Nonparametric Tests Exercises in R: 18 Practice Problems

Explanation: The one-sample Wilcoxon ranks the absolute differences $|x_i - \mu_0|$, then sums the ranks for the positive differences to form $V$. A p-value of 0.64 means there is no evidence that median mpg differs from 20. The continuity correction is the default for $n \geq 50$ and small-$n$ situations with ties; suppress it with correct = FALSE if you want the raw approximation. Common mistake: passing mu as a vector of two values; it must be a single scalar.

Explanation: Every htest object is just a list, so you reach into it with $statistic and $p.value. The unname() strip is important because res$statistic keeps the name "V" attached, and concatenating it under your own V = label would otherwise produce the double name V.V. Building a tidy two-element vector like this is the idiom for stuffing test results into a summarise() output or a knitted table without juggling list columns.

Explanation: A one-sided test redistributes the rejection region into a single tail, so the relationship between the two-sided and one-sided p-values is $p_{one} = p_{two}/2$ when the observed direction matches the alternative, and $p_{one} = 1 - p_{two}/2$ when it does not. Here the sample median is below 9, so testing for "greater" gives a p-value near 1. Always choose the alternative before peeking at the data; flipping it after seeing the sign inflates your Type I error.

Explanation: wilcox.test(x, y) returns the same Mann-Whitney $U$ statistic that older textbooks tabulate, only shifted by a constant: R reports it as $W = U$ for the first sample. A p-value of 0.0019 strongly rejects the null that the two distributions are interchangeable. The output says "location shift", not "median", because the test is sensitive to any stochastic dominance, not strictly to a difference in medians. For a true median comparison you need stronger assumptions (symmetric, identically shaped distributions).

Explanation: The formula interface determines group ordering by sort(unique(am)), which here is c(0, 1), so the "first sample" is automatic transmissions: the same order as the two-vector call in 2.1. That ordering matters because $W$ is computed from the first sample's ranks. If your grouping variable is a factor with relabelled levels, double-check the level order with levels(factor(am)) before interpreting the sign of the rank-biserial effect size.

Explanation: The Hodges-Lehmann estimator is the median of all pairwise differences $x_i - y_j$ across the two samples, and the CI is built by inverting the rank-sum test. It is the natural location-shift summary to pair with a Mann-Whitney p-value because both come from the same rank structure. With ties present R uses an asymptotic CI and emits a warning; for tie-free small samples ask for exact = TRUE to get the discrete exact CI.

Explanation: A paired Wilcoxon is just a one-sample signed-rank test on the within-subject differences day9 - day0, with $\mu_0 = 0$. Pairing strips out between-subject variability, so it almost always has more power than the two-sample Mann-Whitney when each subject contributes both measurements. Common mistake: passing the two vectors in the wrong subject order so that pairs do not align; always sanity-check length(x) == length(y) and that row 1 of each vector belongs to the same subject.

Explanation: Choosing alternative = "greater" here means the test asks whether bp_pre > bp_post typically, i.e. whether the intervention lowered pressure. The direction is x minus y, so flipping the argument order flips the alternative. With 12 paired observations and clearly directional differences, the test reaches $p < 0.005$. If you forgot paired = TRUE, the test would default to Mann-Whitney and treat the 24 values as 12 independent pairs of strangers, throwing away the within-subject correlation and severely losing power.

Explanation: The trick is filtering to chicks that have BOTH a Day 0 and a Day 21 record: some chicks died mid-study and only the survivors are paired. table(diet1$Chick) == 2 selects exactly those IDs. Sorting by Chick then Time is what aligns the two vectors so position $i$ in start and end come from the same chick. Without this filter you would pass mismatched-length vectors and wilcox.test() would error out, or worse pair the wrong subjects and silently return a wrong p-value if the lengths happened to match.

Explanation: Kruskal-Wallis is the rank-based analogue of one-way ANOVA: it pools all observations, ranks them, then asks whether the average ranks differ across groups. The statistic is approximately $\chi^2$-distributed with $k - 1$ degrees of freedom, where $k$ is the number of groups (so 2 here, with three species). A p-value of $\sim 10^{-14}$ rejects the null that all three species share the same Sepal.Width distribution, but Kruskal-Wallis is an omnibus test: it does NOT tell you which pair drives the difference. That is what exercise 5.1 will handle.

