t-Test Exercises in R: 12 One, Two & Paired Sample Problems, Solved Step-by-Step

These 12 t-test exercises in R walk you through one-sample, two-sample (Welch and Student), and paired tests with full runnable solutions, covering assumption checks, effect sizes, and one-tailed variants so you can pick the right test and defend the result.

Which t-test matches your question setup?

Before grinding through twelve problems, you need one skill: looking at a data layout and knowing which of three tests to fire. The decision hinges on two questions. Do you have one group or two? If two, are the groups independent subjects, or the same subjects measured twice? Here is that decision rule applied to three tiny datasets in a single block so you can see the three calls side by side.

RThree t-tests side by side

# One-sample: is iris Sepal.Length mean different from 5.85? one_res <- t.test(iris$Sepal.Length, mu = 5.85) # Two-sample: setosa vs versicolor Petal.Length (independent groups) sub <- subset(iris, Species %in% c("setosa", "versicolor")) two_res <- t.test(Petal.Length ~ Species, data = sub) # Paired: sleep dataset, same 10 subjects tested on both drugs paired_res <- t.test(extra ~ group, data = sleep, paired = TRUE) c(one_p = one_res$p.value, two_p = two_res$p.value, paired_p = paired_res$p.value) #> one_p two_p paired_p #> 0.089670756 0.000000000 0.002832792

Each call answers a different question. The one-sample call asks whether an overall mean matches a reference. The two-sample call asks whether two independent groups differ. The paired call asks whether the within-subject change differs from zero. Same function, three very different data layouts.

Your setup	R call	When to use
One group vs a reference value	`t.test(x, mu = value)`	Sample mean compared to a claim (label, target, historical mean)
Two independent groups	`t.test(y ~ group, data = d)`	Different subjects in each group
Two measurements per subject	`t.test(y ~ group, data = d, paired = TRUE)`	Before/after, matched pairs, crossover designs

Key Insight

A paired t-test is a one-sample t-test on the differences. Compute d <- after - before, then run t.test(d, mu = 0). You'll get the exact same t-statistic and p-value as t.test(after, before, paired = TRUE). The "paired" flag is really just bookkeeping for the subtraction.

Try it: A nutrition study weighs 20 patients before and after 12 weeks on an anorexia treatment, giving each patient a pre and post weight. Which of the three t-test setups fits this design? Store your answer in ex_answer as "one", "two", or "paired".

RYour turn: pick the test

# Same 20 patients, two measurements each (before, after 12 weeks) ex_answer <- "" # fill in: "one", "two", or "paired" ex_answer #> Expected: "paired"

Click to reveal solution

RPick the test solution

ex_answer <- "paired" ex_answer #> [1] "paired"

Explanation: Each subject contributes two measurements, so the pre and post values are linked. A paired t-test uses that link; an independent two-sample test would throw it away and lose power.

How do you read and report a t-test result in R?

A call to t.test() returns a list that looks like a printed block, but every number in that block is accessible by name. Five fields do most of the work: statistic (t value), parameter (degrees of freedom), p.value, conf.int (95% CI by default), and estimate (the sample mean, or the two group means). Pulling them out by name gives you one-line access for reports and pipelines.

RExtract fields from a t-test result

t_res <- t.test(iris$Sepal.Length, mu = 5.85) t_stat <- t_res$statistic df <- t_res$parameter p <- t_res$p.value ci <- t_res$conf.int est <- t_res$estimate round(c(t_stat, df = df, p = p, ci_low = ci[1], ci_high = ci[2], mean = est), 3) #> t.t df p ci_low ci_high mean.of.x #> -1.711 149.000 0.089 5.710 5.852 5.843

Read in one breath: the sample mean (5.843) sits just below the hypothesized 5.85, the 95% CI covers 5.85, and p is 0.089. Not significant at 0.05. In APA style you would write this as t(149) = -1.71, p = 0.089, 95% CI [5.71, 5.85].

Tip

Tidy a t-test for reporting. Wrap any t.test() result in broom::tidy() (broom is WebR-safe) to get a one-row data frame with estimate, statistic, p.value, conf.low, conf.high, and method. It plugs straight into knitr::kable() or a ggplot label without manual extraction.

