| Plot | Before | After | Difference |
|---|---|---|---|
| 1 | 20 | 26 | 6 |
| 2 | 15 | 21 | 6 |
| 3 | 22 | 21 | -1 |
| 4 | 18 | 22 | 4 |
| 5 | 16 | 23 | 7 |
| 6 | 24 | 23 | -1 |
| 7 | 19 | 18 | -1 |
| 8 | 14 | 22 | 8 |
| 9 | 21 | 21 | 0 |
| 10 | 17 | 17 | 0 |
| 11 | 25 | 25 | 0 |
| 12 | 20 | 23 | 3 |
7 Nonparametric and Categorical Data
7.1 Parametric vs. nonparametric tests
Chapter 6 covered three forms of the \(t\)-test (one-sample, paired, and independent-samples, also called two-sample). All three need continuous, roughly normal data; the independent-samples test additionally assumes similar variance across the two groups. This chapter covers what to do when those assumptions aren’t met. We’ll also cover what to do if your data aren’t continuous at all.
So far, we’ve been looking at parametric tests, without calling them that. Every test needs a known null distribution for its test statistic, since that’s the shape we compare our result against to get a \(p\)-value. A test is parametric when we can only know that shape by assuming something specific about the population the data came from.
For the \(t\)-test, that assumption is normality. The \(t\)-distribution isn’t the shape of your raw data, it’s the shape the \(t\)-statistic takes across repeated samples, and we can only derive it because we assumed the population itself was normal, or relied on the Central Limit Theorem for a large enough sample. Nonparametric tests also end up with a known distribution for their own test statistics. They do not need to assume anything about the population’s shape to derive it.
When the distributional assumptions are met, parametric tests are generally the most powerful available. However, when distributional assumptions are not met, parametric tests can become unreliable, and it may be difficult to understand how they become unreliable without extensive simulations. The worse the departures from assumed distributions, the worse the errors that may result.
Nonparametric tests make no assumption about the shape of the population’s distribution. Nonparametric does not mean assumption-free. But they usually have assumptions that are easier to satisfy than those of parametric tests.
Many nonparametric tests are run on the ranks of the data (first, second, third … last), rather than on the raw data. The median is a familiar example: it’s a statistic based on ranks, representing the middle rank in a data set.
This chapter covers two different kinds of tools:
Nonparametric alternatives (to t-tests). Your outcome is still continuous, but it’s skewed, has outliers, or your sample is too small to trust normality. The Wilcoxon signed-rank test (Wilcoxon 1945) and Mann-Whitney-Wilcoxon test (Mann and Whitney 1947) answer the same questions as the one-sample/paired and independent-samples \(t\)-tests, respectively, without assuming normality.
Categorical data. Your outcome isn’t a number at all: a site either has an invasive species or it doesn’t, a sample either exceeds a threshold or it doesn’t. None of the tests above apply here; you need the chi-square (\(\chi^2\)) tests.
We’ll state the assumptions explicitly for each test as we go through them.
A practical bonus of rank-based tests: ordinal data. Because the Wilcoxon and Mann-Whitney-Wilcoxon tests only need a rank ordering, neither is limited to interval- or ratio-scale measurements the way a \(t\)-test is. A \(t\)-test needs a mean, which requires numbers where the distance between values means something. Plenty of environmental data doesn’t work that way: a water-quality rating (poor, fair, good, excellent), a disturbance-severity class, a Likert-style survey response. You can put those in order, but “the average disturbance class was 2.4” isn’t meaningful the way “the average pH was 6.8” is.
7.2 Nonparametric alternative to one-sample & paired t-test: the Wilcoxon signed-rank test
Chapter 6 built the paired \(t\)-test around this same example, wetland restoration and soil moisture, with a full 20-plot dataset. Let’s run a smaller version of that design, and this time check whether we should trust the result before reporting it.
