The Conventional 80–125% Acceptance Limits

Why the 80–125% Limits Matter

The conventional average bioequivalence decision is made by comparing an interval estimate for the test-to-reference geometric mean ratio with prespecified acceptance limits. The limits are usually written as 80% to 125%, or \(0.80\) to \(1.25\) on a ratio scale. This is a decision rule for the estimated relative exposure and its uncertainty; it is not a claim that every individual has the same exposure under the two products.

The earlier material on average bioequivalence establishes the regulatory question: whether the population average exposure under the test product is sufficiently similar to that under the reference product. Here the focus is the conventional numerical interval and the way it is applied. The ratio is oriented as test divided by reference, so a value of 1 indicates equal geometric means, a value below 1 indicates a lower test geometric mean, and a value above 1 indicates a higher test geometric mean.

Definition: Let \(\theta\) denote the test-to-reference geometric mean ratio for the pharmacokinetic endpoint under consideration. The conventional acceptance interval is \(0.80 \leq \theta \leq 1.25\), commonly expressed as 80%–125%. In the standard average bioequivalence assessment, the estimated ratio is accompanied by a 90% confidence interval, and the criterion is met when that entire interval lies within the acceptance limits.

Reading the Limits on the Ratio and Log Scales

The endpoints are not equally distant from 1 when measured by ordinary subtraction: \(1-0.80=0.20\), whereas \(1.25-1=0.25\). They are reciprocal on the ratio scale: \(1/0.80=1.25\). This reciprocal relationship makes them symmetric after taking natural logarithms.

\ln(0.80)=-0.22314,\qquad \ln(1.25)=0.22314

Thus, the ratio-scale acceptance interval corresponds to \([-0.22314,\,0.22314]\) on the log scale. A model is commonly fitted to log-transformed exposure data, and the treatment comparison is estimated as a difference on that scale. Exponentiating the estimated log difference gives the test-to-reference geometric mean ratio. Exponentiating the confidence limits converts them back to the ratio scale, where they can be compared directly with 0.80 and 1.25.

For an estimated log ratio \(\hat{\delta}\), its standard error \(SE\), and the relevant \(df\), the conventional two-sided 90% confidence interval on the log scale has the form:

\hat{\delta} \pm t_{0.95,df}SE

The corresponding ratio-scale interval is:

\left[\exp\!\left(\hat{\delta}-t_{0.95,df}SE\right),\; \exp\!\left(\hat{\delta}+t_{0.95,df}SE\right)\right]

Here \(t_{0.95,df}\) is the 95th percentile of the \(t\) distribution with the analysis degrees of freedom. The calculation presumes the standard log-scale analysis supplies an appropriate estimate, standard error, and degrees of freedom. The particular study design and analysis model determine those quantities; they should not be replaced by an unrelated standard error or by a calculation on the untransformed exposure scale.

Why the Entire 90% Confidence Interval Is Compared

The conventional interval rule is equivalent to two one-sided tests at significance level \(\alpha=0.05\). The null hypothesis is that the ratio is at or beyond at least one unacceptable boundary; the alternative is that it is strictly between the boundaries:

H_0:\ \theta \leq 0.80\ \text{or}\ \theta \geq 1.25 \qquad\text{versus}\qquad H_a:\ 0.80 < \theta < 1.25

Both one-sided null components must be rejected. On the log scale, this requires evidence that the log ratio is above \(\ln(0.80)\) and below \(\ln(1.25)\). Inverting the two one-sided tests gives the requirement that the 90% confidence interval be fully inside the log-scale limits. Exponentiation preserves the ordering of the limits, so this is exactly the requirement that the ratio-scale 90% confidence interval be fully inside \([0.80,1.25]\).

Result: Under the conventional standard average bioequivalence rule, the result meets the criterion only if the lower limit of the 90% confidence interval is at least 0.80 and its upper limit is at most 1.25. If either limit extends beyond the corresponding acceptance boundary, the criterion is not met.

This is an interval decision, not a point-estimate decision. A point estimate inside the bounds does not guarantee that the interval fits inside them: substantial uncertainty can leave a confidence limit beyond a boundary. Conversely, an interval fully inside the bounds necessarily has its center—the estimated ratio—inside them.

Worked Example: A Confidence Interval Fully Inside the Limits

Suppose a hypothetical analysis estimates a test-to-reference geometric mean ratio of 0.9600. On the log scale, \(\hat{\delta}=\ln(0.9600)=-0.04082\). Suppose \(SE=0.0500\), \(df=24\), and \(t_{0.95,24}=1.711\). The assumptions for this calculation are that these are the appropriate log-scale estimates and degrees of freedom from the specified average bioequivalence analysis.

The margin around the log estimate is:

$$ 1.711(0.0500)=0.08555 $$

The 90% confidence interval on the log scale is:

-0.04082 \pm 0.08555=[-0.12637,\ 0.04473]

Exponentiating each endpoint gives the ratio-scale interval:

[\exp(-0.12637),\exp(0.04473)]=[0.8813,\ 1.0457]

The entire interval, from 0.8813 to 1.0457, lies within 0.80 to 1.25. This hypothetical result therefore meets the conventional interval criterion. The conclusion is about the standard average bioequivalence rule for this endpoint; it is not proof that the products are identical.