Explanation: interaction() collapses two factors into a single factor whose levels are every combination, which is what Kruskal-Wallis needs because it cannot represent two crossed factors directly. The downside is you lose the ability to disentangle a main supp effect from a dose effect, so a significant omnibus result here means "some cell differs from some other cell" and nothing more. For a proper two-factor rank-based analysis use the aligned rank transform via the ARTool package, or fit lm() on ranks and test interaction terms with anova().

Explanation: kruskal.test() silently drops rows with NA in either the response or the grouping factor: an attractive default for daily air-quality measurements that have gaps. Always sanity check sum(is.na(airquality$Ozone)) so you know how much data the test actually used; here 37 of 153 days drop out. With 4 degrees of freedom (5 months, $k - 1 = 4$) the chi-squared of 29.27 yields $p < 10^{-5}$, confirming that ozone distributions are not exchangeable across summer months.

Explanation: Bonferroni multiplies each raw p-value by the number of comparisons (3 here for 3 species), capping at 1. It is the strictest of the common adjustments and the safest choice when you genuinely care about family-wise error rate. If you have many groups and want more power, switch to "holm" (the default), which is uniformly more powerful than Bonferroni and still controls the family-wise rate. Reading the matrix: every off-diagonal cell is a pairwise adjusted p-value, and the - slots are the upper triangle (omitted because the matrix is symmetric).

Explanation: The rank-biserial correlation ranges from $-1$ to $+1$ and encodes the probability that a random observation from group 1 exceeds a random observation from group 2, rescaled. The sign convention here means "negative implies group 1 (cyl4) tends to be LOWER ranked than group 2 (cyl8)", though in this dataset the formula's sign output is opposite to that intuition because $W$ is huge: cyl4 cars dominate cyl8 cars on mpg, so $W$ is close to $n_1 n_2$ and $r_{rb}$ approaches $-1$. Pair the effect size with the p-value; the rstatix package wraps this calculation in wilcox_effsize() if you prefer not to write it out.

Explanation: Epsilon-squared is the rank-based analogue of $\eta^2$ from one-way ANOVA: it estimates the proportion of variance in the ranks explained by the grouping factor. The 0.43 value here is a large effect by Cohen's rough conventions ($\geq 0.26$). Unlike eta-squared on raw values, epsilon-squared is unaffected by extreme outliers because it works on ranks. For an unbiased version that adjusts for sample size, use $\varepsilon^2_{adj} = H \cdot (n + 1) / (n^2 - 1)$, which is what the rcompanion package returns by default.

Explanation: Ties (the repeated value 4 across both vectors) break the exact null distribution because the rank sums are no longer uniquely defined, so even with exact = TRUE R falls back to the asymptotic normal approximation and warns you. In practice the two p-values are essentially identical for moderate samples; the warning is a paper trail, not an alarm. If your data have many ties, prefer the permutation version from the coin package via coin::wilcox_test(..., distribution = "exact"), which uses a tie-aware exact distribution.

Explanation: Under the null of exchangeable distributions, every relabelling of group membership is equally likely, so the permutation distribution of $W$ centred at $n_1 n_2 / 2$ is the exact reference distribution. Two-sidedness comes from measuring distance from that null centre on either side. With 5,000 permutations the resolution of p_perm is $1/5000 = 2 \times 10^{-4}$, which is why an "exact-zero" tail rounds up to that floor here. The two p-values agree to two significant figures, validating the asymptotic approximation built into wilcox.test() for this sample size.

Explanation: Friedman's test is the rank-based analogue of repeated-measures ANOVA: within each block (chick), it ranks the response across the treatments (time points), then asks whether the average rank differs by treatment. Restricting to chicks with all 5 timepoints is essential because Friedman requires a complete block design, unlike a mixed-effects model which tolerates unbalanced panels. The huge chi-squared statistic just reflects that weight rises monotonically with time for nearly every chick, so within-block ranks are nearly identical across chicks. For post-hoc pairwise comparisons across timepoints use pairwise.wilcox.test(weight, Time, paired = TRUE, p.adjust.method = "holm") on the same subset.