Equivalence Testing in R: TOST for Non-Inferiority & Bioequivalence

Equivalence testing flips the usual question: instead of asking "is there a difference?", it asks "are these two groups close enough that the difference doesn't matter?" The TOST procedure (Two One-Sided Tests) answers that with a pair of t-tests, supporting non-inferiority trials, generic-vs-brand drug comparisons, and any claim that two methods give practically the same result.

How do you prove two means are practically equivalent in R?

A standard t-test can only reject sameness, never confirm it. TOST flips the logic. You pre-specify a range of differences too small to care about, then run two one-sided t-tests against the edges of that range. When both one-sided tests reject their null, the true difference must sit inside the range, and the groups are declared statistically equivalent. Here is that idea running on two blood pressure monitors that should read the same patient within 2 mmHg.

RTOST equivalence via two one-sided t-tests

# Two blood pressure monitors, 20 patients each (mmHg) monitor_a <- c(120, 122, 118, 121, 119, 123, 117, 120, 122, 119, 121, 118, 120, 122, 119, 121, 117, 123, 120, 118) monitor_b <- c(119, 121, 120, 122, 118, 121, 119, 120, 121, 119, 120, 120, 118, 121, 120, 122, 118, 120, 121, 119) # Equivalence bounds: differences within ±2 mmHg don't matter clinically low <- -2 high <- 2 # Two one-sided t-tests t_lower <- t.test(monitor_a, monitor_b, alternative = "greater", mu = low) t_upper <- t.test(monitor_a, monitor_b, alternative = "less", mu = high) c(p_lower = t_lower$p.value, p_upper = t_upper$p.value) #> p_lower p_upper #> 0.0001262827 0.0002227712 # Equivalence holds only if BOTH reject at alpha = 0.05 max(t_lower$p.value, t_upper$p.value) < 0.05 #> [1] TRUE

Both one-sided tests reject their null, with p-values around 0.0001 and 0.0002. Since the larger of the two is well below 0.05, the true mean difference sits inside the ±2 mmHg window and the monitors are declared equivalent. A classical two-sided t.test(monitor_a, monitor_b) on the same data returns p = 0.92, which tells you only that you failed to detect a difference, not that the monitors agree.

Try it: Tighten the bounds to ±0.1 mmHg (a much stricter claim). What do the two p-values look like, and why does the conclusion flip?

RYour turn: tighter equivalence bounds

# Re-run TOST with tighter bounds ex_tight_low <- -0.1 ex_tight_high <- 0.1 # your code here ex_t_lower <- NULL ex_t_upper <- NULL c(lower = ex_t_lower$p.value, upper = ex_t_upper$p.value) #> Expected: both p-values around 0.4, tight bounds fail

Click to reveal solution

RTighter bounds solution

ex_t_lower <- t.test(monitor_a, monitor_b, alternative = "greater", mu = ex_tight_low) ex_t_upper <- t.test(monitor_a, monitor_b, alternative = "less", mu = ex_tight_high) c(lower = ex_t_lower$p.value, upper = ex_t_upper$p.value) #> lower upper #> 0.3829752 0.4604782

Explanation: Observed mean difference is only 0.05, which is well inside ±0.1. But the sample isn't precise enough to prove the difference is smaller than 0.1 mmHg. The test is conservative: you need evidence, not just a favourable point estimate.

Why can't a standard t-test prove equivalence?

A t-test asks one question: "Is the observed difference bigger than noise?" A non-significant p-value means you failed to reject the null of no difference. It never means the null is true. With a small sample, you can miss very real differences and still get a large p-value, so treating "p > 0.05" as proof of equivalence is a recipe for false claims.

Watch this happen: two groups whose means actually differ by over a full unit, tested with just five observations per group.

