7. t-Tests

One-Sample, Two-Sample, and Paired Mean Comparisons

Author

Affiliation

A. J. Smit

University of the Western Cape

Published

2026/04/05

In This Chapter

One-sample t-tests
Two-sample t-tests
Paired t-tests
Directional tests
Rank-based alternatives within the t-test family

Tasks to Complete in This Chapter

Self-Assessment Task 7-1 (/15)
Self-Assessment Task 7-2 (/10)
Self-Assessment instructions and full task overview

1 Introduction

t-tests compare means. The correct t-test is determined by the structure of the data:

one sample and no grouping variable;
two independent groups defined by a categorical variable with two levels;
paired observations from the same sampling units.

The statistical question is always something like, is the estimated mean, or the estimated mean difference, large enough relative to sampling variation to justify rejecting the null hypothesis? In practice, the main quantity of interest is the estimated effect itself, together with its confidence interval. The p-value supports that interpretation and should not replace it.

Below, I follow the same order throughout: identify the sampling or experimental design, inspect the data, check the assumptions, run the test, and interpret the effect size, uncertainty, and p-value together. With only two groups, the two-sample t-test is the two-group special case of one-way ANOVA. In Chapter 8, I extend the same idea to more than two groups and explain why repeated t-tests inflate Type I error.

2 Design Determines the Test

Consideration of the design comes first:

One-sample design: one continuous variable, no grouping factor, and a biologically meaningful reference value $\mu_0$.
Two-sample design: one continuous response and one categorical grouping variable with exactly two independent levels.
Paired design: two measurements linked within the same sampling unit, so the analysis is done on the within-pair differences.

Checking the assumptions follows:

The response should be continuous.
Observations should be independent in the way implied by the sampling or experimental design.
The relevant distribution should be reasonably well behaved.
In the two-sample case, the spread in the two groups should be assessed before deciding whether Student’s or Welch’s formulation is appropriate.

Approximate normality depends on what is being analysed. In a one-sample test, it concerns the sample values. In a two-sample test, it concerns the distribution within each group. In a paired test, it concerns the paired differences. Chapter 6 developed those checks directly on raw data.

Large samples make t-tests fairly robust to mild non-normality, especially when no single value dominates the result and the group spreads are not extreme. Independence is different. A large sample does not repair pseudoreplication, repeated use of the same individuals, or a poorly designed sampling scheme.

3 R Functions

The main function in this chapter is t.test(). It handles:

one-sample tests with mu = ...;
two independent groups using either x, y or a formula such as response ~ group;
paired tests with paired = TRUE.

The most useful arguments are:

mu for the one-sample reference mean;
alternative for "two.sided", "less", or "greater";
paired for matched observations;
var.equal to force Student’s two-sample t-test instead of Welch’s default.

When assumptions fail badly, wilcox.test() provides rank-based alternatives for one-sample, two-sample, and paired designs. The design still determines the correct procedure.

Mathematical Detail

For a one-sample t-test, the test statistic is:

\[t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \tag{1}\]

In Equation 1, $\bar{x}$ is the sample mean, $\mu_0$ is the hypothesised mean under the null hypothesis, $s$ is the sample standard deviation, and $n$ is the sample size.

For Student’s two-sample t-test, where a common variance is assumed:

\[t = \frac{\bar{A} - \bar{B}}{s_p\sqrt{\frac{1}{n_A} + \frac{1}{n_B}}} \tag{2}\]

with pooled variance

\[s_p^2 = \frac{(n_A - 1)s_A^2 + (n_B - 1)s_B^2}{n_A + n_B - 2} \tag{3}\]

For Welch’s two-sample t-test, which does not assume equal variances:

\[t = \frac{\bar{A} - \bar{B}}{\sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}} \tag{4}\]

with Welch-Satterthwaite degrees of freedom

\[\text{d.f.} = \frac{\left(\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}\right)^2}{\frac{(s_A^2 / n_A)^2}{n_A - 1} + \frac{(s_B^2 / n_B)^2}{n_B - 1}} \tag{5}\]

For a paired t-test, the data are reduced to the within-pair differences and the test becomes:

\[t = \frac{\bar{d}}{s_d/\sqrt{n}} \tag{6}\]

The formulas explain how the statistic is constructed. In practice, R does the arithmetic and we focus on whether the design, assumptions, and interpretation are appropriate.

4 A Simple Experiment

Do This Now!

Work with two friends and devise a real experiment, or a designed sampling exercise, in which you demonstrate the following two principles:

the effect of sample size on the estimate of the mean and variance (use SD),
the effect of repeated sampling on the value of the mean and variance.

In both cases, apply the correct t-test from this chapter to test whether sample size and repeat sampling have a statistically significant effect. Report back comprehensively with Task F.

Many biological questions reduce to a mean comparison. A treatment is compared with a control, a measured trait is compared with a reference value, or the same individuals are measured before and after an intervention. The sampling or experimental design may be determined by scientific question, but the inference always concerns comparison of means or mean differences.

5 One-Sample t-Tests

In a one-sample design there is no grouping variable. We have one continuous sample and a reference value, $\mu_0$, that comes from a threshold, a historical mean, a theoretical expectation, or some other biologically meaningful reference level.

5.1 When to Use a One-Sample t-Test

Use a one-sample t-test when:

you have one sample of observations;
the response variable is continuous;
the observations are independent;
you want to compare the population mean with a known or hypothesised value, $\mu_0$;
the sample is reasonably well behaved, or the sample size is large enough that the test is robust to small departures from normality.

For the standard two-sided form, the hypotheses are:

\[H_{0}: \mu = \mu_{0}\] \[H_{a}: \mu \ne \mu_{0}\]

The one-sample t-test then asks whether the observed sample mean is far enough from $\mu_0$, relative to sampling variation, to justify rejecting $H_0$.

6 Example 1: Tree Height Against Two Reference Means

The built-in trees dataset records black cherry trees, with height given in feet. Here I convert height to metres and evaluate two rounded reference means, 23.0 m and 24.5 m. The example shows how the same sample can lead to different inferences depending on the reference value being tested.

tree_one <- as_tibble(trees) |>
  transmute(height_m = Height * 0.3048, sample = "Black cherry")

6.1 Do an Exploratory Data Analysis (EDA)

This is a one-sample problem, so there is no grouping variable. I inspect the sample itself.

summary(tree_one$height_m)

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  19.20   21.95   23.16   23.16   24.38   26.52

sd(tree_one$height_m)

[1] 1.942129

shapiro.test(tree_one$height_m)


    Shapiro-Wilk normality test

data:  tree_one$height_m
W = 0.96545, p-value = 0.4034

I apply the assumption check to the sample values themselves because this is a one-sample design. The Shapiro-Wilk test shows no strong evidence against normality for tree height, and the summary statistics do not suggest any extreme outlier dominating the sample. That visual comparison is shown in Figure 1. That visual comparison is shown in Figure 1.

Code

ggplot(tree_one, aes(x = sample, y = height_m)) +
  geom_boxplot(fill = "indianred", alpha = 0.3, colour = "black") +
  geom_hline(yintercept = 23.0, colour = "dodgerblue2", linewidth = 0.9) +
  geom_hline(yintercept = 24.5, colour = "indianred2", linewidth = 0.9) +
  labs(x = NULL, y = "Height (m)") +
  coord_flip()

Figure 1: One-sample tree-height data in metres. The blue line marks a reference mean of 23.0 m and the red line marks 24.5 m.

The sample is centred close to 23.16 m. Visually, 23.0 m looks plausible and 24.5 m looks rather high. The assumption check supports proceeding with a one-sample t-test because the sample is reasonably well behaved and the sample size is moderate.

6.2 State the Hypotheses

For the reference value of 23.0 m:

\[H_{0}: \mu = 23.0\] \[H_{a}: \mu \ne 23.0\]

For the reference value of 24.5 m:

\[H_{0}: \mu = 24.5\] \[H_{a}: \mu \ne 24.5\]

6.3 Apply the Test

t_tree_75 <- t.test(tree_one$height_m, mu = 23.0)
t_tree_80 <- t.test(tree_one$height_m, mu = 24.5)

t_tree_75


    One Sample t-test

data:  tree_one$height_m
t = 0.47245, df = 30, p-value = 0.64
alternative hypothesis: true mean is not equal to 23
95 percent confidence interval:
 22.45242 23.87718
sample estimates:
mean of x 
  23.1648

t_tree_80


    One Sample t-test

data:  tree_one$height_m
t = -3.8278, df = 30, p-value = 0.0006115
alternative hypothesis: true mean is not equal to 24.5
95 percent confidence interval:
 22.45242 23.87718
sample estimates:
mean of x 
  23.1648

6.4 Interpret the Results

The mean tree height is 23.16 m. Relative to 23.0 m, that difference is small and the confidence interval includes the reference value. Relative to 24.5 m, the difference is larger and the confidence interval excludes the reference value. I therefore fail to reject $H_0$ when $\mu_0 = 23.0$, but reject $H_0$ when $\mu_0 = 24.5$.

This example shows why the null value must have biological meaning. The data have remained constant, but the inferential conclusion changes because the reference value has shifted.

6.5 Reporting

Write-Up

Methods

A one-sample t-test was used to compare the mean height of black cherry trees with two possible reference means, 23.0 m and 24.5 m. The original data were recorded in feet and converted to metres before analysis. The distribution of tree heights was inspected before the test was applied.

Results

The sample mean tree height was 23.16 m (SD = 1.94, $n = 31$). When tested against a reference mean of 23.0 m, the one-sample t-test provided no strong evidence for a difference ($t_{30} = 0.47$, $p > 0.05$), and the 95% confidence interval for the mean (22.45 to 23.88 m) included 23.0 m. When tested against 24.5 m, the same sample differed from the reference mean ($t_{30} = -3.83$, $p < 0.001$), and the same confidence interval excluded 24.5 m.

Discussion

The inference depends on the biological relevance of the reference value. These data are compatible with a mean tree height near 23.0 m and they reject a mean as high as 24.5 m.

Do It Now!

The built-in sleep dataset (type ?sleep) contains extra hours of sleep gained by patients under two drug treatments. Treat it as a one-sample test: for Drug 1, test whether the mean extra sleep is different from 0 (i.e., whether the drug has any effect on sleep duration).