Step 1: run the paired \(t\)-test, the way Chapter 6 taught it.
| Mean difference | t | df | p-value |
|---|---|---|---|
| 2.583333 | 2.57 | 11 | 0.026 |
\(t(11) = 2.57\), \(p = .026\). Significant. Restoration looks like it increased soil moisture.
Step 2: wait, did we check assumptions first? Let’s look at the differences before trusting that \(p\)-value.
| Statistic | p-value |
|---|---|
| 0.85 | 0.039 |
The Shapiro-Wilk test agrees with the picture (\(p=.039\)). Three plots didn’t change at all, three dropped slightly, and the rest rose by 3 to 8 points. That is not symmetric, let alone normal. The \(t\)-test’s “significant” result depends on an assumption the data don’t meet, so we turn to the Wilcoxon signed-rank test.
The question: did soil moisture systematically change after restoration, or is any change we see just plot-to-plot noise?
\(H_0\): no systematic shift; the differences are symmetric around a median of 0 (positive and negative ranks are balanced).
\(H_a\): there is a systematic shift; the median difference is not 0.
How it works, conceptually. Instead of using the actual sizes of the differences, the test converts them to ranks: it ranks the absolute differences from smallest to largest, then checks whether the positive-signed ranks and negative-signed ranks are roughly balanced. Balanced ranks are the null case, meaning a plot’s rank (how extreme its change was) has nothing to do with whether that change was an increase or a decrease. If soil moisture really increased at most plots, the larger ranks should belong mostly to positive differences instead. Because it works with ranks rather than raw values, one wildly extreme plot can’t dominate the result the way it could with a mean.
Its assumption. The Wilcoxon signed-rank test does not require normality, but it assumes the differences come from a symmetric distribution around their median. That’s a milder assumption than normality (symmetric-but-not-normal distributions are common), but our histogram above strains it: the positive side stretches to 8, the negative side barely reaches -1. We’ll run the test anyway, since it’s still the better-justified choice here, and note the problem when we interpret the result.
| Statistic (V) | p-value | Method | Tails |
|---|---|---|---|
| 39 | 0.057 | Wilcoxon signed rank test with continuity correction | two.sided |
“Two.sided” in the Tails column just means we tested for a shift in either direction, the same one-vs-two-tailed distinction from Chapter 6.
R falls back to a normal approximation twice over here: once for the tied differences (two plots at \(+6\), three at \(-1\)), and once because three plots showed no change at all. Zero differences can’t be ranked as positive or negative, so the Wilcoxon test drops them and runs on the remaining 9 plots instead of all 12. So the rank test is less sensitive to magnitude than the \(t\)-test, and it can also end up with less data.
How \(V\) is calculated. Drop the zero differences, take the absolute value of what’s left, rank those absolute values from smallest to largest (tied absolute values share the average of their ranks), then sum the ranks belonging to the positive differences. That sum is \(V\).
| Difference | Abs. difference | Rank | Sign |
|---|---|---|---|
| 6 | 6 | 6.5 | + |
| 6 | 6 | 6.5 | + |
| -1 | 1 | 2.0 | - |
| 4 | 4 | 5.0 | + |
| 7 | 7 | 8.0 | + |
| -1 | 1 | 2.0 | - |
| -1 | 1 | 2.0 | - |
| 8 | 8 | 9.0 | + |
| 3 | 3 | 4.0 | + |
\(V\) is the sum of the ranks where Sign is “+”: \(6.5+6.5+5+8+9+4 = 39\), matching the table R gave us above.
How the null distribution of \(V\) is built. Under \(H_0\), each plot’s rank has an independent 50/50 chance of being positive or negative. That gives \(2^n\) equally likely sign patterns for \(n\) nonzero differences (here \(n=9\), not 12, since the zeros dropped out). Summing the positive ranks for each of those patterns builds the null distribution of \(V\). It has a known mean and variance:
\[E[V] = \frac{n(n+1)}{4}, \qquad \text{Var}(V) = \frac{n(n+1)(2n+1)}{24}\]
Plugging in our \(n=9\) nonzero plots:
\[E[V] = \frac{9 \times 10}{4} = 22.5, \qquad \text{Var}(V) = \frac{9 \times 10 \times 19}{24} = 71.25\]
That’s an SD of about 8.4. Our observed \(V=39\) sits about 1.96 SDs above that expectation (\(\frac{39-22.5}{8.4} \approx 1.96\)). R’s \(p=.057\) comes from this normal approximation, slightly larger than the .05 a \(z\) of 1.96 would give, because R makes two small corrections: one for tied ranks, and one because \(V\) moves in steps rather than smoothly. For small, tie-free samples with no zeros, R can instead read the exact null distribution directly.