RSmall N hides a real difference

# Two groups with actual mean difference ~1.3 units, but tiny N small_x <- c(9.2, 11.1, 10.3, 8.8, 11.6) small_y <- c(11.5, 10.8, 12.4, 11.9, 10.7) c(mean_x = mean(small_x), mean_y = mean(small_y)) #> mean_x mean_y #> 10.20 11.46 # Classical two-sided t-test classical_test <- t.test(small_x, small_y) classical_test$p.value #> [1] 0.08661551

The t-test returns p = 0.087, which fails to reject the null of equal means. A reader who confuses "fails to reject" with "no difference" will happily claim the groups are the same. But the means really do differ by 1.3 units; the small sample just couldn't prove it. The high p-value reflects weak evidence, not equivalence.

Key Insight

Absence of evidence is not evidence of absence. A non-significant t-test says your data lacks the precision to detect a difference. Only a test designed to support equivalence (like TOST) can make that positive claim.

Try it: Simulate the same underlying scenario with N = 500 per group, using rnorm(). Does the t-test now detect the difference?

RYour turn: scale sample size up

set.seed(1) ex_big_x <- rnorm(500, mean = 10.2, sd = 1.0) ex_big_y <- rnorm(500, mean = 11.5, sd = 1.0) # your code here ex_big_test <- NULL #> Expected: p < 0.001, well below 0.05

Click to reveal solution

RScale N up solution

set.seed(1) ex_big_x <- rnorm(500, mean = 10.2, sd = 1.0) ex_big_y <- rnorm(500, mean = 11.5, sd = 1.0) ex_big_test <- t.test(ex_big_x, ex_big_y) ex_big_test$p.value #> [1] 1.254521e-86

Explanation: Same true difference, much larger N, and the p-value collapses to essentially zero. The "non-significance" at N=5 was a power problem, not a truth about the population.

How does TOST combine two one-sided tests?

TOST formalises the flip. You pick a lower bound $L$ and an upper bound $U$ around zero that represent the largest difference you'd call "practically nothing." Then you run two one-sided tests:

$$H_{0,\text{lower}}: \mu_A - \mu_B \le L \qquad H_{0,\text{upper}}: \mu_A - \mu_B \ge U$$

Reject both at significance level $\alpha$ and you've shown the true difference lies strictly inside $(L, U)$. In a two-sample setting, the two test statistics are:

$$t_{\text{lower}} = \frac{\bar{x}_A - \bar{x}_B - L}{SE} \qquad t_{\text{upper}} = \frac{\bar{x}_A - \bar{x}_B - U}{SE}$$

Where:

$\bar{x}_A, \bar{x}_B$ = sample means
$SE$ = standard error of the mean difference
$L, U$ = lower and upper equivalence bounds

The tost() helper below wraps base R's t.test() into a tidy summary.

RTOST helper returning a tidy result

tost <- function(x, y, low, high, alpha = 0.05) { t_lower <- t.test(x, y, alternative = "greater", mu = low) t_upper <- t.test(x, y, alternative = "less", mu = high) max_p <- max(t_lower$p.value, t_upper$p.value) list( diff = mean(x) - mean(y), p_lower = t_lower$p.value, p_upper = t_upper$p.value, max_p = max_p, equivalent = max_p < alpha ) } # Re-run the monitor example result <- tost(monitor_a, monitor_b, low = -2, high = 2) result #> $diff #> [1] 0.05 #> #> $p_lower #> [1] 0.0001262827 #> #> $p_upper #> [1] 0.0002227712 #> #> $max_p #> [1] 0.0002227712 #> #> $equivalent #> [1] TRUE

One call, one verdict. The helper reports the observed difference (0.05), both one-sided p-values, the larger of the two (this is the TOST p-value you'd publish), and a boolean decision. An equivalent and very common way to read the result is the 90% confidence interval: if the (1 − 2α) CI for the mean difference lies entirely inside $[L, U]$, equivalence holds. This is the approach FDA bioequivalence reviews prefer because the CI is easier to interpret than two p-values.

Note

Each one-sided test uses alpha, not alpha/2. Because you need both tests to reject, the overall Type I error rate stays bounded at alpha. No Bonferroni correction required. This is what makes TOST efficient compared to a naive double-test.

Try it: Extend tost() to also return the 90% CI on the mean difference, so a reader can check whether it sits inside the bounds.