Run the EDA: make a histogram and note whether the distribution looks roughly normal.
State $H_0$ and $H_a$.
Apply t.test() with mu = 0.
Report the result in one sentence following the format from the Write-Up example above.

7 Two-Sample t-Tests

In a two-sample design, the grouping variable has two levels and the observations in one group are independent of those in the other. Now I ask if the two population means are equal.

7.1 When to Use a Two-Sample t-Test

Use a two-sample t-test when:

the response variable is continuous;
the grouping variable has exactly two independent levels;
the observations in one group are independent of those in the other;
the within-group distributions are reasonably well behaved, or the sample sizes are large enough for mild departures from normality to be unimportant;
the biological question concerns the mean difference between the two groups.

For the standard two-sided form, the hypotheses are:

\[H_{0}: \mu_{A} = \mu_{B}\] \[H_{a}: \mu_{A} \ne \mu_{B}\]

or, equivalently,

\[H_{0}: \mu_{A} - \mu_{B} = 0\] \[H_{a}: \mu_{A} - \mu_{B} \ne 0\]

7.2 Choosing Between Student and Welch

Welch’s two-sample t-test should be the default. Use Student’s t-test only when the group spreads are similar and the sample sizes are also similar enough that the pooled-variance assumption is defensible.

Use this rule:

start with plots and grouped summaries;
if the spreads are clearly different, use Welch;
if the spreads are similar, Welch is still acceptable and often simplest;
use Student only when you want the pooled-variance form for a clear reason.

Welch’s test is robust, widely used, and already the default in t.test(). In most routine analyses there is no penalty for choosing it.

7.3 Other Tests for Homogeneity

Formal tests such as Levene’s test or Bartlett’s test provide supporting evidence about group spread. They should not become the main decision step. Visual comparison of spread comes first, and Welch remains the safer choice when the spreads differ.

If the apparent variance difference is tied to strong skewness or a long upper tail, a transformation may be preferable to a raw-scale t-test. In Chapter 6 I developed those diagnostic decisions, and the ozone example below demonstrates this in practice in the final test.

8 Example 2: Penguin Flipper Length in Two Species

The penguins dataset provides a realistic two-group comparison. Here I compare flipper length between Adelie and Chinstrap penguins. The grouping variable is species, and it has two independent levels in this subset.

peng_two <- penguins |>
  select(species, flipper_length_mm) |>
  drop_na() |>
  filter(species %in% c("Adelie", "Chinstrap")) |>
  mutate(species = factor(species, levels = c("Adelie", "Chinstrap")))

8.1 Do an Exploratory Data Analysis (EDA)

I first inspect the grouped summaries, the within-group normality checks, and the spread across groups.

peng_two |>
  group_by(species) |>
  summarise(
    n = n(),
    mean = mean(flipper_length_mm),
    sd = sd(flipper_length_mm),
    shapiro_p = shapiro.test(flipper_length_mm)$p.value,
    .groups = "drop"
  )

# A tibble: 2 × 5
  species       n  mean    sd shapiro_p
  <fct>     <int> <dbl> <dbl>     <dbl>
1 Adelie      151  190.  6.54     0.720
2 Chinstrap    68  196.  7.13     0.811

car::leveneTest(flipper_length_mm ~ species, data = peng_two)

Levene's Test for Homogeneity of Variance (center = median)
       Df F value Pr(>F)
group   1  0.6238 0.4305
      217

I apply the assumption checks within each species because the two-sample design compares group means. The Shapiro-Wilk tests show no strong evidence against normality within either species, and Levene’s test provides no strong evidence that the group spreads differ. Together with the boxplots in Figure 2, these checks support a standard two-group mean comparison.

Code

ggplot(peng_two, aes(x = species, y = flipper_length_mm, fill = species)) +
  geom_boxplot(alpha = 0.35, colour = "black", outlier.shape = NA) +
  geom_jitter(width = 0.08, alpha = 0.6, size = 1.5) +
  labs(x = "Species", y = "Flipper length (mm)") +
  theme(legend.position = "none")

Figure 2: Flipper length in Adelie and Chinstrap penguins. The group centres differ, while the group spreads are similar.

The boxplots show a clear shift in centre, with Chinstrap penguins tending to have longer flippers. The spreads are similar. The implication is that the assumptions do not force a change of method here, so a Welch two-sample t-test is a defensible default.

8.2 State the Hypotheses

\[H_{0}: \mu_{\text{Adelie}} = \mu_{\text{Chinstrap}}\] \[H_{a}: \mu_{\text{Adelie}} \ne \mu_{\text{Chinstrap}}\]

8.3 Apply the Test

peng_welch <- t.test(flipper_length_mm ~ species, data = peng_two)
peng_welch


    Welch Two Sample t-test

data:  flipper_length_mm by species
t = -5.7804, df = 119.68, p-value = 6.049e-08
alternative hypothesis: true difference in means between group Adelie and group Chinstrap is not equal to 0
95 percent confidence interval:
 -7.880530 -3.859244
sample estimates:
   mean in group Adelie mean in group Chinstrap 
               189.9536                195.8235

The formula flipper_length_mm ~ species tells R to compare the flipper lengths across the levels of the grouping variable species. This same formula notation appears again in ANOVA and regression chapters.

8.4 Interpret the Results

The estimated difference in mean flipper length is about 5.9 mm, with Chinstrap penguins having the larger mean. The 95% confidence interval excludes zero, and the p-value is very small. We therefore reject $H_0$ and conclude that the two species differ in mean flipper length.

The confidence interval shows the plausible range for the size of the difference, which is the biologically useful part of the result.

8.5 Reporting

Write-Up

Methods

Mean flipper length was compared between Adelie and Chinstrap penguins using a two-sample Welch t-test. Species was treated as a two-level grouping factor, and flipper length was the continuous response variable. Grouped summaries and graphical checks were used to inspect the data before the test was applied.

Results

Adelie penguins had a mean flipper length of 190.0 mm (SD = 6.54, $n = 151$), whereas Chinstrap penguins had a mean flipper length of 195.8 mm (SD = 7.13, $n = 68$). A two-sample Welch t-test indicated that mean flipper length differed between species ($t_{119.68} = -5.78$, $p < 0.001$). The estimated mean difference was 5.87 mm, and the 95% confidence interval ranged from 3.86 to 7.88 mm, with Chinstrap penguins having the larger mean.

Discussion

The estimated difference is several millimetres and the confidence interval remains well away from zero, so the species difference is biologically substantial as well. This is the two-group case that Chapter 8 generalises to more than two groups.

Do This Now!

Create a visualisation that makes the mean difference and the spread difference in this penguin example easy to compare at a glance.

Self-Assessment Task 7-1

Please refer to this two-sided two-sample t-test:

# random normal data
set.seed(666)
r_two <- data.frame(dat = c(rnorm(n = 20, mean = 4, sd = 1),
                            rnorm(n = 20, mean = 5, sd = 1)),
                    sample = c(rep("A", 20), rep("B", 20)))

# perform t-test
# note how we set the `var.equal` argument to TRUE because we know 
# our data has the same SD (they are simulated as such!)
t.test(dat ~ sample, data = r_two, var.equal = TRUE)


    Two Sample t-test

data:  dat by sample
t = -1.9544, df = 38, p-value = 0.05805
alternative hypothesis: true difference in means between group A and group B is not equal to 0
95 percent confidence interval:
 -1.51699175  0.02670136
sample estimates:
mean in group A mean in group B 
       4.001438        4.746584

Repeat this analyses using the Welch’s t.test(). (/5)
Repeat your analysis, above, using the even more old-fashioned Equation 4 in the lecture. Show the code, talk us through the step you followed to read the p-values off the table of t-statistics. (/10)

9 Example 3: A Two-Sample Comparison After Log Transformation

The ozone example from Chapter 6 can now be carried through to a t-test for inference. I compare ozone concentrations in May-June with those in August-September. In the airquality dataset, ozone is recorded in parts per billion (ppb). On the raw scale, the data are strongly right-skewed and the group variances differ. I apply a log transformation, which improves both features. So the two-sample comparison is done on the log scale and then back-transformed for reporting.

aq <- na.omit(airquality[, c("Ozone", "Month")]) |>
  as_tibble() |>
  filter(Month %in% c(5, 6, 8, 9)) |>
  mutate(
    season = factor(
      if_else(Month %in% c(5, 6), "May-June", "August-September")
    ),
    log_ozone = log(Ozone)
  )

9.1 Do an Exploratory Data Analysis (EDA)

First, I inspect the grouped distributions on the raw and transformed scales in Figure 3.

Code

plt_raw <- ggplot(aq, aes(season, Ozone, fill = season)) +
  geom_boxplot(alpha = 0.35, colour = "black", outlier.shape = NA) +
  geom_jitter(width = 0.08, alpha = 0.6, size = 1.5) +
  labs(x = NULL, y = "Ozone (ppb)") +
  theme(legend.position = "none")

plt_log <- ggplot(aq, aes(season, log_ozone, fill = season)) +
  geom_boxplot(alpha = 0.35, colour = "black", outlier.shape = NA) +
  geom_jitter(width = 0.08, alpha = 0.6, size = 1.5) +
  labs(x = NULL, y = "log(Ozone)") +
  theme(legend.position = "none")

ggarrange(plt_raw, plt_log, ncol = 2)

Figure 3: Ozone concentrations before and after log transformation. A) Raw ozone values (ppb) by season. B) Log-transformed ozone values by season.