To get the critical values, we build the null distribution ourselves by simulating 20,000 experiments. In each one, every one of the 9 nonzero ranks gets a random \(+\) or \(-\) sign, as \(H_0\) says it should, and we add up the positive ranks.
Our observed \(V=39\) lands on the upper boundary but not inside the 5% region. \(V\) can only take certain values, so the boundary can’t fall exactly at 2.5%: a \(V\) of 39 or more comes up just over 2.5% of the time under \(H_0\), which puts the two-tailed \(p\) just over .05. The result is not significant, though it is close to the cutoff.
So far: \(t\)-test \(p=.026\) (significant), Wilcoxon \(p=.057\) (not quite). The two tests reach different verdicts on the same 12 plots because the \(t\)-test’s numerator and denominator both used the exact magnitudes the Wilcoxon test deliberately ignores.
The sign test. There’s an even simpler nonparametric option, the sign test, which looks only at whether each difference is positive or negative. It discards the magnitude entirely and asks whether pluses outnumber minuses more than chance would suggest. It has essentially no distributional assumption beyond the universal one (independence of observations, from a sound study design; see Chapter 6), not even symmetry.
That’s useful here: we already flagged the symmetry assumption as shaky for this data (the positive side stretching to 8, the negative side barely reaching -1), so the sign test sidesteps the one assumption the Wilcoxon test still needed. That robustness comes at a big cost in power, though, so it’s usually a last resort.
| Plots increased | Plots total | p-value |
|---|---|---|
| 6 | 9 | 0.508 |
Six of the 9 nonzero plots increased. \(p=.508\): nowhere close to significant.
The full picture. \(t\)-test: \(p=.026\), significant. Wilcoxon signed-rank: \(p=.057\), not quite. Sign test: \(p=.508\), not close. Same 12 plots, but the three tests don’t agree with each other.
Each test uses less of the data than the one before it. The \(t\)-test uses the full magnitude of every difference, Wilcoxon keeps the ranks but drops the zeros, and the sign test keeps only direction.
That’s not an argument that nonparametric tests are worse. The \(t\)-test gave the smallest \(p\)-value here because it relied most on the normality assumption, which these data do not meet. Check your assumptions before you report a \(p\)-value, not after you’ve already seen one you like.
7.3 Nonparametric alternative to the independent-samples t-test: the Mann-Whitney-Wilcoxon test
Now let’s run through the same kind of example for the independent-samples case. Researchers surveyed plant species richness, the number of distinct plant species found, at restored and unrestored wetland plots, to see whether restoration increased plant diversity. Figure 7.3 gives the design.
| Plot | Restored | Unrestored |
|---|---|---|
| 1 | 21 | 17 |
| 2 | 16 | 18 |
| 3 | 22 | 18 |
| 4 | 16 | 12 |
| 5 | 17 | 11 |
| 6 | 16 | 11 |
| 7 | 17 | 11 |
| 8 | 34 | 10 |
Step 1: run the \(t\)-test, the way Chapter 6 taught it.
| Mean difference | t | df | p-value |
|---|---|---|---|
| 6.375 | 2.54 | 14 | 0.024 |
\(t(14) = 2.54\), \(p = .024\). Significant. Restoration looks like it increased richness.
Step 2: wait, did we check assumptions first? We didn’t, and Chapter 6 was explicit that you’re supposed to. Let’s actually look at the data before trusting that \(p\)-value.
