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Confidence Intervals and the TOST Procedure · Tutorial 217 of 399

TOST With Asymmetric Difference Margins

You will learn to test equivalence with asymmetric difference margins and verify the decision using the corresponding confidence interval.

Intermediate 10 min read

What You'll Learn

  • State the equivalence hypotheses when the lower and upper difference margins are unequal.
  • Calculate and interpret each one-sided test statistic using its own boundary.
  • Connect the two TOST decisions to a confidence interval with the appropriate confidence level.
  • Work through examples where the point estimate is inside the margins but equivalence is not established.
  • Check the assumptions and identify common errors in asymmetric-margin analyses.

Why Unequal Difference Margins Need Care

Equivalence is a claim that a population difference is sufficiently small in either direction. “Sufficiently small” need not mean equally far from zero on both sides. A scientific or clinical justification may set a lower acceptable difference at one value and an upper acceptable difference at another. In that case, the equivalence region is off-center or has unequal widths, and each boundary must be assessed on its own terms.

Let \(\delta\) denote the population treatment difference, defined here as treatment minus reference on the original difference scale. Let \(L\) be the prespecified lower margin and \(U\) the prespecified upper margin, with \(L<U\). The estimates and margins must use the same difference definition and units. For example, if \(\delta\) is measured in milligrams, both margins are also in milligrams.

Definition: With asymmetric difference margins \(L\) and \(U\), equivalence means \(L<\delta<U\). The null hypothesis is \(H_0:\delta\leq L\) or \(\delta\geq U\), and the alternative is \(H_a:L<\delta<U\). The null is the union of two ways equivalence can fail: the difference is too low, or it is too high.

The margins need not be centered at zero, and the distances \(U-\delta\) and \(\delta-L\) are not interchangeable. TOST therefore uses one one-sided test at \(L\) and another at \(U\); both must reject their respective null hypotheses. This is the intersection-union logic of TOST, applied with unequal boundaries.

Model, Assumptions, and Test Calculations

Suppose an analysis produces an estimate \(\hat{\delta}\), its standard error \(SE\), and degrees of freedom \(df\). For the usual t-based procedure, assume the estimate is unbiased for \(\delta\), its standard error is valid under the sampling design and fitted model, and its standardized sampling error follows a t distribution with the stated degrees of freedom. In a small-sample model, this commonly relies on an appropriate normal-error model and on the model’s covariance assumptions. The design and analysis must also correctly represent the units of analysis and any pairing or clustering.

The margins must be chosen before examining the results and justified in context. TOST does not establish that the margins are scientifically acceptable; it assesses the estimate against the supplied margins. If the model or standard error is not credible, the resulting tests and interval are not repaired merely by applying the TOST formulas.

At significance level \(\alpha\), test the lower boundary with \(H_{0L}:\delta\leq L\) against \(H_{aL}:\delta>L\). Then test the upper boundary with \(H_{0U}:\delta\geq U\) against \(H_{aU}:\delta<U\). Using \(t_{1-\alpha,df}\), the \(1-\alpha\) quantile of the t distribution:

$$ t_L=\frac{\hat{\delta}-L}{SE}, \qquad t_U=\frac{\hat{\delta}-U}{SE}. $$

Reject the lower-boundary null if \(t_L>t_{1-\alpha,df}\), and reject the upper-boundary null if \(t_U<-t_{1-\alpha,df}\). Equivalently, the one-sided p-values are \(1-F_{t,df}(t_L)\) and \(F_{t,df}(t_U)\), where \(F_{t,df}\) is the t-distribution cumulative distribution function. Declare equivalence only if both p-values are at most \(\alpha\), using the same rejection convention for both tests.

The two statistics have different boundary values in their numerators. Replacing \(L\) and \(U\) with a symmetric pair around zero, or using the same distance from the estimate for both tests, changes the hypotheses and is not a valid shortcut.

