> Black 68 20 15 5 #> Brown 119 84 54 29 #> Red 26 17 14 14 #> Blond 7 94 10 16 # Step 2 - run the test he_chi <- chisq.test(he_full) # Step 3 - assumption check: every expected count >= 5? min(he_chi$expected) #> [1] 5.524 # Step 4 - effect size (Cramér's V) he_v <- sqrt(unname(he_chi$statistic) / (sum(he_full) * (min(dim(he_full)) - 1))) round(he_v, 3) #> [1] 0.279 # Step 5 - hotspots: |stdres| > 2 he_hot <- which(abs(he_chi$stdres) > 2, arr.ind = TRUE) head(data.frame(hair = rownames(he_full)[he_hot[, 1]], eye = colnames(he_full)[he_hot[, 2]], stdres = round(he_chi$stdres[he_hot], 2)), 4) #> hair eye stdres #> 1 Black Brown 6.21 #> 2 Brown Brown 2.99 #> 3 Black Blue -4.54 #> 4 Blond Blue 11.16

Warning

Always check expected counts before trusting the p-value. Here every expected count is ≥ 5.5, so the asymptotic chi-square is safe. Skipping this step is the most common reporting error in introductory categorical analyses.

A one-paragraph APA-style write-up: A chi-square test of independence examined whether hair colour and eye colour were associated in the 592-person HairEyeColor dataset. The relationship was highly significant, $\chi^2(9, N = 592) = 138.3, p < .001$, with a large effect size, $V = 0.28$. Examination of standardised residuals indicated that the Blond/Blue combination was strongly over-represented (stdres = +11.2) while Blond/Brown was under-represented (stdres = −9.4). That sentence answers three questions at once: is the effect real, is it big, and where does it live.

Summary

Ask	R call	Key field
Goodness of fit	`chisq.test(x, p = expected)`	`$p.value`
Independence / homogeneity	`chisq.test(two_way_table)`	`$p.value`
2×2 without Yates	`chisq.test(tab, correct = FALSE)`	`$statistic`
Small expected counts	`chisq.test(..., simulate.p.value = TRUE, B = 10000)`	`$p.value`
Effect size	`sqrt(chisq / (n * (min(dim) - 1)))`	Cramér's V
Cell-level pattern	`$stdres`, flag `abs(stdres) > 2`	standardised residuals

References

R Core Team. chisq.test, R Stats Reference. Link
Agresti, A. Categorical Data Analysis, 3rd edition. Wiley (2013).
Sheskin, D. J. Handbook of Parametric and Nonparametric Statistical Procedures, 5th edition. Chapman & Hall/CRC (2011).
Mendel, G. Experiments in Plant Hybridisation (1866), source of the classic 9:3:3:1 phenotype data.
Yates, F. "Contingency Tables Involving Small Numbers and the χ² Test." Journal of the Royal Statistical Society Supplement 1(2), 1934.
Cramér, H. Mathematical Methods of Statistics. Princeton University Press (1946).
Fisher, R. A. Statistical Methods for Research Workers, 14th edition. Oliver & Boyd (1970).

Continue Learning

Chi-Square Tests in R: Independence, Goodness-of-Fit & Effect Size covers the parent theory, assumption derivations, residual analysis, and Cramér's V in detail.
Hypothesis Testing Exercises in R gives broader inference practice across t-tests, proportions, non-parametrics, and power analysis.
t-Test Exercises in R uses the same drill-sheet format for comparing means instead of proportions.

Chi-Square Test Exercises in R: 10 Independence & Goodness-of-Fit Problems, Solved Step-by-Step

Which chi-square test matches your data layout?

How do you read chi-square output in R?

Practice Exercises

Exercise 1: Goodness-of-fit with equal expected counts

Exercise 2: Goodness-of-fit with unequal expected (Mendelian 9:3:3:1)

Exercise 3: Goodness-of-fit with a Monte Carlo p-value

Exercise 4: 2×2 independence with Yates' correction (default)

Exercise 5: Same 2×2 without Yates' correction

Exercise 6: R×C independence on mtcars (cyl × gear)

Exercise 7: Extract observed, expected, residuals into a tidy data frame

Exercise 8: Find the residual hotspots in HairEyeColor

Exercise 9: Effect size, Cramér's V on HairEyeColor

Exercise 10: Raw data → xtabs() → chi-square

Complete Example: Hair × Eye Colour Survey

Summary

References

Continue Learning

Related Tutorials

Exercise 10: Raw data → `xtabs()` → chi-square