The circled plot has 34 species, nearly double any other plot in either group. We don’t just throw away outliers; maybe that plot sits in an unusually favorable microhabitat. Either way, it violates the normality assumption:
| Group | Statistic | p-value |
|---|---|---|
| Restored | 0.70 | 0.002 |
| Unrestored | 0.77 | 0.015 |
Both groups fail the Shapiro-Wilk test, the restored group badly (\(p=.002\)), driven almost entirely by that one plot. The \(t\)-test’s “significant” result depends on an assumption the data don’t meet. The Mann-Whitney-Wilcoxon test is designed for this situation.
The question: did restoration increase plant species richness, or is the difference we see consistent with random plot-to-plot variation?
\(H_0\): no difference in richness between restored and unrestored plots; ranks are well-mixed between the groups.
\(H_a\): richness tends to rank systematically higher in one group than the other.
How it works, conceptually. Pool all the observations from both groups together, rank them from smallest to largest, then compare the sum of ranks in Group A to the sum of ranks in Group B. If the groups truly don’t differ, ranks should be well mixed between them; if one group tends to have higher values, its ranks should be systematically higher too. That outlier plot still gets a rank, but only one rank, the highest one, no matter how far above the other plots it is. Its magnitude, which is what distorted the \(t\)-test, can’t distort the Mann-Whitney-Wilcoxon test the same way.
Its assumption. The Mann-Whitney-Wilcoxon test assumes both groups’ parent distributions have the same shape: it does not require normality, and the assumption is not “equal variance,” even though same-shape distributions happen to have equal variance as a side effect. If the two groups have differently-shaped distributions (say, one skewed and one symmetric), the test still runs, but interpreting a significant result purely as “the medians differ” becomes murkier, since it could be a difference in location, shape, or both. Our two groups don’t quite share a shape: the restored group’s skew comes almost entirely from the one 34-species plot, while the unrestored group is only mildly skewed. So read a result from this test as “richness tends to be higher in one group,” not strictly as a difference in medians.
| Statistic (U) | p-value | Method | Tails |
|---|---|---|---|
| 50 | 0.063 | Wilcoxon rank sum test with continuity correction | two.sided |
R labels this statistic \(W\); for two groups it is the Mann-Whitney \(U\). R falls back to the normal approximation again, because both groups share some tied richness values (two plots at 16, two at 11, and so on).
How the null distribution of \(U\) is built. The idea is the same as for the Wilcoxon signed-rank test, but the null distribution is built from group membership instead of signs. Under \(H_0\), both groups are draws from the same population, so once you pool and rank all \(N = n_A + n_B = 16\) values, which ranks end up in the restored group is just a matter of chance: any of the \(\binom{16}{8}\) ways of splitting those 16 ranks into two groups of 8 is equally likely. That’s the null distribution of \(U\), and its mean and variance are known:
\[E[U] = \frac{n_A n_B}{2}, \qquad \text{Var}(U) = \frac{n_A n_B (n_A+n_B+1)}{12}\]
Plugging in our two groups of 8:
\[E[U] = \frac{8 \times 8}{2} = 32, \qquad \text{Var}(U) = \frac{8 \times 8 \times 17}{12} \approx 90.7\]
That’s an SD of about 9.5. Our observed \(U=50\) sits about 1.9 SDs above that expectation (\(\frac{50-32}{9.5} \approx 1.9\)). R’s \(p=.063\) comes from this normal approximation, slightly larger than a \(z\) of 1.9 alone would give, because of the same two small corrections as in the Wilcoxon case: tied ranks, and \(U\) moving in steps rather than smoothly. For small samples without ties, R can instead read the exact null distribution directly, no approximation needed.
The critical values come from the null distribution just described, the \(\binom{16}{8}\) equally likely ways to split 16 pooled ranks into two groups of 8. Rather than enumerate every one of those splits by hand, we approximate that distribution the same way as before, by simulation: pool the richness values, then repeatedly shuffle which 8 ranks land in the restored group:
Our observed \(U=50\) lands just inside the critical value of \(50.5\), matching R’s \(p=.063\): close to significant, but not quite.