Why the Confidence-Interval Rule Works

The two one-sided tests have a direct confidence-interval interpretation. Let \(c=t_{1-\alpha,df}\). The lower test rejects exactly when

$$ \frac{\hat{\delta}-L}{SE}>c \quad\Longleftrightarrow\quad \hat{\delta}-cSE>L. $$

The upper test rejects exactly when

$$ \frac{\hat{\delta}-U}{SE}<-c \quad\Longleftrightarrow\quad \hat{\delta}+cSE<U. $$

Thus both tests reject precisely when the interval from \(\hat{\delta}-cSE\) to \(\hat{\delta}+cSE\) lies inside the equivalence region. This is a two-sided \(100(1-2\alpha)\%\) t confidence interval: its two tails each have probability \(\alpha\). At \(\alpha=0.05\), the matching interval is a 90% confidence interval, not a 95% interval.

$$ \left[ \hat{\delta}-t_{1-\alpha,df}SE,\; \hat{\delta}+t_{1-\alpha,df}SE \right] \subset (L,U). $$

This containment rule is valid for unequal margins just as it is for equal-width margins. The interval itself is centered at the estimate; the allowed region need not be centered there or at zero. For decisions exactly at a boundary, apply the test’s stated rejection convention consistently; conclusions near equality can be sensitive to rounding.

Worked Example: Both Tests Pass With Unequal Margins

Suppose a hypothetical crossover analysis estimates a treatment-minus-reference difference of \(\hat{\delta}=1.2\) mg, with \(SE=2.0\) mg and \(df=24\). Before analysis, the acceptable difference was specified as \(L=-4\) mg and \(U=6\) mg. Test at \(\alpha=0.05\). Assume the analysis model and its standard error are appropriate for these data.

For \(df=24\), \(t_{0.95,24}\approx1.711\). The lower-boundary statistic is

$$ t_L=\frac{1.2-(-4)}{2.0} =\frac{5.2}{2.0} =2.60. $$

Because \(2.60>1.711\), reject \(H_{0L}\). For the upper boundary,

$$ t_U=\frac{1.2-6}{2.0} =\frac{-4.8}{2.0} =-2.40. $$

Because \(-2.40<-1.711\), reject \(H_{0U}\) as well. Both one-sided tests reject, so the result supports equivalence at the 5% level.

The corresponding 90% interval is

$$ 1.2\pm(1.711)(2.0) =1.2\pm3.422 =[-2.222,\;4.622]\text{ mg}. $$

Both endpoints fall inside \((-4,6)\) mg, confirming the test decision. This conclusion concerns the population difference under the model and prespecified margins; it does not mean that the probability the true difference is in this particular interval is 90%.

Examples Where the Boundaries Matter

Worked Example: Estimate Inside the Margins, but Equivalence Is Not Established

In another hypothetical analysis, let \(\hat{\delta}=3.0\) mg, \(SE=1.8\) mg, and \(df=24\), with prespecified margins \(L=-4\) mg and \(U=5\) mg. The estimate lies inside the margins, but that fact alone is not enough to establish equivalence. Use \(t_{0.95,24}\approx1.711\).

At the lower boundary,

$$ t_L=\frac{3.0-(-4)}{1.8} =\frac{7.0}{1.8} \approx3.889. $$

This exceeds 1.711, so the lower-boundary null is rejected. At the upper boundary,

$$ t_U=\frac{3.0-5}{1.8} =\frac{-2.0}{1.8} \approx-1.111. $$

This is not less than \(-1.711\), so the upper-boundary null is not rejected. Since both tests must reject, equivalence is not established. The 90% interval makes the reason visible:

$$ 3.0\pm(1.711)(1.8) =3.0\pm3.0798 =[-0.0798,\;6.0798]\text{ mg}. $$

The upper endpoint exceeds the upper margin of 5 mg. The data are compatible with a difference above that limit, despite the point estimate being within the allowed region.

Worked Example: An Off-Center Equivalence Region

Suppose a third hypothetical analysis gives \(\hat{\delta}=-0.5\) mg, \(SE=1.1\) mg, and \(df=24\). The prespecified margins are \(L=-3\) mg and \(U=2\) mg. These margins are unequal in their distances from zero. Assume again that the model-based t assumptions are adequate.