My formal checks support the same conclusion:

aq |>
  group_by(season) |>
  summarise(
    raw_w = unname(shapiro.test(Ozone)$statistic),
    raw_p = shapiro.test(Ozone)$p.value,
    log_w = unname(shapiro.test(log_ozone)$statistic),
    log_p = shapiro.test(log_ozone)$p.value,
    raw_mean = mean(Ozone),
    geo_mean = exp(mean(log_ozone)),
    .groups = "drop"
  )

# A tibble: 2 × 7
  season           raw_w      raw_p log_w  log_p raw_mean geo_mean
  <fct>            <dbl>      <dbl> <dbl>  <dbl>    <dbl>    <dbl>
1 August-September 0.858 0.0000112  0.973 0.260      44.9     33.6
2 May-June         0.765 0.00000448 0.939 0.0537     25.1     18.6

car::leveneTest(Ozone ~ season, data = aq)

Levene's Test for Homogeneity of Variance (center = median)
      Df F value  Pr(>F)  
group  1  6.3393 0.01362 *
      88                  
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

car::leveneTest(log_ozone ~ season, data = aq)

Levene's Test for Homogeneity of Variance (center = median)
      Df F value Pr(>F)
group  1  0.0964 0.7569
      88

I apply the assumption checks within season on both the raw and transformed scales because this is still a two-group comparison. On the raw scale, both groups show strong evidence against normality and the variances differ. On the log scale, the group distributions are much closer to normal and the variance difference largely disappears. The implication is that the raw-scale comparison is a poor choice, while the transformed comparison is defensible.

9.2 State the Hypotheses

Because the analysis is now on the log scale, the hypotheses concern the mean log ozone concentration:

\[H_{0}: \mu_{\log,\text{May-June}} = \mu_{\log,\text{August-September}}\] \[H_{a}: \mu_{\log,\text{May-June}} \ne \mu_{\log,\text{August-September}}\]

Back-transformation changes the way the result is expressed. Equality on the log scale corresponds to a ratio of 1 on the original scale, so the null hypothesis becomes:

\[H_{0}: \frac{\text{geometric mean ozone in August-September}}{\text{geometric mean ozone in May-June}} = 1\]

The alternative hypothesis is that this ratio is not 1. A ratio of 1 means that the two periods have the same geometric mean ozone concentration in ppb. A ratio greater than 1 means that ozone is higher in August-September. A ratio less than 1 means that ozone is lower in August-September.

9.3 Apply the Test

Welch’s two-sample t-test remains the default choice in R and is appropriate here.

ozone_t <- t.test(log_ozone ~ season, data = aq)
ozone_t


    Welch Two Sample t-test

data:  log_ozone by season
t = 3.3038, df = 67.271, p-value = 0.001531
alternative hypothesis: true difference in means between group August-September and group May-June is not equal to 0
95 percent confidence interval:
 0.2350506 0.9523976
sample estimates:
mean in group August-September         mean in group May-June 
                      3.514982                       2.921258

aq |>
  group_by(season) |>
  summarise(
    geometric_mean = exp(mean(log_ozone)),
    .groups = "drop"
  )

# A tibble: 2 × 2
  season           geometric_mean
  <fct>                     <dbl>
1 August-September           33.6
2 May-June                   18.6

ratio_est <- exp(ozone_t$estimate[[1]] - ozone_t$estimate[[2]])
ratio_ci <- exp(ozone_t$conf.int)

tibble(
  comparison = "August-September / May-June",
  ratio = ratio_est,
  lower = ratio_ci[1],
  upper = ratio_ci[2]
)

# A tibble: 1 × 4
  comparison                  ratio lower upper
  <chr>                       <dbl> <dbl> <dbl>
1 August-September / May-June  1.81  1.26  2.59

9.4 Interpret the Results

The test on the log scale compares mean log ozone concentrations between the two periods. After back-transformation, the result is easier to interpret as a ratio. August-September ozone concentrations are estimated to be about 1.81 times those in May-June, with a 95% confidence interval from 1.26 to 2.59.

The geometric mean is the back-transformed mean from the log scale. It differs from the arithmetic mean in that it summarises multiplicative variation rather than additive variation. The arithmetic mean is the ordinary average obtained by adding values and dividing by the sample size. The geometric mean is obtained by averaging the logged values and then back-transforming. For right-skewed data, it is often less influenced by the upper tail than the arithmetic mean.

In this example, I show the complete sequence of decision steps. I start with diagnosing the assumption violation on the raw scale, then I transform the data for a clear reason, I re-assess the assumptions, apply the test, and then I return the result to the original scale for reporting.

9.5 Reporting

Write-Up

Methods

Ozone concentrations (ppb) were compared between May-June and August-September using a two-sample t-test. Exploratory plots and assumption checks showed strong right-skew and unequal variances on the raw scale, so ozone was log-transformed before analysis. A Welch two-sample t-test was then applied to log-transformed ozone concentrations. Results were back-transformed and reported as geometric means and their ratio.

Results

On the transformed scale, ozone concentrations differed between the two periods (Welch two-sample t-test: $t_{67.27} = 3.30$, $p < 0.01$). The geometric mean ozone concentration was 18.6 ppb in May-June and 33.6 ppb in August-September. After back-transformation, August-September ozone concentrations were estimated to be 1.81 times those in May-June (95% CI 1.26 to 2.59).

Discussion

The transformed analysis supports higher ozone concentrations in August-September than in May-June. Because the model was fitted on the log scale, the clearest interpretation is multiplicative: the later period has ozone concentrations about 1.8 times higher.

10 Paired t-Tests

In a paired design, the observations are linked. The pairing may come from repeated measurements on the same individuals, left-right comparisons on the same organism, or deliberately matched experimental units. The analysis is done on the within-pair differences.

10.1 When to Use a Paired t-Test

Use a paired t-test when:

the response variable is continuous;
the two measurements are meaningfully paired within the same sampling unit;
the pairs are independent of one another;
the paired differences are reasonably well behaved.

For the standard two-sided form, the hypotheses are:

\[H_{0}: \mu_{d} = 0\] \[H_{a}: \mu_{d} \ne 0\]

where $\mu_d$ is the population mean of the paired differences.

11 Example 4: Before-and-After Measurements on Piglets

I will use a small example in which piglets are weighed before and after a three-week feeding period. The inferential question is whether body mass increases over time within the same individuals.

set.seed(666)
piglets <- data.frame(
  id = 1:12,
  before = round(rnorm(12, mean = 8.2, sd = 0.7), 2),
  after = round(rnorm(12, mean = 9.0, sd = 0.7), 2)
)

piglets$diff <- piglets$after - piglets$before
piglets

   id before after  diff
1   1   8.73  9.61  0.88
2   2   9.61  7.80 -1.81
3   3   7.95  9.09  1.14
4   4   9.62  8.95 -0.67
5   5   6.65  9.60  2.95
6   6   8.73  9.24  0.51
7   7   7.29  8.59  1.30
8   8   7.64  9.55  1.91
9   9   6.95  8.52  1.57
10 10   8.17  8.17  0.00
11 11   9.71  9.89  0.18
12 12   6.96  8.78  1.82

11.1 Do an Exploratory Data Analysis (EDA)

Before running the paired t-test, I inspect the paired differences because these are the values used directly by the test.

summary(piglets$diff)

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 -1.810   0.135   1.010   0.815   1.633   2.950

sd(piglets$diff)

[1] 1.277024

shapiro.test(piglets$diff)


    Shapiro-Wilk normality test

data:  piglets$diff
W = 0.9798, p-value = 0.9828

I apply the assumption check to the paired differences not to the raw before and after measurements. The Shapiro-Wilk test shows no strong evidence against normality in those differences, so I can use the paired t-test without changing method. The distribution of differences is shown in Figure 4. The distribution of differences is shown in Figure 4.

Code

ggplot(data = piglets, aes(x = diff)) +
  geom_histogram(binwidth = 0.5, fill = "indianred", colour = "black", alpha = 0.5) +
  geom_vline(xintercept = 0, colour = "dodgerblue2", linewidth = 0.9) +
  labs(x = "Paired difference in mass (after - before)", y = "Frequency")

Figure 4: Distribution of paired differences in piglet body mass after minus before the feeding period.

The paired differences are centred above zero, which already suggests that the piglets tended to gain mass over the feeding period. The distribution is well behaved enough for a paired t-test, so the next step is to test whether the mean difference is different from zero.

11.2 State the Hypotheses

\[H_{0}: \mu_{d} = 0\] \[H_{a}: \mu_{d} \ne 0\]

If the biological question had specified direction in advance, such as whether mass increased after feeding, the same paired design could be analysed with a one-sided alternative.

11.3 Apply the Test

t.test(piglets$after, piglets$before, paired = TRUE)


    Paired t-test

data:  piglets$after and piglets$before
t = 2.2108, df = 11, p-value = 0.04915
alternative hypothesis: true mean difference is not equal to 0
95 percent confidence interval:
 0.003617508 1.626382492
sample estimates:
mean difference 
          0.815

The paired t-test is a one-sample test applied to the set of paired differences. That is why the raw before and after values are not analysed as though they were independent groups.

11.4 Interpret the Results

The mean paired difference is positive, so the piglets tended to be heavier after the feeding period. The confidence interval for the mean difference lies just above zero and the p-value is slightly below 0.05. We therefore reject $H_0$ and conclude that body mass increased over the feeding period in this example.

11.5 Reporting

Write-Up

Methods

Piglet body mass was measured before and after a three-week feeding period, and the resulting paired observations were analysed with a paired t-test. The analysis was based on the within-individual differences in mass.

Results

Piglet body mass increased over the three-week feeding period. A paired t-test showed that the mean within-individual change in body mass differed from zero ($t_{11} = 2.21$, $p < 0.05$). The mean paired difference was 0.815 kg, with a 95% confidence interval from 0.004 to 1.626 kg.

Discussion

The paired design strengthens the inference because each piglet acts as its own control. The biological conclusion therefore concerns change within individuals rather than a difference between independent groups.

12 When Assumptions Are Difficult

The response to assumption problems depends on the sampling and experimental design and on the source of the problem.

Situation	Typical response
One sample, small $n$, strong skew or extreme outliers	Consider transformation or a one-sample Wilcoxon procedure
Two groups, similar shape but different spreads	Use Welch’s t-test
Two groups, skewness linked to spread	Transform, re-check, then analyse if the new scale is defensible
Two groups, strong outliers or badly distorted shapes	Consider a Wilcoxon rank-sum procedure
Paired data, non-normal paired differences	Consider a paired Wilcoxon signed-rank procedure

Use Welch when the main problem is unequal spread. Use a transformation when the problem is scale, such as a long right tail that also drives a variance difference. Use a rank-based alternative when the mean-based comparison itself is no longer a good description of the biological question.

