Applying Confidence-Interval Rules Under Tight NTI Limits

Why the Whole Interval Matters

For a narrow therapeutic index (NTI) drug, a small difference in exposure may be clinically important. The bioequivalence decision therefore uses acceptance limits specified for the relevant endpoint and regulatory framework. As established in the earlier material on why tight NTI limits are not variance-scaled limits, these limits are prespecified fixed bounds: they do not widen because a study estimates greater variability.

The decision is not made by checking only whether the estimated test-to-reference ratio falls between the bounds. A ratio estimate is uncertain. The confidence interval describes the range of values compatible with the estimated contrast at the stated confidence level, under the analysis assumptions. For equivalence, the entire interval must fit within both limits. An interval that crosses either bound does not demonstrate equivalence, even if its point estimate is comfortably inside.

Definition: Let \(\delta\) be the population mean difference between test and reference log exposure for the endpoint being assessed. The test-to-reference geometric mean ratio is \(\exp(\delta)\). For prespecified ratio limits \(L\) and \(U\), equivalence is demonstrated by the confidence-interval rule when the chosen confidence interval for \(\exp(\delta)\) lies entirely within \([L,U]\). The limits and confidence level must be those specified for the applicable analysis; they are not selected after observing the results.

Constructing and Applying the Interval

In a crossover comparison, the analysis commonly estimates a test-minus-reference contrast on the log-transformed endpoint. Denote its estimate by \(\hat{\delta}\), its standard error by \(SE\), and its degrees of freedom by \(df\). Under the specified analysis, a two-sided \(100(1-2\alpha)\%\) confidence interval for \(\delta\) is

\hat{\delta} \pm t_{1-\alpha,df}SE

Here \(t_{1-\alpha,df}\) is the \(1-\alpha\) quantile of the \(t\) distribution with \(df\) degrees of freedom. Exponentiating the two endpoints gives the confidence interval for the geometric mean ratio:

\left[ \exp\!\left(\hat{\delta}-t_{1-\alpha,df}SE\right), \exp\!\left(\hat{\delta}+t_{1-\alpha,df}SE\right) \right].

For example, when the protocol uses two one-sided tests at \(\alpha=0.05\), the corresponding interval is a 90% confidence interval. This is a conventional rule, not a universal confidence level for every jurisdiction or endpoint. Always use the level and acceptance limits prespecified for the particular assessment.

For a ratio interval, the lower and upper acceptance limits can equivalently be compared on the log scale. The condition is

\log(L) \leq \hat{\delta}-t_{1-\alpha,df}SE \quad\text{and}\quad \hat{\delta}+t_{1-\alpha,df}SE \leq \log(U).

For illustration below, the hypothetical analysis uses ratio limits \(L=0.90\) and \(U=1.1111\) and a 90% confidence interval. These numbers are examples only; they are not a substitute for the limits applicable to a real assessment.

Decision rule: Under the conventional two-one-sided-test framework at level \(\alpha\), equivalence is demonstrated when the corresponding \(100(1-2\alpha)\%\) confidence interval lies within the prespecified bounds. If either confidence limit extends beyond its acceptance boundary, equivalence is not demonstrated. Use the exact endpoint and boundary convention stated in the applicable protocol or guidance.

Why the Interval Rule Matches Two One-Sided Tests

Let the prespecified log-scale limits be \(\log(L)\) and \(\log(U)\). The equivalence alternative is that the true contrast lies between those boundaries. The null is the union of the two non-equivalence possibilities: the contrast is at or below the lower boundary, or at or above the upper boundary. Equivalence requires evidence against both possibilities.

\begin{aligned} H_0 &: \delta \leq \log(L)\ \text{or}\ \delta \geq \log(U),\\ H_a &: \log(L) < \delta < \log(U). \end{aligned}

For the lower-bound test, the test statistic is \((\hat{\delta}-\log(L))/SE\). Rejecting its one-sided null at level \(\alpha\) requires

\frac{\hat{\delta}-\log(L)}{SE}>t_{1-\alpha,df} \quad\Longleftrightarrow\quad \hat{\delta}-t_{1-\alpha,df}SE>\log(L).

