Activity: Non-Parametric T-Tests

Rank-based tests

Hands-on activity: effect size and rank-based tests on our own pine needle data, where the assumptions hold and the tests cost power, then assumption checking, transformations, Welch’s t-test, Mann-Whitney, and permutation tests on the lake trout dataset.

Author

Bill Perry

Worksheet: Non-Parametric T-Tests

📥 Pine needle data 📥 Lake trout data 📥 Sagebrush cricket data

📥 Tools script (.R) 📥 Script skeleton (.R)

Part 1 · Our Pine Needles — Start With Data You Trust

Column names in pine_data.csv / pine_data.xlsx

Both files hold the same six columns, with exactly these names. Type them exactly as written — R is case-sensitive, and Needle_Length or needle_length will not work:

Column	What it is
`source_file`	the raw team file the row came from
`team`	field team — one team = one tree
`replicate`	needle number within a tree side (1-8)
`side`	sun exposure — `"shady"` or `"sunny"`
`needle_length_mm`	needle length, millimetres
`needle_width_mm`	needle width, millimetres

After you read the file in, click pine_df in the Variables pane in Positron and check the names against this table before you write any other code.

Last week you ran a paired t-test on our four trees and got p = 0.072. Before we reach for anything non-parametric, we have to establish that we actually need it — so we start by checking the assumptions on the data you collected yourselves.

▶ Run this:

# ── Part 1.1 · Setup and our pine needles ─────────────────────────────
library(tidyverse)
library(janitor)     # round_half_up(), used inside summary_stats()
library(car)         # leveneTest()
library(patchwork)   # combining plots
library(perm)        # permutation tests
library(effectsize)  # cohens_d()

source("themes_functions/r_themes_and_functions.R")

pine_df <- read_csv("data/pine_data.csv")

# same two-stage summary as the t-test lecture
p_df <- pine_df %>%
  group_by(team, side) %>%
  summarise(needle_length_mm = mean(needle_length_mm, na.rm = TRUE),
            .groups = "drop")

p_wide_df <- p_df %>%
  pivot_wider(names_from = side, values_from = needle_length_mm)

p_wide_df

▶ Run this — do the assumptions actually hold?

# ── Part 1.2 · Checking the assumptions first ─────────────────────────
# normality, one side at a time -- all 32 needles per side
shapiro.test(pine_df$needle_length_mm[pine_df$side == "sunny"])
shapiro.test(pine_df$needle_length_mm[pine_df$side == "shady"])

# equal variance
leveneTest(needle_length_mm ~ factor(side), data = pine_df)

✏️ Your turn: Record the three p-values, and remember which way these tests point — H₀ is that the assumption holds.

Shapiro-Wilk, sunny:   p = ______
Shapiro-Wilk, shady:   p = ______
Levene's test:         p = ______

✏️ Your turn: Based on those three numbers, is there any assumption of the paired t-test you ran last week that you can reject?

________________________

✏️ Your turn: So — do these data need a non-parametric test?

________________________

Effect size — a demonstration

Note

👀 Read this one, don’t sweat it

This section is here so you have seen effect size before we need it in the power lecture. Run the two blocks, read the output, answer the two short questions. There is nothing to get stuck on.

Your paired t-test last week gave p = 0.072 — “not significant.” But p never said how much shorter the shady needles were. Cohen’s d does: it is the difference between groups measured in standard deviations, so it carries no units and can be compared across studies.

For a paired design, d is the mean of the paired differences divided by the standard deviation of those differences.

▶ Run this:

# ── Part 1.3 · Effect size (demonstration) ────────────────────────────
diffs <- p_wide_df$shady - p_wide_df$sunny

mean(diffs) / sd(diffs)     # Cohen's d, by hand

# and with a confidence interval
cohens_d(p_wide_df$shady, p_wide_df$sunny, paired = TRUE)

✏️ Your turn: Write the value of d and its confidence interval here, then say in one plain sentence what the d value means.

d  = ______        95% CI = [ ______ , ______ ]

In words: the shady side averaged ______ standard deviations
          ____________ the sunny side.

▶ Run this — why it matters:

# ── Part 1.4 · Effect size (demonstration) ────────────────────────────
2.7227 / sqrt(4)      # our paired t statistic, divided by the square root of n

Compare that number to your d. They are the same, because t = d × √n — the t-statistic is the effect size multiplied by the square root of your sample size.