The full picture. The \(t\)-test said \(p=.024\): significant, restoration works. The Mann-Whitney-Wilcoxon test on the exact same data says \(p=.063\): not significant at \(\alpha=.05\), right on the boundary. The two tests give different answers to the same question on the same 16 plots. One outlier plot was doing a lot of the work behind that “significant” \(t\)-test result, and that’s exactly the kind of influence the normality assumption is supposed to rule out. That’s not a reason to always prefer the nonparametric test; if the data really had been roughly normal, the \(t\)-test’s extra power would have been the right tool. It’s why you check assumptions before reporting a \(p\)-value, not after.
7.4 Categorical data: the chi-square tests
Every test so far (the \(t\)-tests in Chapter 6, and the Wilcoxon and Mann-Whitney-Wilcoxon tests above) needs a continuous outcome variable. But a lot of environmental data isn’t continuous at all: a site either has an invasive species or it doesn’t; a stream sample either exceeds a safety threshold or it doesn’t; a plot falls into one land-cover category, not a number line. For categorical data, none of the tests above apply: you need a different tool entirely, the chi-square (\(\chi^2\)) test (Pearson 1900).
Chapter 3 introduced the shape of the chi-square distribution: always non-negative, right-skewed, naturally one-tailed, with its exact shape depending on degrees of freedom. Only the upper tail counts against \(H_0\): \(\chi^2\) adds up squared gaps between observed and expected, so a mismatch in either direction makes it bigger, and a \(\chi^2\) near 0 just means the counts match the expectation closely. Figure 7.6 shows what the rejection region looks like in practice, using the goodness-of-fit example we’ll run below, which has \(df = 3\). The critical value at \(\alpha = .05\) is 7.81, and any statistic beyond that point (the shaded region) occurs less than 5% of the time by chance.
How can a nonparametric test have its own named distribution? “Nonparametric” describes what you assume about the data: you don’t assume they came from a particular family like the normal. Every test still needs a null distribution for its test statistic: the distribution that statistic would have if \(H_0\) were true. For the rank tests, that distribution is exact: it comes from every possible arrangement of the ranks under \(H_0\). For the chi-square tests, the chi-square curve is an approximation to the statistic’s distribution under \(H_0\), and it is a good one only when expected counts are large enough. That’s why each chi-square test below comes with an expected-count rule.
There are two versions of the chi-square test, depending on the question you’re asking.
7.4.1 Chi-square goodness-of-fit: does one categorical variable match an expected pattern?
The name comes straight from what the test does: it checks how well (“goodness”) your observed counts match (“fit”) a specific pattern you’d expect, whether that expectation comes from a historical baseline, a theory, or an assumption of “equally likely.” A small chi-square statistic means a good fit (observed counts close to expected); a large one means a poor fit (observed counts far from expected).
A state wildlife agency has historically found fish evenly distributed across four zones of a lake, 25% in each. This year’s survey of 200 fish gives a different picture. Figure 7.7 lays out the structure, which differs from every design so far in this book.
Everywhere else in this book we have had an independent variable splitting units into groups and a dependent variable measured on each unit. Here there is only one variable, lake zone, and nothing is measured on the fish beyond which zone they turned up in. The counts are the data, and the comparison is against an expectation rather than against another group.
| Zone | Observed |
|---|---|
| A | 38 |
| B | 61 |
| C | 52 |
| D | 49 |
The question: does this year’s distribution across zones differ from the historical 25%-each pattern, or is this the kind of unevenness you’d expect from random sampling alone?
\(H_0\): fish are still distributed 25% per zone; the pattern matches the historical baseline.
