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I ran into this presentation on Statistical aspects of two-way cross-over studies by Ing. Helmut Schütz (http://bebac.at). He presented some code and referred to the bear package. The bear package is menu driven, which is not my thing. I had to try and do that in R via other packages. The aim is to estimate if the treatments are bioequivalent and to estimate intra subject variation.

### Data

This seems like a funny way to create data, but the presentation contains the data in these four lines. The code after that, putting it in a data.frame, is mine.
T1 <- c(28.39,33.36,24.29,28.61,59.49,42.30)
T2 <- c(49.42,36.78,34.81,45.54,28.23,25.71)
R1 <- c(39.86,32.75,34.97,45.44,27.87,24.26)
R2 <- c(35.44,33.40,24.65,31.77,65.29,37.01)
datain <- data.frame(
aval=c(T1,R2,R1,T2),
subjid=factor(c(
rep(c(1,4,6,7,9,12),2),
rep(c(2,3,5,8,10,11),2))),
seqa=rep(c(‘TR’,’RT’),each=12),
trta=rep(c(‘T’,’R’,’R’,’T’),each=6),
aperiod=factor(rep(c(1,2,1,2),each=6))
)
datain <- datain[order(datain$subjid,datain$aperiod),]
datain$logAval <- log(datain$aval)

### Anova

With these balanced data Anova gives no problems
A1 <- aov(logAval ~  aperiod +seqa+ trta + Error(subjid),
data=datain)
summary(A1)

Error: subjid
Df Sum Sq Mean Sq F value Pr(>F)
seqa       1 0.0023  0.0023   0.014  0.907
Residuals 10 1.5943  0.1594

Error: Within
Df  Sum Sq  Mean Sq F value Pr(>F)
aperiod    1 0.02050 0.020501   3.784 0.0804 .
trta       1 0.00040 0.000397   0.073 0.7921
Residuals 10 0.05417 0.005417
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ‘ 1

#### Extraction of parameters of interest.

The degrees of freedom are calculated outside the anova. This particular bit should be correct even with unbalanced data, but that is untested.
wi <- summary(A1)$Error: Within[] countn <- rowSums(table(datain$seqa,datain$subjid)>0) sn <- sqrt(1/(2*countn) + 1/(2*countn)) SE <- sqrt(wi$Mean Sq[rownames(wi)=='Residuals'])*sn
td <- qt(.05,wi$Df[rownames(wi)=='Residuals']) cat(‘Point estimate (%)’,100*exp(coef(A1)$Within[‘trtaT’]),’\n’,
‘90% interval (%)  ‘  ,
100*exp(coef(A1)$Within[‘trtaT’]+td*SE),’to ‘, 100*exp(coef(A1)$Within[‘trtaT’]-td*SE),
‘\n’,
‘CVIntra (%)       ‘,
100*sqrt(exp(wi$Mean Sq[rownames(wi)==’Residuals’])-1), ‘\n’) Point estimate (%) 100.8168 90% interval (%) 95.47312 to 106.4596 CVIntra (%) 7.370138 ### nlme nlme gave no problems. the parameters are the same. library(nlme) L1 <- lme(logAval ~ aperiod + seqa + trta, random=~1|subjid, data=datain) summary(L1) Linear mixed-effects model fit by REML Data: datain AIC BIC logLik 6.767892 12.74229 2.616054 Random effects: Formula: ~1 | subjid (Intercept) Residual StdDev: 0.2775045 0.07360159 Fixed effects: logAval ~ aperiod + seqa + trta Value Std.Error DF t-value p-value (Intercept) 3.510257 0.11720776 10 29.949011 0.0000 aperiod2 0.058454 0.03004772 10 1.945360 0.0804 seqaTR 0.019577 0.16301059 10 0.120097 0.9068 trtaT 0.008135 0.03004772 10 0.270732 0.7921 Correlation: (Intr) aperd2 seqaTR aperiod2 -0.128 seqaTR -0.695 0.000 trtaT -0.128 0.000 0.000 Standardized Within-Group Residuals: Min Q1 Med Q3 Max -1.21397837 -0.41862241 -0.05987476 0.37505780 1.30243785 Number of Observations: 24 Number of Groups: 12 #### Extraction of parameters of interest One noticeable things here is that VarCorr() resulted in a character variable, so this is transformed by as.numerical. coL1 <- fixef(L1) SEL1 <- sqrt(vcov(L1)['trtaT','trtaT']) dfL1 <- L1$fixDF$terms[names(L1$fixDF$terms)=='trta'] tdL1 <- qt(.05,dfL1) cat(‘Point estimate (%)’,100*exp(coL1[‘trtaT’]),’\n’, ‘90% interval (%) ‘ , 100*exp(coL1[‘trtaT’]+tdL1*SEL1),’to ‘, 100*exp(coL1[‘trtaT’]-tdL1*SEL1), ‘\n’, ‘CVIntra (%) ‘, 100*sqrt(exp(as.numeric(VarCorr(L1)[‘Residual’,’Variance’]))-1), ‘\n’) Point estimate (%) 100.8168 90% interval (%) 95.47312 to 106.4596 CVIntra (%) 7.370138 ### MCMCglmm Definitely not the standard or expected calculation, but the answers are really close to the expected values. M1 <- MCMCglmm(logAval ~ aperiod + seqa + trta, random= ~ subjid , data=datain,family=’gaussian’,nitt=500000,thin=20, burnin=50000,verbose=FALSE) summary(M1) Iterations = 50001:499981 Thinning interval = 20 Sample size = 22500 DIC: -41.6716 G-structure: ~subjid post.mean l-95% CI u-95% CI eff.samp subjid 0.09517 0.02514 0.1989 22500 R-structure: ~units post.mean l-95% CI u-95% CI eff.samp units 0.006832 0.00187 0.01401 22500 Location effects: logAval ~ aperiod + seqa + trta post.mean l-95% CI u-95% CI eff.samp pMCMC (Intercept) 3.510768 3.245017 3.764446 22518 <4e-05 *** aperiod2 0.058452 -0.008879 0.125118 22500 0.0838 . seqaTR 0.018367 -0.326948 0.397998 22500 0.9148 trtaT 0.007888 -0.056976 0.077410 22500 0.7985 Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ‘ 1 #### Extraction of parameters of interest Fairly straightforward, it is all in the samples. Notice the CV Intra is larger than expected, this was observed before in previous REML exercises. By and large, getting decent (point) estimates for variances is a major weakness for MCMCglmm. M1est <- 100*exp(quantile(M1$Sol[,'trtaT'],c(.05,.5,.95)))
cat(‘Point estimate (%)’,M1est,’\n’,
‘90% interval (%)  ‘  ,
M1est,’to ‘,M1est,
‘\n’,
‘CVIntra (%)       ‘,
100*sqrt(exp(mean(M1\$VCV[,’units’]))-1),
‘\n’)

