Simulated dataset bank

Generating truth, models,
and analysis results

63 methods · simulated data · saved analysis results

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Use this guide to choose a biological example, see exactly how it was generated, and connect its scientific question to the null being tested. Every record is simulated.

The quick index links to each method’s CSV, full R script, plot, and saved output. The figures and numerical results come from those module files.

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Generating truth and sample evidence are different.

A nonzero generating effect can yield a large p-value, and a true null can yield a small one. Read the stated model and truth separately from the observed result.

Rows count CSV records. For paired, repeated, clustered, or aggregated data, the design identifies the actual independent unit.

Quick index

Rows and column previews are read directly from the saved CSVs. Result summaries are read from the outputs generated by each module’s R script.

Method	Biological scenario	Rows	Seed	Generating truth	Saved result	Data
01 · Means One-sample t test	Cell biologists measure oxygen consumption in independently cultured cell lines.	24	1101	The population mean is 10.8 nmol/min, so the reference-mean null of 10 is false; the true mean difference is +0.8.	Mean = 10.75 nmol/min; n = 24 t = 2.48; df = 23; p = 0.02107 Full result ↓	CSV
02 · Means Pooled two-sample t test	Evolutionary biologists compare adult beetle body length from two rearing environments.	56	1118	The true Cool minus Warm mean difference is -0.7 mm, and the two population variances are both 0.4225 mm squared.	Cool - Warm mean difference = -1.02 mm Pooled t = -4.94; df = 54.00; p = 7.927e-06 Full result ↓	CSV
03 · Means Welch two-sample t test	Evolutionary biologists compare adult beetle body length from two rearing environments.	56	1102	The true Cool minus Warm mean difference is -0.7 mm. Variances differ (0.4225 versus 1.3225 mm squared).	Cool - Warm mean difference = -1.01 mm Welch t = -4.21; df = 53.33; p = 9.973e-05 Full result ↓	CSV
04 · Means Paired t test	Plant physiologists measure photosynthesis in the same leaves before and after heat stress.	20	1103	The mean after-minus-before difference is -1.3 photosynthesis units, even if this sample provides limited evidence against zero.	Mean change (after - before) = -0.73 Paired t = -1.91; df = 19; p = 0.07128 Full result ↓	CSV
05 · Variance and diagnostics F test of two variances	Evolutionary developmental biologists compare variation in wing length between two inbred fly lines.	60	1109	Population means are equal, but the true variance ratio A/B is 0.11 squared / 0.21 squared, approximately 0.274.	Variance ratio (A / B) = 0.376 F = 0.376; df = 29, 29; p = 0.01048 Full result ↓	CSV
06 · Variance and diagnostics Levene and Brown–Forsythe tests	Microbiologists compare the variability of colony diameters across three nutrient media.	84	1110	Group geometric means are 4 mm, but variances differ. Arithmetic means also vary slightly because log-normal means depend on log-scale spread.	Brown–Forsythe F = 5.55; df = 2, 81; p = 0.005524 Original Levene F = 9.72; p = 0.0001645 Full result ↓	CSV
07 · Variance and diagnostics Shapiro–Wilk and graphical checks	Cell biologists examine the distribution of time to lysis among independent cultured cells.	36	1111	The population is log-normal and positively skewed, not normal. The population geometric mean is 45 minutes.	Shapiro–Wilk W = 0.937; p = 0.04064; n = 36 Curvature in the Q–Q plot indicates a shape mismatch. Full result ↓	CSV
08 · Nonparametric alternatives Sign test	Behavioral biologists record the change in refuge-use time after a predator cue.	30	1104	Positive changes are more likely than negative changes. The unrounded mean is 4 seconds, but the sign test is about direction rather than that mean.	Positive changes: 18 of 30 nonzero differences Exact two-sided p = 0.3616 Full result ↓	CSV
09 · Nonparametric alternatives Wilcoxon signed-rank test	Microbiologists compare log bacterial density before and after a treatment in matched cultures.	28	1105	The symmetric difference distribution is centered at +0.28 log10 units; its zero-center null is false.	Pseudomedian change estimate = 0.249 V = 328.0; p = 0.004582 Full result ↓	CSV
10 · Nonparametric alternatives Wilcoxon rank-sum test	Ecologists compare parasite burdens among fish from two independent lakes.	64	1106	The lake distributions differ, including their mean and spread. The same-distribution null is false; there is not a simple equal-shape location shift.	Observed medians: Clear 3.0; Reedy 8.0 parasites W = 362.5; asymptotic p = 0.04452 Full result ↓	CSV
11 · ANOVA One-way ANOVA	Ecologists compare seedling biomass under three independent soil treatments.	75	1117	The equal-means null is false. All three populations share variance 0.85 squared; Loam minus Sand is truly 0.9 g.	Ordinary ANOVA F(2, 72) = 11.59; p = 4.335e-05 Welch ANOVA F = 8.89; denominator df = 46.37; p = 0.0005408 Full result ↓	CSV
12 · ANOVA Welch one-way ANOVA	Ecologists compare seedling biomass under three independent soil treatments.	78	1119	The true means differ, and variances differ (0.36, 1.21, and 2.25 g squared). Welch addresses the equal-means null without a shared variance.	Welch ANOVA F = 13.65; df = 2, 49.85; p = 1.873e-05 Group sample sizes: Sand 20; Loam 26; Clay 32 Full result ↓	CSV
13 · Nonparametric alternatives Kruskal–Wallis test	Plant pathologists score leaf damage in three independently assigned pathogen treatments.	72	1107	The score distributions differ across all three generating probabilities; expected scores are 2.0, 3.2, and 4.64. The equal-distributions null is false.	Kruskal–Wallis chi-square = 35.75; df = 2; p = 1.73e-08 Observed medians: Control 2; Strain A 3; Strain B 4.5 Full result ↓	CSV
14 · ANOVA and contrasts Factorial ANOVA: interactions	Beetle larvae from two genotypes are independently reared at two temperatures.	88	2101	The temperature effect is 2 mg/week in genotype A and 4.3 mg/week in B, so the interaction is 2.3 mg/week. The interaction null is false.	Interaction = 4.30 mg/week t(84) = 5.69; p = 1.78e-07 Full result ↓	CSV
15 · ANOVA and contrasts ANCOVA: an adjusted group comparison	Plant ecologists compare nutrient treatment while accounting for initial plant height.	90	2102	The nutrient treatment increases the population mean biomass by 2.4 g at any shared initial height. Both groups have slope 1.5 g/cm. The adjusted treatment null is false.	Adjusted nutrient minus control = 2.65 g t(87) = 6.55; p = 3.797e-09 Full result ↓	CSV
16 · Repeated and clustered data Repeated-measures ANOVA	Physiologists measure the same animals at four assay temperatures.	96	2103	Population means are 8, 9.75, 11.5 and 13.25 µmol/hour. The endpoint difference is 5.25 µmol/hour and the equal-means null is false.	Condition F(3, 69) = 84.32 Omnibus p = 4.966e-23 Full result ↓	CSV
17 · Nonparametric alternatives Friedman test	Sensory biologists record activity scores from each insect under three light conditions.	54	1108	Condition effects exist before rounding (0, +0.8, +1.6), and condition score distributions differ after rounding. Repeated observations share an insect baseline.	Independent blocks = 18; conditions = 3 Friedman chi-square = 17.63; df = 2; p = 0.0001482 Full result ↓	CSV
18 · ANOVA and contrasts Tukey / Tukey–Kramer comparisons	Plant growth is compared across four nutrient regimes.	80	2104	Population means for A–D are 8, 9, 11.3 and 12.1 g; all distinct pairwise mean-equality nulls are false.	Family: all six pairwise nutrient comparisons Tukey-adjusted p = 0.0415 Full result ↓	CSV
19 · ANOVA and contrasts Fisher least significant difference	Plant growth is compared across four nutrient regimes.	80	2104	Population means for A–D are 8, 9, 11.3 and 12.1 g; all distinct pairwise mean-equality nulls are false.	Omnibus p = 5.683e-12 Unadjusted p = 0.05163 Full result ↓	CSV
20 · ANOVA and contrasts Scheffé contrasts	Plant growth is compared across four nutrient regimes.	80	2104	Population means for A–D are 8, 9, 11.3 and 12.1 g; all distinct pairwise mean-equality nulls are false. The specified (C+D)/2 − (A+B)/2 contrast equals 3.2 g.	Contrast: (C + D)/2 − (A + B)/2 Scheffé-adjusted p = 9.177e-11 Full result ↓	CSV
21 · Rank methods Dunn rank comparisons	Ecologists compare skewed leaf-damage measurements among four treatments.	84	2105	Group distributions differ; their population medians are exp(1.2), exp(1.3), exp(1.7) and exp(1.9) mm². Dunn tests pooled-rank tendencies, not arithmetic mean equality. The distributions also differ in original-scale spread.	Kruskal–Wallis p = 1.352e-05 Two-sided Holm-adjusted p = 0.0001018 Full result ↓	CSV
22 · Association Pearson correlation	Evolutionary biologists measure wing and tail lengths in adult birds from a single population.	45	1112	The generating population Pearson correlation is 0.555 / sqrt((0.555)^2 + 3.5^2), approximately 0.618. It is not the observed sample correlation.	Pearson r = 0.700 t = 6.43; df = 43; p = 8.606e-08 Full result ↓	CSV
23 · Nonparametric alternatives Spearman rank correlation	Microbial ecologists compare soil salinity with bacterial richness across independent soil cores.	45	1113	The underlying mean response decreases nonlinearly with salinity, creating a negative monotonic association. No fixed numeric population Spearman coefficient was specified.	Spearman rho = -0.752; n = 45 S = 26599.8; asymptotic two-sided p = 2.549e-09 Full result ↓	CSV
24 · Regression Simple linear regression	Plant ecologists grow seedlings under a gradient of nitrogen supply.	36	1114	The conditional mean intercept is 2.2 g and slope is +0.42 g per mg nitrogen. The zero-slope null is false.	Slope = 0.414 g biomass per mg nitrogen Slope t = 9.90; df = 34; p = 1.525e-11 Full result ↓	CSV
25 · Regression Multiple linear regression	Evolutionary biologists relate beetle horn length to body size and larval nutrition.	60	1115	The true conditional body slope is 0.35 mm/mm and nutrition slope 0.16 mm/mg, with no interaction. Both zero-coefficient nulls are false.	Body slope, adjusted for nutrition = 0.299 mm/mm Body coefficient p = 7.658e-05; nested-model F = 18.17 Full result ↓	CSV
26 · Regression Theil–Sen robust slope	Physiologists measure enzyme activity along a temperature gradient with occasional unusually high assays.	40	1116	The underlying uncontaminated linear trend has slope 1.4 U/mL per degree C. Two deliberately elevated assays illustrate robustness; the bootstrap interval comes from the saved contaminated sample.	Theil–Sen slope = 1.408 U/mL per degree C 95% pairs-bootstrap interval: 1.184 to 1.618 Full result ↓	CSV
27 · Proportions and categories Exact binomial test	Host choice in a parasitoid wasp	80	3101	The true native-host probability is 0.68, so the null probability 0.50 is false. The population difference from the null is +0.18.	Native-host choices: 51 / 80 Exact two-sided p = 0.01832 Full result ↓	CSV
28 · Proportions and categories One-proportion test	Germination of salt-tolerant seeds	120	3102	The true germination probability is 0.67 rather than the null benchmark 0.50; the population difference is +0.17.	Germination: 70 / 120 Two-sided p = 0.06789 Full result ↓	CSV
29 · Proportions and categories Two-proportion test	Heat survival in evolved and ancestral yeast	220	3103	Evolved minus ancestral survival probability is 0.20, so the equal-probabilities null is false. The corresponding population odds ratio is approximately 2.37.	Evolved: 78 / 110; ancestral: 61 / 110 Two-sided p = 0.01748 Full result ↓	CSV
30 · Proportions and categories Chi-square goodness-of-fit	Segregation in an F2 cross	240	3104	The simulated probabilities differ from 1:2:1 by -0.07, +0.04, and +0.03. The goodness-of-fit null is false.	Observed: 43, 128, 69 p = 0.03508 Full result ↓	CSV
31 · Proportions and categories Chi-square test of independence	Infection across three lizard habitats	270	3105	Infection probability depends on habitat, so the independence null is false. Urban minus forest infection probability is 0.30; the simulation does not assign causality to habitat.	Infection probabilities: Forest 0.14; Grassland 0.41; Urban 0.54 p = 1.042e-07; Cramer V = 0.345 Full result ↓	CSV
32 · Proportions and categories Fisher exact test	Survival of rare resistant and susceptible clones	24	3106	The resistant-to-susceptible population survival odds ratio is (0.65/0.35)/(0.20/0.80), approximately 7.43. The odds-ratio-one null is false.	Survivors: susceptible 2/12; resistant 7/12 Two-sided exact p = 0.08938 Full result ↓	CSV
33 · Paired binary outcomes McNemar test	Antibody detection before and after an exposure season	80	3107	Marginal after positivity is 0.45×0.85 + 0.55×0.35 = 0.575, a +0.125 change from before. Population gain and loss probabilities are 0.1925 and 0.0675, so the McNemar null is false.	Gained detection: 10; lost detection: 5 McNemar p = 0.1967; exact paired p = 0.3018 Full result ↓	CSV
34 · Paired binary outcomes Cochran’s Q test	Bee visits to three floral odors	180	3108	Increasing odor effects produce strictly increasing marginal visit probabilities across A, B, and C. The equal-probabilities null is false; the latent log-odds differences are not themselves the marginal probability differences.	Matched bees: 60; conditions: 3 Omnibus p = 0.07643 Full result ↓	CSV
35 · Generalized linear models Logistic regression	Hybrid viability across parental divergence	160	3110	The true divergence coefficient is -0.5; viability odds are multiplied by exp(-0.5) ≈ 0.607 per divergence percentage point. The zero-slope null is false. The probability decline is nonlinear.	Divergence coefficient: -0.464 Likelihood-ratio p = 6.573e-09 Full result ↓	CSV
36 · Counts and exposure Exact Poisson rate test	De novo mutations in sequenced microbial lineages	60	3109	The true rate is 0.40 mutations/Mb rather than the null rate 0.25/Mb. The population rate ratio to the null is 1.6.	Events: 345; exposure: 845.0 Mb Exact p = 3.246e-17 Full result ↓	CSV
37 · Generalized linear models Poisson regression	Coral recruit counts and live coral cover	120	3111	The true cover coefficient is 0.016. A 10-point cover increase multiplies recruit density by exp(0.16) ≈ 1.174. The zero-slope null is false, and the exposure coefficient is fixed at one.	Rate ratio per 10 cover points: 1.200 Likelihood-ratio p = 2.64e-30 Full result ↓	CSV
38 · Generalized linear models Quasi-Poisson regression	Bacterial colony counts with clumping	120	3113	High medium multiplies mean colony density by exp(0.5) ≈ 1.649; the equal-rate null is false. The true dispersion multiplier is 4, so Poisson uncertainty would be too small.	High / low mean rate ratio: 1.351 Quasi-Poisson t-test p = 0.002166 Full result ↓	CSV
39 · Generalized linear models Negative binomial regression	Parasite abundance and fish body length	130	3112	The length coefficient is 0.085, so 5 cm multiplies expected parasite abundance by exp(0.425) ≈ 1.530. The slope-zero null is false; the Poisson variance assumption is intentionally violated.	Mean count ratio per 5 cm: 1.602 Wald p = 1.634e-22 Full result ↓	CSV
40 · Generalized linear models Multinomial logistic regression	Fish use of three unordered feeding microhabitats	240	3114	The outcome distributions differ, so the joint no-food-effect null is false. Relative to Open, the food coefficients are log(0.30/0.50)-log(0.50/0.25) ≈ -1.204 for Shelter and log(0.20/0.50)-log(0.25/0.25) ≈ -0.916 for Surface.	Low-food probabilities: 0.217, 0.508, 0.275 Joint p = 6.115e-05 Full result ↓	CSV
41 · Generalized linear models Ordinal logistic regression	Coral bleaching severity along a temperature gradient	180	3115	The common temperature coefficient is 0.8, so each °C multiplies odds of higher versus lower severity by exp(0.8) ≈ 2.226 at every cumulative threshold. The zero-effect null is false and proportional odds holds by construction.	Common odds ratio per 1°C: 2.613 p = 4.544e-23 Full result ↓	CSV
42 · Repeated and clustered data Linear mixed model	Evolutionary biologists follow larval growth under two diets.	144	2106	The control slope is 2 mg/week; enrichment adds 0.9 mg/week. The diet-by-week null is false. Simulated random-intercept and slope SDs are 2.3 mg and 0.7 mg/week.	Growth-slope difference = 0.79 mg/week Satterthwaite df = 34.0; t = 3.20; p = 0.002966 Full result ↓	CSV
43 · Repeated and clustered data Binary mixed model	Immunologists record repeated infection outcomes in vaccinated and control hosts.	288	2107	The conditional vaccine odds ratio is exp(−1) ≈ 0.368; the week odds ratio is exp(0.25) ≈ 1.284. The vaccine null is false. These are conditional effects, not population-average odds ratios.	Conditional vaccine odds ratio = 0.24 Wald z = -3.99; p = 6.63e-05 Full result ↓	CSV
44 · Repeated and clustered data Count mixed model: Poisson and negative binomial	Pollinator visits are repeatedly counted on plants in two habitats.	192	2114	The conditional sheltered/open rate ratio is exp(0.55) ≈ 1.733; the habitat null is false. The generating distribution has extra variation beyond Poisson even after conditioning on the plant intercept.	Conditional sheltered/open rate ratio = 1.50 Wald z = 2.62; p = 0.008759 Full result ↓	CSV
45 · Repeated and clustered data Ordinal mixed model	Fish behavior is scored repeatedly using ordered stress categories.	210	2109	The conditional common odds ratio for a higher score under stress is exp(0.8) ≈ 2.226. The treatment null is false. The simulated thresholds obey proportional odds.	Conditional higher-score odds ratio = 2.51 Wald z = 2.78; p = 0.005491 Full result ↓	CSV
46 · Dependent observations Binary generalized estimating equations	Repeated infection screening after vaccination	320	3116	The true population-average vaccination coefficient is -0.9 and odds ratio exp(-0.9) ≈ 0.407, adjusted for week. The no-group-effect null is false. The true marginal week coefficient is 0.35. A working exchangeable correlation is an approximation, while the marginal mean model is exact.	Independent animals: 80; observations: 320 Robust Wald p = 0.06864 Full result ↓	CSV
47 · Time to event Kaplan–Meier survival estimation	Ecologists follow seedling survival until death or final observation.	120	2110	The population survival function is exp(−0.018 × days); true day-40 survival is exp(−0.72) ≈ 0.487. This is an estimation target, not a null hypothesis.	120 seedlings; 79 events; 41 censored Estimated survival at day 40 = 0.455 Full result ↓	CSV
48 · Time to event Log-rank survival comparison	Plant pathologists compare time to infection under three protective treatments.	150	2111	Equal survival is false: day-30 infection-free probabilities are exp(−1.35) ≈ 0.259, exp(−0.9) ≈ 0.407 and exp(−0.54) ≈ 0.583. The high/control hazard ratio is 0.4, though the log-rank test itself does not estimate it.	Log-rank chi-square(2) = 17.18 Omnibus p = 0.0001863 Full result ↓	CSV
49 · Survival Stratified log-rank test	Larval survival under stress is compared within 12 rearing blocks.	120	240501	Stress has a true hazard ratio of 1.6 within each block; the within-block equality null is false.	Stratified chi-square = 13.433 p = 0.0002472 Full result ↓	CSV
50 · Time to event Cox proportional-hazards regression	Seedling death times are related to protective dose and temperature.	180	2112	The adjusted hazard ratio per dose unit is exp(−0.45) ≈ 0.638; per °C it is exp(0.12) ≈ 1.127. The dose null is false and proportional hazards holds in the generating model.	Dose hazard ratio per unit = 0.69 Wald z = -3.72; p = 0.0001969 Full result ↓	CSV
51 · Time to event Parametric survival: accelerated failure time	Botanists compare germination timing after control or cold treatment.	140	2113	The population cold/control time ratio is exp(0.4) ≈ 1.492, so the equal-time null is false. The generating Weibull distribution matches the fitted AFT family.	Cold/control time ratio = 1.70 Wald z = 4.54; p = 5.745e-06 Full result ↓	CSV
52 · Survival Shared-frailty survival model	Larvae from 24 populations experience a control or stress condition.	240	240502	The conditional stress hazard ratio is exp(0.5) ≈ 1.65; shared population variance is 0.5.	Conditional hazard ratio = 1.61 Wald p = 0.002373 Full result ↓	CSV
53 · Simulation and multivariate methods Independent-sample permutation test	Independent insect larvae are randomly assigned to ambient or warm rearing.	50	240503	The simulated warm-minus-control mean difference is 1.1 mm; the same-distribution null is false.	Warm − control = 0.72 mm Two-sided Monte Carlo p = 0.187 Full result ↓	CSV
54 · Simulation and multivariate methods Paired permutation test	Two comparable leaves per plant receive randomized control or shade treatment.	28	240504	The true mean shaded-minus-control difference is −1.2; plants also differ in their shared baseline.	Shaded − control = -0.97 units Two-sided permutation p = 0.0006 Full result ↓	CSV
55 · Simulation and multivariate methods Monte Carlo null-model test	Test whether candidate sex-determination genes are unusually concentrated on the X chromosome.	80	240505	The true occupancy probability is 0.32, so the tested 0.20 null is false.	Observed on X: 32 of 80 Two-sided Monte Carlo p = 0.0002 Full result ↓	CSV
56 · Simulation and multivariate methods Bootstrap confidence interval	Estimate average biomass from independent bacterial colonies with a right-skewed distribution.	40	240507	The population arithmetic mean is approximately 13.28 mg; the interval may or may not cover it in one sample.	Sample mean = 11.54 mg Percentile 95% CI = 10.15 to 13.07 mg Full result ↓	CSV
57 · Simulation and multivariate methods Sampling variation and CI coverage	Independent cell cultures differ in the waiting time to a developmental transition.	25000	240506	The population mean is exactly 5 days; nominal 95% coverage is a procedure property to evaluate, not an imposed simulation result.	Mean estimated waiting time: 5.03 days Observed 95% t-interval coverage: 92.6% Full result ↓	CSV
58 · Multiple testing Bonferroni correction	Compare gene expression between two conditions across 40 genes.	960	240508	The first eight genes have true mean effects of 1.1 log2 units; the other 32 nulls are true.	Unadjusted p<.05: 8 genes Adjusted p<.05: 2 genes Full result ↓	CSV
59 · Multiple testing Benjamini–Hochberg FDR	Compare gene expression between two conditions across 40 genes.	960	240509	The first eight genes have true mean effects of 1.1 log2 units; the other 32 nulls are true.	Unadjusted p<.05: 7 genes Adjusted p<.05: 3 genes Full result ↓	CSV
60 · Multivariate exploration Principal component analysis	Four correlated morphological traits are measured in beetles from three ecotypes.	75	240510	Groups differ in body-size means and antenna shifts; all traits share a common body-size component.	PC1 accounts for 84.3% of variance PC2 accounts for 8.6% Full result ↓	CSV
61 · Multivariate exploration Classical multidimensional scaling	Four correlated morphological traits are measured in beetles from three ecotypes.	75	240511	Groups differ in body-size means and antenna shifts; all traits share a common body-size component.	Two-dimensional goodness of fit: 0.920 Euclidean distances between standardized traits Full result ↓	CSV
62 · Multivariate exploration Linear discriminant analysis	Four correlated morphological traits are measured in beetles from three ecotypes.	75	240512	Groups differ in body-size means and antenna shifts; all traits share a common body-size component.	Leave-one-out classification accuracy: 82.7% Three known ecotype labels used during training Full result ↓	CSV
63 · Dependent observations Count and rate GEE	Microbial ecologists count bacterial colonies in repeated water samples from experimental pond mesocosms.	360	1401	The exact population-average Warm / Ambient rate ratio is exp(0.45), approximately 1.568; the equal-rate null is false. Each week multiplies the marginal rate by exp(0.1), approximately 1.105. Marginal count variance is mu + 0.5 × mu squared. The working exchangeable correlation is approximate, while the specified marginal mean is exact.	Independent mesocosms: 90; observations: 360 Robust Wald p = 0.00243 Full result ↓	CSV

Dataset 01 · Means

One-sample t test

CSV rows: 24Seed: 1101Module: one_sample_t

Saved CSV Full runnable R script Full-size plot

Cell biologists measure oxygen consumption in independently cultured cell lines.

Question and null

Scientific question

Is mean oxygen consumption different from 10 nmol/min?

Null being tested / estimation target

The population mean is 10 nmol/min.

Generating truth

The population mean is 10.8 nmol/min, so the reference-mean null of 10 is false; the true mean difference is +0.8.

Exact generating model

24 independent oxygen-consumption values are drawn from Normal(mean 10.8, SD 1.5). Each row is one culture.

Design and independent unit: One measurement from each of 24 independent cultures; compare with a prespecified reference of 10 nmol/min.

Columns in the saved CSV

Column	Meaning	Unit / type
`culture`	Independent culture identifier	none
`oxygen`	Oxygen consumption	nmol/min

Preview the actual saved observations

culture	oxygen
1	9.36365439317128
2	10.4629957911243
3	9.49364187398395
4	6.73001021715413
5	12.1475731965392

Analysis from the saved dataset

One-sample t test plot. Cell biologists measure oxygen consumption in independently cultured cell lines. — Simulated observations; generated by this module’s R script.

Mean = 10.75 nmol/min; n = 24
95% CI for mean: 10.12 to 11.37
t = 2.48; df = 23; p = 0.02107

Mean oxygen consumption was 10.75 nmol/min (95% CI 10.12 to 11.37); against the reference mean of 10, t(23) = 2.48, p = 0.02107. This tests a population mean, not whether every culture exceeds 10.

Teaching pointThe reference must express the biological question; the test is about a population mean.

Analysis code excerpt

d <- read.csv("modules/one_sample_t/generated/data.csv")
fit <- t.test(d$oxygen, mu = 10,
              alternative = "two.sided", conf.level = .95)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


	One Sample t-test

data:  d$oxygen
t = 2.4762, df = 23, p-value = 0.02107
alternative hypothesis: true mean is not equal to 10
95 percent confidence interval:
 10.12316 11.37336
sample estimates:
mean of x 
 10.74826

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/one_sample_t/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(1101)
d <- data.frame(culture = 1:24,
                oxygen = rnorm(24, mean = 10.8, sd = 1.5))
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/one_sample_t/analysis.R modules/one_sample_t/generated

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Dataset 02 · Means

Pooled two-sample t test

CSV rows: 56Seed: 1118Module: pooled_t

Saved CSV Full runnable R script Full-size plot

Evolutionary biologists compare adult beetle body length from two rearing environments.

Question and null

Scientific question

Do mean body lengths differ between cool and warm environments?

Null being tested / estimation target

The two population means are equal.

Generating truth

The true Cool minus Warm mean difference is -0.7 mm, and the two population variances are both 0.4225 mm squared.

Exact generating model

26 independent Cool beetles are Normal(10, 0.65 SD), and 30 independent Warm beetles are Normal(10.7, 0.65 SD).

Design and independent unit: Independent beetles in two rearing environments; a common population variance is a substantive model assumption.

Columns in the saved CSV

Column	Meaning	Unit / type
`environment`	Independent rearing group	category
`length_mm`	Adult beetle body length	mm

Preview the actual saved observations

environment	length_mm
Cool	9.94949373653932
Cool	10.1150886481904
Cool	10.0064592350897
Cool	11.6353521543521
Cool	9.48477688877161

Analysis from the saved dataset

Pooled two-sample t test plot. Evolutionary biologists compare adult beetle body length from two rearing environments. — Simulated observations; generated by this module’s R script.

Cool - Warm mean difference = -1.02 mm
95% CI: -1.44 to -0.61 mm
Pooled t = -4.94; df = 54.00; p = 7.927e-06

Mean body length differed by -1.02 mm (Cool minus Warm; 95% CI -1.44 to -0.61). Pooled t = -4.94, df = 54.00, p = 7.927e-06. This analysis assumes equal population variances. Welch tests the same null without that assumption.

Teaching pointPooled and Welch t tests share the mean-difference null; their variance assumptions differ.

Analysis code excerpt

d <- read.csv("modules/pooled_t/generated/data.csv")
d$environment <- factor(d$environment, levels = c("Cool", "Warm"))
fit <- t.test(length_mm ~ environment, data = d,
              var.equal = TRUE, conf.level = .95)
difference <- mean(d$length_mm[d$environment == "Cool"]) -
              mean(d$length_mm[d$environment == "Warm"])

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


	Two Sample t-test

data:  length_mm by environment
t = -4.9399, df = 54, p-value = 7.927e-06
alternative hypothesis: true difference in means between group Cool and group Warm is not equal to 0
95 percent confidence interval:
 -1.4367878 -0.6072203
sample estimates:
mean in group Cool mean in group Warm 
          9.833382          10.855386

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/pooled_t/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(1118)
d <- data.frame(environment = rep(c("Cool", "Warm"), c(26, 30)),
                length_mm = c(rnorm(26, 10, .65), rnorm(30, 10.7, .65)))
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/pooled_t/analysis.R modules/pooled_t/generated

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Dataset 03 · Means

Welch two-sample t test

CSV rows: 56Seed: 1102Module: welch_t

Saved CSV Full runnable R script Full-size plot

Evolutionary biologists compare adult beetle body length from two rearing environments.

Question and null

Scientific question

Do mean body lengths differ between cool and warm environments?

Null being tested / estimation target

The two population means are equal.

Generating truth

The true Cool minus Warm mean difference is -0.7 mm. Variances differ (0.4225 versus 1.3225 mm squared).

Exact generating model

26 independent Cool lengths come from Normal(10, 0.65 SD), and 30 independent Warm lengths from Normal(10.7, 1.15 SD).

Design and independent unit: Independent beetles assigned to cool or warm rearing; each beetle measured once.

Columns in the saved CSV

Column	Meaning	Unit / type
`environment`	Independent rearing group	category
`length_mm`	Adult beetle body length	mm

Preview the actual saved observations

environment	length_mm
Cool	9.71117587142829
Cool	10.2447524393685
Cool	10.7359369601148
Cool	9.85099491058948
Cool	8.43873459705919

Analysis from the saved dataset

Welch two-sample t test plot. Evolutionary biologists compare adult beetle body length from two rearing environments. — Simulated observations; generated by this module’s R script.

Cool - Warm mean difference = -1.01 mm
95% CI: -1.49 to -0.53 mm
Welch t = -4.21; df = 53.33; p = 9.973e-05

Mean body length differed by -1.01 mm (Cool minus Warm; 95% CI -1.49 to -0.53). Welch's t = -4.21, df = 53.33, p = 9.973e-05. Group ordering determines the sign; unequal group variances do not require switching to ranks.

Teaching pointWelch answers the equal-means question while allowing unequal group variances.

Analysis code excerpt

d <- read.csv("modules/welch_t/generated/data.csv")
d$environment <- factor(d$environment, levels = c("Cool", "Warm"))
fit <- t.test(length_mm ~ environment, data = d,
              var.equal = FALSE, conf.level = .95)
difference <- mean(d$length_mm[d$environment == "Cool"]) -
              mean(d$length_mm[d$environment == "Warm"])

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


	Welch Two Sample t-test

data:  length_mm by environment
t = -4.2066, df = 53.33, p-value = 9.973e-05
alternative hypothesis: true difference in means between group Cool and group Warm is not equal to 0
95 percent confidence interval:
 -1.4937879 -0.5292964
sample estimates:
mean in group Cool mean in group Warm 
          9.763781          10.775323

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/welch_t/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(1102)
d <- data.frame(environment = rep(c("Cool", "Warm"), c(26, 30)),
                length_mm = c(rnorm(26, 10, .65), rnorm(30, 10.7, 1.15)))
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/welch_t/analysis.R modules/welch_t/generated

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Dataset 04 · Means

Paired t test

CSV rows: 20Seed: 1103Module: paired_t

Saved CSV Full runnable R script Full-size plot

Plant physiologists measure photosynthesis in the same leaves before and after heat stress.

Question and null

Scientific question

Is the mean within-plant change in photosynthesis zero?

Null being tested / estimation target

The population mean of after minus before differences is zero.

Generating truth

The mean after-minus-before difference is -1.3 photosynthesis units, even if this sample provides limited evidence against zero.

Exact generating model

20 independent baseline values come from Normal(18, 2.2 SD). Each after value is its own baseline plus an independent Normal(-1.3, 1.7 SD) change.

Design and independent unit: Twenty independent plants contribute one before/after pair each.

Columns in the saved CSV

Column	Meaning	Unit / type
`plant`	Independent plant and pair identifier	none
`before`	Photosynthesis before heat	micromol CO2 / m2 / s
`after`	Photosynthesis after heat	micromol CO2 / m2 / s

Preview the actual saved observations

plant	before	after
1	18.1287374581788	16.5908885265036
2	19.7004842068794	17.1879623862451
3	17.0935988722271	15.9246713928148
4	19.1601148465114	18.1799555013732
5	17.2033077635387	19.5839635129769

Analysis from the saved dataset

Paired t test plot. Plant physiologists measure photosynthesis in the same leaves before and after heat stress. — Simulated observations; generated by this module’s R script.

Mean change (after - before) = -0.73
95% CI for mean change: -1.52 to 0.07
Paired t = -1.91; df = 19; p = 0.07128

Photosynthesis changed by a mean of -0.73 units after heat stress (95% CI -1.52 to 0.07); paired t(19) = -1.91, p = 0.07128. Pairing makes the biological unit the plant and the analyzed response its change.

Teaching pointA large p-value does not make the simulated nonzero effect disappear; paired differences define the target.

Analysis code excerpt

d <- read.csv("modules/paired_t/generated/data.csv")
change <- d$after - d$before
fit <- t.test(d$after, d$before, paired = TRUE,
              alternative = "two.sided")

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


	Paired t-test

data:  d$after and d$before
t = -1.9105, df = 19, p-value = 0.07128
alternative hypothesis: true mean difference is not equal to 0
95 percent confidence interval:
 -1.52238792  0.06942308
sample estimates:
mean difference 
     -0.7264824

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/paired_t/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(1103)
before <- rnorm(20, 18, 2.2)
after <- before + rnorm(20, -1.3, 1.7)
d <- data.frame(plant = 1:20, before = before, after = after)
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/paired_t/analysis.R modules/paired_t/generated

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Dataset 05 · Variance and diagnostics

F test of two variances

CSV rows: 60Seed: 1109Module: f_variance

Saved CSV Full runnable R script Full-size plot

Evolutionary developmental biologists compare variation in wing length between two inbred fly lines.

Question and null

Scientific question

Is wing-length variance the same in the two fly lines?

Null being tested / estimation target

The population variance ratio (Line A / Line B) is 1.