Keep the following rules in mind:

do not switch to Student’s t-test merely because Levene’s test is non-significant;
do not treat a formal normality test as the only decision criterion;
do not transform data automatically; transform for a stated reason and then re-check;
do not forget that a large sample helps with mild non-normality but does not fix non-independence.

13 Optional: One-Sided t-Tests

One-sided tests should be used only when the direction of the effect is specified before the data are analysed. If the real question is whether two means differ, use the standard two-sided form.

13.1 One-Sided One-Sample t-Test

Suppose the forestry question is directional: are the sampled trees shorter than 24.5 m on average?

The hypotheses are:

\[H_{0}: \mu \ge 24.5\] \[H_{a}: \mu < 24.5\]

t.test(tree_one$height_m, mu = 24.5, alternative = "less")


    One Sample t-test

data:  tree_one$height_m
t = -3.8278, df = 30, p-value = 0.0003058
alternative hypothesis: true mean is less than 24.5
95 percent confidence interval:
     -Inf 23.75683
sample estimates:
mean of x 
  23.1648

t.test(tree_one$height_m, mu = 24.5, alternative = "greater")


    One Sample t-test

data:  tree_one$height_m
t = -3.8278, df = 30, p-value = 0.9997
alternative hypothesis: true mean is greater than 24.5
95 percent confidence interval:
 22.57277      Inf
sample estimates:
mean of x 
  23.1648

The first test matches the direction implied by the data and the question, but the second does not. Direction changes the hypothesis, the tail area, and the interpretation.

13.2 One-Sided Two-Sample t-Test

Suppose the biological question is directional: are Adelie penguin flippers shorter than Chinstrap penguin flippers?

The hypotheses are:

\[H_{0}: \mu_{\text{Adelie}} \ge \mu_{\text{Chinstrap}}\] \[H_{a}: \mu_{\text{Adelie}} < \mu_{\text{Chinstrap}}\]

t.test(flipper_length_mm ~ species, data = peng_two, alternative = "less")


    Welch Two Sample t-test

data:  flipper_length_mm by species
t = -5.7804, df = 119.68, p-value = 3.025e-08
alternative hypothesis: true difference in means between group Adelie and group Chinstrap is less than 0
95 percent confidence interval:
      -Inf -4.186534
sample estimates:
   mean in group Adelie mean in group Chinstrap 
               189.9536                195.8235

t.test(flipper_length_mm ~ species, data = peng_two, alternative = "greater")


    Welch Two Sample t-test

data:  flipper_length_mm by species
t = -5.7804, df = 119.68, p-value = 1
alternative hypothesis: true difference in means between group Adelie and group Chinstrap is greater than 0
95 percent confidence interval:
 -7.55324      Inf
sample estimates:
   mean in group Adelie mean in group Chinstrap 
               189.9536                195.8235

Again, the test only makes sense when the direction was specified before the analysis. One-sided tests are not a rescue strategy for a non-significant two-sided result.

14 Rank-Based Alternatives Within the t-Test Family

The parametric t-tests in this chapter are mean-based procedures. When the response is badly non-normal in a small sample, strongly skewed, or dominated by influential outliers, a rank-based alternative may be more defensible. These methods belong to the same inferential family as the t-tests because they serve the same one-sample, two-sample, or paired designs.

The three main rank-based counterparts are:

one-sample Wilcoxon signed-rank test for a single sample compared with a reference value;
Wilcoxon rank-sum test for two independent groups;
paired Wilcoxon signed-rank test for matched or before-after observations.

All three are handled by wilcox.test() in R:

wilcox.test(x, mu = ...) for a one-sample problem;
wilcox.test(x, y) or wilcox.test(response ~ group, data = ...) for two independent groups;
wilcox.test(x, y, paired = TRUE) for paired data.

These procedures are often introduced as alternatives when the mean is a poor summary of the data. They should not be described as direct tests of medians in every case. Their interpretation depends on the design and on the structure of the distributions being compared.

Use the Design, Not Just the Function Name

The one-sample and paired Wilcoxon procedures are often confused because they use the same function. The distinction is exactly the same as for t-tests:

one sample against a reference value → one-sample signed-rank test;
two related measurements on the same units → paired signed-rank test;
two independent groups → rank-sum test.

15 Decision Steps for Any t-Test

For most problems in this chapter, the workflow is:

identify the design: one sample, two independent groups, or paired data;
visualise the data;
check the relevant assumptions for shape, spread, and independence;
choose the test, with Welch as the default for two groups;
run the test;
interpret the estimated effect, its confidence interval, and the p-value together.

If you follow these steps you’ll prevent a large number of common mistakes.

Do It Now!

Work through the six decision steps above using the built-in mtcars dataset. Compare fuel economy (mpg) between cars with automatic (am = 0) and manual (am = 1) transmissions.

Step 1: What is the design?
Step 2: Make a grouped box plot of mpg ~ factor(am).
Step 3: Apply Shapiro-Wilk within each group and Levene’s test across groups.
Step 4: Choose Welch’s or Student’s t-test based on step 3.
Step 5: Run t.test(mpg ~ factor(am), data = mtcars, var.equal = ...).
Step 6: Write a one-sentence results statement with the test statistic, df, p-value, and 95% CI.

16 A t-Test Workflow

Now that the individual cases are in place, I can walk through a complete analysis in R. For this example we use the ecklonia data from Intro R Workshop: Data Manipulation, Analysis, Graphing.

16.1 Loading Data

Before I analyse anything, I need to load the data and reshape them into long format.

ecklonia <- read_csv(here::here("data", "BCB744", "ecklonia.csv")) |>
  gather(key = "variable", value = "value", -species, -site, -ID)

16.2 Visualising Data

I visualise the data first in Figure 5.

Code

ggplot(data = ecklonia, aes(x = variable, y = value, fill = site)) +
  geom_boxplot(colour = "black", fill = "dodgerblue4", alpha = 0.4) +
  coord_flip()

Figure 5: Boxplots showing differences in morphometric properties of the kelp *Ecklonia maxima* at two sites in False Bay.

The measurements occupy very different scales, which makes the full figure difficult to compare directly between sites. But, nevertheless, the figure’s value is orientating me to the variables in the dataset and showing me that a more focused comparison is needed before I can state a hypothesis clearly.

16.3 Formulating a Hypothesis

Let me narrow the question to stipe mass and focus only on two sites, Batsata Rock and Boulders Beach, as shown in Figure 6.

Code

ecklonia_sub <- ecklonia |>
  filter(variable == "stipe_mass")

ggplot(data = ecklonia_sub, aes(x = variable, y = value, fill = site)) +
  geom_boxplot(colour = "black", alpha = 0.4) +
  coord_flip() +
  labs(y = "Stipe mass (kg)", x = "") +
  theme(axis.text.y = element_blank(),
        axis.ticks.y = element_blank())

Figure 6: Boxplots showing the difference in stipe mass (kg) of the kelp *Ecklonia maxima* at two sites in False Bay.

I can state the directional question as: are kelp stipes heavier at Batsata Rock than at Boulders Beach?

The corresponding hypotheses are:

\[H_{0}: \mu_{\text{Batsata}} \le \mu_{\text{Boulders}}\] \[H_{a}: \mu_{\text{Batsata}} > \mu_{\text{Boulders}}\]

16.4 Choosing a Test

There are two independent groups, one continuous response, and a directional hypothesis specified in advance. That combination points to a one-sided two-sample t-test. Because the group spreads appear similar, Student’s formulation is acceptable here, although Welch would also be defensible.

16.5 Checking Assumptions

The response is continuous and the sites define the two independent groups. I still need to inspect spread and approximate normality within groups.

ecklonia_sub |>
  group_by(site) |>
  summarise(
    stipe_mass_var = var(value),
    stipe_mass_norm = shapiro.test(value)$p.value,
    .groups = "drop"
  )

# A tibble: 2 × 3
  site           stipe_mass_var stipe_mass_norm
  <chr>                   <dbl>           <dbl>
1 Batsata Rock             2.00           0.813
2 Boulders Beach           2.64           0.527

The grouped summaries suggest similar spread and no obvious normality problem. If these checks had failed badly, Chapter 6 provides the next steps.

16.6 Running an Analysis

t.test(value ~ site, data = ecklonia_sub,
       var.equal = TRUE, alternative = "greater")


    Two Sample t-test

data:  value by site
t = 1.8741, df = 24, p-value = 0.03657
alternative hypothesis: true difference in means between group Batsata Rock and group Boulders Beach is greater than 0
95 percent confidence interval:
 0.09752735        Inf
sample estimates:
  mean in group Batsata Rock mean in group Boulders Beach 
                    6.116154                     4.996154

16.7 Interpreting the Results

The output gives the test statistic, degrees of freedom, p-value, and confidence interval for the mean difference. If the one-sided p-value is small enough, I reject $H_0$ and conclude that mean stipe mass is greater at Batsata Rock. The confidence interval gives the plausible size of that difference.

16.8 Drawing Conclusions

The scientific conclusion should consider what local ecological or hydrodynamic conditions could account for the difference in stipe mass between the two sites.

16.9 Reporting

Write-Up

Methods

Stipe mass of Ecklonia maxima was compared between Batsata Rock and Boulders Beach using a one-sided two-sample Student’s t-test. The analysis treated site as the grouping factor and stipe mass as the continuous response variable, after checking that the group spreads were similar and that the data were approximately normal within each site.

Results

Mean stipe mass of Ecklonia maxima was higher at Batsata Rock ($6.12 \pm 1.41$ kg SD, $n = 13$) than at Boulders Beach ($5.00 \pm 1.63$ kg SD, $n = 13$). A one-sided two-sample Student’s t-test indicated that stipe mass at Batsata Rock was greater than at Boulders Beach ($t_{24} = 1.87$, $p < 0.05$). The one-sided 95% confidence interval for the mean difference ranged from 0.10 kg to $\infty$.

Discussion

The biological conclusion is that kelp at Batsata Rock had heavier stipes than kelp at Boulders Beach in this comparison. The next step would be to explain why that difference exists.

