similar to: null hypothesis for two-way anova

Hi! I would like to do Cox regression using the available routines in the survival package BUT I want to use an arbitrary link function, i.e. want to use the model h(t)=h_0(t)r(beta'z) with arbitrary function r, instead of h(t)=h_0(t)exp(beta'z) Grateful for any comment on this, Dragi ----------------------------------------------------------- Dragi Anevski, PhD Mathematical

Hi, In the following results I interpret exp(coef) as the factor that multiplies the base hazard rate if the corresponding variable is TRUE. For example, when the bucket is ks008 and fidelity <= 3, then the rate, compared to the base rate h_0(t), is h(t) = 0.200 h_0(t). My question is then, to what case does the base hazard rate correspond to? I would expect the reference to be the first

My question is, how to compute hazard function(H(t)) after building the coxph model. I even aware of the terminology that differs from hazard function(H(t)) and the hazard rate(h(t)). Here onward I wish to calculate both. Here what I have done in two different methods; ##########################################################################################

Dear All, I am currently working with the coxph function within the package survival. I have the model h_ij = h_0(t) exp(b1x1 + b2x2) where the indicator variables are as follows: x1 x2 VPS 0 0 LTG 1 0 TPM 0 1 [[alternative HTML version deleted]]

Hello again, I m trying to use timereg package as you suggested (R2.7.1 on XP Pro). here is my script based on the example from timereg for a fine & gray model in which relt = time to event, rels = status 0/1/2 2=competing, 1=event of interest, 0=censored random = covariate I want to test library(timereg) rel<-read.csv("relapse2.csv", header = TRUE, sep = ",",

I have a question and it's only relation to R is that I probably need R after I understand what to do. Both models are delta y_t = Beta + epslion and suppose I have a null hypothesis and alternative hypothesis H_0 : delta y_t = zero + epsilon epsilon is normal ( 0, sigmazero^2 ) H_1 delta y_t = beta + epsilon epsilon is normal ( sigmabeta^2 )

Hi, I'm using the ks.test() on two vectors. I looked up the reference and also coded up a version of the two sample Smirnov test. My question is that how can I decide from the output of R that the two vectors x & y come from the same distribution? Am I correct in assuming that smaller D values indicate that they come from the same distribution? In addition how can I use the p value that

Dear R-helpers, I was testing the truncgof CRAN package, found something that looked like a bug, and did my job: contacted the maintainer. But he did not reply, so I am resending my query here. I installed package truncgof and run the example for function ad.test. I got the following output: set.seed(123) treshold <- 10 xc <- rlnorm(100, 2, 2) # complete sample xt <- xc[xc >=

I don't know from the nortest package, but it should ***always*** be the case that you test hypotheses H_0: The data have a normal distribution. vs. H_a: The data do not have a normal distribution. So if you get a p-value < 0.05 you can say that ***there is evidence*** (at the 0.05 significance level) that the data are not from a normal distribution. If the nortest package does

Hello, I ran the coxph model and everything worked fine. When I extract the output from the basehaz(y) function and was wondering if that baseline is cumulative or not. I then do: F(t,t+1)=1-exp(h_0(t+1) exp(coeff(1)*covariate(1)+...)) Does this give me the probability of death from t to t+1 (in which case the baseline is not cumulative) or the probability of death before t+1 (in which case the

> Subject: [R] LRT in arima models > Date: Mon, 17 Feb 2003 11:53:04 +0100 > From: "vito muggeo" <vito.muggeo at giustizia.it> > To: <r-help at stat.math.ethz.ch> > > Dear all, > > For some reason I'm evaluating the size of the LRT testing for the effect of > some explanatory variable in arima models. > I performed three different simulations

I'd like to compare tests based on the mixed model representation of additive models, testing among others y=f(x1)+f(x2) vs y=f(x1)+f(x2)+f(x1,x2) (testing for additivity) In mixed model representation, where X represents the unpenalized part of the spline functions and Z the "wiggly" parts, this would be: y=X%*%beta+ Z_1%*%b_1+ Z_2%*%b_2 vs y=X%*%beta+ Z_1%*%b_1+ Z_2%*%b_2 + Z_12

Full_Name: Yves Rosseel Version: 2.2.1 OS: Submission from: (NULL) (157.193.116.152) The code below sketches a possible implementation of a function 'mlh.mlm' which I think would be a good complement to the 'anova.mlm' function in the stats package. It tests a single linear hypothesis of the form H_0: LBT'= 0 where B is the matrix of regression coefficients; L is a matrix

Dear Carlos, Duncan and everyone You may have already sorted the matter by now, but since I have not seen anything posted since Duncan's reply, here I go. I apologize in advance for the spam, if it turns out I've missed some post. I think the test and the implementation of the truncgof package are just fine. I've done Carlos' experiment (repeatedly generating samples and testing

Dear R help list, I am modeling some survival data with coxph and survreg (dist='weibull') using package survival. I have 2 problems: 1) I do not understand how to interpret the regression coefficients in the survreg output and it is not clear, for me, from ?survreg.objects how to. Here is an example of the codes that points out my problem: - data is stc1 - the factor is dichotomous

1. The weibull is the only distribution that can be written in both a proportional hazazrds for and an accelerated failure time form. Survreg uses the latter. In an ACF model, we model the time to failure. Positive coefficients are good (longer time to death). In a PH model, we model the death rate. Positive coefficients are bad (higher death rate). You are not the first to be confused

Thanks for sharing the questions and responses! Is it possible to appreciate how much the coefficients matter in one or the other model? Say, using Biau's example, using coxph, as.factor(grade2 == "high")TRUE gives hazard ratio 1.27 (rounded). As clinician I can grasp this HR as 27% relative increase. I can relate with other published results. With survreg the Weibull model gives a

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

Dear All I am trying to replicate a numerical application (not computed on R) from an article. Using, ks.test() I computed the exact D value shown in the article but the p-values I obtain are quite different from the one shown in the article. The tests are performed on a sample of 37 values (please see "[0] DATA" below) for truncated Exponential, Pareto and truncated LogNormal

Dear All, I am using wilcox.test to test two samples, data_a and data_b, earch sample has 3 replicates, suppose data_a and data_b are 20*3 matrix. Then I used the following to test the null hypothesis (they are from same distribution.): wilcox.test(x=data_a, y=data_b, alternative="g") I got pvalue = 1.90806170863311e-09. When I switched data_a and data_b by doing the following: