How this calculation works
A small estimated difference is not enough to establish practical equivalence. TOST asks whether the population mean is supported inside two prespecified limits, rather than merely failing to detect a difference from zero. This one-sample calculation evaluates both one-sided t tests and their matching confidence interval. The lower and upper bounds remain explicit raw-unit assumptions instead of being chosen automatically from the data.
tL = (mean−lower)/SE; tU = (mean−upper)/SE; TOST p = max(pL,pU)A worked example
For observations −1, zero and one, sample mean is zero and sample standard deviation is one. Standard error is one divided by the square root of three. With bounds −5 and +5, both boundary t statistics have magnitude 8.66025 and the combined p value is about 0.00653623. Narrowing the bounds to ±0.1 fails to show equivalence with the same observations.
Good to know
The observations must support the independent one-sample t model. No paired or independent two-group equivalence analysis, multiplicity correction or sample-size planning is included. This page does not select acceptable engineering, financial or clinical limits. The statistical calculation cannot supply the substantive criterion that gives practical equivalence its meaning.
Reference: JMP — one sample mean equivalence TOST (opens in a new tab) ↗
What this calculation answers
A small estimated difference is not enough to establish practical equivalence. TOST asks whether the population mean is supported inside two prespecified limits, rather than merely failing to detect a difference from zero.
This one-sample calculation evaluates both one-sided t tests and their matching confidence interval. The lower and upper bounds remain explicit raw-unit assumptions instead of being chosen automatically from the data.
The Welch two sample t test calculator asks about a mean difference against a point null. That ordinary test cannot establish practical equivalence inside two prespecified bounds.
Use the right input conventions
Enter independent observations and bounds in the same measurement unit. Each bound should be selected for a substantive reason before looking at the outcome. Upper bound must exceed lower bound.
Alpha is the significance level of each one-sided test. At alpha 0.05, the matching two-sided interval has ninety-percent coverage, not ninety-five percent. Positive sample variation is required for the finite t model.
Interpret the result
Equivalence is supported only when both one-sided p values are below alpha. Taking their maximum gives the combined TOST p value. A conventional nonsignificant difference test cannot replace that intersection of conditions.
A failure to demonstrate equivalence does not prove a practically important difference. The data may simply be too imprecise, so the estimated mean and interval need to be read alongside the decision.
Work through the example
For observations −1, zero and one, sample mean is zero and sample standard deviation is one. Standard error is one divided by the square root of three.
With bounds −5 and +5, both boundary t statistics have magnitude 8.66025 and the combined p value is about 0.00653623. Narrowing the bounds to ±0.1 fails to show equivalence with the same observations.
Where the model stops
The observations must support the independent one-sample t model. No paired or independent two-group equivalence analysis, multiplicity correction or sample-size planning is included.
This page does not select acceptable engineering, financial or clinical limits. The statistical calculation cannot supply the substantive criterion that gives practical equivalence its meaning.
A few questions, answered
Does a nonsignificant ordinary t test prove equivalence?
No. Equivalence requires evidence inside prespecified bounds, not merely insufficient evidence against zero.
What changes in the comparison?
Narrow the practical bounds without changing the sample; the uncertainty then prevents equivalence.