## Strange result [Regulatives / Guidelines]

Hi John,

» […] I read some report […] and the one sided 90% CI was reported as lower = -18.33%, upper = -6.36%; T/R Ratio = 93.86, n=24. I found it strange to report CI in such a way so I was wondering if the reporting was based on some old rules.

Strange!
As the name tells, TOST are interval (significance) tests giving two p-values – one for the 1st (left) test where

H01: PE < ln(1–0.2)
H11: PE ≥ ln(1–0.2)

and one for the 2nd (right) test where

H02: PE > ln(1+0.2)
H12: PE ≤ ln(1+0.2).

Only if both H01 and H02 are rejected at the level α, BE is concluded. In my entire career I haven’t seen a single report where this was done1 – though most claim that TOST was applied. The confidence interval inclusion test is two-sided (though OTST is difficult to pronounce) and “operationally equivalent” to TOST. Some call that just “an algebraic coincidence”.2 There is no CI in TOST.3 The FDA’s 1992 guidance4 mixes both approaches up:

The two one-sided hypotheses at the α = 0.05 level of significance should be tested for AUC and Cmax by con- structing the 90% confidence interval for the ratio between the test and reference averages.

However, conclusions of TOST and the confidence interval inclusion method are the same.

Coming back to your case. I’ve never seen results reported in such a way.
If we would report the CI as usual we would give (93.86 – 18.33)% = 75.53% and (93.86 – 6.36)% = 87.53% and the study failed (like nobody assumed). But then we have another problem with the reported T/R-ratio which transforms to (93.86 – 100)% = –6.14%. Then –18.33% < –6.14% < –6.36%, or what? Based on 100√0.7553 × 0.8753 we get 81.29% ≠ 93.86%.
Or are the boundaries given relative to 100%? Bizarre. However, then the study would pass since (100 – 18.33)% = 81.67% and (100 – 6.36)% = 93.64%. Note that the reported PE is outside the CI…

» […] (dated 1990; the synopsis only)

Was the study performed for Health Canada? In the 1989 draft 80–120% (untransformed data) were recommended and changed to 80–125% (log-transformed) in 1991.
Then the study would have passed again cause –18.33% > –20% and –6.36% < +20%. However, the problem with the PE persists cause 100(–0.1833 + (–0.0636)) / 2 = –12.35% ≠ –6.14%. I don’t get it.

1. I did it myself in a few studies upon sponsor’s wish. In all cases assessors asked for the 90% CI later…
For ages I do it the other way ’round. Describe the confidence interval inclusion approach according to the guidelines and – to make newbies happy – state that it is “operationally equivalent” to TOST.
2. Brown LD, Casella G, Hwang JTG. Optimal Confidence Sets, Bioequivalence, and the Limaçon of Pascal. J Amer Statist Assoc. 1995;90(431):880–9. doi:10.1080/01621459.1995.10476587. free resource.
3. Schuirmann DJ. A comparison of the Two One-Sided Tests Procedure and the Power Approach for Assessing the Equivalence of Average Bioavailability. J Pharmacokin Biopharm. 1987;15(6):657–80. doi:10.1007/BF01068419.
4. FDA/CDER. Guidance for Industry. Statistical Procedures for Bioequivalence Studies Using a Standard Two-Treatment Crossover Design. Jul 1992.

Cheers,
Helmut Schütz

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