ElMaestro
★★★

Denmark,
2013-04-08 22:23
(4419 d 00:37 ago)

Posting: # 10375
Views: 6,968
 

 EMA dataset 1 trouble [RSABE / ABEL]

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Hi all,

I am fiddling a bit around with scaled BE and trying to use R for same.
Cannot get to 107.11% - 124.89% for EMA dataset 1 / method A. My PE is correct so the trouble I guess is in somewhere in one of these parts:

1. sqrt(0.5*(1/nSeq1 + 1/nSeq2)
2. df and the critical T-value.

Ad 1:
Looks to me like the lm allows effect estimates for 75 subjects; this means subjects with a missing value are not tossed out. So I guess all is fine with 77 in total; seemingly 38 and 39. But I get the wrong CI anyway.

Ad 2:
For df I tried both the df form the full model and from the Ref-only model (let me add: I think it is a valid argument that the df from the latter should be used but I am sure we can have a good fight over that some other time).

Any ideas?
Here's some code bits:
A=read.table("EMAset1b.csv", header=T)
Model1=lm(log(A$DATA)~0+factor(A$FORMULATION)+
factor(A$PERIOD)+factor(A$SEQUENCE)+factor(A$SUBJECT))
lnPE=(coef(Model1)[2]-coef(Model1)[1]) ## works!
B=subset(A, FORMULATION=='R')
Model2=lm(log(B$DATA)~
factor(B$PERIOD)+factor(B$SEQUENCE)+factor(B$SUBJECT))
s=summary(Model2)$sigma ## works!
CV=sqrt(exp(s*s)-1.0) ## works!
DF=Model1$df.residual ## or use Model2's residual. Discuss some other time.
CritT=qt(1-0.05, DF)

(blah blah code for extraction of n per sequence)

lnU=lnPE+s*CritT*sqrt(0.5*(1/nBABA+1/nABAB)) ##can't get it right.
lnL=lnPE-s*CritT*sqrt(0.5*(1/nBABA+1/nABAB)) ##can't get it right.

Thanks in advance.
—
Pass or fail!
ElMaestro
 
Helmut
★★★
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Vienna, Austria,
2013-04-09 04:37
(4418 d 18:23 ago)

@ ElMaestro
Posting: # 10376
Views: 5,919
 

 complete (RR) only

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Hi ElMaestro!

❝ I am fiddling a bit around with scaled BE and trying to use R for same.


Welcome to the crippled model!

❝ Ad 1:

❝ Looks to me like the lm allows effect estimates for 75 subjects; this means subjects with a missing value are not tossed out. So I guess all is fine with 77 in total; seemingly 38 and 39.


Wonderful. Phoenix/WinNonlin by default uses all data as well (REML…). Since we have to mimick SAS we have to get rid of subjects with incomplete data (#24, 31, 67, 71). The complete data set contains 73 subjects with two administrations of R.

❝ Ad 2:

❝ For df I tried both the df form the full model and from the Ref-only model (let me add: I think it is a valid argument that the df from the latter should be used but I am sure we can have a good fight over that some other time).


Yep. 71.

❝ A=read.table("EMAset1b.csv", header=T)


Can you post your EMAset1b.csv? Since my crystal ball is at the laundry I can’t check your code.
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d_labes
★★★

Berlin, Germany,
2013-04-09 11:01
(4418 d 11:59 ago)

@ ElMaestro
Posting: # 10379
Views: 5,926
 

 EMA dataset 1 - programmers trouble

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Hi ElMaestro,

❝ Here's some code bits:

❝

❝ A=read.table("EMAset1b.csv", header=T)

❝ ....

❝ lnU=lnPE+s*CritT*sqrt(0.5*(1/nBABA+1/nABAB)) ##can't get it right.

❝ lnL=lnPE-s*CritT*sqrt(0.5*(1/nBABA+1/nABAB)) ##can't get it right.


Try it with:
lnU=lnPE+s*CritT*sqrt(0.25*(1/nBABA+1/nABAB))
lnL=lnPE-s*CritT*sqrt(0.25*(1/nBABA+1/nABAB))

or even easier fit a model with intercept and get the confidence interval via call of confint(). No need to fiddle with n in sequences:
all <- lm(log(Data) ~ as.factor(formulation)+ as.factor(period) +as.factor(subject), data=A)
lnPE <- coef(all)[2]
CI   <- confint(all,level=0.9)[2,]

BTW: I have omitted sequence since it doesn't contribute to the mse to have such a term. Moreover I have all effects in lower letters (caps lock not arrested :-D).

My SAS gives (without back transformation):
                              Standard
  Parameter    Estimate          Error   t Value   Pr > |t|     90% Confidence Limits

  test-ref   0.14547372     0.04650869      3.13     0.0020     0.06864574   0.22230170

The latter R code gives:
> print(lnPE)
as.factor(tmt)T
      0.1454737
> print(CI)
       5 %       95 %
0.06864569 0.22230165
—
Regards,

Detlew
 
ElMaestro
★★★

Denmark,
2013-04-09 12:12
(4418 d 10:48 ago)

@ d_labes
Posting: # 10380
Views: 5,846
 

 Still not exactly sure

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Thank you, Helmut and d_labes, for your input.

I am not planning to via the in-built confidence interval function because I can port it directly to C when I know exactly what's going on.

Helmut, I downloaded your spreadsheet, saved the dataset as csv with tab delimiter and adjusted the code to fit your format.