Using \(t_{0.95,24}\approx1.711\), the lower test statistic is

$$ t_L=\frac{-0.5-(-3)}{1.1} =\frac{2.5}{1.1} \approx2.273, $$

which exceeds 1.711. The upper statistic is

$$ t_U=\frac{-0.5-2}{1.1} =\frac{-2.5}{1.1} \approx-2.273, $$

which is less than \(-1.711\). Both tests reject their null hypotheses. The interval check gives

$$ -0.5\pm(1.711)(1.1) =-0.5\pm1.8821 =[-2.3821,\;1.3821]\text{ mg}. $$

The interval lies within \((-3,2)\) mg, so equivalence is supported at \(\alpha=0.05\). Notice that the region’s midpoint is \(-0.5\) mg, not zero. TOST does not require the acceptable range to be symmetric about zero; it requires the interval to fit within the justified range.

Diagnostics, Alternatives, and Limitations

Before interpreting a result, check that the margins were prespecified, use the same sign convention for \(\hat{\delta}\), \(L\), and \(U\), and verify that all three use the same units. Then inspect the interval endpoint nearest each margin. The lower endpoint is compared with \(L\); the upper endpoint is compared with \(U\). A plot can help reveal which boundary is limiting, but it does not replace the numerical test or a check of the model assumptions.

Warning signs include a standard error taken from a model that does not reflect the design, an incorrect degrees-of-freedom value, a sign reversal in the difference definition, or margins selected after seeing the estimate. An interval that extends beyond a boundary means the corresponding one-sided test has not established equivalence. A point estimate inside both margins is not a diagnostic substitute for this interval check.

The main alternative is a symmetric-difference analysis, which uses margins \(-M\) and \(M\). That is appropriate only when a symmetric acceptable range is scientifically justified; imposing it when the prespecified limits are unequal tests a different claim. A conventional test of whether \(\delta=0\) asks whether there is evidence of a difference from zero, not whether the difference is small enough to be equivalent. Likewise, a nonsignificant test of zero difference does not establish equivalence.

TOST is a decision procedure relative to chosen margins, a significance level, and a sampling model. It does not show that every individual response is within the margins, prove exact equality, or compensate for poor precision. When one test fails, report which boundary was not cleared and the interval in context rather than describing the result as proof of non-equivalence.

Common pitfall: “The estimate is between the margins, so the treatments are equivalent.” This ignores sampling uncertainty. The correction is to test both boundaries, or equivalently verify that the matching \(100(1-2\alpha)\%\) confidence interval lies inside both margins.

Check Your Understanding

Use the standard TOST convention that both one-sided tests must reject. For numerical calculations, take \(t_{0.95,24}=1.711\) where needed.

  1. Write the equivalence alternative and the two one-sided null hypotheses for margins \(L=-2\) and \(U=5\) mg.
  2. For \(\hat{\delta}=1.0\) mg, \(SE=1.0\) mg, and \(df=24\), calculate both test statistics and decide whether equivalence is established for those margins at \(\alpha=0.05\).
  3. For an estimate and standard error, what confidence level of two-sided interval corresponds to TOST at \(\alpha=0.025\)?
  4. Explain why a point estimate between \(L\) and \(U\) does not by itself establish equivalence.

Solutions. (1) \(H_a:-2<\delta<5\) mg; the null is \(H_0:\delta\leq-2\) mg or \(\delta\geq5\) mg. The separate nulls are \(H_{0L}:\delta\leq-2\) mg and \(H_{0U}:\delta\geq5\) mg. (2) \(t_L=(1-(-2))/1=3.00\), and \(t_U=(1-5)/1=-4.00\). Since \(3.00>1.711\) and \(-4.00<-1.711\), both tests reject and equivalence is established. (3) The matching interval is \(100(1-2\alpha)\%=95\%\). (4) The estimate does not show how uncertain the population difference is. The interval may extend beyond either margin, so both boundary tests must be passed.