Helmut
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Vienna, Austria,
2020-10-01 18:41
(1446 d 09:15 ago)

Posting: # 21957
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 Adjusted indirect comparisons: Algebra [General Sta­tis­tics]

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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?

  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; 2nd ed. 2016. \(\small{(3.8)}\) p. 37.
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d_labes
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Berlin, Germany,
2020-10-01 19:03
(1446 d 08:52 ago)

@ Helmut
Posting: # 21961
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 Adjusted indirect comparisons: Algebra

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Dear Helmut,

❝ ... The error term in the 2×2×2 crossover is given by $$SE_\textrm{(d)}=SE_\Delta=\widehat{\sigma}_\textrm{w}\sqrt{\frac{1}{n_\textrm{1}}+\frac{1}{n_\textrm{2}}},\tag{2}$$where \(\small{\widehat{\sigma}_\textrm{w}=SD_\textrm{w}=\sqrt{MSE}}\) from ANOVA. Alternatively we can write $$SE_\Delta=\sqrt{\frac{SD_{\textrm{w}}^{2}}{2}\left (\frac{1}{n_\textrm{1}}+\frac{1}{n_\textrm{2}}\right )}\tag{3}$$

Here I can't follow you. From where arises the 2 in formula (3)
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Detlew
 
Helmut
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2020-10-01 19:13
(1446 d 08:43 ago)

@ d_labes
Posting: # 21962
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 Adjusted indirect comparisons: Typo

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Dear Detlew,

❝ ❝ … Alternatively we can write $$SE_\Delta=\sqrt{\frac{SD_{\textrm{w}}^{2}}{2}\left (\frac{1}{n_\textrm{1}}+\frac{1}{n_\textrm{2}}\right )}\tag{3}$$

❝ Here I can't follow you. From where arises the 2 in formula (3)


Sorry, bloody copy & paste error!
\((2)\) in my OP was wrong. Correct: $$SE_\textrm{(d)}=SE_\Delta=\widehat{\sigma}_\textrm{w}\sqrt{\frac{1}{2}\left (\frac{1}{n_\textrm{1}}+\frac{1}{n_\textrm{2}}\right )}\tag{2}$$
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Helmut
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2020-10-02 16:56
(1445 d 10:59 ago)

@ Helmut
Posting: # 21964
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 SE of ∆?

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Dear all,

now I’m confused.
In Chow & Liu* (\(\small{(3.3.1)}\) p.62, Table 3.4.1. p.65, and \(\small{(4.2.2)}\) p.83) the standard error of the difference is given as $$\hat{\sigma}_\textrm{d}\sqrt{\frac{1}{n_\textrm{1}}+\frac{1}{n_\textrm{2}}}$$ I beg your pardon?

  • Chow SC, Liu JP. Design and Analysis of Bioavailability and Bioequivalence Studies. Boca Raton: CRC Press; 3rd ed. 2009.
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d_labes
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Berlin, Germany,
2020-10-02 20:54
(1445 d 07:01 ago)

@ Helmut
Posting: # 21965
Views: 2,632
 

 SE of ∆? or what?

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Dear Helmut,

❝ now I’m confused.


I think all the confusion comes from that sigmaw, sigmad, sigmadelta values including their estimates
which are used by all the authors cited within this thread in a different meaning.

I'm not able to figure out who is who, what is what. Sorry.

The only thing I'm convinced of is that your formula (2) above is correct.
If you write the confidence interval for the BE decision as
PE(T-R) +- SD(d)*tval(0.95, df)

The rest of your algebra is straight forward.
And correct if you ask me ;-).

BTW: the formula (2) is not the error term in the 2×2×2 crossover.
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Detlew
 
Helmut
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2020-10-03 00:29
(1445 d 03:26 ago)

@ d_labes
Posting: # 21966
Views: 2,710
 

 SE of ∆

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Dear Detlew,

❝ I think all the confusion comes from that sigmaw, sigmad, sigmadelta values including their estimates which are used by all the authors cited within this thread in a different meaning.


Quite possible.

❝ I'm not able to figure out who is who, what is what. Sorry.


F**ing terminology.

❝ The only thing I'm convinced of is that your formula (2) above is correct.

❝ If you write the confidence interval for the BE decision as

❝ PE(T-R) +- SD(d)*tval(0.95, df)


Exactly.

❝ The rest of your algebra is straight forward.

❝ And correct if you ask me ;-).


THX. Now three people agree. I even didn’t trust my rusty algebra and asked Maxima for help:

[image]


❝ BTW: the formula (2) is not the error term in the 2×2×2 crossover.


How would you call it? We use \((3)\) in PowerTOST’s BE_CI.R line 30:

sqrt(mse*ades$bkni*nc)

where for design = "2x2" ades$bkni is 0.5 and nc is sum(1/n[1]+1/n[2]).
If we agree that \(\small{\widehat{\sigma}_\textrm{w}=\sqrt{MSE}}\) * we end up with \((2)\):

      For the special case of a two-treatment, two-period crossover study in which \(\small{n_\textrm{1}}\) subjects receive the test formulation in period one and the reference formulation in period two, while \(\small{n_\textrm{2}}\) subjects receive the reference formulation in period one and the reference formulation in period two, the unbiased estimator is given by $$\small{Est.=\frac{\left(\bar{X}_\textrm{T1}+\bar{X}_\textrm{T2} \right)}{2}-\frac{\left(\bar{X}_\textrm{R1}+\bar{X}_\textrm{R2} \right)}{2}}$$ where
      \(\small{\bar{X}_\textrm{T1}=}\) the observed mean of the \(\small{n_\textrm{1}}\) observations of the test formulation in period one.
      \(\small{\bar{X}_\textrm{T2}=}\) the observed mean of the \(\small{n_\textrm{2}}\) observations of the test formulation in period two.
      \(\small{\bar{X}_\textrm{R1}=}\) the observed mean of the \(\small{n_\textrm{2}}\) observations of the reference formulation in period one.
      \(\small{\bar{X}_\textrm{R2}=}\) the observed mean of the \(\small{n_\textrm{1}}\) observations of the reference formulation in period two.
The standard error of this estimator is $$\small{SE=s\sqrt{\frac{1}{2}\left (\frac{1}{n_\textrm{1}}+\frac{1}{n_\textrm{2}} \right)}}$$ where […] \(\small{s}\) is the square root of the “error” mean square from the crossover analysis of variance, based on \(\small{\nu}\) degrees of freedom.


  • 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.
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d_labes
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Berlin, Germany,
2020-10-04 12:23
(1443 d 15:32 ago)

@ Helmut
Posting: # 21969
Views: 2,490
 

 SE of ∆

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Dear Helmut,

❝ ❝ BTW: the formula (2) is not the error term in the 2×2×2 crossover.

❝

❝ How would you call it?


See the subject of your post.
The error term in the 2×2×2 crossover is MSE (aka mean squared error). Or the square root of MSE.

❝ We use \((3)\) in PowerTOST’s BE_CI.R line 30:

sqrt(mse*ades$bkni*nc)

where for design = "2x2" ades$bkni is 0.5 and nc is sum(1/n[1]+1/n[2]).

❝ If we agree that \(\small{\widehat{\sigma}_\textrm{w}=\sqrt{MSE}}\) * we end up with \((2)\)


Correct.
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Regards,

Detlew
 
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