Generating truth

Population means are equal, but the true variance ratio A/B is 0.11 squared / 0.21 squared, approximately 0.274.

Exact generating model

30 wing lengths per independent fly line come from normal distributions with shared mean 2.6 mm, SD 0.11 for Line A and SD 0.21 for Line B.

Design and independent unit: Thirty independent flies per line are reared under a common environment.

Columns in the saved CSV

Column	Meaning	Unit / type
`line`	Fly line	category
`wing_mm`	Wing length of one independent fly	mm

Preview the actual saved observations

line	wing_mm
Line A	2.4658158961294
Line A	2.52823113418399
Line A	2.70352610924267
Line A	2.70954325259251
Line A	2.58877954006514

Analysis from the saved dataset

F test of two variances plot. Evolutionary developmental biologists compare variation in wing length between two inbred fly lines. — Simulated observations; generated by this module’s R script.

Variance ratio (A / B) = 0.376
95% CI for variance ratio: 0.179 to 0.791
F = 0.376; df = 29, 29; p = 0.01048

Wing-length variance in Line A was 0.376 times that in Line B (95% CI 0.179 to 0.791); F(29, 29) = 0.376, p = 0.01048. The target is variability, and this exact F reference relies strongly on normal populations.

Teaching pointA variance question can have a different answer from a mean question; the exact F test depends strongly on normality.

Analysis code excerpt

d <- read.csv("modules/f_variance/generated/data.csv")
d$line <- factor(d$line, levels = c("Line A", "Line B"))
fit <- var.test(wing_mm ~ line, data = d, ratio = 1,
                alternative = "two.sided", conf.level = .95)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


	F test to compare two variances

data:  wing_mm by line
F = 0.37637, num df = 29, denom df = 29, p-value = 0.01048
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
 0.1791387 0.7907512
sample estimates:
ratio of variances 
         0.3763697

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/f_variance/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(1109)
d <- data.frame(line = rep(c("Line A", "Line B"), each = 30),
                wing_mm = c(rnorm(30, 2.6, .11), rnorm(30, 2.6, .21)))
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/f_variance/analysis.R modules/f_variance/generated

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Dataset 06 · Variance and diagnostics

Levene and Brown–Forsythe tests

CSV rows: 84Seed: 1110Module: levene

Saved CSV Full runnable R script Full-size plot

Microbiologists compare the variability of colony diameters across three nutrient media.

Question and null

Scientific question

Do population variances differ among nutrient media?

Null being tested / estimation target

The groups have equal variances; the test compares absolute deviations from their group centers.

Generating truth

Group geometric means are 4 mm, but variances differ. Arithmetic means also vary slightly because log-normal means depend on log-scale spread.

Exact generating model

28 independently cultured colonies per medium have log-normal diameters. All have meanlog log(4); log-scale SD is 0.12, 0.20, or 0.35 for Low, Medium, and High medium.

Design and independent unit: Independent colonies are grown in separately inoculated culture wells; one colony measured per well.

Columns in the saved CSV

Column	Meaning	Unit / type
`medium`	Nutrient medium	category
`diameter_mm`	Diameter of a colony from one independent well	mm

Preview the actual saved observations

medium	diameter_mm
Low	3.9804700059381
Low	4.6128778865365
Low	3.64741655289966
Low	3.98306846588256
Low	4.07284163093935

Analysis from the saved dataset

Levene and Brown–Forsythe tests plot. Microbiologists compare the variability of colony diameters across three nutrient media. — Simulated observations; generated by this module’s R script.

Brown–Forsythe F = 5.55; df = 2, 81; p = 0.005524
Original Levene F = 9.72; p = 0.0001645
Observed group SDs: Low 0.59; Medium 0.84; High 1.64

The median-centered Brown–Forsythe comparison gave F(2, 81) = 5.55, p = 0.005524. Inspect the distributions as well as the p-value: the procedure compares absolute deviations, which can reflect both spread and shape.

Teaching pointMedian-centered Brown–Forsythe reduces sensitivity to long tails; it is not a required gate before Welch mean comparisons.

Analysis code excerpt

d <- read.csv("modules/levene/generated/data.csv")
d$medium <- factor(d$medium, levels = c("Low", "Medium", "High"))
brown_forsythe <- leveneTest(diameter_mm ~ medium, data = d,
                            center = median)
original_levene <- leveneTest(diameter_mm ~ medium, data = d,
                              center = mean)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output

Levene's Test for Homogeneity of Variance (center = median)
      Df F value   Pr(>F)   
group  2  5.5471 0.005524 **
      81                    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Levene's Test for Homogeneity of Variance (center = mean)
      Df F value    Pr(>F)    
group  2  9.7209 0.0001645 ***
      81                      
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/levene/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(1110)
d <- data.frame(medium = rep(c("Low", "Medium", "High"), each = 28),
                diameter_mm = c(rlnorm(28, log(4), .12),
                rlnorm(28, log(4), .20), rlnorm(28, log(4), .35)))
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/levene/analysis.R modules/levene/generated

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Dataset 07 · Variance and diagnostics

Shapiro–Wilk and graphical checks

CSV rows: 36Seed: 1111Module: shapiro

Saved CSV Full runnable R script Full-size plot

Cell biologists examine the distribution of time to lysis among independent cultured cells.

Question and null

Scientific question

Is a normal model a reasonable working description of these lysis times?

Null being tested / estimation target

The independent observations come from a normal distribution.

Generating truth

The population is log-normal and positively skewed, not normal. The population geometric mean is 45 minutes.

Exact generating model

36 independent lysis times follow a log-normal distribution with meanlog log(45) and log-scale SD 0.5.

Design and independent unit: One lysis time for each independently treated culture; use a diagnostic to investigate shape.

Columns in the saved CSV

Column	Meaning	Unit / type
`culture`	Independent culture identifier	none
`lysis_minutes`	Time until lysis	minutes

Preview the actual saved observations

culture	lysis_minutes
1	43.0935112245607
2	87.1756199933112
3	61.9615176128064
4	80.9681422942562
5	47.6940973120447

Analysis from the saved dataset

Shapiro–Wilk and graphical checks plot. Cell biologists examine the distribution of time to lysis among independent cultured cells. — Simulated observations; generated by this module’s R script.

Shapiro–Wilk W = 0.937; p = 0.04064; n = 36
Curvature in the Q–Q plot indicates a shape mismatch.
A nonsignificant result would not establish normality.

For the 36 lysis times, Shapiro–Wilk W = 0.937, p = 0.04064. Use the Q–Q plot to understand the mismatch; this diagnostic alone does not choose the scientific test or show that a mean comparison is invalid.

Teaching pointA diagnostic explains how a working model misses the data; its p-value should not automatically select the scientific analysis.

Analysis code excerpt

d <- read.csv("modules/shapiro/generated/data.csv")
fit <- shapiro.test(d$lysis_minutes)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


	Shapiro-Wilk normality test

data:  d$lysis_minutes
W = 0.93687, p-value = 0.04064

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/shapiro/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(1111)
d <- data.frame(culture = 1:36,
                lysis_minutes = rlnorm(36, log(45), .5))
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/shapiro/analysis.R modules/shapiro/generated

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Dataset 08 · Nonparametric alternatives

Sign test

CSV rows: 30Seed: 1104Module: sign_test

Saved CSV Full runnable R script Full-size plot

Behavioral biologists record the change in refuge-use time after a predator cue.

Question and null

Scientific question

Is an increase just as likely as a decrease in refuge-use time?

Null being tested / estimation target

Among nonzero changes, positive and negative changes are equally likely; for continuous differences, the median is zero.

Generating truth

Positive changes are more likely than negative changes. The unrounded mean is 4 seconds, but the sign test is about direction rather than that mean.

Exact generating model

30 independent changes are Gamma(shape 1.4, scale 5) minus 3 seconds, rounded to 0.1 seconds. Exact zeros, if present, are excluded from the sign comparison.

Design and independent unit: Each independent fish is measured twice; the changes are skewed.

Columns in the saved CSV

Column	Meaning	Unit / type
`fish`	Independent fish and pair identifier	none
`change_seconds`	After-minus-before refuge-use time	seconds

Preview the actual saved observations

fish	change_seconds
1	1.3
2	-0.6
3	10
4	8.4
5	2.8

Analysis from the saved dataset

Sign test plot. Behavioral biologists record the change in refuge-use time after a predator cue. — Simulated observations; generated by this module’s R script.

Positive changes: 18 of 30 nonzero differences
Zero changes excluded: 0
Median observed change = 1.25 seconds
95% CI for positive-change probability: 0.41 to 0.77
Exact two-sided p = 0.3616

18 of 30 nonzero changes were positive (exact sign-test p = 0.3616; 95% CI for the positive-change probability 0.41 to 0.77). The test asks about direction, not mean change; 0 exact zero differences were excluded.

Teaching pointA sign test handles asymmetric differences by asking about the probability of a positive change.

Analysis code excerpt

d <- read.csv("modules/sign_test/generated/data.csv")
nonzero <- d$change_seconds[d$change_seconds != 0]
positive <- sum(nonzero > 0)
fit <- binom.test(positive, length(nonzero), p = .5,
                 alternative = "two.sided")

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


	Exact binomial test

data:  positive and length(nonzero)
number of successes = 18, number of trials = 30, p-value = 0.3616
alternative hypothesis: true probability of success is not equal to 0.5
95 percent confidence interval:
 0.4060349 0.7734424
sample estimates:
probability of success 
                   0.6

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/sign_test/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(1104)
change <- rgamma(30, shape = 1.4, scale = 5) - 3
d <- data.frame(fish = 1:30, change_seconds = round(change, 1))
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/sign_test/analysis.R modules/sign_test/generated

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Dataset 09 · Nonparametric alternatives

Wilcoxon signed-rank test

CSV rows: 28Seed: 1105Module: signed_rank

Saved CSV Full runnable R script Full-size plot

Microbiologists compare log bacterial density before and after a treatment in matched cultures.

Question and null

Scientific question

Is the distribution of within-culture changes centered at zero?

Null being tested / estimation target

The difference distribution is symmetric about zero.

Generating truth

The symmetric difference distribution is centered at +0.28 log10 units; its zero-center null is false.

Exact generating model

28 baselines come from Normal(6, 0.4 SD). Paired changes equal 0.28 + 0.28 times a Student t variable with 4 degrees of freedom; after equals before plus change.

Design and independent unit: Each independent culture contributes one difference; occasional large differences make ranks useful.

Columns in the saved CSV

Column	Meaning	Unit / type
`culture`	Independent culture and pair identifier	none
`before`	Log10 bacterial density before treatment	log10 CFU/mL
`after`	Log10 bacterial density after treatment	log10 CFU/mL

Preview the actual saved observations

culture	before	after
1	6.78593799756038	7.1849263353085
2	6.07893070261203	5.8610915500608
3	5.47851842742343	5.65918500394574
4	6.1870424999375	6.51968430410169
5	6.23750914625024	6.52507806739053

Analysis from the saved dataset

Wilcoxon signed-rank test plot. Microbiologists compare log bacterial density before and after a treatment in matched cultures. — Simulated observations; generated by this module’s R script.

Pseudomedian change estimate = 0.249
95% CI: 0.094 to 0.341
V = 328.0; p = 0.004582

The estimated pseudomedian log10 density change was 0.249 (95% CI 0.094 to 0.341); signed-rank V = 328.0, p = 0.004582. Under a symmetric location-shift model this describes the common center of the differences; the procedure does not generally test a mean.

Teaching pointSigned-rank is useful for symmetric heavy-tailed differences; sign is the simpler alternative when symmetry is doubtful.

Analysis code excerpt

d <- read.csv("modules/signed_rank/generated/data.csv")
change <- d$after - d$before
fit <- wilcox.test(d$after, d$before, paired = TRUE,
                   exact = FALSE, conf.int = TRUE)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


	Wilcoxon signed rank test with continuity correction

data:  d$after and d$before
V = 328, p-value = 0.004582
alternative hypothesis: true location shift is not equal to 0
95 percent confidence interval:
 0.09375924 0.34088634
sample estimates:
(pseudo)median 
     0.2486103

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/signed_rank/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(1105)
before <- rnorm(28, 6, .4)
after <- before + .28 + .28 * rt(28, df = 4)
d <- data.frame(culture = 1:28, before = before, after = after)
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/signed_rank/analysis.R modules/signed_rank/generated

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Dataset 10 · Nonparametric alternatives

Wilcoxon rank-sum test

CSV rows: 64Seed: 1106Module: rank_sum

Saved CSV Full runnable R script Full-size plot

Ecologists compare parasite burdens among fish from two independent lakes.

Question and null

Scientific question

Do parasite-burden distributions differ between the two lakes?

Null being tested / estimation target

The two groups have the same response distribution.

Generating truth

The lake distributions differ, including their mean and spread. The same-distribution null is false; there is not a simple equal-shape location shift.

Exact generating model

32 independent fish per lake have negative-binomial parasite counts. Clear has mean 5 and Reedy mean 10, both with size parameter 1.5; variances are mean + mean squared / 1.5.

Design and independent unit: Each fish is sampled once; counts are strongly skewed and tied.

Columns in the saved CSV

Column	Meaning	Unit / type
`lake`	Lake of independently sampled fish	category
`parasites`	Parasite burden of one fish	count

Preview the actual saved observations

lake	parasites
Clear	7
Clear	4
Clear	0
Clear	6
Clear	8

Analysis from the saved dataset

Wilcoxon rank-sum test plot. Ecologists compare parasite burdens among fish from two independent lakes. — Simulated observations; generated by this module’s R script.

Observed medians: Clear 3.0; Reedy 8.0 parasites
W = 362.5; asymptotic p = 0.04452
Medians describe these data; the general null concerns distributions.

Parasite burdens had observed medians 3.0 (Clear) and 8.0 (Reedy). The rank-sum comparison gave W = 362.5, p = 0.04452. This is evidence about the distributions; calling it a median test requires additional shape assumptions.

Teaching pointA rank-sum result can reflect distributional differences; the mean-count question also has a direct count-model route.

Analysis code excerpt

d <- read.csv("modules/rank_sum/generated/data.csv")
fit <- wilcox.test(parasites ~ lake, data = d,
                   exact = FALSE, correct = TRUE)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


	Wilcoxon rank sum test with continuity correction

data:  parasites by lake
W = 362.5, p-value = 0.04452
alternative hypothesis: true location shift is not equal to 0

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/rank_sum/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(1106)
d <- data.frame(lake = rep(c("Clear", "Reedy"), each = 32),
                parasites = c(rnbinom(32, mu = 5, size = 1.5),
                              rnbinom(32, mu = 10, size = 1.5)))
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/rank_sum/analysis.R modules/rank_sum/generated

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Dataset 11 · ANOVA

One-way ANOVA

CSV rows: 75Seed: 1117Module: one_way_anova

Saved CSV Full runnable R script Full-size plot

Ecologists compare seedling biomass under three independent soil treatments.

Question and null

Scientific question

Are mean seedling biomasses equal across the three soils?

Null being tested / estimation target

All soil-treatment population means are equal.

Generating truth

The equal-means null is false. All three populations share variance 0.85 squared; Loam minus Sand is truly 0.9 g.

Exact generating model

25 independent biomass values per soil come from normal distributions: Sand mean 4.5, Loam mean 5.4, Clay mean 5.1 g; every group has SD 0.85 g.

Design and independent unit: Twenty-five independently assigned pots per soil treatment; one biomass value per pot.

Columns in the saved CSV

Column	Meaning	Unit / type
`soil`	Assigned soil treatment	category
`biomass_g`	Seedling biomass from one independent pot	g

Preview the actual saved observations

soil	biomass_g
Sand	1.98665566013475
Sand	3.19457483907652
Sand	3.81605028387267
Sand	5.40523457987103
Sand	4.40294379235544

Analysis from the saved dataset

One-way ANOVA plot. Ecologists compare seedling biomass under three independent soil treatments. — Simulated observations; generated by this module’s R script.

Ordinary ANOVA F(2, 72) = 11.59; p = 4.335e-05
Welch ANOVA F = 8.89; denominator df = 46.37; p = 0.0005408
Observed means (g): Sand 4.28; Loam 5.28; Clay 5.08
Neither omnibus test identifies a specific differing pair.

Mean seedling biomass was compared across soils with ordinary ANOVA: F(2, 72) = 11.59, p = 4.335e-05. Welch ANOVA, which relaxes equal variances, gave p = 0.0005408. Use a planned contrast or adjusted pairwise comparisons to answer which means differ.

Teaching pointThe omnibus equal-means question differs from a scientific question about one particular contrast.

Analysis code excerpt

d <- read.csv("modules/one_way_anova/generated/data.csv")
d$soil <- factor(d$soil, levels = c("Sand", "Loam", "Clay"))
fit <- aov(biomass_g ~ soil, data = d)
anova_table <- summary(fit)[[1]]
welch <- oneway.test(biomass_g ~ soil, data = d,
                    var.equal = FALSE)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output

            Df Sum Sq Mean Sq F value   Pr(>F)    
soil         2  14.14   7.068   11.59 4.34e-05 ***
Residuals   72  43.92   0.610                     
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

	One-way analysis of means (not assuming equal variances)

data:  biomass_g and soil
F = 8.8862, num df = 2.000, denom df = 46.371, p-value = 0.0005408

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/one_way_anova/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(1117)
d <- data.frame(soil = rep(c("Sand", "Loam", "Clay"), each = 25),
                biomass_g = c(rnorm(25, 4.5, .85), rnorm(25, 5.4, .85),
                              rnorm(25, 5.1, .85)))
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/one_way_anova/analysis.R modules/one_way_anova/generated

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Dataset 12 · ANOVA

Welch one-way ANOVA

CSV rows: 78Seed: 1119Module: welch_anova

Saved CSV Full runnable R script Full-size plot

Ecologists compare seedling biomass under three independent soil treatments.

Question and null

Scientific question

Are mean seedling biomasses equal across the three soils?

Null being tested / estimation target

All soil-treatment population means are equal.

Generating truth

The true means differ, and variances differ (0.36, 1.21, and 2.25 g squared). Welch addresses the equal-means null without a shared variance.

Exact generating model

Independent soil groups have different sample sizes and normal biomass distributions: Sand n=20, mean 4.5, SD 0.6; Loam n=26, mean 5.4, SD 1.1; Clay n=32, mean 5.1, SD 1.5.

Design and independent unit: Independent pots in three soil treatments; sample sizes and group variances differ.

Columns in the saved CSV

Column	Meaning	Unit / type
`soil`	Assigned soil treatment	category
`biomass_g`	Seedling biomass from one independent pot	g

Preview the actual saved observations

soil	biomass_g
Sand	4.65612728606543
Sand	4.69014305099215
Sand	4.86568502600326
Sand	4.2595795115919
Sand	3.62311763376286

Analysis from the saved dataset

Welch one-way ANOVA plot. Ecologists compare seedling biomass under three independent soil treatments. — Simulated observations; generated by this module’s R script.

Welch ANOVA F = 13.65; df = 2, 49.85; p = 1.873e-05
Group sample sizes: Sand 20; Loam 26; Clay 32
Observed means (g): Sand 4.4; Loam 5.55; Clay 4.87
Unequal variance is modeled; the omnibus test does not select a pair.

Welch ANOVA compared mean biomass across soils while allowing unequal variances: F(2, 49.85) = 13.65, p = 1.873e-05. This answers the same equal-means question as ordinary ANOVA; a specific pair requires an appropriate adjusted follow-up.

Teaching pointUnequal variances have a parametric solution that preserves the scientific question about means.

Analysis code excerpt

d <- read.csv("modules/welch_anova/generated/data.csv")
d$soil <- factor(d$soil, levels = c("Sand", "Loam", "Clay"))
fit <- oneway.test(biomass_g ~ soil, data = d,
                  var.equal = FALSE)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


	One-way analysis of means (not assuming equal variances)

data:  biomass_g and soil
F = 13.65, num df = 2.000, denom df = 49.854, p-value = 1.873e-05

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/welch_anova/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(1119)
d <- data.frame(soil = rep(c("Sand", "Loam", "Clay"), c(20, 26, 32)),
                biomass_g = c(rnorm(20, 4.5, .6), rnorm(26, 5.4, 1.1),
                              rnorm(32, 5.1, 1.5)))
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/welch_anova/analysis.R modules/welch_anova/generated

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Dataset 13 · Nonparametric alternatives

Kruskal–Wallis test

CSV rows: 72Seed: 1107Module: kruskal_wallis

Saved CSV Full runnable R script Full-size plot

Plant pathologists score leaf damage in three independently assigned pathogen treatments.

Question and null

Scientific question

Do damage-score distributions differ among pathogen treatments?

Null being tested / estimation target

All groups share the same response distribution.

Generating truth

The score distributions differ across all three generating probabilities; expected scores are 2.0, 3.2, and 4.64. The equal-distributions null is false.

Exact generating model

24 independent plants per group receive binomial damage scores with 8 trials and probabilities 0.25 (Control), 0.40 (Strain A), and 0.58 (Strain B). The score is used as an ordered severity scale.

Design and independent unit: Twenty-four independent plants per treatment; ordinal damage scores run from 0 to 8.

Columns in the saved CSV

Column	Meaning	Unit / type
`treatment`	Assigned pathogen treatment	category
`damage`	Ordered leaf-damage score	score 0 to 8

Preview the actual saved observations

treatment	damage
Control	2
Control	2
Control	4
Control	0
Control	3

Analysis from the saved dataset

Kruskal–Wallis test plot. Plant pathologists score leaf damage in three independently assigned pathogen treatments. — Simulated observations; generated by this module’s R script.

Kruskal–Wallis chi-square = 35.75; df = 2; p = 1.73e-08
Observed medians: Control 2; Strain A 3; Strain B 4.5
The omnibus result does not identify which pairs differ.

Damage-score distributions were compared using Kruskal–Wallis: chi-square(2) = 35.75, p = 1.73e-08. An omnibus result concerns at least one group difference; it does not identify a particular pair or generally establish different medians.

Teaching pointAn omnibus rank result identifies a group-level difference without choosing a pair.

Analysis code excerpt

d <- read.csv("modules/kruskal_wallis/generated/data.csv")
d$treatment <- factor(d$treatment, levels = c("Control", "Strain A", "Strain B"))
fit <- kruskal.test(damage ~ treatment, data = d)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


	Kruskal-Wallis rank sum test

data:  damage by treatment
Kruskal-Wallis chi-squared = 35.745, df = 2, p-value = 1.73e-08

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/kruskal_wallis/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(1107)
d <- data.frame(treatment = rep(c("Control", "Strain A", "Strain B"), each = 24),
                damage = c(rbinom(24, 8, .25), rbinom(24, 8, .40),
                           rbinom(24, 8, .58)))
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/kruskal_wallis/analysis.R modules/kruskal_wallis/generated

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Dataset 14 · ANOVA and contrasts

Factorial ANOVA: interactions

CSV rows: 88Seed: 2101Module: factorial_anova

Saved CSV Full runnable R script Full-size plot

Beetle larvae from two genotypes are independently reared at two temperatures.

Question and null

Scientific question

Does the temperature effect on mean growth depend on genotype?

Null being tested / estimation target

The warm-minus-cool mean growth difference is the same for genotypes A and B.

Generating truth

The temperature effect is 2 mg/week in genotype A and 4.3 mg/week in B, so the interaction is 2.3 mg/week. The interaction null is false.

Exact generating model

There are 22 larvae in each of four genotype-by-temperature combinations. Growth = 14 + 1.4 if genotype B + 2 if warm + 2.3 if both genotype B and warm + independent Normal(0, 2.1²) error, in mg/week. Values are not rounded.

Design and independent unit: 22 independent larvae per genotype × temperature combination.

Columns in the saved CSV

Column	Meaning	Unit / type
`replicate`	Larva number within a genotype/temperature combination	ID
`genotype`	Genotype A or B	category
`heat`	Cool or warm rearing condition	category
`growth`	Larval growth rate	mg/week

Preview the actual saved observations

replicate	genotype	heat	growth
1	A	Cool	16.7089844789053
2	A	Cool	13.5201815473616
3	A	Cool	15.5885877638367
4	A	Cool	14.1910320609535
5	A	Cool	14.1142847985379

Analysis from the saved dataset

Factorial ANOVA: interactions plot. Beetle larvae from two genotypes are independently reared at two temperatures. — Simulated observations; generated by this module’s R script.

Interaction = 4.30 mg/week
95% CI 2.80 to 5.81
t(84) = 5.69; p = 1.78e-07

The warm-minus-cool growth effect was 4.30 mg/week greater in genotype B than A (95% CI 2.80 to 5.81; p = 1.78e-07). The interaction asks whether the temperature effect depends on genotype; a single average temperature effect would hide this difference.

Teaching pointAn interaction is a difference of differences: a main-effect summary can conceal it.

Analysis code excerpt

d <- read.csv("modules/factorial_anova/generated/data.csv")
fit <- lm(growth ~ genotype * heat, data = d)
summary(fit)
interaction_ci <- confint(fit)["genotypeB:heatWarm", ]

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


Call:
lm(formula = growth ~ genotype * heat, data = d)

Residuals:
    Min      1Q  Median      3Q     Max 
-3.8521 -1.1843 -0.0363  1.0850  4.6476 

Coefficients:
                   Estimate Std. Error t value Pr(>|t|)    
(Intercept)         14.9526     0.3779  39.567  < 2e-16 ***
genotypeB            0.2676     0.5344   0.501    0.618    
heatWarm             0.7240     0.5344   1.355    0.179    
genotypeB:heatWarm   4.3035     0.7558   5.694 1.78e-07 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1.773 on 84 degrees of freedom
Multiple R-squared:  0.6099,	Adjusted R-squared:  0.5959 
F-statistic: 43.77 on 3 and 84 DF,  p-value: < 2.2e-16

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/factorial_anova/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(2101)
d <- expand.grid(replicate = 1:22, genotype = c("A", "B"),
                 heat = c("Cool", "Warm"))
d$growth <- with(d, 14 + 1.4 * (genotype == "B") +
  2 * (heat == "Warm") + 2.3 * (genotype == "B" & heat == "Warm") +
  rnorm(nrow(d), 0, 2.1))
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/factorial_anova/analysis.R modules/factorial_anova/generated

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Dataset 15 · ANOVA and contrasts

ANCOVA: an adjusted group comparison

CSV rows: 90Seed: 2102Module: ancova

Saved CSV Full runnable R script Full-size plot

Plant ecologists compare nutrient treatment while accounting for initial plant height.

Question and null

Scientific question

Do treatments differ in mean biomass at the same initial height?

Null being tested / estimation target

The treatment mean difference is zero at a common initial height in the stated model.

Generating truth

The nutrient treatment increases the population mean biomass by 2.4 g at any shared initial height. Both groups have slope 1.5 g/cm. The adjusted treatment null is false.

Exact generating model

45 independent plants are assigned to each treatment. Initial height is Uniform(3, 9) cm. Biomass = 2 + 1.5 × initial height + 2.4 if nutrient-treated + independent Normal(0, 1.8²) error, in g. No rounding is applied.

Design and independent unit: 90 independent plants; treatment groups overlap in initial height.

Columns in the saved CSV

Column	Meaning	Unit / type
`plant`	Independent plant identifier	ID
`treatment`	Control or nutrient addition	category
`initial`	Height before treatment	cm
`biomass`	Final dry biomass	g

Preview the actual saved observations

plant	treatment	initial	biomass
1	Control	5.70954237738624	11.3170776975508
2	Control	7.87733154091984	12.8563258031163
3	Control	4.50338011421263	11.1223644431843
4	Control	5.56966270180419	12.574935834342
5	Control	6.05019523808733	9.98673653862866

Analysis from the saved dataset

ANCOVA: an adjusted group comparison plot. Plant ecologists compare nutrient treatment while accounting for initial plant height. — Simulated observations; generated by this module’s R script.

Adjusted nutrient minus control = 2.65 g
95% CI 1.84 to 3.45
t(87) = 6.55; p = 3.797e-09

At the same initial height, nutrient-treated plants had 2.65 g greater mean biomass (95% CI 1.84 to 3.45; p = 3.797e-09). This fitted model assumes a common height slope; the adjusted treatment comparison is supported by overlapping height ranges.

Teaching pointThe scientific comparison holds the covariate fixed; the raw mean comparison is different.

Analysis code excerpt

d <- read.csv("modules/ancova/generated/data.csv")
fit <- lm(biomass ~ treatment + initial, data = d)
summary(fit)
adjusted_ci <- confint(fit)["treatmentNutrient", ]

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


Call:
lm(formula = biomass ~ treatment + initial, data = d)

Residuals:
   Min     1Q Median     3Q    Max 
-3.692 -1.195  0.030  1.050  4.634 

Coefficients:
                  Estimate Std. Error t value Pr(>|t|)    
(Intercept)         2.1147     0.7830   2.701  0.00831 ** 
treatmentNutrient   2.6479     0.4040   6.555  3.8e-09 ***
initial             1.4427     0.1212  11.900  < 2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1.899 on 87 degrees of freedom
Multiple R-squared:  0.6571,	Adjusted R-squared:  0.6492 
F-statistic: 83.35 on 2 and 87 DF,  p-value: < 2.2e-16

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/ancova/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(2102)
d <- data.frame(plant = 1:90, treatment = rep(c("Control", "Nutrient"), each = 45),
                initial = runif(90, 3, 9))
d$biomass <- with(d, 2 + 1.5 * initial +
  2.4 * (treatment == "Nutrient") + rnorm(90, 0, 1.8))
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/ancova/analysis.R modules/ancova/generated

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Dataset 16 · Repeated and clustered data

Repeated-measures ANOVA

CSV rows: 96Seed: 2103Module: repeated_anova

Saved CSV Full runnable R script Full-size plot

Physiologists measure the same animals at four assay temperatures.

Question and null

Scientific question

Are mean oxygen-use rates equal across temperatures within animals?

Null being tested / estimation target

All four population temperature means are equal.

Generating truth

Population means are 8, 9.75, 11.5 and 13.25 µmol/hour. The endpoint difference is 5.25 µmol/hour and the equal-means null is false.

Exact generating model

24 animals are measured at 15, 20, 25 and 30°C. Oxygen use = 8 + 0.35 × (temperature − 15) + an animal-specific Normal(0, 2²) intercept + independent Normal(0, 1.3²) observation error. Intercepts are shared within animals, producing repeated-measure dependence and a spherical covariance structure. No rounding is applied.

Design and independent unit: Each of 24 animals provides all four measurements.

Columns in the saved CSV

Column	Meaning	Unit / type
`temperature`	Assay temperature	°C
`subject`	Animal measured at all four temperatures	ID
`oxygen`	Measured oxygen consumption	µmol/hour

Preview the actual saved observations

temperature	subject	oxygen
15	1	8.10836399815
20	1	7.98710181120176
25	1	13.7872320434406
30	1	14.0568662705288
15	2	9.8833785773146

Analysis from the saved dataset

Repeated-measures ANOVA plot. Physiologists measure the same animals at four assay temperatures. — Simulated observations; generated by this module’s R script.

Condition F(3, 69) = 84.32
Omnibus p = 4.966e-23
Planned 30−15°C mean difference = 5.68 µmol/hour
95% CI 4.99 to 6.38

Within the same 24 animals, temperature means differed (F(3, 69) = 84.32, p = 4.966e-23). The prespecified 30−15°C mean increase was 5.68 µmol/hour (95% CI 4.99 to 6.38). The repeated-measures F test assumes sphericity; the experiment has 24 independent animals, not 96 independent rows.

Teaching pointThe subject ID, not the row, defines independent replication.

Analysis code excerpt

d <- read.csv("modules/repeated_anova/generated/data.csv")
fit <- aov(oxygen ~ condition + Error(subject/condition), data = d)
summary(fit)
endpoints <- t.test(d$oxygen[d$temperature == 30],
                   d$oxygen[d$temperature == 15], paired = TRUE)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


Error: subject
          Df Sum Sq Mean Sq F value Pr(>F)
Residuals 23  285.9   12.43               

Error: subject:condition
          Df Sum Sq Mean Sq F value Pr(>F)    
condition  3  467.9  155.98   84.32 <2e-16 ***
Residuals 69  127.6    1.85                   
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/repeated_anova/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(2103)
d <- expand.grid(temperature = c(15, 20, 25, 30), subject = 1:24)
individual <- rnorm(24, 0, 2)
d$oxygen <- 8 + .35 * (d$temperature - 15) +
  individual[d$subject] + rnorm(nrow(d), 0, 1.3)
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/repeated_anova/analysis.R modules/repeated_anova/generated

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Dataset 17 · Nonparametric alternatives

Friedman test

CSV rows: 54Seed: 1108Module: friedman

Saved CSV Full runnable R script Full-size plot

Sensory biologists record activity scores from each insect under three light conditions.

Question and null

Scientific question

Do light conditions differ after accounting for insect-to-insect variation?

Null being tested / estimation target

Within blocks, condition labels are exchangeable: no systematic condition effect.

Generating truth

Condition effects exist before rounding (0, +0.8, +1.6), and condition score distributions differ after rounding. Repeated observations share an insect baseline.

Exact generating model

18 insects each receive all 3 conditions. Each insect has a Normal(5, 1.3 SD) baseline; Dim, Moderate, and Bright add 0, 0.8, and 1.6. Independent Normal(0, 1 SD) noise is added, then scores are rounded and restricted to 0–10.

Design and independent unit: Eighteen independent insects each complete all three conditions in randomized order.

Columns in the saved CSV

Column	Meaning	Unit / type
`insect`	Independent insect/block identifier	none
`condition`	Light condition	category
`activity_score`	Rounded and bounded activity measure	score 0 to 10

Preview the actual saved observations

insect	condition	activity_score
1	Dim	6
1	Moderate	7
1	Bright	7
2	Dim	7
2	Moderate	7

Analysis from the saved dataset

Friedman test plot. Sensory biologists record activity scores from each insect under three light conditions. — Simulated observations; generated by this module’s R script.

Independent blocks = 18; conditions = 3
Friedman chi-square = 17.63; df = 2; p = 0.0001482
Condition medians: Dim 5; Moderate 6; Bright 7

For 18 insects measured under all three conditions, the Friedman comparison gave chi-square(2) = 17.63, p = 0.0001482. This tests a systematic condition effect within insects; it does not identify which conditions differ.

Teaching pointRank within each insect so that between-insect differences do not masquerade as condition effects.

Analysis code excerpt

d <- read.csv("modules/friedman/generated/data.csv")
d$condition <- factor(d$condition, levels = c("Dim", "Moderate", "Bright"))
d$insect <- factor(d$insect)
fit <- friedman.test(activity_score ~ condition | insect, data = d)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


	Friedman rank sum test

data:  activity_score and condition and insect
Friedman chi-squared = 17.633, df = 2, p-value = 0.0001482

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/friedman/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(1108)
insect <- rep(1:18, each = 3)
condition <- rep(c("Dim", "Moderate", "Bright"), 18)
score <- pmax(0, pmin(10, round(rep(rnorm(18, 5, 1.3), each = 3) +
             rep(c(0, .8, 1.6), 18) + rnorm(54, 0, 1))))
d <- data.frame(insect, condition, activity_score = score)
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/friedman/analysis.R modules/friedman/generated

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Dataset 18 · ANOVA and contrasts

Tukey / Tukey–Kramer comparisons

CSV rows: 80Seed: 2104Module: tukey

Saved CSV Full runnable R script Full-size plot

Plant growth is compared across four nutrient regimes.