16.10 Going Further

In this example, I isolated two sites so that the comparison could fit a two-sample t-test. The full dataset contains more than two sites, however, and that changes the analysis. Once the question becomes whether mean stipe mass differs across several (i.e., > 2) sites, the correct next step is ANOVA rather than a bunch of pairwise t-tests. In Chapter 8, I develop that extension and explain why repeated two-group testing inflates Type I error.

The biological questions also inflates at that point. If stipe mass differs among several sites, what combination of exposure, nutrient regime, competition, or age structure produced that pattern?

Self-Assessment Task 7-2

For each scenario below, identify the correct t-test design (one-sample, two-sample, or paired) and state which assumption needs to be checked, and where:

You measure the resting heart rates of 25 athletes and compare them with the published population mean of 72 bpm. (/2)
You compare the nitrogen content of leaves from trees in two adjacent forest types (20 trees per forest). (/2)
You weigh 15 juvenile tortoises in March and again in October and ask whether they gained mass. (/2)
You measure the length of left and right tibiae from 18 rabbits. (/2)

Discuss with a partner whether (b) and (d) are structurally similar or different. What changes between them? (/2)

17 Conclusion

t-tests are among the first inferential tools used in biostatistics because they connect directly to common biological questions. The main decision is design: one sample, two independent groups, or paired observations. From there, the workflow simply requires that you inspect the data, check the assumptions, choose the appropriate test, and interpret the estimated effect with its uncertainty.

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Citation

BibTeX citation:

@online{smit2026,
  author = {Smit, A. J.},
  title = {7. {*t*-Tests}},
  date = {2026-04-05},
  url = {https://tangledbank.netlify.app/BCB744/basic_stats/07-t_tests.html},
  langid = {en}
}

For attribution, please cite this work as:

Smit AJ (2026) 7. *t*-Tests. https://tangledbank.netlify.app/BCB744/basic_stats/07-t_tests.html.

