similar to: LM intercept

Hi, I am not a statistics expert, so I have this question. A linear model gives me the following summary: Call: lm(formula = N ~ N_alt) Residuals: Min 1Q Median 3Q Max -110.30 -35.80 -22.77 38.07 122.76 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 13.5177 229.0764 0.059 0.9535 N_alt 0.2832 0.1501 1.886 0.0739

I have some data with two categorises plus/minus (p53) and a particular time (Time) and the outcome is a continuous vairable (Result). I set up a maximum model. ancova <- lm(Result~Time*p53) > summary(ancova) .. Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 0.05919 0.55646 0.106 0.916 Time -0.02134 0.01785 -1.195 0.241 p53plus

A colleague asked me if there is a way to specify with a ***variable*** (say ``cflag'') whether there is an intercept in a linear model. She had in mind something like lm(y ~ x - cflag) where cflag could be 0 or 1; if it's 0 an intercept should be fitted, if it's 1 then no intercept. This doesn't work ``of course''. The cflag just gets treated as another predictor

Dear R users, I need help with translating these SAS codes into R with lme()? I have a longitudinal data with repeated measures (measurements are equally spaced in time, subjects are measured several times a year). I need to allow slope and intercept vary. SAS codes are: proc mixed data = survey method=reml; class subject var1 var3 var2 time; model score = var2 score_base var4 var5 var3

Dear all, I am trying to estimate the prediction from a fixed effects model and their confidence intervals as well. Though I do not want to include in the prediction and at the confidence intervals the intercept. For that reason I used the argument incl.non.slopes=FALSE. But either if it is TRUE or FALSE it does not have any difference and also the system does not provide any warning. I really

Dear R-help list members, I am currently trying to fit a generalized linear model using a binomial with the canonical link. The usual solution is to use the R function glm() in the package "stats". However, I run into problem when I want to fit a glm without an intercept. It is indicated that the solution is in changing the function glm.fit (also in "stats"), by specifying

When one is doing simple regression and needs to force a zero intercept ( for whatever reason. I realize it's a controversial issue ), then subtracting the means of the left hand side and the right hand side from themselves does the trick. Does anyone know if there is a similar trick when the RHS has two variables ? Thanks. -------------------------------------------------------- This is

Hi, The below very strange: # a) aov function av <- aov(Sepal.Length ~ Species, data=iris) # Error in parse(text = x) : # unexpected symbol in "Sepal(Sepal.Length+Species)Length" av <- aov(iris[, 1] ~ iris[, 5]) # summary(av) # Df Sum Sq Mean Sq F value Pr(>F) # iris[, 5] 2 63.2 31.6 119 <2e-16 *** # Residuals 147 39.0 0.3 # ---

Dear R-people ... I'm a new user. I can't get predict.lm() to produce predictions for new independent data. There are some messages in archived help about this problem, but I still don't see my error after reviewing those. I understand that the new independent data must have the same name(s) as used when the model was made. In the example below, predict.lm produces the

I keep getting repeated copies of ``R-help post acknowledgements'' in respect of a couple of postings that I made to the list this morning (my time). I only posted each posting *once*. I hope that others are not getting repeated copies of my postings .... I mean I *know* my postings are so wonderful they merit re-reading, but one does not need another copy in order to re-read! :-)

I feel bad even asking, but: Rgames> data(OrchardSprays) Rgames> model<-lm(decrease~.,data=OrchardSprays) Rgames> model Call: lm(formula = decrease ~ ., data = OrchardSprays) Coefficients: (Intercept) rowpos colpos treatmentB treatmentC 22.705 -2.784 -1.234 3.000 20.625 treatmentD treatmentE treatmentF treatmentG treatmentH

Dear Everyone in list: I am writing some codes to automate the process of fitting linear models where the names of variables of models are produced and stored in character vectors. But I have problems to pass the vectors to the lm( ) because I don't know how to strip the quotation marks automatically. Here are the codes of the example of lm( ): ## Annette Dobson (1990) "An Introduction

Hello, I have conducted an SEM in which the resultant standardized path coefficients are much higher than would be expected from the raw correlation matrix. To explore further, I stripped the model down to a simple bivariate relationship between two variables (NDVI, and species richness), where it's my understanding that the SEM's standardized path coefficient should equal the correlation

Peter Dalgaard writes (in response to a question about 2-way ANOVA with imbalance): > ... There are various > boneheaded ways in which people try to use to assign some kind of > SumSq to main effects in the presence of interaction, and they are all > wrong - although maybe not very wrong if the unbalance is slight. People keep saying this

Hi I would like to run a multi-level hierarchical logistic regression model with sampling weight? Is this possible with R? Thanks a lot, Qian Guo --------------------------------- [[alternative HTML version deleted]]

I had occasion recently to read in a one-line *.csv file that looked like: "CandidateName","NSN","Ethnicity","dob","gender" "Smith, Mary Jane",111222333,"E","2/25/1989","F" That "F" (for female) in the last field got transformed to FALSE. Apparently read.csv (and hence read.table) are inferring

Hi, I am running a simple linear model with (say) 5 independent variables. Is there a simple way of getting the variance-covariance matrix of the coeffcient estimates? None of the values of the lm() seem to provide this. Thanks in advance, Ritwik Sinha rsinha@darwin.cwru.edu Grad Student Case Western Reserve University [[alternative HTML version deleted]]

Hello, I have different results from these two softwares for a simple binomial GLM problem. >From Genmod in SAS: LogLikelihood=-4.75, coeff(intercept)=-3.59, coeff(x)=0.95 >From glm in R: LogLikelihood=-0.94, coeff(intercept)=-3.99, coeff(x)=1.36 Is there anyone tell me what I did wrong? Here are the code and results, 1) SAS Genmod: % r: # of failure % k: size of a risk set data

For reasons best known only to myself ( :-) ) I wish to create a data frame with 0 rows and 9 columns. The best I've been able to come up with is: junk <- as.data.frame(matrix(0,nrow=0,ncol=9)) Is there a sexier way? cheers, Rolf ###################################################################### Attention:\ This e-mail message is privileged and confid...{{dropped:9}}

Hi, I would like to fit my data with a 4th order polynomial. Now I have only 5 data point, I should have a polynomial that exactly pass the five point Then I would like to compute the "fitted" or "predict" value with a relatively large x dataset. How can I do it? BTW, I thought the model "prodfn" should pass by (0,0), but I just wonder why the const is