Nonparametric Tests Exercises in R: 10 Practice Problems, Solved Step-by-Step

These 10 nonparametric tests exercises in R cover one-sample Wilcoxon signed-rank, two-sample Mann-Whitney U, paired signed-rank, Kruskal-Wallis for three or more groups, post-hoc pairwise comparisons, Hodges-Lehmann confidence intervals, tie handling, and rank-based effect sizes, each one runnable in your browser.

Which non-parametric test matches your design?

One of three R calls covers 90% of rank-based problems. The choice depends only on how your data is laid out: one group, two independent groups, two paired groups, or three-plus groups. Here are all three tests run against the same built-in datasets so you can see the output shape before starting the exercises.

RThree rank-based tests, three data layouts

# Layout 1: one sample, test median vs a claimed value one_res <- wilcox.test(mtcars$mpg, mu = 20) # Layout 2: two independent groups (Mann-Whitney U) set4 <- mtcars$mpg[mtcars$cyl == 4] set8 <- mtcars$mpg[mtcars$cyl == 8] mw_res <- wilcox.test(set4, set8) # Layout 3: three or more groups (Kruskal-Wallis) kw_res <- kruskal.test(Sepal.Width ~ Species, data = iris) c(one_sample = one_res$p.value, mann_whitney = mw_res$p.value, kruskal_wallis = kw_res$p.value) #> one_sample mann_whitney kruskal_wallis #> 4.027e-01 4.192e-06 9.514e-15

Three calls, three p-values, same $p.value field on every result. The one-sample test fails to reject that median mpg equals 20 (p = 0.40). The Mann-Whitney test strongly rejects equal mpg across 4-cyl and 8-cyl cars (p = 4e-06). The Kruskal-Wallis test crushes the null that Sepal.Width is interchangeable across iris species (p = 1e-14). Same function family, same result shape, just different input layouts.

Here is the one-line decision rule:

You have...	Are measurements paired?	R call
One sample + a claimed centre	N/A	`wilcox.test(x, mu = m)`
Two independent groups	No	`wilcox.test(y ~ g)` or `wilcox.test(x, y)`
Two groups, same subjects	Yes	`wilcox.test(x, y, paired = TRUE)`
Three or more groups	No	`kruskal.test(y ~ g)`

Key Insight

Mann-Whitney U and Wilcoxon rank-sum are the same test under two names. R only ships wilcox.test(), which returns a W statistic equal to the Wilcoxon rank sum shifted by a constant. Whether a paper calls the result "U" or "W", the p-value is identical. "Non-parametric" means the test does not assume a specific distribution shape (like normal), not that it is assumption-free. It still assumes independence and, for the paired test, a symmetric distribution of differences.

Try it: A trainer measures blood pressure for 15 patients before and after an intervention. Same patients, two measurements each. Which flag in wilcox.test() must you set? Save "paired" or "mu" to ex_flag.

RYour turn: pick the wilcox.test flag

# Try it: before/after on the same 15 patients ex_flag <- "___" # replace with "paired" or "mu" ex_flag #> Expected: "paired"

Click to reveal solution

RPaired design solution

ex_flag <- "paired" ex_flag #> [1] "paired"

Explanation: Same subjects measured twice is a paired design. Set paired = TRUE and pass the two vectors, wilcox.test(before, after, paired = TRUE). mu is the one-sample null centre, irrelevant when you have two measurements per subject.

How do you read wilcox.test() and kruskal.test() output?

Both functions return a htest list with everything you need for a write-up: the test statistic, p-value, alternative hypothesis, and (for Wilcoxon with conf.int = TRUE) a confidence interval plus Hodges-Lehmann location estimate. Pulling fields by name gives one-line access for report tables, assumption checks, and downstream plots.