--- date: last-modified date-format: "YYYY/MM/DD" title: "7. *t*-Tests" subtitle: "One-Sample, Two-Sample, and Paired Mean Comparisons" reference-location: margin --- ```{r code-brewing-opts, echo=FALSE} knitr::opts_chunk$set( comment = "R>", warning = FALSE, message = FALSE, fig.asp = NULL, fig.align = "center", fig.retina = 2, dpi = 300 ) ggplot2::theme_set( ggplot2::theme_grey(base_size = 8) ) ``` ```{r code-knitr-opts-chunk-set, echo=FALSE} library(tidyverse) library(ggpubr) library(palmerpenguins) ``` ::: {.callout-note appearance="simple"} ## In This Chapter - One-sample *t*-tests - Two-sample *t*-tests - Paired *t*-tests - Directional tests - Rank-based alternatives within the *t*-test family ::: ::: {.callout-important appearance="simple"} ## Tasks to Complete in This Chapter - Self-Assessment Task 7-1 **(/15)** - Self-Assessment Task 7-2 **(/10)** - [Self-Assessment instructions and full task overview](../tasks/BCB744_Biostats_Self-Assessment.qmd) ::: # Introduction *t*-tests compare means. The correct *t*-test is determined by the structure of the data: - one sample and no grouping variable; - two independent groups defined by a categorical variable with two levels; - paired observations from the same sampling units. The statistical question is always something like, is the estimated mean, or the estimated mean difference, large enough relative to sampling variation to justify rejecting the null hypothesis? In practice, the main quantity of interest is the estimated effect itself, together with its confidence interval. The *p*-value supports that interpretation and should not replace it. Below, I follow the same order throughout: identify the sampling or experimental design, inspect the data, check the assumptions, run the test, and interpret the effect size, uncertainty, and *p*-value together. With only two groups, the two-sample *t*-test is the two-group special case of one-way ANOVA. In [Chapter 8](08-anova.qmd), I extend the same idea to more than two groups and explain why repeated *t*-tests inflate Type I error. # Design Determines the Test Consideration of the design comes first: - **One-sample design:** one continuous variable, no grouping factor, and a biologically meaningful reference value $\mu_0$. - **Two-sample design:** one continuous response and one categorical grouping variable with exactly two independent levels. - **Paired design:** two measurements linked within the same sampling unit, so the analysis is done on the within-pair differences. Checking the assumptions follows: - The response should be continuous. - Observations should be independent in the way implied by the sampling or experimental design. - The relevant distribution should be reasonably well behaved. - In the two-sample case, the spread in the two groups should be assessed before deciding whether Student's or Welch's formulation is appropriate. Approximate normality depends on what is being analysed. In a one-sample test, it concerns the sample values. In a two-sample test, it concerns the distribution within each group. In a paired test, it concerns the paired differences. [Chapter 6](06-assumptions-and-transformations.qmd) developed those checks directly on raw data. Large samples make *t*-tests fairly robust to mild non-normality, especially when no single value dominates the result and the group spreads are not extreme. Independence is different. A large sample does not repair pseudoreplication, repeated use of the same individuals, or a poorly designed sampling scheme. # R Functions The main function in this chapter is `t.test()`. It handles: - one-sample tests with `mu = ...`; - two independent groups using either `x, y` or a formula such as `response ~ group`; - paired tests with `paired = TRUE`. The most useful arguments are: - `mu` for the one-sample reference mean; - `alternative` for `"two.sided"`, `"less"`, or `"greater"`; - `paired` for matched observations; - `var.equal` to force Student's two-sample *t*-test instead of Welch's default. When assumptions fail badly, `wilcox.test()` provides rank-based alternatives for one-sample, two-sample, and paired designs. The design still determines the correct procedure. ::: {.callout-note collapse="true" appearance="simple"} ## Mathematical Detail For a one-sample *t*-test, the test statistic is: $$t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$$ {#eq-onesample} In @eq-onesample, $\bar{x}$ is the sample mean, $\mu_0$ is the hypothesised mean under the null hypothesis, $s$ is the sample standard deviation, and $n$ is the sample size. For Student's two-sample *t*-test, where a common variance is assumed: $$t = \frac{\bar{A} - \bar{B}}{s_p\sqrt{\frac{1}{n_A} + \frac{1}{n_B}}}$$ {#eq-student-t} with pooled variance $$s_p^2 = \frac{(n_A - 1)s_A^2 + (n_B - 1)s_B^2}{n_A + n_B - 2}$$ {#eq-pooledvar} For Welch's two-sample *t*-test, which does not assume equal variances: $$t = \frac{\bar{A} - \bar{B}}{\sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}}$$ {#eq-welch} with Welch-Satterthwaite degrees of freedom $$\text{d.f.} = \frac{\left(\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}\right)^2}{\frac{(s_A^2 / n_A)^2}{n_A - 1} + \frac{(s_B^2 / n_B)^2}{n_B - 1}}$$ {#eq-satterthwaite} For a paired *t*-test, the data are reduced to the within-pair differences and the test becomes: $$t = \frac{\bar{d}}{s_d/\sqrt{n}}$$ {#eq-paired-t} The formulas explain how the statistic is constructed. In practice, R does the arithmetic and we focus on whether the design, assumptions, and interpretation are appropriate. ::: # A Simple Experiment ::: callout-important ## Do This Now! Work with two friends and devise a real experiment, or a designed sampling exercise, in which you demonstrate the following two principles: a. the effect of sample size on the estimate of the mean and variance (use SD), b. the effect of repeated sampling on the value of the mean and variance. In both cases, apply the correct *t*-test from this chapter to test whether sample size and repeat sampling have a statistically significant effect. Report back comprehensively with Task F. ::: Many biological questions reduce to a mean comparison. A treatment is compared with a control, a measured trait is compared with a reference value, or the same individuals are measured before and after an intervention. The sampling or experimental design may be determined by scientific question, but the inference always concerns comparison of means or mean differences. # One-Sample *t*-Tests In a one-sample design there is no grouping variable. We have one continuous sample and a reference value, $\mu_0$, that comes from a threshold, a historical mean, a theoretical expectation, or some other biologically meaningful reference level. ## When to Use a One-Sample *t*-Test Use a one-sample *t*-test when: - you have one sample of observations; - the response variable is continuous; - the observations are independent; - you want to compare the population mean with a known or hypothesised value, $\mu_0$; - the sample is reasonably well behaved, or the sample size is large enough that the test is robust to small departures from normality. For the standard two-sided form, the hypotheses are: $$H_{0}: \mu = \mu_{0}$$ $$H_{a}: \mu \ne \mu_{0}$$ The one-sample *t*-test then asks whether the observed sample mean is far enough from $\mu_0$, relative to sampling variation, to justify rejecting $H_0$. # Example 1: Tree Height Against Two Reference Means The built-in `trees` dataset records black cherry trees, with height given in feet. Here I convert height to metres and evaluate two rounded reference means, 23.0 m and 24.5 m. The example shows how the same sample can lead to different inferences depending on the reference value being tested. ```{r code-tree-one} tree_one <- as_tibble(trees) |> transmute(height_m = Height * 0.3048, sample = "Black cherry") ``` ## Do an Exploratory Data Analysis (EDA) This is a one-sample problem, so there is no grouping variable. I inspect the sample itself. ```{r code-tree-one-summary} summary(tree_one$height_m) sd(tree_one$height_m) shapiro.test(tree_one$height_m) ``` I apply the assumption check to the sample values themselves because this is a one-sample design. The Shapiro-Wilk test shows no strong evidence against normality for tree height, and the summary statistics do not suggest any extreme outlier dominating the sample. That visual comparison is shown in @fig-tree-one. That visual comparison is shown in @fig-tree-one. ```{r fig-tree-one} #| fig-cap: "One-sample tree-height data in metres. The blue line marks a reference mean of 23.0 m and the red line marks 24.5 m." #| fig-width: 4 #| fig-height: 3 #| code-fold: true ggplot(tree_one, aes(x = sample, y = height_m)) + geom_boxplot(fill = "indianred", alpha = 0.3, colour = "black") + geom_hline(yintercept = 23.0, colour = "dodgerblue2", linewidth = 0.9) + geom_hline(yintercept = 24.5, colour = "indianred2", linewidth = 0.9) + labs(x = NULL, y = "Height (m)") + coord_flip() ``` The sample is centred close to 23.16 m. Visually, 23.0 m looks plausible and 24.5 m looks rather high. The assumption check supports proceeding with a one-sample *t*-test because the sample is reasonably well behaved and the sample size is moderate. ## State the Hypotheses For the reference value of 23.0 m: $$H_{0}: \mu = 23.0$$ $$H_{a}: \mu \ne 23.0$$ For the reference value of 24.5 m: $$H_{0}: \mu = 24.5$$ $$H_{a}: \mu \ne 24.5$$ ## Apply the Test ```{r code-tree-one-test} t_tree_75 <- t.test(tree_one$height_m, mu = 23.0) t_tree_80 <- t.test(tree_one$height_m, mu = 24.5) t_tree_75 t_tree_80 ``` ## Interpret the Results The mean tree height is 23.16 m. Relative to 23.0 m, that difference is small and the confidence interval includes the reference value. Relative to 24.5 m, the difference is larger and the confidence interval excludes the reference value. I therefore fail to reject $H_0$ when $\mu_0 = 23.0$, but reject $H_0$ when $\mu_0 = 24.5$. This example shows why the null value must have biological meaning. The data have remained constant, but the inferential conclusion changes because the reference value has shifted. ## Reporting ::: {.callout-note appearance="simple"} ## Write-Up **Methods** A one-sample *t*-test was used to compare the mean height of black cherry trees with two possible reference means, 23.0 m and 24.5 m. The original data were recorded in feet and converted to metres before analysis. The distribution of tree heights was inspected before the test was applied. **Results** The sample mean tree height was 23.16 m (SD = 1.94, $n = 31$). When tested against a reference mean of 23.0 m, the one-sample *t*-test provided no strong evidence for a difference ($t_{30} = 0.47$, $p > 0.05$), and the 95% confidence interval for the mean (22.45 to 23.88 m) included 23.0 m. When tested against 24.5 m, the same sample differed from the reference mean ($t_{30} = -3.83$, $p < 0.001$), and the same confidence interval excluded 24.5 m. **Discussion** The inference depends on the biological relevance of the reference value. These data are compatible with a mean tree height near 23.0 m and they reject a mean as high as 24.5 m. ::: ::: callout-important ## Do It Now! The built-in `sleep` dataset (type `?sleep`) contains extra hours of sleep gained by patients under two drug treatments. Treat it as a one-sample test: for Drug 1, test whether the mean extra sleep is different from 0 (*i.e.*, whether the drug has any effect on sleep duration). a. Run the EDA: make a histogram and note whether the distribution looks roughly normal. b. State $H_0$ and $H_a$. c. Apply `t.test()` with `mu = 0`. d. Report the result in one sentence following the format from the Write-Up example above. ````{=html}  ```` ::: # Two-Sample *t*-Tests In a two-sample design, the grouping variable has two levels and the observations in one group are independent of those in the other. Now I ask if the two population means are equal. ## When to Use a Two-Sample *t*-Test Use a two-sample *t*-test when: - the response variable is continuous; - the grouping variable has exactly two independent levels; - the observations in one group are independent of those in the other; - the within-group distributions are reasonably well behaved, or the sample sizes are large enough for mild departures from normality to be unimportant; - the biological question concerns the mean difference between the two groups. For the standard two-sided form, the hypotheses are: $$H_{0}: \mu_{A} = \mu_{B}$$ $$H_{a}: \mu_{A} \ne \mu_{B}$$ or, equivalently, $$H_{0}: \mu_{A} - \mu_{B} = 0$$ $$H_{a}: \mu_{A} - \mu_{B} \ne 0$$ ## Choosing Between Student and Welch Welch's two-sample *t*-test should be the default. Use Student's *t*-test only when the group spreads are similar and the sample sizes are also similar enough that the pooled-variance assumption is defensible. Use this rule: 1. start with plots and grouped summaries; 2. if the spreads are clearly different, use Welch; 3. if the spreads are similar, Welch is still acceptable and often simplest; 4. use Student only when you want the pooled-variance form for a clear reason. Welch's test is robust, widely used, and already the default in `t.test()`. In most routine analyses there is no penalty for choosing it. ## Other Tests for Homogeneity Formal tests such as Levene's test or Bartlett's test provide supporting evidence about group spread. They should not become the main decision step. Visual comparison of spread comes first, and Welch remains the safer choice when the spreads differ. If the apparent variance difference is tied to strong skewness or a long upper tail, a transformation may be preferable to a raw-scale *t*-test. In [Chapter 6](06-assumptions-and-transformations.qmd) I developed those diagnostic decisions, and the ozone example below demonstrates this in practice in the final test. # Example 