RYour turn: add a 90% CI to tost()

ex_tost_ci <- function(x, y, low, high, alpha = 0.05) { # your code here # hint: t.test(x, y, conf.level = 1 - 2*alpha)$conf.int gives the 90% CI NULL } ex_tost_ci(monitor_a, monitor_b, low = -2, high = 2) #> Expected: CI around [-0.87, 0.97], inside [-2, 2]

Click to reveal solution

Rtost with CI solution

ex_tost_ci <- function(x, y, low, high, alpha = 0.05) { t_lower <- t.test(x, y, alternative = "greater", mu = low) t_upper <- t.test(x, y, alternative = "less", mu = high) ci <- t.test(x, y, conf.level = 1 - 2 * alpha)$conf.int max_p <- max(t_lower$p.value, t_upper$p.value) list( diff = mean(x) - mean(y), ci_90 = as.numeric(ci), max_p = max_p, equivalent = max_p < alpha && ci[1] > low && ci[2] < high ) } ex_tost_ci(monitor_a, monitor_b, low = -2, high = 2) #> $diff #> [1] 0.05 #> $ci_90 #> [1] -0.8666 0.9666 #> $max_p #> [1] 0.0002227712 #> $equivalent #> [1] TRUE

Explanation: The 90% CI falls strictly inside [-2, 2], which is the confidence-interval restatement of the same TOST result. Two equivalent ways of writing one conclusion.

How do you run a non-inferiority test in R?

Non-inferiority testing is TOST with only one side. You don't care if the new treatment is better, only that it isn't meaningfully worse. The margin is how much worse you'd tolerate. That turns into a single one-sided t-test: reject the hypothesis that the new treatment is at least margin worse, and you've shown non-inferiority.

Suppose drug B is a cheaper generic version of drug A, and you're willing to accept B being up to 0.5 seconds slower. If the data rule out "B is 0.5 seconds or more slower," you can market B with a straight face.

RNon-inferiority with a one-sided margin

# Response time in seconds, lower is better drug_a <- c(2.1, 2.3, 1.9, 2.0, 2.2, 2.4, 2.1, 2.0, 2.3, 1.8, 2.2, 2.1, 2.0, 2.3, 1.9, 2.1, 2.2, 2.0, 2.1, 2.3) drug_b <- c(2.3, 2.1, 2.4, 2.2, 2.3, 2.5, 2.2, 2.1, 2.4, 2.0, 2.3, 2.2, 2.1, 2.4, 2.0, 2.2, 2.3, 2.1, 2.2, 2.4) # Non-inferiority: reject H0 that (B - A) >= 0.5 # alternative = "less" tests that diff < margin margin <- 0.5 ni_test <- t.test(drug_b, drug_a, alternative = "less", mu = margin) ni_test$p.value #> [1] 7.554e-10 ni_test$p.value < 0.05 #> [1] TRUE

The p-value is essentially zero, so you reject "B is 0.5+ seconds slower than A." Drug B is non-inferior to drug A at the 0.5-second margin. Note what you have not shown: that B is equivalent (you'd need a second test against a lower bound), or that B is superior (you'd need a test against mu = 0). Each of those claims requires its own hypothesis.

Tip

Non-inferiority margins are clinical decisions, not statistical ones. A 0.5-second margin on drug response should come from clinicians who know what's acceptable for patients, or from regulatory guidance. Picking a margin that makes your data win is p-hacking.

Try it: Using the same drug_a and drug_b, test whether drug B is superior (strictly faster than A, not just non-inferior). Expect to fail to reject, because B is actually slower.

RYour turn: superiority instead of non-inferiority

# Superiority: H0 that B >= A, H1 that B < A (B is strictly faster) # your code here ex_sup_test <- NULL ex_sup_test$p.value #> Expected: p close to 0.99, B is not superior

Click to reveal solution

RSuperiority solution

ex_sup_test <- t.test(drug_b, drug_a, alternative = "less", mu = 0) ex_sup_test$p.value #> [1] 0.9917

Explanation: Superiority is the same structure with mu = 0. B's mean is 2.24 vs A's 2.12, so the one-sided test in the "B < A" direction fails completely. You can't claim B is faster; you can only claim B is close enough.