\(H_a\): fish are not evenly distributed across zones; the pattern has changed.
| Statistic (<U+03C7><U+00B2>) | df | p-value | Method |
|---|---|---|---|
| 5.4 | 3 | 0.145 | Chi-squared test for given probabilities |
Assumption. The goodness-of-fit test doesn’t assume anything about the shape of your data (categories aren’t numbers, so there’s no shape to assume). It does have a practical requirement instead: expected counts should generally be 5 or more in each category. That “5” is a conventional rule of thumb (Cochran 1954; Larntz 1978), not something calculated from this data or compared to a critical value; it’s a check on whether the chi-square curve is a trustworthy approximation, done before the test itself runs. Here, expected count per zone is \(200 \times 0.25 = 50\) (total fish times the 25%-per-zone expectation under \(H_0\)), comfortably above that minimum. This is the same expected-value logic from Chapter 3, \(E[X]=\sum x_i P(X=x_i)\), just applied to a count instead of a single random variable: multiply the total by the probability each category gets under \(H_0\). When expected counts are too small, the chi-square approximation breaks down and a different method (like Fisher’s exact test) is more appropriate.
Where \(\chi^2\) actually comes from. For each category, take the squared difference between observed and expected, scaled by expected: \(\chi^2 = \sum \frac{(O-E)^2}{E}\).
| Zone | Observed | Expected | Contribution |
|---|---|---|---|
| A | 38 | 50 | 2.88 |
| B | 61 | 50 | 2.42 |
| C | 52 | 50 | 0.08 |
| D | 49 | 50 | 0.02 |
Summing the last column: \(2.88+2.42+0.08+0.02 = 5.4\), matching R’s statistic above. For goodness-of-fit, \(df = k-1\), where \(k\) is the number of categories: \(4-1=3\). With \(df=3\), the critical value at \(\alpha=.05\) is 7.81 (the same one marked on the distribution figure earlier in this chapter); our \(5.4\) falls well short of it.
Interpretation. \(\chi^2(3) = 5.4\), \(p = .145\): not significant. This year’s zone counts are consistent with the historical 25%-each pattern; the unevenness we see is within the range chance alone would produce.
7.4.2 Chi-square test of independence: are two categorical variables related?
Now suppose researchers classify 150 stream sites by land use and by whether each site’s E. coli level exceeds a recreational safety threshold. Every site falls into exactly one cell of Figure 7.8.
In that table, nobody assigned anything. The sites already had their land use and their E. coli level before the researchers arrived, and the cells simply count how many turned out each way. That makes this an observational design, and it is why the cells hold counts of sites rather than measurements taken on them.
| No | Yes | |
|---|---|---|
| Forested | 42 | 8 |
| Agricultural | 25 | 25 |
| Urban | 20 | 30 |
The question: is E. coli exceedance independent of land use, or are they related?
\(H_0\): land use and exceedance are independent; knowing one tells you nothing about the other.
\(H_a\): land use and exceedance are related; they are not independent.
| Statistic (<U+03C7>^2) | df | p-value | Method |
|---|---|---|---|
| 21.84 | 2 | < .001 | Pearson's Chi-squared test |
Assumption. Same rule as goodness-of-fit: expected counts should generally be 5 or more in every cell of the table. Unlike the goodness-of-fit test, where expected counts came from a stated proportion (25% per zone), here they come from the table’s own row and column totals: the expected count for a cell is \(\frac{\text{row total} \times \text{column total}}{\text{grand total}}\), the count you’d see in that cell if land use and exceedance really were unrelated. For Forested/No, that’s \(\frac{50 \times 87}{150} = 29\) (50 forested sites total, 87 sites total with no exceedance, 150 sites overall).
| No | Yes | |
|---|---|---|
| Forested | 29 | 21 |
| Agricultural | 29 | 21 |
| Urban | 29 | 21 |
All six expected counts are between 21 and 29, well above the 5-count minimum. This check does not tell us whether the test is significant. It only tells us whether the chi-square approximation is trustworthy here.