2: Penguin Flipper Length in Two Species The `penguins` dataset provides a realistic two-group comparison. Here I compare flipper length between Adelie and Chinstrap penguins. The grouping variable is `species`, and it has two independent levels in this subset. ```{r code-peng-two} peng_two <- penguins |> select(species, flipper_length_mm) |> drop_na() |> filter(species %in% c("Adelie", "Chinstrap")) |> mutate(species = factor(species, levels = c("Adelie", "Chinstrap"))) ``` ## Do an Exploratory Data Analysis (EDA) I first inspect the grouped summaries, the within-group normality checks, and the spread across groups. ```{r code-peng-two-summary} peng_two |> group_by(species) |> summarise( n = n(), mean = mean(flipper_length_mm), sd = sd(flipper_length_mm), shapiro_p = shapiro.test(flipper_length_mm)$p.value, .groups = "drop" ) car::leveneTest(flipper_length_mm ~ species, data = peng_two) ``` I apply the assumption checks within each species because the two-sample design compares group means. The Shapiro-Wilk tests show no strong evidence against normality within either species, and Levene's test provides no strong evidence that the group spreads differ. Together with the boxplots in @fig-peng-two, these checks support a standard two-group mean comparison. ```{r fig-peng-two} #| fig-cap: "Flipper length in Adelie and Chinstrap penguins. The group centres differ, while the group spreads are similar." #| fig-width: 4.5 #| fig-height: 3.5 #| code-fold: true ggplot(peng_two, aes(x = species, y = flipper_length_mm, fill = species)) + geom_boxplot(alpha = 0.35, colour = "black", outlier.shape = NA) + geom_jitter(width = 0.08, alpha = 0.6, size = 1.5) + labs(x = "Species", y = "Flipper length (mm)") + theme(legend.position = "none") ``` The boxplots show a clear shift in centre, with Chinstrap penguins tending to have longer flippers. The spreads are similar. The implication is that the assumptions do not force a change of method here, so a Welch two-sample *t*-test is a defensible default. ## State the Hypotheses $$H_{0}: \mu_{\text{Adelie}} = \mu_{\text{Chinstrap}}$$ $$H_{a}: \mu_{\text{Adelie}} \ne \mu_{\text{Chinstrap}}$$ ## Apply the Test ```{r code-peng-two-test} peng_welch <- t.test(flipper_length_mm ~ species, data = peng_two) peng_welch ``` The formula `flipper_length_mm ~ species` tells R to compare the flipper lengths across the levels of the grouping variable `species`. This same formula notation appears again in ANOVA and regression chapters. ## Interpret the Results The estimated difference in mean flipper length is about 5.9 mm, with Chinstrap penguins having the larger mean. The 95% confidence interval excludes zero, and the *p*-value is very small. We therefore reject $H_0$ and conclude that the two species differ in mean flipper length. The confidence interval shows the plausible range for the size of the difference, which is the biologically useful part of the result. ## Reporting ::: {.callout-note appearance="simple"} ## Write-Up **Methods** Mean flipper length was compared between Adelie and Chinstrap penguins using a two-sample Welch *t*-test. Species was treated as a two-level grouping factor, and flipper length was the continuous response variable. Grouped summaries and graphical checks were used to inspect the data before the test was applied. **Results** Adelie penguins had a mean flipper length of 190.0 mm (SD = 6.54, $n = 151$), whereas Chinstrap penguins had a mean flipper length of 195.8 mm (SD = 7.13, $n = 68$). A two-sample Welch *t*-test indicated that mean flipper length differed between species ($t_{119.68} = -5.78$, $p < 0.001$). The estimated mean difference was 5.87 mm, and the 95% confidence interval ranged from 3.86 to 7.88 mm, with Chinstrap penguins having the larger mean. **Discussion** The estimated difference is several millimetres and the confidence interval remains well away from zero, so the species difference is biologically substantial as well. This is the two-group case that Chapter 8 generalises to more than two groups. ::: ::: callout-important ## Do This Now! Create a visualisation that makes the mean difference and the spread difference in this penguin example easy to compare at a glance. ::: ::: {#self-assessment-task-7-1 .callout-important appearance="simple"} ## Self-Assessment Task 7-1 Please refer to this two-sided two-sample *t*-test: ```{r} # random normal data set.seed(666) r_two <- data.frame(dat = c(rnorm(n = 20, mean = 4, sd = 1), rnorm(n = 20, mean = 5, sd = 1)), sample = c(rep("A", 20), rep("B", 20))) # perform t-test # note how we set the `var.equal` argument to TRUE because we know # our data has the same SD (they are simulated as such!) t.test(dat ~ sample, data = r_two, var.equal = TRUE) ``` a. Repeat this analyses using the Welch's `t.test()`. **(/5)** b. Repeat your analysis, above, using the even more old-fashioned Equation 4 in the lecture. Show the code, talk us through the step you followed to read the *p*-values off the table of *t*-statistics. **(/10)** ::: # Example 3: A Two-Sample Comparison After Log Transformation The ozone example from [Chapter 6](06-assumptions-and-transformations.qmd) can now be carried through to a *t*-test for inference. I compare ozone concentrations in **May-June** with those in **August-September**. In the `airquality` dataset, ozone is recorded in parts per billion (`ppb`). On the raw scale, the data are strongly right-skewed and the group variances differ. I apply a log transformation, which improves both features. So the two-sample comparison is done on the log scale and then back-transformed for reporting. ```{r code-ozone-ttest-setup} aq <- na.omit(airquality[, c("Ozone", "Month")]) |> as_tibble() |> filter(Month %in% c(5, 6, 8, 9)) |> mutate( season = factor( if_else(Month %in% c(5, 6), "May-June", "August-September") ), log_ozone = log(Ozone) ) ``` ## Do an Exploratory Data Analysis (EDA) First, I inspect the grouped distributions on the raw and transformed scales in @fig-ozone-transform-ttest. ```{r fig-ozone-transform-ttest} #| fig-cap: "Ozone concentrations before and after log transformation. A) Raw ozone values (ppb) by season. B) Log-transformed ozone values by season." #| fig-width: 8 #| fig-height: 3.5 #| code-fold: true plt_raw <- ggplot(aq, aes(season, Ozone, fill = season)) + geom_boxplot(alpha = 0.35, colour = "black", outlier.shape = NA) + geom_jitter(width = 0.08, alpha = 0.6, size = 1.5) + labs(x = NULL, y = "Ozone (ppb)") + theme(legend.position = "none") plt_log <- ggplot(aq, aes(season, log_ozone, fill = season)) + geom_boxplot(alpha = 0.35, colour = "black", outlier.shape = NA) + geom_jitter(width = 0.08, alpha = 0.6, size = 1.5) + labs(x = NULL, y = "log(Ozone)") + theme(legend.position = "none") ggarrange(plt_raw, plt_log, ncol = 2) ``` My formal checks support the same conclusion: ```{r code-ozone-checks} aq |> group_by(season) |> summarise( raw_w = unname(shapiro.test(Ozone)$statistic), raw_p = shapiro.test(Ozone)$p.value, log_w = unname(shapiro.test(log_ozone)$statistic), log_p = shapiro.test(log_ozone)$p.value, raw_mean = mean(Ozone), geo_mean = exp(mean(log_ozone)), .groups = "drop" ) car::leveneTest(Ozone ~ season, data = aq) car::leveneTest(log_ozone ~ season, data = aq) ``` I apply the assumption checks within season on both the raw and transformed scales because this is still a two-group comparison. On the raw scale, both groups show strong evidence against normality and the variances differ. On the log scale, the group distributions are much closer to normal and the variance difference largely disappears. The implication is that the raw-scale comparison is a poor choice, while the transformed comparison is defensible. ## State the Hypotheses Because the analysis is now on the log scale, the hypotheses concern the mean log ozone concentration: $$H_{0}: \mu_{\log,\text{May-June}} = \mu_{\log,\text{August-September}}$$ $$H_{a}: \mu_{\log,\text{May-June}} \ne \mu_{\log,\text{August-September}}$$ Back-transformation changes the way the result is expressed. Equality on the log scale corresponds to a ratio of 1 on the original scale, so the null hypothesis becomes: $$H_{0}: \frac{\text{geometric mean ozone in August-September}}{\text{geometric mean ozone in May-June}} = 1$$ The alternative hypothesis is that this ratio is not 1. A ratio of 1 means that the two periods have the same geometric mean ozone concentration in `ppb`. A ratio greater than 1 means that ozone is higher in August-September. A ratio less than 1 means that ozone is lower in August-September. ## Apply the Test Welch's two-sample *t*-test remains the default choice in R and is appropriate here. ```{r code-ozone-log-ttest} ozone_t <- t.test(log_ozone ~ season, data = aq) ozone_t aq |> group_by(season) |> summarise( geometric_mean = exp(mean(log_ozone)), .groups = "drop" ) ratio_est <- exp(ozone_t$estimate[[1]] - ozone_t$estimate[[2]]) ratio_ci <- exp(ozone_t$conf.int) tibble( comparison = "August-September / May-June", ratio = ratio_est, lower = ratio_ci[1], upper = ratio_ci[2] ) ``` ## Interpret the Results The test on the log scale compares mean log ozone concentrations between the two periods. After back-transformation, the result is easier to interpret as a ratio. August-September ozone concentrations are estimated to be about 1.81 times those in May-June, with a 95% confidence interval from 1.26 to 2.59. The geometric mean is the back-transformed mean from the log scale. It differs from the arithmetic mean in that it summarises multiplicative variation rather than additive variation. The arithmetic mean is the ordinary average obtained by adding values and dividing by the sample size. The geometric mean is obtained by averaging the logged values and then back-transforming. For right-skewed data, it is often less influenced by the upper tail than the arithmetic mean. In this example, I show the complete sequence of decision steps. I start with diagnosing the assumption violation on the raw scale, then I transform the data for a clear reason, I re-assess the assumptions, apply the test, and then I return the result to the original scale for reporting. ## Reporting ::: {.callout-note appearance="simple"} ## Write-Up **Methods** Ozone concentrations (`ppb`) were compared between May-June and August-September using a two-sample *t*-test. Exploratory plots and assumption checks showed strong right-skew and unequal variances on the raw scale, so ozone was log-transformed before analysis. A Welch two-sample *t*-test was then applied to log-transformed ozone concentrations. Results were back-transformed and reported as geometric means and their ratio. **Results** On the transformed scale, ozone concentrations differed between the two periods (Welch two-sample *t*-test: $t_{67.27} = 3.30$, $p < 0.01$). The geometric mean ozone concentration was 18.6 ppb in May-June and 33.6 ppb in August-September. After back-transformation, August-September ozone concentrations were estimated to be 1.81 times those in May-June (95% CI 1.26 to 2.59). **Discussion** The transformed analysis supports higher ozone concentrations in August-September than in May-June. Because the model was fitted on the log scale, the clearest interpretation is multiplicative: the later period has ozone concentrations about 1.8 times higher. ::: # Paired *t*-Tests In a paired design, the observations are linked. The pairing may come from repeated measurements on the same individuals, left-right comparisons on the same organism, or deliberately matched experimental units. The analysis is done on the within-pair differences. ## When to Use a Paired *t*-Test Use a paired *t*-test when: - the response variable is continuous; - the two measurements are meaningfully paired within the same sampling unit; - the pairs are independent of one another; - the paired differences are reasonably well behaved. For the standard two-sided form, the hypotheses are: $$H_{0}: \mu_{d} = 0$$ $$H_{a}: \mu_{d} \ne 0$$ where $\mu_d$ is the population mean of the paired differences. # Example 4: Before-and-After Measurements on Piglets I will use a small example in which piglets are weighed before and after a three-week feeding period. The inferential question is whether body mass increases over time within the same individuals. ```{r code-paired-piglets} set.seed(666) piglets <- data.frame( id = 1:12, before = round(rnorm(12, mean = 8.2, sd = 0.7), 2), after = round(rnorm(12, mean = 9.0, sd = 0.7), 2) ) piglets$diff <- piglets$after - piglets$before piglets ``` ## Do an Exploratory Data Analysis (EDA) Before running the paired *t*-test, I inspect the paired differences because these are the values used directly by the test. ```{r code-paired-summary} summary(piglets$diff) sd(piglets$diff) shapiro.test(piglets$diff) ``` I apply the assumption check to the paired differences not to the raw before and after measurements. The Shapiro-Wilk test shows no strong evidence against normality in those differences, so I can use the paired *t*-test without changing method. The distribution of differences is shown in @fig-paired-differences. The distribution of differences is shown in @fig-paired-differences. ```{r fig-paired-differences} #| fig-cap: "Distribution of paired differences in piglet body mass after minus before the feeding period." #| fig-width: 4 #| fig-height: 3 #| code-fold: true ggplot(data = piglets, aes(x = diff)) + geom_histogram(binwidth = 0.5, fill = "indianred", colour = "black", alpha = 0.5) + geom_vline(xintercept = 0, colour = "dodgerblue2", linewidth = 0.9) + labs(x = "Paired difference in mass (after - before)", y = "Frequency") ``` The paired differences are centred above zero, which already suggests that the piglets tended to gain mass over the feeding period. The distribution is well behaved enough for a paired *t*-test, so the next step is to test whether the mean difference is different