✏️ Your turn: If a p-value depends on both the size of the effect and how many samples you took, why is reporting d alongside p more informative than reporting p on its own?

________________________

Important

⚠️ The sentence worth remembering

“No significant difference” sounds like we found nothing. Our four trees actually gave a large effect (d = −1.36) that we simply could not pin down (the CI runs from −2.74 to 0.10). Those are very different messages, and only the second one is honest.

Now back to the rank tests. Run them even though the assumptions held — the point of this part is to find out what that choice would have cost you.

▶ Run this:

# ── Part 1.5 · The two rank-based tests ───────────────────────────────
# unpaired: Mann-Whitney U
wilcox.test(needle_length_mm ~ side, data = p_df)

# paired: Wilcoxon signed-rank
wilcox.test(p_wide_df$shady, p_wide_df$sunny, paired = TRUE)

✏️ Your turn: Fill in the comparison table.

                        parametric        non-parametric
unpaired:               p = 0.425         p = ______
paired:                 p = 0.072         p = ______

✏️ Your turn: Did the two approaches lead you to the same conclusion? Did they give the same p-value?

________________________

✏️ Your turn: Look at the signed-rank statistic, V. What value did you get, and what does it say about how consistent the four trees were?

________________________

▶ Run this — the floor:

# ── Part 1.6 · The floor ──────────────────────────────────────────────
2 / 2^4     # smallest two-sided p a signed-rank test can give with 4 pairs

✏️ Your turn: Compare that number to the p-value you got from the signed-rank test. What does that tell you about whether the test could have found our effect significant, even in principle?

________________________

✏️ Your turn: Based on all of this, which test would you report for the pine needle data, and why?

________________________

Tip

🚀 If you finish early: How many pairs would you need before a signed-rank test could possibly reach p < 0.05? Try 2 / 2^5, 2 / 2^6, and so on until you find it.

Part 2 · A Dataset Where the Assumptions Really Do Fail

Part 1 showed you what non-parametric tests cost. It could not show you what they buy you, because nothing about the pine data was broken.

The lake trout data is a genuine mess: six lakes, wildly unequal sample sizes, and mass distributions that fail every check we just ran. Everything from here on uses mass_g.

Let’s start by exploring only lake NE 12, as if doing a single-sample t-test, then compare NE 12 to Island Lake.

We want to test if the mass of lake trout in NE 12 differs from a mean of 500g.

✏️ Your turn: Write out the hypotheses:

H0: mu = _____ (the mean mass of lake trout in NE 12 is _____ g)
H1: mu ≠ _____ (the mean mass of lake trout in NE 12 is not _____ g)

▶ Run this in your Script:

# ── Part 2.1 · Reading in the lake trout ────────────────────────────────
# libraries and source() are already loaded from Part 1.1
lt_df <- read_csv("data/lake_trout.csv")

✏️ Your turn: How many lakes are in this dataset?

________________________

▶ Run this — calculate the mode (R has no built-in mode() for this):

# ── Part 2.2 · Setup and the Single-Sample Question ─────────────────────
lt_df %>%
  filter(!is.na(mass_g)) %>%
  group_by(lake, mass_g) %>%
  summarise(count = n(), .groups = "drop_last") %>%
  arrange(desc(count)) %>%
  slice(1) %>%
  select(-count) %>%
  rename(mode_mass = mass_g)

▶ Run this — create a data frame with just NE 12:

# ── Part 2.3 · Setup and the Single-Sample Question ─────────────────────
ne12_df <- lt_df %>%
  filter(lake == "NE 12") %>%
  filter(!is.na(mass_g))  # Remove any NA values

Tip

🚀 If you finish early: Compute n, mean, sd and se for mass_g across all lakes at once — lt_df %>% group_by(lake) %>% reframe(summary_stats(mass_g)).

# ── Part 2.4 · Setup and the Single-Sample Question ─────────────────────
# Write your code in the R script:

Part 3 · Testing Assumptions on NE 12

Before conducting our t-test, we need to verify our data meets the necessary assumptions. Methods to test normality: visual (QQ plots, histograms) and statistical (Shapiro-Wilk test).