Question and null

Scientific question

Which pairs of nutrient means differ?

Null being tested / estimation target

For each pair, the two population means are equal.

Generating truth

Population means for A–D are 8, 9, 11.3 and 12.1 g; all distinct pairwise mean-equality nulls are false.

Exact generating model

20 independent plants per nutrient regime have normal biomass with means 8, 9, 11.3 and 12.1 g for A–D and a common SD of 1.7 g. Values are not rounded. The identical dataset is regenerated separately for each comparison procedure.

Design and independent unit: 20 independent plants per regime; all procedures use the same simulated dataset.

Columns in the saved CSV

Column	Meaning	Unit / type
`plant`	Independent plant identifier	ID
`nutrient`	Nutrient regime A–D	category
`biomass`	Dry biomass	g

Preview the actual saved observations

plant	nutrient	biomass
1	A	6.6708047417735
2	A	9.4212660941788
3	A	6.28560846867295
4	A	7.79904241740413
5	A	6.52604285760043

Analysis from the saved dataset

Tukey / Tukey–Kramer comparisons plot. Plant growth is compared across four nutrient regimes. — Simulated observations; generated by this module’s R script.

Family: all six pairwise nutrient comparisons
D minus C = 1.42 g
Simultaneous 95% CI 0.04 to 2.80
Tukey-adjusted p = 0.0415

Regime D exceeded C by 1.42 g (simultaneous 95% CI 0.04 to 2.80; Tukey-adjusted p = 0.0415). The interval and p-value account for all six pairwise comparisons; they do not protect every possible weighted contrast.

Teaching pointProtect the family of all six pairwise mean comparisons.

Analysis code excerpt

d <- read.csv("modules/tukey/generated/data.csv")
fit <- aov(biomass ~ nutrient, data = d)
comparisons <- TukeyHSD(fit, "nutrient", conf.level = .95)
comparisons

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output

            Df Sum Sq Mean Sq F value   Pr(>F)    
nutrient     3  222.4   74.14   26.94 5.68e-12 ***
Residuals   76  209.2    2.75                     
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
  Tukey multiple comparisons of means
    95% family-wise confidence level

Fit: aov(formula = biomass ~ nutrient, data = d)

$nutrient
        diff         lwr      upr     p adj
B-A 1.037343 -0.34073261 2.415418 0.2056314
C-A 2.905036  1.52696022 4.283111 0.0000025
D-A 4.322102  2.94402599 5.700177 0.0000000
C-B 1.867693  0.48961729 3.245768 0.0035291
D-B 3.284759  1.90668306 4.662834 0.0000001
D-C 1.417066  0.03899023 2.795141 0.0415003

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/tukey/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(2104)
d <- data.frame(plant = 1:80,
                nutrient = rep(c("A", "B", "C", "D"), each = 20))
d$biomass <- rnorm(80, rep(c(8, 9, 11.3, 12.1), each = 20), 1.7)
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/tukey/analysis.R modules/tukey/generated

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Dataset 19 · ANOVA and contrasts

Fisher least significant difference

CSV rows: 80Seed: 2104Module: fisher_lsd

Saved CSV Full runnable R script Full-size plot

Plant growth is compared across four nutrient regimes.

Question and null

Scientific question

What does an unadjusted pairwise comparison after the ANOVA gate test?

Null being tested / estimation target

For a chosen pair, the two population means are equal.

Generating truth

Population means for A–D are 8, 9, 11.3 and 12.1 g; all distinct pairwise mean-equality nulls are false.

Exact generating model

Design and independent unit: 20 independent plants per regime; all procedures use the same simulated dataset.

Columns in the saved CSV

Column	Meaning	Unit / type
`plant`	Independent plant identifier	ID
`nutrient`	Nutrient regime A–D	category
`biomass`	Dry biomass	g

Preview the actual saved observations

plant	nutrient	biomass
1	A	6.6708047417735
2	A	9.4212660941788
3	A	6.28560846867295
4	A	7.79904241740413
5	A	6.52604285760043

Analysis from the saved dataset

Fisher least significant difference plot. Plant growth is compared across four nutrient regimes. — Simulated observations; generated by this module’s R script.

Omnibus p = 5.683e-12
A - B = -1.04 g
Individual 95% CI -2.08 to 0.01
Unadjusted p = 0.05163
Protected LSD first requires a significant omnibus test.

The omnibus test had p = 5.683e-12, so the protected LSD gate is passed in this example. The A - B difference was -1.04 g (individual 95% CI -2.08 to 0.01; unadjusted p = 0.05163). This gate does not generally protect all pairwise decisions when some of four or more means differ; use Tukey for the all-pairs family.

Teaching pointAn omnibus gate plus unadjusted comparisons is not general all-pairs protection.

Analysis code excerpt

d <- read.csv("modules/fisher_lsd/generated/data.csv")
fit <- lm(biomass ~ nutrient, data = d)
omnibus_p <- anova(fit)$`Pr(>F)`[1]
means <- emmeans::emmeans(fit, ~ nutrient)
comparisons <- NULL
if (omnibus_p < .05) {
  comparisons <- summary(emmeans::contrast(means, "pairwise"),
                         infer = c(TRUE, TRUE), adjust = "none")
}

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


Call:
lm(formula = biomass ~ nutrient, data = d)

Residuals:
    Min      1Q  Median      3Q     Max 
-3.6874 -1.0377  0.1064  1.2669  3.2888 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)   8.2277     0.3710  22.179  < 2e-16 ***
nutrientB     1.0373     0.5246   1.977   0.0516 .  
nutrientC     2.9050     0.5246   5.537 4.21e-07 ***
nutrientD     4.3221     0.5246   8.239 3.85e-12 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1.659 on 76 degrees of freedom
Multiple R-squared:  0.5153,	Adjusted R-squared:  0.4962 
F-statistic: 26.94 on 3 and 76 DF,  p-value: 5.683e-12

 contrast estimate    SE df lower.CL upper.CL t.ratio p.value
 A - B       -1.04 0.525 76    -2.08  0.00753  -1.977  0.0516
 A - C       -2.91 0.525 76    -3.95 -1.86016  -5.537 <0.0001
 A - D       -4.32 0.525 76    -5.37 -3.27723  -8.239 <0.0001
 B - C       -1.87 0.525 76    -2.91 -0.82282  -3.560  0.0006
 B - D       -3.28 0.525 76    -4.33 -2.23988  -6.261 <0.0001
 C - D       -1.42 0.525 76    -2.46 -0.37219  -2.701  0.0085

Confidence level used: 0.95

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/fisher_lsd/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(2104)
d <- data.frame(plant = 1:80,
                nutrient = rep(c("A", "B", "C", "D"), each = 20))
d$biomass <- rnorm(80, rep(c(8, 9, 11.3, 12.1), each = 20), 1.7)
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/fisher_lsd/analysis.R modules/fisher_lsd/generated

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Dataset 20 · ANOVA and contrasts

Scheffé contrasts

CSV rows: 80Seed: 2104Module: scheffe

Saved CSV Full runnable R script Full-size plot

Plant growth is compared across four nutrient regimes.

Question and null

Scientific question

Does the average of regimes C and D differ from the average of A and B?

Null being tested / estimation target

The contrast (μC + μD)/2 − (μA + μB)/2 is zero.

Generating truth

Population means for A–D are 8, 9, 11.3 and 12.1 g; all distinct pairwise mean-equality nulls are false. The specified (C+D)/2 − (A+B)/2 contrast equals 3.2 g.

Exact generating model

Design and independent unit: 20 independent plants per regime; all procedures use the same simulated dataset.

Columns in the saved CSV

Column	Meaning	Unit / type
`plant`	Independent plant identifier	ID
`nutrient`	Nutrient regime A–D	category
`biomass`	Dry biomass	g

Preview the actual saved observations

plant	nutrient	biomass
1	A	6.6708047417735
2	A	9.4212660941788
3	A	6.28560846867295
4	A	7.79904241740413
5	A	6.52604285760043

Analysis from the saved dataset

Scheffé contrasts plot. Plant growth is compared across four nutrient regimes. — Simulated observations; generated by this module’s R script.

Contrast: (C + D)/2 − (A + B)/2
Estimated difference = 3.09 g
Scheffé 95% CI 2.03 to 4.16
Scheffé-adjusted p = 9.177e-11

The average of regimes C and D exceeded the average of A and B by 3.09 g (Scheffé simultaneous 95% CI 2.03 to 4.16; adjusted p = 9.177e-11). Using rank 3 protects all mean contrasts among these four groups, a larger family than the pairwise comparisons protected by Tukey.

Teaching pointA weighted contrast asks a specific biological question; Scheffé protects the larger all-contrast family.

Analysis code excerpt

d <- read.csv("modules/scheffe/generated/data.csv")
fit <- lm(biomass ~ nutrient, data = d)
means <- emmeans::emmeans(fit, ~ nutrient)
contrast <- emmeans::contrast(means,
  list("CD average minus AB average" = c(-.5, -.5, .5, .5)))
comparison <- summary(contrast, infer = c(TRUE, TRUE),
                      adjust = "scheffe", scheffe.rank = 3)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


Call:
lm(formula = biomass ~ nutrient, data = d)

Residuals:
    Min      1Q  Median      3Q     Max 
-3.6874 -1.0377  0.1064  1.2669  3.2888 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)   8.2277     0.3710  22.179  < 2e-16 ***
nutrientB     1.0373     0.5246   1.977   0.0516 .  
nutrientC     2.9050     0.5246   5.537 4.21e-07 ***
nutrientD     4.3221     0.5246   8.239 3.85e-12 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1.659 on 76 degrees of freedom
Multiple R-squared:  0.5153,	Adjusted R-squared:  0.4962 
F-statistic: 26.94 on 3 and 76 DF,  p-value: 5.683e-12

 contrast                    estimate    SE df lower.CL upper.CL t.ratio
 CD average minus AB average     3.09 0.371 76     2.03     4.16   8.343
 p.value
 <0.0001

Confidence level used: 0.95 
Conf-level adjustment: scheffe method with rank 3 
P value adjustment: scheffe method with rank 3

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/scheffe/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(2104)
d <- data.frame(plant = 1:80,
                nutrient = rep(c("A", "B", "C", "D"), each = 20))
d$biomass <- rnorm(80, rep(c(8, 9, 11.3, 12.1), each = 20), 1.7)
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/scheffe/analysis.R modules/scheffe/generated

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Dataset 21 · Rank methods

Dunn rank comparisons

CSV rows: 84Seed: 2105Module: dunn

Saved CSV Full runnable R script Full-size plot

Ecologists compare skewed leaf-damage measurements among four treatments.

Question and null

Scientific question

Which pairs differ in their rank tendencies?

Null being tested / estimation target

For a chosen pair, pooled-rank tendencies are equal under the common-distribution null.

Generating truth

Group distributions differ; their population medians are exp(1.2), exp(1.3), exp(1.7) and exp(1.9) mm². Dunn tests pooled-rank tendencies, not arithmetic mean equality. The distributions also differ in original-scale spread.

Exact generating model

21 independent leaves in each of four groups have log-normal damage. Log damage is Normal(mean, 0.48²), with means 1.2, 1.3, 1.7 and 1.9 for A–D. Damage remains positive and unrounded. All leaves come from different plants.

Design and independent unit: 21 independent leaves from different plants per treatment.

Columns in the saved CSV

Column	Meaning	Unit / type
`leaf`	Independent leaf from a distinct plant	ID
`treatment`	Treatment A–D	category
`damage`	Area removed from a leaf	mm²

Preview the actual saved observations

leaf	treatment	damage
1	A	2.38558409407124
2	A	5.56560940971648
3	A	2.18275187368594
4	A	6.04565914498524
5	A	5.86565211470618

Analysis from the saved dataset

Dunn rank comparisons plot. Ecologists compare skewed leaf-damage measurements among four treatments. — Simulated observations; generated by this module’s R script.

Kruskal–Wallis p = 1.352e-05
Highlighted pair: B - D
Dunn z = -4.30
Two-sided Holm-adjusted p = 0.0001018

Across the six rank comparisons, the highlighted B - D comparison had Dunn z = -4.30 and two-sided Holm-adjusted p = 0.0001018. This compares pooled rank tendencies; with unequal distribution shapes, it is not simply a test of equal medians. The smallest adjusted p-value is highlighted after protecting the full six-pair family.

Teaching pointAfter choosing a rank question, retain multiplicity protection for pairwise follow-ups.

Analysis code excerpt

d <- read.csv("modules/dunn/generated/data.csv")
overall <- kruskal.test(damage ~ treatment, data = d)
comparisons <- dunn.test::dunn.test(d$damage, d$treatment,
  method = "holm", altp = TRUE, kw = FALSE, list = FALSE)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


	Kruskal-Wallis rank sum test

data:  damage by treatment
Kruskal-Wallis chi-squared = 25.276, df = 3, p-value = 1.352e-05

$chi2
[1] 25.27627

$Z
[1]  0.5693239 -2.5682832 -3.1376070 -3.7322342 -4.3015580 -1.1639510

$altP
[1] 5.691364e-01 1.022036e-02 1.703331e-03 1.897889e-04 1.696014e-05
[6] 2.444439e-01

$altP.adjusted
[1] 0.5691363809 0.0306610890 0.0068133231 0.0009489447 0.0001017608
[6] 0.4888877458

$comparisons
[1] "A - B" "A - C" "B - C" "A - D" "B - D" "C - D"

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/dunn/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(2105)
d <- data.frame(leaf = 1:84,
                treatment = rep(c("A", "B", "C", "D"), each = 21))
d$damage <- rlnorm(84, rep(c(1.2, 1.3, 1.7, 1.9), each = 21), .48)
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/dunn/analysis.R modules/dunn/generated

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Dataset 22 · Association

Pearson correlation

CSV rows: 45Seed: 1112Module: pearson

Saved CSV Full runnable R script Full-size plot

Evolutionary biologists measure wing and tail lengths in adult birds from a single population.

Question and null

Scientific question

Are wing and tail lengths linearly associated in this population?

Null being tested / estimation target

The population Pearson correlation is zero.

Generating truth

The generating population Pearson correlation is 0.55*5 / sqrt((0.55*5)^2 + 3.5^2), approximately 0.618. It is not the observed sample correlation.

Exact generating model

45 independent wing lengths are Normal(75, 5 SD). Tail length is 15 + 0.55 times wing length + independent Normal(0, 3.5 SD) noise.

Design and independent unit: Each independent bird contributes one pair of continuous traits.

Columns in the saved CSV

Column	Meaning	Unit / type
`bird`	Independent bird identifier	none
`wing_mm`	Wing length	mm
`tail_mm`	Tail length	mm

Preview the actual saved observations

bird	wing_mm	tail_mm
1	78.9384301858733	53.9216466189698
2	72.889876095192	57.8349867119479
3	75.2845108566451	58.1155862248892
4	78.5528715633195	59.64706614043
5	67.062689624881	47.1987321119575

Analysis from the saved dataset

Pearson correlation plot. Evolutionary biologists measure wing and tail lengths in adult birds from a single population. — Simulated observations; generated by this module’s R script.

Pearson r = 0.700
95% CI for population correlation: 0.512 to 0.824
t = 6.43; df = 43; p = 8.606e-08

Wing and tail lengths had Pearson r = 0.700 (95% CI 0.512 to 0.824), p = 8.606e-08. This describes linear association within the sampled population; it does not establish that one trait causes the other.

Teaching pointCorrelation describes linear association symmetrically; causal direction and a slope are separate questions.

Analysis code excerpt

d <- read.csv("modules/pearson/generated/data.csv")
fit <- cor.test(d$wing_mm, d$tail_mm, method = "pearson",
                alternative = "two.sided", conf.level = .95)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


	Pearson's product-moment correlation

data:  d$wing_mm and d$tail_mm
t = 6.4317, df = 43, p-value = 8.606e-08
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
 0.5119136 0.8243291
sample estimates:
      cor 
0.7002284

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/pearson/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(1112)
wing <- rnorm(45, 75, 5)
tail <- 15 + .55 * wing + rnorm(45, 0, 3.5)
d <- data.frame(bird = 1:45, wing_mm = wing, tail_mm = tail)
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/pearson/analysis.R modules/pearson/generated

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Dataset 23 · Nonparametric alternatives

Spearman rank correlation

CSV rows: 45Seed: 1113Module: spearman

Saved CSV Full runnable R script Full-size plot

Microbial ecologists compare soil salinity with bacterial richness across independent soil cores.

Question and null

Scientific question

Does bacterial richness tend to decline as salinity increases?

Null being tested / estimation target

The population Spearman rank correlation is zero.

Generating truth

The underlying mean response decreases nonlinearly with salinity, creating a negative monotonic association. No fixed numeric population Spearman coefficient was specified.

Exact generating model

45 independent site salinities are Uniform(0.5, 8). Richness is 90 - 25*log(salinity + 1) + Normal(0, 7 SD) noise, rounded to the nearest taxon.

Design and independent unit: Forty-five independent sites contribute one salinity and richness measurement each.

Columns in the saved CSV

Column	Meaning	Unit / type
`site`	Independent site identifier	none
`salinity_ppt`	Soil salinity	parts per thousand
`richness`	Observed bacterial taxon richness	taxa

Preview the actual saved observations

site	salinity_ppt	richness
1	2.09540565568022	60
2	7.34450018976349	42
3	3.97358826256823	46
4	1.95299733732827	62
5	4.00734839797951	45

Analysis from the saved dataset

Spearman rank correlation plot. Microbial ecologists compare soil salinity with bacterial richness across independent soil cores. — Simulated observations; generated by this module’s R script.

Spearman rho = -0.752; n = 45
S = 26599.8; asymptotic two-sided p = 2.549e-09
The coefficient describes rank association, not a change in taxa per ppt.

Salinity and bacterial richness had Spearman rho = -0.752 (two-sided p = 2.549e-09). This quantifies their monotonic rank association. It does not estimate a change in mean richness per unit salinity.

Teaching pointRanks can summarize a monotonic curve without claiming a constant change in the original response units.

Analysis code excerpt

d <- read.csv("modules/spearman/generated/data.csv")
fit <- cor.test(d$salinity_ppt, d$richness,
                method = "spearman", exact = FALSE,
                alternative = "two.sided")

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


	Spearman's rank correlation rho

data:  d$salinity_ppt and d$richness
S = 26600, p-value = 2.549e-09
alternative hypothesis: true rho is not equal to 0
sample estimates:
       rho 
-0.7522922

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/spearman/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(1113)
salinity <- runif(45, .5, 8)
richness <- round(90 - 25 * log(salinity + 1) + rnorm(45, 0, 7))
d <- data.frame(site = 1:45, salinity_ppt = salinity, richness = richness)
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/spearman/analysis.R modules/spearman/generated

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Dataset 24 · Regression

Simple linear regression

CSV rows: 36Seed: 1114Module: simple_lm

Saved CSV Full runnable R script Full-size plot

Plant ecologists grow seedlings under a gradient of nitrogen supply.

Question and null

Scientific question

How much does mean seedling biomass change per additional mg of nitrogen?

Null being tested / estimation target

The population slope of mean biomass on nitrogen is zero.

Generating truth

The conditional mean intercept is 2.2 g and slope is +0.42 g per mg nitrogen. The zero-slope null is false.

Exact generating model

36 nitrogen doses are Uniform(0, 12) mg. Independent biomass values follow 2.2 + 0.42*nitrogen + Normal(0, 0.8 SD).

Design and independent unit: Thirty-six independently assigned pots receive one nitrogen level and yield one biomass measurement.

Columns in the saved CSV

Column	Meaning	Unit / type
`pot`	Independent pot identifier	none
`nitrogen_mg`	Assigned nitrogen dose	mg
`biomass_g`	Seedling biomass	g

Preview the actual saved observations

pot	nitrogen_mg	biomass_g
1	9.54561813455075	5.99222564791988
2	2.35906611476094	4.64012466744188
3	7.31420148164034	6.1187426715042
4	2.36318408232182	3.58543334210748
5	0.470603071153164	2.45911362156326

Analysis from the saved dataset

Simple linear regression plot. Plant ecologists grow seedlings under a gradient of nitrogen supply. — Simulated observations; generated by this module’s R script.

Slope = 0.414 g biomass per mg nitrogen
95% slope CI: 0.329 to 0.499
Slope t = 9.90; df = 34; p = 1.525e-11
R-squared = 0.742

Each additional mg of nitrogen was associated with 0.414 g higher mean biomass (95% CI 0.329 to 0.499); slope t(34) = 9.90, p = 1.525e-11. The narrow band concerns mean biomass; the wider prediction limits concern a new individual pot.

Teaching pointChoose confidence intervals for a mean response and prediction intervals for a new individual pot.

Analysis code excerpt

d <- read.csv("modules/simple_lm/generated/data.csv")
fit <- lm(biomass_g ~ nitrogen_mg, data = d)
coefficients <- summary(fit)$coefficients
ci <- confint(fit)
grid <- data.frame(nitrogen_mg = seq(min(d$nitrogen_mg), max(d$nitrogen_mg), length.out = 100))
mean_ci <- predict(fit, grid, interval = "confidence")
new_pot <- predict(fit, grid, interval = "prediction")

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


Call:
lm(formula = biomass_g ~ nitrogen_mg, data = d)

Residuals:
    Min      1Q  Median      3Q     Max 
-1.8507 -0.4546 -0.1079  0.4972  1.9486 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  2.21791    0.29034   7.639 7.05e-09 ***
nitrogen_mg  0.41409    0.04185   9.896 1.52e-11 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.811 on 34 degrees of freedom
Multiple R-squared:  0.7423,	Adjusted R-squared:  0.7347 
F-statistic: 97.93 on 1 and 34 DF,  p-value: 1.525e-11

                2.5 %    97.5 %
(Intercept) 1.6278685 2.8079514
nitrogen_mg 0.3290528 0.4991331

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/simple_lm/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(1114)
nitrogen <- runif(36, 0, 12)
biomass <- 2.2 + .42 * nitrogen + rnorm(36, 0, .8)
d <- data.frame(pot = 1:36, nitrogen_mg = nitrogen, biomass_g = biomass)
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/simple_lm/analysis.R modules/simple_lm/generated

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Dataset 25 · Regression

Multiple linear regression

CSV rows: 60Seed: 1115Module: multiple_lm

Saved CSV Full runnable R script Full-size plot

Evolutionary biologists relate beetle horn length to body size and larval nutrition.

Question and null

Scientific question

Does body size predict mean horn length after accounting for larval nutrition?

Null being tested / estimation target

The body-size coefficient is zero conditional on nutrition in the specified model.

Generating truth

The true conditional body slope is 0.35 mm/mm and nutrition slope 0.16 mm/mg, with no interaction. Both zero-coefficient nulls are false.

Exact generating model

60 independent body lengths are Uniform(7, 12) mm and nutrition values independently Uniform(2, 10) mg. Horn length = 0.4 + 0.35*body + 0.16*nutrition + Normal(0, 0.65 SD).

Design and independent unit: Sixty independent beetles are raised across a nutrient gradient and measured as adults.

Columns in the saved CSV

Column	Meaning	Unit / type
`beetle`	Independent beetle identifier	none
`body_mm`	Adult body length	mm
`nutrition_mg`	Larval nutrition amount	mg
`horn_mm`	Adult horn length	mm

Preview the actual saved observations

beetle	body_mm	nutrition_mg	horn_mm
1	9.02821977436543	3.09207137487829	2.49057035137996
2	11.0083169168793	8.33163026347756	6.74357274274548
3	10.5993990674615	3.7476684320718	4.57263497441412
4	10.5076748183928	9.84045981802046	5.36436749727661
5	7.96044226549566	6.83275267481804	4.62986831024042

Analysis from the saved dataset

Multiple linear regression plot. Evolutionary biologists relate beetle horn length to body size and larval nutrition. — Simulated observations; generated by this module’s R script.

Body slope, adjusted for nutrition = 0.299 mm/mm
95% CI: 0.159 to 0.440
Body coefficient p = 7.658e-05; nested-model F = 18.17
Nutrition slope = 0.181 mm/mg; p = 1.112e-05
Adjusted R-squared = 0.411

Holding larval nutrition fixed in the additive model, each additional mm of body length was associated with 0.299 mm higher mean horn length (95% CI 0.159 to 0.440; p = 7.658e-05). This conditional association is not the same as the unadjusted relationship or proof of causation.

Teaching pointRead each coefficient conditionally on the other variables in the stated model.

Analysis code excerpt

d <- read.csv("modules/multiple_lm/generated/data.csv")
fit <- lm(horn_mm ~ body_mm + nutrition_mg, data = d)
reduced <- lm(horn_mm ~ nutrition_mg, data = d)
comparison <- anova(reduced, fit)
coefficients <- summary(fit)$coefficients
ci <- confint(fit)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


Call:
lm(formula = horn_mm ~ body_mm + nutrition_mg, data = d)

Residuals:
     Min       1Q   Median       3Q      Max 
-1.51574 -0.38170 -0.00496  0.19296  1.60870 

Coefficients:
             Estimate Std. Error t value Pr(>|t|)    
(Intercept)   0.69688    0.70981   0.982     0.33    
body_mm       0.29930    0.07021   4.263 7.66e-05 ***
nutrition_mg  0.18104    0.03757   4.818 1.11e-05 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.7242 on 57 degrees of freedom
Multiple R-squared:  0.4306,	Adjusted R-squared:  0.4106 
F-statistic: 21.55 on 2 and 57 DF,  p-value: 1.07e-07

                  2.5 %    97.5 %
(Intercept)  -0.7244990 2.1182587
body_mm       0.1587143 0.4398833
nutrition_mg  0.1057975 0.2562805

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/multiple_lm/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(1115)
body <- runif(60, 7, 12)
nutrition <- runif(60, 2, 10)
horn <- .4 + .35 * body + .16 * nutrition + rnorm(60, 0, .65)
d <- data.frame(beetle = 1:60, body_mm = body, nutrition_mg = nutrition, horn_mm = horn)
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/multiple_lm/analysis.R modules/multiple_lm/generated

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Dataset 26 · Regression

Theil–Sen robust slope

CSV rows: 40Seed: 1116Module: theil_sen

Saved CSV Full runnable R script Full-size plot

Physiologists measure enzyme activity along a temperature gradient with occasional unusually high assays.

Question and null

Scientific question

What is the robust linear change in enzyme activity per degree?

Null being tested / estimation target

The target is the robust linear activity–temperature slope; zero means no linear location trend.

Generating truth

The underlying uncontaminated linear trend has slope 1.4 U/mL per degree C. Two deliberately elevated assays illustrate robustness; the bootstrap interval comes from the saved contaminated sample.

Exact generating model

40 temperatures are Uniform(16, 30) degrees C. Core activity = 8 + 1.4*temperature + 2 times a Student t variable with 4 degrees of freedom. Assays 5 and 18 then receive an extra +25 U/mL.

Design and independent unit: Forty independent assay preparations contribute one temperature/activity pair each.

Columns in the saved CSV

Column	Meaning	Unit / type
`assay`	Independent assay identifier	none
`temp_c`	Assay temperature	degrees C
`activity`	Enzyme activity	U/mL

Preview the actual saved observations

assay	temp_c	activity
1	25.6160467932932	45.0424989221144
2	19.7068037008867	35.5608470863034
3	23.0981907695532	38.7956486383163
4	18.9486378859729	37.2015015674194
5	24.699953973759	65.372094333161

Analysis from the saved dataset

Theil–Sen robust slope plot. Physiologists measure enzyme activity along a temperature gradient with occasional unusually high assays. — Simulated observations; generated by this module’s R script.

Theil–Sen slope = 1.408 U/mL per degree C
95% pairs-bootstrap interval: 1.184 to 1.618
Finite original pairwise slopes = 780; bootstrap replicates = 999
The interval is approximate; no exact test p-value is reported.

The robust slope was 1.408 U/mL per degree C, with a 95% percentile pairs-bootstrap interval of 1.184 to 1.618 (999 replicates). This estimates a linear location trend resistant to extreme responses; its interval is approximate and is not an exact slope test.

Teaching pointA robust slope retains interpretable units while limiting extreme-response influence; uncertainty here is an approximate pairs-bootstrap interval.

Analysis code excerpt

d <- read.csv("modules/theil_sen/generated/data.csv")
pairs <- combn(seq_len(nrow(d)), 2)
slopes <- (d$activity[pairs[2, ]] - d$activity[pairs[1, ]]) /
          (d$temp_c[pairs[2, ]] - d$temp_c[pairs[1, ]])
slope <- median(slopes)
intercept <- median(d$activity - slope * d$temp_c)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output

intercept     slope 
 7.832577  1.407600 
    2.5%    97.5% 
1.183998 1.618227 
[1] 999

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/theil_sen/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(1116)
temperature <- runif(40, 16, 30)
activity <- 8 + 1.4 * temperature + rt(40, df = 4) * 2
activity[c(5, 18)] <- activity[c(5, 18)] + 25
d <- data.frame(assay = 1:40, temp_c = temperature, activity = activity)
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/theil_sen/analysis.R modules/theil_sen/generated

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Dataset 27 · Proportions and categories

Exact binomial test

CSV rows: 80Seed: 3101Module: binomial

Saved CSV Full runnable R script Full-size plot

Host choice in a parasitoid wasp

Question and null

Scientific question

Do wasps choose the native host with probability 0.50?

Null being tested / estimation target

The population probability of choosing the native host is 0.50.

Generating truth

The true native-host probability is 0.68, so the null probability 0.50 is false. The population difference from the null is +0.18.

Exact generating model

Generate 80 independent Bernoulli choices, each with probability 0.68 of a native-host choice. Each wasp appears once. No rounding or dependence is introduced.

Design and independent unit: One independent choice per wasp; two equally available host species.

Columns in the saved CSV

Column	Meaning	Unit / type
`wasp`	Unique independent wasp identifier	none
`native`	1 = native host chosen; 0 = introduced host	binary

Preview the actual saved observations

wasp	native
1	0
2	0
3	1
4	1
5	0

Analysis from the saved dataset

Exact binomial test plot. Host choice in a parasitoid wasp — Simulated observations; generated by this module’s R script.

Native-host choices: 51 / 80
Estimated probability: 0.637
95% exact CI: 0.522 to 0.742
Exact two-sided p = 0.01832

Native-host choice was 63.7% (95% exact CI 52.2–74.2%; p = 0.0183 against 50%). This estimates preference under the simulated choice conditions.

Teaching pointThe null probability comes from the scientific prediction, not automatically from the fact that there are two outcomes.

Analysis code excerpt

d <- read.csv("modules/binomial/generated/data.csv")
x <- sum(d$native)
fit <- binom.test(x, nrow(d), p = 0.5,
                  alternative = "two.sided")
estimate <- x / nrow(d)
ci <- fit$conf.int

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


	Exact binomial test

data:  x and nrow(d)
number of successes = 51, number of trials = 80, p-value = 0.01832
alternative hypothesis: true probability of success is not equal to 0.5
95 percent confidence interval:
 0.5223873 0.7421051
sample estimates:
probability of success 
                0.6375

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

out_dir <- "modules/binomial/generated"
dir.create(out_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(3101)
d <- data.frame(wasp = 1:80,
                native = rbinom(80, 1, 0.68))
write.csv(d, file.path(out_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/binomial/analysis.R modules/binomial/generated

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Dataset 28 · Proportions and categories

One-proportion test

CSV rows: 120Seed: 3102Module: one_proportion

Saved CSV Full runnable R script Full-size plot

Germination of salt-tolerant seeds

Question and null

Scientific question

Does the germination probability differ from the 50% benchmark?

Null being tested / estimation target

The population germination probability is 0.50.

Generating truth

The true germination probability is 0.67 rather than the null benchmark 0.50; the population difference is +0.17.

Exact generating model

Generate 120 independent seeds with Bernoulli germination probability 0.67. Every seed has the same probability. No rounding or tray dependence is simulated.

Design and independent unit: One germination outcome per seed under a fixed salt treatment.

Columns in the saved CSV

Column	Meaning	Unit / type
`seed`	Unique independent seed identifier	none
`germinated`	1 = germinated; 0 = did not germinate	binary

Preview the actual saved observations

seed	germinated
1	1
2	1
3	0
4	0
5	1

Analysis from the saved dataset

One-proportion test plot. Germination of salt-tolerant seeds — Simulated observations; generated by this module’s R script.

Germination: 70 / 120
Estimated probability: 0.583
95% score CI: 0.494 to 0.668
Two-sided p = 0.06789

Germination was 58.3% (95% score CI 49.4–66.8%; p = 0.0679 against 50%). The test concerns this germination probability, not the mean of arbitrary continuous percentages.

Teaching pointA binomial proportion is successes out of a known number of trials; arbitrary continuous percentages are different data.

Analysis code excerpt

d <- read.csv("modules/one_proportion/generated/data.csv")
x <- sum(d$germinated)
fit <- prop.test(x, nrow(d), p = 0.5,
                 alternative = "two.sided", correct = FALSE)
estimate <- x / nrow(d)
ci <- fit$conf.int

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


	1-sample proportions test without continuity correction

data:  x out of nrow(d), null probability 0.5
X-squared = 3.3333, df = 1, p-value = 0.06789
alternative hypothesis: true p is not equal to 0.5
95 percent confidence interval:
 0.4938803 0.6676165
sample estimates:
        p 
0.5833333

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

out_dir <- "modules/one_proportion/generated"
dir.create(out_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(3102)
d <- data.frame(seed = 1:120,
                germinated = rbinom(120, 1, 0.67))
write.csv(d, file.path(out_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/one_proportion/analysis.R modules/one_proportion/generated

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Dataset 29 · Proportions and categories

Two-proportion test

CSV rows: 220Seed: 3103Module: two_proportions

Saved CSV Full runnable R script Full-size plot

Heat survival in evolved and ancestral yeast

Question and null

Scientific question

Does survival probability differ between evolved and ancestral yeast?

Null being tested / estimation target

Evolved and ancestral cultures have equal survival probabilities.

Generating truth

Evolved minus ancestral survival probability is 0.20, so the equal-probabilities null is false. The corresponding population odds ratio is approximately 2.37.

Exact generating model

Generate 110 independent evolved cultures and 110 independent ancestral cultures. Survival is Bernoulli with probabilities 0.72 and 0.52, respectively. No batching or pairing is introduced.

Design and independent unit: Independent cultures assigned to one genotype; survival is scored once.