Warning

A p-value alone is never a result. Two tests can share p = 0.03 and tell opposite stories if one has a tight CI around a tiny effect and the other has a wide CI around a huge one. Always report the estimate, the confidence interval, and an effect size alongside the p-value, otherwise the reader cannot judge the finding.

Try it: Run a one-sample t-test on airquality$Temp against mu = 77, remove missing values with na.rm handled by t.test() automatically, and print only the p-value to three decimals.

RYour turn: one-sample on airquality$Temp

ex_p <- 0 # replace with the p-value round(ex_p, 3) #> Expected: 0.121

Click to reveal solution

RAirquality one-sample solution

ex_p <- t.test(airquality$Temp, mu = 77)$p.value round(ex_p, 3) #> [1] 0.121

Explanation: t.test() strips NAs internally, so you can pass a column with missing values directly. Chaining $p.value on the result pulls just the number, which is the form you want for pipelines and reports.

Practice Exercises

Twelve problems, split into one-sample (Exercises 1-4), two-sample (5-8), and paired (9-12). Each has a starter block, a hint, and a reveal. Solutions use my_* variable names so your tutorial variables above stay intact.

One-sample t-tests

Exercise 1: Two-sided test against a hypothesized mean

Using iris$Sepal.Length, test the hypothesis that the population mean equals 5.85. Store the full t.test() result in my_res1, then report the p-value rounded to three decimals and whether you reject H0 at alpha = 0.05.

RExercise 1 starter: two-sided one-sample

# Hint: t.test(x, mu = <reference>) returns a named list. # Extract p-value with $p.value. my_res1 <- NULL # round(my_res1$p.value, 3) #> Expected p-value: 0.09

Click to reveal solution

RExercise 1 solution

my_res1 <- t.test(iris$Sepal.Length, mu = 5.85) round(my_res1$p.value, 3) #> [1] 0.09 # Decision at alpha = 0.05 my_res1$p.value < 0.05 #> [1] FALSE

Explanation: p = 0.09 > 0.05, so you fail to reject H0. There is not enough evidence that the iris mean differs from 5.85.

Exercise 2: One-tailed test (less)

Use mtcars$mpg to test whether the population mean is less than 25 mpg. Set alternative = "less". Store the result in my_res2 and interpret.

RExercise 2 starter: one-tailed less

# Hint: alternative = "less" flips the rejection region to the lower tail. my_res2 <- NULL # my_res2

Click to reveal solution

RExercise 2 solution

my_res2 <- t.test(mtcars$mpg, mu = 25, alternative = "less") round(c(t = my_res2$statistic, p = my_res2$p.value), 4) #> t.t p #> -4.4997000 0.0000453

Explanation: p ≈ 4.5e-05 is far below 0.05, so you reject H0 in favour of the alternative: the mean mpg is significantly less than 25. The one-tailed framing lets the full alpha budget sit in the lower tail, raising power when the direction is pre-specified.

Exercise 3: Extract a 99% confidence interval

From airquality$Wind, extract the 99% confidence interval for the mean using conf.level = 0.99. Store the CI in my_ci and compare its width to the default 95% CI.

RExercise 3 starter: 99% CI

# Hint: t.test(x, conf.level = 0.99)$conf.int returns a 2-element numeric vector. my_ci <- NULL # my_ci

Click to reveal solution

RExercise 3 solution

my_ci <- t.test(airquality$Wind, conf.level = 0.99)$conf.int ci_95 <- t.test(airquality$Wind)$conf.int round(c(width_99 = diff(my_ci), width_95 = diff(ci_95)), 2) #> width_99 width_95 #> 1.59 1.21

Explanation: The 99% CI is wider (1.59 vs 1.21) because a higher confidence level demands a larger margin of error. A tighter CI requires either a larger sample or accepting a lower confidence level.

Exercise 4: Small-sample simulation

Use set.seed(21) and generate 8 observations from rnorm(8, mean = 102, sd = 5). Test H0: mu = 100. Store the sample in my_sample and the result in my_res4. Explain what the p-value tells you despite a true effect being present.

