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Confidence Intervals and the TOST Procedure · Tutorial 228 of 399

Worked Example: Ratio Confidence Interval Crossing One Limit

Use a back-transformed ratio interval to identify which equivalence-boundary test fails and explain what the result does—and does not—establish.

Intermediate 12 min read

What You'll Learn

  • Identify the equivalence limit crossed by a back-transformed ratio confidence interval.
  • Match a crossed lower or upper limit to its corresponding one-sided TOST.
  • Calculate log-scale confidence limits and exponentiate them correctly.
  • Check one-sided statistics against the appropriate rejection direction.
  • Diagnose common inconsistencies between interval and test results.
  • Distinguish failure to establish equivalence from evidence that the ratio is exactly one.

Why a Crossed Limit Identifies a Failed Test

For a ratio-scale treatment comparison, the equivalence region is usually specified by a lower ratio limit \(L\) and an upper ratio limit \(U\), where \(0<L<U\). A confidence interval that extends past one of these limits does not establish equivalence, even if the point estimate is close to 1. The location of the crossed endpoint tells us which one-sided test in the two one-sided tests procedure (TOST) fails.

The calculations are performed on the log scale because a ratio becomes a difference after taking logarithms. Let \(\theta\) denote the population ratio of interest, such as a treatment-to-reference geometric mean ratio, and let \(\eta=\log(\theta)\). Write \(\hat{\eta}\) for the estimate of \(\eta\), with standard error \(SE\). The ratio limits \(L\) and \(U\) correspond to log-scale limits \(\log(L)\) and \(\log(U)\).

Definition: For equivalence limits \(L\) and \(U\), the TOST consists of a lower-bound test, \(H_0:\eta\leq\log(L)\) against \(H_a:\eta>\log(L)\), and an upper-bound test, \(H_0:\eta\geq\log(U)\) against \(H_a:\eta<\log(U)\). Equivalence is established only when both null hypotheses are rejected at the chosen significance level \(\alpha\).

For the common limits \(L=0.80\) and \(U=1.25\), the log limits are approximately \(-0.22314\) and \(0.22314\). They are symmetric around zero on the log scale, although they are not equally far from 1 as additive differences on the ratio scale. Exponentiation is increasing, so a log-scale interval crosses a log limit if and only if its back-transformed ratio interval crosses the corresponding ratio limit.

How the Interval and the Two Tests Correspond

Suppose the estimate follows a normal model with known \(SE\), or a \(t\) model with an appropriate degrees of freedom \(df\). Let \(c\) be the relevant one-sided critical value: \(z_{1-\alpha}\) for a standard normal test, or \(t_{1-\alpha,df}\) for a \(t\) test. The two one-sided statistics are

$$ T_L=\frac{\hat{\eta}-\log(L)}{SE}, \qquad T_U=\frac{\hat{\eta}-\log(U)}{SE}. $$

The lower-bound test rejects when \(T_L>c\); the upper-bound test rejects when \(T_U<-c\). Equivalently, for the matching \(100(1-2\alpha)\%\) two-sided confidence interval, both endpoints must fall strictly inside the log-scale equivalence limits.

$$ \left[\hat{\eta}-cSE,\ \hat{\eta}+cSE\right] \subset \left(\log(L),\ \log(U)\right). $$

The connection follows directly from rearranging the test inequalities. The lower test rejects exactly when

$$ T_L>c \quad\Longleftrightarrow\quad \frac{\hat{\eta}-\log(L)}{SE}>c \quad\Longleftrightarrow\quad \hat{\eta}-cSE>\log(L). $$

Thus, if the lower confidence endpoint is below \(\log(L)\), the lower-bound test does not reject. For the upper test,

$$ T_U<-c \quad\Longleftrightarrow\quad \frac{\hat{\eta}-\log(U)}{SE}<-c \quad\Longleftrightarrow\quad \hat{\eta}+cSE<\log(U). $$

So if the upper confidence endpoint is above \(\log(U)\), the upper-bound test does not reject. Exponentiating the interval endpoints gives the same checks on the ratio scale. This relationship assumes the interval and tests use the same estimate, standard error, degrees of freedom or reference distribution, and significance level.