Uncertainty Can Change the Decision

How close the estimate is to 1 is only part of the decision. The width of the confidence interval also matters. An imprecise estimate may fail the criterion even when its point estimate is within the acceptance limits, because the data do not locate the ratio precisely enough to place the whole interval inside them.

Worked Example: Point Estimate Inside, Criterion Not Met

Consider a second hypothetical analysis with geometric mean ratio 1.040. Its log estimate is \(\hat{\delta}=\ln(1.040)=0.03922\). Let \(SE=0.130\), \(df=24\), and \(t_{0.95,24}=1.711\). As in the previous calculation, assume the estimate, standard error, and degrees of freedom come from the appropriate log-scale analysis.

The log-scale margin is:

$$ 1.711(0.130)=0.22243 $$

Therefore, the 90% confidence interval on the log scale is:

0.03922 \pm 0.22243=[-0.18321,\ 0.26165]

After exponentiation:

[\exp(-0.18321),\exp(0.26165)]=[0.8326,\ 1.2991]

The point estimate, 1.040, is inside the acceptance limits, but the upper confidence limit, 1.2991, exceeds 1.25. The interval is not wholly inside the limits, so this hypothetical result does not meet the conventional criterion. It would be incorrect to conclude that it passes based only on the point estimate.

The reverse check is also useful. If the estimated ratio is outside the limits, a conventional confidence interval centered on its log estimate cannot be wholly inside them. For example, an estimate of 0.7900 is already below 0.80. The confidence interval still must be calculated and reported, but the estimate itself shows that the full-interval rule cannot be satisfied.

Worked Example: Point Estimate Below the Lower Limit

Suppose a third hypothetical analysis gives a geometric mean ratio of 0.7900, so \(\hat{\delta}=\ln(0.7900)=-0.23572\). Let \(SE=0.0350\), \(df=24\), and \(t_{0.95,24}=1.711\). Assume these are the relevant analysis quantities for the endpoint.

The margin and log-scale interval are:

1.711(0.0350)=0.05989 $$ $$ -0.23572 \pm 0.05989=[-0.29561,\ -0.17583]

Converting the limits to the ratio scale gives:

[\exp(-0.29561),\exp(-0.17583)]=[0.7441,\ 0.8388]

The interval crosses the lower acceptance boundary: its lower limit is below 0.80, while its upper limit is above 0.80. It is therefore not fully within the limits. The point estimate is also below 0.80, so the result does not meet the conventional criterion.

Diagnostics, Comparisons, and Limitations

Before interpreting a result, check the calculation and its context. Confirm that the ratio is oriented as test divided by reference, that the confidence limits are exponentiated from the log-scale interval, and that the appropriate standard error and degrees of freedom were used. Also verify that the endpoint and analysis correspond to the regulatory question being assessed. A reversed ratio changes the interpretation: for instance, \(1/0.80=1.25\), so a test/reference result should not be compared with limits after silently switching to reference/test.

The interval rule is stricter than simply checking whether the point estimate lies between 0.80 and 1.25. A point estimate alone ignores precision. It is also different from asking whether a conventional 95% confidence interval contains 1: that is a different question and does not implement the standard average bioequivalence rule. The 90% confidence interval is used because it corresponds to the two one-sided tests at \(\alpha=0.05\), not because a 90% interval says there is a 90% probability that the fixed ratio lies within it.

A failed criterion does not, by itself, establish that the products differ by a clinically important amount. It says the data did not satisfy this specified interval rule. Failure may arise because the estimated ratio is outside the bounds, because uncertainty is too large, or both. Likewise, a passing result is evidence that the interval criterion has been met for the endpoint and analysis under consideration; it is not a universal guarantee of identical response in every patient or of equivalence on every possible measure.

The numerical limits are regulatory acceptance limits, not direct statements that an 80% exposure is necessarily clinically adequate or that a 25% increase is harmless for every medicine. Their applicability depends on the governing regulatory framework and the product-specific criteria. The present rule describes the conventional interval; it does not determine whether that interval is suitable for a particular product.

Common mistake: “The estimated ratio is 1.04, which is between 0.80 and 1.25, so the result passes.” The correction is to compare both endpoints of the 90% confidence interval with the limits. In Worked Example 2, the estimate is within bounds but the upper confidence limit is not, so the conventional criterion is not met.

Check Your Understanding

Answer each question using the ratio-scale interval rule. The solutions follow the questions.

What are the conventional acceptance limits on the ratio scale, and what are their approximate logarithms?
A hypothetical 90% confidence interval is \([0.82,1.24]\). Does it meet the conventional interval criterion?
A hypothetical estimate is 0.98, but its 90% confidence interval is \([0.78,1.10]\). Does the point estimate determine a pass?
Why are 80% and 125% symmetric on the log scale even though their distances from 100% are not equal on the percentage scale?

Solutions. (1) The limits are 0.80 and 1.25; their natural logarithms are approximately \(-0.22314\) and \(0.22314\). (2) Yes. Both confidence limits lie within 0.80 and 1.25. (3) No. The lower confidence limit is below 0.80, so the full interval is not inside the bounds, even though the point estimate is. (4) The limits are reciprocals: \(1/0.80=1.25\). Taking natural logarithms changes reciprocals into opposite signs, so \(\ln(0.80)=-\ln(1.25)\).