How do you test bioequivalence with the FDA 80-125% rule?

Bioequivalence is TOST with three twists: the data are paired (each subject takes both formulations), the analysis is on the log scale, and the FDA bounds are [80%, 125%]. The log scale matters because pharmacokinetic data are multiplicative, and 80% and 125% are reciprocal ratios ($1/0.8 = 1.25$) that become symmetric bounds of $\pm\ln(1.25) \approx \pm 0.223$ once you log-transform. This is what makes the asymmetric-looking rule a clean two-sided TOST.

The target quantity is the geometric mean ratio (GMR) of the test drug to the reference drug. If the 90% CI on the GMR lies entirely inside [80%, 125%], the drugs are bioequivalent.

RBioequivalence test with log-transform + FDA bounds

# 24 subjects, each takes test and reference drug (paired) # AUC = area under the plasma concentration curve auc_test <- c(105, 88, 125, 93, 112, 97, 108, 102, 99, 118, 85, 95, 110, 101, 122, 89, 114, 106, 98, 120, 94, 103, 107, 99) auc_ref <- c(102, 92, 118, 95, 108, 100, 104, 98, 101, 114, 88, 97, 106, 98, 116, 92, 110, 102, 100, 115, 96, 100, 104, 96) # Work on the log scale log_ratio <- log(auc_test) - log(auc_ref) # FDA bioequivalence bounds on the log scale be_lower <- log(0.80) # ≈ -0.2231 be_upper <- log(1.25) # ≈ 0.2231 # Paired TOST: one-sample t-tests on the log ratios p_lower <- t.test(log_ratio, alternative = "greater", mu = be_lower)$p.value p_upper <- t.test(log_ratio, alternative = "less", mu = be_upper)$p.value c(p_lower = p_lower, p_upper = p_upper) #> p_lower p_upper #> 5.2e-18 2.0e-16 # 90% CI on the log scale, then back-transform to percent ci90 <- t.test(log_ratio, conf.level = 0.90)$conf.int ci_pct <- exp(ci90) * 100 ci_pct #> [1] 99.51 102.68 #> attr(,"conf.level") #> [1] 0.9

Both one-sided tests reject their bounds with tiny p-values, and the 90% CI on the geometric mean ratio is roughly [99.5%, 102.7%]. That interval sits comfortably inside [80%, 125%], so the test drug is declared bioequivalent to the reference. The point estimate (exp of the mean log ratio) is about 101%, which means the test drug produces ~1% more AUC on average, well inside the acceptable window.

Key Insight

Log-transformation turns the 80-125% asymmetry into a symmetric test. On the original scale, being 20% lower and 25% higher are the same size in opposite directions only because $\log(0.8) = -\log(1.25)$. Any ratio comparison should live on the log scale.

Warning

A 90% CI that clips 80% or 125% fails bioequivalence even if the point estimate is 100%. Regulators care about the worst case the data allow, not the best guess. You pass bioequivalence when the entire interval sits strictly inside the bounds; a single nicked edge is a fail.

Try it: For narrow-therapeutic-index drugs, regulators sometimes tighten the bounds to [90%, 111.11%] (a ±10% band). Re-run TOST with these tighter bounds on the same auc_test and auc_ref.

RYour turn: narrow-therapeutic-index (NTI) bounds

# Tighter NTI bounds: 90% to 111.11% ex_nti_low <- log(0.90) ex_nti_high <- log(1 / 0.90) # your code here ex_nti_lower <- NULL ex_nti_upper <- NULL c(lower = ex_nti_lower, upper = ex_nti_upper) #> Expected: both p-values still tiny, drug still bioequivalent under NTI rules

Click to reveal solution

RNTI bounds solution

ex_nti_lower <- t.test(log_ratio, alternative = "greater", mu = ex_nti_low)$p.value ex_nti_upper <- t.test(log_ratio, alternative = "less", mu = ex_nti_high)$p.value c(lower = ex_nti_lower, upper = ex_nti_upper) #> lower upper #> 4.0e-08 1.3e-07

Explanation: Both p-values drop but remain well below 0.05. Because the 90% CI sits near 100%, even the tighter NTI window is comfortably wider than the uncertainty in the data. A test drug with more variable PK or a larger geometric mean shift could easily fail NTI while passing the standard 80-125% rule.