Columns in the saved CSV

Column	Meaning	Unit / type
`group`	Culture genotype: Evolved or Ancestral	none
`survived`	1 = survived heat; 0 = died	binary

Preview the actual saved observations

group	survived
Evolved	1
Evolved	0
Evolved	1
Evolved	1
Evolved	0

Analysis from the saved dataset

Two-proportion test plot. Heat survival in evolved and ancestral yeast — Simulated observations; generated by this module’s R script.

Evolved: 78 / 110; ancestral: 61 / 110
Difference (evolved − ancestral): 0.155
95% CI for difference: 0.029 to 0.280
Two-sided p = 0.01748

Evolved minus ancestral survival was 15.5 percentage points (95% CI 2.9 to 28.0; p = 0.0175). The interval for this difference answers the comparison directly.

Teaching pointKeep group order explicit: the contrast is evolved minus ancestral, and its confidence interval addresses the actual comparison.

Analysis code excerpt

d <- read.csv("modules/two_proportions/generated/data.csv")
groups <- c("Evolved", "Ancestral")
x <- sapply(groups, function(g) sum(d$survived[d$group == g]))
n <- sapply(groups, function(g) sum(d$group == g))
fit <- prop.test(x, n, correct = FALSE)
estimate <- x / n
difference <- estimate[1] - estimate[2]
ci <- fit$conf.int
group_ci <- sapply(1:2, function(i) prop.test(x[i], n[i],
                                            correct = FALSE)$conf.int)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


	2-sample test for equality of proportions without continuity correction

data:  x out of n
X-squared = 5.647, df = 1, p-value = 0.01748
alternative hypothesis: two.sided
95 percent confidence interval:
 0.02872612 0.28036479
sample estimates:
   prop 1    prop 2 
0.7090909 0.5545455

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

out_dir <- "modules/two_proportions/generated"
dir.create(out_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(3103)
d <- data.frame(group = rep(c("Evolved", "Ancestral"), each = 110))
d$survived <- rbinom(220, 1, rep(c(.72, .52), each = 110))
write.csv(d, file.path(out_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/two_proportions/analysis.R modules/two_proportions/generated

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Dataset 30 · Proportions and categories

Chi-square goodness-of-fit

CSV rows: 240Seed: 3104Module: chi_gof

Saved CSV Full runnable R script Full-size plot

Segregation in an F2 cross

Question and null

Scientific question

Do the genotype frequencies follow the expected 1:2:1 segregation ratio?

Null being tested / estimation target

Genotype probabilities are AA = 0.25, Aa = 0.50, and aa = 0.25.

Generating truth

The simulated probabilities differ from 1:2:1 by -0.07, +0.04, and +0.03. The goodness-of-fit null is false.

Exact generating model

Independently sample 240 offspring genotypes with probabilities AA = 0.18, Aa = 0.54, aa = 0.28. Test against the fixed Mendelian probabilities 0.25, 0.50, 0.25; no probabilities are estimated for the null.

Design and independent unit: Each offspring is assigned to one of three genotypes.

Columns in the saved CSV

Column	Meaning	Unit / type
`offspring`	Unique independent offspring identifier	none
`genotype`	AA, Aa, or aa genotype category	none

Preview the actual saved observations

offspring	genotype
1	aa
2	aa
3	Aa
4	AA
5	Aa

Analysis from the saved dataset

Chi-square goodness-of-fit plot. Segregation in an F2 cross — Simulated observations; generated by this module’s R script.

Observed: 43, 128, 69
Expected: 60, 120, 60
Chi-square = 6.700; df = 2
p = 0.03508

Observed genotype proportions were 0.179, 0.533, 0.287. The goodness-of-fit test gives chi-square(2) = 6.70, p = 0.0351 against the prespecified 1:2:1 ratio; the test does not identify a biological mechanism for the departure.

Teaching pointExpected counts encode the biological hypothesis. Equal expected counts would test a different hypothesis.

Analysis code excerpt

d <- read.csv("modules/chi_gof/generated/data.csv")
observed <- table(factor(d$genotype, levels = c("AA", "Aa", "aa")))
fit <- chisq.test(observed, p = c(.25, .50, .25))
expected <- as.numeric(fit$expected)
departures <- observed / sum(observed) - c(.25, .50, .25)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


	Chi-squared test for given probabilities

data:  observed
X-squared = 6.7, df = 2, p-value = 0.03508

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

out_dir <- "modules/chi_gof/generated"
dir.create(out_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(3104)
d <- data.frame(offspring = 1:240,
  genotype = sample(c("AA", "Aa", "aa"), 240, replace = TRUE,
                    prob = c(.18, .54, .28)))
write.csv(d, file.path(out_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/chi_gof/analysis.R modules/chi_gof/generated

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Dataset 31 · Proportions and categories

Chi-square test of independence

CSV rows: 270Seed: 3105Module: chi_independence

Saved CSV Full runnable R script Full-size plot

Infection across three lizard habitats

Question and null

Scientific question

Is infection status associated with habitat?

Null being tested / estimation target

Infection status and habitat are independent; infection probabilities are equal across habitats.

Generating truth

Infection probability depends on habitat, so the independence null is false. Urban minus forest infection probability is 0.30; the simulation does not assign causality to habitat.

Exact generating model

Generate 90 independent lizards from each of Forest, Grassland, and Urban habitats. Infection is Bernoulli with probabilities 0.25, 0.42, and 0.55, respectively. No site-level clustering is simulated.

Design and independent unit: Independent lizards sampled once in one of three habitats.

Columns in the saved CSV

Column	Meaning	Unit / type
`habitat`	Sampling habitat: Forest, Grassland, or Urban	none
`infection`	Infected or Uninfected status	none

Preview the actual saved observations

habitat	infection
Forest	Uninfected
Forest	Uninfected
Forest	Uninfected
Forest	Uninfected
Forest	Infected

Analysis from the saved dataset

Chi-square test of independence plot. Infection across three lizard habitats — Simulated observations; generated by this module’s R script.

Infection probabilities: Forest 0.14; Grassland 0.41; Urban 0.54
Minimum expected count: 33.0
Chi-square(2) = 32.153
p = 1.042e-07; Cramer V = 0.345

Infection probabilities differed descriptively across habitats (Forest 14%, Grassland 41%, Urban 54%). The independence test gives chi-square(2) = 32.15, p = 1.04e-07; Cramer V = 0.35. This observational association does not establish a habitat effect.

Teaching pointA significant omnibus association does not establish a causal habitat effect or identify every differing pair.

Analysis code excerpt

d <- read.csv("modules/chi_independence/generated/data.csv")
tab <- table(d$habitat, d$infection)
fit <- chisq.test(tab, correct = FALSE)
cramers_v <- sqrt(as.numeric(fit$statistic) /
                  (sum(tab) * min(nrow(tab)-1, ncol(tab)-1)))
rates <- prop.table(tab, 1)[, "Infected"]

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


	Pearson's Chi-squared test

data:  tab
X-squared = 32.153, df = 2, p-value = 1.042e-07

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

out_dir <- "modules/chi_independence/generated"
dir.create(out_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(3105)
d <- data.frame(habitat = rep(c("Forest", "Grassland", "Urban"), each = 90))
d$infection <- ifelse(rbinom(270, 1, rep(c(.25, .42, .55), each = 90)),
                       "Infected", "Uninfected")
write.csv(d, file.path(out_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/chi_independence/analysis.R modules/chi_independence/generated

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Dataset 32 · Proportions and categories

Fisher exact test

CSV rows: 24Seed: 3106Module: fisher_exact

Saved CSV Full runnable R script Full-size plot

Survival of rare resistant and susceptible clones

Question and null

Scientific question

Are survival odds associated with clone type?

Null being tested / estimation target

The population odds ratio for clone type and survival is 1.

Generating truth

The resistant-to-susceptible population survival odds ratio is (0.65/0.35)/(0.20/0.80), approximately 7.43. The odds-ratio-one null is false.

Exact generating model

Generate 12 independent susceptible clones with survival probability 0.20 and 12 independent resistant clones with probability 0.65. Each contributes one Bernoulli outcome; small table cells are intentional.

Design and independent unit: Two small independent clone groups; binary survival after exposure.

Columns in the saved CSV

Column	Meaning	Unit / type
`clone`	Unique independent clone identifier	none
`type`	Susceptible or Resistant clone type	none
`survived`	1 = survived exposure; 0 = died	binary

Preview the actual saved observations

clone	type	survived
1	Susceptible	0
2	Susceptible	0
3	Susceptible	0
4	Susceptible	1
5	Susceptible	0

Analysis from the saved dataset

Fisher exact test plot. Survival of rare resistant and susceptible clones — Simulated observations; generated by this module’s R script.

Survivors: susceptible 2/12; resistant 7/12
Survival odds ratio (resistant / susceptible): 6.395
95% exact CI: 0.810 to 85.671
Two-sided exact p = 0.08938

The resistant-to-susceptible survival odds ratio was 6.39 (95% exact CI 0.81–85.67; p = 0.0894). A wide interval reflects limited information from these small groups; odds are not probabilities.

Teaching pointExact inference handles sparse counts, but it cannot create the precision missing from a small sample.

Analysis code excerpt

d <- read.csv("modules/fisher_exact/generated/data.csv")
tab <- table(factor(d$type, levels = c("Susceptible", "Resistant")),
             factor(d$survived, levels = c(0, 1)))
fit <- fisher.test(tab, alternative = "two.sided")
odds_ratio <- unname(fit$estimate)
ci <- fit$conf.int

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


	Fisher's Exact Test for Count Data

data:  tab
p-value = 0.08938
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
  0.8098091 85.6709653
sample estimates:
odds ratio 
  6.394505

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

out_dir <- "modules/fisher_exact/generated"
dir.create(out_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(3106)
d <- data.frame(clone = 1:24,
  type = rep(c("Susceptible", "Resistant"), each = 12))
d$survived <- rbinom(24, 1, rep(c(.20, .65), each = 12))
write.csv(d, file.path(out_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/fisher_exact/analysis.R modules/fisher_exact/generated

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Dataset 33 · Paired binary outcomes

McNemar test

CSV rows: 80Seed: 3107Module: mcnemar

Saved CSV Full runnable R script Full-size plot

Antibody detection before and after an exposure season

Question and null

Scientific question

Does the marginal probability of antibody detection change between visits?

Null being tested / estimation target

The probability of changing from negative to positive equals that of changing from positive to negative.

Generating truth

Marginal after positivity is 0.45×0.85 + 0.55×0.35 = 0.575, a +0.125 change from before. Population gain and loss probabilities are 0.1925 and 0.0675, so the McNemar null is false.

Exact generating model

Generate 80 independent mice. Before status is Bernoulli(0.45). Conditional after positivity is 0.85 for a previously positive mouse and 0.35 for a previously negative mouse, producing correlated within-mouse measurements.

Design and independent unit: The same 80 wild mice are sampled twice.

Columns in the saved CSV

Column	Meaning	Unit / type
`mouse`	Matched mouse identifier	none
`before`	Antibody detected at the first visit (1 = yes)	binary
`after`	Antibody detected at the second visit (1 = yes)	binary

Preview the actual saved observations

mouse	before	after
1	1	1
2	1	1
3	1	0
4	1	0
5	0	0

Analysis from the saved dataset

McNemar test plot. Antibody detection before and after an exposure season — Simulated observations; generated by this module’s R script.

Gained detection: 10; lost detection: 5
Before: 0.438; after: 0.500
Change: 0.062
McNemar p = 0.1967; exact paired p = 0.3018

Detection changed from 43.8% to 50.0% (change 6.2 percentage points). There were 10 gains and 5 losses; McNemar p = 0.197 and exact paired p = 0.302. The comparison preserves mouse identity.

Teaching pointThe information about a change is in discordant pairs. Keeping the mouse identity changes the appropriate test.

Analysis code excerpt

d <- read.csv("modules/mcnemar/generated/data.csv")
tab <- table(factor(d$before, levels = 0:1),
             factor(d$after, levels = 0:1))
fit <- mcnemar.test(tab, correct = FALSE)
gained <- tab[1, 2]
lost <- tab[2, 1]
exact <- binom.test(gained, gained + lost, p = .5)
change <- mean(d$after) - mean(d$before)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output

$McNemar

	McNemar's Chi-squared test

data:  tab
McNemar's chi-squared = 1.6667, df = 1, p-value = 0.1967


$exact_discordant_pairs

	Exact binomial test

data:  gained and gained + lost
number of successes = 10, number of trials = 15, p-value = 0.3018
alternative hypothesis: true probability of success is not equal to 0.5
95 percent confidence interval:
 0.3838037 0.8817589
sample estimates:
probability of success 
             0.6666667

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

out_dir <- "modules/mcnemar/generated"
dir.create(out_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(3107)
d <- data.frame(mouse = 1:80, before = rbinom(80, 1, .45))
d$after <- rbinom(80, 1, ifelse(d$before == 1, .85, .35))
write.csv(d, file.path(out_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/mcnemar/analysis.R modules/mcnemar/generated

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Dataset 34 · Paired binary outcomes

Cochran’s Q test

CSV rows: 180Seed: 3108Module: cochran_q

Saved CSV Full runnable R script Full-size plot

Bee visits to three floral odors

Question and null

Scientific question

Are marginal visit probabilities equal for all three odors?

Null being tested / estimation target

The three odor conditions have equal marginal probabilities of a visit.

Generating truth

Increasing odor effects produce strictly increasing marginal visit probabilities across A, B, and C. The equal-probabilities null is false; the latent log-odds differences are not themselves the marginal probability differences.

Exact generating model

Generate 60 independent bees, each with a Normal(0, 0.8²) latent effect. Each bee encounters odors A, B, and C. Visit logits are -0.7 + bee effect + odor effect, with odor effects 0, 0.5, and 1. Outcomes are independent conditional on the bee effect but correlated within a bee marginally.

Design and independent unit: Each bee experiences all odors in randomized sessions with a washout.

Columns in the saved CSV

Column	Meaning	Unit / type
`bee`	Independent bee and repeated-measures identifier	none
`odor`	Repeated odor condition: A, B, or C	none
`visit`	1 = visit; 0 = no visit	binary

Preview the actual saved observations

bee	odor	visit
1	A	0
1	B	1
1	C	1
2	A	0
2	B	1

Analysis from the saved dataset

Cochran’s Q test plot. Bee visits to three floral odors — Simulated observations; generated by this module’s R script.

Matched bees: 60; conditions: 3
Visit probabilities: 0.317, 0.417, 0.517
Q = 5.143; df = 2
Omnibus p = 0.07643

Visit probabilities were 31.7%, 41.7%, 51.7%. Cochran Q(2) = 5.14, p = 0.0764 tests equality across all odors while retaining matching. It does not identify which odor pairs differ.

Teaching pointA matched omnibus binary test uses complete sets of responses; a significant result does not identify which conditions differ.

Analysis code excerpt

d <- read.csv("modules/cochran_q/generated/data.csv")
wide <- xtabs(visit ~ bee + odor, data = d)
k <- ncol(wide)
C <- colSums(wide)
R <- rowSums(wide)
Q <- (k - 1) * (k * sum(C^2) - sum(C)^2) /
     (k * sum(R) - sum(R^2))
p_value <- pchisq(Q, df = k - 1, lower.tail = FALSE)
rates <- C / nrow(wide)
ci <- sapply(1:k, function(j) binom.test(C[j], nrow(wide))$conf.int)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output

$statistic
[1] 5.142857

$df
[1] 2

$p.value
[1] 0.07642629

$visit_probabilities
        A         B         C 
0.3166667 0.4166667 0.5166667

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

out_dir <- "modules/cochran_q/generated"
dir.create(out_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(3108)
bee_effect <- rnorm(60, 0, .8)
d <- data.frame(bee = rep(1:60, each = 3),
                odor = rep(c("A", "B", "C"), 60))
d$visit <- rbinom(180, 1,
  plogis(-.7 + rep(bee_effect, each = 3) + rep(c(0, .5, 1), 60)))
write.csv(d, file.path(out_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/cochran_q/analysis.R modules/cochran_q/generated

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Dataset 35 · Generalized linear models

Logistic regression

CSV rows: 160Seed: 3110Module: logistic

Saved CSV Full runnable R script Full-size plot

Hybrid viability across parental divergence

Question and null

Scientific question

Does viability probability change with parental divergence?

Null being tested / estimation target

The divergence coefficient is zero: modeled viability odds do not change with divergence.

Generating truth

The true divergence coefficient is -0.5; viability odds are multiplied by exp(-0.5) ≈ 0.607 per divergence percentage point. The zero-slope null is false. The probability decline is nonlinear.

Exact generating model

Generate 160 independent crosses with divergence Uniform(0,8) percentage points. One hybrid viability outcome per cross is Bernoulli with probability plogis(1.9 - 0.5 × divergence). Predictors are not rounded.

Design and independent unit: One independent hybrid offspring per cross.

Columns in the saved CSV

Column	Meaning	Unit / type
`cross`	Unique independent cross identifier	none
`divergence`	Parental divergence	percentage points
`viable`	1 = viable hybrid; 0 = nonviable	binary

Preview the actual saved observations

cross	divergence	viable
1	5.03366027772427	0
2	4.56829151697457	0
3	6.65665970183909	0
4	5.46810280717909	0
5	7.20037802867591	0

Analysis from the saved dataset

Logistic regression plot. Hybrid viability across parental divergence — Simulated observations; generated by this module’s R script.

Divergence coefficient: -0.464
Odds ratio per percentage point: 0.629
95% Wald CI for odds ratio: 0.528 to 0.749
Likelihood-ratio p = 6.573e-09

Each additional percentage point of divergence multiplied modeled viability odds by 0.63 (95% Wald CI 0.53–0.75; likelihood-ratio p = 6.57e-09). The curve translates odds into probabilities; this association alone does not establish causation.

Teaching pointA coefficient describes log odds; translate to odds ratios or plotted probabilities to answer the biological question.

Analysis code excerpt

d <- read.csv("modules/logistic/generated/data.csv")
fit <- glm(viable ~ divergence, data = d, family = binomial)
test <- drop1(fit, test = "Chisq")
p_value <- test["divergence", "Pr(>Chi)"]
b <- coef(fit)["divergence"]
se <- sqrt(vcov(fit)["divergence", "divergence"])
odds_ratio <- exp(b)
ci <- exp(b + c(-1, 1) * 1.96 * se)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


Call:
glm(formula = viable ~ divergence, family = binomial, data = d)

Coefficients:
            Estimate Std. Error z value Pr(>|z|)    
(Intercept)  1.99600    0.41934   4.760 1.94e-06 ***
divergence  -0.46382    0.08915  -5.203 1.96e-07 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 221.78  on 159  degrees of freedom
Residual deviance: 188.12  on 158  degrees of freedom
AIC: 192.12

Number of Fisher Scoring iterations: 4

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

out_dir <- "modules/logistic/generated"
dir.create(out_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(3110)
d <- data.frame(cross = 1:160, divergence = runif(160, 0, 8))
d$viable <- rbinom(160, 1, plogis(1.9 - .5 * d$divergence))
write.csv(d, file.path(out_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/logistic/analysis.R modules/logistic/generated

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Dataset 36 · Counts and exposure

Exact Poisson rate test

CSV rows: 60Seed: 3109Module: poisson_exact

Saved CSV Full runnable R script Full-size plot

De novo mutations in sequenced microbial lineages

Question and null

Scientific question

Does the mutation rate differ from 0.25 mutations per callable megabase?

Null being tested / estimation target

The common mutation rate is 0.25 per callable megabase.

Generating truth

The true rate is 0.40 mutations/Mb rather than the null rate 0.25/Mb. The population rate ratio to the null is 1.6.

Exact generating model

Generate 60 independent lineages with callable sequence Uniform(8,20) megabases. Each count is Poisson with mean 0.40 × callable megabases. The population rate is constant across lineages and exposure is unrounded.

Design and independent unit: Independent lineages have different amounts of callable sequence.

Columns in the saved CSV

Column	Meaning	Unit / type
`lineage`	Unique independent microbial lineage	none
`megabases`	Callable sequence exposure	Mb
`mutations`	Observed de novo mutations	count

Preview the actual saved observations

lineage	megabases	mutations
1	14.7936320584267	7
2	19.4386634835973	6
3	19.1721002459526	3
4	12.0031125238165	4
5	18.0816876189783	5

Analysis from the saved dataset

Exact Poisson rate test plot. De novo mutations in sequenced microbial lineages — Simulated observations; generated by this module’s R script.

Events: 345; exposure: 845.0 Mb
Rate: 0.408 mutations/Mb
95% exact CI: 0.366 to 0.454
Exact p = 3.246e-17

The mutation rate was 0.408 per callable Mb (95% exact CI 0.366–0.454; p = 3.25e-17 against 0.25/Mb). Pooling counts is justified here by the common-rate Poisson model.

Teaching pointA count becomes a rate only after the exposure and its units are supplied.

Analysis code excerpt

d <- read.csv("modules/poisson_exact/generated/data.csv")
fit <- poisson.test(sum(d$mutations), T = sum(d$megabases),
                    r = .25, alternative = "two.sided")
rate <- sum(d$mutations) / sum(d$megabases)
ci <- fit$conf.int

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


	Exact Poisson test

data:  sum(d$mutations) time base: sum(d$megabases)
number of events = 345, time base = 844.99, p-value < 2.2e-16
alternative hypothesis: true event rate is not equal to 0.25
95 percent confidence interval:
 0.3663373 0.4537273
sample estimates:
event rate 
 0.4082887

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

out_dir <- "modules/poisson_exact/generated"
dir.create(out_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(3109)
d <- data.frame(lineage = 1:60, megabases = runif(60, 8, 20))
d$mutations <- rpois(60, .40 * d$megabases)
write.csv(d, file.path(out_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/poisson_exact/analysis.R modules/poisson_exact/generated

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Dataset 37 · Generalized linear models

Poisson regression

CSV rows: 120Seed: 3111Module: poisson_glm

Saved CSV Full runnable R script Full-size plot

Coral recruit counts and live coral cover

Question and null

Scientific question

Does recruit density change with live coral cover after accounting for surveyed area?

Null being tested / estimation target

The coral-cover coefficient is zero: recruit rate per square meter is unrelated to cover in this model.

Generating truth

The true cover coefficient is 0.016. A 10-point cover increase multiplies recruit density by exp(0.16) ≈ 1.174. The zero-slope null is false, and the exposure coefficient is fixed at one.

Exact generating model

Generate 120 independent transects. Cover is Uniform(0,80) percentage points and area is independently Uniform(2,8) square meters. Recruit counts are Poisson(area × exp(-0.5 + 0.016 × cover)). No clustering or rounding is introduced.

Design and independent unit: Independent transects differ in surveyed area and live coral cover.

Columns in the saved CSV

Column	Meaning	Unit / type
`transect`	Unique independent transect identifier	none
`cover`	Live coral cover	percentage points
`area`	Surveyed transect area	m²
`recruits`	Observed coral recruits	count

Preview the actual saved observations

transect	cover	area	recruits
1	56.6454520076513	7.21462797746062	15
2	58.0549041554332	6.60326516302302	9
3	23.0353081598878	7.21808331925422	5
4	9.47735395282507	4.16359867714345	3
5	75.7575953565538	6.12687354674563	12

Analysis from the saved dataset

Poisson regression plot. Coral recruit counts and live coral cover — Simulated observations; generated by this module’s R script.

Rate ratio per 10 cover points: 1.200
95% Wald CI: 1.162 to 1.240
Likelihood-ratio p = 2.64e-30
Pearson dispersion diagnostic: 1.111

A 10-percentage-point increase in coral cover multiplied modeled recruit density by 1.20 (95% Wald CI 1.16–1.24; p = 2.64e-30). Area is handled as exposure; the observational association is not a causal effect.

Teaching pointThe exposure offset separates a biological density difference from simply looking over a larger area.

Analysis code excerpt

d <- read.csv("modules/poisson_glm/generated/data.csv")
fit <- glm(recruits ~ cover + offset(log(area)), data = d,
           family = poisson)
p_value <- drop1(fit, test = "Chisq")["cover", "Pr(>Chi)"]
b <- coef(fit)["cover"]
se <- sqrt(vcov(fit)["cover", "cover"])
rate_ratio <- exp(10 * b)
ci <- exp(10 * (b + c(-1, 1) * 1.96 * se))
dispersion <- sum(residuals(fit, type = "pearson")^2) / df.residual(fit)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


Call:
glm(formula = recruits ~ cover + offset(log(area)), family = poisson, 
    data = d)

Coefficients:
             Estimate Std. Error z value Pr(>|z|)    
(Intercept) -0.587917   0.092543  -6.353 2.11e-10 ***
cover        0.018256   0.001649  11.071  < 2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for poisson family taken to be 1)

    Null deviance: 258.31  on 119  degrees of freedom
Residual deviance: 127.43  on 118  degrees of freedom
AIC: 541.72

Number of Fisher Scoring iterations: 4

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

out_dir <- "modules/poisson_glm/generated"
dir.create(out_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(3111)
d <- data.frame(transect = 1:120, cover = runif(120, 0, 80),
                area = runif(120, 2, 8))
d$recruits <- rpois(120, d$area * exp(-.5 + .016 * d$cover))
write.csv(d, file.path(out_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/poisson_glm/analysis.R modules/poisson_glm/generated

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Dataset 38 · Generalized linear models

Quasi-Poisson regression

CSV rows: 120Seed: 3113Module: quasipoisson

Saved CSV Full runnable R script Full-size plot

Bacterial colony counts with clumping

Question and null

Scientific question

Does mean colony density differ between media after accounting for plated volume?

Null being tested / estimation target

The high-medium coefficient is zero: mean colony rates are equal between media.

Generating truth

High medium multiplies mean colony density by exp(0.5) ≈ 1.649; the equal-rate null is false. The true dispersion multiplier is 4, so Poisson uncertainty would be too small.

Exact generating model

Generate 60 low-medium and 60 high-medium independent plates, with volume Uniform(0.7,1.3) ml. Set mean count μ = volume × exp(2.4 + 0.5 × high). Generate clumps ~ Poisson(μ/3), then colonies ~ Poisson(3 × clumps). Marginal mean is μ and variance is 4μ.

Design and independent unit: Independent plated samples vary in medium and plated volume.

Columns in the saved CSV

Column	Meaning	Unit / type
`medium`	Low or High nutrient medium	none
`volume_ml`	Plated volume exposure	ml
`colonies`	Observed colony count after clumping	count

Preview the actual saved observations

medium	volume_ml	colonies
Low	1.18410207144916	7
Low	0.738803575234488	13
Low	0.832271371362731	8
Low	0.969846675405279	13
Low	0.774454141547903	10

Analysis from the saved dataset

Quasi-Poisson regression plot. Bacterial colony counts with clumping — Simulated observations; generated by this module’s R script.

High / low mean rate ratio: 1.351
95% t-based CI: 1.117 to 1.633
Dispersion multiplier: 3.953
Quasi-Poisson t-test p = 0.002166

High-nutrient medium had 1.35 times the modeled colony rate of low-nutrient medium (95% t-based CI 1.12–1.63; p = 0.00217). Estimated dispersion 3.95 adjusts uncertainty for extra-Poisson variation.

Teaching pointQuasi-Poisson retains the log-mean question while changing the uncertainty model; ordinary likelihood AIC is unavailable.

Analysis code excerpt

d <- read.csv("modules/quasipoisson/generated/data.csv")
d$medium <- factor(d$medium, levels = c("Low", "High"))
fit <- glm(colonies ~ medium + offset(log(volume_ml)),
           data = d, family = quasipoisson)
b <- coef(fit)["mediumHigh"]
se <- sqrt(vcov(fit)["mediumHigh", "mediumHigh"])
critical <- qt(.975, df.residual(fit))
rate_ratio <- exp(b)
ci <- exp(b + c(-1, 1) * critical * se)
p_value <- summary(fit)$coefficients["mediumHigh", "Pr(>|t|)"]

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


Call:
glm(formula = colonies ~ medium + offset(log(volume_ml)), family = quasipoisson, 
    data = d)

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  2.52196    0.07284  34.622  < 2e-16 ***
mediumHigh   0.30075    0.09592   3.135  0.00217 ** 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for quasipoisson family taken to be 3.953259)

    Null deviance: 489.45  on 119  degrees of freedom
Residual deviance: 450.13  on 118  degrees of freedom
AIC: NA

Number of Fisher Scoring iterations: 4

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

out_dir <- "modules/quasipoisson/generated"
dir.create(out_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(3113)
d <- data.frame(medium = rep(c("Low", "High"), each = 60),
                volume_ml = runif(120, .7, 1.3))
mu <- d$volume_ml * exp(2.4 + .5 * (d$medium == "High"))
clumps <- rpois(120, mu / 3)
d$colonies <- rpois(120, 3 * clumps)
write.csv(d, file.path(out_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/quasipoisson/analysis.R modules/quasipoisson/generated

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Dataset 39 · Generalized linear models

Negative binomial regression

CSV rows: 130Seed: 3112Module: negbin_glm

Saved CSV Full runnable R script Full-size plot

Parasite abundance and fish body length

Question and null

Scientific question

Does expected parasite burden change with fish length?

Null being tested / estimation target

The length coefficient is zero: expected parasite abundance does not change with length in this model.

Generating truth

The length coefficient is 0.085, so 5 cm multiplies expected parasite abundance by exp(0.425) ≈ 1.530. The slope-zero null is false; the Poisson variance assumption is intentionally violated.

Exact generating model

Generate 130 independent fish with length Uniform(18,42) cm. Parasite count follows a negative binomial with mean exp(-0.2 + 0.085 × length) and size/theta 2. Conditional variance is mean + mean²/2.

Design and independent unit: Independent fish vary in body length; parasite burdens are highly variable.

Columns in the saved CSV

Column	Meaning	Unit / type
`fish`	Unique independent fish identifier	none
`length_cm`	Fish body length	cm
`parasites`	Observed parasite burden	count

Preview the actual saved observations

fish	length_cm	parasites
1	27.863254705444	7
2	27.5214403253049	1
3	29.7717737685889	4
4	20.5594320036471	2
5	21.0298875980079	2

Analysis from the saved dataset

Negative binomial regression plot. Parasite abundance and fish body length — Simulated observations; generated by this module’s R script.

Mean count ratio per 5 cm: 1.602
95% Wald CI: 1.458 to 1.761
Wald p = 1.634e-22
Estimated negative-binomial theta: 2.292

A 5-cm increase in fish length multiplied expected parasite count by 1.60 (95% Wald CI 1.46–1.76; Wald p = 1.63e-22). Negative binomial variation allows more spread than a Poisson model.

Teaching pointOverdispersion can be addressed with a parametric count model; it does not automatically require replacing the scientific question with a rank test.

Analysis code excerpt

d <- read.csv("modules/negbin_glm/generated/data.csv")
fit <- MASS::glm.nb(parasites ~ length_cm, data = d)
b <- coef(fit)["length_cm"]
se <- sqrt(vcov(fit)["length_cm", "length_cm"])
p_value <- summary(fit)$coefficients["length_cm", "Pr(>|z|)"]
rate_ratio <- exp(5 * b)
ci <- exp(5 * (b + c(-1, 1) * 1.96 * se))

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


Call:
MASS::glm.nb(formula = parasites ~ length_cm, data = d, init.theta = 2.292042228, 
    link = log)

Coefficients:
             Estimate Std. Error z value Pr(>|z|)    
(Intercept) -0.617706   0.299553  -2.062   0.0392 *  
length_cm    0.094281   0.009658   9.762   <2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for Negative Binomial(2.292) family taken to be 1)

    Null deviance: 234.38  on 129  degrees of freedom
Residual deviance: 140.90  on 128  degrees of freedom
AIC: 814.62

Number of Fisher Scoring iterations: 1


              Theta:  2.292 
          Std. Err.:  0.374 

 2 x log-likelihood:  -808.615

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

out_dir <- "modules/negbin_glm/generated"
dir.create(out_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(3112)
d <- data.frame(fish = 1:130, length_cm = runif(130, 18, 42))
mu <- exp(-.2 + .085 * d$length_cm)
d$parasites <- rnbinom(130, mu = mu, size = 2)
write.csv(d, file.path(out_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/negbin_glm/analysis.R modules/negbin_glm/generated

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Dataset 40 · Generalized linear models

Multinomial logistic regression

CSV rows: 240Seed: 3114Module: multinomial

Saved CSV Full runnable R script Full-size plot

Fish use of three unordered feeding microhabitats

Question and null

Scientific question

Does food availability change the distribution of feeding microhabitat choices?

Null being tested / estimation target

All nonreference food coefficients are zero; microhabitat probabilities are the same under low and high food.

Generating truth

The outcome distributions differ, so the joint no-food-effect null is false. Relative to Open, the food coefficients are log(0.30/0.50)-log(0.50/0.25) ≈ -1.204 for Shelter and log(0.20/0.50)-log(0.25/0.25) ≈ -0.916 for Surface.

Exact generating model

Generate 120 independent fish choices under Low food with Open/Shelter/Surface probabilities 0.25/0.50/0.25 and 120 under High food with probabilities 0.50/0.30/0.20. Choices are unordered, mutually exclusive, and independent.

Design and independent unit: Independent fish encounter either low or high food availability.

Columns in the saved CSV

Column	Meaning	Unit / type
`food`	Low or High food availability	none
`choice`	Open, Shelter, or Surface microhabitat	none

Preview the actual saved observations

food	choice
Low	Surface
Low	Shelter
Low	Shelter
Low	Shelter
Low	Surface

Analysis from the saved dataset

Multinomial logistic regression plot. Fish use of three unordered feeding microhabitats — Simulated observations; generated by this module’s R script.

Low-food probabilities: 0.217, 0.508, 0.275
High-food probabilities: 0.475, 0.292, 0.233
Likelihood ratio = 19.404; df = 2
Joint p = 6.115e-05

Predicted probabilities for open/shelter/surface were 0.22/0.51/0.28 under low food and 0.48/0.29/0.23 under high food. The joint likelihood-ratio test gives chi-square(2) = 19.40, p = 6.11e-05; it tests the whole distribution of choices.

Teaching pointA categorical response with more than two levels requires a joint question across several coefficients.