Worked Example: The Ratio Interval Crosses the Upper Limit

Suppose a hypothetical analysis estimates \(\hat{\eta}=0.18232\), with \(SE=0.0800\) and \(df=30\). Use equivalence limits 0.80 and 1.25 and a 90% two-sided \(t\) confidence interval. For these degrees of freedom, take \(t_{0.95,30}\approx1.697\). Determine which one-sided test fails.

Calculate the log-scale interval. The margin of error is \(1.697(0.0800)=0.13576\). Therefore,

$$ [0.18232-0.13576,\ 0.18232+0.13576] =[0.04656,\ 0.31808]. $$

Back-transform. Exponentiating both endpoints gives

$$ [\exp(0.04656),\ \exp(0.31808)] \approx[1.0477,\ 1.3745]. $$

The lower endpoint is above 0.80, but the upper endpoint exceeds 1.25. The interval therefore crosses the upper equivalence limit.

Check the lower-bound test. Its statistic is

$$ T_L=\frac{0.18232-\log(0.80)}{0.0800} =\frac{0.18232+0.22314}{0.0800} =\frac{0.40546}{0.0800} \approx5.068. $$

This exceeds the critical value 1.697, so the lower-bound test rejects. For a \(t\) distribution with 30 degrees of freedom, its one-sided upper-tail p-value is approximately \(9.606\times10^{-6}\).

Check the upper-bound test. Its statistic is

$$ T_U=\frac{0.18232-\log(1.25)}{0.0800} =\frac{0.18232-0.22314}{0.0800} \approx-0.5103. $$

The upper test requires a statistic below \(-1.697\) to reject. Since \(-0.5103\) is not that small, this test fails to reject its null hypothesis. The upper endpoint crossing 1.25 and the failed upper-bound test are two equivalent ways to express the same result.

Check the Corresponding Boundary, Not Just the Estimate

A confidence interval can fail the equivalence criterion at either boundary. Crossing the lower limit points to the lower-bound test; crossing the upper limit points to the upper-bound test. The test direction matters: the lower test looks for evidence that the parameter is above the lower limit, whereas the upper test looks for evidence that it is below the upper limit. A statistic's sign alone does not determine the decision; compare it with the appropriate critical value.

Worked Example: A Second Interval Crossing the Upper Limit

Consider hypothetical results \(\hat{\eta}=0.14842\), \(SE=0.1000\), and \(df=18\), using \(t_{0.95,18}\approx1.734\). The log-scale 90% interval is

$$ [0.14842-1.734(0.1000),\ 0.14842+1.734(0.1000)] =[-0.02498,\ 0.32182]. $$

After exponentiation, the ratio interval is

$$ [\exp(-0.02498),\ \exp(0.32182)] \approx[0.9753,\ 1.3796]. $$

The lower endpoint is above 0.80, but the upper endpoint is above 1.25. Thus the interval crosses the upper limit. The lower-bound statistic is

$$ T_L=\frac{0.14842-\log(0.80)}{0.1000} =\frac{0.14842+0.22314}{0.1000} \approx3.716. $$

This exceeds the critical value 1.734, so the lower-bound test rejects; its one-sided upper-tail p-value is about 0.0008. In contrast, the upper-bound statistic is

$$ T_U=\frac{0.14842-\log(1.25)}{0.1000} =\frac{-0.07472}{0.1000} =-0.7472. $$

It is not below \(-1.734\), so the upper-bound test fails. The point estimate itself is inside the ratio limits, but that fact does not overcome the uncertainty represented by the interval.

Worked Example: The Interval Crosses the Lower Limit Instead

For a contrasting hypothetical case, let \(\hat{\eta}=-0.17435\), \(SE=0.0900\), and \(df=20\), with \(t_{0.95,20}\approx1.725\). The log-scale 90% interval is

$$ [-0.17435-1.725(0.0900),\ -0.17435+1.725(0.0900)] =[-0.32960,\ -0.01910]. $$

The corresponding ratio interval is approximately

$$ [\exp(-0.32960),\ \exp(-0.01910)] \approx[0.7193,\ 0.9811]. $$

Its lower endpoint is below 0.80, while its upper endpoint is below 1.25. The lower-bound test is the one that fails. Its statistic is

$$ T_L=\frac{-0.17435-\log(0.80)}{0.0900} =\frac{-0.17435+0.22314}{0.0900} \approx0.5421. $$