▶ Run this — histogram:

# ── Part 3.1 · Testing Assumptions on NE 12 ─────────────────────────────
ne12_histo_plot <- ggplot(ne12_df, aes(x = mass_g)) +
  geom_histogram(binwidth = 200) +
  labs(x = "mass (g)", y = "Frequency")
ne12_histo_plot

▶ Run this — dotplot, boxplot, and QQ plot:

# ── Part 3.2 · Testing Assumptions on NE 12 ─────────────────────────────
ne12_dot_plot <- ggplot(ne12_df, aes(x = mass_g, y = "")) +
  geom_dotplot(binwidth = 60, stackdir = "center", fill = "darkblue", dotsize = 0.5) +
  labs(title = "Dotplot", x = "Mass (g)", y = "")
ne12_dot_plot

ne12_box_plot <- ggplot(ne12_df, aes(y = mass_g)) +
  geom_boxplot(fill = "darkblue") +
  labs(y = "Mass (g)", x = "") +
  coord_flip()
ne12_box_plot

ne12_qq_plot <- ggplot(ne12_df, aes(sample = mass_g)) +
  stat_qq(color = "darkblue") +
  stat_qq_line() +
  labs(title = "QQ Plot", x = "Theoretical Quantiles", y = "Sample Quantiles") +
  theme_regular(base_size = 10) +
  theme(plot.title = element_text(hjust = 0.5))
ne12_qq_plot

▶ Run this — combine all four with patchwork:

# ── Part 3.3 · Testing Assumptions on NE 12 ─────────────────────────────
combined_stats_plot <- (ne12_histo_plot + ne12_dot_plot) / (ne12_box_plot + ne12_qq_plot) +
  plot_annotation(
    theme = theme(plot.title = element_text(hjust = 0.5),
                  plot.subtitle = element_text(hjust = 0.5))
  )
combined_stats_plot

▶ Run this — the formal test (note: really you want to do this on residuals, but here we test the raw values directly):

# ── Part 3.4 · Testing Assumptions on NE 12 ─────────────────────────────
shapiro_test <- shapiro.test(ne12_df$mass_g)
shapiro_test

✏️ Your turn: Does the QQ plot agree with the Shapiro-Wilk result? Is NE 12’s mass normally distributed?

________________________

Tip

🚀 If you finish early: Run the same histogram, QQ plot and Shapiro-Wilk on a different lake — try Toolik or N 01. Is any lake’s mass close to normal?

# ── Part 3.5 · Testing Assumptions on NE 12 ─────────────────────────────
# Write your code in the R script:

Part 4 · Comparing Two Lakes — NE 12 vs. Island Lake

Now that we’ve shown NE 12’s mass fails (or passes) normality, let’s explore a comparison of NE 12 and Island Lake mass_g.

▶ Run this — build the two-lake data frame:

# ── Part 4.1 · Comparing Two Lakes ──────────────────────────────────────
in_df <- lt_df %>%
  filter(lake %in% c("NE 12", "Island Lake")) %>%
  filter(!is.na(mass_g))
head(in_df)

▶ Run this — summary stats by lake:

# ── Part 4.2 · Comparing Two Lakes ──────────────────────────────────────
summary_by_lake <- in_df %>% group_by(lake) %>% reframe(summary_stats(mass_g))

summary_by_lake

▶ Run this — histograms by lake:

# ── Part 4.3 · Comparing Two Lakes ──────────────────────────────────────
hist_plot <- in_df %>%
  ggplot(aes(x = mass_g, fill = lake)) +
  geom_histogram(bins = 20, alpha = 0.7) +
  labs(x = "Mass (g)", y = "Count") +
  theme_regular(base_size = 10) +
  facet_wrap(~lake, scales = "free_y")
hist_plot

▶ Run this — QQ plots by lake (in a QQ plot, points that follow the line indicate normally distributed data; deviations suggest non-normality):

# ── Part 4.4 · Comparing Two Lakes ──────────────────────────────────────
qq_plot <- in_df %>%
  ggplot(aes(sample = mass_g, color = lake)) +
  stat_qq() +
  stat_qq_line() +
  labs(title = "QQ Plot for Normality Check",
       x = "Theoretical Quantiles", y = "Sample Quantiles") +
  theme_regular(base_size = 10) +
  facet_wrap(~lake)
qq_plot

▶ Run this — the formal test, done for each lake separately (note: Island Lake can look non-normal in the QQ plot but come out close to normal in the Shapiro-Wilk test):

