Displaying 20 results from an estimated 4000 matches similar to: "Plotting two regression lines on one graph"
2005 Apr 06
2
two methods for regression, two different results
Please forgive a straight stats question, and the informal notation.
let us say we wish to perform a liner regression:
y=b0 + b1*x + b2*z
There are two ways this can be done, the usual way, as a single
regression,
fit1<-lm(y~x+z)
or by doing two regressions. In the first regression we could have y as
the dependent variable and x as the independent variable
fit2<-lm(y~x).
The second
2009 Apr 03
1
Trouble extracting graphic results from a bootstrap
Hi,
I'm trying to extract a histogram over the results from a bootstrap. However
I keep receiving the error message "Error in hist.default(boot.lrtest$ll,
breaks = "scott") : 'x' must be numeric".
The bootstrap I'm running looks like:
> boot.test <- function(data, indeces, maxit=20) {
+ y1 <- fit1+e1[indeces]
+ mod1 <- glm(y1 ~ X1-1, maxit=maxit)
+
2010 Sep 21
1
package gbm, predict.gbm with offset
Dear all,
the help file for predict.gbm states that "The predictions from gbm do not
include the offset term. The user may add the value of the offset to the
predicted value if desired." I am just not sure how exactly, especially for
a Poisson model, where I believe the offset is multiplicative ?
For example:
library(MASS)
fit1 <- glm(Claims ~ District + Group + Age +
2018 Jan 17
1
Assessing calibration of Cox model with time-dependent coefficients
I am trying to find methods for testing and visualizing calibration to Cox
models with time-depended coefficients. I have read this nice article
<http://journals.sagepub.com/doi/10.1177/0962280213497434>. In this paper,
we can fit three models:
fit0 <- coxph(Surv(futime, status) ~ x1 + x2 + x3, data = data0) p <-
log(predict(fit0, newdata = data1, type = "expected")) lp
2012 Nov 08
2
Comparing nonlinear, non-nested models
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,
2020 Sep 29
5
2 KM curves on the same plot
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
2011 Mar 25
2
A question on glmnet analysis
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
2011 Jan 26
2
Extracting the terms from an rpart object
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 <-
2017 Dec 20
1
Nonlinear regression
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)
2009 Apr 08
2
Null-Hypothesis
Hello R users,
I've used the following help two compare two regression line slopes.
Wanted to test if they differ significantly:
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
2018 Jan 18
1
Time-dependent coefficients in a Cox model with categorical variants
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
2004 Dec 20
2
problems with limma
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 <-
2011 Sep 07
2
reporting ANOVA for nested models
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
2008 Jul 02
1
survival package test stats
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
2008 Jan 05
1
Likelihood ratio test for proportional odds logistic regression
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
2009 May 04
1
Nelson-Aalen estimator of cumulative hazard
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
2004 Mar 09
3
update forgets about offset() (PR#6656)
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,
2011 Oct 06
1
anova.rq {quantreg) - Why do different level of nesting changes the P values?!
Hello dear R help members.
I am trying to understand the anova.rq, and I am finding something which I
can not explain (is it a bug?!):
The example is for when we have 3 nested models. I run the anova once on
the two models, and again on the three models. I expect that the p.value
for the comparison of model 1 and model 2 would remain the same, whether or
not I add a third model to be compared
2011 Apr 02
3
Plotting MDS (multidimensional scaling)
Hi,
I just encountered what I thought was strange behavior in MDS. However, it
turned out that the mistake was mine. The lesson learned from my mistake is
that one should plot on a square pane when plotting results of an MDS. Not
doing so can be very misleading. Follow the example of an equilateral
triangle below to see what I mean. I hope this helps others to avoid this
kind of headache.
2009 Jul 28
2
A hiccup when using anova on gam() fits.
I stumbled across a mild glitch when trying to compare the
result of gam() fitting with the result of lm() fitting.
The following code demonstrates the problem:
library(gam)
x <- rep(1:10,10)
set.seed(42)
y <- rnorm(100)
fit1 <- lm(y~x)
fit2 <- gam(y~lo(x))
fit3 <- lm(y~factor(x))
print(anova(fit1,fit2)) # No worries.
print(anova(fit1,fit3)) # Likewise.
print(anova(fit2,fit3)) #