Worked Example: Equivalence for a Ratio Estimate

From a Ratio Estimate to an Equivalence Decision

A ratio estimate can be close to 1 and still be too imprecise to support equivalence. Conversely, an estimate not exactly equal to 1 may be sufficiently precise to show that the ratio lies within acceptable bounds. The decision therefore depends on both the estimate and its uncertainty, evaluated against equivalence limits chosen for the application.

For a ratio parameter, the log scale is useful because ratios become differences after taking logarithms. This lets us apply the two one-sided tests (TOST) to a log-ratio estimate, then express the final interval and conclusion on the original ratio scale.

Definition: Let \(\rho\) be the population ratio of interest, with \(\rho>0\), and let \(L\) and \(U\) be its lower and upper equivalence limits, where \(0<L<U\). Define the log-ratio parameter \(\delta=\ln(\rho)\), the estimate \(\hat{\delta}\), and its standard error \(SE\). The equivalence limits on the log scale are \(\ln(L)\) and \(\ln(U)\). Equivalence means that \(\rho\) lies between \(L\) and \(U\), or equivalently that \(\delta\) lies between \(\ln(L)\) and \(\ln(U)\).

The ratio and its log are dimensionless when they are formed from quantities with the same units. In an exposure comparison, for example, \(\rho\) might be the test-to-reference ratio of geometric means for an exposure measure. The analysis below assumes that the estimate and standard error refer to the log ratio, not to the ratio itself.

Assumptions and Test Setup

The calculations require a valid estimate of the intended log-ratio contrast. The model must account for the study design and dependence in the data; its standard error and degrees of freedom must be appropriate for that contrast. For the t-based calculations here, assume that the sampling distribution of \(\hat{\delta}\), standardized by its estimated standard error, follows a t distribution with the stated \(df\), under the relevant boundary value. A sufficiently accurate large-sample normal approximation may be used in other settings, but the critical values and p-values must then follow that approximation consistently.

The limits \(L\) and \(U\) are scientific or regulatory criteria, not values selected to make an observed result pass. The ratio must be positive for its logarithm to be defined. The usual TOST also presumes that the reported contrast, estimand, and limits describe the same comparison and population.

At significance level \(\alpha\), the two component tests are:

\begin{aligned} H_{0L}: \delta \le \ln(L) &\quad \text{versus} \quad H_{aL}: \delta > \ln(L),\\ H_{0U}: \delta \ge \ln(U) &\quad \text{versus} \quad H_{aU}: \delta < \ln(U). \end{aligned}

For the lower-bound test, the statistic is \(t_L=(\hat{\delta}-\ln(L))/SE\); its one-sided p-value is the upper-tail probability under the t distribution with the specified degrees of freedom. For the upper-bound test, \(t_U=(\hat{\delta}-\ln(U))/SE\); its p-value is the lower-tail probability. Equivalence is concluded only when both null hypotheses are rejected at level \(\alpha\), or, equivalently, both one-sided p-values are at most \(\alpha\).

Why the 90% Interval Matches TOST at 5%

When the two tests use the same \(SE\) and \(df\), their joint decision can be checked with a two-sided confidence interval. This is not a separate test with a different conclusion: it is an equivalent way to express the two TOST inequalities.

Result: For TOST at level \(\alpha\), form the \(100(1-2\alpha)\%\) t confidence interval for \(\delta\). Equivalence is supported when the entire interval lies within \((\ln(L),\ln(U))\). Exponentiating both endpoints gives the corresponding interval for \(\rho\), which must lie within \((L,U)\).

To justify the result, let \(c=t_{1-\alpha,df}\), the t critical value with cumulative probability \(1-\alpha\). Rejection of the lower-bound null requires \(t_L>c\), which rearranges to \(\hat{\delta}-cSE>\ln(L)\). Rejection of the upper-bound null requires \(t_U<-c\), which rearranges to \(\hat{\delta}+cSE<\ln(U)\). Both conditions hold exactly when the interval \([\hat{\delta}-cSE,\hat{\delta}+cSE]\) is inside the two log limits. Since \(c\) is the critical value for a \(100(1-2\alpha)\%\) two-sided interval, the result follows. Exponentiation is increasing, so it preserves the ordering of the endpoints.