# ── Part 4.5 · Comparing Two Lakes ──────────────────────────────────────
normality_results <- in_df %>%
  group_by(lake) %>%
  summarize(
    shapiro_stat = shapiro.test(mass_g)$statistic,
    shapiro_p_value = shapiro.test(mass_g)$p.value,
    normal_distribution = if_else(shapiro_p_value > 0.05, "Normal", "Non-normal")
  )
normality_results

▶ Run this — Levene’s test for equal variances (H₀: variances are equal across groups; we want this non-significant):

# ── Part 4.6 · Comparing Two Lakes ──────────────────────────────────────
levene_result <- leveneTest(mass_g ~ lake, data = in_df)
levene_result

✏️ Your turn: Based on your normality and Levene’s test results, do these two lakes meet the assumptions for a standard t-test?

________________________

Tip

🚀 If you finish early: Rerun Levene’s test on a different pair of lakes — say Toolik and N 01, which have far more similar sample sizes. Same conclusion?

# ── Part 4.7 · Comparing Two Lakes ──────────────────────────────────────
# Write your code in the R script:

Part 5 · Data Transformations

A log₁₀ transformation commonly helps right-skewed data like this pass the normality assumption.

▶ Run this:

# ── Part 5.1 · Data Transformations ─────────────────────────────────────
in_df <- in_df %>%
  mutate(log_mass = log10(mass_g))
head(in_df)

▶ Run this — check if it worked, with a histogram:

# ── Part 5.2 · Data Transformations ─────────────────────────────────────
log_hist_plot <- in_df %>%
  ggplot(aes(x = log_mass, fill = lake)) +
  geom_histogram(bins = 20, alpha = 0.7) +
  labs(title = "Distribution of Log-Transformed Lake Trout Mass",
       x = "Log10 Mass", y = "Count") +
  theme_regular(base_size = 10) +
  facet_wrap(~lake, scales = "free_y")
log_hist_plot

▶ Run this — a QQ plot of the transformed data (we’ll skip Shapiro-Wilk for a moment):

# ── Part 5.3 · Data Transformations ─────────────────────────────────────
log_qq_plot <- in_df %>%
  ggplot(aes(sample = log_mass, color = lake)) +
  stat_qq() +
  stat_qq_line() +
  labs(title = "QQ Plot for Log-Transformed Data",
       x = "Theoretical Quantiles", y = "Sample Quantiles") +
  theme_regular(base_size = 10) +
  facet_wrap(~lake)
log_qq_plot

▶ Run this — now the formal tests on the transformed data:

# ── Part 5.4 · Data Transformations ─────────────────────────────────────
log_normality_results <- in_df %>%
  group_by(lake) %>%
  summarize(
    shapiro_stat = shapiro.test(log10(mass_g))$statistic,
    shapiro_p_value = shapiro.test(log10(mass_g))$p.value,
    normal_distribution = if_else(shapiro_p_value > 0.05, "Normal", "Non-normal")
  )
log_normality_results

levene_log_result <- leveneTest(log_mass ~ lake, data = in_df)
levene_log_result

✏️ Your turn: Did the log transformation fix normality, equal variance, both, or neither?

________________________

Tip

🚀 If you finish early: Try a square-root transformation (sqrt(mass_g)) instead of log₁₀. Build a histogram and rerun the Shapiro-Wilk test — does it do better or worse than the log transform?

# ── Part 5.5 · Data Transformations ─────────────────────────────────────
# Write your code in the R script:

Part 6 · Running the Tests

Even if the transformation doesn’t fully fix things, let’s run the full set of tests so we can compare them side by side.

▶ Run this — standard two-sample t-test on the raw data:

# ── Part 6.1 · Running the Tests ────────────────────────────────────────
t_test_result <- t.test(
  mass_g ~ lake,
  data = in_df,
  var.equal = TRUE,
  alternative = "two.sided"
)
t_test_result

▶ Run this — the same t-test on the log-transformed data:

# ── Part 6.2 · Running the Tests ────────────────────────────────────────
log_t_test_result <- t.test(
  log_mass ~ lake,
  data = in_df,
  var.equal = TRUE,
  alternative = "two.sided"
)
log_t_test_result

Note

When analyzing log-transformed data:

The mean of log-transformed data, back-transformed, gives the geometric mean (not the arithmetic mean).
The back-transformed confidence intervals represent the CI for the geometric mean, and must be calculated carefully.
Report results like: “The geometric mean mass of lake trout in NE 12 was X g (95% CI: Y–Z).”
You can’t just take 10^SE to get the standard error — instead, back-transform mean - se and mean + se separately.

