similar to: interaction terms in regression analysis

Displaying 20 results from an estimated 200 matches similar to: "interaction terms in regression analysis"

2006 Jun 15
1
Repost: Estimation when interaction is present: How do I get get the parameters from nlme?
Gday, This is a repost since I only had one direct reply and I remain mystified- This may be stupidity on my part but it may not be so simple. In brief, my problem is I'm not sure how to extract parameter values/effect sizes from a nonlinear regression model with a significant interaction term. My data sets are dose response curves (force and dose) for muscle that also have two
2011 Jul 29
2
Multifactor boxplots
Dear All I would like to produce interaction boxplots and this seems to work: par(mfrow=c(2,2)) A=sample(rnorm(50,50,10)) B=sample(rnorm(50,100,10)) Test=merge(A,B,by=0)#by=0 where 0 is the row.names TreatA=(gl(2,50,100,labels=c("High","Low"))) TreatB=rep(gl(2,25,50,labels=c("High","Low")),2) Newdata=data.frame(TreatA,TreatB,Test)
2005 Sep 04
1
specification for glmmPQL
Hello All, I have a question regarding how glmmPQL should be specified. Which of these two is correct? summary(fm.3 <- glmmPQL(cbind(response, 100 - response) ~ expt, data = data.1, random = ~ 1 | subject, family = binomial)) summary(fm.4 <- glmmPQL(response ~ expt, data = data.2, random = ~ 1 | subject, family =
2010 Jan 22
1
Estimate Slope from Boltzmann Model (package: DRC)
Dear R Community, I am using the package DRC ( to fit a boltzman model to my data. I can fit the model and extract the lower limit, upper limit, and ED50 (aka V50), but I cannot figure out how to get the slope of the curve at ED50. Is there a simple way to do this? I've searched the mailing list and looked through the package documentation, but could not find anything. I am new to r, and
2008 Jan 15
1
how to fit model to split data and get residual plots
I have a data set with the following structure (with many more obs.): var1 expt day diameter 1 1 2 0.5 1 1 3 0.9 1 1 4 1.3 1 1 5 1.7 1 2 2 0.3 1 2 3 0.5 1 2 4 0.9 1 2 5 1.6 2 1 2 0.7 2 1 3 1.2 2 1 4 1.6 2 1 5 2.3 2 2 2 0.4 2 2 3 0.8 2 2 4 1.6 2 2 5 3.2 I can get separate regression analysis for each level of var1 and expt with the command: by(data3, data3$var1:data3$expt, function(x)
2005 Nov 14
0
Trouble with aovlist and Tukey test
I am having what I think is a strange problem with applying TukeyHSD to an aov fit with error strata. TukeyHSD is supposed to take "A fitted model object, usually an 'aov' fit." aov (with error strata) is supposed to generate an object of type aovlist, which is a list of objects of type aov. But I can't seem to feed components of my aovlist to TukeyHSD. I guess I
2012 May 04
0
oddsratio and some basic help on epitools
Here is a working snippet. library(epitools) mat <- matrix(c(10,15,60,25,98, 12,10,70,28,14, 9,11,68,10,12 ,8,13,20,11,58) ,ncol=2) colnames(mat) <- c("treatmentA","treatmentB") row.names(mat) <- paste("Cond",rep(1:10,1)) dimnames(mat) <- list("Condition" = row.names(mat), "instrument" = colnames(mat)) > mat instrument
2012 May 04
0
oddsratio epitool and chi-square
Here is a working snippet. library(epitools) mat <- matrix(c(10,15,60,25,98, 12,10,70,28,14, 9,11,68,10,12 ,8,13,20,11,58) ,ncol=2) colnames(mat) <- c("treatmentA","treatmentB") row.names(mat) <- paste("Cond",rep(1:10,1)) dimnames(mat) <- list("Condition" = row.names(mat), "instrument" = colnames(mat)) > mat instrument
2012 May 04
0
epitools question
Here is a working snippet. library(epitools) mat <- matrix(c(10,15,60,25,98, 12,10,70,28,14, 9,11,68,10,12 ,8,13,20,11,58) ,ncol=2) colnames(mat) <- c("treatmentA","treatmentB") row.names(mat) <- paste("Cond",rep(1:10,1)) dimnames(mat) <- list("Condition" = row.names(mat), "instrument" = colnames(mat)) > mat instrument
2010 Aug 12
0
DRC: Effective doses versus Predicted values
Hi! I want to use the DRC package in order to calculate the IC50 value of an enzyme inhibition assay. The problem is that the estimated ED50, is always out of the fitted curve. In the example below, I had a ED50 value of 2.2896, But when I predict the response level for this concentration I get a value of 45.71 instead of the expected value of 50. This is my data: #Dose unit is concentration
2008 Jul 03
3
apply with a division
Hi, I'd like to normalize a dataset by dividing each row by the first row. Very simple, right? I tried this: > expt.fluor X1 X2 X3 1 124 120 134 2 165 163 174 3 52 51 43 4 179 171 166 5 239 238 235 >first.row <- expt.fluor[1,] > normed <- apply(expt.fluor, 1, function(r) {r / first.row}) >normed [[1]] X1 X2 X3 1 1 1 1 [[2]] X1 X2 X3 1
2010 Jul 12
1
ed50
I am using semiparametric Model  library(mgcv) sm1=gam(y~x1+s(x2),family=binomial, f) How should I  find out standard error for ed50 for the above model ED50 =( -sm1$coef[1]-f(x2)) / sm1$coef [2]   f(x2) is estimated value for non parametric term.   Thanks [[alternative HTML version deleted]]
2012 Jan 03
1
ED50 calculation in drc package
Hi, I am trying to use drc package to calculate IC50 value. The ED50 calculated in some models (LL4 for example) as a response half-way between the upper and lower limit, which is the definition of the relative IC50 value. Does that mean the ED50 in drc package is IC50? How the ED function in drc package distinguish to estimate ED or IC values? Thanks a lot [[alternative HTML version
2006 Oct 18
1
conversion of LL coordenates to UTM problems (ED50-WGS84 format)
Hi R-Users, I have plotted a region whose polygon coordinates are given in shp format ED50 UTM (zone=30) ) using "readShapePoly" in library(maptools). Now I need to plot a set of points in that region (my.dataframe, with X and Y geographic coordinates), which have been read using GPS in Longitud-Latitud form (using WGS84 system), so I first need to convert these Longitud-Latitud data
2009 Oct 26
1
explalinig the output of my linear model analysis
Hi, I am new in statistics and i manage to make the linear model analysis but i have some difficulties in explaining the results. Can someone help me explalinig the output of my linear model analysis ? My data are with 2 variables habitat (e,s) and treatment (a,c,p) with multiple trials within. Thank you in advance Call: lm(formula = a$wild ~ a$habitat/a$treatment/a$trial) Residuals: Min
2009 Jul 30
3
What is the best method to produce means by categorical factors?
I am attempting to replicate some of my experience from SAS in R and assume there are best methods for using a combination of summary(), subset, and which() to produce a subset of mean values by categorical or ordinal factors. within sas I would write proc means mean data=dataset; class factor1 factor2 var variable1 variable2; RUN; producing an output with means for each variable by factor
2009 Apr 16
0
incorrect handling of NAs by na.action with lmList (package nlme) (PR#13658)
Greetings, I just found out a bug in the function lmList of the package nlme with R 2.8.1 running under windows XP 32-bits. I have a data table with various columns corresponding to continuous variables as well as treatment variables taken on several years and several sites. Here is an example : Id year treatment A treatment B variable1 variable2 variable3 1
2008 Jul 23
1
Calling LISP programs in R
I have written some programs in Common Lisp and I have been using SAS to pipe those programs to my lisp compiler in batch mode by using the %xlog and %xlst SAS commands. I wonder if there is in R a similar way to pipe commands to LISP so that all my work would be concentrated in R even when I have to call a LISP program? I have looked at the foreign library but this seems to adjust data types not
2013 Nov 23
0
Re: [R-SIG-Mac] morley object?
Hi Ann Looks like a typo - I see "moreley.tab" that should be "morley.tab". Works for me after correcting that. > filepath <- system.file("data", "moreley.tab", package="datasets") > filepath [1] "" > > filepath <- system.file("data", "morley.tab", package="datasets") > filepath [1]
2005 Jan 24
4
lme and varFunc()
Dear R users, I am currently analyzing a dataset using lme(). The model I use has the following structure: model<-lme(response~Covariate+TreatmentA+TreatmentB,random=~1|Block/Plot,method="ML") When I plot the residuals against the fitted values, I see a clear positive trend (meaning that the variance increases with the mean). I tried to solve this issue using weights=varPower(),