RExtract every useful field from wilcox.test and kruskal.test

# Mann-Whitney with confidence interval switched on mw_full <- wilcox.test(set4, set8, conf.int = TRUE) mw_full$statistic #> W #> 117 mw_full$p.value #> [1] 4.192e-06 mw_full$estimate #> difference in location #> 10.50 mw_full$conf.int #> [1] 6.300 14.200 #> attr(,"conf.level") #> [1] 0.95 # Kruskal-Wallis fields (reuse kw_res from earlier) kw_res$statistic #> Kruskal-Wallis chi-squared #> 63.57 kw_res$parameter #> df #> 2 kw_res$p.value #> [1] 9.514e-15

The Wilcoxon $estimate is the Hodges-Lehmann estimator, the median of all pairwise differences between the two samples. It is a rank-based analogue of a mean difference, and its 95% CI of 6.3 to 14.2 mpg is the interval you report instead of mean(set4) - mean(set8). For Kruskal-Wallis, $statistic is H and $parameter is df = groups minus 1. Those three fields plus the p-value are everything a reviewer asks for.

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

Field	wilcox.test	kruskal.test	When to use
`$statistic`	W (rank-sum / signed-rank)	H (chi-sq approximation)	Always report
`$parameter`	not present	df = k − 1	KW write-ups
`$p.value`	Tail probability	Tail probability	The decision number
`$estimate`	Hodges-Lehmann (with `conf.int`)	not present	Effect magnitude
`$conf.int`	95% CI on HL (with `conf.int`)	not present	Uncertainty range

Tip

broom::tidy() turns any R test into a one-row data frame. broom::tidy(mw_full) returns a clean estimate / statistic / p.value / method / alternative row that drops into markdown tables or ggplot panels. Useful when you run the same test across many subsets with purrr::map().

Warning

If you see "cannot compute exact p-value with ties", the asymptotic approximation was used. Rank-based tests assume continuous data so ties shouldn't occur, but real data has them. R silently falls back to a normal approximation (with continuity correction unless you set correct = FALSE). For small samples with many ties, the fallback is inaccurate, consider coin::wilcox_test() for an exact permutation-based alternative.

Try it: Extract just the p-value from a one-sample Wilcoxon test on mtcars$mpg with null median mu = 20. Save it to ex_p, rounded to 4 decimals.

RYour turn: extract the one-sample p-value

# Try it: one-sample wilcox on mtcars$mpg against mu = 20, keep only p, rounded ex_p <- round(wilcox.test(___, mu = ___)$p.value, 4) # fill in the vector and null ex_p #> Expected: 0.4027

Click to reveal solution

ROne-sample p-value solution

ex_p <- round(wilcox.test(mtcars$mpg, mu = 20)$p.value, 4) ex_p #> [1] 0.4027

Explanation: wilcox.test(x, mu = 20) runs the one-sample signed-rank test of whether the pseudo-median of x equals 20. The observed median of mtcars$mpg is 19.2, close enough to 20 that with n = 32 the test cannot reject, p = 0.40.

Practice Exercises

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

Exercise 1: One-sample Wilcoxon signed-rank against a claimed median

Test whether the median weight in mtcars$wt differs from 3.2 (thousand pounds). Save the full result to ex1_res. State the decision at α = 0.05 and explain why the pseudo-median is a better target than the simple median when the distribution is skewed.

RExercise 1 starter: one-sample signed-rank

# Exercise 1: is median mtcars$wt different from 3.2? # Hint: wilcox.test(vec, mu = 3.2, conf.int = TRUE) ex1_vec <- mtcars$wt median(ex1_vec) #> [1] 3.325 # Write your code below:

Click to reveal solution

RExercise 1 solution

ex1_res <- wilcox.test(ex1_vec, mu = 3.2, conf.int = TRUE) ex1_res #> #> Wilcoxon signed rank test with continuity correction #> #> data: ex1_vec #> V = 269, p-value = 0.3073 #> alternative hypothesis: true location is not equal to 3.2 #> 95 percent confidence interval: #> 2.9975 3.6100 #> sample estimates: #> (pseudo)median #> 3.2325

Explanation: The signed-rank test converts each x - 3.2 difference to a signed rank and asks whether the positive and negative ranks balance. V = 269, p = 0.31, so we fail to reject. The 95% CI on the pseudo-median (2.998 to 3.610) contains 3.2, telling the same story from a location-estimate angle. Use the pseudo-median instead of median() when you want a signed-rank-compatible centre for a potentially asymmetric distribution.

Exercise 2: Mann-Whitney U on two independent groups

Using the iris dataset, compare Petal.Length between setosa and versicolor. Save the result to ex2_res. Report W and the p-value.

RExercise 2 starter: Mann-Whitney on iris

# Exercise 2: Petal.Length setosa vs versicolor # Hint: wilcox.test(x, y) when you have two independent numeric vectors ex2_g_set <- iris$Petal.Length[iris$Species == "setosa"] ex2_g_ver <- iris$Petal.Length[iris$Species == "versicolor"] # Write your code below:

Click to reveal solution

RExercise 2 solution

ex2_res <- wilcox.test(ex2_g_set, ex2_g_ver) ex2_res #> #> Wilcoxon rank sum test with continuity correction #> #> data: ex2_g_set and ex2_g_ver #> W = 0, p-value < 2.2e-16 #> alternative hypothesis: true location shift is not equal to 0

Explanation: Setosa petals range 1.0 to 1.9 cm and versicolor petals range 3.0 to 5.1 cm. Zero overlap means every setosa value ranks below every versicolor value, so W = 0 (the minimum possible). The p-value is below .Machine$double.eps, R prints < 2.2e-16. When two samples are perfectly separable, the rank-sum test returns the most extreme p-value it can produce.

Exercise 3: One-tailed Mann-Whitney (direction matters)

Re-run Exercise 2 with the alternative hypothesis "versicolor petals are longer than setosa petals". Save the result to ex3_res. Explain why a one-tailed test halves the p-value when the effect is in the hypothesised direction.

RExercise 3 starter: one-tailed Mann-Whitney

# Exercise 3: one-tailed (versicolor > setosa) # Hint: the first vector is what the alternative says is larger # Write your code below:

Click to reveal solution

RExercise 3 solution

ex3_res <- wilcox.test(ex2_g_ver, ex2_g_set, alternative = "greater") ex3_res #> #> Wilcoxon rank sum test with continuity correction #> #> data: ex2_g_ver and ex2_g_set #> W = 2500, p-value < 2.2e-16 #> alternative hypothesis: true location shift is greater than 0

Explanation: Putting versicolor first in wilcox.test(ver, set, alternative = "greater") tests whether the first sample is stochastically larger than the second. W = 2500 (the maximum possible for 50 × 50 samples), and the one-tailed p-value equals the right-tail area only. With perfect separation it saturates at the floating-point limit. For a less extreme effect, the one-tailed p would be exactly half the two-tailed p when the observed direction matches the alternative.

Tip

**alternative = "two.sided" is the safe default, change it only when the direction is pre-specified.* Picking one-tailed after* peeking at group means inflates false positives and is a common reviewer red flag. Lock in one-tailed from your pre-registration or hypothesis, not from the data.

Exercise 4: Paired Wilcoxon signed-rank on the sleep dataset

R's built-in sleep dataset records extra hours of sleep from the same 10 subjects taking two soporific drugs (groups 1 and 2). Run a paired Wilcoxon signed-rank test and save the result to ex4_res.

RExercise 4 starter: paired sleep data

# Exercise 4: paired signed-rank on sleep # Hint: each ID appears in both groups, pass paired = TRUE ex4_d1 <- sleep$extra[sleep$group == 1] ex4_d2 <- sleep$extra[sleep$group == 2] head(cbind(ex4_d1, ex4_d2)) #> ex4_d1 ex4_d2 #> [1,] 0.7 1.9 #> [2,] -1.6 0.8 #> [3,] -0.2 1.1 #> [4,] -1.2 0.1 #> [5,] -0.1 -0.1 #> [6,] 3.4 4.4 # Write your code below:

Click to reveal solution

RExercise 4 solution

ex4_res <- suppressWarnings(wilcox.test(ex4_d1, ex4_d2, paired = TRUE)) ex4_res #> #> Wilcoxon signed rank test with continuity correction #> #> data: ex4_d1 and ex4_d2 #> V = 0, p-value = 0.009091 #> alternative hypothesis: true location shift is not equal to 0

Explanation: With paired = TRUE, R computes d <- ex4_d1 - ex4_d2 per subject, ranks the |d| values, signs them, and sums the positive ranks into V. Here every subject slept longer on drug 2 than drug 1, so every signed difference is negative, positive ranks sum to V = 0 (the minimum). The asymptotic p-value of 0.0091 rejects "no difference" at α = 0.05, drug 2 outperforms drug 1 in this sample.