The lower test requires a statistic greater than 1.725 to reject, and \(0.5421\) does not meet that condition. For comparison, the upper-bound statistic is

$$ T_U=\frac{-0.17435-\log(1.25)}{0.0900} =\frac{-0.39749}{0.0900} \approx-4.417. $$

This is below \(-1.725\), so the upper-bound test rejects. The interval and tests agree: the lower boundary is crossed, and the lower-bound test fails.

Assumptions, Diagnostics, and Alternatives

These calculations are only as reliable as the estimate and uncertainty model that produced them. The contrast must estimate the intended log ratio, the standard error must represent its sampling variability, and the stated reference distribution and degrees of freedom must be appropriate. For a \(t\)-based calculation, the standard error and degrees of freedom must come from the model or analysis method supporting that \(t\) inference. The equivalence limits should be specified independently of the observed results.

Useful checks include confirming that both endpoints were calculated from the same \(\hat{\eta}\) and \(SE\), that the confidence level matches the TOST significance level, and that the ratio interval was obtained by exponentiating—not by adding—the log-scale endpoints. Because exponentiation is monotone, the ratio interval must preserve endpoint order. A reversed interval or a reported endpoint that does not exponentiate from its stated log value signals a calculation or transcription error.

The confidence-interval presentation and the two-test presentation are alternatives for communicating the same decision when they use matching assumptions and settings. The interval makes the estimated range and the crossed boundary visible; the tests name the one-sided null hypothesis that was not rejected and can provide a p-value. Either way, crossing one limit means the evidence does not establish that the full interval lies within the prespecified equivalence region. It does not, by itself, show that the ratio equals 1 or that the treatments differ by a particular amount.

  • Pitfall: “The point estimate is inside 0.80 to 1.25, so equivalence has been shown.” Correction: Equivalence requires both one-sided tests to reject; check the interval endpoints, not only the point estimate.
  • Pitfall: “The interval crosses 1, so the upper-bound test fails.” Correction: Crossing 1 is not the test criterion. Compare the endpoint with 0.80 and 1.25 and identify which equivalence limit it crosses.
  • Pitfall: “A statistic of \(-0.51\) is negative, so the upper test rejects.” Correction: The upper test rejects only when its statistic is sufficiently negative—below the negative critical value.
Key takeaway: On the ratio scale, an interval extending above \(U\) identifies the upper-bound test as the one that fails; an interval extending below \(L\) identifies the lower-bound test. Confirm the correspondence using the appropriate one-sided statistic and critical value.

Check Your Understanding

Use limits 0.80 and 1.25 and a 90% confidence interval unless a question specifies otherwise.

  1. A back-transformed interval is [0.91, 1.28]. Which boundary is crossed, and which one-sided test fails?
  2. A log-scale interval is [-0.18, 0.20]. Does it cross either log equivalence limit? State the corresponding decision about the two tests.
  3. For \(\hat{\eta}=0.10\), \(SE=0.05\), and normal critical value 1.645, calculate the log-scale confidence interval and identify whether it crosses a limit.
  4. Explain why an interval crossing 1 does not, by itself, identify a failed equivalence test.

Solutions. (1) The upper endpoint 1.28 exceeds 1.25, so the interval crosses the upper limit and the upper-bound test fails to reject. The interval's lower endpoint is above 0.80. (2) The log limits are approximately \(-0.22314\) and \(0.22314\). Both endpoints, \(-0.18\) and \(0.20\), lie between them, so neither limit is crossed and both tests reject under the matching interval-test assumptions. (3) The margin of error is \(1.645(0.05)=0.08225\); the interval is \([0.10-0.08225,\ 0.10+0.08225]=[0.01775,\ 0.18225]\). Both endpoints lie inside the log limits, so neither equivalence boundary is crossed. (4) The equivalence tests compare the interval with 0.80 and 1.25, not with 1. An interval can include 1 while remaining inside both limits, or include 1 and extend past one of them; the crossed equivalence limit determines which test fails.