Here's some code:
A=read.table("EMAset1c.csv", header=T)

Model1=lm(log(A$Data)~0+factor(A$Formulation)+
factor(A$Period)+factor(A$Sequence)+factor(A$Subject))

lnPE=(coef(Model1)[2]-coef(Model1)[1])


B=subset(A, Formulation=='R')
Model2=lm(log(B$Data)~factor(B$Period)+factor(B$Sequence)+factor(B$Subject))
s=summary(Model2)$sigma
CV=sqrt(exp(s*s)-1.0)
DF=Model2$df.residual
CritT=qt(1-0.05, DF)

nTRTR=0
nRTRT=0

for (i in min(B$Subject):max(B$Subject))
if (i %in% B$Subject) ##if the subject exists
{
  C=subset(B, Subject==i)
  r=subset(C, Formulation=='R')
  if(length(r$Formulation)==2) ##if the subject ate pocus two times.
   {
    if (C$Sequence[1]=='TRTR') nTRTR=nTRTR+1
    if (C$Sequence[1]=='RTRT') nRTRT=nRTRT+1
   }
} ## there's prolly a more efficient way of doing it.

lnU=lnPE+s*CritT*sqrt(0.25*(1/nTRTR+1/nRTRT))
lnL=lnPE-s*CritT*sqrt(0.25*(1/nTRTR+1/nRTRT))
cat("PE=", exp(lnPE),"\n")
cat("Upper Limit=", exp(lnU),"\n")
cat("Lower Limit=", exp(lnL),"\n")
cat("CV=", CV, "\n")
cat("nTRTR , nRTRT:", nTRTR, ",", nRTRT, "\n")
 

I am still not really there :-D
—
Pass or fail!
ElMaestro
 
d_labes
★★★

Berlin, Germany,
2013-04-09 17:57
(4418 d 05:03 ago)

@ ElMaestro
Posting: # 10383
Views: 5,814
 

 Still not exactly sure - why

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Dear ElMaestro,

could you kindly give a more elaborate description of what is your problem?
Without charging me with the burden to run your code and figure out what's going on!
—
Regards,

Detlew
 
ElMaestro
★★★

Denmark,
2013-04-09 18:39
(4418 d 04:21 ago)

@ d_labes
Posting: # 10384
Views: 5,716
 

 My problem(s)

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Hi d_labes,

❝ could you kindly give a more elaborate description of what is your problem?

❝ Without charging me with the burden to run your code and figure out what's going on!


I would like to reproduce the result of EMA's Q&A with method A: 107.11% - 124.89% for the CI using R.
The code was an attempt at that and it was of course a miserable failure which can probably be explained by my lack of skills. :vomit:
—
Pass or fail!
ElMaestro
 
ElMaestro
★★★

Denmark,
2013-04-10 01:29
(4417 d 21:31 ago)

@ d_labes
Posting: # 10386
Views: 5,793
 

 More confusion

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Gentlemen...

I have the point estimate and s right, this much is certain.
Since something is wrong, I just need to figure out which n-per-sequence that would result in limits such as the lower one of 107.11:

for (n1 in 30:80)
for (n2 in 30:80)
 {
  df=n1+n2-2
  ct = qt(1-0.05, df)
   lnL=lnPE-s*ct*sqrt(0.25*(1/n1+1/n2))
  if (exp(lnL)>1.0710)
  if (exp(lnL)<1.0712) cat("n1=", n1, "n2=",n2, "lower limit=", exp(lnL), "\n")
 }


The output is:
n1= 34 n2= 77 lower limit= 1.071095
n1= 35 n2= 72 lower limit= 1.071009
n1= 35 n2= 73 lower limit= 1.0712
n1= 36 n2= 69 lower limit= 1.071178
n1= 37 n2= 65 lower limit= 1.07102
n1= 39 n2= 60 lower limit= 1.071096
n1= 40 n2= 58 lower limit= 1.071152
n1= 41 n2= 56 lower limit= 1.071138
n1= 42 n2= 54 lower limit= 1.071052
n1= 44 n2= 51 lower limit= 1.071036
n1= 45 n2= 50 lower limit= 1.071145
n1= 50 n2= 45 lower limit= 1.071145
n1= 51 n2= 44 lower limit= 1.071036
n1= 54 n2= 42 lower limit= 1.071052
n1= 56 n2= 41 lower limit= 1.071138
n1= 58 n2= 40 lower limit= 1.071152
n1= 60 n2= 39 lower limit= 1.071096
n1= 65 n2= 37 lower limit= 1.07102
n1= 69 n2= 36 lower limit= 1.071178
n1= 72 n2= 35 lower limit= 1.071009
n1= 73 n2= 35 lower limit= 1.0712
n1= 77 n2= 34 lower limit= 1.071095

So, realistic n1 and n2 add up to irrelevant values, therefore I might be inclined to believe the problem must be one of these lines:
df=n1+n2-2
ct = qt(1-0.05, df)
lnL=lnPE-s*ct*sqrt(0.25*(1/n1+1/n2))

:confused::confused::confused:
—
Pass or fail!
ElMaestro
 
d_labes
★★★

Berlin, Germany,
2013-04-10 12:51
(4417 d 10:10 ago)

@ ElMaestro
Posting: # 10396
Views: 5,714
 

 Confusion help

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My Dear,

don't puzzle in one of your mentioned directions.
The formulas used are obviously only correct in case of no missings. And the EMA set I has missings.

In case of missings you have to use a complicated formula involving the design matrix only ElMaestro can understand :-D. Have a look at the explanation given for the ESTIMATE statement of SAS Proc GLM.
But don't ask me what the vectors and matrices are there. I myself could it only understand after some pills of Schützomycin.

Hope this helps against confusion seizures.
—
Regards,

Detlew
 
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