Bebac user
☆

Egypt,
2022-01-23 11:55
(368 d 18:04 ago)

Posting: # 22738
Views: 1,412

## intrasubject variability [Power / Sample Size]

How can I calculate the intrasubject variability (CV) after my study was done ?
Using MSE or (POE, U, L) ?

I tried to compare them for Partially replicated study:
CV for Cmax by MSE = 51.59%
CV for Cmax by (POE, U, L) = 43.27%
So, which one is the correct ?
ElMaestro
★★★

Denmark,
2022-01-23 13:24
(368 d 16:36 ago)

@ Bebac user
Posting: # 22739
Views: 1,205

## intrasubject variability

Hi Bebac_user,

❝ How can I calculate the intrasubject variability (CV) after my study was done ?

❝ Using MSE or (POE, U, L) ?

Both should work and both will give the same result within the meaning of rounding.

❝ I tried to compare them for Partially replicated study:

❝ CV for Cmax by MSE = 51.59%

❝ CV for Cmax by (POE, U, L) = 43.27%

❝ So, which one is the correct ?

Well, I think you may have made a tiny mistake somewhere since you did not get the same result. But I can't tell from your post where the hickup is.
Probably the MSE is the most direct method for getting a CV, but if the design was semirepl. then the intrasubject var. associated with ref might not have come from an MSE. It really depends on the steps you took. The devil is in the detail. So kindly provide more of it Pass or fail!
ElMaestro
Helmut
★★★  Vienna, Austria,
2022-01-23 14:12
(368 d 15:47 ago)

@ Bebac user
Posting: # 22740
Views: 1,224

## replicate designs: intrasubject variability

Hi Bebac user,

❝ How can I calculate the intrasubject variability (CV) after my study was done ?

❝ Using MSE or (POE, U, L) ?

❝ I tried to compare them for Partially replicated study:

❝ CV for Cmax by MSE = 51.59%

❝ CV for Cmax by (POE, U, L) = 43.27%

❝ So, which one is the correct ?

It depends on what you want and what for. From the ANOVA of replicate designs (partial and full) you get the with-subject $$\small{CV_\textrm{w}}$$ by $$\small{CV_\textrm{w}=\sqrt{\exp(MSE)-1}}\tag{1}$$ To get the pooled $$\small{CV_\textrm{w}}$$ from $$\small{CV_\textrm{wT}}$$ and $$\small{CV_\textrm{wR}}$$ you have to convert them to their associated variances $$\small{\begin{matrix}\tag{2} s_\textrm{wT}^2=\log_{e}(CV_\textrm{wT}^2+1)\\ s_\textrm{wR}^2=\log_{e}(CV_\textrm{wR}^2+1) \end{matrix}}$$ calculated their mean $$\small{s_\textrm{w}^2=(s_\textrm{wT}^2+s_\textrm{wR}^2)/2}\tag{3}$$ and back-transform to the $$\small{CV_\textrm{w}}$$ $$\small{CV_\textrm{w}= \sqrt{\exp(s_\textrm{w}^2)-1}}\tag{4}$$ This is exact only in a full replicate design because in a partial replicate $$\small{CV_\textrm{wT}}$$ is unknown. $$\small{CV_\textrm{w}}$$ obtained by $$\small{(4)}$$ is identical to the one obtained by $$\small{(1)}$$ only if sequences are balanced. Let’s try that:

library(replicateBE) library(PowerTOST) # Patterson & Jones https://doi.org/10.1002/pst.498 (Table II) # partial replicate, balanced, complete x   <- method.A(data = rds04, verbose = TRUE, print = FALSE,                 details = TRUE) # get subjects / sequence n   <- as.integer(strsplit(x[["Sub/seq"]], "\\|")[]) # CVw from the CI CVw <- CI2CV(lower = x[["CL.lo(%)"]], upper = x[["CL.hi(%)"]],              design = "2x3x3", n = n) cat(paste0("Design: ", x[["Design"]],     ", subjects / sequence: ", x[["Sub/seq"]], "\n"),     sprintf("CVwR from Method A  = %.2f%%", x[["CVwR(%)"]]), "\n",     sprintf("CVwT from Method A  = %.2f%%", x[["CVwT(%)"]]), "\n",     sprintf("CVw  from CI        = %.2f%%", 100 * CVw), "\n",     sprintf("CVw  from ANOVA MSE = %.2f%%", 100 * mse2CV(0.28377)), "\n") Data set DS04: Method A by lm() ─────────────────────────────────── Type III Analysis of Variance Table Response: log(PK)                  Df  Sum Sq Mean Sq  F value     Pr(>F) sequence          2  4.2619 2.13095  2.06562 0.13785915 period            2  0.0657 0.03287  0.11582 0.89075874 treatment         1  3.4031 3.40306 11.99223 0.00079058 sequence:subject 48 49.5180 1.03162  3.63540 2.9213e-08 Residuals        99 28.0935 0.28377                    treatment T – R:    Estimate  Std. Error     t value    Pr(>|t|) 0.316370000 0.091357800 3.462980000 0.000790582 99 Degrees of Freedom Design: TRR|RTR|RRT, subjects / sequence: 17|17|17  CVwR from Method A  = 61.22%  CVwT from Method A  = NA%  CVw  from CI        = 57.28%  CVw  from ANOVA MSE = 57.28%

Of course, we don’t get $$\small{CV_\textrm{wT}}$$ in the partial replicate design. Since the study was balanced, $$\small{CV_\textrm{w}}$$ calculated from the CI is identical to the one obtained from the ANOVA’s $$\small{MSE}$$.