Analysis code excerpt

d <- read.csv("modules/multinomial/generated/data.csv")
d$food <- factor(d$food, levels = c("Low", "High"))
d$choice <- factor(d$choice, levels = c("Open", "Shelter", "Surface"))
fit <- nnet::multinom(choice ~ food, data = d, trace = FALSE)
fit0 <- nnet::multinom(choice ~ 1, data = d, trace = FALSE)
LR <- 2 * as.numeric(logLik(fit) - logLik(fit0))
df <- attr(logLik(fit), "df") - attr(logLik(fit0), "df")
p_value <- pchisq(LR, df, lower.tail = FALSE)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output

Call:
nnet::multinom(formula = choice ~ food, data = d, trace = FALSE)

Coefficients:
        (Intercept)   foodHigh
Shelter   0.8527665 -1.3404733
Surface   0.2384121 -0.9492706

Std. Errors:
        (Intercept)  foodHigh
Shelter   0.2342112 0.3177579
Surface   0.2622292 0.3493172

Residual Deviance: 499.8932 
AIC: 507.8932

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

out_dir <- "modules/multinomial/generated"
dir.create(out_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(3114)
d <- data.frame(food = rep(c("Low", "High"), each = 120))
d$choice <- c(sample(c("Open", "Shelter", "Surface"), 120, TRUE,
                      prob = c(.25, .50, .25)),
              sample(c("Open", "Shelter", "Surface"), 120, TRUE,
                      prob = c(.50, .30, .20)))
write.csv(d, file.path(out_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/multinomial/analysis.R modules/multinomial/generated

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Dataset 41 · Generalized linear models

Ordinal logistic regression

CSV rows: 180Seed: 3115Module: ordinal_logistic

Saved CSV Full runnable R script Full-size plot

Coral bleaching severity along a temperature gradient

Question and null

Scientific question

Does temperature shift bleaching severity toward higher ordered categories?

Null being tested / estimation target

The common temperature coefficient is zero; the modeled severity distribution does not change with temperature.

Generating truth

The common temperature coefficient is 0.8, so each °C multiplies odds of higher versus lower severity by exp(0.8) ≈ 2.226 at every cumulative threshold. The zero-effect null is false and proportional odds holds by construction.

Exact generating model

Generate 180 independent fragments with temperature Uniform(26,32) °C. Latent severity is 0.8 × (temperature - 28) plus a standard logistic error. Cut at -1, 0.7, and 2.2 to obtain None, Mild, Moderate, Severe. Do not treat category spacing as numeric distance.

Design and independent unit: Independent coral fragments are assigned a four-level ordered bleaching score.

Columns in the saved CSV

Column	Meaning	Unit / type
`fragment`	Unique independent coral fragment identifier	none
`temperature`	Experimental temperature	°C
`severity`	Ordered bleaching level: None < Mild < Moderate < Severe	none

Preview the actual saved observations

fragment	temperature	severity
1	28.1531253228895	Moderate
2	28.3028749451041	None
3	27.6260669585317	None
4	27.5612253323197	Mild
5	31.1645849784836	Moderate

Analysis from the saved dataset

Ordinal logistic regression plot. Coral bleaching severity along a temperature gradient — Simulated observations; generated by this module’s R script.

Common odds ratio per 1°C: 2.613
95% Wald CI: 2.105 to 3.244
Likelihood ratio = 97.836; df = 1
p = 4.544e-23

Each 1°C increase multiplied the modeled odds of higher rather than lower bleaching severity by 2.61 (95% Wald CI 2.11–3.24; likelihood-ratio p = 4.54e-23), assuming the same temperature effect across thresholds.

Teaching pointOrdering carries information without assuming equal distances; the common effect across thresholds is the key model assumption.

Analysis code excerpt

d <- read.csv("modules/ordinal_logistic/generated/data.csv")
d$severity <- ordered(d$severity, levels = c("None", "Mild", "Moderate", "Severe"))
fit <- ordinal::clm(severity ~ temperature, data = d, link = "logit")
fit0 <- ordinal::clm(severity ~ 1, data = d, link = "logit")
LR <- 2 * as.numeric(logLik(fit) - logLik(fit0))
p_value <- pchisq(LR, df = 1, lower.tail = FALSE)
b <- fit$beta["temperature"]
se <- sqrt(vcov(fit)["temperature", "temperature"])
odds_ratio <- exp(b)
ci <- exp(b + c(-1, 1) * 1.96 * se)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output

formula: severity ~ temperature
data:    d

 link  threshold nobs logLik  AIC    niter max.grad cond.H 
 logit flexible  180  -200.34 408.69 5(0)  2.85e-11 1.3e+06

Coefficients:
            Estimate Std. Error z value Pr(>|z|)    
temperature   0.9606     0.1103    8.71   <2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Threshold coefficients:
                Estimate Std. Error z value
None|Mild         25.935      3.098   8.373
Mild|Moderate     27.589      3.178   8.682
Moderate|Severe   29.316      3.273   8.957

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

out_dir <- "modules/ordinal_logistic/generated"
dir.create(out_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(3115)
d <- data.frame(fragment = 1:180, temperature = runif(180, 26, 32))
latent <- .8 * (d$temperature - 28) + rlogis(180)
d$severity <- cut(latent, c(-Inf, -1, .7, 2.2, Inf),
                  labels = c("None", "Mild", "Moderate", "Severe"))
write.csv(d, file.path(out_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/ordinal_logistic/analysis.R modules/ordinal_logistic/generated

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Dataset 42 · Repeated and clustered data

Linear mixed model

CSV rows: 144Seed: 2106Module: linear_mixed

Saved CSV Full runnable R script Full-size plot

Evolutionary biologists follow larval growth under two diets.

Question and null

Scientific question

Does diet change the mean growth slope?

Null being tested / estimation target

The population growth-slope difference between diets is zero.

Generating truth

The control slope is 2 mg/week; enrichment adds 0.9 mg/week. The diet-by-week null is false. Simulated random-intercept and slope SDs are 2.3 mg and 0.7 mg/week.

Exact generating model

36 larvae (18 per diet) are measured at weeks 0–3. Mass = 12 + 2 × week + 0.8 if enriched + 0.9 × week if enriched + a larval Normal(0, 2.3²) intercept + a larval Normal(0, 0.7²) slope × week + independent Normal(0, 1.1²) error. Simulated intercepts and slopes are independent. Repeated values share both effects; no rounding.

Design and independent unit: 36 larvae measured at weeks 0–3; diet assigned to larvae.

Columns in the saved CSV

Column	Meaning	Unit / type
`week`	Time since baseline	week
`larva`	Repeated biological individual	ID
`treatment`	Control or enriched diet	category
`mass`	Larval mass	mg

Preview the actual saved observations

week	larva	treatment	mass
0	1	Control	11.1027584930035
1	1	Control	11.9429528919379
2	1	Control	15.1949825881597
3	1	Control	16.07766272148
0	2	Control	12.3917285881197

Analysis from the saved dataset

Linear mixed model plot. Evolutionary biologists follow larval growth under two diets. — Simulated observations; generated by this module’s R script.

Growth-slope difference = 0.79 mg/week
Approximate 95% Wald CI 0.31 to 1.27
Satterthwaite df = 34.0; t = 3.20; p = 0.002966
36 larvae; 144 rows; singular fit: FALSE

Enrichment increased the fitted growth slope by 0.79 mg/week (approximate 95% Wald CI 0.31 to 1.27; Satterthwaite t with 34.0 df, p = 0.002966). Individual intercepts and slopes model repeated larvae; 144 rows do not represent 144 independent animals.

Teaching pointRandom intercepts describe starting differences; random slopes describe trajectory differences.

Analysis code excerpt

d <- read.csv("modules/linear_mixed/generated/data.csv")
fit <- lmerTest::lmer(mass ~ treatment * week + (week | larva), data = d)
summary(fit, ddf = "Satterthwaite")
trend_ci <- confint(fit, parm = "treatmentEnriched:week", method = "Wald")

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output

Linear mixed model fit by REML. t-tests use Satterthwaite's method [
lmerModLmerTest]
Formula: mass ~ treatment * week + (week | larva)
   Data: d

REML criterion at convergence: 570

Scaled residuals: 
     Min       1Q   Median       3Q      Max 
-1.64582 -0.56247  0.05774  0.54128  1.60492 

Random effects:
 Groups   Name        Variance Std.Dev. Corr  
 larva    (Intercept) 5.7165   2.3909         
          week        0.3129   0.5594   -0.01 
 Residual             1.1701   1.0817         
Number of obs: 144, groups:  larva, 36

Fixed effects:
                       Estimate Std. Error      df t value Pr(>|t|)    
(Intercept)             12.2501     0.6026 33.9995  20.330  < 2e-16 ***
treatmentEnriched        1.3465     0.8522 33.9995   1.580  0.12335    
week                     2.0463     0.1743 34.0002  11.740 1.65e-13 ***
treatmentEnriched:week   0.7891     0.2465 34.0002   3.201  0.00297 ** 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Correlation of Fixed Effects:
            (Intr) trtmnE week  
trtmntEnrch -0.707              
week        -0.193  0.137       
trtmntEnrc:  0.137 -0.193 -0.707

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/linear_mixed/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(2106)
d <- expand.grid(week = 0:3, larva = 1:36)
d$treatment <- rep(rep(c("Control", "Enriched"), each = 18), each = 4)
intercepts <- rnorm(36, 0, 2.3); slopes <- rnorm(36, 0, .7)
d$mass <- 12 + 2 * d$week + .8 * (d$treatment == "Enriched") +
  .9 * d$week * (d$treatment == "Enriched") + intercepts[d$larva] +
  slopes[d$larva] * d$week + rnorm(nrow(d), 0, 1.1)
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/linear_mixed/analysis.R modules/linear_mixed/generated

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Dataset 43 · Repeated and clustered data

Binary mixed model

CSV rows: 288Seed: 2107Module: binary_glmm

Saved CSV Full runnable R script Full-size plot

Immunologists record repeated infection outcomes in vaccinated and control hosts.

Question and null

Scientific question

Does vaccination affect infection odds while accounting for repeated hosts?

Null being tested / estimation target

The vaccination coefficient is zero: its conditional odds ratio is one.

Generating truth

The conditional vaccine odds ratio is exp(−1) ≈ 0.368; the week odds ratio is exp(0.25) ≈ 1.284. The vaccine null is false. These are conditional effects, not population-average odds ratios.

Exact generating model

72 hosts (36 per group) are scored at weeks 0–3. Host intercepts are Normal(0, 0.8²). The infection log odds equal −0.4 + 0.25 × week − 1.0 if vaccinated + host intercept. Independent Bernoulli outcomes are sampled conditional on those probabilities. Host intercepts create dependence across visits; latent probabilities are removed from the saved student dataset.

Design and independent unit: 72 hosts with four weekly binary responses.

Columns in the saved CSV

Column	Meaning	Unit / type
`week`	Scheduled visit relative to baseline	week
`host`	Repeated host identifier	ID
`treatment`	Control or vaccine assignment	category
`infected`	Infection outcome: 1 infected, 0 not infected	binary

Preview the actual saved observations

week	host	treatment	infected
0	1	Control	1
1	1	Control	1
2	1	Control	1
3	1	Control	1
0	2	Control	0

Analysis from the saved dataset

Binary mixed model plot. Immunologists record repeated infection outcomes in vaccinated and control hosts. — Simulated observations; generated by this module’s R script.

Conditional vaccine odds ratio = 0.24
Approximate 95% CI 0.12 to 0.48
Wald z = -3.99; p = 6.63e-05
72 independent hosts; four repeated observations each

At the same week and host random effect, vaccination multiplied infection odds by 0.24 (approximate 95% Wald CI 0.12 to 0.48; p = 6.63e-05). This is a conditional odds ratio, not a probability ratio or the population-average odds ratio estimated by a GEE.

Teaching pointRepeated binary outcomes need both a response distribution and a dependence model.

Analysis code excerpt

d <- read.csv("modules/binary_glmm/generated/data.csv")
fit <- lme4::glmer(infected ~ treatment + week + (1 | host),
                   family = binomial, data = d,
                   control = lme4::glmerControl(optimizer = "bobyqa"))
summary(fit)
log_or_ci <- confint(fit, parm = "treatmentVaccine", method = "Wald")

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output

Generalized linear mixed model fit by maximum likelihood (Laplace
  Approximation) [glmerMod]
 Family: binomial  ( logit )
Formula: infected ~ treatment + week + (1 | host)
   Data: d
Control: lme4::glmerControl(optimizer = "bobyqa")

      AIC       BIC    logLik -2*log(L)  df.resid 
    366.0     380.7    -179.0     358.0       284 

Scaled residuals: 
    Min      1Q  Median      3Q     Max 
-1.4049 -0.6888 -0.4268  0.7494  1.8362 

Random effects:
 Groups Name        Variance Std.Dev.
 host   (Intercept) 0.8177   0.9043  
Number of obs: 288, groups:  host, 72

Fixed effects:
                 Estimate Std. Error z value Pr(>|z|)    
(Intercept)       0.34030    0.29953   1.136    0.256    
treatmentVaccine -1.42502    0.35722  -3.989 6.63e-05 ***
week             -0.05141    0.12047  -0.427    0.670    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Correlation of Fixed Effects:
            (Intr) trtmnV
tretmntVccn -0.550       
week        -0.607  0.016

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/binary_glmm/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(2107)
d <- expand.grid(week = 0:3, host = 1:72)
d$treatment <- rep(rep(c("Control", "Vaccine"), each = 36), each = 4)
individual <- rnorm(72, 0, .8)
d$probability <- plogis(-.4 + .25 * d$week -
  1.0 * (d$treatment == "Vaccine") + individual[d$host])
d$infected <- rbinom(nrow(d), 1, d$probability)
d$probability <- NULL
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/binary_glmm/analysis.R modules/binary_glmm/generated

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Dataset 44 · Repeated and clustered data

Count mixed model: Poisson and negative binomial

CSV rows: 192Seed: 2114Module: count_glmm

Saved CSV Full runnable R script Full-size plot

Pollinator visits are repeatedly counted on plants in two habitats.

Question and null

Scientific question

Does habitat change visit rate while accounting for repeated plants?

Null being tested / estimation target

The habitat coefficient is zero: its conditional visit-rate ratio is one.

Generating truth

The conditional sheltered/open rate ratio is exp(0.55) ≈ 1.733; the habitat null is false. The generating distribution has extra variation beyond Poisson even after conditioning on the plant intercept.

Exact generating model

48 plants (24 per habitat) are observed on days 1–4. Observation duration is sampled uniformly from the integers 20–60 minutes. Plant intercepts are Normal(0, 0.45²). Log rate = −2 + 0.55 if sheltered + 0.06 × day + plant intercept. Counts are negative binomial with mean = minutes × rate and size = 5; conditional variance is mean + mean²/5. Repeated plants share the intercept.

Design and independent unit: 48 plants observed on four days with unequal observation durations.

Columns in the saved CSV

Column	Meaning	Unit / type
`day`	Observation day	day
`plant`	Repeated plant identifier	ID
`habitat`	Open or sheltered habitat	category
`minutes`	Observation exposure	minute
`visits`	Number of pollinator visits	count

Preview the actual saved observations

day	plant	habitat	minutes	visits
1	1	Open	32	22
2	1	Open	58	10
3	1	Open	25	9
4	1	Open	48	23
1	2	Open	43	5

Analysis from the saved dataset

Count mixed model: Poisson and negative binomial plot. Pollinator visits are repeatedly counted on plants in two habitats. — Simulated observations; generated by this module’s R script.

Conditional sheltered/open rate ratio = 1.50
Approximate 95% CI 1.11 to 2.02
Wald z = 2.62; p = 0.008759
Poisson AIC 1281.6; negative-binomial AIC 1165.5

In the negative-binomial mixed model, sheltered plants had 1.50 times the visit rate of open plants at the same day and random effect (approximate 95% Wald CI 1.11 to 2.02; p = 0.008759). Minutes are an exposure offset; plant intercepts model repeated observations. Extra conditional variation motivates checking the negative-binomial variant, not automatically selecting it from a single cutoff.

Teaching pointOffsets retain count information while accounting for exposure; negative binomial and random effects address different variation sources.

Analysis code excerpt

d <- read.csv("modules/count_glmm/generated/data.csv")
poisson_fit <- lme4::glmer(visits ~ habitat + day +
  offset(log(minutes)) + (1 | plant), family = poisson, data = d,
  control = lme4::glmerControl(optimizer = "bobyqa"))
fit <- lme4::glmer.nb(visits ~ habitat + day +
  offset(log(minutes)) + (1 | plant), data = d,
  control = lme4::glmerControl(optimizer = "bobyqa"))
summary(fit)
log_ratio_ci <- confint(fit, parm = "habitatSheltered", method = "Wald")

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output

Generalized linear mixed model fit by maximum likelihood (Laplace
  Approximation) [glmerMod]
 Family: Negative Binomial(5.8365)  ( log )
Formula: visits ~ habitat + day + offset(log(minutes)) + (1 | plant)
   Data: d
Control: lme4::glmerControl(optimizer = "bobyqa")

      AIC       BIC    logLik -2*log(L)  df.resid 
   1165.5    1181.8    -577.8    1155.5       187 

Scaled residuals: 
    Min      1Q  Median      3Q     Max 
-1.6584 -0.6798 -0.1409  0.4246  3.2452 

Random effects:
 Groups Name        Variance Std.Dev.
 plant  (Intercept) 0.2057   0.4536  
Number of obs: 192, groups:  plant, 48

Fixed effects:
                 Estimate Std. Error z value Pr(>|z|)    
(Intercept)      -1.90884    0.14304 -13.345  < 2e-16 ***
habitatSheltered  0.40249    0.15355   2.621  0.00876 ** 
day               0.05234    0.03605   1.452  0.14658    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Correlation of Fixed Effects:
            (Intr) hbttSh
habttShltrd -0.547       
day         -0.638 -0.005
Generalized linear mixed model fit by maximum likelihood (Laplace
  Approximation) [glmerMod]
 Family: poisson  ( log )
Formula: visits ~ habitat + day + offset(log(minutes)) + (1 | plant)
   Data: d
Control: lme4::glmerControl(optimizer = "bobyqa")

      AIC       BIC    logLik -2*log(L)  df.resid 
   1281.6    1294.7    -636.8    1273.6       188 

Scaled residuals: 
    Min      1Q  Median      3Q     Max 
-3.0928 -0.9462 -0.1606  0.6940  5.4685 

Random effects:
 Groups Name        Variance Std.Dev.
 plant  (Intercept) 0.2447   0.4947  
Number of obs: 192, groups:  plant, 48

Fixed effects:
                 Estimate Std. Error z value Pr(>|z|)    
(Intercept)      -1.90548    0.12177 -15.649   <2e-16 ***
habitatSheltered  0.40022    0.15209   2.631   0.0085 ** 
day               0.04840    0.02086   2.321   0.0203 *  
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Correlation of Fixed Effects:
            (Intr) hbttSh
habttShltrd -0.644       
day         -0.447  0.011

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/count_glmm/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(2114)
d <- expand.grid(day = 1:4, plant = 1:48)
d$habitat <- rep(rep(c("Open", "Sheltered"), each = 24), each = 4)
d$minutes <- sample(20:60, nrow(d), replace = TRUE)
individual <- rnorm(48, 0, .45)
rate <- exp(-2.0 + .55 * (d$habitat == "Sheltered") +
              .06 * d$day + individual[d$plant])
d$visits <- rnbinom(nrow(d), mu = d$minutes * rate, size = 5)
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/count_glmm/analysis.R modules/count_glmm/generated

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Dataset 45 · Repeated and clustered data

Ordinal mixed model

CSV rows: 210Seed: 2109Module: ordinal_mixed

Saved CSV Full runnable R script Full-size plot

Fish behavior is scored repeatedly using ordered stress categories.

Question and null

Scientific question

Does treatment shift repeated scores toward higher categories?

Null being tested / estimation target

The treatment coefficient is zero: its conditional cumulative odds ratio is one.

Generating truth

The conditional common odds ratio for a higher score under stress is exp(0.8) ≈ 2.226. The treatment null is false. The simulated thresholds obey proportional odds.

Exact generating model

70 fish (35 per treatment) are scored at visits 1–3. Fish intercepts are Normal(0, 0.8²). Latent severity = 0.8 if stressed + 0.15 × visit + fish intercept + standard logistic error. Thresholds −0.8, 0.3 and 1.3 produce None, Mild, Moderate and High. The latent score is not saved; repeated scores share the fish intercept.

Design and independent unit: 70 fish measured at three visits; treatment assigned between fish.

Columns in the saved CSV

Column	Meaning	Unit / type
`visit`	Scheduled repeated assessment	visit
`fish`	Repeated fish identifier	ID
`treatment`	Control or stress assignment	category
`score`	Ordered behavior severity	ordered category

Preview the actual saved observations

visit	fish	treatment	score
1	1	Control	Mild
2	1	Control	Mild
3	1	Control	High
1	2	Control	None
2	2	Control	High

Analysis from the saved dataset

Ordinal mixed model plot. Fish behavior is scored repeatedly using ordered stress categories. — Simulated observations; generated by this module’s R script.

Conditional higher-score odds ratio = 2.51
Approximate 95% CI 1.31 to 4.82
Wald z = 2.78; p = 0.005491
70 fish; three repeated ordinal scores each

At the same visit and fish random effect, stress treatment multiplied the odds of being above any score threshold by 2.51 (approximate 95% Wald CI 1.31 to 4.82; p = 0.005491). The proportional-odds model uses a common effect across thresholds; category labels are ordered, not equally spaced measurements.

Teaching pointUse ordered category probabilities; arbitrary numerical category spacings are unnecessary.

Analysis code excerpt

d <- read.csv("modules/ordinal_mixed/generated/data.csv")
fit <- ordinal::clmm(score ~ treatment + visit + (1 | fish),
                     data = d, link = "logit", Hess = TRUE)
summary(fit)
b <- coef(summary(fit))["treatmentStress", ]
log_or_ci <- b[1] + c(-1, 1) * qnorm(.975) * b[2]

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output

Cumulative Link Mixed Model fitted with the Laplace approximation

formula: score ~ treatment + visit + (1 | fish)
data:    d

 link  threshold nobs logLik  AIC    niter    max.grad cond.H 
 logit flexible  210  -274.11 560.22 240(723) 5.81e-05 1.0e+02

Random effects:
 Groups Name        Variance Std.Dev.
 fish   (Intercept) 0.6701   0.8186  
Number of groups:  fish 70 

Coefficients:
                Estimate Std. Error z value Pr(>|z|)   
treatmentStress   0.9221     0.3321   2.777  0.00549 **
visit             0.2399     0.1608   1.492  0.13578   
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Threshold coefficients:
              Estimate Std. Error z value
None|Mild      -0.6966     0.4029  -1.729
Mild|Moderate   0.6908     0.4025   1.716
Moderate|High   1.5974     0.4207   3.797

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/ordinal_mixed/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(2109)
d <- expand.grid(visit = 1:3, fish = 1:70)
d$treatment <- rep(rep(c("Control", "Stress"), each = 35), each = 3)
individual <- rnorm(70, 0, .8)
latent <- .8 * (d$treatment == "Stress") + .15 * d$visit +
  individual[d$fish] + rlogis(nrow(d))
d$score <- cut(latent, c(-Inf, -.8, .3, 1.3, Inf),
               labels = c("None", "Mild", "Moderate", "High"))
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/ordinal_mixed/analysis.R modules/ordinal_mixed/generated

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Dataset 46 · Dependent observations

Binary generalized estimating equations

CSV rows: 320Seed: 3116Module: binary_gee

Saved CSV Full runnable R script Full-size plot

Repeated infection screening after vaccination

Question and null

Scientific question

Does population-average infection probability differ by vaccination after accounting for week?

Null being tested / estimation target

The population-average vaccination coefficient is zero, adjusting for week.

Generating truth

The true population-average vaccination coefficient is -0.9 and odds ratio exp(-0.9) ≈ 0.407, adjusted for week. The no-group-effect null is false. The true marginal week coefficient is 0.35. A working exchangeable correlation is an approximation, while the marginal mean model is exact.

Exact generating model

Generate 80 independent animals, 40 per vaccination group, measured at weeks 0,1,2,3. Set the marginal infection probability to plogis(-0.6 + 0.35 × week - 0.9 × vaccinated). For each row, z = sqrt(0.3) × shared animal Normal(0,1) + sqrt(0.7) × independent Normal(0,1). Because each z is standard normal, pnorm(z) is marginally uniform. Infection is 1 when pnorm(z) is below the specified probability. The shared component creates dependence while preserving the exact marginal probabilities.

Design and independent unit: 80 independent animals are screened at four scheduled visits.

Columns in the saved CSV

Column	Meaning	Unit / type
`animal`	Independent animal and clustering identifier	none
`week`	Scheduled visit time	weeks
`vaccine`	Control or Vaccinated group	none
`infected`	1 = infection detected; 0 = not detected	binary

Preview the actual saved observations

animal	week	vaccine	infected
1	0	Control	0
1	1	Control	0
1	2	Control	0
1	3	Control	1
2	0	Control	0

Analysis from the saved dataset

Binary generalized estimating equations plot. Repeated infection screening after vaccination — Simulated observations; generated by this module’s R script.

Independent animals: 80; observations: 320
Marginal vaccination odds ratio: 0.555
95% robust CI: 0.294 to 1.046
Robust Wald p = 0.06864
Working correlation estimate: 0.265

The estimated population-average infection odds ratio for vaccination was 0.55 (95% robust CI 0.29–1.05; Wald p = 0.0686), adjusted for week. The model treats 80 animals, rather than 320 rows, as independent units.

Teaching pointChoose between a population-average question and an individual-conditional question; repeated rows are not independent animals.

Analysis code excerpt

d <- read.csv("modules/binary_gee/generated/data.csv")
d$vaccine <- factor(d$vaccine, levels = c("Control", "Vaccinated"))
d <- d[order(d$animal, d$week), ]
fit <- geepack::geeglm(infected ~ vaccine + week, id = animal,
  data = d, family = binomial, corstr = "exchangeable")
b <- coef(fit)["vaccineVaccinated"]
se <- sqrt(vcov(fit)["vaccineVaccinated", "vaccineVaccinated"])
odds_ratio <- exp(b)
ci <- exp(b + c(-1, 1) * 1.96 * se)
p_value <- 2 * pnorm(abs(b / se), lower.tail = FALSE)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


Call:
geepack::geeglm(formula = infected ~ vaccine + week, family = binomial, 
    data = d, id = animal, corstr = "exchangeable")

 Coefficients:
                  Estimate  Std.err   Wald Pr(>|W|)    
(Intercept)       -0.98785  0.25570 14.926 0.000112 ***
vaccineVaccinated -0.58896  0.32346  3.315 0.068637 .  
week               0.46535  0.08719 28.486 9.44e-08 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Correlation structure = exchangeable 
Estimated Scale Parameters:

            Estimate Std.err
(Intercept)        1 0.08157
  Link = identity 

Estimated Correlation Parameters:
      Estimate Std.err
alpha   0.2655  0.0796
Number of clusters:   80  Maximum cluster size: 4

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

out_dir <- "modules/binary_gee/generated"
dir.create(out_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(3116)
shared <- rep(rnorm(80), each = 4)
d <- data.frame(animal = rep(1:80, each = 4),
  week = rep(0:3, 80), vaccine = rep(rep(c("Control", "Vaccinated"), each = 40), each = 4))
z <- sqrt(.3) * shared + sqrt(.7) * rnorm(320)
p <- plogis(-.6 + .35 * d$week - .9 * (d$vaccine == "Vaccinated"))
d$infected <- as.integer(pnorm(z) < p)
write.csv(d, file.path(out_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/binary_gee/analysis.R modules/binary_gee/generated

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Dataset 47 · Time to event

Kaplan–Meier survival estimation

CSV rows: 120Seed: 2110Module: kaplan_meier

Saved CSV Full runnable R script Full-size plot

Ecologists follow seedling survival until death or final observation.

Question and null

Scientific question

What fraction of seedlings remain alive at day 40?

Null being tested / estimation target

No null hypothesis: the target is the survival probability over time.

Generating truth

The population survival function is exp(−0.018 × days); true day-40 survival is exp(−0.72) ≈ 0.487. This is an estimation target, not a null hypothesis.

Exact generating model

120 independent event times are drawn from an exponential distribution with death rate 0.018/day. Each independent final follow-up time is Uniform(35, 100) days. Observed days = minimum(event time, follow-up); event = 1 when the event occurs by follow-up and 0 otherwise. Times remain unrounded. Censoring is independent by construction.

Design and independent unit: 120 independent seedlings; follow-up ends at different times.

Columns in the saved CSV

Column	Meaning	Unit / type
`seedling`	Independent seedling identifier	ID
`days`	Observed event or censoring time	day
`event`	1 death; 0 right-censored	binary

Preview the actual saved observations

seedling	days	event
1	48.0300462433041	1
2	77.7376460412052	0
3	12.8673076185368	1
4	43.8299575634301	0
5	38.0694387066695	1

Analysis from the saved dataset

Kaplan–Meier survival estimation plot. Ecologists follow seedling survival until death or final observation. — Simulated observations; generated by this module’s R script.

120 seedlings; 79 events; 41 censored
Estimated survival at day 40 = 0.455
95% CI 0.364 to 0.542
Still at risk at day 40: 48

Estimated day-40 survival was 45.5% (95% CI 36.4% to 54.2%), with 48 seedlings still at risk. Kaplan–Meier estimates survival rather than testing a null. Censored seedlings contribute information until their final follow-up; they are not assumed to survive forever.

Teaching pointCensored observations contribute until follow-up ends; no invented p-value is needed for estimation.

Analysis code excerpt

d <- read.csv("modules/kaplan_meier/generated/data.csv")
fit <- survival::survfit(survival::Surv(days, event) ~ 1,
                         data = d, conf.type = "log-log")
at_day40 <- summary(fit, times = 40)
print(at_day40)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output

Call: survfit(formula = survival::Surv(days, event) ~ 1, data = d, 
    conf.type = "log-log")

   time n.risk n.event survival std.err lower 95% CI upper 95% CI
  0.106    120       1    0.992  0.0083        0.942        0.999
  0.126    119       1    0.983  0.0117        0.935        0.996
  2.720    118       1    0.975  0.0143        0.925        0.992
  2.786    117       1    0.967  0.0164        0.914        0.987
  3.710    116       1    0.958  0.0182        0.903        0.982
  3.869    115       1    0.950  0.0199        0.892        0.977
  3.900    114       1    0.942  0.0214        0.882        0.972
  3.983    113       1    0.933  0.0228        0.871        0.966
  4.104    112       1    0.925  0.0240        0.861        0.960
  4.377    111       1    0.917  0.0252        0.851        0.954
  4.446    110       1    0.908  0.0263        0.841        0.948
  4.531    109       1    0.900  0.0274        0.831        0.942
  4.822    108       1    0.892  0.0284        0.821        0.936
  5.181    107       1    0.883  0.0293        0.811        0.929
  6.574    106       1    0.875  0.0302        0.801        0.923
  6.788    105       1    0.867  0.0310        0.792        0.916
  7.888    104       1    0.858  0.0318        0.782        0.909
  8.457    103       1    0.850  0.0326        0.773        0.903
  8.532    102       1    0.842  0.0333        0.763        0.896
  9.082    101       1    0.833  0.0340        0.754        0.889
  9.103    100       1    0.825  0.0347        0.744        0.882
 12.133     99       1    0.817  0.0353        0.735        0.875
 12.867     98       1    0.808  0.0359        0.726        0.868
 13.038     97       1    0.800  0.0365        0.717        0.861
 13.059     96       1    0.792  0.0371        0.707        0.854
 13.633     95       1    0.783  0.0376        0.698        0.847
 15.020     94       1    0.775  0.0381        0.689        0.840
 15.281     93       1    0.767  0.0386        0.680        0.833
 15.370     92       1    0.758  0.0391        0.671        0.825
 15.590     91       1    0.750  0.0395        0.662        0.818
 15.865     90       1    0.742  0.0400        0.653        0.811
 18.267     89       1    0.733  0.0404        0.645        0.803
 18.634     88       1    0.725  0.0408        0.636        0.796
 18.744     87       1    0.717  0.0411        0.627        0.788
 18.800     86       1    0.708  0.0415        0.618        0.781
 18.822     85       1    0.700  0.0418        0.609        0.773
 19.473     84       1    0.692  0.0422        0.601        0.766
 20.529     83       1    0.683  0.0425        0.592        0.758
 22.516     82       1    0.675  0.0428        0.583        0.751
 23.080     81       1    0.667  0.0430        0.575        0.743
 23.526     80       1    0.658  0.0433        0.566        0.736
 24.822     79       1    0.650  0.0435        0.558        0.728
 24.903     78       1    0.642  0.0438        0.549        0.720
 26.414     77       1    0.633  0.0440        0.540        0.712
 26.537     76       1    0.625  0.0442        0.532        0.705
 27.108     75       1    0.617  0.0444        0.523        0.697
 30.038     74       1    0.608  0.0446        0.515        0.689
 31.827     73       1    0.600  0.0447        0.507        0.681
 31.855     72       1    0.592  0.0449        0.498        0.673
 32.051     71       1    0.583  0.0450        0.490        0.666
 32.881     70       1    0.575  0.0451        0.482        0.658
 33.146     69       1    0.567  0.0452        0.473        0.650
 33.241     68       1    0.558  0.0453        0.465        0.642
 34.796     67       1    0.550  0.0454        0.457        0.634
 34.815     66       1    0.542  0.0455        0.448        0.626
 35.056     65       1    0.533  0.0455        0.440        0.618
 35.379     64       1    0.525  0.0456        0.432        0.610
 35.787     63       1    0.517  0.0456        0.424        0.602
 35.834     62       1    0.508  0.0456        0.416        0.594
 35.927     61       1    0.500  0.0456        0.408        0.585
 35.995     60       1    0.492  0.0456        0.400        0.577
 36.970     54       1    0.483  0.0457        0.391        0.568
 38.069     53       1    0.473  0.0457        0.382        0.560
 38.550     52       1    0.464  0.0457        0.373        0.551
 39.659     49       1    0.455  0.0458        0.364        0.542
 40.280     48       1    0.445  0.0458        0.354        0.532
 40.896     47       1    0.436  0.0458        0.345        0.523
 43.820     45       1    0.426  0.0458        0.336        0.514
 45.663     40       1    0.416  0.0459        0.325        0.503
 47.372     39       1    0.405  0.0459        0.315        0.493
 47.792     37       1    0.394  0.0460        0.304        0.483
 48.030     36       1    0.383  0.0460        0.293        0.472
 52.841     29       1    0.370  0.0462        0.280        0.460
 58.048     22       1    0.353  0.0471        0.262        0.445
 59.211     21       1    0.336  0.0478        0.245        0.430
 61.635     19       1    0.319  0.0484        0.227        0.414
 65.496     16       1    0.299  0.0493        0.206        0.397
 81.042      9       1    0.265  0.0538        0.167        0.374
 90.357      4       1    0.199  0.0702        0.084        0.349

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/kaplan_meier/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(2110)
d <- data.frame(seedling = 1:120)
event_time <- rexp(120, rate = .018)
follow_up <- runif(120, 35, 100)
d$days <- pmin(event_time, follow_up)
d$event <- as.integer(event_time <= follow_up)
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/kaplan_meier/analysis.R modules/kaplan_meier/generated

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Dataset 48 · Time to event

Log-rank survival comparison

CSV rows: 150Seed: 2111Module: log_rank

Saved CSV Full runnable R script Full-size plot

Plant pathologists compare time to infection under three protective treatments.

Question and null

Scientific question

Are the infection-free survival curves equal across treatments?

Null being tested / estimation target

The groups have the same survival function over follow-up.

