Framework I — t-test and ANOVA
When to use it: Your response variable is continuous (a measurement), and your predictor is a categorical group variable.
- Two groups → use a t-test
- Three or more groups → use one-way ANOVA (or its non-parametric equivalent, the Kruskal-Wallis test, when assumptions are violated)
The core question
Is the difference in group means large enough — relative to the within-group variability — that it is unlikely to have arisen by chance?
\[t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]
For ANOVA, the same logic is generalised using the F-ratio: variance between groups divided by variance within groups.
\[F = \frac{MS_{\text{between}}}{MS_{\text{within}}}\]
A large F means the group means are more spread out than individual observations within groups — evidence against the null hypothesis that all group means are equal.
Environmental example: soil nitrogen across land-use types
set.seed(42)
# Soil total nitrogen (g/kg) at three land-use types
nitrogen <- data.frame(
landuse = rep(c("Forest", "Grassland", "Cropland"), each = 25),
N_gkg = c(
rnorm(25, mean = 4.8, sd = 0.7), # Forest — highest
rnorm(25, mean = 3.2, sd = 0.9), # Grassland
rnorm(25, mean = 1.9, sd = 0.6) # Cropland — lowest
)
)Step 1 — Visualise the groups
Always plot your data before running any test. Boxplots show the median, interquartile range, and any outliers simultaneously.
ggplot(nitrogen, aes(x = landuse, y = N_gkg, fill = landuse)) +
geom_boxplot(width = 0.5, outlier.shape = 21, outlier.size = 1.8,
colour = "#1e3a2f", alpha = 0.85) +
geom_jitter(width = 0.15, size = 0.9,
colour = "#1e3a2f", alpha = 0.35) +
scale_fill_manual(values = c("#4a7c5e", "#8aad96", "#c8832a")) +
labs(x = NULL, y = "Soil nitrogen (g/kg)",
title = "Soil Nitrogen by Land-Use Type") +
theme_minimal(base_size = 13) +
theme(legend.position = "none",
panel.grid.major.x = element_blank(),
plot.title = element_text(family = "serif", face = "bold",
colour = "#1e3a2f"))
Soil nitrogen (g/kg) across three land-use types. The horizontal bar is the median; the box spans the IQR; whiskers extend to 1.5×IQR. Cropland shows consistently lower nitrogen than forest.
Step 2 — Run the ANOVA
model_aov <- aov(N_gkg ~ landuse, data = nitrogen)
summary(model_aov)## Df Sum Sq Mean Sq F value Pr(>F)
## landuse 2 112.48 56.24 88.7 <2e-16 ***
## Residuals 72 45.65 0.63
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1The F-statistic and its p-value tell you whether any groups differ. A significant ANOVA only tells you that the group means are not all equal — it does not tell you which pairs differ.
Step 3 — Post-hoc comparisons
When ANOVA is significant, use Tukey’s Honest Significant Difference (HSD) test to identify which specific pairs of groups differ, while controlling the overall Type I error rate.
TukeyHSD(model_aov)## Tukey multiple comparisons of means
## 95% family-wise confidence level
##
## Fit: aov(formula = N_gkg ~ landuse, data = nitrogen)
##
## $landuse
## diff lwr upr p adj
## Forest-Cropland 2.9456297 2.4066573 3.484602 0.000000
## Grassland-Cropland 0.9813626 0.4423903 1.520335 0.000126
## Grassland-Forest -1.9642671 -2.5032394 -1.425295 0.000000
Tukey HSD 95% confidence intervals for all pairwise differences. Intervals that do not cross zero indicate a statistically significant difference between those groups.
Checking assumptions
ANOVA rests on three assumptions. Always check them with diagnostic plots.
par(mfrow = c(1, 2), mar = c(4, 4, 3, 1), las = 1)
plot(model_aov, which = 1, main = "Residuals vs Fitted")
plot(model_aov, which = 2, main = "Normal Q-Q")
Left: residuals should scatter randomly around zero (no pattern = good). Right: residuals on a Q-Q plot should follow the diagonal line (normality check).
par(mfrow = c(1, 1))⚠️ When ANOVA assumptions are violated
If residuals are strongly non-normal or variances are very unequal
across groups, consider: (1) log-transforming the
response variable (common for skewed environmental data like
concentrations), (2) using Welch’s ANOVA
(oneway.test()) which does not assume equal variances, or
(3) the non-parametric Kruskal-Wallis
test (kruskal.test()), which tests whether group
medians differ without assuming normality.
⚙️ Two-group case: the t-test
When you have
exactly two groups, t.test(y ~ group, data = df) is
equivalent to a two-group ANOVA (F = t²). Use
var.equal = TRUE if Levene’s test confirms equal variances;
otherwise the default Welch correction is safer.
