Helmut ★★★ Vienna, Austria, 20201001 18:41 (1295 d 03:27 ago) Posting: # 21957 Views: 2,681 

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 crossover^{2} 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. Alternatively^{3} 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?
— Diftor heh smusma 🖖🏼 Довге життя Україна! _{} Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes 
d_labes ★★★ Berlin, Germany, 20201001 19:03 (1295 d 03:05 ago) @ Helmut Posting: # 21961 Views: 2,243 

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 $$SE_\Delta=\sqrt{\frac{SD_{\textrm{w}}^{2}}{2}\left (\frac{1}{n_\textrm{1}}+\frac{1}{n_\textrm{2}}\right )}\tag{3}$$ — Regards, Detlew 
Helmut ★★★ Vienna, Austria, 20201001 19:13 (1295 d 02:55 ago) @ d_labes Posting: # 21962 Views: 2,227 

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: $$SE_\textrm{(d)}=SE_\Delta=\widehat{\sigma}_\textrm{w}\sqrt{\frac{1}{2}\left (\frac{1}{n_\textrm{1}}+\frac{1}{n_\textrm{2}}\right )}\tag{2}$$ — Diftor heh smusma 🖖🏼 Довге життя Україна! _{} Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes 
Helmut ★★★ Vienna, Austria, 20201002 16:56 (1294 d 05:12 ago) @ Helmut Posting: # 21964 Views: 2,164 

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?
— Diftor heh smusma 🖖🏼 Довге життя Україна! _{} Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes 
d_labes ★★★ Berlin, Germany, 20201002 20:54 (1294 d 01:13 ago) @ Helmut Posting: # 21965 Views: 2,124 

Dear Helmut, ❝ now I’m confused. I think all the confusion comes from that sigma_{w}, sigma_{d}, sigma_{delta} 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(TR) + 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. — Regards, Detlew 
Helmut ★★★ Vienna, Austria, 20201003 00:29 (1293 d 21:39 ago) @ d_labes Posting: # 21966 Views: 2,192 

Dear Detlew, ❝ I think all the confusion comes from that sigma_{w}, sigma_{d}, sigma_{delta} 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(TR) + 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: ❝ 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:
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 \((2)\):
— Diftor heh smusma 🖖🏼 Довге життя Україна! _{} Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes 
d_labes ★★★ Berlin, Germany, 20201004 12:23 (1292 d 09:45 ago) @ Helmut Posting: # 21969 Views: 1,977 

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
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. — Regards, Detlew 