similar to: Bootstrap CI of Slope in a Weighted Simple Linear Regression

Dear all I am intrested in making iterative estimation (thro' loop statements) of, say, linear regression model. For this purpose, I have written the following programme and that I have made use of a sample data (viz., exp.txt): ? Programme: ? # Linear regression modelling with sample data (try5.txt) # Repeated estimation through loop statement

From: Martin Schwidefsky <schwidefsky@de.ibm.com> From: Hubertus Franke <frankeh@watson.ibm.com> From: Himanshu Raj <rhim@cc.gatech.edu> [patch 2/9] Guest page hinting: unused / free pages on s390. s390 uses the milli-coded ESSA instruction to set the page state. The page state is formed by four guest page states called block usage states and three host page states called block

From: Martin Schwidefsky <schwidefsky@de.ibm.com> From: Hubertus Franke <frankeh@watson.ibm.com> From: Himanshu Raj <rhim@cc.gatech.edu> [patch 2/9] Guest page hinting: unused / free pages on s390. s390 uses the milli-coded ESSA instruction to set the page state. The page state is formed by four guest page states called block usage states and three host page states called block

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

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

Hello, I want to know if two quadratic regressions are significantly different. I was advised to make the test using step 1 bootstrapping both quadratic regressions and get their slope coefficients. (Let's call the slope coefficient *â*^1 and *â*^2) step 2 use the slope difference *â*^1-*â*^2 and bootstrap the slope coefficent step 3 find out the sampling distribution above and

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

Dear List: Below is the validation output of a fitted ordinal logistic model using the bootstrap in the rms package. My interpretation is that most of the corrected indices indicate little overfitting, however the slope seems to indicate that the model is too optimistic. Given that most of the corrected indices seem reasonable, would it be appropriate to use this model on future data if the

On 10/25/2017 4:38 AM, Ista Zahn wrote: > On Tue, Oct 24, 2017 at 3:05 PM, BooBoo <booboo at gforcecable.com> wrote: >> This has every appearance of being a bug. If it is not a bug, can someone >> tell me what I am asking for when I ask for "x[x[,2]==0,]". Thanks. > You are asking for elements of x where the second column is equal to zero. > >

This has every appearance of being a bug. If it is not a bug, can someone tell me what I am asking for when I ask for "x[x[,2]==0,]". Thanks. > #here is the toy dataset > x <- rbind(c(1,1),c(2,2),c(3,3),c(4,0),c(5,0),c(6,NA), + c(7,NA),c(8,NA),c(9,NA),c(10,NA) + ) > x [,1] [,2] [1,] 1 1 [2,] 2 2 [3,] 3 3 [4,] 4 0 [5,] 5 0

> On Oct 25, 2017, at 6:57 AM, BooBoo <booboo at gforcecable.com> wrote: > > On 10/25/2017 4:38 AM, Ista Zahn wrote: >> On Tue, Oct 24, 2017 at 3:05 PM, BooBoo <booboo at gforcecable.com> wrote: >>> This has every appearance of being a bug. If it is not a bug, can someone >>> tell me what I am asking for when I ask for "x[x[,2]==0,]". Thanks.

On Tue, Oct 24, 2017 at 3:05 PM, BooBoo <booboo at gforcecable.com> wrote: > This has every appearance of being a bug. If it is not a bug, can someone > tell me what I am asking for when I ask for "x[x[,2]==0,]". Thanks. You are asking for elements of x where the second column is equal to zero. help("==") and help("[") explain what happens when

I know this is easy, but I am stumped: > gsub("0","K","8.00+00") [1] "8.KK+KK" > gsub("+","K","8.00+00") Error in gsub("+", "K", "8.00+00") : invalid regular expression '+' In addition: Warning message: In gsub("+", "K", "8.00+00") : regcomp error:

I have this problem: my screen seems not wide enough in R. My true computer monitor is very wide, so I wish that I can line up the text when I stretch it wide. i.e.: This is typical R returns: > summary(a) V1 V2 V3 V4 F:625 Min. :20020103 EMC : 34 Min. : 9300300 L:944 1st Qu.:20020530 BRW : 27 1st Qu.: 9323865

Hi R-Users,   Thanks in advance.   I am using R-2.12.0 on Windows XP.   I am trying to produce an n X m matrix from text data stored in different files. Where n = number of words (say w1, w2, …, wn). M is the number of documents (say d1, d2, …, dm)   A. Using package tm   I am using package tm to do the job. I have provided the code below:   > my.corpus <- Corpus(DirSource(my.path),

Dear R-Users, I am new to R, so please excuse the ignorance. I have data: x (2.14, 2.41, 1.09, 0.17, 8.18) y (3.81, 5.13, 0.63, 0.75, 6.35) I would like to use simple linear regression and test 2 things: 1) slope of line of best fit is statistically different from 1 2) y-intercept is statically different from 0 Could anyone provide me with the R terminology to do this? Thanks so much, Dan

Is there a way to do a linear regression with lm (having one predictor variable) and constrain the slope of the line to equal 1? Tom -- View this message in context: http://www.nabble.com/Linear-Regression-with-lm-Forcing-the-Slope-to-Equal-1-tf4828804.html#a13815432 Sent from the R help mailing list archive at Nabble.com.

I want to fit a linear model with fixed slope e.g. y = x + b (instead of general: y = a*x + b) Is it possible to do with lm()? Regards, Ryszard -------------------------------------------------------------------------- Confidentiality Notice: This message is private and may ...{{dropped:11}}

I am wondering if it possible to normalize the slope of a linear regression to its intercept to allow for valid between-group comparisons. Here is the scenario: I need to compare the slopes of biomass increase among NAFO divisions of Northwest Atlantic cod. However, the initial division biomass is a confounding factor that may influence the slope of the regression model. How can I normalize the

Dear All, First of all I would like to say I do not have much knowledge about this subject, so most of you can find it really easy. I am doing a linear regression and I want to test if the slope of the curve is 0. R gives the summary statistics: Call: lm(formula = x ~ s) Residuals: Min 1Q Median 3Q Max -0.025096 -0.020316 -0.001203 0.011658 0.044970 Coefficients: