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:  LC_COLLATE=French_France.1252 LC_CTYPE=French_France.1252 LC_MONETARY=French_France.1252  LC_NUMERIC=C LC_TIME=French_France.1252 attached base packages:  stats graphics grDevices utils datasets methods base other attached packages:  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):  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  nloptr_1.0.4 svMisc_0.9-70 splines_3.4.4 tools_3.4.4 compiler_3.4.4 tcltk_3.4.4 [[alternative HTML version deleted]] ______________________________________________ Remail@example.com mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.