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Non-Inferiority and Equivalence

Equivalence Test for Means (Two One-Sided Tests)

A two-sample equivalence test for independent means using Schuirmann’s two one-sided tests (TOST). Enter summary statistics and pre-specified equivalence limits to determine whether the confidence interval for the mean difference lies completely within the equivalence region. Runs entirely in your browser.

Two Independent Groups

This implementation uses the equal-variance two-sample t-test, matching the PASS procedure for equivalence tests of two means using differences.

TOST Results

Equivalence is established only when both one-sided null hypotheses are rejected.
Enter summary statistics and click Run Equivalence Test.

Methodology

The calculator implements the two one-sided tests (TOST) procedure for equivalence of two independent means, using the difference in means as the parameter of interest and a common pooled variance. The equivalence region is the interval from the lower equivalence limit ΔL to the upper equivalence limit ΔU.

Hypotheses

H01: μ1 − μ2 ≤ ΔL
H02: μ1 − μ2 ≥ ΔU
HA: ΔL < μ1 − μ2 < ΔU

Each null hypothesis is tested at the specified one-sided significance level α. Equivalence is concluded only if both null hypotheses are rejected.

Test Statistics

For the equal-variance two-sample t-test, the pooled standard deviation and standard error are

sp = √[((n1−1)s12 + (n2−1)s22) / (n1+n2−2)]

SE = sp √(1/n1 + 1/n2)

With D = x̄1 − x̄2, the two TOST statistics are tL = (D − ΔL)/SE and tU = (D − ΔU)/SE, with n1 + n2 − 2 degrees of freedom. The lower-side p-value is P(T ≥ tL), while the upper-side p-value is P(T ≤ tU).

Confidence-Interval Interpretation

The TOST decision is equivalent to checking whether the two-sided 100(1−2α)% confidence interval for the mean difference is contained completely inside the equivalence limits. With α = 0.05, this is a 90% confidence interval.

Validation Example

A numerical verification example uses Group 1 mean = 5.2, SD = 1.5, n = 50; Group 2 mean = 5.0, SD = 1.4, n = 50; equivalence limits = −0.5 and +0.5; and α = 0.05. The expected calculations are: pooled SD = 1.4508618, SE = 0.2901724, df = 98, mean difference = 0.2000, tL = 2.4123593 with p = 0.0088550, tU = −1.0338683 with p = 0.1518712, TOST p = 0.1518712, and 90% CI = [−0.2818461, 0.6818461]. Because the interval extends above +0.5, equivalence is not established.

References

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, 657–680.

Phillips, K. F. (1990). Power of the two one-sided tests procedure in bioequivalence. International Journal of Clinical Pharmacology, Therapy and Toxicology, 28, 261–266.

NCSS, LLC. PASS User’s Guide II, Chapter 460: Equivalence Tests for Two Means using Differences. PASS documents the parallel-group TOST procedure and its calculation/validation examples. The current PASS documentation lists the corresponding procedure under Equivalence → Means → Two Independent Means.

Statistical Solutions / nQuery Advisor. nQuery Advisor User’s Guide, Chapter 11: Demonstrating Equivalence. The guide describes the two-group TOST for differences in means and identifies the MTE1tg procedure.