Point estimate (%) 100.7712
90% interval (%)   95.45551 to  106.4557
CVIntra (%)        8.279958

### lme4

lme4 was non-cooperating. I expected to use mcmcsamp(), but that function has disappeared in my version of lme4. Since I was a minor version behind on R I updated, which lead to updating Eclipse to from Juno to Kepler, which lead to updating Java while I was at it, but mcmcsamp() was gone. As a note, I had an old version of R/ & lme4 sitting around, the values out of mcmcsamp were quite different from expected. I am  using version 0.999999-2 for these calculations.
library(lme4)
R1 <- lmer(logAval ~ aperiod + seqa + trta + (1|subjid),
data=datain)
summary(R1)
Linear mixed model fit by REML [‘lmerMod’]
Formula: logAval ~ aperiod + seqa + trta + (1 | subjid)
Data: datain

REML criterion at convergence: -5.2321

Random effects:
Groups   Name        Variance Std.Dev.
subjid   (Intercept) 0.077009 0.2775
Residual             0.005417 0.0736
Number of obs: 24, groups: subjid, 12

Fixed effects:
Estimate Std. Error t value
(Intercept) 3.510257   0.117208  29.949
aperiod2    0.058454   0.030048   1.945
seqaTR      0.019577   0.163011   0.120
trtaT       0.008135   0.030048   0.271

Correlation of Fixed Effects:
(Intr) aperd2 seqaTR
aperiod2 -0.128
seqaTR   -0.695  0.000
trtaT    -0.128  0.000  0.000

#### Extraction of parameters of interest

The easy way out as detailed in http://glmm.wikidot.com/faq would be to impute the df value from another program, such as aov(). This would mean 10 df, and all the values would be equal to the presentation. I this blog I want to try something else. I wonder of this combination of simulate() and refit() is legit? The 90% interval is slightly narrower than the one in the presentation.
simeff <- function(m0) {
s <- simulate(m0)
fixef(refit(m0,s))[‘trtaT’]
}
ff <- replicate(1000,simeff(R1))
R1est <- 100*exp(quantile(ff,c(.05,.5,.95)))
cat(‘Point estimate (%)’,R1est,’\n’,
‘90% interval (%)  ‘  ,
R1est,’to ‘,R1est,
‘\n’,
‘CVIntra (%)       ‘,
100*sqrt(exp(sigma(R1)^2)-1),
‘\n’)
Point estimate (%) 100.7082
90% interval (%)   95.75779 to  105.7563
CVIntra (%)        7.370137