CV from CI [Power / Sample Size]
Hi Louis,
I don’t know which presentation you are referring to. Link?
I’m referring to the one held in Moscow 2014 (slides 24–28).
No. See slide 27: The confidence interval (CI) is given with 0.91–1.15. Hence, the lower confidence limit (CLlo) was 0.91 and the upper one (CLhi) 1.15. The point estimate (PE) aka geometric means ratio (GMR) was not given. It can be calculated by \(\sqrt{CL_{lo} \times CL_{hi}}\), i.e., \(\sqrt{0.91 \times 1.15} = 1.023\).
Yes. This entire game is only necessary if the MSE is not given.
As I already wrote, all calculations are done in the logarithmic* domain. Did you ever wonder why the acceptance range of 80–125% is not symmetric around 100%? Most biologic variables follow a log-normal distribution. Makes sense because negative and even zero values are not possible (what’s a negative concentration?). It means also that log-transformed values follow a normal distribution which is one of the assumptions in the ANOVA. In BE we accept a deviation (Δ) of 20% as to be clinically not relevant. Consequently the lower limit of the acceptance range is 100(1–Δ)=80% and the upper one its reciprocal 100(1–Δ)–1=125%. Transformed to logs we get –0.2231 and +0.2231 which are symmetric around zero. Bingo. We could apply the ANOVA.
Remember some useful rules for working with logs:
\(CI=e^{\log(PE) \mp t \cdot SE_\Delta )}\). SEΔ is calculated from the mean square error MSE and the total sample size (N) as \(\sqrt{2 \cdot MSE / N}\) (slide 28). See slide 27 how the MSE is calculated from the difference between one of the CLs and PE, the subjects in both sequences (n1, n2), and the t-value. Note that \(\log(CL_{hi}) - \log(PE), \log(PE) - \log(CL_{lo})\), or the logs of ratios give the same result: \(\log(1.15) - \log(1.024) = \log(1.023) - \log(0.91) = \log((1.15/1.023)) = \log((1.023/0.91)) = 0.11702\)
Once we have the MSE, calculation of the CV is easy. If you do it by hand (I hope not), I suggest to calculate the CI afterwards. If you don’t get the original values, something went wrong.
I don’t know which presentation you are referring to. Link?
I’m referring to the one held in Moscow 2014 (slides 24–28).
❝ If I am to look on slide 26 when the PE is derived it seems that the CLlo and CLhi are for the ratio for AUCs or Cmaxs as Test vs Ref from previous study.
No. See slide 27: The confidence interval (CI) is given with 0.91–1.15. Hence, the lower confidence limit (CLlo) was 0.91 and the upper one (CLhi) 1.15. The point estimate (PE) aka geometric means ratio (GMR) was not given. It can be calculated by \(\sqrt{CL_{lo} \times CL_{hi}}\), i.e., \(\sqrt{0.91 \times 1.15} = 1.023\).
❝ So maybe this is case when there is no other information about a previous study other than that. Because otherwise if we know the MSE for AUC or Cmax then it it easy to go get directly Cvintra. Am I correct on this?
Yes. This entire game is only necessary if the MSE is not given.
❝ I'm asking because when delta(CL) is derived it seems that we take the logs. Not sure why...
As I already wrote, all calculations are done in the logarithmic* domain. Did you ever wonder why the acceptance range of 80–125% is not symmetric around 100%? Most biologic variables follow a log-normal distribution. Makes sense because negative and even zero values are not possible (what’s a negative concentration?). It means also that log-transformed values follow a normal distribution which is one of the assumptions in the ANOVA. In BE we accept a deviation (Δ) of 20% as to be clinically not relevant. Consequently the lower limit of the acceptance range is 100(1–Δ)=80% and the upper one its reciprocal 100(1–Δ)–1=125%. Transformed to logs we get –0.2231 and +0.2231 which are symmetric around zero. Bingo. We could apply the ANOVA.
Remember some useful rules for working with logs:
- \(\log {(x \cdot y)} = \log {(x)} + \log {(y)}, \log {(x / y)} = \log {(x)} - \log {(y)}\)
- \(\log {(x^y)} = y \cdot \log {(x)}, \sqrt[y]{x} = \log {(x)} / y\)
- The back-transformed arithmetic mean of logs is the geometric mean of original values:
\(e^{1/n \cdot \sum {\log {(x_1)}, \ldots, \log {(x_i)}})} =\sqrt[n]{\prod {x_1, \ldots, x_n}}\)
Before the age of computers the first formula was used. With the latter even pocket calculators reach their limit soon. Not to speak of paper/pencil/brain.
\(CI=e^{\log(PE) \mp t \cdot SE_\Delta )}\). SEΔ is calculated from the mean square error MSE and the total sample size (N) as \(\sqrt{2 \cdot MSE / N}\) (slide 28). See slide 27 how the MSE is calculated from the difference between one of the CLs and PE, the subjects in both sequences (n1, n2), and the t-value. Note that \(\log(CL_{hi}) - \log(PE), \log(PE) - \log(CL_{lo})\), or the logs of ratios give the same result: \(\log(1.15) - \log(1.024) = \log(1.023) - \log(0.91) = \log((1.15/1.023)) = \log((1.023/0.91)) = 0.11702\)
Once we have the MSE, calculation of the CV is easy. If you do it by hand (I hope not), I suggest to calculate the CI afterwards. If you don’t get the original values, something went wrong.
- logℯ(x), the natural of Napierian logarithm (base ℯ≈2.71828…). If ever possible, avoid the common logarithm log10(x).
If your analysis was done on log10-transformed data: \(CV_{intra} = \sqrt{10^{\log_e(10) \cdot MSE} - 1}\).
—
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Helmut Schütz
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Dif-tor heh smusma 🖖🏼 Довге життя Україна!
![[image]](https://static.bebac.at/pics/Blue_and_yellow_ribbon_UA.png)
Helmut Schütz
![[image]](https://static.bebac.at/img/CC by.png)
The quality of responses received is directly proportional to the quality of the question asked. 🚮
Science Quotes
Complete thread:
- Finding the CI limits Louis52 2018-02-01 19:57 [Power / Sample Size]
- CV from CI Helmut 2018-02-02 00:56
- CV from CI Louis52 2018-02-03 17:12
- CV from CIHelmut 2018-02-03 21:48
- CV from CI Louis52 2018-02-04 14:38
- CV from CI Helmut 2018-02-05 11:39
- CV from CI Louis52 2018-02-04 14:38
- CV from CIHelmut 2018-02-03 21:48
- CV from CI Louis52 2018-02-03 17:12
- CV from CI Helmut 2018-02-02 00:56