Worked Example: A Ratio Near the Lower Equivalence Limit

Hypothetical data. Suppose a study gives a test-to-reference ratio estimate of \(1.08\), so \(\hat{\delta}=\ln(1.08)=0.076961\). The estimated standard error on the log scale is \(0.03377\), with \(df=24\). The prespecified ratio limits are \(L=0.90\) and \(U=1.25\). Use \(\alpha=0.05\).

Transform the limits. The log-scale boundaries are \(\ln(0.90)=-0.105361\) and \(\ln(1.25)=0.223144\). The estimated log ratio is between them.

Calculate both tests. For the lower boundary,

t_L=\frac{0.076961-(-0.105361)}{0.03377} =\frac{0.182322}{0.03377} \approx 5.399.

With 24 degrees of freedom, the lower one-sided p-value is approximately \(7.593\times10^{-6}\). For the upper boundary,

t_U=\frac{0.076961-0.223144}{0.03377} =\frac{-0.146183}{0.03377} \approx -4.329.

The upper one-sided p-value is the lower-tail probability at \(-4.329\) with 24 degrees of freedom; it is less than \(0.001\). Both p-values are below \(0.05\), so both component null hypotheses are rejected.

Check with the interval. The 90% t critical value for 24 degrees of freedom is approximately \(1.7109\). Thus the log-scale interval is

0.076961\pm 1.7109(0.03377) =0.076961\pm0.057777 =[0.019184,\,0.134738].

Exponentiating its endpoints gives the ratio-scale interval \([e^{0.019184},e^{0.134738}]\), approximately \([1.0194,1.1442]\). This interval is wholly within \([0.90,1.25]\), agreeing with the TOST decision. In context, the estimated ratio is \(1.08\), and the 90% confidence interval for the population ratio is approximately \(1.019\) to \(1.144\); under the stated model, the TOST supports equivalence at the specified limits and 5% level.

Two Further Ratio Calculations

Worked Example: Equivalence With Limits of 0.80 and 1.25

Hypothetical data. A second analysis estimates a ratio of \(1.04\), giving \(\hat{\delta}=\ln(1.04)=0.039221\), with \(SE=0.04000\) and \(df=30\). Use limits \(L=0.80\), \(U=1.25\), and \(\alpha=0.05\).

The transformed boundaries are \(\ln(0.80)=-0.223144\) and \(\ln(1.25)=0.223144\). The test statistics are

t_L=\frac{0.039221-(-0.223144)}{0.04000}=6.559, \qquad t_U=\frac{0.039221-0.223144}{0.04000}=-4.598.

Each corresponding one-sided p-value is below \(0.001\), so both tests reject their null hypotheses at the 5% level. For a cross-check, the 90% t critical value for 30 degrees of freedom is approximately \(1.6973\). The log interval is

0.039221\pm1.6973(0.04000) =[-0.028671,\,0.107113].

After exponentiation, the interval is approximately \([0.9717,1.1131]\). It lies within \([0.80,1.25]\), so the interval and TOST give the same equivalence conclusion. These limits are asymmetric around 1 on the ratio scale, even though their log limits are symmetric around 0.

Worked Example: A Model-Provided Log-Ratio Contrast

Hypothetical data. A fitted model reports a log-ratio contrast \(\hat{\delta}=0.019803\), corresponding to a ratio estimate of \(e^{0.019803}\approx1.020\). Its standard error is \(0.02500\), with \(df=40\). Suppose the prespecified limits are \(0.90\) and \(1.25\), and \(\alpha=0.05\).