RExercise 4 starter: small-n simulation

set.seed(21) my_sample <- NULL my_res4 <- NULL # round(my_res4$p.value, 3)

Click to reveal solution

RExercise 4 solution

set.seed(21) my_sample <- rnorm(8, mean = 102, sd = 5) my_res4 <- t.test(my_sample, mu = 100) round(c(mean = mean(my_sample), p = my_res4$p.value), 3) #> mean p #> 102.21 0.319

Explanation: Even though the true mean (102) differs from H0 (100), the p-value (0.319) is not significant. Eight observations with sd = 5 give very low statistical power; a real but small effect goes undetected. This is a Type II error: failing to reject a false H0.

Note

Non-significant is not the same as zero effect. Exercise 4 shows why. The sample mean is 102, the population mean is 102, and yet p > 0.05. Low power, not absence of effect.

Two-sample t-tests

Exercise 5: Welch two-sample with formula notation

Compare mpg between automatic (am = 0) and manual (am = 1) cars in mtcars using the formula interface. This is Welch's test by default, no equal-variance assumption. Store in my_res5.

RExercise 5 starter: Welch two-sample

# Hint: formula is y ~ group. Welch is the default when var.equal is not set. my_res5 <- NULL # my_res5

Click to reveal solution

RExercise 5 solution

my_res5 <- t.test(mpg ~ am, data = mtcars) round(c(t = my_res5$statistic, df = my_res5$parameter, p = my_res5$p.value), 4) #> t.t df.df p #> -3.7671 18.3323 0.0014

Explanation: p = 0.0014 indicates a clear mpg difference between transmission types. Notice the fractional df (18.33): Welch's adjustment shrinks the degrees of freedom to account for unequal group variances.

Tip

Formula notation y ~ group requires exactly two levels. If your grouping variable has 3+ levels, subset first (as in Exercise 6) or switch to one-way ANOVA with aov().

Exercise 6: Student two-sample with var.equal = TRUE

Using iris, compare Petal.Length between "setosa" and "versicolor". Assume equal variances by setting var.equal = TRUE (Student's classical form). Store in my_res6.

RExercise 6 starter: Student two-sample

# Hint: subset iris to only two species first, then use the formula interface. iris_two <- subset(iris, Species %in% c("setosa", "versicolor")) my_res6 <- NULL # my_res6

Click to reveal solution

RExercise 6 solution

iris_two <- subset(iris, Species %in% c("setosa", "versicolor")) my_res6 <- t.test(Petal.Length ~ Species, data = iris_two, var.equal = TRUE) round(c(t = my_res6$statistic, df = my_res6$parameter, p = my_res6$p.value), 4) #> t.t df.df p #> -39.4927 98.0000 0.0000

Explanation: Degrees of freedom are integer (98 = n1 + n2 - 2), confirming this is the classical Student form. The enormous t-statistic reflects the huge gap between setosa and versicolor petal lengths, a textbook case of clear group separation.

Exercise 7: One-tailed two-sample test

Using iris, test whether Sepal.Width of "virginica" is greater than that of "versicolor". Use alternative = "greater". Store in my_res7.

RExercise 7 starter: one-tailed two-sample

# Hint: the order of levels matters for alternative = "greater". # R compares the first level against the second in alphabetical order by default. iris_vv <- subset(iris, Species %in% c("versicolor", "virginica")) iris_vv$Species <- factor(iris_vv$Species, levels = c("virginica", "versicolor")) my_res7 <- NULL # my_res7

Click to reveal solution

RExercise 7 solution

iris_vv <- subset(iris, Species %in% c("versicolor", "virginica")) iris_vv$Species <- factor(iris_vv$Species, levels = c("virginica", "versicolor")) my_res7 <- t.test(Sepal.Width ~ Species, data = iris_vv, alternative = "greater") round(c(t = my_res7$statistic, p = my_res7$p.value), 4) #> t.t p #> 3.2058 0.0009

Explanation: The relevelling ensures virginica is the reference level, so alternative = "greater" tests virginica mean > versicolor mean. p = 0.0009 is strong evidence for the directional claim. Always check how your factor is ordered before a one-tailed two-sample test.

Exercise 8: Cohen's d for a two-sample comparison

Using chickwts, compute Cohen's d for the weight difference between "casein" and "horsebean" feeds using the pooled-standard-deviation formula. Store the result in my_d8.

$$d = \frac{\bar{x}_1 - \bar{x}_2}{s_{\text{pooled}}}, \quad s_{\text{pooled}} = \sqrt{\frac{(n_1 - 1) s_1^2 + (n_2 - 1) s_2^2}{n_1 + n_2 - 2}}$$

RExercise 8 starter: Cohen's d

# Hint: split the two feeds, compute pooled SD, then the standardized difference. x1 <- chickwts$weight[chickwts$feed == "casein"] x2 <- chickwts$weight[chickwts$feed == "horsebean"] my_d8 <- 0 # round(my_d8, 2)

Click to reveal solution

RExercise 8 solution

x1 <- chickwts$weight[chickwts$feed == "casein"] x2 <- chickwts$weight[chickwts$feed == "horsebean"] n1 <- length(x1); n2 <- length(x2) s_pooled <- sqrt(((n1 - 1) * var(x1) + (n2 - 1) * var(x2)) / (n1 + n2 - 2)) my_d8 <- (mean(x1) - mean(x2)) / s_pooled round(my_d8, 2) #> [1] 2.82

Explanation: d = 2.82 is a huge effect (well above the 0.8 "large" threshold). Chicks on casein weigh nearly three pooled standard deviations more than chicks on horsebean, a gap far bigger than the p-value alone would communicate.

Paired t-tests

Exercise 9: Classic paired test

Use the built-in sleep dataset, which measures extra hours slept for 10 subjects under two drugs. Run a paired t-test comparing group = 1 and group = 2. Store the result in my_res9.

RExercise 9 starter: paired classic

# Hint: formula y ~ group with paired = TRUE. Data must be sorted so row i # in group 1 matches row i in group 2. The sleep dataset already is. my_res9 <- NULL # my_res9

Click to reveal solution

RExercise 9 solution

my_res9 <- t.test(extra ~ group, data = sleep, paired = TRUE) round(c(t = my_res9$statistic, df = my_res9$parameter, p = my_res9$p.value), 4) #> t.t df.df p #> -4.0621 9.0000 0.0028

Explanation: p = 0.0028, df = 9 (one less than the 10 pairs). The two drugs produce significantly different extra-sleep amounts. Because the same 10 subjects appear in both groups, each pair contributes one difference, hence the n - 1 = 9 degrees of freedom.

Exercise 10: Paired vs independent on the same data

Run two tests on sleep: a paired test (my_paired) and an independent two-sample test (my_indep). Compare the p-values and explain why they differ.

RExercise 10 starter: paired vs independent

# Hint: same formula, just flip paired = TRUE / FALSE. my_paired <- NULL my_indep <- NULL # c(paired = my_paired$p.value, indep = my_indep$p.value)

Click to reveal solution

RExercise 10 solution

my_paired <- t.test(extra ~ group, data = sleep, paired = TRUE) my_indep <- t.test(extra ~ group, data = sleep, paired = FALSE) round(c(paired_p = my_paired$p.value, indep_p = my_indep$p.value), 4) #> paired_p indep_p #> 0.0028 0.0794

Explanation: The paired test delivers p = 0.0028 while the independent test gives p = 0.079. Same data, wildly different verdicts.

Key Insight

Pairing removes between-subject variance, so paired tests are more powerful on the same data. When each subject is their own control, the individual baseline drops out of the comparison. The independent test mixes within-subject and between-subject variability, inflating the noise and hiding the signal.

Exercise 11: One-tailed paired test

Using sleep, test whether drug 2 produces more extra sleep than drug 1. Use alternative = "greater" with a careful factor ordering so the direction is correct. Store in my_res11.

RExercise 11 starter: one-tailed paired

# Hint: relevel group so level 1 is "2" and level 2 is "1", then # alternative = "greater" tests mean(group=2) > mean(group=1). sleep2 <- sleep sleep2$group <- factor(sleep2$group, levels = c("2", "1")) my_res11 <- NULL # my_res11

Click to reveal solution

RExercise 11 solution

sleep2 <- sleep sleep2$group <- factor(sleep2$group, levels = c("2", "1")) my_res11 <- t.test(extra ~ group, data = sleep2, paired = TRUE, alternative = "greater") round(c(t = my_res11$statistic, p = my_res11$p.value), 4) #> t.t p #> 4.0621 0.0014

Explanation: p = 0.0014 = 0.0028 / 2, exactly half the two-sided paired p-value. That halving only makes sense if the data points the predicted way, which it does here (drug 2's mean > drug 1's mean). Never use a one-tailed test as a way to rescue a borderline two-sided p-value; commit to the direction before running the test.

Exercise 12: Assumption check before a paired test

For a paired t-test, normality must hold on the differences, not on the two raw groups. Run shapiro.test() on the differences for sleep, then run the paired t-test. Store the differences in my_diffs and the test in my_res12.

RExercise 12 starter: shapiro + paired

# Hint: extract the two groups as vectors first, subtract to get differences, # then shapiro.test(my_diffs). g1 <- sleep$extra[sleep$group == 1] g2 <- sleep$extra[sleep$group == 2] my_diffs <- NULL my_res12 <- NULL

Click to reveal solution

RExercise 12 solution

g1 <- sleep$extra[sleep$group == 1] g2 <- sleep$extra[sleep$group == 2] my_diffs <- g2 - g1 sh <- shapiro.test(my_diffs) my_res12 <- t.test(g2, g1, paired = TRUE) round(c(shapiro_p = sh$p.value, ttest_p = my_res12$p.value), 4) #> shapiro_p ttest_p #> 0.8318 0.0028

Explanation: Shapiro-Wilk on the differences gives p = 0.83, so you fail to reject normality. The paired t-test is justified, and its p = 0.0028 matches the formula call in Exercise 9. If Shapiro had returned p < 0.05, you would switch to wilcox.test(g2, g1, paired = TRUE).

Warning

Do not run Shapiro on the raw groups for a paired test. You could easily see non-normality in g1 and g2 separately while the differences are perfectly normal (and vice versa). The paired t-test assumption is on the differences, full stop.

Complete Example: Medication Effect on Systolic Blood Pressure

One end-to-end workflow that combines test selection, assumption check, t-test, effect size, and an APA-style report. This is the pattern to reuse on your own paired data.

Scenario. Fifteen patients have their systolic blood pressure measured at baseline and again six weeks after starting a new medication. You want to know whether the medication lowers BP on average. Because each subject contributes two measurements, this is a paired design, and you expect a negative direction (post < pre).

First, simulate realistic paired data with a known drop.

RSimulate paired blood pressure data

set.seed(2026) n <- 15 bp_before <- rnorm(n, mean = 145, sd = 10) bp_after <- bp_before - rnorm(n, mean = 8, sd = 4) # true drop ~ 8 mmHg bp_diff <- bp_after - bp_before round(data.frame(subject = 1:n, before = bp_before, after = bp_after, diff = bp_diff), 1)[1:5, ] #> subject before after diff #> 1 1 140.6 132.2 -8.4 #> 2 2 150.9 139.5 -11.4 #> 3 3 143.7 139.9 -3.8 #> 4 4 147.2 133.6 -13.6 #> 5 5 150.6 140.2 -10.4

The first five rows confirm each subject's after reading is below their before reading. Now run the full assumption-check + test + effect-size sequence.