▶ Run this — back-transform to the geometric mean:

# ── Part 6.3 · Running the Tests ────────────────────────────────────────
back_transformed <- in_df %>%
  group_by(lake) %>%
  summarise(
    n = n(),
    mean_log = mean(log_mass),
    sd_log = sd(log_mass),
    se_log = sd_log / sqrt(n),
    geometric_mean = 10^mean_log,
    lower_se = 10^(mean_log - se_log),
    upper_se = 10^(mean_log + se_log),
    arithmetic_mean = mean(mass_g)
  )

back_transformed %>%
  select(lake, mean_log, geometric_mean, arithmetic_mean)

▶ Run this — plot the back-transformed geometric means (note the error bars are not symmetrical!):

# ── Part 6.4 · Running the Tests ────────────────────────────────────────
geo_mean_plot <- back_transformed %>%
  ggplot(aes(x = lake, y = geometric_mean, fill = lake)) +
  geom_bar(stat = "identity", width = 0.5, alpha = 0.7) +
  geom_errorbar(aes(ymin = lower_se, ymax = upper_se), width = 0.2, linewidth = 1) +
  labs(title = "Geometric Mean Lake Trout Mass with Standard Error",
       subtitle = "Back-transformed from log10 scale",
       x = "Lake", y = "Geometric Mean Mass (g)") +
  theme_regular(base_size = 10) +
  theme(legend.position = "none")
geo_mean_plot

▶ Run this — Welch’s t-test (doesn’t assume equal variances):

# ── Part 6.5 · Running the Tests ────────────────────────────────────────
welch_test_result <- t.test(
  mass_g ~ lake,
  data = in_df,
  var.equal = FALSE,
  alternative = "two.sided"
)
welch_test_result

💡 Welch’s t-test is preferred when group variances are unequal (per Levene’s test), sample sizes differ between groups, or you just want a more robust default.

▶ Run this — the Mann-Whitney Wilcoxon test (non-parametric):

# ── Part 6.6 · Running the Tests ────────────────────────────────────────
wilcox_test_result <- wilcox.test(
  mass_g ~ lake,
  data = in_df,
  alternative = "two.sided"
)
wilcox_test_result

💡 Mann-Whitney is preferred when data isn’t normal even after transformation, you’re comparing medians rather than means, or outliers might distort a t-test — it compares ranks, not raw values.

▶ Run this — a permutation test (balance the sample sizes first):

# ── Part 6.7 · Running the Tests ────────────────────────────────────────
set.seed(123)  # For reproducibility

island_size <- sum(in_df$lake == "Island Lake")

ne12_sample <- in_df %>%
  filter(lake == "NE 12") %>%
  slice_sample(n = island_size)

balanced_df <- bind_rows(
  ne12_sample,
  in_df %>% filter(lake == "Island Lake")
)

perm_test_result <- permTS(
  mass_g ~ lake,
  data = balanced_df,
  alternative = "two.sided",
  method = "exact.mc",
  control = permControl(nmc = 10000)
)
perm_test_result

💡 Permutation tests are useful for small samples, data that doesn’t meet parametric assumptions, or when you want a test that makes minimal distributional assumptions — and they can test statistics other than the mean.

Tip

🚀 If you finish early: Rerun the permutation test with nmc = 1000 instead of 10000. Does the p-value change meaningfully? What’s the tradeoff of using fewer Monte Carlo replications?

# ── Part 6.8 · Running the Tests ────────────────────────────────────────
# Write your code in the R script:

Part 7 · Comparing All Results

▶ Run this — build a comparison table:

# ── Part 7.1 · Comparing All Results ────────────────────────────────────
test_results <- data.frame(
  Test = c("Standard t-test",
           "Log-transformed t-test",
           "Welch's t-test",
           "Mann-Whitney Wilcoxon test"),
  Statistic = c(paste("t =", round(t_test_result$statistic, 2)),
               paste("t =", round(log_t_test_result$statistic, 2)),
               paste("t =", round(welch_test_result$statistic, 2)),
               paste("W =", wilcox_test_result$statistic)),
  p_value = c(t_test_result$p.value,
             log_t_test_result$p.value,
             welch_test_result$p.value,
             wilcox_test_result$p.value),
  Significant = c(t_test_result$p.value < 0.05,
                 log_t_test_result$p.value < 0.05,
                 welch_test_result$p.value < 0.05,
                 wilcox_test_result$p.value < 0.05)
)
test_results