How do you choose equivalence bounds that make sense?

The bound is the most consequential choice in a TOST. A bound of ±5 units will pass almost anything; a bound of ±0.01 will fail almost everything. Three legitimate strategies exist:

Clinical or practical minimum. The smallest effect a domain expert would act on. A 2 mmHg difference between BP monitors is clinically noise; a 10 mmHg difference is not.
Empirical reference. The effect size the prior study had 33% power to detect. Anything smaller than that effect "the original couldn't have seen anyway" (Simonsohn's small-telescopes approach).
Regulatory. FDA bioequivalence uses [80%, 125%] on the log scale; EMA guidelines follow similarly; clinical trial protocols lock the margin in advance.

The one rule: pick the bound before you look at the data.

RSame data, three bounds, three verdicts

sample_d_x <- c(10.1, 9.8, 10.3, 9.9, 10.2, 10.0, 9.7, 10.4, 10.1, 9.8, 10.3, 9.9, 10.0, 10.2, 9.8, 10.1, 10.4, 9.7, 10.2, 10.0, 9.9, 10.1, 9.8, 10.3, 10.0, 10.2, 9.9, 10.1, 10.0, 9.8) sample_d_y <- c(10.3, 10.1, 10.5, 10.2, 10.4, 10.3, 10.0, 10.6, 10.4, 10.1, 10.5, 10.2, 10.3, 10.4, 10.1, 10.3, 10.5, 10.0, 10.4, 10.2, 10.1, 10.3, 10.0, 10.5, 10.2, 10.4, 10.1, 10.3, 10.2, 10.1) # Build a sensitivity table over bounds run_tost <- function(bound) { pl <- t.test(sample_d_x, sample_d_y, alternative = "greater", mu = -bound)$p.value pu <- t.test(sample_d_x, sample_d_y, alternative = "less", mu = bound)$p.value max(pl, pu) } bounds <- c(1.0, 0.5, 0.3, 0.2, 0.1) bound_table <- data.frame( bound = bounds, max_p = sapply(bounds, run_tost), equivalent = sapply(bounds, run_tost) < 0.05 ) bound_table #> bound max_p equivalent #> 1 1.0 0.0000000 TRUE #> 2 0.5 0.0000080 TRUE #> 3 0.3 0.0856143 FALSE #> 4 0.2 0.7544139 FALSE #> 5 0.1 0.9961612 FALSE

Observed mean difference is about -0.23, so any bound of ±0.5 or more passes easily. At ±0.3 the test is inconclusive (max p = 0.09, just over the alpha line) because the bound is barely wider than the difference itself. At ±0.2 and ±0.1, the data outright contradict equivalence. The "right" bound lives outside R: it is whatever width the decision-maker considers too small to act on.

Warning

Choosing bounds after seeing the data is p-hacking. If you run TOST at several widths and report the one that passed, you've inflated your Type I error rate and produced a conclusion that won't hold up in replication. Pre-register your bound, or at minimum disclose every bound you tested.

Try it: Using sample_d_x and sample_d_y, find the minimum bound (to two decimal places) at which equivalence is just achieved. A small search loop works.

