9
$\begingroup$

From the paper Equivalence and Noninferiority Testing in Regression Models and Repeated-Measures Designs the following graphic is taken.

enter image description here

I am interested in a non-inferiority test between two proportions (difference), say $p_{T}$ and $p_{S}$ where T is the new treatment and S is the standard. I am willing to accept that the new treatment is not more than 0.05 worse than the standard. I also have a variable I need to control for and thus want to use regression. Call this variable $C$.

So, I believe according to this graphic that if I fit a logistic regression model :

$logit(p) = \beta_{0} + \beta_{1}C + \beta_{2}Treatment$

where Treatment is coded 1 for treatment and 0 for standard, will an exponentiation of the lower side of the confidence interval for $\beta_{2}$ , if larger than -0.05 conclude that treatment is non-inferior to the standard?

$\endgroup$
1
  • $\begingroup$ I have added another citation to my answer. The paper contains an example of recalculation of noninferiority bounds for proportions to noninferiority bounds for odds ratios. $\endgroup$
    – Viktor
    May 6, 2020 at 1:47

1 Answer 1

3
$\begingroup$

No. The exponentiation of the coefficients of logistic regression yields odds ratios. So you have to compare the confidence interval (CI) for the odds ratio to 1. If the outcome in the regression is adverse events count, then the upper side of the CI has to be less than 1.05 to imply noninferiority.

However, the common practice seems to impose noninferiority bounds not on the odds ratio, but on the difference of two proportions. For instance, if we wish that the two proportions' difference does not exceed 15%, then the noninfriority bound on the odds ratio will be 0.55 (or 1/0.55). See Tunes da Silva G, Logan BR, Klein JP. Methods for equivalence and noninferiority testing. Biol Blood Marrow Transplant. 2009;15(1 Suppl):120‐127. doi:10.1016/j.bbmt.2008.10.004 https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2701110/

Analogously, for the Cox regression, noninferiority bound 10% for difference of survival probabilities yields the upper bound 1.31 for exponentiated coefficients of the regression. See the mentioned paper.

Althunian et al. write:

"There are several applications of the margin in the analysis of noninferiority trials, but the recommended approach by regulators, such as the US Food and Drug Administration (FDA), is to compare the estimated 95% confidence interval (CI) of the new drug vs. the active comparator from the noninferiority trial to a predefined margin. If the CI lies entirely below the margin (e.g. for effect measures where the larger the effect the worse the outcome), noninferiority of the new drug to the active comparator can be concluded." https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5510081/

Some authors suggest using 90% CI instead of 95% CI. E.g., see the mentioned paper by Tunes da Silva et al. or https://www.ncss.com/wp-content/themes/ncss/pdf/Procedures/NCSS/Two-Sample_Non-Inferiority_Tests_for_Survival_Data_using_Cox_Regression.pdf

That corresponds to one-sided test with $\alpha=5\%$. But many methodologists support using $\alpha=2.5\%$ for one-sided tests.

$\endgroup$
4
  • $\begingroup$ Do you have a source to cite? $\endgroup$
    – B_Miner
    May 8, 2019 at 23:07
  • 1
    $\begingroup$ An odds ratio of 1.05 isn't the same as a difference in proportions of 0.05 $\endgroup$
    – B_Miner
    May 8, 2019 at 23:08
  • $\begingroup$ @B_Miner Certainly, noninferiority for proportions and noninferiority for odds ratios do differ. $\endgroup$
    – Viktor
    May 12, 2019 at 5:25
  • $\begingroup$ @D_Miner It is sufficient just to google "noninferiority odds ratios" to get some references. E.g., ncbi.nlm.nih.gov/pubmed/20191595 where logarithms of odds ratios are considered or ncss-wpengine.netdna-ssl.com/wp-content/themes/ncss/pdf/… $\endgroup$
    – Viktor
    May 12, 2019 at 5:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.