similar to: Null-Hypothesis

Hi, I've made a research about how to compare two regression line slopes (of y versus x for 2 groups, "group" being a factor ) using R. I knew the method based on the following statement : t = (b1 - b2) / sb1,b2 where b1 and b2 are the two slope coefficients and sb1,b2 the pooled standard error of the slope (b) which can be calculated in R this way: > df1 <-

Dear List, I used rlm to calculate two regression models for two data sets (rlm due to two outlying values in one of the data sets). Now I want to compare the two regression slopes. I came across some R-code of Spencer Graves in reply to a similar problem: http://www.mail-archive.com/r-help at stat.math.ethz.ch/msg06666.html The code was: > df1 <- data.frame(x=1:10, y=1:10+rnorm(10))

Hi, I am computing the Nelson-Aalen (NA) estimate of baseline cumulative hazard in two different ways using the "survival" package. I am expecting that they should be identical. However, they are not. Their difference is a monotonically increasing with time. This difference is probably not large to make any impact in the application, but is annoyingly non-trivial for me to just

I would suggest the method of Sokal and Rholf (1995) S. 498, using the F test. Below I repeat the analysis by Spencer Graves: Spencer: > df1 <- data.frame(x=1:3, y=1:3+rnorm(3)) > df2 <- data.frame(x=1:3, y=1:3+rnorm(3)) > fit1 <- lm(y~x, df1) > s1 <- summary(fit1)$coefficients > fit2 <- lm(y~x, df2) > s2 <- summary(fit2)$coefficients > db <-

You also need to reply-all so the mailing list stays in the loop. -- Sent from my phone. Please excuse my brevity. On December 19, 2017 4:00:29 PM PST, Timothy Axberg <axbergtimothy at gmail.com> wrote: >Sorry about that. Here is the code typed directly on the email. > >qe = (Qmax * Kl * ce) / (1 + Kl * ce) > >##The data >ce <- c(15.17, 42.15, 69.12, 237.7, 419.77)

Hello, I've written a simple (although probably overly roundabout) function to test whether two regression slope coefficients from two linear models on independent data sets are significantly different. I'm a bit concerned, because when I test it on simulated data with different sample sizes and variances, the function seems to be extremely sensitive both of these. I am wondering if

Dear All, I have a rather unusual problem. I have a set of data for a class in subsurface processes. From that dataset, I must calculate the slope of the best-fit line (which is the parameter of interest). The problem I have is twofold: 1) for the purposes of the exercise, I must force my best-fit line to go through the origin (0,0), and 2) the line must be linear, even though the data is

First, as others have said please obey the mailing list rules and turn of First, as others have said please obey the mailing list rules and turn off html, not everyone uses an html email client. Here is your code, formatted and with line numbers added. I also fixed one error: "y" should be "status". 1. fit0 <- coxph(Surv(futime, status) ~ x1 + x2 + x3, data = data0) 2. p

Dear R-users, I have noticed small discrepencies in the reported estimate of the variance of the frailty by the print method for survreg() and the 'theta' component included in the object fit: # Examples in R-2.6.2 for Windows library(survival) # version 2.34-1 (2008-03-31) # discrepancy fit1 <- survreg(Surv(time, status) ~ rx + frailty(litter), rats) fit1 fit1$history[[1]]$theta

quantreg package is used. *fit1 results are* Call: rq(formula = op ~ inp1 + inp2 + inp3 + inp4 + inp5 + inp6 + inp7 + inp8 + inp9, tau = 0.15, data = wbc) Coefficients: (Intercept) inp1 inp2 inp3 inp4 inp5 -0.191528450 0.005276347 0.021414032 0.016034803 0.007510343 0.005276347 inp6 inp7 inp8 inp9 0.058708544

Hi, I noticed a failure of mksh built with klibc on mips64el. The failing test, on a high level, is this: :>a sleep 2 :>b test a -nt b echo $? This is supposed to echo 1 (false) because a is not newer than b. The test code is roughly: // const char *opnd1 = "a"; // const char *opnd2 = "b"; // struct stat b1, b2; // int s; return (test_stat(opnd1, &b1) ==

Dear R users, Could somebody please help me to find a way of comparing nonlinear, non-nested models in R, where the number of parameters is not necessarily different? Here is a sample (growth rates, y, as a function of internal substrate concentration, x): x <- c(0.52, 1.21, 1.45, 1.64, 1.89, 2.14, 2.47, 3.20, 4.47, 5.31, 6.48) y <- c(0.00, 0.35, 0.41, 0.49, 0.58, 0.61, 0.71, 0.83, 0.98,

Hello, Can anyone suggest a simple way to generate a Kaplan-Meier plot with 2 survfit objects, just like this one:? https://drive.google.com/file/d/1fEcpdIdE2xYtA6LBQN9ck3JkL6-goabX/view?usp=sharing Suppose I have 2 survfit objects: fit1 is for the curve on the left (survtime has been truncated to the cutoff line: year 5), fit2 is for the curve on the right (minimum survival time is at the

Hi, I am trying to do logistic regression for data of 104 patients, which have one outcome (yes or no) and 15 variables (9 categorical factors [yes or no] and 6 continuous variables). Number of yes outcome is 25. Twenty-five events and 15 variables mean events per variable is much less than 10. Therefore, I tried to analyze the data with penalized regression method. I would like please some of the

Hello all, I wish to extract the terms from an rpart object. Specifically, I would like to be able to know what is the response variable (so I could do some manipulation on it). But in general, such a method for rpart will also need to handle a "." case (see fit2) Here are two simple examples: fit1 <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis) fit1$call fit2 <-

I try to send this message To Gordon Smyth at smyth at vehi,edu.au but it bounced back, so here it is to r-help I am trying to use limma, just downloaded it from CRAN. I use R 2.0.1 on Win XP see the following: > library(RODBC) > chan1 <- odbcConnectExcel("D:/Data/mgc/Chips/Chips4.xls") > dd <- sqlFetch(chan1,"Raw") # all data 12000 > # > nzw <-

I have the following results for an ANOVA comparing two nested models. I wasn't sure how I am supposed to report this result in the area of psychology. Specifically, am I supposed to report the DF's or just the F ratio? I could manually calculate the degrees of freedoms, but there must be a reason why R does not give this information, i.e. those are not conventionally used in the

Hello, Is there a function in the survival package that will allow me to test a subset of independent variables for joint significance? I am thinking along the lines of a Wald, likelihood ratio, or F-test. I am using the survreg procedure to estimate my parameters. Thank you. Geoff Geoffrey Smith Visiting Assistant Professor Department of Finance University of Illinois at Urbana-Champaign

Hi, I want to do a global likelihood ratio test for the proportional odds logistic regression model and am unsure how to go about it. I am using the polr() function in library(MASS). 1. Is the p-value from the likelihood ratio test obtained by anova(fit1,fit2), where fit1 is the polr model with only the intercept and fit2 is the full polr model (refer to example below)? So in the case of the

In R1.7 and above (including R 1.9 alpha), 'update.formula' forgets to copy any offset(...) term in the original '.' formula: test> df <- data.frame( x=1:4, y=sqrt( 1:4), z=c(2:4,1)) test> fit1 <- glm( y~offset(x)+z, data=df) test> fit1$call glm(formula = y ~ offset(x) + z, data = df) test> fit1u <- update( fit1, ~.) test> fit1u$call glm(formula = y ~ z,