similar to: Why can't Anove (car package) see the data?

If one estimates a model using multinom, is it possible to perform the omnibus LR test ( the analogue to omnibus F in linear models ) using the output from multinom ? The residual deviance is there but I was hoping I could somehow pull out the deviance based on just using an intercept ? Sample code is below from the CAR book but I wasn't sure how to do it based on that example. Thanks

Hi, P-values in Durbin-Watson test obtained through the use of functions available in packages "lmtest" and "car" are different. The difference is quite significant. function "dwtest" in "lmtest" is much faster than "burbinwatson" in "car". Actually, you can take a nap while the latter trying to calculated Durbin-Watson test. My question

I have design with two repeated-measures factor, and no grouping factor. I can analyze the dataset successfully in other software, including my legacy DOS version BMDP, and R's 'aov' function. I would like to use 'Anova' in 'car' in order to obtain the sphericity tests and the H-F corrected p-values. I do not believe the data are truly deficient in rank. I

I am getting an error message from scatterplot: > library(car) > scatterplot(Prestige$income~Prestige$type) Error in Summary.factor(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, : range not meaningful for factors In addition: Warning message: In Ops.factor(x[floor(d)], x[ceiling(d)]) : + not meaningful for factors > The command does output the kind of graph that I want (boxplots).

Greetings all The help file for GAMM in mgcv indicates that the log likelihood for a GAMM reported using summary(my.gamm$lme) (as an example) is not correct. However, in a past R-help post (included below), there is some indication that the likelihood ratio test in anova.lme(mygamm$lme, mygamm1$lme) is valid. How can I tell if anova.lme results are meaningful (are AIC, BIC, and logLik

Dear Rlist, Using an example in box.cox.powers() help, I have the following warning message. example: library(car) >attach(Prestige) > box.cox.powers(income) Box-Cox Transformation to Normality Est.Power Std.Err. Wald(Power=0) Wald(Power=1) 0.1793 0.1108 1.6179 -7.4062 L.R. test, power = 0: 2.7103 df = 1 p = 0.0997 L.R. test, power = 1: 47.261 df = 1 p = 0

Dear list users How is it possible to visualise both a linear trend line and a quadratic trend line on a plot of two variables? Here my almost working exsample. data(Duncan) attach(Duncan) plot(prestige ~ income) abline(lm(prestige ~ income), col=2, lwd=2) Now I would like to add yet another trend line, but this time a quadratic one. So I have two trend lines. One linear trend line

Dear R users, Suppose I have two different response variables y1, y2 that I regress separately on the same explanatory variable, x; sample sizes are n1=n2. Is it legitimate to compare the regression slopes (equal variances assumed) by using lm(y~x*FACTOR), where FACTOR gets "y1" if y1 is the response, and "y2" if y2 is the response? The problem I see here is that the

Dear all, I know that my question is somewhat special but I tried several times to solve the problems on my own but I am unfortunately not able to compute the following test statistic using the quantreg package. Well, here we go, I appreciate every little comment or help as I really do not know how to tell R what I want it to do^^ My situation is as follows: I have a data set containing a

I am estimating a multinomial model with two quantitative predictors, X1 and X2, and 3 responses. The responses are called neutral, positive and negative with neutral being the baseline. There are actually many models being estimated because I estimate the model over time and also for various parameter sets but that's not important. When I estimate a model, since neutral is the baseline

Dear expeRts, there is obviously a general trend to use model comparisons, LRT and AIC instead of Wald-test-based significance, at least in the R community. I personally like this approach. And, when using LME's, it seems to be the preferred way (concluded from postings of Brian Ripley and Douglas Bates' article in R-News 5(2005)1), esp. because of problems with the d.f. approximation.

Hello R-Users,   I have one binary dependent variable and a set of independent variables (glm(formula,…,family=”binomial”) ) and I am using the function stepAIC (“MASS”) for choosing an optimal model. However I am not sure if stepAIC considers significance properties like Likelihood ratio test and Wald test (see example below).     > y <- rbinom(30,1,0.4) > x1 <- rnorm(30) > x2

Dear R users, Which test is most appropriate in glm when the family is Gamma? In the help page of anova.glm, I found the following ?For models with known dispersion (e.g., binomial and Poisson fits) the chi-squared test is most appropriate, and for those with dispersion estimated by moments (e.g., gaussian, quasibinomial and quasipoisson fits) the F test is most appropriate.? My questions :

Hello All, Am having some trouble computing Type III SS in a Cox Regression using either drop1 or Anova from the car package. Am hoping that people will take a look to see if they can tell what's going on. Here is my R code: cox3grp <- subset(survData, Treatment %in% c("DC", "DA", "DO"), c("PTNO", "Treatment", "PFS_CENSORED",

I have two models, controlled by dummy variables to see if the models can be combined into one model with similar intercepts and slopes. Has anyone tried to conduct this type of test in R. I am utilizing the econometric idea of hypothesis testing through the hypothesis of coincidence. I have tried to run an anova with test of Chisq, but I am not sure what the results are telling. In addition, I

I have a doubt about the calculation of tests for fixed effects in mixed-effects models. I have read that, except in well-balanced designs, the F statistic that is usually calculated for ANOVA tables may be far from being distributed as an exact F distribution, and that's the reason why the anova method on "mer" objects (calculated by lmer) do not calculate the denominator df nor a

Hi all, It seems that I can get White's (HC3) test using MASS. The syntax I used for the particular problem is anova(scireg3, white.adjust="hc3") where scireg3 is an object from the lm function. But, the anova summary table is all I get. I don't get the new estimates or standard errors correcting for heteroskedasticity. Is there a way to get that information? Thanks

Dear all, I am trying to compare the estimated coefficients of a quantile regression model between two different samples. It is a Wald test, but I cannot find one way to do that in R.The samples are collected conditional on a specific characteristic and I would like to test whether such characteristic indeed affect the estimators. The problem in the test anova.rq is that the response variable

Dear R list members, I'm encountering a problem with expand.model.frame(): Suppose that I define the following simple function (meant just to illustrate the problem): > fun <- function(model){ + expand.model.frame(model, all.vars(formula(model))) + } > and I have the following model, created with an explicit data argument: > mod Call:

hello, I work since a few time on R and i wanted to know how to obtain the Wald chi square value when you make a binary logistic regression. In fact, i have the z value and the signification but is there a script to see what is the value of Wald chi square. You can see my model below, Best regards, S??verine Erhel [Previously saved workspace restored] > m3 = glm(reponse2 ~ form +