Generating truth

Equal survival is false: day-30 infection-free probabilities are exp(−1.35) ≈ 0.259, exp(−0.9) ≈ 0.407 and exp(−0.54) ≈ 0.583. The high/control hazard ratio is 0.4, though the log-rank test itself does not estimate it.

Exact generating model

50 independent plants per group have exponential infection times with rates 0.045, 0.030 and 0.018/day in Control, Low and High treatment. Independent follow-up is Uniform(40, 80) days. Save the minimum time and an event indicator; no rounding. The construction gives proportional hazards and noninformative censoring.

Design and independent unit: 50 independent plants per treatment; some plants remain uninfected at final follow-up.

Columns in the saved CSV

Column	Meaning	Unit / type
`plant`	Independent plant identifier	ID
`treatment`	Protective treatment level	category
`days`	Time to infection or last follow-up	day
`event`	1 infection; 0 right-censored	binary

Preview the actual saved observations

plant	treatment	days	event
1	Control	33.2739301421118	1
2	Control	15.4534463243951	1
3	Control	9.15904902956552	1
4	Control	65.3983728960156	0
5	Control	10.1598302523295	1

Analysis from the saved dataset

Log-rank survival comparison plot. Plant pathologists compare time to infection under three protective treatments. — Simulated observations; generated by this module’s R script.

Log-rank chi-square(2) = 17.18
Omnibus p = 0.0001863
Control day-30 survival 0.24 (95% CI 0.15–0.39)
High day-30 survival 0.62 (95% CI 0.50–0.77)

The three infection-free survival curves differed (log-rank chi-square(2) = 17.18, p = 0.0001863). Estimated day-30 infection-free fractions were 0.24 in control and 0.62 in high treatment. The omnibus test does not identify each differing pair and does not estimate a hazard ratio.

Teaching pointAn omnibus survival comparison is not a pairwise effect estimate.

Analysis code excerpt

d <- read.csv("modules/log_rank/generated/data.csv")
test <- survival::survdiff(survival::Surv(days, event) ~ treatment,
                           data = d, rho = 0)
fit <- survival::survfit(survival::Surv(days, event) ~ treatment, data = d)
p_value <- pchisq(test$chisq, df = length(test$n) - 1, lower.tail = FALSE)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output

Call: survfit(formula = survival::Surv(days, event) ~ treatment, data = d)

                treatment=Control 
   time n.risk n.event survival std.err lower 95% CI upper 95% CI
  0.813     50       1   0.9800  0.0198       0.9420        1.000
  0.992     49       1   0.9600  0.0277       0.9072        1.000
  1.344     48       1   0.9400  0.0336       0.8764        1.000
  2.481     47       1   0.9200  0.0384       0.8478        0.998
  2.498     46       1   0.9000  0.0424       0.8206        0.987
  2.584     45       1   0.8800  0.0460       0.7944        0.975
  4.478     44       1   0.8600  0.0491       0.7690        0.962
  4.495     43       1   0.8400  0.0518       0.7443        0.948
  5.943     42       1   0.8200  0.0543       0.7201        0.934
  6.715     41       1   0.8000  0.0566       0.6965        0.919
  7.222     40       1   0.7800  0.0586       0.6732        0.904
  7.687     39       1   0.7600  0.0604       0.6504        0.888
  7.875     38       1   0.7400  0.0620       0.6279        0.872
  8.784     37       1   0.7200  0.0635       0.6057        0.856
  9.159     36       1   0.7000  0.0648       0.5838        0.839
  9.416     35       1   0.6800  0.0660       0.5623        0.822
 10.160     34       1   0.6600  0.0670       0.5409        0.805
 10.390     33       1   0.6400  0.0679       0.5199        0.788
 10.392     32       1   0.6200  0.0686       0.4991        0.770
 10.587     31       1   0.6000  0.0693       0.4785        0.752
 10.749     30       1   0.5800  0.0698       0.4581        0.734
 11.233     29       1   0.5600  0.0702       0.4380        0.716
 11.492     28       1   0.5400  0.0705       0.4181        0.697
 11.969     27       1   0.5200  0.0707       0.3984        0.679
 13.215     26       1   0.5000  0.0707       0.3790        0.660
 14.101     25       1   0.4800  0.0707       0.3597        0.641
 15.453     24       1   0.4600  0.0705       0.3407        0.621
 15.867     23       1   0.4400  0.0702       0.3218        0.602
 20.316     22       1   0.4200  0.0698       0.3032        0.582
 20.978     21       1   0.4000  0.0693       0.2849        0.562
 21.332     20       1   0.3800  0.0686       0.2667        0.541
 21.591     19       1   0.3600  0.0679       0.2488        0.521
 21.986     18       1   0.3400  0.0670       0.2311        0.500
 25.038     17       1   0.3200  0.0660       0.2136        0.479
 27.135     16       1   0.3000  0.0648       0.1964        0.458
 28.011     15       1   0.2800  0.0635       0.1795        0.437
 28.269     14       1   0.2600  0.0620       0.1629        0.415
 29.354     13       1   0.2400  0.0604       0.1466        0.393
 31.881     12       1   0.2200  0.0586       0.1305        0.371
 33.274     11       1   0.2000  0.0566       0.1149        0.348
 34.181     10       1   0.1800  0.0543       0.0996        0.325
 37.319      9       1   0.1600  0.0518       0.0848        0.302
 42.870      7       1   0.1371  0.0492       0.0679        0.277
 52.549      3       1   0.0914  0.0497       0.0315        0.265

                treatment=Low 
   time n.risk n.event survival std.err lower 95% CI upper 95% CI
  0.336     50       1    0.980  0.0198       0.9420        1.000
  0.962     49       1    0.960  0.0277       0.9072        1.000
  1.650     48       1    0.940  0.0336       0.8764        1.000
  2.796     47       1    0.920  0.0384       0.8478        0.998
  3.411     46       1    0.900  0.0424       0.8206        0.987
  3.787     45       1    0.880  0.0460       0.7944        0.975
  4.355     44       1    0.860  0.0491       0.7690        0.962
  4.363     43       1    0.840  0.0518       0.7443        0.948
  5.133     42       1    0.820  0.0543       0.7201        0.934
  5.437     41       1    0.800  0.0566       0.6965        0.919
  5.899     40       1    0.780  0.0586       0.6732        0.904
  6.838     39       1    0.760  0.0604       0.6504        0.888
  7.224     38       1    0.740  0.0620       0.6279        0.872
  7.538     37       1    0.720  0.0635       0.6057        0.856
  7.675     36       1    0.700  0.0648       0.5838        0.839
  8.909     35       1    0.680  0.0660       0.5623        0.822
  9.707     34       1    0.660  0.0670       0.5409        0.805
  9.890     33       1    0.640  0.0679       0.5199        0.788
 10.080     32       1    0.620  0.0686       0.4991        0.770
 13.278     31       1    0.600  0.0693       0.4785        0.752
 14.667     30       1    0.580  0.0698       0.4581        0.734
 15.481     29       1    0.560  0.0702       0.4380        0.716
 16.421     28       1    0.540  0.0705       0.4181        0.697
 21.299     27       1    0.520  0.0707       0.3984        0.679
 22.051     26       1    0.500  0.0707       0.3790        0.660
 23.496     25       1    0.480  0.0707       0.3597        0.641
 23.714     24       1    0.460  0.0705       0.3407        0.621
 28.097     23       1    0.440  0.0702       0.3218        0.602
 29.766     22       1    0.420  0.0698       0.3032        0.582
 30.795     21       1    0.400  0.0693       0.2849        0.562
 32.531     20       1    0.380  0.0686       0.2667        0.541
 34.979     19       1    0.360  0.0679       0.2488        0.521
 36.550     18       1    0.340  0.0670       0.2311        0.500
 37.875     17       1    0.320  0.0660       0.2136        0.479
 39.991     16       1    0.300  0.0648       0.1964        0.458
 51.134     10       1    0.270  0.0649       0.1686        0.432
 52.005      8       1    0.236  0.0650       0.1378        0.405
 55.730      7       1    0.202  0.0639       0.1091        0.376
 63.940      6       1    0.169  0.0615       0.0826        0.345
 69.234      5       1    0.135  0.0577       0.0584        0.312
 77.609      1       1    0.000     NaN           NA           NA

                treatment=High 
   time n.risk n.event survival std.err lower 95% CI upper 95% CI
  0.198     50       1    0.980  0.0198        0.942        1.000
  0.954     49       1    0.960  0.0277        0.907        1.000
  2.322     48       1    0.940  0.0336        0.876        1.000
  3.432     47       1    0.920  0.0384        0.848        0.998
  4.231     46       1    0.900  0.0424        0.821        0.987
  5.337     45       1    0.880  0.0460        0.794        0.975
  6.714     44       1    0.860  0.0491        0.769        0.962
  7.608     43       1    0.840  0.0518        0.744        0.948
  7.978     42       1    0.820  0.0543        0.720        0.934
  8.346     41       1    0.800  0.0566        0.696        0.919
  8.408     40       1    0.780  0.0586        0.673        0.904
 10.973     39       1    0.760  0.0604        0.650        0.888
 14.870     38       1    0.740  0.0620        0.628        0.872
 15.989     37       1    0.720  0.0635        0.606        0.856
 16.848     36       1    0.700  0.0648        0.584        0.839
 18.024     35       1    0.680  0.0660        0.562        0.822
 18.038     34       1    0.660  0.0670        0.541        0.805
 22.455     33       1    0.640  0.0679        0.520        0.788
 26.415     32       1    0.620  0.0686        0.499        0.770
 35.083     31       1    0.600  0.0693        0.478        0.752
 39.551     30       1    0.580  0.0698        0.458        0.734
 40.646     29       1    0.560  0.0702        0.438        0.716
 41.403     26       1    0.538  0.0707        0.416        0.697
 48.025     21       1    0.513  0.0719        0.390        0.675
 56.342     16       1    0.481  0.0742        0.355        0.651
 59.986     15       1    0.449  0.0758        0.322        0.625
 62.376     11       1    0.408  0.0792        0.279        0.597
 65.583      7       1    0.350  0.0867        0.215        0.568
 69.705      6       1    0.291  0.0897        0.159        0.533

Call:
survival::survdiff(formula = survival::Surv(days, event) ~ treatment, 
    data = d, rho = 0)

                   N Observed Expected (O-E)^2/E (O-E)^2/V
treatment=Control 50       44     28.8     8.000     11.16
treatment=Low     50       41     36.7     0.515      0.76
treatment=High    50       29     48.5     7.857     14.15

 Chisq= 17.2  on 2 degrees of freedom, p= 2e-04

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/log_rank/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(2111)
d <- data.frame(plant = 1:150,
                treatment = rep(c("Control", "Low", "High"), each = 50))
event_time <- rexp(150, rate = rep(c(.045, .03, .018), each = 50))
follow_up <- runif(150, 40, 80)
d$days <- pmin(event_time, follow_up)
d$event <- as.integer(event_time <= follow_up)
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/log_rank/analysis.R modules/log_rank/generated

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Dataset 49 · Survival

Stratified log-rank test

CSV rows: 120Seed: 240501Module: stratified_logrank

Saved CSV Full runnable R script Full-size plot

Larval survival under stress is compared within 12 rearing blocks.

Question and null

Scientific question

Do matched groups differ in survival after accounting for the matching strata?

Null being tested / estimation target

Within the specified strata, groups have the same survival function.

Generating truth

Stress has a true hazard ratio of 1.6 within each block; the within-block equality null is false.

Exact generating model

120 independent exponential event times conditional on 12 fixed block rates evenly spaced from 0.025 to 0.08 per day. Stress multiplies every block rate by 1.6. Follow-up stops at 30 days.

Design and independent unit: Independent larvae within known rearing blocks; administrative censoring at day 30.

Columns in the saved CSV

Column	Meaning	Unit / type
`stratum`	Rearing block	ID
`treatment`	Assigned condition	category
`time`	Observed event or censoring time	days
`event`	Event observed by day 30	TRUE/FALSE

Preview the actual saved observations

stratum	treatment	time	event
1	Control	30	FALSE
1	Control	27.4622309207916	TRUE
1	Control	30	FALSE
1	Control	3.6188505589962	TRUE
1	Control	30	FALSE

Analysis from the saved dataset

Stratified log-rank test plot. Larval survival under stress is compared within 12 rearing blocks. — Simulated observations; generated by this module’s R script.

Stratified chi-square = 13.433
p = 0.0002472
12 rearing blocks; 120 larvae

After comparison within rearing blocks, chi-square = 13.43 (1 df), p = 0.000247. The test compares survival; estimate the direction and magnitude with survival curves or a suitable model.

Teaching pointStratification permits different baseline survival across known blocks.

Analysis code excerpt

d <- read.csv("modules/stratified_logrank/generated/data.csv")
result <- survival::survdiff(survival::Surv(time, event) ~ treatment + strata(stratum), data=d)
p_value <- pchisq(result$chisq, df=1, lower.tail=FALSE)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output

Call:
survival::survdiff(formula = survival::Surv(time, event) ~ treatment + 
    strata(stratum), data = d)

                   N Observed Expected (O-E)^2/E (O-E)^2/V
treatment=Control 60       42     58.5      4.64      13.4
treatment=Stress  60       54     37.5      7.23      13.4

 Chisq= 13.4  on 1 degrees of freedom, p= 2e-04

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

out_dir <- "modules/stratified_logrank/generated"
dir.create(out_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(240501)
stratum <- rep(1:12, each=10)
treatment <- rep(rep(c("Control", "Stress"), each=5), 12)
rate <- rep(seq(.025, .08, length.out=12), each=10)
event_time <- rexp(120, rate * ifelse(treatment=="Stress", 1.6, 1))
d <- data.frame(stratum, treatment, time=pmin(event_time, 30), event=event_time<=30)
write.csv(d, file.path(out_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/stratified_logrank/analysis.R modules/stratified_logrank/generated

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Dataset 50 · Time to event

Cox proportional-hazards regression

CSV rows: 180Seed: 2112Module: cox_ph

Saved CSV Full runnable R script Full-size plot

Seedling death times are related to protective dose and temperature.

Question and null

Scientific question

Does dose alter death hazard after accounting for temperature?

Null being tested / estimation target

The dose coefficient is zero: the adjusted hazard ratio per dose unit is one.

Generating truth

The adjusted hazard ratio per dose unit is exp(−0.45) ≈ 0.638; per °C it is exp(0.12) ≈ 1.127. The dose null is false and proportional hazards holds in the generating model.

Exact generating model

180 seedlings have dose Uniform(0, 3) and temperature Normal(25, 2²)°C. Conditional event times are exponential with rate exp(log(0.035) − 0.45 × dose + 0.12 × (temperature − 25)). Independent censoring times are Uniform(45, 100) days. Save the minimum time and event indicator without rounding.

Design and independent unit: 180 independent seedlings with right-censoring.

Columns in the saved CSV

Column	Meaning	Unit / type
`seedling`	Independent seedling identifier	ID
`dose`	Protective dose	dose unit
`temperature`	Rearing temperature	°C
`days`	Observed death or censoring time	day
`event`	1 death; 0 right-censored	binary

Preview the actual saved observations

seedling	dose	temperature	days	event
1	2.46703382371925	23.6347555251591	86.6266853827983	0
2	1.43185360520147	27.5634497970171	33.0444991868693	1
3	1.13769101304933	24.9126378926086	62.699396465905	0
4	0.799748866818845	22.4205103275966	15.3950857075481	1
5	0.796565959230065	26.1267323197537	5.6205234633493	1

Analysis from the saved dataset

Cox proportional-hazards regression plot. Seedling death times are related to protective dose and temperature. — Simulated observations; generated by this module’s R script.

Dose hazard ratio per unit = 0.69
95% CI 0.56 to 0.84
Wald z = -3.72; p = 0.0001969
PH diagnostic global p = 0.01934

Each dose unit multiplied the death hazard by 0.69 at the same temperature (95% CI 0.56 to 0.84; Wald p = 0.0001969). This is an instantaneous-risk comparison among seedlings still alive, not a survival probability ratio. The fitted constant hazard ratio must be checked over time.

Teaching pointA hazard ratio compares instantaneous event risk among individuals still at risk.

Analysis code excerpt

d <- read.csv("modules/cox_ph/generated/data.csv")
fit <- survival::coxph(survival::Surv(days, event) ~ dose + temperature,
                       data = d, x = TRUE)
summary(fit)
ph_check <- survival::cox.zph(fit)
hazard_ci <- exp(confint(fit)["dose", ])

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output

Call:
survival::coxph(formula = survival::Surv(days, event) ~ dose + 
    temperature, data = d, x = TRUE)

  n= 180, number of events= 126 

                coef exp(coef) se(coef)      z Pr(>|z|)    
dose        -0.37705   0.68588  0.10128 -3.723 0.000197 ***
temperature  0.16013   1.17367  0.04852  3.300 0.000966 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

            exp(coef) exp(-coef) lower .95 upper .95
dose           0.6859      1.458    0.5624    0.8365
temperature    1.1737      0.852    1.0672    1.2908

Concordance= 0.628  (se = 0.027 )
Likelihood ratio test= 21.6  on 2 df,   p=2e-05
Wald test            = 21.43  on 2 df,   p=2e-05
Score (logrank) test = 21.27  on 2 df,   p=2e-05

            chisq df      p
dose        7.726  1 0.0054
temperature 0.942  1 0.3317
GLOBAL      7.891  2 0.0193

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/cox_ph/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(2112)
d <- data.frame(seedling = 1:180, dose = runif(180, 0, 3),
                temperature = rnorm(180, 25, 2))
rate <- exp(log(.035) - .45 * d$dose + .12 * (d$temperature - 25))
event_time <- rexp(180, rate = rate)
follow_up <- runif(180, 45, 100)
d$days <- pmin(event_time, follow_up)
d$event <- as.integer(event_time <= follow_up)
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/cox_ph/analysis.R modules/cox_ph/generated

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Dataset 51 · Time to event

Parametric survival: accelerated failure time

CSV rows: 140Seed: 2113Module: parametric_survival

Saved CSV Full runnable R script Full-size plot

Botanists compare germination timing after control or cold treatment.

Question and null

Scientific question

Does cold treatment stretch the germination-time distribution?

Null being tested / estimation target

The treatment coefficient is zero: the cold/control time ratio is one.

Generating truth

The population cold/control time ratio is exp(0.4) ≈ 1.492, so the equal-time null is false. The generating Weibull distribution matches the fitted AFT family.

Exact generating model

140 independent seeds (70 per group) have Weibull germination times with shape 1.8 and scale exp(log(20) + 0.4 if Cold). Thus control scale is 20 days and cold scale is about 29.84 days. Independent censoring times are Uniform(22, 50) days. Save minimum time and event indicator without rounding.

Design and independent unit: 140 independent seeds; some remain ungerminated at final follow-up.

Columns in the saved CSV

Column	Meaning	Unit / type
`seed`	Independent seed identifier	ID
`treatment`	Control or cold pretreatment	category
`days`	Observed germination or censoring time	day
`event`	1 germinated; 0 right-censored	binary

Preview the actual saved observations

seed	treatment	days	event
1	Control	11.058161895054	1
2	Control	26.1480823980189	1
3	Control	5.19415270853795	1
4	Control	22.3929165285081	0
5	Control	8.92397164253965	1

Analysis from the saved dataset

Parametric survival: accelerated failure time plot. Botanists compare germination timing after control or cold treatment. — Simulated observations; generated by this module’s R script.

Cold/control time ratio = 1.70
95% CI 1.35 to 2.13
Wald z = 4.54; p = 5.745e-06
Fitted Weibull shape = 1.64

Under the Weibull accelerated-failure-time model, cold-treated seeds had 1.70 times the germination time of controls (95% CI 1.35 to 2.13; p = 5.745e-06). This is a time ratio, not a Cox hazard ratio. The distributional fit and censoring assumptions remain part of the interpretation.

Teaching pointAn AFT coefficient becomes a time ratio; it is not interpreted as a Cox hazard ratio.

Analysis code excerpt

d <- read.csv("modules/parametric_survival/generated/data.csv")
fit <- survival::survreg(survival::Surv(days, event) ~ treatment,
                         data = d, dist = "weibull")
summary(fit)
time_ci <- exp(confint(fit)["treatmentCold", ])

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


Call:
survival::survreg(formula = survival::Surv(days, event) ~ treatment, 
    data = d, dist = "weibull")
                Value Std. Error     z       p
(Intercept)    2.8854     0.0773 37.32 < 2e-16
treatmentCold  0.5288     0.1166  4.54 5.7e-06
Log(scale)    -0.4935     0.0759 -6.50 8.1e-11

Scale= 0.61 

Weibull distribution
Loglik(model)= -454.6   Loglik(intercept only)= -465
	Chisq= 20.72 on 1 degrees of freedom, p= 5.3e-06 
Number of Newton-Raphson Iterations: 5 
n= 140

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/parametric_survival/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(2113)
d <- data.frame(seed = 1:140,
                treatment = rep(c("Control", "Cold"), each = 70))
log_scale <- log(20) + .4 * (d$treatment == "Cold")
event_time <- rweibull(140, shape = 1.8, scale = exp(log_scale))
follow_up <- runif(140, 22, 50)
d$days <- pmin(event_time, follow_up)
d$event <- as.integer(event_time <= follow_up)
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/parametric_survival/analysis.R modules/parametric_survival/generated

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Dataset 52 · Survival

Shared-frailty survival model

CSV rows: 240Seed: 240502Module: shared_frailty

Saved CSV Full runnable R script Full-size plot

Larvae from 24 populations experience a control or stress condition.

Question and null

Scientific question

Does stress change larval death hazard after accounting for shared population conditions?

Null being tested / estimation target

The stress coefficient is zero: the hazard ratio conditional on population frailty is one.

Generating truth

The conditional stress hazard ratio is exp(0.5) ≈ 1.65; shared population variance is 0.5.

Exact generating model

24 population frailties follow Gamma(shape=2, rate=2), mean 1. Conditional event times are exponential with rate 0.04 × frailty × exp(0.5 × stress). Ten larvae per population; censoring at 35 days.

Design and independent unit: Larvae share an unobserved population risk; time-to-death is censored at day 35.

Columns in the saved CSV

Column	Meaning	Unit / type
`population`	Shared risk group	ID
`stress`	Control=0, stress=1	indicator
`time`	Observed follow-up	days
`event`	Death observed	TRUE/FALSE

Preview the actual saved observations

population	time	event
1	35	FALSE
1	35	FALSE
1	35	FALSE
1	33.9286072057958	TRUE
1	21.937562494658	TRUE

Analysis from the saved dataset

Shared-frailty survival model plot. Larvae from 24 populations experience a control or stress condition. — Simulated observations; generated by this module’s R script.

Conditional hazard ratio = 1.61
Wald 95% CI 1.18 to 2.18
Wald p = 0.002373

Stress has an estimated hazard ratio of 1.61 conditional on population frailty (95% Wald CI 1.18 to 2.18; p = 0.00237). Population frailty represents shared differences in baseline risk.

Teaching pointFrailty accounts for shared latent risk; it is different from giving each stratum its own baseline curve.

Analysis code excerpt

d <- read.csv("modules/shared_frailty/generated/data.csv")
result <- survival::coxph(survival::Surv(time, event) ~ stress + frailty(population), data=d)
b <- coef(result)["stress"]
se <- sqrt(vcov(result)["stress", "stress"])
hr <- exp(b); ci <- exp(b+c(-1,1)*1.96*se)
p_value <- 2*pnorm(-abs(b/se))

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output

Call:
survival::coxph(formula = survival::Surv(time, event) ~ stress + 
    frailty(population), data = d)

                      coef se(coef)    se2  Chisq   DF       p
stress               0.475    0.156  0.155  9.236  1.0  0.0024
frailty(population)                        55.288 17.2 6.8e-06

Iterations: 8 outer, 31 Newton-Raphson
     Variance of random effect= 0.337   I-likelihood = -848 
Degrees of freedom for terms=  1.0 17.2 
Likelihood ratio test=82.8  on 18.1 df, p=3e-10
n= 240, number of events= 174

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

out_dir <- "modules/shared_frailty/generated"
dir.create(out_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(240502)
population <- rep(1:24, each=10)
stress <- rep(rep(0:1, each=5), 24)
frailty <- rgamma(24, shape=2, rate=2)
event_time <- rexp(240, .04 * frailty[population] * exp(.5*stress))
d <- data.frame(population, stress, time=pmin(event_time, 35), event=event_time<=35)
write.csv(d, file.path(out_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/shared_frailty/analysis.R modules/shared_frailty/generated

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Dataset 53 · Simulation and multivariate methods

Independent-sample permutation test

CSV rows: 50Seed: 240503Module: permutation_independent

Saved CSV Full runnable R script Full-size plot

Independent insect larvae are randomly assigned to ambient or warm rearing.

Question and null

Scientific question

Would the observed group statistic be unusual under assignments permitted by the design?

Null being tested / estimation target

For ordinary group-label permutation, group labels are exchangeable under the null; in a randomized experiment, the sharp null is no treatment effect on any unit.

Generating truth

The simulated warm-minus-control mean difference is 1.1 mm; the same-distribution null is false.

Exact generating model

25 independent values per group. Control growth is Normal(8, SD=1.8) mm; warm growth is Normal(9.1, SD=1.8) mm. No rounding or clipping.

Design and independent unit: Two independent randomized groups; statistic is the mean difference.

Columns in the saved CSV

Column	Meaning	Unit / type
`group`	Randomized treatment	category
`growth_mm`	Larval growth	mm

Preview the actual saved observations

group	growth_mm
Control	7.97504548497602
Control	4.89425790226894
Control	9.02612575667418
Control	8.05558482342149
Control	5.14803845685443

Analysis from the saved dataset

Independent-sample permutation test plot. Independent insect larvae are randomly assigned to ambient or warm rearing. — Simulated observations; generated by this module’s R script.

Warm − control = 0.72 mm
Two-sided Monte Carlo p = 0.187
4999 permitted label shuffles

Mean growth was 0.72 mm higher under warming; the two-sided permutation p-value is 0.187 from 4999 random assignments. The shuffle must match the design.

Teaching pointChoose the statistic for the scientific question, then shuffle in a way the design permits.

Analysis code excerpt

d <- read.csv("modules/permutation_independent/generated/data.csv")
observed <- mean(d$growth_mm[d$group=="Warm"])-mean(d$growth_mm[d$group=="Control"])
B <- 4999
null <- numeric(B)
for (i in 1:B) {
  shuffled <- sample(d$group)
  null[i] <- mean(d$growth_mm[shuffled=="Warm"])-mean(d$growth_mm[shuffled=="Control"])
}
p_value <- (1+sum(abs(null)>=abs(observed)))/(B+1)
result <- data.frame(mean_difference=observed, permutations=B, p=p_value)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output

  mean_difference permutations     p
1       0.7207885         4999 0.187

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

out_dir <- "modules/permutation_independent/generated"
dir.create(out_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(240503)
group <- rep(c("Control", "Warm"), each=25)
d <- data.frame(group, growth_mm=rnorm(50, 8+1.1*(group=="Warm"), 1.8))
write.csv(d, file.path(out_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/permutation_independent/analysis.R modules/permutation_independent/generated

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Dataset 54 · Simulation and multivariate methods

Paired permutation test

CSV rows: 28Seed: 240504Module: permutation_paired

Saved CSV Full runnable R script Full-size plot

Two comparable leaves per plant receive randomized control or shade treatment.

Question and null

Scientific question

Is the within-pair or within-block contrast unusual under the actual randomization?

Null being tested / estimation target

Under a paired randomized design, treatment assignment has no effect on any unit; permitted within-pair swaps are equally plausible.

Generating truth

The true mean shaded-minus-control difference is −1.2; plants also differ in their shared baseline.

Exact generating model

Plant baselines are Normal(15, SD=2). Independent leaf errors are Normal(0, SD=1). Shading subtracts 1.2 photosynthesis units.

Design and independent unit: Matched leaves within 28 independent plants; treatment labels are randomized within each plant.

Columns in the saved CSV

Column	Meaning	Unit / type
`plant`	Independent matched unit	ID
`control`	Control-leaf rate	photosynthesis units
`shaded`	Shaded-leaf rate	photosynthesis units

Preview the actual saved observations

plant	control	shaded
1	15.2814577712788	12.6839981994452
2	12.9668676942456	11.611469977333
3	19.0552980882258	19.7852502878579
4	14.1404402330399	12.4842119069942
5	12.8207158930959	12.7003378907753

Analysis from the saved dataset

Paired permutation test plot. Two comparable leaves per plant receive randomized control or shade treatment. — Simulated observations; generated by this module’s R script.

Shaded − control = -0.97 units
Two-sided permutation p = 0.0006
28 pairs; labels swap only within plants

Within plants, shading changed mean photosynthesis by -0.97 units (paired permutation p = 0.0006). Swaps preserve plant pairing.

Teaching pointSwapping labels within a pair is equivalent to reversing the sign of that pair’s difference.

Analysis code excerpt

d <- read.csv("modules/permutation_paired/generated/data.csv")
difference <- d$shaded-d$control
observed <- mean(difference)
B <- 4999
null <- numeric(B)
for (i in 1:B) null[i] <- mean(difference * sample(c(-1,1), nrow(d), replace=TRUE))
p_value <- (1+sum(abs(null)>=abs(observed)))/(B+1)
result <- data.frame(mean_difference=observed, permutations=B, p=p_value)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output

  mean_difference permutations     p
1      -0.9724543         4999 6e-04

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

out_dir <- "modules/permutation_paired/generated"
dir.create(out_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(240504)
baseline <- rnorm(28, 15, 2)
d <- data.frame(plant=1:28, control=baseline+rnorm(28,0,1),
                shaded=baseline-1.2+rnorm(28,0,1))
write.csv(d, file.path(out_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/permutation_paired/analysis.R modules/permutation_paired/generated

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Dataset 55 · Simulation and multivariate methods

Monte Carlo null-model test

CSV rows: 80Seed: 240505Module: monte_carlo

Saved CSV Full runnable R script Full-size plot

Test whether candidate sex-determination genes are unusually concentrated on the X chromosome.

Question and null

Scientific question

Is chromosome occupancy inconsistent with the specified opportunity model?

Null being tested / estimation target

Each independent candidate gene has probability 0.20 of being on X.

Generating truth

The true occupancy probability is 0.32, so the tested 0.20 null is false.

Exact generating model

80 independent Bernoulli indicators with true X probability 0.32. The tested null probability is 0.20; null simulation uses 9999 binomial draws.

Design and independent unit: A teaching null assigns independent genes to X with probability equal to its opportunity share.

Columns in the saved CSV

Column	Meaning	Unit / type
`gene`	Candidate gene	ID
`on_X`	X chromosome=1, other=0	indicator

Preview the actual saved observations

gene	on_X
1	0
2	0
3	1
4	0
5	1

Analysis from the saved dataset

Monte Carlo null-model test plot. Test whether candidate sex-determination genes are unusually concentrated on the X chromosome. — Simulated observations; generated by this module’s R script.

Observed on X: 32 of 80
Expected under null: 16.0
Two-sided Monte Carlo p = 0.0002

32 of 80 genes were on X, compared with null expectation 16.0. The absolute count-deviation Monte Carlo test gives p = 0.0002. Here the null assigns each independent gene probability 0.20 of being on X.

Teaching pointA simulation test is only as useful as its biological null model.

Analysis code excerpt

d <- read.csv("modules/monte_carlo/generated/data.csv")
observed <- sum(d$on_X)
p0 <- .20; n <- nrow(d); B <- 9999
null <- rbinom(B, size=n, prob=p0)
p_value <- (1+sum(abs(null-n*p0)>=abs(observed-n*p0)))/(B+1)
result <- data.frame(observed_X=observed, expected_X=n*p0, monte_carlo_p=p_value)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output

  observed_X expected_X monte_carlo_p
1         32         16         2e-04

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

out_dir <- "modules/monte_carlo/generated"
dir.create(out_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(240505)
d <- data.frame(gene=1:80, on_X=rbinom(80, 1, .32))
write.csv(d, file.path(out_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/monte_carlo/analysis.R modules/monte_carlo/generated

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Dataset 56 · Simulation and multivariate methods

Bootstrap confidence interval

CSV rows: 40Seed: 240507Module: bootstrap

Saved CSV Full runnable R script Full-size plot

Estimate average biomass from independent bacterial colonies with a right-skewed distribution.

Question and null

Scientific question

How uncertain is the estimated mean biomass?

Null being tested / estimation target

No null is required for this confidence interval; the target is the population mean.

Generating truth

The population arithmetic mean is approximately 13.28 mg; the interval may or may not cover it in one sample.

Exact generating model

40 independent Lognormal(meanlog=log(12), sdlog=0.45) colony biomasses. The generating arithmetic mean is exp(log(12)+0.45^2/2), approximately 13.28 mg.

Design and independent unit: One biomass measurement per independent colony.

Columns in the saved CSV

Column	Meaning	Unit / type
`colony`	Independent colony	ID
`biomass_mg`	Measured biomass	mg

Preview the actual saved observations

colony	biomass_mg
1	5.62201495443718
2	12.4819548811077
3	11.1032813991177
4	10.5493765124805
5	6.30828473813244

Analysis from the saved dataset

Bootstrap confidence interval plot. Estimate average biomass from independent bacterial colonies with a right-skewed distribution. — Simulated observations; generated by this module’s R script.

Sample mean = 11.54 mg
Percentile 95% CI = 10.15 to 13.07 mg
4999 resamples of independent colonies

Mean colony biomass is 11.54 mg, with a percentile bootstrap 95% interval of 10.15 to 13.07 mg. The interval estimates uncertainty in the mean, not the spread of individual colonies.

Teaching pointResampling changes the computational procedure, not what counts as an independent replicate.

Analysis code excerpt

d <- read.csv("modules/bootstrap/generated/data.csv")
B <- 4999
boot_mean <- numeric(B)
for (i in 1:B) boot_mean[i] <- mean(sample(d$biomass_mg, replace=TRUE))
ci <- quantile(boot_mean, c(.025,.975))
result <- data.frame(mean=mean(d$biomass_mg), lower=ci[1], upper=ci[2])

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output

         mean    lower    upper
2.5% 11.53829 10.14616 13.06527

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

out_dir <- "modules/bootstrap/generated"
dir.create(out_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(240507)
d <- data.frame(colony=1:40, biomass_mg=rlnorm(40, log(12), .45))
write.csv(d, file.path(out_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/bootstrap/analysis.R modules/bootstrap/generated

↑ Back to the quick index

Dataset 57 · Simulation and multivariate methods

Sampling variation and CI coverage

CSV rows: 25000Seed: 240506Module: sampling_coverage

Saved CSV Full runnable R script Full-size plot

Independent cell cultures differ in the waiting time to a developmental transition.

Question and null

Scientific question

How often does a stated interval procedure cover the true population mean?

Null being tested / estimation target

This is a performance simulation, not a single hypothesis test. The known population mean is 5 days.