Note

Paired vs independent is a design decision, not a data decision. If the same subject contributes to both groups (before/after, left/right, rater/re-rater), it is paired. If subjects are independently sampled into group A vs group B, it is unpaired. Running the wrong variant can flip a significant result either way because pairing removes between-subject variance.

Exercise 5: Tie handling, exact vs approximate p-values

Here is a small dataset with deliberate ties. Run wilcox.test() on it first with defaults (watch for the ties warning), then suppress the exact calculation with exact = FALSE. Save the final result to ex5_res.

RExercise 5 starter: ties in the data

# Exercise 5: ties + exact p-value # Hint: first use tryCatch to capture the warning message, then set exact = FALSE ex5_tied <- c(3, 5, 7, 7, 9, 12, 15) ex5_base <- c(4, 6, 8, 10, 11, 13) # Write your code below:

Click to reveal solution

RExercise 5 solution

# First pass: see the warning ex5_warn <- tryCatch( wilcox.test(ex5_tied, ex5_base), warning = function(w) conditionMessage(w) ) ex5_warn #> [1] "cannot compute exact p-value with ties" # Second pass: switch to asymptotic approximation ex5_res <- wilcox.test(ex5_tied, ex5_base, exact = FALSE) ex5_res #> #> Wilcoxon rank sum test with continuity correction #> #> data: ex5_tied and ex5_base #> W = 22.5, p-value = 0.9362 #> alternative hypothesis: true location shift is not equal to 0

Explanation: The repeated 7 in ex5_tied blocks the exact computation. R falls back to a normal approximation with continuity correction and tie-corrected variance, W = 22.5, p = 0.94. Setting exact = FALSE explicitly makes the fallback deterministic and silences the warning. For small n with heavy ties consider coin::wilcox_test() which runs an exact permutation test.

Exercise 6: Kruskal-Wallis on three groups

Use the built-in PlantGrowth dataset (yields under a control and two treatments, n = 10 each). Save a Kruskal-Wallis test of weight ~ group to ex6_res.

RExercise 6 starter: Kruskal-Wallis

# Exercise 6: weight across three plant growth groups # Hint: kruskal.test(y ~ group, data = ...) ex6_df <- PlantGrowth head(ex6_df, 3) #> weight group #> 1 4.17 ctrl #> 2 5.58 ctrl #> 3 5.18 ctrl # Write your code below:

Click to reveal solution

RExercise 6 solution

ex6_res <- kruskal.test(weight ~ group, data = ex6_df) ex6_res #> #> Kruskal-Wallis rank sum test #> #> data: weight by group #> Kruskal-Wallis chi-squared = 7.9882, df = 2, p-value = 0.01842

Explanation: H = 7.99 on df = 2 gives p = 0.018. At α = 0.05 we reject the null that all three groups share the same location. Kruskal-Wallis tells us somewhere in the three groups there is a difference, but it does not tell us where, that is the job of the post-hoc comparison in Exercise 7.

Exercise 7: Post-hoc pairwise comparisons with BH adjustment

After a significant Kruskal-Wallis, use pairwise.wilcox.test() with p.adjust.method = "BH" (Benjamini-Hochberg) to identify which pairs of plant-growth groups differ. Save the result to ex7_res.

RExercise 7 starter: pairwise wilcox with BH

# Exercise 7: pairwise.wilcox.test across the three groups # Hint: pairwise.wilcox.test(y, g, p.adjust.method = "BH") # Write your code below:

Click to reveal solution

RExercise 7 solution

ex7_res <- suppressWarnings(pairwise.wilcox.test(ex6_df$weight, ex6_df$group, p.adjust.method = "BH")) ex7_res #> #> Pairwise comparisons using Wilcoxon rank sum test with continuity correction #> #> data: ex6_df$weight and ex6_df$group #> #> ctrl trt1 #> trt1 0.310 - #> trt2 0.095 0.027 #> #> P value adjustment method: BH

Explanation: Of the three pairs, only trt1 vs trt2 clears α = 0.05 after BH adjustment (adjusted p = 0.027). The ctrl vs trt2 contrast is borderline (p = 0.095) and ctrl vs trt1 is not significant. BH controls the expected false discovery rate, less conservative than Bonferroni (which would multiply raw p-values by 3). Always report the adjustment method, unadjusted pairwise p-values inflate the false-positive risk.