from zero. ## State the Hypotheses $$H_{0}: \mu_{d} = 0$$ $$H_{a}: \mu_{d} \ne 0$$ If the biological question had specified direction in advance, such as whether mass increased after feeding, the same paired design could be analysed with a one-sided alternative. ## Apply the Test ```{r code-paired-test} t.test(piglets$after, piglets$before, paired = TRUE) ``` The paired *t*-test is a one-sample test applied to the set of paired differences. That is why the raw before and after values are not analysed as though they were independent groups. ## Interpret the Results The mean paired difference is positive, so the piglets tended to be heavier after the feeding period. The confidence interval for the mean difference lies just above zero and the *p*-value is slightly below 0.05. We therefore reject $H_0$ and conclude that body mass increased over the feeding period in this example. ## Reporting ::: {.callout-note appearance="simple"} ## Write-Up **Methods** Piglet body mass was measured before and after a three-week feeding period, and the resulting paired observations were analysed with a paired *t*-test. The analysis was based on the within-individual differences in mass. **Results** Piglet body mass increased over the three-week feeding period. A paired *t*-test showed that the mean within-individual change in body mass differed from zero ($t_{11} = 2.21$, $p < 0.05$). The mean paired difference was 0.815 kg, with a 95% confidence interval from 0.004 to 1.626 kg. **Discussion** The paired design strengthens the inference because each piglet acts as its own control. The biological conclusion therefore concerns change within individuals rather than a difference between independent groups. ::: # When Assumptions Are Difficult The response to assumption problems depends on the sampling and experimental design and on the source of the problem. | Situation | Typical response | |:-----------------------------------|:-----------------------------------| | One sample, small $n$, strong skew or extreme outliers | Consider transformation or a one-sample Wilcoxon procedure | | Two groups, similar shape but different spreads | Use Welch's *t*-test | | Two groups, skewness linked to spread | Transform, re-check, then analyse if the new scale is defensible | | Two groups, strong outliers or badly distorted shapes | Consider a Wilcoxon rank-sum procedure | | Paired data, non-normal paired differences | Consider a paired Wilcoxon signed-rank procedure | Use Welch when the main problem is unequal spread. Use a transformation when the problem is scale, such as a long right tail that also drives a variance difference. Use a rank-based alternative when the mean-based comparison itself is no longer a good description of the biological question. Keep the following rules in mind: - do not switch to Student's *t*-test merely because Levene's test is non-significant; - do not treat a formal normality test as the only decision criterion; - do not transform data automatically; transform for a stated reason and then re-check; - do not forget that a large sample helps with mild non-normality but does not fix non-independence. # Optional: One-Sided *t*-Tests One-sided tests should be used only when the direction of the effect is specified before the data are analysed. If the real question is whether two means differ, use the standard two-sided form. ## One-Sided One-Sample *t*-Test Suppose the forestry question is directional: are the sampled trees shorter than 24.5 m on average? The hypotheses are: $$H_{0}: \mu \ge 24.5$$ $$H_{a}: \mu < 24.5$$ ```{r code-tree-one-sided} t.test(tree_one$height_m, mu = 24.5, alternative = "less") t.test(tree_one$height_m, mu = 24.5, alternative = "greater") ``` The first test matches the direction implied by the data and the question, but the second does not. Direction changes the hypothesis, the tail area, and the interpretation. ## One-Sided Two-Sample *t*-Test Suppose the biological question is directional: are Adelie penguin flippers shorter than Chinstrap penguin flippers? The hypotheses are: $$H_{0}: \mu_{\text{Adelie}} \ge \mu_{\text{Chinstrap}}$$ $$H_{a}: \mu_{\text{Adelie}} < \mu_{\text{Chinstrap}}$$ ```{r code-peng-one-sided} t.test(flipper_length_mm ~ species, data = peng_two, alternative = "less") t.test(flipper_length_mm ~ species, data = peng_two, alternative = "greater") ``` Again, the test only makes sense when the direction was specified before the analysis. One-sided tests are not a rescue strategy for a non-significant two-sided result. # Rank-Based Alternatives Within the *t*-Test Family The parametric *t*-tests in this chapter are mean-based procedures. When the response is badly non-normal in a small sample, strongly skewed, or dominated by influential outliers, a rank-based alternative may be more defensible. These methods belong to the same inferential family as the *t*-tests because they serve the same one-sample, two-sample, or paired designs. The three main rank-based counterparts are: - **one-sample Wilcoxon signed-rank test** for a single sample compared with a reference value; - **Wilcoxon rank-sum test** for two independent groups; - **paired Wilcoxon signed-rank test** for matched or before-after observations. All three are handled by `wilcox.test()` in R: - `wilcox.test(x, mu = ...)` for a one-sample problem; - `wilcox.test(x, y)` or `wilcox.test(response ~ group, data = ...)` for two independent groups; - `wilcox.test(x, y, paired = TRUE)` for paired data. These procedures are often introduced as alternatives when the mean is a poor summary of the data. They should not be described as direct tests of medians in every case. Their interpretation depends on the design and on the structure of the distributions being compared. ::: {.callout-note appearance="simple"} ## Use the Design, Not Just the Function Name The one-sample and paired Wilcoxon procedures are often confused because they use the same function. The distinction is exactly the same as for *t*-tests: - one sample against a reference value → one-sample signed-rank test; - two related measurements on the same units → paired signed-rank test; - two independent groups → rank-sum test. ::: # Decision Steps for Any *t*-Test For most problems in this chapter, the workflow is: 1. identify the design: one sample, two independent groups, or paired data; 2. visualise the data; 3. check the relevant assumptions for shape, spread, and independence; 4. choose the test, with Welch as the default for two groups; 5. run the test; 6. interpret the estimated effect, its confidence interval, and the *p*-value together. If you follow these steps you'll prevent a large number of common mistakes. ::: callout-important ## Do It Now! Work through the six decision steps above using the built-in `mtcars` dataset. Compare fuel economy (`mpg`) between cars with automatic (`am = 0`) and manual (`am = 1`) transmissions. Step 1: What is the design?\ Step 2: Make a grouped box plot of `mpg ~ factor(am)`.\ Step 3: Apply Shapiro-Wilk within each group and Levene's test across groups.\ Step 4: Choose Welch's or Student's *t*-test based on step 3.\ Step 5: Run `t.test(mpg ~ factor(am), data = mtcars, var.equal = ...)`.\ Step 6: Write a one-sentence results statement with the test statistic, df, p-value, and 95% CI. ````{=html}  ```` ::: # A *t*-Test Workflow Now that the individual cases are in place, I can walk through a complete analysis in R. For this example we use the `ecklonia` data from [Intro R Workshop: Data Manipulation, Analysis, Graphing](https://robwschlegel.github.io/Intro_R_Workshop/). ## Loading Data Before I analyse anything, I need to load the data and reshape them into long format. ```{r code-ecklonia-read-csv-here-here} ecklonia <- read_csv(here::here("data", "BCB744", "ecklonia.csv")) |> gather(key = "variable", value = "value", -species, -site, -ID) ``` ## Visualising Data I visualise the data first in @fig-ecklonia1. ```{r fig-ecklonia1} #| fig-cap: "Boxplots showing differences in morphometric properties of the kelp _Ecklonia maxima_ at two sites in False Bay." #| code-fold: true ggplot(data = ecklonia, aes(x = variable, y = value, fill = site)) + geom_boxplot(colour = "black", fill = "dodgerblue4", alpha = 0.4) + coord_flip() ``` The measurements occupy very different scales, which makes the full figure difficult to compare directly between sites. But, nevertheless, the figure's value is orientating me to the variables in the dataset and showing me that a more focused comparison is needed before I can state a hypothesis clearly. ## Formulating a Hypothesis Let me narrow the question to stipe mass and focus only on two sites, Batsata Rock and Boulders Beach, as shown in @fig-ecklonia2. ```{r fig-ecklonia2} #| fig-cap: "Boxplots showing the difference in stipe mass (kg) of the kelp _Ecklonia maxima_ at two sites in False Bay." #| code-fold: true ecklonia_sub <- ecklonia |> filter(variable == "stipe_mass") ggplot(data = ecklonia_sub, aes(x = variable, y = value, fill = site)) + geom_boxplot(colour = "black", alpha = 0.4) + coord_flip() + labs(y = "Stipe mass (kg)", x = "") + theme(axis.text.y = element_blank(), axis.ticks.y = element_blank()) ``` I can state the directional question as: are kelp stipes heavier at Batsata Rock than at Boulders Beach? The corresponding hypotheses are: $$H_{0}: \mu_{\text{Batsata}} \le \mu_{\text{Boulders}}$$ $$H_{a}: \mu_{\text{Batsata}} > \mu_{\text{Boulders}}$$ ## Choosing a Test There are two independent groups, one continuous response, and a directional hypothesis specified in advance. That combination points to a one-sided two-sample *t*-test. Because the group spreads appear similar, Student's formulation is acceptable here, although Welch would also be defensible. ## Checking Assumptions The response is continuous and the sites define the two independent groups. I still need to inspect spread and approximate normality within groups. ```{r code-ecklonia-sub} ecklonia_sub |> group_by(site) |> summarise( stipe_mass_var = var(value), stipe_mass_norm = shapiro.test(value)$p.value, .groups = "drop" ) ``` The grouped summaries suggest similar spread and no obvious normality problem. If these checks had failed badly, [Chapter 6](06-assumptions-and-transformations.qmd) provides the next steps. ## Running an Analysis ```{r code-t-test-value-site} t.test(value ~ site, data = ecklonia_sub, var.equal = TRUE, alternative = "greater") ``` ## Interpreting the Results The output gives the test statistic, degrees of freedom, *p*-value, and confidence interval for the mean difference. If the one-sided *p*-value is small enough, I reject $H_0$ and conclude that mean stipe mass is greater at Batsata Rock. The confidence interval gives the plausible size of that difference. ## Drawing Conclusions The scientific conclusion should consider what local ecological or hydrodynamic conditions could account for the difference in stipe mass between the two sites. ## Reporting ::: {.callout-note appearance="simple"} ## Write-Up **Methods** Stipe mass of *Ecklonia maxima* was compared between Batsata Rock and Boulders Beach using a one-sided two-sample Student's *t*-test. The analysis treated site as the grouping factor and stipe mass as the continuous response variable, after checking that the group spreads were similar and that the data were approximately normal within each site. **Results** Mean stipe mass of *Ecklonia maxima* was higher at Batsata Rock ($6.12 \pm 1.41$ kg SD, $n = 13$) than at Boulders Beach ($5.00 \pm 1.63$ kg SD, $n = 13$). A one-sided two-sample Student's *t*-test indicated that stipe mass at Batsata Rock was greater than at Boulders Beach ($t_{24} = 1.87$, $p < 0.05$). The one-sided 95% confidence interval for the mean difference ranged from 0.10 kg to $\infty$. **Discussion** The biological conclusion is that kelp at Batsata Rock had heavier stipes than kelp at Boulders Beach in this comparison. The next step would be to explain why that difference exists. ::: ## Going Further In this example, I isolated two sites so that the comparison could fit a two-sample *t*-test. The full dataset contains more than two sites, however, and that changes the analysis. Once the question becomes whether mean stipe mass differs across several (*i.e.*, \> 2) sites, the correct next step is ANOVA rather than a bunch of pairwise *t*-tests. In [Chapter 8](08-anova.qmd), I develop that extension and explain why repeated two-group testing inflates Type I error. The biological questions also inflates at that point. If stipe mass differs among several sites, what combination of exposure, nutrient regime, competition, or age structure produced that pattern? ::: {#self-assessment-task-7-2 .callout-important appearance="simple"} ## Self-Assessment Task 7-2 For each scenario below, identify the correct *t*-test design (one-sample, two-sample, or paired) and state which assumption needs to be checked, and where: a. You measure the resting heart rates of 25 athletes and compare them with the published population mean of 72 bpm. **(/2)** b. You compare the nitrogen content of leaves from trees in two adjacent forest types (20 trees per forest). **(/2)** c. You weigh 15 juvenile tortoises in March and again in October and ask whether they gained mass. **(/2)** d. You measure the length of left and right tibiae from 18 rabbits. **(/2)** Discuss with a partner whether (b) and (d) are structurally similar or different. What changes between them? **(/2)** ::: # Conclusion *t*-tests are among the first inferential tools used in biostatistics because they connect directly to common biological questions. The main decision is design: one sample, two independent groups, or paired observations. From there, the workflow simply requires that you inspect the data, check the assumptions, choose the appropriate test, and interpret the estimated effect with its uncertainty.