Imph ★ 20220620 18:16 (666 d 22:57 ago) Posting: # 23069 Views: 2,003 

Hello, I would like to know if in the parallel design, we should not assume the equality of the variances. If the equality of variances is not assumed, how do the statistical model, the statistical analysis and the formula for the confidence interval become? Best regards. 
Helmut ★★★ Vienna, Austria, 20220620 18:47 (666 d 22:26 ago) @ Imph Posting: # 23070 Views: 1,824 

Hi Imph, ❝ I would like to know if in the parallel design, we should not assume the equality of the variances. Correct. Specifically stated by the FDA For parallel designs […] equal variances should not be assumed. and in between the lines by the EMAThe precise model to be used for the analysis should be prespecified in the protocol. The statistical analysis should take into account sources of variation that can be reasonably assumed to have an effect on the response variable. ❝ […] how do the statistical model, the statistical analysis and the formula for the confidence interval become? Perform Welch’s ttest, i.e., with Satterthwaite’s approximation of the degrees of freedom. $$\nu=\frac{\big{(}\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\big{)}^2}{\frac{s_1^4}{n_1^2(n_11)}+\frac{s_2^4}{n_2^2(n_21)}}$$where \(\small{n_1,n_2}\) are the number of subjects under treatments T and R, respectively. \(\small{s_1,s_2}\) are the standard deviations of treatment arms. Not by any chance it is the default in and SAS , whereas SPSS calculates always the conventional ttest (assuming equal variances) and the Welchtest. In the output table, use the second row.Calculate the confidence interval as usual with the tvalue for given \(\small{\alpha,\,\nu}\). It’s also easy in Phoenix WinNonlin and PKanalix. Consult the respective manual. You should check whether your setup is correct.*
— 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 
Imph ★ 20220623 15:02 (664 d 02:11 ago) @ Helmut Posting: # 23074 Views: 1,743 

Hello, Thank you for your valuable response. Is it possible to conduct an analysis of variance according to the following table without assuming equality of variances? Does Welch's correction only concern the calculation of confidence intervals? Thank's a lot. 
Helmut ★★★ Vienna, Austria, 20220623 16:19 (664 d 00:55 ago) @ Imph Posting: # 23075 Views: 1,718 

Hi Imph, ❝ Thank you for your valuable response. Welcome. ❝ Is it possible to conduct an analysis of variance according to the following table without assuming equality of variances? You could but it would be wrong. Since you must not assume equal variances and in your table the residual degrees of freedom n_{1}+n_{2}–2 are the ones of the conventional ttest. ❝ Does Welch's correction only concern the calculation of confidence intervals? You obtain also the pvalue of the treatment difference (which is, however, not relevant in bioequivalence). Crossvalidated (‼) example codes are given as supplementary information to the publication mentioned in my last post:The first one should work also in Excel. There is no excuse to use a flawed method. — 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 
Imph ★ 20220626 12:06 (661 d 05:07 ago) @ Helmut Posting: # 23089 Views: 1,659 

Hello, How can we analyze the treatment effect if we cannot use the analysis of variance? 
Helmut ★★★ Vienna, Austria, 20220626 14:12 (661 d 03:02 ago) @ Imph Posting: # 23090 Views: 1,649 

Hi Imph, ❝ How can we analyze the treatment effect if we cannot use the analysis of variance? Why didn’t you bother reading the paper given there and trying one of the codes given there? Again: Why are you interested in the – irrelevant – treatment effect (a pvalue)? Statistical significance ≠ clinical relevance. BE is solely based on the latter. All formulas for logtransformed data. Testing for a statistically significant difference $$H_0:\mu_\text{T}\mu_\text{R}=0\;vs\;H_1:\mu_\text{T}\mu_\text{R}\neq0\tag{1}$$ was abandoned more than thirty years ago by the Two OneSided Tests [1] $$\eqalign{ H_{01}:\mu_\text{T}\mu_\text{R}\leq\theta_1&vs\;H_{11}:\mu_\text{T}\mu_\text{R}>\theta_1\\ &\text{and}\\ H_{02}:\mu_\text{T}\mu_\text{R}\geq\theta_2&vs\;H_{12}:\mu_\text{T}\mu_\text{R}<\theta_2}\tag{2}$$ or the operationally identical – and preferred in guidelines – confidence interval inclusion approach $$H_0:\mu_\text{T}\mu_\text{R}\ni\left\{\theta_1,\,\theta_2\right\}\;vs\;H_1:\theta_1<\mu_\text{T}\mu_\text{R}<\theta_2\tag{3}$$ The limits \(\small{\left\{\theta_1,\,\theta_2\right\}}\) are based on the clinically not relevant difference \(\small{\Delta}\). In other words, nobody is interested in a pvalue. See also this article. An script to evaluate the reference data set #6 [2] (unequal groups sizes and unequal variances):
By setting eq < TRUE in the script you will get at the end (only for comparison):
Results are similar because there is no large difference in group sizes and variances are similar as well. However, the conventional ttest is liberal (anticonservative, inflated patient’s risk) because its confidence interval is always narrower than the one of Welch’s test. If you have extremely different sample sizes and variances, the outcome may differ substantially. The reference data set #7 [3] is actually a complete failure with a 90% CI of 97.38–138.5% (Welch’s test: 201.164 degrees of freedom, standard error of the difference 0.106596) but almost passes BE with 106.86–126.2% (ttest: 1198 df, SE 0.0505922). Why is the CI obtained by the ttest narrower than by Welch’s test though the point estimates are equal? We have more degrees of freedom (≈ 6×) and the standard error of the difference is smaller (≈ ½).
— 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 