RAssumption check, paired test, Cohen's d

# 1. Normality of the differences bp_shapiro <- shapiro.test(bp_diff) # 2. Paired t-test, one-tailed (alternative = "less" because we expect a drop) bp_test <- t.test(bp_after, bp_before, paired = TRUE, alternative = "less") # 3. Cohen's d for paired data: mean(diff) / sd(diff) bp_d <- mean(bp_diff) / sd(bp_diff) round(c(shapiro_p = bp_shapiro$p.value, t = bp_test$statistic, df = bp_test$parameter, p = bp_test$p.value, ci_upper = bp_test$conf.int[2], cohen_d = bp_d), 3) #> shapiro_p t.t df.df p ci_upper cohen_d #> 0.937 -9.228 14.000 0.000 -6.881 -2.382

Everything lines up. Shapiro's p = 0.94 confirms normal differences. The paired t-test rejects H0 overwhelmingly with t(14) = -9.23, p < 0.001, and the one-sided 95% CI rules out any drop smaller than 6.88 mmHg. Cohen's d is -2.38, far beyond "large." The APA-style report writes itself:

A paired-samples t-test showed a significant drop in systolic blood pressure after six weeks of medication (M_before = 147.6, M_after = 139.4), t(14) = -9.23, p < .001, one-sided 95% CI [-∞, -6.88], Cohen's d = -2.38.

This template, simulate or load → check assumption → test → effect size → one-sentence APA report, is the exact sequence to reuse when writing up any of the twelve exercises above against your own data.

Summary

Test	R call	Assumption check	Effect size
One-sample	`t.test(x, mu = m0)`	Shapiro on `x`	`(mean(x) - m0) / sd(x)`
Welch two-sample	`t.test(y ~ g, data = d)`	Shapiro per group	Cohen's d with pooled SD
Student two-sample	`t.test(y ~ g, data = d, var.equal = TRUE)`	Shapiro per group + `var.test()`	Cohen's d with pooled SD
Paired	`t.test(y ~ g, data = d, paired = TRUE)`	Shapiro on the differences	`mean(diff) / sd(diff)`

Key habits you should now have locked in:

Pick the test from the data layout, not from what looks convenient.
Always extract the named fields ($p.value, $conf.int, $estimate) rather than eyeballing the printed block.
Run the assumption check on the correct quantity (raw groups for two-sample, differences for paired).
Report p-value, 95% CI, and an effect size together, never the p-value alone.
Commit to one-sided direction before seeing the data; do not use it as a p-hacking rescue.

References

R Core Team, t.test() reference documentation. Link
Student, "The Probable Error of a Mean." Biometrika 6(1), 1908. The original t-test paper. Link
Welch, B. L., "The generalization of Student's problem when several different population variances are involved." Biometrika 34, 1947. Link
Navarro, D., Learning Statistics with R, Chapter 13: Comparing two means. Link
Diez, Barr, Çetinkaya-Rundel, OpenIntro Statistics, 4th ed., Chapter 7. Link
Cohen, J., Statistical Power Analysis for the Behavioral Sciences, 2nd ed., Routledge, 1988. Effect-size thresholds for d.
Wickham, H. & Grolemund, G., R for Data Science. Link

Continue Learning

t-Tests in R, the canonical reference that covers the decision rule, assumption checks, and effect sizes in depth; use it when any exercise above leaves you wanting more theory.
Hypothesis Testing in R, situates the t-test inside the broader hypothesis-testing framework, alongside chi-square, proportion tests, and non-parametric alternatives.
Effect Size in R, deep-dive on Cohen's d, Hedges' g, and Glass's delta, with the pooled-SD formulas that Exercise 8 and the Complete Example use.

t-Test Exercises in R: 12 One, Two & Paired Sample Problems, Solved Step-by-Step

Which t-test matches your question setup?

How do you read and report a t-test result in R?

Practice Exercises

Exercise 1: Two-sided test against a hypothesized mean

Exercise 2: One-tailed test (less)

Exercise 3: Extract a 99% confidence interval

Exercise 4: Small-sample simulation

Exercise 5: Welch two-sample with formula notation

Exercise 6: Student two-sample with var.equal = TRUE

Exercise 7: One-tailed two-sample test

Exercise 8: Cohen's d for a two-sample comparison

Exercise 9: Classic paired test

Exercise 10: Paired vs independent on the same data

Exercise 11: One-tailed paired test

Exercise 12: Assumption check before a paired test

Complete Example: Medication Effect on Systolic Blood Pressure

Summary

References

Continue Learning

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