▶ Run this — visualize the raw comparison:

# ── Part 7.2 · Comparing All Results ────────────────────────────────────
combined_plot <- in_df %>%
  ggplot(aes(x = lake, y = mass_g, fill = lake)) +
  geom_boxplot(alpha = 0.7, outlier.shape = NA) +
  geom_jitter(width = 0.2, alpha = 0.5, size = 2) +
  labs(x = "Lake", y = "Mass (g)") +
  theme_regular(base_size = 10) +
  theme(legend.position = "none")
combined_plot

✏️ Your turn: Do all four tests agree on whether the difference is significant? If any disagree, what does that tell you about how sensitive the conclusion is to test choice?

________________________

When reporting results, include:

Standard t-test: “Lake trout from NE 12 had significantly different mass (M = [mean], SD = [SD]) compared to Island Lake (M = [mean], SD = [SD]), t([df]) = [t-value], p = [p-value].”
Log-transformed t-test: “After log transformation to meet normality assumptions, lake trout from NE 12 had significantly different mass (geometric mean = [value], 95% CI [lower–upper]) compared to Island Lake (geometric mean = [value], 95% CI [lower–upper]), t([df]) = [t-value], p = [p-value].”
Welch’s t-test: “Assuming unequal variances, lake trout from NE 12 had significantly different mass (M = [mean], SD = [SD]) compared to Island Lake (M = [mean], SD = [SD]), Welch’s t([df]) = [t-value], p = [p-value].”
Mann-Whitney Wilcoxon test: “Lake trout mass differed significantly between NE 12 (Mdn = [median]) and Island Lake (Mdn = [median]), W = [W-value], p = [p-value].”
Permutation test: “Permutation testing (10,000 iterations) revealed significant differences in lake trout mass between NE 12 and Island Lake, p = [p-value].”

✏️ Your turn: Write the standard t-test and Mann-Whitney reporting sentences using your own numbers from Parts 5–6.

________________________________________________________
________________________________________________________

Tip

🚀 If you finish early: Add the permutation test’s p-value as a fifth row to test_results. (Hint: perm_test_result$p.value.)

# ── Part 7.3 · Comparing All Results ────────────────────────────────────
# Write your code in the R script:

Part 8 · Take-Home Extension — Sagebrush Crickets

Due Wednesday, September 30 — before the Study Design & Sampling class.

📥 Sagebrush cricket data

This time the test itself is the question. You now know five ways to compare two groups — standard t-test, log-transformed t-test, Welch’s t-test, Mann-Whitney, and permutation test. Given a new dataset, you decide which one is appropriate and defend your choice.

Background

Male sagebrush crickets (Cyphoderris strepitans) have an unusual mating system: during copulation the female eats the tips of the male’s hind wings and drinks his haemolymph. A male who has already mated has damaged wings, and is therefore visibly “starved” relative to a virgin male.

The question is whether that costs him. Johnson et al. measured time to mating — how many hours a male waited before securing a mate — for males that had already mated (starved) and males that had not (fed).

Important

⚠️ This is Example 13.5 in Whitlock & Schluter

These are the data the textbook uses to introduce the Mann-Whitney U test in Chapter 13. That is deliberate: after you have made your own decision and written it up, you can read the chapter and see how the authors justified theirs.

Do the work first. The point of this assignment is your reasoning, not matching the book.

Column names in sage_brush_crickets_feeding_mating.csv

Column	What it is
`feeding_status`	`"fed"` (virgin male) or `"starved"` (already mated)
`time_to_mating_h`	hours the male waited before mating

▶ Run this — read the data in and look at it:

# ── Part 8.1 · Take-Home Extension ──────────────────────────────────────
cricket_df <- read_csv("data/sage_brush_crickets_feeding_mating.csv")

cricket_df %>% group_by(feeding_status) %>% reframe(summary_stats(time_to_mating_h))

✏️ Your turn: Look at the n, mean, and sd for each group before you do anything else. What do you notice about the two standard deviations relative to the two means?

________________________

Your Task

Question: Does time to mating differ between starved and fed male sagebrush crickets?

