Dear list,
this seemed to me like a very trivial question, but finally I haven't found
any similar postings with suitable solutions on the net ...
Basically, instead of regressing two simple series of measures 'a' and 'b'
(like b ~ a), I would like to use independent replicate measurements for
each variable at each level (ie, instead of having just one 'a' and one 'b'
I have independent replicates for all measures of 'a' and 'b', 'a1' could
be as well compared to 'b1' as to 'b2' etc.)
In analogy one could think of a procedure claiming to act and increase a
given output value by eg 30% (compared to not emplying this procedure).
Now I have indepedent repeated measures (since the measures themselves are
considered not very precise) for a (large) number of individuals with and
without the treatment.
Basically, I want to test the hypthesis that applying the procedure
increases values in a linear way by a given factor, thus test the
parameters of a linear regression (eg slope=1.3, offset may be different to
0). In extension to this, how could I make a confidence-interval for the
estimated slop (due to the treatment) to check if the claimed value is
indeed inside ?
# Here some toy data, my real data are much larger and might ressemble
somehow to this.
# Lines are for subjects and columns for 2 groups and repeat-measurements.
# in this case I introduce a toy-factor of 1.3 to the 2nd part of my data
(in the real data such a factor is just a hypothesis), which I would like
to investigate/confirm
dat <- matrix(rep(1:12,7)+rnorm(84),nc=7)
dat[,4:7] <- 1.3*dat[,4:7]+runif(48)
# some individual measures may be missing :
dat[2:3,4] <- NA
colnames(dat) <- paste(rep(c("a","b"),3:4),c(1:3,1:4),sep="")
# In analogy to the ample documentation of lm() :
datMean <- cbind(aM=rowMeans(dat[,1:3]),bM=rowMeans(dat[,4:7]))
(lmMean <- lm(bM ~ aM,data=as.data.frame(datMean)))
# I suppose the estimated parameters (intercet & slope) may be correct but
sice the degrees of freedom are not made of means I am convinced they are
incorrect and thus any statistics using them will be so, too ...
df.residual(lmMean)
summary(lmMean)
# I also thought about a workaround reorganizing the data into a 'simple'
two-column setup using somthing like stack() and allowing b ~ a, but again,
I suppose the degrees of freedom won't be correct neither.
# 1) should I simply correct the degrees of freedom in my lm-object, would
this be the correct number of degrees of freedom
lmMean$df.residual <- nrow(dat)*5-2
# then I suppose I would need to change the standard errors, I'm shur what
is the best way to do so
# or 2) is there a package allowing to do these steps, thus returning
correct DF, Std Err and Pr(>|t) ?
Thanks in advance,
Wolfgang Raffelsberger
for completeness :
sessionInfo()
> sessionInfo()
R version 3.4.4 (2018-03-15)
Platform: x86_64-w64-mingw32/x64 (64-bit)
Running under: Windows 7 x64 (build 7601) Service Pack 1
Matrix products: default
locale:
[1] LC_COLLATE=French_France.1252 LC_CTYPE=French_France.1252
LC_MONETARY=French_France.1252
[4] LC_NUMERIC=C LC_TIME=French_France.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] limma_3.34.9 lme4_1.1-15 Matrix_1.2-12 TinnRcom_1.0.20
formatR_1.4 svSocket_0.9-57
loaded via a namespace (and not attached):
[1] Rcpp_0.12.16 lattice_0.20-35 MASS_7.3-49 grid_3.4.4
nlme_3.1-131.1 minqa_1.2.4
[7] nloptr_1.0.4 svMisc_0.9-70 splines_3.4.4 tools_3.4.4
compiler_3.4.4 tcltk_3.4.4
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