The log boundaries are \(-0.105361\) and \(0.223144\). The statistics are

t_L=\frac{0.019803-(-0.105361)}{0.02500}\approx5.007, \qquad t_U=\frac{0.019803-0.223144}{0.02500}\approx-8.134.

Both one-sided p-values are below \(0.001\), supporting rejection of both null hypotheses. With \(df=40\), the 90% t critical value is approximately \(1.6839\). The log interval is

0.019803\pm1.6839(0.02500) =0.019803\pm0.042098 =[-0.022295,\,0.061901].

Exponentiating gives a ratio interval of approximately \([0.9780,1.0639]\), which lies within \([0.90,1.25]\). This example uses a model-provided contrast: the calculation is valid only if its estimate, standard error, degrees of freedom, and defined estimand are all on the appropriate log-ratio scale.

Diagnostics, Alternatives, and Limitations

Before reporting a result, check that the estimate and its standard error are on the log scale; that both margins have been transformed using natural logarithms; and that the same \(df\) and \(SE\) are used for the tests and matching interval. Confirm that the back-transformation exponentiates both interval endpoints, rather than taking the logarithm of an interval already on the ratio scale. Also verify that the limits were fixed independently of the observed estimate.

TOST p-values and the matching confidence interval are alternative presentations of the same decision when their assumptions and inputs agree. The p-values show the evidence against each boundary null separately. The interval makes the range of values compatible with the estimate and its uncertainty easier to compare with both limits. A point estimate inside the limits alone does not establish equivalence; uncertainty must also fit inside them.

A difference-scale analysis answers a different question from a ratio-scale analysis. For example, an absolute difference of a given size can have very different implications at different exposure levels, whereas a ratio describes a relative change. The scale and margins should therefore follow the scientific estimand, not whichever calculation gives a more favorable result.

The conclusion is conditional on the model, estimand, and chosen margins. A t-based interval does not repair a misspecified model, an inappropriate standard error, or a study design that fails to support the intended comparison. Reporting only the back-transformed interval also hides the scale on which the tests were performed; include the ratio estimate, limits, confidence level, and TOST decision.

Common mistake: testing the ratio estimate directly against log limits. A ratio such as \(1.08\) cannot be compared numerically with \(\ln(0.90)\). Transform the estimate to \(\ln(1.08)\), or transform the limits back to the ratio scale and use a method defined for that scale.
Common mistake: treating an in-bounds point estimate as proof of equivalence. The point estimate in Example 1 is inside the margins, but that fact alone says nothing about whether its uncertainty is also inside them. Use both one-sided tests or the matching interval.
Common mistake: back-transforming only the estimate. Exponentiate each log-interval endpoint separately. This produces an interval on the ratio scale with the correct ordering and preserves the TOST comparison.

Key takeaway: For ratio equivalence, transform the estimate and both limits to the log scale, perform both one-sided tests using the correct uncertainty model, and confirm the result with the matching confidence interval. Exponentiate the interval endpoints for interpretation in ratio units.

Check Your Understanding

Answer each question, then check the brief solutions below.

A ratio estimate is \(1.05\). What estimate is used on the log scale?
For limits \(0.80\) and \(1.25\), what are the two log-scale boundaries?
At \(\alpha=0.05\), what confidence level corresponds to the TOST interval?
A 90% ratio-scale interval is \([0.92,1.18]\), with prespecified limits \([0.90,1.25]\). What does the interval criterion imply at the corresponding TOST level?
Why is it not sufficient that a ratio estimate itself lies between the equivalence limits?

Solutions. (1) \(\ln(1.05)\approx0.04879\). (2) \(\ln(0.80)\approx-0.22314\) and \(\ln(1.25)\approx0.22314\). (3) The matching interval is 90%, since \(100(1-2(0.05))=90\). (4) The full interval lies inside both limits, so the interval criterion supports equivalence at the corresponding 5% TOST level, subject to the stated model assumptions. (5) The estimate does not convey its uncertainty; the interval may extend beyond a margin even when the point estimate is inside.