Where \(\chi^2\) actually comes from. Same formula as the goodness-of-fit test, applied cell by cell instead of category by category: \(\chi^2 = \sum \frac{(O-E)^2}{E}\), using the observed table and the expected table above.
| No | Yes | |
|---|---|---|
| Forested | 5.83 | 8.05 |
| Agricultural | 0.55 | 0.76 |
| Urban | 2.79 | 3.86 |
Summing all six cells: \(5.83+8.05+0.55+0.76+2.79+3.86 = 21.84\), matching R’s statistic above. For a test of independence, \(df=(r-1)(c-1)\) for \(r\) rows and \(c\) columns: \((3-1)(2-1)=2\). The number of sites never enters. With \(df=2\), the critical value at \(\alpha=.05\) is 5.99; our \(21.84\) is more than three times that, which is where \(p<.001\) comes from.
Interpretation. \(\chi^2(2) = 21.84\), \(p < .001\): significant. E. coli exceedance is not independent of land use. To see how the two variables are related, look at the table. Divide within each row to get the exceedance rate for each land use:
| Land use | Exceeds / total | Share |
|---|---|---|
| Forested | 8 / 50 | 16% |
| Agricultural | 25 / 50 | 50% |
| Urban | 30 / 50 | 60% |
Dividing within a row matters because raw counts only compare fairly when the row totals match. Agricultural and urban sites exceed the threshold three to four times as often as forested ones. The test says the two variables are related, but, like correlation, it does not by itself establish that land use causes the exceedance.
ImportantResults reporting: nonparametric and chi-square tests
These tests report differently from a \(t\)-test. There is no mean difference to quote, because none of them used the means. Give the test’s own statistic, the \(p\)-value, and a median rather than a mean.
Wilcoxon signed-rank, on the restoration differences:
Soil moisture did not change significantly after restoration, Wilcoxon signed-rank V = 39, p = .057 (median change +1.5 percentage points across all 12 plots; 9 non-zero differences).
Mann-Whitney-Wilcoxon, on the richness comparison:
Restored plots did not differ significantly from unrestored plots in species richness, U = 50, p = .063 (medians 17 vs. 11.5).
Chi-square, where the counts themselves are the result:
E. coli exceedance was not independent of land use, \(\chi^2(2) = 21.84\), p < .001. Forested sites exceeded the threshold in 8 of 50 cases, compared with 25 of 50 agricultural and 30 of 50 urban sites.
Report the number of non-zero differences for a signed-rank test, since dropped zeros change the effective sample size. And for chi-square, give the actual counts, not just the statistic: the counts are what a reader can interpret.
7.5 Chapter Summary
Why it matters. Real environmental data often fails the assumptions a \(t\)-test needs: small samples, skewed distributions, or outcomes that are counts and categories rather than measurements. This chapter covers the tests for those cases and how to tell when you need one.
Core ideas
- Rank-based tests trade power for weaker assumptions. The Wilcoxon signed-rank and Mann-Whitney-Wilcoxon tests ask the same questions as the paired and independent-samples \(t\)-tests, but work from ranks instead of raw magnitudes. That buys a milder distributional assumption at the cost of some sensitivity.
- The trade-off is a spectrum. The sign test goes further still, using only the direction of each difference. On the same restoration data, the \(t\)-test, Wilcoxon, and sign test gave progressively weaker verdicts, because each one throws away more information than the last.
- Nonparametric does not mean assumption-free. Wilcoxon assumes the differences are symmetric about their median. Mann-Whitney-Wilcoxon assumes the two distributions have the same shape, which is not the same as equal variance. State the assumption you are actually making.
- Check assumptions before you trust a \(p\)-value, not after. Both worked examples in this chapter ran the parametric test first, then went back and found the violation. That order is deliberate: it shows what goes wrong when you test first and check assumptions afterward.
- Categorical data needs a different tool entirely. The chi-square tests handle counts: goodness-of-fit asks whether one variable’s counts match an expected pattern, and the test of independence asks whether two categorical variables are related.
- The testing logic never changes. State \(H_0\) and \(H_a\), compute a statistic, and ask how surprising that statistic would be if \(H_0\) were true. Only the statistic and its null distribution differ from test to test.