Tip

Choose p.adjust.method by the error you want to control. "bonferroni" controls family-wise error rate (reject zero true nulls), strict but stable. "holm" does the same with a step-down gain in power. "BH" (Benjamini-Hochberg) controls the false discovery rate, the expected fraction of rejections that are wrong, and is the default in most modern biological and psychological reporting.

Exercise 8: Rank-biserial effect size for Mann-Whitney

The p-value from Exercise 2 said "the two groups differ". Quantify how much with the rank-biserial correlation $r$, computed as $r = 1 - \frac{2U}{n_1 n_2}$ where $U$ is the smaller of the two U statistics. Save ex8_r and classify the magnitude.

RExercise 8 starter: rank-biserial effect size

# Exercise 8: rank-biserial r from the Exercise 2 W stat # Hint: U1 = W; U2 = n1*n2 - U1; use the smaller, then r = 1 - 2*U/(n1*n2) ex8_n1 <- length(ex2_g_set) ex8_n2 <- length(ex2_g_ver) # Write your code below:

Click to reveal solution

RExercise 8 solution

ex8_U1 <- unname(ex2_res$statistic) # W from Exercise 2 ex8_U2 <- ex8_n1 * ex8_n2 - ex8_U1 ex8_U <- min(ex8_U1, ex8_U2) ex8_r <- 1 - 2 * ex8_U / (ex8_n1 * ex8_n2) round(ex8_r, 3) #> [1] 1

Explanation: With U1 = 0, U2 = 2500, the smaller is 0. Plug into the formula: $r = 1 - (2 \cdot 0)/(50 \cdot 50) = 1.0$. Perfect separation maps to $r = 1$, and that lines up with what we saw in Exercise 2. Cohen's thresholds on rank-biserial: 0.1 small, 0.3 medium, 0.5 large. An $r$ of 1.0 is as large as it goes.

Exercise 9: Epsilon-squared effect size for Kruskal-Wallis

A Kruskal-Wallis p-value says there is some difference, but $\varepsilon^2$ quantifies it. Use the formula $\varepsilon^2 = \frac{H (n + 1)}{n^2 - 1}$ on the iris Sepal.Width vs Species Kruskal-Wallis from the opening section. Classify by the convention 0.01 small, 0.08 medium, 0.26 large.

RExercise 9 starter: epsilon-squared for KW

# Exercise 9: epsilon^2 from the iris KW statistic # Hint: reuse kw_res$statistic; n = nrow(iris) ex9_H <- unname(kw_res$statistic) ex9_n <- nrow(iris) # Write your code below:

Click to reveal solution

RExercise 9 solution

ex9_eps <- ex9_H * (ex9_n + 1) / (ex9_n^2 - 1) round(ex9_eps, 3) #> [1] 0.427

Explanation: With $H = 63.57$ and $n = 150$, $\varepsilon^2 = 63.57 \cdot 151 / (150^2 - 1) = 0.427$. That sits well above the 0.26 "large" threshold: Species explains a meaningful share of the rank variance in Sepal.Width. Always pair a KW p-value with $\varepsilon^2$ (or $\eta^2_H$, which is nearly identical); a highly significant KW on a huge n can have a trivial effect.

Key Insight

P-values scale with sample size, effect sizes do not. In a study with 10,000 observations, a Kruskal-Wallis can flag a tiny $\varepsilon^2 = 0.002$ as $p < 10^{-10}$. That is "statistically significant but practically irrelevant". Reporting the rank-biserial $r$ or $\varepsilon^2$ alongside the p-value keeps the reader grounded in how big the effect actually is.

Exercise 10: Full pipeline on chickwts, test → post-hoc → effect size

Real data usually arrives as one row per observation with a categorical grouping. R's built-in chickwts has the weight of 71 chicks across 6 feed types. Put the full workflow together: Kruskal-Wallis, pairwise post-hoc with Holm adjustment, and $\varepsilon^2$.

