Bad science (statistics is taboo) [Study As­sess­ment]

posted by Helmut Homepage – Vienna, Austria, 2020-07-17 01:01 (27 d 03:13 ago) – Posting: # 21727
Views: 618

Hi Loky do,

» » ... it is considered acceptable that low-lier profiles can be excluded from statistical analysis […] if they occur with the same or lower frequency in the test product compared to the reference product.
»
» Is this case-specific for this product, and its formula?

Yes. It’s based on numerous replicate design studies reviewed by the PKWP. AFAIK, in all more outliers were seen after the reference. It’s a terrible product showing also funky batch-to-batch variability.

» or it could be followed for other products with the same cases?

If it’s a delayed release product, see my previous post. However, note the subtle different wording “number” vs. “frequency”.

In the draft MR-GL “comparable frequency” was used as well. From a statistical point of view that calls for a two-sided test. When I showed an example at the GL-meeting in Bonn (2013) I stirred up a hornets’ nest. Imagine: 64 subjects, 4 outliers after T and 3 after R. Is the frequency comparable? Yes, the p-value is 0.7139!1 The members of the PKWP did not like that at all. Oops, one (!) more. Changed in the final GL. Not acceptable. You can have it even more extreme. A 4-period full replicate study, one outlier (0.78%) after T and none after R → p-value 0.5.2 Not acceptable because the bloody number is higher!
Lesson learned: Bad science (statistics is taboo). If you have one more outlier after T than after R – independent of the sample size – you’re dead.


  1. Two-sided test (is T = R?)
    subj <- 64
    per  <- 2
    seq  <- 2
    OL.R <- 3
    OL.T <- 4
    n.T  <- n.R <- subj*per/seq
    outliers  <- matrix(c(OL.T, n.T-OL.T, OL.R, n.R), nrow = 2,
                        dimnames=list(Guess = c("T", "R"),
                                      Truth = c("T", "R")))
    fisher.test(outliers, alternative = "two.sided")

            Fisher's Exact Test for Count Data
    data:  outliers

    p-value = 0.7139
    alternative hypothesis: true odds ratio is not equal to 1
    95 percent confidence interval:
      0.2296229 10.0829801
    sample estimates:
    odds ratio
      1.418408


  2. One-sided test (is T > R?)
    subj <- 64
    per  <- 4
    seq  <- 2
    OL.R <- 0
    OL.T <- OL.R + 1
    n.T  <- n.R <- subj*per/seq
    outliers  <- matrix(c(OL.T, n.T-OL.T, OL.R, n.R), nrow = 2,
                        dimnames=list(Guess = c("T", "R"),
                                      Truth = c("T", "R")))
    fisher.test(outliers, alternative = "greater")

            Fisher's Exact Test for Count Data

    data:  outliers

    p-value = 0.5
    alternative hypothesis: true odds ratio is greater than 1
    95 percent confidence interval:
     0.05262625        Inf
    sample estimates:
    odds ratio
           Inf

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