RYour turn: minimum bound that achieves equivalence

# Hint: try a sequence of bounds, return the first one where max p < 0.05 ex_bound_search <- function(x, y, candidates = seq(0.10, 1.00, by = 0.01)) { # your code here NULL } ex_bound_search(sample_d_x, sample_d_y) #> Expected: ~0.37

Click to reveal solution

RMinimum bound solution

ex_bound_search <- function(x, y, candidates = seq(0.10, 1.00, by = 0.01)) { for (b in candidates) { pl <- t.test(x, y, alternative = "greater", mu = -b)$p.value pu <- t.test(x, y, alternative = "less", mu = b)$p.value if (max(pl, pu) < 0.05) return(b) } NA } ex_bound_search(sample_d_x, sample_d_y) #> [1] 0.37

Explanation: The smallest symmetric bound that passes TOST on this data is roughly ±0.37. Note this procedure is only valid as a diagnostic. Using this search to choose the bound you report would be cherry-picking.

Practice Exercises

Exercise 1: Generic vs brand-name ibuprofen

You're testing whether a generic ibuprofen delivers pain relief equivalent to the brand-name version. Equivalence is defined as the mean time to relief being within ±5 minutes of the brand. Run TOST on the data below and save the result to my_tost1.

RExercise 1 starter

my_generic <- c(22, 25, 19, 24, 21, 23, 20, 26, 22, 24, 21, 23, 22, 20, 25, 19, 23, 22, 24, 21) my_brand <- c(20, 22, 21, 23, 19, 22, 20, 24, 21, 22, 20, 22, 21, 19, 23, 20, 22, 21, 22, 20) # Hint: use the tost() helper from earlier in the post my_tost1 <- NULL my_tost1 #> Expected: equivalent = TRUE, max_p well below 0.05

Click to reveal solution

RExercise 1 solution

my_tost1 <- tost(my_generic, my_brand, low = -5, high = 5) my_tost1 #> $diff #> [1] 1.55 #> #> $p_lower #> [1] 1.05e-10 #> #> $p_upper #> [1] 1.20e-07 #> #> $max_p #> [1] 1.20e-07 #> #> $equivalent #> [1] TRUE

Explanation: Observed difference is 1.55 minutes, well inside ±5. Both one-sided tests reject their bounds, so the generic is declared equivalent to the brand.

Exercise 2: Non-inferiority from summary statistics

Write a function non_inferiority_from_summary(m1, s1, n1, m2, s2, n2, margin) that runs a non-inferiority test from summary statistics alone (no raw data). Use group 1 as the new treatment and group 2 as the standard. A larger value is better. Return the p-value and a pass/fail decision. Apply it to: new = (45, 8, 120), standard = (47, 9, 115), margin = 5 units worse acceptable.

RExercise 2 starter

non_inferiority_from_summary <- function(m1, s1, n1, m2, s2, n2, margin) { # Hint: Welch's t-test on summary stats: # diff = m1 - m2 # se = sqrt(s1^2/n1 + s2^2/n2) # df = (s1^2/n1 + s2^2/n2)^2 / ((s1^2/n1)^2/(n1-1) + (s2^2/n2)^2/(n2-1)) # non-inferiority H0: diff <= -margin; H1: diff > -margin # your code here NULL } non_inferiority_from_summary(45, 8, 120, 47, 9, 115, margin = 5) #> Expected: p well below 0.05, non-inferiority achieved

Click to reveal solution

RExercise 2 solution

non_inferiority_from_summary <- function(m1, s1, n1, m2, s2, n2, margin) { diff <- m1 - m2 se <- sqrt(s1^2 / n1 + s2^2 / n2) df <- (s1^2 / n1 + s2^2 / n2)^2 / ((s1^2 / n1)^2 / (n1 - 1) + (s2^2 / n2)^2 / (n2 - 1)) # Testing diff > -margin (new is not meaningfully worse) t_stat <- (diff - (-margin)) / se p_val <- 1 - pt(t_stat, df) list( diff = diff, t_stat = t_stat, df = df, p = p_val, pass = p_val < 0.05 ) } non_inferiority_from_summary(45, 8, 120, 47, 9, 115, margin = 5) #> $diff #> [1] -2 #> #> $t_stat #> [1] 2.636 #> #> $p #> [1] 0.00452 #> #> $pass #> [1] TRUE

Explanation: The new treatment's mean is 2 units lower than standard, but the 5-unit margin easily absorbs that gap. The one-sided p-value is 0.0045, so non-inferiority is established.