RExercise 10 starter: chickwts end-to-end

# Exercise 10: full KW pipeline on chickwts # Hint: kruskal.test -> pairwise.wilcox.test (holm) -> epsilon-squared ex10_df <- chickwts table(ex10_df$feed) #> casein horsebean linseed meatmeal soybean sunflower #> 12 10 12 11 14 12 # Write your code below:

Click to reveal solution

RExercise 10 solution

# Step 1: Kruskal-Wallis ex10_res <- kruskal.test(weight ~ feed, data = ex10_df) ex10_res$statistic #> Kruskal-Wallis chi-squared #> 37.343 ex10_res$p.value #> [1] 5.113e-07 # Step 2: pairwise.wilcox.test with Holm ex10_pw <- suppressWarnings( pairwise.wilcox.test(ex10_df$weight, ex10_df$feed, p.adjust.method = "holm") ) round(ex10_pw$p.value, 3) #> casein horsebean linseed meatmeal soybean #> horsebean 0.001 NA NA NA NA #> linseed 0.004 0.096 NA NA NA #> meatmeal 0.674 0.008 0.237 NA NA #> soybean 0.097 0.024 0.674 0.674 NA #> sunflower 0.823 0.001 0.004 0.674 0.097 # Step 3: epsilon-squared ex10_n <- nrow(ex10_df) ex10_H <- unname(ex10_res$statistic) ex10_eps <- ex10_H * (ex10_n + 1) / (ex10_n^2 - 1) round(ex10_eps, 3) #> [1] 0.527

Explanation: Three steps capture the whole story. Kruskal-Wallis says weights differ across feeds ($p = 5 \times 10^{-7}$). Holm-adjusted pairwise comparisons say casein beats horsebean and linseed but ties with meatmeal / soybean / sunflower, and horsebean is clearly the weakest feed. Epsilon-squared of 0.53 flags this as a very large effect, much of the weight variance is explained by feed choice. That three-line sketch, test / post-hoc / effect size, is the template for any categorical comparison in real data.

Complete Example: end-to-end chickwts analysis with reporting

Let's stitch the full workflow, including an assumption check, into one block that mirrors how a real report would read.

REnd-to-end nonparametric analysis on chickwts

# Step 1 - quick sanity check: n, median, IQR per group cw_tab <- aggregate(weight ~ feed, data = chickwts, FUN = function(v) round(c(n = length(v), median = median(v), IQR = IQR(v)), 1)) cw_tab #> feed weight.n weight.median weight.IQR #> 1 casein 12 342.0 80.8 #> 2 horsebean 10 151.5 60.5 #> 3 linseed 12 221.0 94.8 #> 4 meatmeal 11 263.0 84.0 #> 5 soybean 14 248.0 105.5 #> 6 sunflower 12 328.0 50.8 # Step 2 - Shapiro-Wilk per group (sample sizes are small, so err on nonparametric) cw_shapiro <- sapply(split(chickwts$weight, chickwts$feed), function(v) shapiro.test(v)$p.value) round(cw_shapiro, 3) #> casein horsebean linseed meatmeal soybean sunflower #> 0.215 0.506 0.923 0.931 0.506 0.365 # Step 3 - Kruskal-Wallis (the right default when any group is non-normal or small n) cw_kw <- kruskal.test(weight ~ feed, data = chickwts) cw_kw$p.value #> [1] 5.113e-07 # Step 4 - pairwise with Holm + effect size cw_pw_min <- min(suppressWarnings( pairwise.wilcox.test(chickwts$weight, chickwts$feed, p.adjust.method = "holm")$p.value), na.rm = TRUE) cw_eps <- unname(cw_kw$statistic) * (nrow(chickwts) + 1) / (nrow(chickwts)^2 - 1) c(min_adj_p = round(cw_pw_min, 4), epsilon2 = round(cw_eps, 3)) #> min_adj_p epsilon2 #> 0.001 0.527

Warning

Median is not mean, do not swap the research question when you swap the test. Kruskal-Wallis tests whether the distributions (usually summarised by medians or pseudo-medians) differ, not whether the means differ. If your client asked about average weight, explain upfront why you are answering a median-based question, and consider a transformed-data t-test / ANOVA or a robust mean like the Hodges-Lehmann estimator as complements.

