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LSM calculation in 2x2 2-stage design

LSM 2-stage design bioequivalence

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#1 Vladimir

Vladimir

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Posted 25 July 2017 - 04:51 PM

Hello all!

 

Could someone kindly explain how LSM is computing in PWNL?

 

I am use this model:

Sequence+Stage+Period(Stage)+Formulation+Sequence*Stage+Subject(Sequence*Stage)

 

All effects is "fixed".

 

And it is clear for me how coefficients of least squares means  is computed, but not clear how to estimate LSM.

 

If I try to calculate LSM like = (Test_mean_seqence1_stage1 + Test_mean_seqence2_stage1 + Test_mean_seqence1_stage2 + Test_mean_seqence1_stage2)/4  - a have result not equal to PWNL.

 

Is there  any analytical expression to do it?

 


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#2 mittyright

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Posted 26 July 2017 - 09:28 PM

Hi Vladimir,

 

Sorry, there's no simple formula as for conventional model due to interaction and nested terms

Citing R help:

 

A special type of linear estimates is the so called least–squares means (or LS–means). Other names for these estimates include population means and marginal means. Consider an imaginary field experiment analyzed with model of the form

> lm( y ~ treat + block + year)

where treat is a treatment factor, block is a blocking factor and year is the year (a factor) where the experiment is repeated over several years. This model specifies the conditional mean E(Y |treat, block, year). One may be interested in predictions of the form E(Y |treat). This quantity can not formally be found from the model. However, it is tempting to average the fitted values of E(Y |treat, block, year) across the levels of block and year and think of this average as E(Y |treat). This average is precisely what is called the LS–means. If the experiment is balanced then this average is identical to the average of the observations when stratified according to treat.

 

So LSmeans are conditional means over other factors. And it's true, it is just an average in case of simple model.

The problem is that 2-stage EMA model is not simple and you need to take into account intrablock information. No analytical ways out

 

hope it helps,

Mittyright



#3 mittyright

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Posted 28 July 2017 - 12:07 PM

Hi again

 

If you are interested in cross-validation with R and in understanding of algo, please review the code below

library(dplyr)
library(nlme)
Subj=as.factor(rep(1:20, each = 2))
Seq=as.factor(c(rep(c("TR","RT"), each = 6), 
                rep(c("RT","TR"), each = 6), 
                rep(c("TR","RT"), each = 4), 
                rep(c("RT","TR"), each = 4)))
Per=as.factor((rep(c(1,2), 20)))
Trt=as.factor(c(rep(c(1,2), 6), 
                rep(c(2,1), 6), 
                rep(c(1,2), 4), 
                rep(c(2,1), 4)))
Stg=as.factor(c(rep(1,24), 
                rep(2,16)))
PK=c(4.8, 4.6, 1.4, 1.3, 3.9, 4.4, 1.5, 2.4, 3.8, 2.9, 3.3, 2.5, 1.7, 1.5, 1.3, 1.9, 2.1, 2.4, 11.4, 13.9, 1.5, 2.3, 2.6, 2.4,
     4.0,4.2,0.7, 0.5, 10.3, 11.9,4.7, 4.8, 5.9,5.5, 3.8, 5.4, 4.3, 2.1, 3.6, 2.4)
M=lm(PK ~ Seq + Stg/Per + Stg:Seq/Subj + Trt)
newdat <- expand.grid(Trt = levels(Trt), Seq = levels(Seq), Subj = levels(Subj), Per = levels(Per), Stg = levels(Stg))
preddata <- cbind(newdat, predict(M, newdat))
dta <- cbind.data.frame(Trt, Seq, Subj, Per,  Stg, PK)
preddatawoNA <- na.omit(left_join(preddata, dta))
preddatawoNA%>%
  group_by(Trt, Seq, Per, Stg) %>%
  summarize(Subjmean = mean(`predict(M, newdat)`)) %>%
  group_by(Trt) %>%
  summarize(lsmean = mean(Subjmean))

All the best,

Mittyright







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