PowerTOST: CVfromCI -> CI.BE [Study As­sess­ment]

posted by d_labes  – Berlin, Germany, 2019-02-20 14:12  – Posting: # 19938
Views: 3,728

Dear Helmut,

» ...
» How to discover which method was used?
» Work backwards, i.e., see with which CV you can reproduce the reported results for each comparison.
» res.1 <- CI.BE(pe=pe, CV=CV.1, n=n, design=des)
» res.2 <- CI.BE(pe=pe, CV=CV.2, n=n, design=eval)
» cat(paste0("\nBack-calculated 90% CI by",
»     "\n  Pooled ANOVA           : ",
»     sprintf("%.2f%%%s", 100*res.1[["lower"]], "\u2013"),
»     sprintf("%.2f%%", 100*res.1[["upper"]]),
»     "\n  Two-at-a-Time Principle: ",
»     sprintf("%.2f%%%s", 100*res.2[["lower"]], "\u2013"),
»     sprintf("%.2f%%", 100*res.2[["upper"]]), "\n"))
» Back-calculated 90% CI by
»   Pooled ANOVA           : 85.00%–106.18%
»   Two-at-a-Time Principle: 85.00%–106.18%

IMHO this suggestion is an orouboros.
Calculating the CV from the CI and using this CV to calculate the CI will give you always the CI used in the starting step. Regardless of the design used in both steps.
As you has demonstrated with your calculations :cool:.



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