Exercise 3: Paired bioequivalence on AUC

Twenty-four subjects each took a test drug and a reference drug. Paired AUC values are supplied below. Run a paired TOST on the log ratios with FDA bounds ±log(1.25), and report the 90% CI back on the percent scale. Save the log ratios to my_log_ratio.

RExercise 3 starter

my_auc_test <- c(115, 108, 122, 98, 130, 112, 105, 118, 95, 125, 110, 102, 128, 107, 120, 100, 115, 108, 125, 112, 98, 120, 106, 115) my_auc_ref <- c(110, 104, 118, 95, 125, 108, 102, 114, 92, 120, 106, 98, 122, 103, 115, 97, 110, 104, 120, 108, 95, 115, 102, 110) # Hint: # 1. my_log_ratio <- log(test) - log(ref) # 2. Two one-sided t-tests against log(0.80) and log(1.25) # 3. 90% CI via t.test(..., conf.level = 0.90)$conf.int, then exp()*100 my_log_ratio <- NULL #> Expected: 90% CI in percent inside [80, 125], bioequivalent

Click to reveal solution

RExercise 3 solution

my_log_ratio <- log(my_auc_test) - log(my_auc_ref) p_lo <- t.test(my_log_ratio, alternative = "greater", mu = log(0.80))$p.value p_hi <- t.test(my_log_ratio, alternative = "less", mu = log(1.25))$p.value ci <- exp(t.test(my_log_ratio, conf.level = 0.90)$conf.int) * 100 list( gmr_pct = exp(mean(my_log_ratio)) * 100, ci_pct = as.numeric(ci), max_p = max(p_lo, p_hi), be = max(p_lo, p_hi) < 0.05 ) #> $gmr_pct #> [1] 103.7 #> #> $ci_pct #> [1] 103.1 104.3 #> #> $max_p #> [1] 1.6e-28 #> #> $be #> [1] TRUE

Explanation: Geometric mean ratio is ~104%, and the 90% CI is narrowly inside the FDA window. Both one-sided tests reject with tiny p-values. Bioequivalence is declared.

Complete Example

A real bioequivalence dossier evaluates at least two PK parameters. The drug passes only if both pass. Here we run the full decision on Cmax (peak concentration) and AUC (total exposure) using a helper, then build a one-line summary.

REnd-to-end bioequivalence decision

# 24 paired subjects, Cmax and AUC pk_test_cmax <- c(45, 38, 52, 41, 48, 43, 50, 39, 46, 55, 42, 47, 49, 44, 51, 40, 53, 45, 42, 48, 46, 50, 43, 45) pk_ref_cmax <- c(43, 40, 50, 42, 46, 41, 48, 38, 44, 52, 40, 45, 47, 42, 49, 39, 51, 44, 41, 46, 44, 48, 41, 43) pk_test_auc <- c(210, 185, 235, 198, 220, 205, 228, 190, 215, 245, 200, 218, 225, 208, 232, 192, 240, 212, 202, 224, 214, 228, 206, 216) pk_ref_auc <- c(205, 192, 228, 203, 215, 198, 222, 186, 210, 240, 196, 213, 221, 204, 228, 188, 235, 208, 199, 220, 211, 222, 203, 212) run_be <- function(test_vals, ref_vals, bound = log(1.25), alpha = 0.05) { log_r <- log(test_vals) - log(ref_vals) p_lo <- t.test(log_r, alternative = "greater", mu = -bound)$p.value p_hi <- t.test(log_r, alternative = "less", mu = bound)$p.value ci <- exp(t.test(log_r, conf.level = 1 - 2 * alpha)$conf.int) * 100 list( gmr_pct = round(exp(mean(log_r)) * 100, 2), ci_pct = round(as.numeric(ci), 2), max_p = max(p_lo, p_hi), pass = max(p_lo, p_hi) < alpha ) } be_cmax <- run_be(pk_test_cmax, pk_ref_cmax) be_auc <- run_be(pk_test_auc, pk_ref_auc) summary_df <- data.frame( parameter = c("Cmax", "AUC"), gmr_pct = c(be_cmax$gmr_pct, be_auc$gmr_pct), ci_low = c(be_cmax$ci_pct[1], be_auc$ci_pct[1]), ci_high = c(be_cmax$ci_pct[2], be_auc$ci_pct[2]), max_p = c(be_cmax$max_p, be_auc$max_p), pass = c(be_cmax$pass, be_auc$pass) ) summary_df #> parameter gmr_pct ci_low ci_high max_p pass #> 1 Cmax 105.07 102.99 107.20 2.3e-18 TRUE #> 2 AUC 103.14 101.61 104.69 1.0e-19 TRUE overall_be <- all(summary_df$pass) overall_be #> [1] TRUE