Generating truth

The population mean is exactly 5 days; nominal 95% coverage is a procedure property to evaluate, not an imposed simulation result.

Exact generating model

1000 samples of 25 independent Exponential(rate=0.2/day) observations. Every sample uses a conventional two-sided 95% t interval for its mean.

Design and independent unit: Repeat a 25-cell sampling experiment 1000 times from a known population.

Columns in the saved CSV

Column	Meaning	Unit / type
`experiment`	Repeated simulated sample	ID
`waiting_days`	Time to transition	days

Preview the actual saved observations

experiment	waiting_days
1	8.52699033772569
1	3.75008905997234
1	6.87738209010518
1	8.81315140942943
1	8.54428097563696

Analysis from the saved dataset

Sampling variation and CI coverage plot. Independent cell cultures differ in the waiting time to a developmental transition. — Simulated observations; generated by this module’s R script.

Mean estimated waiting time: 5.03 days
Observed 95% t-interval coverage: 92.6%
True population mean: 5 days

Across 1000 experiments of 25 cells, 92.6% of nominal 95% t intervals contain the true mean of 5 days. Coverage is evaluated against that truth, not each interval’s own sample mean.

Teaching pointLarger B stabilizes the estimated coverage; larger n changes the experiment.

Analysis code excerpt

d <- read.csv("modules/sampling_coverage/generated/data.csv")
low <- high <- estimate <- numeric(1000)
for (i in 1:1000) {
  x <- d$waiting_days[d$experiment==i]
  estimate[i] <- mean(x)
  margin <- qt(.975, length(x)-1)*sd(x)/sqrt(length(x))
  low[i] <- mean(x)-margin; high[i] <- mean(x)+margin
}
covered <- low<=5 & high>=5
result <- data.frame(experiment=1:1000, estimate, low, high, covered)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output

     experiment estimate      low      high covered
1             1 5.522029 4.072284  6.971773    TRUE
2             2 5.204283 2.985148  7.423418    TRUE
3             3 4.368782 2.517560  6.220003    TRUE
4             4 5.642785 3.234415  8.051154    TRUE
5             5 4.238518 2.021749  6.455288    TRUE
6             6 5.643793 2.979302  8.308284    TRUE
7             7 5.469335 3.573552  7.365119    TRUE
8             8 5.008316 3.136312  6.880320    TRUE
9             9 7.688537 4.535211 10.841862    TRUE
10           10 4.479982 2.489549  6.470416    TRUE
11           11 5.263651 1.711425  8.815877    TRUE
12           12 3.318445 1.975755  4.661135   FALSE
13           13 5.013864 2.746185  7.281543    TRUE
14           14 5.850578 3.541759  8.159396    TRUE
15           15 4.512140 2.771186  6.253094    TRUE
16           16 5.354784 3.188556  7.521012    TRUE
17           17 6.757512 3.640167  9.874857    TRUE
18           18 4.132858 2.769656  5.496059    TRUE
19           19 5.497705 3.455908  7.539502    TRUE
20           20 6.413697 4.247548  8.579845    TRUE
21           21 4.688334 3.063164  6.313503    TRUE
22           22 5.239921 3.067297  7.412546    TRUE
23           23 6.752626 3.853531  9.651721    TRUE
24           24 4.809608 3.067157  6.552059    TRUE
25           25 5.655483 4.247544  7.063422    TRUE
26           26 5.876048 3.895268  7.856828    TRUE
27           27 4.618555 2.384767  6.852343    TRUE
28           28 3.658729 2.417663  4.899794   FALSE
29           29 4.338001 2.857263  5.818739    TRUE
30           30 5.506953 3.072848  7.941058    TRUE
31           31 5.687501 3.054839  8.320163    TRUE
32           32 5.066542 3.191820  6.941265    TRUE
33           33 4.692163 2.311953  7.072373    TRUE
34           34 4.805327 3.165974  6.444680    TRUE
35           35 5.233745 3.158254  7.309237    TRUE
36           36 4.789032 3.003878  6.574186    TRUE
37           37 4.012812 1.996415  6.029208    TRUE
38           38 4.326758 2.382734  6.270782    TRUE
39           39 3.964481 2.574601  5.354360    TRUE
40           40 6.382165 4.232534  8.531796    TRUE
41           41 3.434084 1.829988  5.038180    TRUE
42           42 5.814381 3.787407  7.841355    TRUE
43           43 6.348211 4.113734  8.582688    TRUE
44           44 6.842268 4.091238  9.593298    TRUE
45           45 5.445202 2.970798  7.919606    TRUE
46           46 4.089058 2.582495  5.595621    TRUE
47           47 6.638345 4.348471  8.928219    TRUE
48           48 4.803594 2.537863  7.069326    TRUE
49           49 4.649075 2.458765  6.839386    TRUE
50           50 6.536089 4.425454  8.646724    TRUE
51           51 4.634365 2.627069  6.641661    TRUE
52           52 4.411398 3.077791  5.745006    TRUE
53           53 3.947442 2.577601  5.317284    TRUE
54           54 5.809957 3.245246  8.374669    TRUE
55           55 4.911161 2.853966  6.968356    TRUE
56           56 2.839206 1.888804  3.789608   FALSE
57           57 4.344700 3.039214  5.650185    TRUE
58           58 4.519626 2.867844  6.171408    TRUE
59           59 5.930605 2.995504  8.865706    TRUE
60           60 4.083792 2.659315  5.508269    TRUE
61           61 4.400565 2.899233  5.901897    TRUE
62           62 4.608569 3.206468  6.010671    TRUE
63           63 6.122087 3.529017  8.715158    TRUE
64           64 4.017382 2.632192  5.402571    TRUE
65           65 3.818379 2.364116  5.272641    TRUE
66           66 3.765858 2.317919  5.213796    TRUE
67           67 4.754093 2.844543  6.663642    TRUE
68           68 6.341148 3.941493  8.740803    TRUE
69           69 4.799054 2.697553  6.900555    TRUE
70           70 3.796780 2.443970  5.149590    TRUE
71           71 4.136329 2.731771  5.540887    TRUE
72           72 4.030844 2.327800  5.733889    TRUE
73           73 6.796037 4.119067  9.473006    TRUE
74           74 5.556472 2.973111  8.139833    TRUE
75           75 6.207926 3.910795  8.505058    TRUE
76           76 4.107061 2.152500  6.061622    TRUE
77           77 4.753323 3.124626  6.382020    TRUE
78           78 2.794476 1.369826  4.219126   FALSE
79           79 4.566371 2.837708  6.295035    TRUE
80           80 5.894649 3.659524  8.129775    TRUE
81           81 4.690335 2.896712  6.483958    TRUE
82           82 4.491655 2.640704  6.342605    TRUE
83           83 5.663687 3.229420  8.097953    TRUE
84           84 4.115975 2.707549  5.524400    TRUE
85           85 5.732207 3.571358  7.893057    TRUE
86           86 4.876916 2.965229  6.788603    TRUE
87           87 5.618509 4.067509  7.169509    TRUE
88           88 4.811652 3.344971  6.278334    TRUE
89           89 3.998192 2.273287  5.723097    TRUE
90           90 4.719024 2.958358  6.479690    TRUE
91           91 3.695699 2.389634  5.001764    TRUE
92           92 5.504377 3.267597  7.741157    TRUE
93           93 4.458405 2.330069  6.586741    TRUE
94           94 3.596332 2.582829  4.609835   FALSE
95           95 3.905252 2.464322  5.346181    TRUE
96           96 4.234980 2.561521  5.908438    TRUE
97           97 5.397633 3.101522  7.693744    TRUE
98           98 3.804190 2.235648  5.372732    TRUE
99           99 4.555470 2.447108  6.663832    TRUE
100         100 3.965643 1.957545  5.973742    TRUE
101         101 3.936298 2.425014  5.447583    TRUE
102         102 4.434617 2.475259  6.393976    TRUE
103         103 5.100318 3.290936  6.909700    TRUE
104         104 4.067811 2.542494  5.593129    TRUE
105         105 3.317712 1.908598  4.726826   FALSE
106         106 6.516611 4.021650  9.011571    TRUE
107         107 4.493778 3.003852  5.983703    TRUE
108         108 3.950111 2.552436  5.347785    TRUE
109         109 5.101214 3.126121  7.076308    TRUE
110         110 5.391561 3.582766  7.200356    TRUE
111         111 5.672264 2.436146  8.908382    TRUE
112         112 5.308363 3.405103  7.211624    TRUE
113         113 4.648292 3.070992  6.225592    TRUE
114         114 4.833650 2.986126  6.681173    TRUE
115         115 6.413949 3.788569  9.039328    TRUE
116         116 3.755544 2.547431  4.963657   FALSE
117         117 5.953093 4.095236  7.810951    TRUE
118         118 5.423245 2.814981  8.031510    TRUE
119         119 8.026679 5.101515 10.951842   FALSE
120         120 4.622307 2.434032  6.810582    TRUE
121         121 4.080753 2.700272  5.461233    TRUE
122         122 4.235137 2.645343  5.824930    TRUE
123         123 5.288931 2.278319  8.299543    TRUE
124         124 5.873955 3.016026  8.731884    TRUE
125         125 5.171276 3.380162  6.962391    TRUE
126         126 3.250377 1.489852  5.010902    TRUE
127         127 4.191097 2.668626  5.713568    TRUE
128         128 6.754089 3.601342  9.906836    TRUE
129         129 4.793879 3.097612  6.490145    TRUE
130         130 6.746923 3.533661  9.960186    TRUE
131         131 6.187196 3.894694  8.479698    TRUE
132         132 4.224100 2.524231  5.923970    TRUE
133         133 3.434999 2.221782  4.648216   FALSE
134         134 6.834802 4.139378  9.530226    TRUE
135         135 4.191357 2.319007  6.063707    TRUE
136         136 3.151103 2.060237  4.241968   FALSE
137         137 5.026459 3.169689  6.883229    TRUE
138         138 3.577657 2.528884  4.626430   FALSE
139         139 3.687805 1.994784  5.380827    TRUE
140         140 5.351641 3.422936  7.280347    TRUE
141         141 3.879084 2.304154  5.454013    TRUE
142         142 5.286025 3.516029  7.056022    TRUE
143         143 4.693587 3.300891  6.086283    TRUE
144         144 5.109116 2.401925  7.816308    TRUE
145         145 5.089690 3.333748  6.845632    TRUE
146         146 5.430569 3.870811  6.990327    TRUE
147         147 6.782960 4.339468  9.226452    TRUE
148         148 5.823611 3.331024  8.316199    TRUE
149         149 4.366575 2.390465  6.342684    TRUE
150         150 4.184145 2.110662  6.257627    TRUE
151         151 7.730201 4.354545 11.105856    TRUE
152         152 6.140021 3.348536  8.931505    TRUE
153         153 4.117216 2.409964  5.824468    TRUE
154         154 4.398973 1.430581  7.367365    TRUE
155         155 4.508656 2.892396  6.124916    TRUE
156         156 4.142960 2.664366  5.621555    TRUE
157         157 5.245062 3.098888  7.391237    TRUE
158         158 5.040628 3.008546  7.072711    TRUE
159         159 4.218798 2.480088  5.957507    TRUE
160         160 5.347144 3.433166  7.261123    TRUE
161         161 6.686552 3.977025  9.396078    TRUE
162         162 4.142875 3.147446  5.138305    TRUE
163         163 3.269228 1.170891  5.367566    TRUE
164         164 4.955636 2.996825  6.914446    TRUE
165         165 4.556690 2.991951  6.121429    TRUE
166         166 5.400123 2.910192  7.890054    TRUE
167         167 4.277913 2.393319  6.162508    TRUE
168         168 3.254105 2.029229  4.478981   FALSE
169         169 4.566055 3.025270  6.106840    TRUE
170         170 5.348849 3.080346  7.617352    TRUE
171         171 2.395981 1.556685  3.235278   FALSE
172         172 4.640689 2.748807  6.532571    TRUE
173         173 3.508499 2.009903  5.007095    TRUE
174         174 5.225143 2.841299  7.608988    TRUE
175         175 4.983477 2.784035  7.182919    TRUE
176         176 3.795989 1.917545  5.674432    TRUE
177         177 5.690686 3.581406  7.799967    TRUE
178         178 4.848927 2.914100  6.783754    TRUE
179         179 5.374085 2.260090  8.488080    TRUE
180         180 6.475772 3.811408  9.140136    TRUE
181         181 4.741831 2.402756  7.080906    TRUE
182         182 5.321358 3.240530  7.402186    TRUE
183         183 5.191675 1.921541  8.461809    TRUE
184         184 4.658280 2.766932  6.549628    TRUE
185         185 5.753889 4.188246  7.319532    TRUE
186         186 6.257938 3.073481  9.442395    TRUE
187         187 5.566719 3.929474  7.203964    TRUE
188         188 2.465539 1.547398  3.383681   FALSE
189         189 4.206448 2.561935  5.850961    TRUE
190         190 3.479337 1.933332  5.025342    TRUE
191         191 3.463501 2.085828  4.841174   FALSE
192         192 3.727588 2.093795  5.361380    TRUE
193         193 6.847350 3.846342  9.848358    TRUE
194         194 5.713059 3.598224  7.827895    TRUE
195         195 3.507900 2.070407  4.945393   FALSE
196         196 6.905772 3.535558 10.275986    TRUE
197         197 7.418474 4.791591 10.045357    TRUE
198         198 5.314492 3.311919  7.317065    TRUE
199         199 5.561324 3.223291  7.899358    TRUE
200         200 5.457266 3.586288  7.328244    TRUE
201         201 5.989511 3.791370  8.187652    TRUE
202         202 3.913985 2.508346  5.319624    TRUE
203         203 5.290439 3.427504  7.153373    TRUE
204         204 4.686989 2.752241  6.621737    TRUE
205         205 4.463568 2.682710  6.244426    TRUE
206         206 5.165704 3.407387  6.924021    TRUE
207         207 5.076360 2.702550  7.450170    TRUE
208         208 6.213807 2.760825  9.666789    TRUE
209         209 4.867417 3.309513  6.425321    TRUE
210         210 3.577422 1.434624  5.720220    TRUE
211         211 5.894050 3.514567  8.273532    TRUE
212         212 4.338063 2.560657  6.115469    TRUE
213         213 6.143999 3.323829  8.964169    TRUE
214         214 5.598588 3.768305  7.428870    TRUE
215         215 6.272771 3.983205  8.562336    TRUE
216         216 7.138216 4.579673  9.696759    TRUE
217         217 6.026648 4.078599  7.974697    TRUE
218         218 3.850279 2.341270  5.359289    TRUE
219         219 4.487893 2.978306  5.997479    TRUE
220         220 5.245262 3.003950  7.486574    TRUE
221         221 4.822648 2.948739  6.696558    TRUE
222         222 8.021933 4.637147 11.406719    TRUE
223         223 5.935316 3.847466  8.023167    TRUE
224         224 4.980738 2.542093  7.419382    TRUE
225         225 4.281605 2.961476  5.601734    TRUE
226         226 5.418768 2.691508  8.146029    TRUE
227         227 4.427298 2.528119  6.326477    TRUE
228         228 6.569612 4.002026  9.137198    TRUE
229         229 5.126956 3.152984  7.100928    TRUE
230         230 3.853325 2.401367  5.305283    TRUE
231         231 4.502664 2.540948  6.464380    TRUE
232         232 3.975073 2.580084  5.370062    TRUE
233         233 3.434649 2.342173  4.527125   FALSE
234         234 5.400147 2.760415  8.039879    TRUE
235         235 4.140376 2.134298  6.146454    TRUE
236         236 5.589214 4.041403  7.137025    TRUE
237         237 3.532313 2.393871  4.670754   FALSE
238         238 4.704814 2.780582  6.629047    TRUE
239         239 6.548639 3.403325  9.693952    TRUE
240         240 6.697136 4.277313  9.116959    TRUE
241         241 6.611054 3.033616 10.188492    TRUE
242         242 3.994481 2.210885  5.778078    TRUE
243         243 4.509383 3.090736  5.928030    TRUE
244         244 5.026530 3.017718  7.035342    TRUE
245         245 4.815224 3.105586  6.524861    TRUE
246         246 5.571338 2.475466  8.667209    TRUE
247         247 3.350048 2.078539  4.621557   FALSE
248         248 4.217988 2.648283  5.787692    TRUE
249         249 4.310275 2.014899  6.605652    TRUE
250         250 4.440112 2.616232  6.263992    TRUE
251         251 3.553821 2.381495  4.726146   FALSE
252         252 5.472761 3.405546  7.539975    TRUE
253         253 4.248571 2.281546  6.215597    TRUE
254         254 5.081565 3.002994  7.160135    TRUE
255         255 7.247705 3.855316 10.640094    TRUE
256         256 6.569047 4.377447  8.760647    TRUE
257         257 7.361497 5.158275  9.564720   FALSE
258         258 4.632382 2.402952  6.861812    TRUE
259         259 4.290517 2.594137  5.986896    TRUE
260         260 6.130597 3.595685  8.665509    TRUE
261         261 6.192151 4.400440  7.983862    TRUE
262         262 3.833823 2.241736  5.425911    TRUE
263         263 5.779415 3.498400  8.060429    TRUE
264         264 4.294562 2.340403  6.248721    TRUE
265         265 5.183574 1.892786  8.474362    TRUE
266         266 4.889187 2.752254  7.026119    TRUE
267         267 5.756816 2.918858  8.594774    TRUE
268         268 6.884198 4.169422  9.598975    TRUE
269         269 4.989537 2.320946  7.658129    TRUE
270         270 5.013138 2.522201  7.504074    TRUE
271         271 6.021809 3.395412  8.648207    TRUE
272         272 5.107057 2.865401  7.348712    TRUE
273         273 4.792391 3.209370  6.375412    TRUE
274         274 4.249709 2.221006  6.278412    TRUE
275         275 5.170267 3.380168  6.960365    TRUE
276         276 5.768764 3.080663  8.456865    TRUE
277         277 6.234953 3.083420  9.386486    TRUE
278         278 5.028424 2.781633  7.275215    TRUE
279         279 5.616455 3.487074  7.745837    TRUE
280         280 6.236102 3.960979  8.511224    TRUE
281         281 4.983969 2.177201  7.790738    TRUE
282         282 4.940238 2.891489  6.988987    TRUE
283         283 5.390487 3.274993  7.505981    TRUE
284         284 3.381895 2.098824  4.664966   FALSE
285         285 5.791524 2.618764  8.964283    TRUE
286         286 4.716518 3.311282  6.121754    TRUE
287         287 4.849553 2.268082  7.431024    TRUE
288         288 6.041211 3.583686  8.498735    TRUE
289         289 5.229720 3.264473  7.194968    TRUE
290         290 7.439641 4.404122 10.475160    TRUE
291         291 6.378421 3.828685  8.928157    TRUE
292         292 4.015060 2.681271  5.348850    TRUE
293         293 4.837685 2.984054  6.691316    TRUE
294         294 4.038949 2.735889  5.342009    TRUE
295         295 5.393227 2.732074  8.054380    TRUE
296         296 3.603210 1.905268  5.301151    TRUE
297         297 4.987438 2.865588  7.109287    TRUE
298         298 3.485912 2.313194  4.658630   FALSE
299         299 5.140612 2.954029  7.327194    TRUE
300         300 3.477965 2.236296  4.719633   FALSE
301         301 4.838179 2.666346  7.010013    TRUE
302         302 4.378519 2.904589  5.852448    TRUE
303         303 6.498538 4.652734  8.344341    TRUE
304         304 7.888626 4.962471 10.814781    TRUE
305         305 5.866072 2.703398  9.028746    TRUE
306         306 4.548210 2.879283  6.217138    TRUE
307         307 5.812011 3.232438  8.391584    TRUE
308         308 4.729734 1.924979  7.534489    TRUE
309         309 4.724547 3.012132  6.436962    TRUE
310         310 5.509638 3.535638  7.483638    TRUE
311         311 4.860572 3.116204  6.604940    TRUE
312         312 6.229047 3.491158  8.966936    TRUE
313         313 7.803910 4.373291 11.234530    TRUE
314         314 4.172664 2.392126  5.953202    TRUE
315         315 5.385459 2.042203  8.728715    TRUE
316         316 7.443384 5.117284  9.769484   FALSE
317         317 3.497343 2.049574  4.945112   FALSE
318         318 4.548129 2.747034  6.349224    TRUE
319         319 3.575881 2.061335  5.090426    TRUE
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692         692 4.796377 2.911562  6.681192    TRUE
693         693 7.166511 3.473505 10.859517    TRUE
694         694 3.737361 2.370923  5.103799    TRUE
695         695 4.946349 2.374817  7.517881    TRUE
696         696 5.439459 3.581471  7.297447    TRUE
697         697 5.315822 3.447856  7.183788    TRUE
698         698 3.741409 2.290181  5.192637    TRUE
699         699 4.774971 3.073936  6.476006    TRUE
700         700 3.402316 2.136310  4.668323   FALSE
701         701 6.020801 3.099626  8.941975    TRUE
702         702 4.449331 1.966136  6.932527    TRUE
703         703 5.102650 2.891447  7.313854    TRUE
704         704 6.881138 2.040926 11.721351    TRUE
705         705 3.850459 2.259788  5.441130    TRUE
706         706 4.413014 2.002804  6.823224    TRUE
707         707 6.850892 4.650988  9.050796    TRUE
708         708 4.725583 3.138983  6.312183    TRUE
709         709 4.881814 3.193956  6.569672    TRUE
710         710 3.611745 2.037187  5.186303    TRUE
711         711 3.327933 1.994595  4.661271   FALSE
712         712 6.106955 3.311340  8.902570    TRUE
713         713 3.416288 2.020751  4.811825   FALSE
714         714 4.884710 2.936531  6.832889    TRUE
715         715 5.180604 2.607237  7.753971    TRUE
716         716 5.281995 2.733991  7.830000    TRUE
717         717 4.175568 2.081826  6.269309    TRUE
718         718 5.017237 2.621908  7.412565    TRUE
719         719 5.467241 3.864656  7.069826    TRUE
720         720 3.153686 1.932826  4.374546   FALSE
721         721 7.155385 4.427355  9.883415    TRUE
722         722 4.341135 2.774704  5.907567    TRUE
723         723 3.037412 1.669276  4.405547   FALSE
724         724 5.474374 3.772444  7.176304    TRUE
725         725 6.450400 3.799504  9.101295    TRUE
726         726 6.750764 3.514467  9.987061    TRUE
727         727 6.845386 4.481699  9.209074    TRUE
728         728 4.667582 2.473775  6.861389    TRUE
729         729 3.362999 1.530116  5.195882    TRUE
730         730 5.378504 3.487381  7.269627    TRUE
731         731 5.766296 3.740023  7.792570    TRUE
732         732 4.098024 2.725293  5.470754    TRUE
733         733 4.005968 2.481660  5.530277    TRUE
734         734 5.245817 2.606822  7.884813    TRUE
735         735 4.533103 2.987374  6.078832    TRUE
736         736 5.538313 3.563316  7.513309    TRUE
737         737 4.860992 3.043456  6.678529    TRUE
738         738 6.143755 3.583157  8.704353    TRUE
739         739 4.551941 2.655501  6.448382    TRUE
740         740 4.473198 2.789070  6.157327    TRUE
741         741 5.453317 2.970865  7.935769    TRUE
742         742 4.773511 3.273516  6.273505    TRUE
743         743 4.269461 2.958314  5.580608    TRUE
744         744 3.756605 2.266408  5.246802    TRUE
745         745 5.240357 3.412053  7.068661    TRUE
746         746 5.540027 3.514813  7.565241    TRUE
747         747 6.274828 2.552602  9.997053    TRUE
748         748 5.829695 3.977654  7.681736    TRUE
749         749 6.163465 3.715670  8.611259    TRUE
750         750 6.687634 4.175748  9.199520    TRUE
751         751 5.688588 3.342611  8.034566    TRUE
752         752 5.748188 4.057759  7.438617    TRUE
753         753 5.199793 3.391007  7.008579    TRUE
754         754 4.378812 2.994439  5.763185    TRUE
755         755 3.754616 2.704143  4.805090   FALSE
756         756 4.932217 2.380414  7.484020    TRUE
757         757 4.050452 2.442749  5.658155    TRUE
758         758 6.332122 3.545771  9.118473    TRUE
759         759 4.303519 2.527341  6.079696    TRUE
760         760 2.960447 2.094130  3.826763   FALSE
761         761 5.997577 3.645953  8.349201    TRUE
762         762 3.706199 1.912164  5.500233    TRUE
763         763 4.375903 2.424948  6.326859    TRUE
764         764 5.380407 2.557484  8.203330    TRUE
765         765 4.623833 2.975591  6.272076    TRUE
766         766 5.731270 3.832596  7.629945    TRUE
767         767 3.936541 1.520358  6.352724    TRUE
768         768 5.630331 3.625201  7.635460    TRUE
769         769 4.858964 3.136736  6.581192    TRUE
770         770 4.676496 3.058747  6.294244    TRUE
771         771 4.408279 2.590964  6.225594    TRUE
772         772 4.756400 2.692028  6.820773    TRUE
773         773 6.340416 4.269570  8.411262    TRUE
774         774 4.193975 2.424266  5.963684    TRUE
775         775 3.795168 2.251851  5.338484    TRUE
776         776 5.300706 3.510871  7.090542    TRUE
777         777 4.364782 3.042306  5.687257    TRUE
778         778 6.294204 3.516965  9.071443    TRUE
779         779 5.999043 3.449786  8.548300    TRUE
780         780 6.136780 3.655391  8.618169    TRUE
781         781 7.005303 4.334360  9.676245    TRUE
782         782 3.163227 1.703438  4.623016   FALSE
783         783 4.364169 2.640040  6.088298    TRUE
784         784 4.599123 3.042723  6.155522    TRUE
785         785 3.837009 2.050359  5.623659    TRUE
786         786 5.653517 3.040694  8.266340    TRUE
787         787 6.824836 4.443580  9.206093    TRUE
788         788 5.531718 3.731936  7.331501    TRUE
789         789 3.376157 2.149207  4.603107   FALSE
790         790 5.154140 3.051966  7.256315    TRUE
791         791 3.730045 2.337028  5.123063    TRUE
792         792 5.159005 2.997285  7.320726    TRUE
793         793 6.109767 3.730349  8.489185    TRUE
794         794 5.310507 1.937030  8.683984    TRUE
795         795 3.821305 2.450911  5.191699    TRUE
796         796 4.503406 2.751958  6.254855    TRUE
797         797 6.215615 3.772428  8.658803    TRUE
798         798 5.963003 4.186761  7.739245    TRUE
799         799 4.547936 2.966518  6.129353    TRUE
800         800 5.935948 3.831795  8.040100    TRUE
801         801 4.580775 3.262942  5.898609    TRUE
802         802 5.288510 3.293573  7.283447    TRUE
803         803 4.003014 2.249933  5.756094    TRUE
804         804 4.578937 2.767465  6.390409    TRUE
805         805 5.208572 2.679992  7.737152    TRUE
806         806 4.510099 2.834455  6.185742    TRUE
807         807 5.541238 3.028892  8.053584    TRUE
808         808 5.587650 3.922296  7.253003    TRUE
809         809 5.868813 3.030253  8.707373    TRUE
810         810 6.811166 3.956606  9.665726    TRUE
811         811 4.709316 2.678808  6.739824    TRUE
812         812 4.609925 2.372038  6.847812    TRUE
813         813 4.616457 2.873843  6.359071    TRUE
814         814 6.079642 3.810205  8.349078    TRUE
815         815 4.919606 3.317186  6.522027    TRUE
816         816 4.729133 2.687433  6.770834    TRUE
817         817 4.503897 2.746692  6.261102    TRUE
818         818 4.966847 2.799646  7.134048    TRUE
819         819 4.129039 2.239144  6.018934    TRUE
820         820 6.426912 3.939504  8.914320    TRUE
821         821 4.147807 2.227563  6.068051    TRUE
822         822 4.291451 2.683235  5.899667    TRUE
823         823 4.312255 2.782270  5.842240    TRUE
824         824 5.449740 3.637872  7.261609    TRUE
825         825 5.059137 3.375397  6.742877    TRUE
826         826 5.734931 2.777250  8.692611    TRUE
827         827 3.832671 1.863564  5.801779    TRUE
828         828 5.752577 3.043486  8.461667    TRUE
829         829 4.141864 2.360535  5.923192    TRUE
830         830 7.076177 4.395425  9.756928    TRUE
831         831 3.144744 2.211525  4.077963   FALSE
832         832 5.077636 3.190094  6.965177    TRUE
833         833 3.399460 2.176792  4.622129   FALSE
834         834 3.481387 2.116464  4.846311   FALSE
835         835 4.274422 2.708855  5.839988    TRUE
836         836 4.662463 3.128964  6.195962    TRUE
837         837 3.014837 1.529162  4.500511   FALSE
838         838 4.790803 2.854306  6.727300    TRUE
839         839 4.339132 2.613838  6.064426    TRUE
840         840 3.702371 2.466085  4.938657   FALSE
841         841 4.896887 2.952496  6.841277    TRUE
842         842 6.039827 3.678936  8.400717    TRUE
843         843 3.734504 1.991932  5.477075    TRUE
844         844 4.153317 1.642083  6.664552    TRUE
845         845 4.611843 2.374738  6.848947    TRUE
846         846 4.794271 3.111052  6.477490    TRUE
847         847 7.798794 3.923451 11.674137    TRUE
848         848 6.130638 3.656101  8.605176    TRUE
849         849 4.226358 2.782900  5.669816    TRUE
850         850 4.228714 3.036633  5.420794    TRUE
851         851 4.464075 3.111874  5.816277    TRUE
852         852 4.676103 2.719248  6.632958    TRUE
853         853 4.480655 2.820526  6.140784    TRUE
854         854 4.470416 2.921734  6.019097    TRUE
855         855 3.911622 2.537023  5.286222    TRUE
856         856 3.940003 2.787154  5.092851    TRUE
857         857 6.795089 3.883985  9.706194    TRUE
858         858 4.961688 3.646689  6.276688    TRUE
859         859 5.467941 3.495495  7.440387    TRUE
860         860 5.532097 2.580928  8.483267    TRUE
861         861 4.301249 3.162170  5.440327    TRUE
862         862 4.519407 2.838443  6.200371    TRUE
863         863 4.562370 3.423932  5.700807    TRUE
864         864 7.015168 5.088167  8.942168   FALSE
865         865 4.450826 3.122603  5.779050    TRUE
866         866 5.715641 3.350611  8.080670    TRUE
867         867 4.443041 2.631183  6.254898    TRUE
868         868 4.290362 2.808323  5.772401    TRUE
869         869 7.454553 4.734805 10.174301    TRUE
870         870 4.640577 2.597255  6.683899    TRUE
871         871 5.670020 3.387079  7.952961    TRUE
872         872 5.482412 3.268457  7.696367    TRUE
873         873 5.342856 3.574643  7.111070    TRUE
874         874 4.285617 2.414433  6.156801    TRUE
875         875 5.142422 3.373784  6.911060    TRUE
876         876 4.420942 2.992030  5.849854    TRUE
877         877 3.598424 2.205445  4.991404   FALSE
878         878 4.806463 3.012270  6.600656    TRUE
879         879 4.783824 2.845521  6.722126    TRUE
880         880 5.191992 2.880083  7.503901    TRUE
881         881 3.418037 2.118500  4.717574   FALSE
882         882 4.897953 2.848338  6.947568    TRUE
883         883 3.398464 1.997536  4.799392   FALSE
884         884 3.386818 2.177829  4.595807   FALSE
885         885 4.340401 2.150313  6.530490    TRUE
886         886 4.118217 2.760658  5.475777    TRUE
887         887 5.153575 2.984333  7.322817    TRUE
888         888 5.510811 3.769060  7.252561    TRUE
889         889 4.602609 2.855847  6.349372    TRUE
890         890 3.905941 2.291495  5.520388    TRUE
891         891 6.774256 4.306045  9.242467    TRUE
892         892 4.317355 2.024624  6.610086    TRUE
893         893 4.723483 2.979927  6.467039    TRUE
894         894 4.097961 2.764166  5.431756    TRUE
895         895 3.331095 2.104561  4.557629   FALSE
896         896 3.764747 2.597417  4.932077   FALSE
897         897 5.044357 2.025798  8.062916    TRUE
898         898 5.104493 2.976858  7.232128    TRUE
899         899 3.882643 2.367587  5.397699    TRUE
900         900 4.726397 3.083887  6.368907    TRUE
901         901 5.481720 2.834072  8.129368    TRUE
902         902 4.482242 2.335264  6.629221    TRUE
903         903 4.535109 1.822728  7.247490    TRUE
904         904 5.196180 2.692429  7.699932    TRUE
905         905 4.492009 2.163685  6.820334    TRUE
906         906 6.678821 3.784831  9.572811    TRUE
907         907 5.884205 3.473011  8.295398    TRUE
908         908 5.687094 2.991138  8.383051    TRUE
909         909 7.412439 4.708236 10.116642    TRUE
910         910 4.303710 2.809604  5.797816    TRUE
911         911 4.964240 2.453755  7.474724    TRUE
912         912 5.801719 3.462369  8.141068    TRUE
913         913 4.950980 3.050451  6.851509    TRUE
914         914 6.001761 3.774318  8.229205    TRUE
915         915 6.528805 3.491221  9.566389    TRUE
916         916 4.288220 2.089472  6.486969    TRUE
917         917 4.956680 3.180271  6.733090    TRUE
918         918 7.170421 3.982702 10.358140    TRUE
919         919 5.185765 3.480158  6.891373    TRUE
920         920 6.595852 3.893265  9.298440    TRUE
921         921 7.114845 4.448619  9.781071    TRUE
922         922 3.623587 1.952435  5.294738    TRUE
923         923 4.543857 2.465439  6.622275    TRUE
924         924 3.258461 1.851580  4.665342   FALSE
925         925 4.014023 2.668096  5.359949    TRUE
926         926 4.276813 2.510167  6.043460    TRUE
927         927 6.814491 3.751824  9.877159    TRUE
928         928 5.772787 3.231194  8.314380    TRUE
929         929 5.447037 2.751048  8.143025    TRUE
930         930 5.931869 2.980780  8.882958    TRUE
931         931 4.041626 1.958424  6.124829    TRUE
932         932 5.316973 3.323518  7.310427    TRUE
933         933 4.688817 2.673399  6.704235    TRUE
934         934 7.467652 4.695951 10.239353    TRUE
935         935 5.564227 2.907007  8.221446    TRUE
936         936 3.423039 2.226523  4.619555   FALSE
937         937 5.734565 3.254354  8.214775    TRUE
938         938 5.020510 2.869966  7.171054    TRUE
939         939 5.100959 3.294427  6.907491    TRUE
940         940 5.657519 3.319490  7.995548    TRUE
941         941 5.606712 2.617362  8.596062    TRUE
942         942 5.931564 3.878500  7.984629    TRUE
943         943 7.327035 4.398522 10.255548    TRUE
944         944 5.762697 3.368771  8.156623    TRUE
945         945 4.435359 2.549444  6.321275    TRUE
946         946 3.444544 2.051309  4.837780   FALSE
947         947 4.634854 2.194194  7.075514    TRUE
948         948 5.300417 3.013622  7.587212    TRUE
949         949 3.831749 2.413875  5.249622    TRUE
950         950 7.433438 4.760416 10.106460    TRUE
951         951 4.202738 2.386948  6.018529    TRUE
952         952 5.257739 3.429020  7.086458    TRUE
953         953 4.035664 2.651484  5.419843    TRUE
954         954 4.220258 2.506679  5.933838    TRUE
955         955 4.543337 3.332731  5.753943    TRUE
956         956 6.475770 4.153527  8.798013    TRUE
957         957 4.484203 2.645032  6.323374    TRUE
958         958 4.818013 2.538188  7.097838    TRUE
959         959 5.850119 3.954613  7.745626    TRUE
960         960 5.065497 3.364905  6.766089    TRUE
961         961 5.269891 2.878345  7.661437    TRUE
962         962 4.138069 2.639614  5.636524    TRUE
963         963 4.647331 2.865612  6.429050    TRUE
964         964 4.934531 3.012154  6.856908    TRUE
965         965 4.033897 2.014314  6.053480    TRUE
966         966 4.862676 2.955659  6.769694    TRUE
967         967 4.686686 2.578652  6.794719    TRUE
968         968 5.974572 3.574119  8.375025    TRUE
969         969 5.134402 2.720357  7.548448    TRUE
970         970 5.522797 4.113178  6.932417    TRUE
971         971 3.967790 2.119553  5.816027    TRUE
972         972 4.745528 3.035956  6.455101    TRUE
973         973 3.799563 2.304391  5.294735    TRUE
974         974 5.458321 3.464534  7.452109    TRUE
975         975 3.848773 2.257088  5.440459    TRUE
976         976 5.323922 3.794530  6.853314    TRUE
977         977 3.833001 1.957141  5.708861    TRUE
978         978 7.722533 5.198700 10.246366   FALSE
979         979 4.958634 2.905588  7.011681    TRUE
980         980 5.787319 3.474365  8.100274    TRUE
981         981 4.950609 2.944975  6.956242    TRUE
982         982 4.231221 2.701450  5.760993    TRUE
983         983 5.810367 3.869281  7.751453    TRUE
984         984 6.648866 4.359803  8.937930    TRUE
985         985 4.933492 3.061508  6.805477    TRUE
986         986 4.763034 2.993199  6.532870    TRUE
987         987 3.640684 1.930856  5.350513    TRUE
988         988 5.302980 2.535678  8.070281    TRUE
989         989 6.424658 3.746124  9.103192    TRUE
990         990 7.601018 4.182932 11.019104    TRUE
991         991 5.223517 3.281845  7.165188    TRUE
992         992 4.524292 1.641465  7.407119    TRUE
993         993 5.903551 3.418944  8.388158    TRUE
994         994 4.888675 3.278416  6.498933    TRUE
995         995 5.960770 3.397590  8.523951    TRUE
996         996 5.388928 2.862317  7.915540    TRUE
997         997 5.236746 2.872016  7.601476    TRUE
998         998 6.589959 4.307956  8.871962    TRUE
999         999 3.836842 1.914978  5.758706    TRUE
1000       1000 2.881929 1.601653  4.162206   FALSE

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

out_dir <- "modules/sampling_coverage/generated"
dir.create(out_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(240506)
d <- data.frame(experiment=rep(1:1000, each=25),
                waiting_days=rexp(25000, rate=.2))
write.csv(d, file.path(out_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/sampling_coverage/analysis.R modules/sampling_coverage/generated

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Dataset 58 · Multiple testing

Bonferroni correction

CSV rows: 960Seed: 240508Module: bonferroni

Saved CSV Full runnable R script Full-size plot

Compare gene expression between two conditions across 40 genes.

