search for: fm3

Hi List, I have to compare a relationship between y and x for two locations. I found logistic regression fits both datasets well, but I am not sure how to test if relationships for both sites are significantly different. I searched the r site, however no answers exactly match the question. I used Tukey's HSD to compare two means, but the relationship in my study was not simply linear. So I

...grateful. Cheers and Thanks in Advance Mike require(MuMIn) data(Cement) # option 1, create model.selection object using dredge fm0 <- lm(y ~ ., data = Cement) print(dd <- dredge(fm0)) fm1 <- lm(formula = y ~ X1 + X2, data = Cement) fm2 <- lm(formula = y ~ X1 + X2 + X4, data = Cement) fm3 <- lm(formula = y ~ X1 + X2 + X3, data = Cement) fm4 <- lm(formula = y ~ X1 + X4, data = Cement) fm5 <- lm(formula = y ~ X1 + X3 + X4, data = Cement) # ranked with AICc by default # obviously this works model.avg(get.models(dd, delta < 4)) # option 2: the aim is to produce a model sel...

...7.5900 0.6003333 6 829.2700 0.5883333 I have fit the data using logistic, Michaelis?Menten, and linear model, they all give significance. > fm1 <- nls(Biomass~SSlogis(P, phi1, phi2, phi3), data=Ca.P.Biomass.A) > fm2 <- nls(Biomass~SSmicmen(P, phi1, phi2), data=Ca.P.Biomass.A) > fm3 <- lm(Biomass~P, data = Ca.P.Biomass.A) I hope to compare the difference among the three models, and I using anova(). As for the example here, the three models seem not have significant difference. However, I am confused by the negative df in the following ANOVA table. And my question is how...

Hi R Users, When I use package lme4 for mixed model analysis, I can't distinguish the significant and insignificant variables from all random independent variables. Here is my data and result: Data: Rice<-data.frame(Yield=c(8,7,4,9,7,6,9,8,8,8,7,5,9,9,5,7,7,8,8,8,4,8,6,4,8,8,9), Variety=rep(rep(c("A1","A2","A3"),each=3),3),

How can I get prediction intervals from a mixed-effects model? Consider the following example: library(nlme) fm3 <- lme(distance ~ age*Sex, data = Orthodont, random = ~ 1) df3.1 <- with(Orthodont, data.frame(age=seq(5, 20, 5), Subject=rep(Subject[1], 4), Sex=rep(Sex[1], 4))) predict(fm3, df3.1, interval='prediction') # M01 M01 M01 M01 #...

...ch extension/context FM2 at default creating local channel [Jan 25 11:34:32] WARNING[23725]: app_followme.c:863 findmeexec: Unable to allocate a channel for Local/FM2/2000 at longdistance cause: Unknown [Jan 25 11:34:32] NOTICE[23725]: chan_local.c:571 local_alloc: No such extension/context FM3 at default creating local channel [Jan 25 11:34:32] WARNING[23725]: app_followme.c:863 findmeexec: Unable to allocate a channel for Local/FM3/2000 at longdistance cause: Unknown (NOTE: This log was taken just prior to making the changes indicated below in my Followme.conf file. But the chang...

...plot(women, xlab = "Height (in)", ylab = "Weight (lb)") lines(ht2,ph2) 3) height <- women$height # 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 weight <- women$weight # 115 117 120 123 126 129 132 135 139 142 146 150 154 159 164 fb3 <- bs(height, df = 5) fm3 <- lm(weight ~ fb3) ht3 <- seq(57, 73, len = 200) ph3 <- predict(fm3, data.frame(height=ht3)) # Error message about newdata. Why ? plot(women, xlab = "Height (in)", ylab = "Weight (lb)") lines(ht3,ph3) # no line Thanks for the reason of this message. Alex Randria...

Hello everyone I am currently trying to conduct analysis of my graduate thesis data using a mixed effects model and I have reached an impass. When I try to conduct a multiple comparison, I get an error (See below): > fm3<- lme(abovegroundbiomass.m.2~medium*amelioration*fertilizer*treatment, random=~1|block/medium/amelioration/fertilizer) > tukeytest<-glht(fm3, linfct=mcp(treatment="Tukey")) Error in contrMat(table(mf[[nm]]), type = types[pm]) : less than two groups I wonder if anyone could...

This is an example from chapter 11 of the 6th edition of Devore's engineering statistics text. It happens to be a balanced data set in two factors but the calculations will also work for unbalanced data. I create a factor called 'cell' from the text representation of the Variety level and the Density level using '/' as the separator character. The coefficients for the linear

...ssumes the data is nominal, not ordinal, using either glm() or loglm(). library(MASS) options(contrasts=c("contr.sum", "contr.poly")) X <- as.integer(gl(4, 4, 16)) - 1 Y <- as.integer(gl(4, 1, 16)) - 1 data.2 <- data.frame(Freq, X = factor(X), Y = factor(Y)) summary(fm3 <- glm(Freq ~ X + Y, data = data.2, family = poisson())) dummy.coef(fm3) fm4 <- loglm(Freq ~ X + Y, data = data.2, param = T, fit = T) fm4; fm4$param My question is this: can glm() or some other function be used in the manner Agresti employed for ordinal count data? Thank you, ANDREW And...