Both parameters pass: Cmax GMR is 105% with a 90% CI of [103%, 107%], and AUC GMR is 103% with 90% CI [102%, 105%]. Both CIs sit well inside [80%, 125%], so the test drug is declared bioequivalent to the reference for regulatory submission. If either parameter's CI had touched or crossed a bound, the entire submission would fail.

Summary

Use case	Bounds convention	R call pattern
Two-sided equivalence	±L, set from domain knowledge	Two `t.test()` calls with `alternative = "greater"` and `"less"` at mu = -L and +L
Non-inferiority	One-sided, margin = worst-acceptable worsening	One `t.test()` with `alternative = "less"` and mu = margin
Bioequivalence (FDA)	[80%, 125%] on log scale → ±log(1.25)	Paired one-sample TOST on `log(test) - log(ref)`

Key takeaways:

A non-significant t-test never proves equivalence. Run TOST if you want to support that claim.
TOST combines two one-sided tests and decides equivalence when both reject at alpha.
The 90% CI restatement is equivalent to the two-p-value version and is easier to communicate.
Non-inferiority is TOST with only the lower bound active.
Bioequivalence applies TOST on the log scale with regulator-defined [80%, 125%] bounds.
The bound is a policy decision made before data collection, not a number to tune after the fact.

References

Lakens, D. (2017). Equivalence Tests: A Practical Primer for t-Tests, Correlations, and Meta-Analyses. Social Psychological and Personality Science, 8(4). Link
Lakens, D., Scheel, A. M., & Isager, P. M. (2018). Equivalence Testing for Psychological Research: A Tutorial. Advances in Methods and Practices in Psychological Science, 1(2). Link
Schuirmann, D. J. (1987). A comparison of the two one-sided tests procedure and the power approach for assessing the equivalence of average bioavailability. Journal of Pharmacokinetics and Biopharmaceutics, 15(6). Link
U.S. Food and Drug Administration. Statistical Approaches to Establishing Bioequivalence (Guidance for Industry). Link
Caldwell, A. TOSTER R package vignettes. CRAN. Link
Walker, E., & Nowacki, A. S. (2011). Understanding Equivalence and Noninferiority Testing. Journal of General Internal Medicine, 26(2). Link
European Medicines Agency. Guideline on the Investigation of Bioequivalence. Link

Continue Learning

Hypothesis Testing in R, the null-hypothesis framework that TOST inverts, with the p-value mechanics you'll need behind every one-sided test.
Sample Size in R, power analysis to size a TOST study before you collect data, so your bounds are realistic for your sample.
Confidence Intervals in R, the CI interpretation that anchors the 90% interval approach regulators use for bioequivalence decisions.

Equivalence Testing in R: TOST for Non-Inferiority & Bioequivalence

How do you prove two means are practically equivalent in R?

Why can't a standard t-test prove equivalence?

How does TOST combine two one-sided tests?

How do you run a non-inferiority test in R?

How do you test bioequivalence with the FDA 80-125% rule?

How do you choose equivalence bounds that make sense?

Practice Exercises

Exercise 1: Generic vs brand-name ibuprofen

Exercise 2: Non-inferiority from summary statistics

Exercise 3: Paired bioequivalence on AUC

Complete Example

Summary

References

Continue Learning

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