A one-paragraph APA-style write-up: A Kruskal-Wallis rank-sum test compared chick weights across six feeds in the chickwts dataset (N = 71). Weight differed significantly across feeds, $H(5) = 37.34$, $p < .001$, with a large effect, $\varepsilon^2 = 0.53$. Holm-adjusted pairwise Wilcoxon rank-sum tests indicated casein- and sunflower-fed chicks weighed significantly more than horsebean-fed chicks (all adjusted $p < .01$), while differences among the remaining feeds were not reliable. That sentence answers three questions at once: is the effect real, is it big, and where does it live.

Summary

Exercise	Test	Key R call	Effect size
1	One-sample signed-rank	`wilcox.test(x, mu)`	n/a
2-3	Mann-Whitney (2- and 1-sided)	`wilcox.test(x, y)`	Rank-biserial $r$
4	Paired signed-rank	`wilcox.test(x, y, paired = TRUE)`	n/a
5	Tie handling	`wilcox.test(..., exact = FALSE)`	n/a
6	Kruskal-Wallis	`kruskal.test(y ~ g)`	Epsilon-squared
7	Post-hoc pairwise	`pairwise.wilcox.test(y, g, "BH")`	n/a
8-9	Effect sizes	Formulas above	$r$, $\varepsilon^2$
10	Full pipeline	KW + post-hoc + $\varepsilon^2$	$\varepsilon^2$

Three principles carry across every exercise: pick the test from the design (one / two / paired / many groups); always pair the p-value with an effect size so the reader knows how big the effect is, not just whether it exists; and adjust pairwise post-hoc p-values so your family-wise error or false discovery rate is under control.

References

R Core Team. wilcox.test, R Stats Reference. Link
R Core Team. kruskal.test, R Stats Reference. Link
Hollander, M., Wolfe, D. A., and Chicken, E. Nonparametric Statistical Methods, 3rd edition. Wiley (2013).
Conover, W. J. Practical Nonparametric Statistics, 3rd edition. Wiley (1999).
Wilcoxon, F. "Individual Comparisons by Ranking Methods." Biometrics Bulletin 1(6), 80-83 (1945).
Mann, H. B. and Whitney, D. R. "On a test of whether one of two random variables is stochastically larger than the other." Annals of Mathematical Statistics 18(1), 50-60 (1947).
Kruskal, W. H. and Wallis, W. A. "Use of Ranks in One-Criterion Variance Analysis." Journal of the American Statistical Association 47(260), 583-621 (1952).
Tomczak, M. and Tomczak, E. "The need to report effect size estimates revisited." Trends in Sport Sciences 21(1), 19-25 (2014).

Continue Learning

Wilcoxon, Mann-Whitney, and Kruskal-Wallis in R: Non-Parametric When Normality Fails covers the parent theory, decision rules, and assumption checks in detail.
t-Test Exercises in R is the parametric counterpart, use it when the normality assumption holds.
Hypothesis Testing Exercises in R gives broader inference practice across proportions, means, and non-parametrics with the same drill-sheet format.

Nonparametric Tests Exercises in R: 10 Practice Problems, Solved Step-by-Step

Which non-parametric test matches your design?

How do you read wilcox.test() and kruskal.test() output?

Practice Exercises

Exercise 1: One-sample Wilcoxon signed-rank against a claimed median

Exercise 2: Mann-Whitney U on two independent groups

Exercise 3: One-tailed Mann-Whitney (direction matters)

Exercise 4: Paired Wilcoxon signed-rank on the sleep dataset

Exercise 5: Tie handling, exact vs approximate p-values

Exercise 6: Kruskal-Wallis on three groups

Exercise 7: Post-hoc pairwise comparisons with BH adjustment

Exercise 8: Rank-biserial effect size for Mann-Whitney

Exercise 9: Epsilon-squared effect size for Kruskal-Wallis

Exercise 10: Full pipeline on chickwts, test → post-hoc → effect size

Complete Example: end-to-end chickwts analysis with reporting

Summary

References

Continue Learning

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