Thus, the lower endpoint of the corresponding confidence interval must exceed the lower boundary. For the upper-bound test, rejection requires

\frac{\hat{\delta}-\log(U)}{SE}<-t_{1-\alpha,df} \quad\Longleftrightarrow\quad \hat{\delta}+t_{1-\alpha,df}SE<\log(U).

So the upper endpoint must be below the upper boundary as well. Both tests must reject their respective null components. This is why a single two-sided interval makes the decision transparent. It is not a test of the null hypothesis that the formulations have exactly equal exposure.

Worked Example: The Interval Fits Within Both Limits

Suppose a hypothetical analysis of log-transformed exposure estimates \(\hat{\delta}=0.0180\), with \(SE=0.0250\) and \(df=24\). Use the example limits \(0.90\) and \(1.1111\), and a 90% confidence interval. Assume the endpoint, analysis population, and contrast were specified in advance, and that the stated standard error and degrees of freedom come from an appropriate analysis.

For \(df=24\), use \(t_{0.95,24}=1.711\). The margin of error on the log scale is

$$ 1.711(0.0250)=0.042775. $$

The interval for the log contrast is therefore

0.0180\pm0.042775=[-0.024775,\ 0.060775].

Exponentiating the estimate gives the geometric mean ratio \(e^{0.0180}=1.018\), or about 101.8%. Exponentiating the interval endpoints gives

[\exp(-0.024775),\ \exp(0.060775)]\approx[0.9755,\ 1.063].

The lower confidence limit, 0.9755, is above 0.90, and the upper limit, 1.063, is below 1.1111. The interval lies within both prespecified limits, so this hypothetical result demonstrates equivalence under the stated rule. The conclusion depends on the stated analysis assumptions; it does not mean the true ratio is known exactly.

Worked Example: The Interval Crosses the Upper Limit

In a second hypothetical analysis, let \(\hat{\delta}=0.0450\), \(SE=0.0400\), and \(df=20\). Use the same example limits and a 90% confidence interval. Assume the log-scale contrast and standard error are valid for the prespecified comparison.

Using \(t_{0.95,20}=1.725\), the margin is

$$ 1.725(0.0400)=0.0690. $$

The log-scale interval is \(0.0450\pm0.0690=[-0.0240,\ 0.1140]\). On the ratio scale, the point estimate is \(\exp(0.0450)=1.046\), and the interval is

[\exp(-0.0240),\ \exp(0.1140)]\approx[0.9763,\ 1.121].

The point estimate lies between 0.90 and 1.1111, but the upper confidence limit, about 1.121, exceeds 1.1111. The interval therefore does not fit within the acceptance limits. Equivalence is not demonstrated. This result alone does not establish that the formulations are inequivalent; the uncertainty is too large to meet the equivalence criterion.

Worked Example: The Interval Crosses the Lower Limit

A third hypothetical analysis gives \(\hat{\delta}=-0.0750\), \(SE=0.0250\), and \(df=24\). Use the same limits and confidence level. Assume the estimate and its uncertainty are appropriate for the specified endpoint.

With \(t_{0.95,24}=1.711\), the margin is \(1.711(0.0250)=0.042775\), so the log-scale interval is

-0.0750\pm0.042775=[-0.117775,\ -0.032225].

The point estimate on the ratio scale is \(\exp(-0.0750)=0.9277\), within the example limits. The interval is

[\exp(-0.117775),\ \exp(-0.032225)]\approx[0.8889,\ 0.9683].

The lower confidence limit, 0.8889, is below 0.90. Even though the upper limit is within range and the point estimate is acceptable, the entire interval is not contained in the limits. Equivalence is not demonstrated under the stated rule.

Assumptions and Diagnostics

The arithmetic of the interval is straightforward; the harder question is whether the estimate, standard error, and reference distribution support it. At a minimum, the interpretation relies on the following conditions:

The estimand and scale are appropriate. The contrast must represent the prespecified test-to-reference comparison for the endpoint, and log transformation must match the analysis plan. Exponentiation then gives a ratio of geometric means, not a difference in arithmetic means.
The study design and data support the contrast. The comparison must be identified by the design, and the analysis must handle the observed subjects and periods as prespecified. Missing or unusable observations must not be treated as if they supplied information.
The standard error and degrees of freedom are valid. The confidence interval assumes that the reported uncertainty calculation and its \(t\)-reference approximation are suitable for the analysis. These quantities should come from the specified statistical model and design, not from an informal calculation that ignores the study structure.
The analysis is not undermined by data or conduct problems. Important deviations, influential observations, unexpected period or sequence patterns, and data-quality concerns may call the estimate or its uncertainty into question. Diagnostics should be interpreted in light of the design and not used as a post hoc reason to choose a more favorable analysis.