Question and null

Scientific question

Which gene differences are supported while controlling the chance of any false positive across the 40 comparisons?

Null being tested / estimation target

For each gene, the treatment and control population means are equal. The adjustment changes the error criterion, not those individual nulls.

Generating truth

The first eight genes have true mean effects of 1.1 log2 units; the other 32 nulls are true.

Exact generating model

40 genes, each with 12 control and 12 treated independent Normal observations, SD=1 log2 unit. Controls have mean 6; the first 8 genes have treatment shift 1.1; the other 32 have shift 0. Genes are independent in this simulation.

Design and independent unit: Each gene comparison uses 12 independent samples per condition; simulation uses independent genes.

Columns in the saved CSV

Column	Meaning	Unit / type
`gene`	Gene comparison	ID
`treated`	Control=0, treated=1	indicator
`expression`	Expression measurement	log2 units

Preview the actual saved observations

gene	treated	expression
1	0	9.3078867236721
1	0	5.03171928240272
1	0	4.51762083093542
1	0	4.35698534771405
1	0	6.10677239406436

Analysis from the saved dataset

Bonferroni correction plot. Compare gene expression between two conditions across 40 genes. — Simulated observations; generated by this module’s R script.

Unadjusted p<.05: 8 genes
Adjusted p<.05: 2 genes
One declared family of 40 gene comparisons

Of 40 gene comparisons, 2 have Bonferroni-adjusted p-values below 0.05. With valid input tests, this rule controls the probability of at least one false positive in the family at 5%. Interpret the gene effect sizes alongside the adjusted evidence.

Teaching pointBonferroni controls the chance of at least one false positive in the declared family; retain the gene effect sizes when interpreting discoveries.

Analysis code excerpt

d <- read.csv("modules/bonferroni/generated/data.csv")
p_value <- difference <- numeric(40)
for (i in 1:40) {
  x <- d$expression[d$gene==i & d$treated==1]
  y <- d$expression[d$gene==i & d$treated==0]
  p_value[i] <- t.test(x,y)$p.value
  difference[i] <- mean(x)-mean(y)
}
adjusted <- p.adjust(p_value, method="bonferroni")
result <- data.frame(gene=1:40, difference, p_value, adjusted)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output

   gene   difference      p_value    adjusted
1     1  1.356069983 2.431656e-02 0.972662302
2     2  1.820104702 3.547859e-03 0.141914361
3     3  0.845346076 1.950008e-02 0.780003216
4     4  1.601922131 2.962924e-04 0.011851697
5     5  0.875891097 1.213253e-02 0.485301058
6     6  0.687357661 7.714434e-02 1.000000000
7     7  1.845249234 9.557971e-05 0.003823188
8     8  1.105882290 1.875913e-02 0.750365364
9     9  0.141888955 7.029532e-01 1.000000000
10   10 -0.525492608 1.433129e-01 1.000000000
11   11 -0.111584860 7.609645e-01 1.000000000
12   12 -0.500318535 1.802215e-01 1.000000000
13   13 -0.174194669 6.714101e-01 1.000000000
14   14 -0.442811468 3.442297e-01 1.000000000
15   15 -0.237449722 4.772690e-01 1.000000000
16   16  0.070743474 8.445295e-01 1.000000000
17   17  1.038355354 2.150182e-02 0.860072962
18   18 -0.478127932 2.694789e-01 1.000000000
19   19 -0.695425512 7.735015e-02 1.000000000
20   20  0.295473084 5.001759e-01 1.000000000
21   21 -0.255778461 5.105560e-01 1.000000000
22   22 -0.052458354 8.819974e-01 1.000000000
23   23  0.019658817 9.659932e-01 1.000000000
24   24  0.428118427 2.948827e-01 1.000000000
25   25  0.078072139 8.409584e-01 1.000000000
26   26  0.329322315 4.137296e-01 1.000000000
27   27  0.115868310 7.837690e-01 1.000000000
28   28 -0.128556430 7.479098e-01 1.000000000
29   29 -0.244565960 5.393038e-01 1.000000000
30   30 -0.444446000 3.158609e-01 1.000000000
31   31 -0.110704620 8.086996e-01 1.000000000
32   32 -0.163111825 7.039041e-01 1.000000000
33   33 -0.460802842 3.168194e-01 1.000000000
34   34 -0.288218696 5.228287e-01 1.000000000
35   35  0.312361668 4.543604e-01 1.000000000
36   36 -0.439299699 3.246300e-01 1.000000000
37   37 -0.001465637 9.964233e-01 1.000000000
38   38  0.588926287 2.193949e-01 1.000000000
39   39 -0.165987262 6.396148e-01 1.000000000
40   40  0.585516005 1.799847e-01 1.000000000

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

out_dir <- "modules/bonferroni/generated"
dir.create(out_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(240508)
gene <- rep(1:40, each=24)
treated <- rep(rep(0:1, each=12), 40)
effect <- rep(c(rep(1.1,8), rep(0,32)), each=24)
d <- data.frame(gene, treated, expression=rnorm(960, 6+effect*treated, 1))
write.csv(d, file.path(out_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/bonferroni/analysis.R modules/bonferroni/generated

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Dataset 59 · Multiple testing

Benjamini–Hochberg FDR

CSV rows: 960Seed: 240509Module: fdr

Saved CSV Full runnable R script Full-size plot

Compare gene expression between two conditions across 40 genes.

Question and null

Scientific question

Which gene differences are supported while controlling the expected fraction of false discoveries among selected genes?

Null being tested / estimation target

For each gene, the treatment and control population means are equal. The adjustment changes the error criterion, not those individual nulls.

Generating truth

The first eight genes have true mean effects of 1.1 log2 units; the other 32 nulls are true.

Exact generating model

Design and independent unit: Each gene comparison uses 12 independent samples per condition; simulation uses independent genes.

Columns in the saved CSV

Column	Meaning	Unit / type
`gene`	Gene comparison	ID
`treated`	Control=0, treated=1	indicator
`expression`	Expression measurement	log2 units

Preview the actual saved observations

gene	treated	expression
1	0	5.77752076269615
1	0	6.30547876318671
1	0	5.97837420004123
1	0	5.55129859688178
1	0	6.62372689168127

Analysis from the saved dataset

Benjamini–Hochberg FDR plot. Compare gene expression between two conditions across 40 genes. — Simulated observations; generated by this module’s R script.

Unadjusted p<.05: 7 genes
Adjusted p<.05: 3 genes
One declared family of 40 gene comparisons

Of 40 gene comparisons, 3 have BH-adjusted p-values below 0.05. Under its assumptions, this rule controls the expected false-discovery proportion among selected genes at 5%. Interpret the gene effect sizes alongside the adjusted evidence.

Teaching pointBH controls the expected false-discovery proportion among selected genes under its assumptions; the fraction in one particular selected set can differ.

Analysis code excerpt

d <- read.csv("modules/fdr/generated/data.csv")
p_value <- difference <- numeric(40)
for (i in 1:40) {
  x <- d$expression[d$gene==i & d$treated==1]
  y <- d$expression[d$gene==i & d$treated==0]
  p_value[i] <- t.test(x,y)$p.value
  difference[i] <- mean(x)-mean(y)
}
adjusted <- p.adjust(p_value, method="BH")
result <- data.frame(gene=1:40, difference, p_value, adjusted)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output

   gene  difference      p_value   adjusted
1     1  0.74361874 0.1104543621 0.40165223
2     2  1.13993835 0.0030532767 0.04071036
3     3  0.63303366 0.1567432146 0.44783776
4     4  1.14326813 0.0073476739 0.06095345
5     5  0.96736406 0.0305870454 0.17478312
6     6  1.09236913 0.0076191816 0.06095345
7     7  1.48722039 0.0005346944 0.02138778
8     8  0.98760270 0.0091921923 0.06128128
9     9  0.51837432 0.1388609646 0.42726451
10   10 -0.14223887 0.7992511250 0.89242739
11   11  0.51406450 0.1902207102 0.50725523
12   12  0.70940140 0.1049898188 0.40165223
13   13 -0.30250964 0.4092528380 0.71174407
14   14  0.42126908 0.2268777905 0.53383010
15   15  0.61198019 0.1050522464 0.40165223
16   16  0.20888941 0.5939132343 0.84844748
17   17 -0.65242027 0.1298588176 0.42726451
18   18 -0.01411687 0.9796145857 0.97961459
19   19  0.24030316 0.6422998740 0.88593086
20   20 -0.64493323 0.1016113193 0.40165223
21   21  0.07513899 0.8145615887 0.89242739
22   22 -0.47464912 0.2974303246 0.66095628
23   23  0.02363298 0.9466910092 0.97096514
24   24  1.02654327 0.0014177568 0.02835514
25   25 -0.29191394 0.4709077705 0.75345243
26   26 -0.10301348 0.8478060177 0.89242739
27   27  0.18143797 0.6983414710 0.89242739
28   28 -0.18717039 0.6687974584 0.89172994
29   29  0.33171934 0.3389648607 0.67310147
30   30  0.15073943 0.7246670673 0.89242739
31   31  0.51380224 0.2236912675 0.53383010
32   32 -0.08888316 0.8389536027 0.89242739
33   33  0.26256464 0.4695329863 0.75345243
34   34 -0.15063432 0.7623551098 0.89242739
35   35 -0.36171374 0.3702058103 0.67310147
36   36  0.45507254 0.3293200199 0.67310147
37   37 -0.33460383 0.5022342610 0.77266809
38   38  0.24260469 0.5607670894 0.83076606
39   39  0.48258110 0.3672953546 0.67310147
40   40  0.10560952 0.7943725893 0.89242739

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

out_dir <- "modules/fdr/generated"
dir.create(out_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(240509)
gene <- rep(1:40, each=24)
treated <- rep(rep(0:1, each=12), 40)
effect <- rep(c(rep(1.1,8), rep(0,32)), each=24)
d <- data.frame(gene, treated, expression=rnorm(960, 6+effect*treated, 1))
write.csv(d, file.path(out_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/fdr/analysis.R modules/fdr/generated

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Dataset 60 · Multivariate exploration

Principal component analysis

CSV rows: 75Seed: 240510Module: pca

Saved CSV Full runnable R script Full-size plot

Four correlated morphological traits are measured in beetles from three ecotypes.

Question and null

Scientific question

Which combinations of traits summarize the most variation?

Null being tested / estimation target

The target is a set of linear trait combinations that explain the most variance; PCA itself does not test a null hypothesis.

Generating truth

Groups differ in body-size means and antenna shifts; all traits share a common body-size component.

Exact generating model

75 beetles: 25 per ecotype. Body size is Normal(mean 12,14,13 mm by Forest,Grassland,Dune; SD 1.2). Wing=1.3×size+Normal(0,1); femur=0.45×size+Normal(0,0.4); antenna=0.7×size+ecotype shift(0,1,−1)+Normal(0,0.5). All errors independent.

Design and independent unit: 25 independent beetles per ecotype; four traits on each individual.

Columns in the saved CSV

Column	Meaning	Unit / type
`ecotype`	Known source habitat	category
`body_mm`	body	mm
`wing_mm`	wing	mm
`femur_mm`	femur	mm
`antenna_mm`	antenna	mm

Preview the actual saved observations

ecotype	body_mm	wing_mm	femur_mm	antenna_mm
Forest	11.2647528949218	15.474078041144	6.01782735146394	7.57321183797288
Forest	12.2959645580009	13.7150827832475	6.00009978323413	8.86446793140073
Forest	10.5848280325085	13.4846115390853	5.152304415982	7.89716867407414
Forest	12.0555880212174	17.3469301269076	5.79020081622552	8.47695353703034
Forest	10.8100740951082	13.554809422668	5.3391807555605	8.15910208119912

Analysis from the saved dataset

Principal component analysis plot. Four correlated morphological traits are measured in beetles from three ecotypes. — Simulated observations; generated by this module’s R script.

PC1 accounts for 84.3% of variance
PC2 accounts for 8.6%
Variables centered and scaled

The first two principal components account for 92.9% of the standardized trait variance. Separation is descriptive; PCA does not test a population difference.

Teaching pointPCA uses the measurements, not class labels, to construct the axes.

Analysis code excerpt

d <- read.csv("modules/pca/generated/data.csv")
z <- d[c("body_mm","wing_mm","femur_mm","antenna_mm")]
result <- prcomp(z, center=TRUE, scale.=TRUE)
variance <- result$sdev^2/sum(result$sdev^2)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output

Standard deviations (1, .., p=4):
[1] 1.8360217 0.5857306 0.4563007 0.2788073

Rotation (n x k) = (4 x 4):
                 PC1         PC2         PC3        PC4
body_mm    0.5283830 -0.09689995  0.05098935 -0.8419156
wing_mm    0.4805746 -0.77094271 -0.17456589  0.3797659
femur_mm   0.5007526  0.30369527  0.74286006  0.3243069
antenna_mm 0.4889849  0.55138637 -0.64427150  0.2044041

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

out_dir <- "modules/pca/generated"
dir.create(out_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(240510)
ecotype <- rep(c("Forest", "Grassland", "Dune"), each=25)
size <- rnorm(75, rep(c(12,14,13),each=25), 1.2)
d <- data.frame(ecotype, body_mm=size,
  wing_mm=1.3*size+rnorm(75,0,1), femur_mm=.45*size+rnorm(75,0,.4),
  antenna_mm=.7*size+rep(c(0,1,-1),each=25)+rnorm(75,0,.5))
write.csv(d, file.path(out_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/pca/analysis.R modules/pca/generated

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Dataset 61 · Multivariate exploration

Classical multidimensional scaling

CSV rows: 75Seed: 240511Module: mds

Saved CSV Full runnable R script Full-size plot

Four correlated morphological traits are measured in beetles from three ecotypes.

Question and null

Scientific question

Can a low-dimensional map preserve the chosen between-individual distances?

Null being tested / estimation target

The target is a low-dimensional configuration that preserves the chosen distances; this map is not itself a hypothesis test.

Generating truth

Groups differ in body-size means and antenna shifts; all traits share a common body-size component.

Exact generating model

Design and independent unit: 25 independent beetles per ecotype; four traits on each individual.

Columns in the saved CSV

Column	Meaning	Unit / type
`ecotype`	Known source habitat	category
`body_mm`	body	mm
`wing_mm`	wing	mm
`femur_mm`	femur	mm
`antenna_mm`	antenna	mm

Preview the actual saved observations

ecotype	body_mm	wing_mm	femur_mm	antenna_mm
Forest	13.7935386870749	17.4922373962112	6.57965527990933	9.90212421085428
Forest	9.75903050159645	11.6742873577423	4.75473118465187	5.61002502406736
Forest	11.6297747472069	16.0650308433011	5.93490554523194	7.92840270593423
Forest	11.5632736390146	15.8686095986522	4.89818858940753	8.76899453836736
Forest	11.4016696440561	13.3131216590948	5.11594644984824	7.07287245338416

Analysis from the saved dataset

Classical multidimensional scaling plot. Four correlated morphological traits are measured in beetles from three ecotypes. — Simulated observations; generated by this module’s R script.

Two-dimensional goodness of fit: 0.920
Euclidean distances between standardized traits
Closer points have more similar trait combinations

The two-dimensional configuration has a goodness-of-fit measure of 0.920 for the supplied distances. Its axes summarize distances; their signs and orientation have no fixed biological meaning.

Teaching pointDistance choice defines what counts as similar.

Analysis code excerpt

d <- read.csv("modules/mds/generated/data.csv")
z <- scale(d[c("body_mm","wing_mm","femur_mm","antenna_mm")])
distance <- dist(z)
result <- cmdscale(distance, k=2, eig=TRUE)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output

$points
             [,1]         [,2]
 [1,]  0.97540338  0.007130556
 [2,] -4.46130829  0.469609156
 [3,] -1.24571197  0.497452164
 [4,] -1.72828046 -0.597133304
 [5,] -2.74594570  0.122395278
 [6,] -2.26815563 -0.447789896
 [7,] -1.12397958 -0.107186606
 [8,] -2.29155329 -0.726311080
 [9,] -1.80227000 -0.471975030
[10,] -0.56171783 -0.334092884
[11,] -1.18275802  0.653631090
[12,] -1.48299577  0.317510437
[13,] -1.62765057 -0.078355363
[14,] -2.98896933 -0.208697033
[15,] -1.52757541 -0.363367937
[16,] -1.24637429 -0.557080003
[17,] -1.68366709 -0.860288254
[18,]  0.75666938  1.285941982
[19,] -2.29549992 -0.667152081
[20,] -2.66289688  0.118922324
[21,]  3.10082130 -0.156332459
[22,] -1.82282364 -0.059656139
[23,]  0.40921175  0.416657106
[24,] -1.32552072 -0.761149165
[25,]  0.44845870  0.200823791
[26,]  0.10324130 -0.589977555
[27,]  1.19923123  0.134419369
[28,]  3.96984785 -0.172448876
[29,]  2.27126118 -0.066311361
[30,]  1.91711754 -0.430709984
[31,]  0.89544237 -0.462383205
[32,] -0.78009775 -0.819192745
[33,]  1.95641125 -0.909672844
[34,] -0.79468191 -0.922655152
[35,] -0.05522046 -0.365612822
[36,]  1.91428844  0.226624485
[37,]  3.20090994 -0.459201925
[38,]  0.21205988 -0.380048608
[39,]  1.02003356 -1.362147284
[40,]  1.24568281 -0.502301932
[41,]  1.57679589 -0.872085050
[42,]  1.81592643 -0.316351601
[43,] -0.27934197 -0.941706918
[44,]  0.71348105 -0.759967619
[45,]  0.38942396 -0.801572292
[46,]  4.47393929 -0.264713577
[47,]  1.98517161 -0.021157208
[48,]  0.36349297 -1.335409653
[49,]  1.15444022 -0.482976416
[50,]  3.46599888 -0.175119031
[51,] -1.53157106 -0.074276676
[52,]  0.78199752  0.930207230
[53,]  1.93515071  0.509660813
[54,] -0.70579706  0.859092352
[55,]  1.93518403  0.658875892
[56,]  0.67278208  0.218760206
[57,]  2.31127845  1.656886824
[58,]  1.38891081  0.771957279
[59,] -3.12887039  0.673693604
[60,]  0.20500793  0.590338474
[61,]  1.08079324  0.219675988
[62,] -1.65195396  0.345045343
[63,] -0.23402694  0.866576022
[64,]  0.63783338  0.691868939
[65,]  0.03377474  0.685551771
[66,] -0.73276347  0.993763279
[67,]  0.38852270  0.645752563
[68,]  1.00849774  0.539426058
[69,] -1.10164072  0.394921013
[70,] -2.75230520  0.157078109
[71,] -1.89821525  0.185367845
[72,]  1.02526505  0.481524596
[73,] -0.55076836  0.062510038
[74,]  0.79132930  1.044617787
[75,] -1.45818095  1.250297811

$eig
 [1]  2.415403e+02  3.065253e+01  1.922729e+01  4.579862e+00  9.815933e-15
 [6]  7.269084e-15  6.521041e-15  6.457920e-15  5.351543e-15  5.024629e-15
[11]  3.536685e-15  3.224538e-15  3.067886e-15  2.784960e-15  2.781320e-15
[16]  2.715877e-15  2.455016e-15  2.311272e-15  2.161329e-15  1.958721e-15
[21]  1.911807e-15  1.878351e-15  1.848735e-15  1.841280e-15  1.815672e-15
[26]  1.404475e-15  1.240902e-15  1.143682e-15  8.747429e-16  6.569501e-16
[31]  5.432930e-16  5.345057e-16  3.332429e-16  2.448202e-16  2.198298e-16
[36]  1.350057e-16  1.320982e-16  7.297038e-17  6.174247e-18 -1.292171e-16
[41] -2.390654e-16 -3.186937e-16 -3.438659e-16 -3.717135e-16 -4.448662e-16
[46] -4.953244e-16 -5.242173e-16 -5.484482e-16 -5.981108e-16 -6.448906e-16
[51] -6.998647e-16 -7.125050e-16 -7.285133e-16 -8.590374e-16 -9.645182e-16
[56] -9.755598e-16 -1.171510e-15 -1.301264e-15 -1.519557e-15 -1.560846e-15
[61] -2.906036e-15 -3.566509e-15 -3.793373e-15 -4.033947e-15 -4.147306e-15
[66] -4.299794e-15 -6.382293e-15 -6.885932e-15 -6.931247e-15 -7.223042e-15
[71] -7.241831e-15 -1.007704e-14 -1.271357e-14 -1.335275e-14 -5.515095e-14

$x
NULL

$ac
[1] 0

$GOF
[1] 0.9195704 0.9195704

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

out_dir <- "modules/mds/generated"
dir.create(out_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(240511)
ecotype <- rep(c("Forest", "Grassland", "Dune"), each=25)
size <- rnorm(75, rep(c(12,14,13),each=25), 1.2)
d <- data.frame(ecotype, body_mm=size,
  wing_mm=1.3*size+rnorm(75,0,1), femur_mm=.45*size+rnorm(75,0,.4),
  antenna_mm=.7*size+rep(c(0,1,-1),each=25)+rnorm(75,0,.5))
write.csv(d, file.path(out_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/mds/analysis.R modules/mds/generated

↑ Back to the quick index

Dataset 62 · Multivariate exploration

Linear discriminant analysis

CSV rows: 75Seed: 240512Module: lda

Saved CSV Full runnable R script Full-size plot

Four correlated morphological traits are measured in beetles from three ecotypes.

Question and null

Scientific question

How well can measured traits predict known group membership?

Null being tested / estimation target

The target is predictive classification of known ecotypes; this example estimates validation accuracy rather than testing a null hypothesis.

Generating truth

Groups differ in body-size means and antenna shifts; all traits share a common body-size component.

Exact generating model

Design and independent unit: 25 independent beetles per ecotype; four traits on each individual.

Columns in the saved CSV

Column	Meaning	Unit / type
`ecotype`	Known source habitat	category
`body_mm`	body	mm
`wing_mm`	wing	mm
`femur_mm`	femur	mm
`antenna_mm`	antenna	mm

Preview the actual saved observations

ecotype	body_mm	wing_mm	femur_mm	antenna_mm
Forest	11.346714570257	13.0609812134225	5.64075711991521	7.47562180433382
Forest	9.94927714151816	14.9876651201588	4.18085927291101	7.39597433384898
Forest	13.1550697258882	19.2382336105822	5.57496613338817	9.55850712775277
Forest	12.1554924312438	15.557648569213	5.68831666441997	8.9100266629901
Forest	11.7423266463407	16.3495097846689	5.10032314912281	7.30273942721568

Analysis from the saved dataset

Linear discriminant analysis plot. Four correlated morphological traits are measured in beetles from three ecotypes. — Simulated observations; generated by this module’s R script.

Leave-one-out classification accuracy: 82.7%
Three known ecotype labels used during training
Equal group sizes in this example

Leave-one-out classification accuracy is 82.7% for these three ecotypes. This is a prediction exercise using known class labels, not an unsupervised discovery or a significance test.

Teaching pointLDA uses the labels to find discrimination; separation on training data is not validation.

Analysis code excerpt

d <- read.csv("modules/lda/generated/data.csv")
result <- MASS::lda(ecotype ~ body_mm + wing_mm + femur_mm + antenna_mm, data=d)
cv <- MASS::lda(ecotype ~ body_mm + wing_mm + femur_mm + antenna_mm, data=d, CV=TRUE)
accuracy <- mean(cv$class==d$ecotype)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output

Call:
lda(ecotype ~ body_mm + wing_mm + femur_mm + antenna_mm, data = d)

Prior probabilities of groups:
     Dune    Forest Grassland 
0.3333333 0.3333333 0.3333333 

Group means:
           body_mm  wing_mm femur_mm antenna_mm
Dune      12.70929 17.02713 5.706581   7.915556
Forest    11.99869 15.71095 5.328634   8.397726
Grassland 13.90283 17.92353 6.288135  10.870551

Coefficients of linear discriminants:
                   LD1         LD2
body_mm     1.01499407 -0.78013116
wing_mm     0.10976514 -0.43360845
femur_mm    0.09092525 -0.08365976
antenna_mm -1.89182813  0.50613401

Proportion of trace:
   LD1    LD2 
0.8823 0.1177

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

out_dir <- "modules/lda/generated"
dir.create(out_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(240512)
ecotype <- rep(c("Forest", "Grassland", "Dune"), each=25)
size <- rnorm(75, rep(c(12,14,13),each=25), 1.2)
d <- data.frame(ecotype, body_mm=size,
  wing_mm=1.3*size+rnorm(75,0,1), femur_mm=.45*size+rnorm(75,0,.4),
  antenna_mm=.7*size+rep(c(0,1,-1),each=25)+rnorm(75,0,.5))
write.csv(d, file.path(out_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/lda/analysis.R modules/lda/generated

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Dataset 63 · Dependent observations

Count and rate GEE

CSV rows: 360Seed: 1401Module: count_gee

Saved CSV Full runnable R script Full-size plot

Microbial ecologists count bacterial colonies in repeated water samples from experimental pond mesocosms.

Question and null

Scientific question

Does warming change the population-average colony rate after accounting for week and sampled volume?

Null being tested / estimation target

The warming coefficient is zero: population-average colony rates are equal at the same week and exposure.

Generating truth

The exact population-average Warm / Ambient rate ratio is exp(0.45), approximately 1.568; the equal-rate null is false. Each week multiplies the marginal rate by exp(0.1), approximately 1.105. Marginal count variance is mu + 0.5 × mu squared. The working exchangeable correlation is approximate, while the specified marginal mean is exact.

Exact generating model

Ninety independent mesocosms (45 Ambient, 45 Warm) are sampled at weeks 0, 1, 2, and 3. Each volume is Uniform(0.5, 1.5) mL. The marginal expected count is mu = volume × exp(2.2 + 0.45 × Warm + 0.1 × week). Each mesocosm has one independent Gamma(shape 2, rate 2) multiplier, with mean 1 and variance 0.5. Conditional counts are independent Poisson(mu × multiplier), so repeated observations share the multiplier. No rounding or censoring is added.

Design and independent unit: Ninety independent mesocosms, 45 per temperature treatment, are sampled at weeks 0–3. Each sample has a known plated volume.

Columns in the saved CSV

Column	Meaning	Unit / type
`mesocosm`	Independent experimental mesocosm and repeated-sample cluster	ID
`week`	Sampling time since experiment began	week
`treatment`	Assigned ambient or warmed temperature	category
`volume_ml`	Sampled/plated water volume: the count exposure	mL
`colonies`	Observed colony count in the plated sample	count

Preview the actual saved observations

mesocosm	week	treatment	volume_ml	colonies
1	0	Ambient	1.28036231640726	9
1	1	Ambient	0.995979028521106	8
1	2	Ambient	1.15856119571254	21
1	3	Ambient	0.592245136154816	5
2	0	Ambient	0.656593127408996	20

Analysis from the saved dataset

Count and rate GEE plot. Microbial ecologists count bacterial colonies in repeated water samples from experimental pond mesocosms. — Simulated observations; generated by this module’s R script.

Independent mesocosms: 90; observations: 360
Population-average Warm / Ambient rate ratio = 1.580
95% robust CI: 1.175 to 2.123
Robust Wald p = 0.00243
Working correlation estimate = 0.816

The population-average colony rate in warmed mesocosms was 1.58 times the ambient rate (95% robust CI 1.18 to 2.12; Wald p = 0.00243), adjusting for week and sampled volume. The 90 mesocosms are the independent units; the model accounts for their repeated samples.

Teaching pointGEE estimates a population-average rate ratio and uses mesocosms as independent units. The offset accounts for exposure; robust uncertainty accommodates the extra variation and repeated observations.

Analysis code excerpt

d <- read.csv("modules/count_gee/generated/data.csv")
d$treatment <- factor(d$treatment, levels = c("Ambient", "Warm"))
d <- d[order(d$mesocosm, d$week), ]
fit <- geepack::geeglm(colonies ~ treatment + week + offset(log(volume_ml)),
  id = mesocosm, data = d, family = poisson(link = "log"),
  corstr = "exchangeable", std.err = "san.se")
b <- coef(fit)["treatmentWarm"]
se <- sqrt(vcov(fit)["treatmentWarm", "treatmentWarm"])
rate_ratio <- exp(b)
ci <- exp(b + c(-1, 1) * 1.96 * se)
p_value <- 2 * pnorm(abs(b / se), lower.tail = FALSE)

The teaching excerpt uses paths relative to the course folder. The full script supplies package loading and any additional preparation.

Full printed R output


Call:
geepack::geeglm(formula = colonies ~ treatment + week + offset(log(volume_ml)), 
    family = poisson(link = "log"), data = d, id = mesocosm, 
    corstr = "exchangeable", std.err = "san.se")

 Coefficients:
              Estimate Std.err    Wald Pr(>|W|)    
(Intercept)    2.20660 0.10491 442.355  < 2e-16 ***
treatmentWarm  0.45715 0.15078   9.193  0.00243 ** 
week           0.10817 0.01322  66.937 3.33e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Correlation structure = exchangeable 
Estimated Scale Parameters:

            Estimate Std.err
(Intercept)    7.602    1.51
  Link = identity 

Estimated Correlation Parameters:
      Estimate Std.err
alpha   0.8161 0.04783
Number of clusters:   90  Maximum cluster size: 4

Recreate the observations

The fixed seed and simulation excerpt are shown below. Run the complete script to regenerate the CSV, figure, and results together.

output_dir <- "modules/count_gee/generated"
dir.create(output_dir, recursive = TRUE, showWarnings = FALSE)
set.seed(1401)
mesocosm <- rep(1:90, each = 4)
week <- rep(0:3, 90)
treatment <- rep(rep(c("Ambient", "Warm"), each = 45), each = 4)
volume_ml <- runif(360, .5, 1.5)
shared_rate <- rgamma(90, shape = 2, rate = 2)
mu <- volume_ml * exp(2.2 + .45 * (treatment == "Warm") + .1 * week)
colonies <- rpois(360, mu * shared_rate[mesocosm])
d <- data.frame(mesocosm, week, treatment, volume_ml, colonies)
write.csv(d, file.path(output_dir, "data.csv"), row.names = FALSE)

Run the complete module from the course folder:

Rscript modules/count_gee/analysis.R modules/count_gee/generated

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Reproducibility notes

Each module owns its fixed seed, simulation, CSV, plot, and analysis output. Identifiers are local to a dataset; repeated IDs in different files do not imply shared organisms.

The model descriptions state any rounding, dependence, censoring, or deliberate unusual observations. The column dictionary and design identify how each record should be interpreted.

This guide reads saved CSVs and result files; it does not resimulate or refit analyses. Run the course builder or the relevant module scripts first after changing a model, then regenerate this guide.

Run source("build_instructor_guide.R") followed by build_instructor_guide(".") from the course folder, or knit Instructor-dataset-guide.Rmd in RStudio. Missing module outputs cause a build error rather than an incomplete guide.

The catalog records the generating model, truth, observed result, file links, and checksums of the CSV and result files used for this build.