Let’s remove the first subject from the data set and see what happens.

rds04 <- rds04[rds04\$subject != 1, ] # keep all but the first -> imbalanced x   <- method.A(data = rds04, verbose = TRUE, print = FALSE,                 details = TRUE) n   <- as.integer(strsplit(x[["Sub/seq"]], "\\|")[]) CVw <- CI2CV(lower = x[["CL.lo(%)"]], upper = x[["CL.hi(%)"]],              design = "2x3x3", n = n) cat(paste0("Design: ", x[["Design"]],     ", subjects / sequence: ", x[["Sub/seq"]], "\n"),     sprintf("CVwR from Method A  = %.2f%%", x[["CVwR(%)"]]), "\n",     sprintf("CVwT from Method A  = %.2f%%", x[["CVwT(%)"]]), "\n",     sprintf("CVw  from CI        = %.2f%%", 100 * CVw), "\n",     sprintf("CVw  from ANOVA MSE = %.2f%%", 100 * mse2CV(0.28825)), "\n") Data set DS04: Method A by lm() ─────────────────────────────────── Type III Analysis of Variance Table Response: log(PK)                  Df  Sum Sq Mean Sq  F value     Pr(>F) sequence          2  2.8418 1.42092  1.43254  0.2489319 period            2  0.0461 0.02303  0.07989  0.9232771 treatment         1  3.2785 3.27853 11.37382  0.0010713 sequence:subject 47 46.6188 0.99189  3.44104 1.3845e-07 Residuals        97 27.9605 0.28825                    treatment T – R:   Estimate Std. Error    t value   Pr(>|t|) 0.31368000 0.09301090 3.37251000 0.00107127 97 Degrees of Freedom Design: TRR|RTR|RRT, subjects / sequence: 17|16|17  CVwR from Method A  = 61.60%  CVwT from Method A  = NA%  CVw  from CI        = 57.79%  CVw  from ANOVA MSE = 57.80%

Not exactly the same but pretty close.

What about an imbalanced and incomplete (periods missing) full replicate design?

# The EMA's (in)famous reference data set I x   <- method.A(data = rds01, verbose = TRUE, print = FALSE,                 details = TRUE) n   <- as.integer(strsplit(x[["Sub/seq"]], "\\|")[]) CVw <- CI2CV(lower = x[["CL.lo(%)"]], upper = x[["CL.hi(%)"]],              design = "2x2x4", n = n) cat(paste0("Design: ", x[["Design"]],     ", subjects / sequence: ", x[["Sub/seq"]], "\n"),     sprintf("CVwR from Method A  = %.2f%%", x[["CVwR(%)"]]), "\n",     sprintf("CVwT from Method A  = %.2f%%", x[["CVwT(%)"]]), "\n",     sprintf("CVw  from CI        = %.2f%%", 100 * CVw), "\n",     sprintf("CVw  from ANOVA MSE = %.2f%%", 100 * mse2CV(0.159995)), "\n") Data set DS01: Method A by lm() ─────────────────────────────────── Type III Analysis of Variance Table Response: log(PK)                   Df   Sum Sq  Mean Sq  F value     Pr(>F) sequence           1   0.0077 0.007652  0.00268  0.9588496 period             3   0.6984 0.232784  1.45494  0.2278285 treatment          1   1.7681 1.768098 11.05095  0.0010405 sequence:subject  75 214.1296 2.855061 17.84467 < 2.22e-16 Residuals        217  34.7190 0.159995              treatment T – R:   Estimate Std. Error    t value   Pr(>|t|) 0.14547400 0.04650870 3.12788000 0.00200215 217 Degrees of Freedom Design: TRTR|RTRT, subjects / sequence: 39|38  CVwR from Method A  = 46.96%  CVwT from Method A  = 35.16%  CVw  from CI        = 42.58%  CVw  from ANOVA MSE = 41.65%

Not so close. $$\small{(3)}$$ is only correct for balanced sequences. Although it’s possible to weigh the variances by the degrees of freedom, I wouldn’t do that.

Conclusion: If you have the ANOVA and are interested in $$\small{CV_\textrm{w}}$$, use $$\small{(1)}$$.

Dif-tor heh smusma 🖖🏼 Довге життя Україна! Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. 🚮
Science Quotes
Bebac user
☆

Egypt,
2022-01-23 15:03
(368 d 14:57 ago)

@ Helmut
Posting: # 22741
Views: 1,172

## replicate designs: intrasubject variability

Thank you my role model for this pretty answer.
Helmut
★★★  Vienna, Austria,
2022-01-23 15:16
(368 d 14:44 ago)

@ Bebac user
Posting: # 22742
Views: 1,211

## OT

Hi Bebac user,

❝ Thank you my role model …

Well. On my way from قنا to سفاجا in August 1979. Note the speargun attached to my backpack.
Was before I got my first diving certification.  ❝ … for this pretty answer.

Welcome!

Dif-tor heh smusma 🖖🏼 Довге життя Україна! Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. 🚮
Science Quotes
Bebac user
☆

Egypt,
2022-01-23 21:47
(368 d 08:13 ago)

@ Helmut
Posting: # 22746
Views: 1,189

## OT

??????????
dshah
★★

India/United Kingdom,
2022-01-24 11:53
(367 d 18:07 ago)

@ Bebac user
Posting: # 22748
Views: 1,210

## intrasubject variability

Dear Bebac User!
You can try FARTSSIE to calculate ISCV.
Regards,
Dshah  Ing. Helmut Schütz 