similar to: lm fitting with a specified slope

I'm trying to manipulate/change a formula prior to passing it to another function. A simplified example: User passes formula to my function: y~x My function does: lm(transform(y)~x) Here, transform() is added to the model's response. What is the best way to accomplish this?

I'd trying to subclass the "lm" class to produce a "mylm" class whose instances behave like lm objects (are accepted by methods like summary.lm) but have additional data or slots of my own design. For starters: setClass("mylm", "lm") produces the somewhat cryptic: Warning message: Old-style (``S3'') class "mylm" supplied as a

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.

Hello. Consider the response-variable of data.frame df is constant, so analytically perfect fit of a linear model is expected. Fitting a regression line using lm result in residuals, slope and std.errors not exactly zero, which is acceptable in some way, but errorneous. But if you use summary.lm it shows inacceptable error propagation in the calculation of the t value and the corresponding

I'm interfacing to C code that uses 1-based indexing on arrays -- it ignores the zeroth element. Thus, input vectors from R must be moved up one, and output arrays must be moved down one. What is the best way to deal with this using R internal code? My current approach is: For an input R vector of length n, allocate a new vector(v) of length n+1 and copy input into v[1] to v[1+n]. Call

In "Modern Applied Statistics with S-Plus" 3rd ed., footnote on page 153 regarding a model lm(Gas~Insul/(Temp+I(Temp^2))-1,whiteside), I read "Notice that when the quadratic terms are present, first degree coefficients mean 'the slope of the curve at temperature zero', so a non-significant value does not mean that the linear term is not needed.

Hello I want to compare the slope (let's say 'b') of a linear model obtained with lm() to a theoretical value (let's say 'th'). To do so, I think I should compute a 't value' using something like : (b - 'th')/standard.deviation(b) and then look at the p-value of this computed t. I don't understand how to do this in a simple way, just using lm()

Hi there, am trying to run a linear regression with a slope of 0. I have a dataset as follows t d 1 303 2 302 3 304 4 306 5 307 6 303 I would like to test the significance that these points would lie on a horizontal straight line. The standard regression lm(d~t) doesn't seem to allow the slope to be set. Any help very welcome. ed

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 learning ggplot2, and need your help. When I try > p <- ggplot(mtcars, aes(x = wt, y=mpg)) + geom_point() > p + geom_abline(slope=5) (from http://had.co.nz/ggplot2/geom_abline.html) the slope of the abline does not change, but this works: > p + geom_abline(intercept=20) In order to have slope work, I have to use > p + geom_abline(aes(slope=5)) Is it a bug, or is there

Hello all, I have 3 time series (tt) that I've fitted segmented regression models to, with 3 breakpoints that are common to all, using code below (requires segmented package). However I wish to specifiy a zero coefficient, a priori, for the last segment of the KW series (green) only. Is this possible to do with segmented? If not, could someone point in a direction? The final goal is to

Dear forum members, How can I force a negative slope in a linear regression even though the slope might be positive? I will need it for the purpose of determining the trend due reasons other than biological because the biological (genetic) trend is not positive for these data. Thanks. Julia Example of the data: [1] 1.254 1.235 1.261 0.952 1.202 1.152 0.801 0.424 0.330 0.251 0.229

Hello, I have a challenge! I have a large dataset with three columns, "date","temp", "location". "date" is in the format %m/%d/%y %H:%M, with a "temp" recorded every 10 minutes. These temperatures of surface temperatures and so fluctuate during the day, heating up and then cooling down, so the data is a series of peaks and troughs. I would like

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

Hi, Is there a neat way of working out the slope of a flat surface in R? Given (x,y,z) co-ordinates of the four corners of a square, is there a function which will allow me to calculate the "mean" slope of the surface in a given direction? Thanks in advance.. Laura Laura Quinn Institute of Atmospheric Science School of Earth and Environment University of Leeds Leeds LS2 9JT tel: +44

[resent, plus small addition; I do not understand why gmail sent a weird charset.] Dear R wizards: I would love to write a general function that matches the slope of a plotted line in an xy-plot at a particular x,y location. something like x<- (1:10)^2; y<- 40:50; plot( x,y, type="l", xlim=c(0,90) ) srt.at5 = text.at.current.plot.with.slope( x, y, 5); text(

Hello, I have a linear model with different slopes for different treatment groups. I need to pairwise compare the different slope estimates for the different treatment groups. Is there a package that does pairwise comparisons of slope coefficients, making the appropriate adjustments in the P values? Thanks, John. -- ========================================== John Janmaat Department of

Hi for some data I working on I am merely plotting time against temperature for a variable named filmclip. So for example, I have volunteers who watched various film clips and have used infared camera to monitor the temperature on their face at every second of the clip. The variable names I have used are Normalised ( for the temperature) and Frame (for the time in seconds). So I have fitted a

Hi all, I try to test a linear slope using offset. I have: > m2 <- glm(Y~X*V) > summary(m2) Call: glm(formula = Y ~ X * V) Deviance Residuals: Min 1Q Median 3Q Max -2.01688 -0.56028 0.05224 0.53213 3.60216 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 1.3673 0.8476 1.613 0.119788 X

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