...ations in which these models are applied. The preferred approach to model-building in R/S-PLUS is to build the model sequentially. I would use fm1 <- lmer(DeathDay ~ Treatment + (1|Clutch:Cup) + (1|Clutch)) fm2 <- lmer(DeathDay ~ 1 + (1|Clutch:Cup) + (1|Clutch)) # equivalent to your model7 fm3 <- lmer(DeathDay ~ 1 + (1|Clutch)) and then compare fm2 and fm3, say with anova(fm2, fm3). As described below the p-value from the likelihood ratio test in this case is not exact because the hypothesis you are testing is on the boundary of the parameter space. However, the value returned wil...

...n ordinary data frame rather than a grouped data object): Assay2 <- as.data.frame(Assay) fm2<-lme(logDens~sample*dilut, data=Assay2, random=list(Block = pdBlocked(list(pdIdent(~1), pdIdent(~sample-1),pdIdent(~dilut-1))) )) Now, block has only 2 levels so I prefer to treat it as fixed: fm3<-lme(logDens~Block+sample*dilut, data=Assay2, random=list(Block = pdBlocked(list(pdIdent(~sample-1),pdIdent(~dilut-1))) )) This works fine until I try a summary() or anova(), e.g. > anova(fm3) numDF denDF F-value p-value (Intercept) 1 29 824.2612 <.0001 Block...

Hello, I am using the party module to estimate the relationship between the probability of being a student and number of siblings (alive). However, I need to include a number of relevant covariates. My code is below: fm3 <- mob(Student ~ age + alive + sex2 + cwa + cha + cym | Religion+Servant + Literacy, control = ctrl, data = samp2, model = glinearModel, family =binomial()) plot(fm3, tp_args = list(which = "alive"),tnex = 2, type = "simple" ) Which works fine. However, the plots generated d...

Hi, this site: http://www.uow.edu.au/~andrewm/linux/ext3/ says "Forbidden" when I try to access .... , where can I download the ext3 patch for 2.4.15 ? Thank you Nikolai

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...mpty model: e.model <- lme(y~1,random=~1|G) With one explanatory variable: fm1.lme <- lme(y~x1,random = ~1|G) With two exp. variables, assuming that there are only maineffects for my variables: fm2.lme <- lme(y ~ x1 + x2+ ... + x7, random = ~1|G) The same, adding an interaction effect: fm3.lme <- lme(y~x1*x2,random=~1|G) This is how far i got. Now i would like to add the z-variable into the model. How do i do this ? Yours sincerly Felix Eschenburg

...ble Model 1: stack.loss ~ Air.Flow + Water.Temp Model 2: stack.loss ~ Air.Flow + Water.Temp + Acid.Conc. Res.Df Res.Sum Sq Df Sum Sq F value Pr(>F) 1 18 188.795 2 17 178.830 1 9.965 0.9473 0.3440 > fm2 <- update(fm1, . ~ . - Water.Temp) > fm3 <- update(fm2, . ~ . - Air.Flow) > anova(fm2, fm1) Analysis of Variance Table Model 1: stack.loss ~ Air.Flow Model 2: stack.loss ~ Air.Flow + Water.Temp Res.Df Res.Sum Sq Df Sum Sq F value Pr(>F) 1 19 319.12 2 18 188.80 1 130.32 12.425 0.00...

Greetings, 1) Is there a nice way of extracting the variance estimates from an lme fit? They don't seem to be part of the lme object. 2) In a series of simulations, I am finding that with ML fitting one of my random effect variances is sometimes being estimated as essentially zero with massive CI instead of the finite value it should have, whilst using REML I get the expected value. I guess

...ance of the variance components I did: fm2 <- aov(fruto~pop+spp+Error(pop:spp), data=p1) Manually, the variance component for the interaction is negative, therefore I turned to lmer() to try to get a better estimate as well as confidence intervals. After invoking the Matrix library, I typed: fm3 <- lmer(fruto~(1|pop) + (1|spp) + (1|pop:spp), data=p1) During calculations I get an error which indicates that the variance components for pop and pop:spp are almost zero. The summary output presents variance components that do not resemble the manually estimated values (eg. sigma_pop = 365...

...andom variation. The egsingle data set in the mlmRev package has repeated measures on students and those students are nested in schools. We can fit a model with random intercepts and slopes for students and students within schools. The first model is fit using lme and the second is fit using lmer. fm3 <- lme(math ~ year, random=~year|schoolid/childid, egsingle) fm4 <- lmer(math ~ year +(year|schoolid:childid) +(year|schoolid), egsingle, control=list(gradient = FALSE, niterEM = 0)) Both result in parameter estimates that are exactly the same. The newest release of lme4 allows for an even e...