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Confidence Intervals and the TOST Procedure · Tutorial 222 of 399

A Worked TOST Calculation From Estimate and Standard Error

Learn to calculate and reconcile the two TOST tests and their confidence interval from a supplied estimate, standard error, degrees of freedom, and margins.

Intermediate 10 min read

What You'll Learn

  • Set up the lower- and upper-margin one-sided tests.
  • Calculate each test statistic and use the correct tail for its p-value.
  • Make the equivalence decision by requiring both tests to pass.
  • Verify the result using the confidence interval corresponding to the test level.
  • Diagnose disagreements caused by tail, sign, or margin errors.

From a Statistical Summary to an Equivalence Decision

A reported estimate and standard error can look reassuring when the estimate is close to zero. But a small estimated difference, by itself, does not establish equivalence: the uncertainty must also be small enough relative to both equivalence margins. The two one-sided tests procedure, or TOST, makes this requirement explicit by testing the lower and upper margins separately.

This tutorial works through TOST calculations when the estimate, standard error, degrees of freedom, and margins are supplied. The calculations use a t reference distribution; they do not determine whether that reference distribution is appropriate. The supplied summary must come from an analysis for which the t-based inference is justified.

Definition: Let \(\theta\) be the parameter being assessed, let \(\hat{\theta}\) be its estimate, and let \(SE\) be the standard error of that estimate. Let \(L\) and \(U\) be the prespecified lower and upper equivalence margins, with \(L<U\). Equivalence means \(L<\theta<U\). At significance level \(\alpha\), TOST tests \(H_{0L}:\theta\leq L\) against \(H_{aL}:\theta>L\), and \(H_{0U}:\theta\geq U\) against \(H_{aU}:\theta<U\). Equivalence is concluded only when both one-sided null hypotheses are rejected.

The lower-margin test asks whether the parameter is sufficiently above \(L\); the upper-margin test asks whether it is sufficiently below \(U\). These are directional tests, so their p-values come from opposite tails. For a t-based analysis with \(df\) degrees of freedom, the statistics and p-values are

$$ t_L=\frac{\hat{\theta}-L}{SE}, \qquad p_L=P(T_{df}\geq t_L), \qquad t_U=\frac{\hat{\theta}-U}{SE}, \qquad p_U=P(T_{df}\leq t_U), $$

where \(T_{df}\) has a t distribution with \(df\) degrees of freedom. Reject \(H_{0L}\) when \(p_L\leq\alpha\), and reject \(H_{0U}\) when \(p_U\leq\alpha\). Equivalence is supported only if both conditions hold. A small p-value for just one test is not enough.

Why the Two Tests Match an Interval

For the t-based procedure, define \(t_{1-\alpha,df}\) as the \(1-\alpha\) quantile of the t distribution with \(df\) degrees of freedom. The two-sided confidence interval corresponding to TOST at level \(\alpha\) is the \(100(1-2\alpha)\%\) interval

$$ \left[ \hat{\theta}-t_{1-\alpha,df}SE,\; \hat{\theta}+t_{1-\alpha,df}SE \right]. $$

The lower test rejects when its statistic is at least the upper-tail critical value:

$$ t_L\geq t_{1-\alpha,df} \quad\Longleftrightarrow\quad \hat{\theta}-t_{1-\alpha,df}SE\geq L. $$

The upper test rejects when its statistic is at most the corresponding lower-tail critical value, \(-t_{1-\alpha,df}\):

$$ t_U\leq -t_{1-\alpha,df} \quad\Longleftrightarrow\quad \hat{\theta}+t_{1-\alpha,df}SE\leq U. $$

Thus, both tests pass exactly when the interval’s lower endpoint is at least \(L\) and its upper endpoint is at most \(U\). With the usual convention of rejecting when \(p\leq\alpha\), equality at an endpoint corresponds to a test p-value equal to \(\alpha\). In practice, calculations should retain adequate precision before rounding the reported endpoints.

