Bioequivalence and Bioavailability Forum

Dear all,

in Gwaza et al.¹ and all following publications this formula is given for the standard deviation of the difference: $$SD=\frac{2\cdot SE_\textrm{(d)}}{\sqrt{\frac{1}{n_\textrm{1}}+\frac{1}{n_\textrm{2}}}}\tag{1}$$ Jiři pointed out that this might not be correct. I checked his algebra and think that he is right.
Let’s do it step by step. The error term in the 2×2×2 crossover² is given by $$SE_\textrm{(d)}=SE_\Delta=\widehat{\sigma}_\textrm{w}\sqrt{\frac{1}{2}\left (\frac{1}{n_\textrm{1}}+\frac{1}{n_\textrm{2}}\right )}\tag{2}$$where \(\small{\widehat{\sigma}_\textrm{w}=SD_\textrm{w}=\sqrt{MSE}}\) from ANOVA. Alternatively³ we can write $$SE_\Delta=\sqrt{\frac{SD_{\textrm{w}}^{2}}{2}\left (\frac{1}{n_\textrm{1}}+\frac{1}{n_\textrm{2}}\right )}\equiv\sqrt{\frac{MSE}{2}\left (\frac{1}{n_\textrm{1}}+\frac{1}{n_\textrm{2}}\right )}\tag{3}$$ Square both sides of \((3)\) $$SE_{\Delta}^{2}=\frac{SD_{\textrm{w}}^{2}}{2}\left (\frac{1}{n_\textrm{1}}+\frac{1}{n_\textrm{2}}\right )\tag{4a}$$ Rearrange $$SD_{\textrm{w}}^{2}=\frac{2\cdot SE_{\Delta}^{2}}{\frac{1}{n_\textrm{1}}+\frac{1}{n_\textrm{2}}}\tag{4b}$$ Square root of both sides $$SD_{\textrm{w}}=\frac{\sqrt{2}\cdot SE_{\Delta}}{\sqrt{\frac{1}{n_\textrm{1}}+\frac{1}{n_\textrm{2}}}}\tag{5}$$ If we apply \((1)\) instead of \((5)\), the confidence interval will be by \(\small{\sqrt{2}}\) too wide. Opinions?

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1. Gwaza L, Gordon J, Welink J, Potthast H, Hansson H, Stahl M, García-Arieta A. Statistical approaches to indirectly compare bioequivalence between generics: a comparison of methodologies employing artemether / lume­fantrine 20/120 mg tablets as prequalified by WHO. Eur J Clin Pharmacol. 2012; 68(12): 1611–8. doi:10.1007/s00228-012-1396-1.
   
2. Hauschke D, Steinijans VW, Pigeot I. Bioequivalence Studies in Drug Development. Chichester: John Wiley; 2007. p. 90.
   
3. Patterson SD, Jones B. Bioequivalence and Statistics in Clinical Pharmacology. Boca Raton: CRC Press; 2^nd ed. 2016. \(\small{(3.8)}\) p. 37.