✏️ Your turn: State your hypotheses. Explore the data yourself — normality, variance, sample size, outliers, whatever you need to check — and decide which test from this unit is the right one for this data. Justify your choice, then run it, interpret the result, and write it up as a proper results sentence.

Warning

⚠️ This dataset will not let you off easily

Unlike the pine needles, the assumptions here are genuinely in trouble — and unlike the lake trout, the answer is not obvious. That is the whole point. Check every assumption, and notice that a log transformation fixes one problem without fixing the other. Your justification is the assignment.

Note

✅ You are graded on the reasoning, not on the p-value

A well-justified test that returns a non-significant result is a complete, full-credit answer. “We found no difference” is a finding. If your p-value is large, do not go hunting through the other four tests looking for a smaller one — that is p-hacking, and it is the one thing that will cost you marks here.

H0 =
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Ha =
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Test I will use and why:
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# ── Part 8.2 · Take-Home Extension ──────────────────────────────────────
# Write your code in the R script:

Interpretation:
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Results sentence:
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Final Figure

Produce one publication-quality figure comparing time to mating between the two feeding statuses — proper axis labels with units, no default ggplot grey background — and export it with ggsave().

# ── Part 8.3 · Take-Home Extension ──────────────────────────────────────
# Write your code in the R script:

Note

📤 What to turn in — due Sept 30

Your write-up should include:

The question
Your hypotheses
Which test you used, and your justification for choosing it
The code and its output
Your interpretation of the result
A final figure showing the result

Tip

🎓 Graduate students — optional extension

Once you have written yours up, read Whitlock & Schluter Chapter 13, Example 13.5 and compare. Did they reach for the same test you did? If not, what did they see in the data that you didn’t — or what did you see that they didn’t?

Then go further: with n = 11 and 13, what effect size would this study have been able to detect? Sketch the connection to the power ideas we will take up shortly.

Conclusion

Two datasets, two different answers to the same question — which test does this data deserve?

On the pine needles the assumptions held, and the rank-based tests cost us power for nothing. On the lake trout the assumptions failed badly, and those same tests were exactly the right tool. Non-parametric is not a safe default; it is a response to a specific problem.

When assumptions are violated:

Try transforming the data (e.g., log transformation)
Use Welch’s t-test if variances are unequal
Use non-parametric tests (Mann-Whitney or permutation tests) if data remains non-normal

All methods have strengths and limitations, and consistency of results across methods can strengthen your conclusions.

Note

When to use each test

Standard t-test: data is normally distributed with equal variances
Log-transformed t-test: raw data is skewed but log-transformation achieves normality
Welch’s t-test: variances are unequal
Mann-Whitney Wilcoxon test: data is not normal and cannot be transformed to normality
Permutation test: sample sizes are small, or assumptions cannot be met

Review and checkpoint

At this point you can:

Test normality with histograms, QQ plots, and the Shapiro-Wilk test
Explain what Cohen’s d measures, and why it belongs next to a p-value
Test equal variance with Levene’s test
Apply and evaluate a log transformation, including back-transforming to a geometric mean
Run and interpret a standard t-test, Welch’s t-test, Mann-Whitney Wilcoxon test, and permutation test
Compare results across test types and explain when to reach for each one

Note

📤 What to turn in before next class

Upload both of these to the course management system:

Your code — the scripts/ folder (or just 07_nonparametric_t_tests.R)
This worksheet, with your written answers

Homework

Homework 4 — Non-Parametric Tests

Rank-based tests on real published data: claw strength in stream-dwelling versus burrowing crayfish, asking whether a big claw is an honest signal of fighting ability or a bluff. You decide whether the parametric assumptions hold before choosing the test.

Data: graham_and_angilletta_claw_data.csv — from Graham & Angilletta (2022), Functional Ecology. Not a textbook dataset, so read the paper rather than a chapter.

Other Formats

Worksheet: Non-Parametric T-Tests

Part 1 · Our Pine Needles — Start With Data You Trust

Effect size — a demonstration

Part 2 · A Dataset Where the Assumptions Really Do Fail

Part 3 · Testing Assumptions on NE 12

Part 4 · Comparing Two Lakes — NE 12 vs. Island Lake

Part 5 · Data Transformations

Part 6 · Running the Tests

Part 7 · Comparing All Results

Part 8 · Take-Home Extension — Sagebrush Crickets

Background

Your Task

Final Figure

Conclusion

Review and checkpoint

Homework