Key result: At significance level \(\alpha\), the t-based TOST concludes equivalence if and only if the \(100(1-2\alpha)\%\) confidence interval is contained within the equivalence limits, using the endpoint convention consistent with the test rule. For example, \(\alpha=0.05\) corresponds to a 90% interval, not a 95% interval.

Worked Calculations

Worked Example: Both Tests Pass

Suppose a hypothetical analysis estimates a treatment difference of \(\hat{\theta}=0.030\) mg·h/L, with \(SE=0.050\) mg·h/L and \(df=24\). The prespecified margins are \(L=-0.150\) mg·h/L and \(U=0.150\) mg·h/L. Use \(\alpha=0.05\).

Lower test. Subtract the lower margin and divide by the standard error:

$$ t_L=\frac{0.030-(-0.150)}{0.050} =\frac{0.180}{0.050} =3.60. $$

The lower test uses the upper tail, so \(p_L=P(T_{24}\geq3.60)\), approximately 0.0007. This is below 0.05, so the lower-margin null is rejected.

Upper test. Compare the estimate with the upper margin:

$$ t_U=\frac{0.030-0.150}{0.050} =\frac{-0.120}{0.050} =-2.40. $$

The upper test uses the lower tail, so \(p_U=P(T_{24}\leq-2.40)\), approximately 0.0123. This too is below 0.05. Both tests pass, so the TOST decision is to conclude equivalence at the 5% level.

Interval check. For \(df=24\), \(t_{0.95,24}\) is approximately 1.711. The 90% interval is

$$ 0.030\pm 1.710882(0.050) \approx0.030\pm0.0855441 \approx[-0.0555,\;0.1155]\ {\rm mg\cdot h/L}. $$

The lower endpoint, \(-0.0555\), is above \(-0.150\), and the upper endpoint, \(0.1155\), is below \(0.150\). The interval confirms the two-test decision.

Worked Example: One Test Fails

In a second hypothetical analysis, \(\hat{\theta}=0.120\) mg·h/L, \(SE=0.080\) mg·h/L, and \(df=18\), with margins \(L=-0.200\) mg·h/L and \(U=0.200\) mg·h/L. Let \(\alpha=0.05\).

The lower statistic is

$$ t_L=\frac{0.120-(-0.200)}{0.080} =\frac{0.320}{0.080} =4.00. $$

The upper-tail probability \(P(T_{18}\geq4.00)\) is approximately 0.0004, so the lower test passes.

For the upper margin,

$$ t_U=\frac{0.120-0.200}{0.080} =\frac{-0.080}{0.080} =-1.00. $$

The correct p-value is the lower-tail probability \(P(T_{18}\leq-1.00)\), approximately 0.1655. Since this exceeds 0.05, the upper test fails. Therefore, equivalence is not established, even though the estimate is inside both margins and the lower test passes.

For a matching interval, \(t_{0.95,18}\) is approximately 1.734. The 90% interval is

$$ 0.120\pm1.734(0.080) =0.120\pm0.1387 =[-0.0187,\;0.2587]\ {\rm mg\cdot h/L}. $$

Its upper endpoint, \(0.2587\) mg·h/L, exceeds the upper margin of \(0.200\) mg·h/L. The interval therefore also fails to establish equivalence.

Worked Example: Asymmetric Margins and a Tail Check

Consider a third hypothetical result with \(\hat{\theta}=0.250\) mg·h/L, \(SE=0.080\) mg·h/L, and \(df=12\). Suppose the limits are asymmetric: \(L=-0.100\) mg·h/L and \(U=0.300\) mg·h/L. Use \(\alpha=0.05\).

For the lower test,

$$ t_L=\frac{0.250-(-0.100)}{0.080} =\frac{0.350}{0.080} =4.375. $$

Thus \(p_L=P(T_{12}\geq4.375)\), approximately 0.0005, and the lower test passes.

For the upper test,

$$ t_U=\frac{0.250-0.300}{0.080} =\frac{-0.050}{0.080} =-0.625. $$

The upper test’s p-value is \(P(T_{12}\leq-0.625)\), approximately 0.272. It fails the 0.05 criterion, so equivalence is not established.