A diagnostic review should include the analysis data and the log-scale outcomes, not just the final ratio interval. Check that the intended subjects, periods, treatments, and endpoint values are present; investigate unusual observations and missingness; and confirm that the model or calculation actually estimates the prespecified contrast. If the confidence interval is near a boundary, small changes in data handling or uncertainty estimation can change the decision, so the analysis trail deserves particular scrutiny. A boundary-crossing interval should not be “repaired” by changing the limits, confidence level, endpoint, or analysis after seeing the result.

Comparisons, Limitations, and a Common Pitfall

The interval rule and the two one-sided tests are two ways of expressing the same conventional equivalence decision when they use the matching confidence level and analysis. The interval is often easier to communicate because it shows both the estimated ratio and its uncertainty relative to each boundary. A p-value-only presentation can obscure which boundary was not met, though the two one-sided test results may be reported alongside the interval.

Do not substitute a familiar acceptance range, such as the conventional 80.00%–125.00% range, for the tighter NTI limits specified for the assessment. Nor should a limit be widened because variability is high: that would change a fixed, prespecified criterion into a different rule. The earlier material on jurisdiction-specific NTI requirements is relevant when identifying the governing limits, confidence level, and endpoint; the numerical limits used in these examples are illustrative only.

The interval also has limitations. It quantifies uncertainty in the specified contrast under the analysis assumptions; it does not establish clinical safety for every patient, demonstrate identical concentration-time profiles, or guarantee that all required endpoints meet their criteria. If the assessment requires separate endpoints, each must be evaluated according to its own prespecified rule. The broader clinical and regulatory interpretation requires more than one ratio interval.

Common mistake: “The estimated ratio is inside the limits, so the formulations are bioequivalent.” This reasoning ignores uncertainty. In the lower-bound example, the point estimate is 0.9277, but the confidence interval extends below 0.90. The correction is to compare both confidence limits with the corresponding prespecified boundaries. If either limit falls outside, equivalence has not been demonstrated; that conclusion is not automatically proof of inequivalence.

Key takeaway: First specify the endpoint, analysis, confidence level, and fixed NTI limits. Then calculate the log-scale interval, exponentiate both endpoints, and compare the entire ratio interval with both bounds. A point estimate inside the limits is not sufficient.

Check Your Understanding

Answer each question using the stated conventional rule. Assume the analysis inputs are valid and the limits are prespecified at 0.90 and 1.1111.

An estimate has \(\hat{\delta}=0\), \(SE=0.02\), and \(df=24\). With \(t_{0.95,24}=1.711\), does its 90% interval demonstrate equivalence?
A ratio confidence interval is 0.91 to 1.12. The point estimate is 1.01. What is the decision, and what does it mean?
Why does a 90% confidence interval correspond to two one-sided tests at \(\alpha=0.05\) in this rule?
A study has a wide interval that crosses the lower limit. Can the limit be widened because the study’s variability was high? Explain.

Solutions. (1) The log interval is \(0\pm1.711(0.02)=[-0.03422,0.03422]\). Exponentiating gives approximately \([0.9664,1.035]\), entirely within the limits, so equivalence is demonstrated under the stated assumptions. (2) The interval crosses the upper limit of 1.1111, so equivalence is not demonstrated. This alone does not prove inequivalence; the point estimate cannot override the interval. (3) Each one-sided test uses \(\alpha=0.05\); the matching two-sided interval has confidence level \(1-2\alpha=0.90\), and its lower and upper endpoints correspond to the two one-sided rejection conditions. (4) No. The limits are prespecified and fixed for this rule. High variability may widen the interval and make it harder to demonstrate equivalence, but it does not authorize changing the acceptance limits.