Using \(t_{0.95,12}\approx1.782\), the 90% interval is

$$ 0.250\pm1.782(0.080) =0.250\pm0.1426 =[0.1074,\;0.3926]\ {\rm mg\cdot h/L}. $$

The lower endpoint is above \(-0.100\) mg·h/L, but the upper endpoint \(0.3926\) mg·h/L exceeds \(0.300\) mg·h/L. The interval agrees with the tests: the lower constraint is met, the upper constraint is not, and equivalence is not concluded.

Assumptions, Checks, and Common Errors

The arithmetic is only as trustworthy as the statistical summary. A t-based calculation assumes that the estimate is centered on the target parameter under the model, that its standard error is valid, and that the stated \(df\) and t reference distribution correctly describe its standardized uncertainty. In a classical normal linear model, for example, these conditions follow from the model and variance estimator when their assumptions hold. A supplied \(SE\) and \(df\) do not by themselves prove those conditions. The estimate, standard error, and degrees of freedom must refer to the same contrast and analysis.

Before interpreting the result, check that \(SE>0\), \(L<U\), the estimate and both margins use the same scale and units, and the degrees of freedom are those associated with the supplied standard error. Then inspect the signs and tails: an estimate above \(L\) should produce a positive \(t_L\); an estimate below \(U\) should produce a negative \(t_U\). If the estimate lies between the margins, these signs are expected, although they do not guarantee that either test passes.

A common mistake is to calculate both p-values from the same tail. The lower-bound test asks for evidence that \(\theta>L\), so its p-value is the upper tail of \(t_L\). The upper-bound test asks for evidence that \(\theta<U\), so its p-value is the lower tail of \(t_U\). For example, in the second worked example, \(t_U=-1.00\) gives a lower-tail p-value near 0.1655—not an upper-tail p-value near 0.8345.

Another error is to declare equivalence because \(\hat{\theta}\) falls between the margins, or because one test passes. Neither is sufficient: the uncertainty must clear both margins. The interval check is a useful independent arithmetic check, but it is not a different evidential standard when it uses the matching confidence level. A 95% interval does not generally correspond to TOST at \(\alpha=0.05\); the matching interval is 90%.

TOST’s principal limitation is that its conclusion depends on the margins, which must be scientifically justified and specified independently of the observed estimate. A decision also does not say that every plausible value within the interval is equally likely, nor does it prove exact equality. It says that, under the specified model and margins, the data provide sufficient evidence to reject both margin-based null hypotheses at the chosen level. In practice, report the estimate, standard error, degrees of freedom, margins, both one-sided p-values, and the matching interval so that the conclusion can be checked.

Check Your Understanding

Try each calculation, then check the solutions below. Use a t reference distribution where degrees of freedom are supplied.

  1. For \(\hat{\theta}=0.20\), \(SE=0.10\), \(df=20\), \(L=-0.10\), and \(U=0.50\), calculate \(t_L\) and \(t_U\). Which tail is used for each p-value?
  2. At \(\alpha=0.05\), what confidence level corresponds to the TOST interval?
  3. Does the interval \([-0.20,0.40]\) establish equivalence for margins \([-0.25,0.35]\)? Explain.
  4. Why is an estimate lying between \(L\) and \(U\) not sufficient by itself to establish equivalence?

Solutions. (1) \(t_L=(0.20-(-0.10))/0.10=3.00\), and \(t_U=(0.20-0.50)/0.10=-3.00\). Use the upper tail for \(p_L\) and the lower tail for \(p_U\). (2) The level is \(100(1-2(0.05))\%=90\%\). (3) No. The lower endpoint, \(-0.20\), is above \(-0.25\), but the upper endpoint, \(0.40\), exceeds \(0.35\); therefore the interval does not establish equivalence. (4) The estimate does not reflect uncertainty. The interval or, equivalently, both one-sided tests must show that the parameter is sufficiently far from both margins.

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