similar to: How to Find the value of r-square change in hierarchical multiple linear regression

I am trying to perform goodness of fit test using R. I am using this website http://wiener.math.csi.cuny.edu/Statistics/R/simpleR/stat013.html for help. However, I am unable to carry out the test successfully. My code follows. It is taken from the website just mentioned. freq=c(22,21,22,27,22,36) # frequencies obtained after rolling the dice 150 times. prob=c(1,1,1,1,1,1)/6 # specify expected

Residuals are stored as a numeric vector. The R software comes with a document "Introduction to R" that discusses basic math functions and logical operators that can create logical vectors: abs( stdresiduals ) > 2.5 It also discusses indexing using logical vectors: stdresiduals[ abs( stdresiduals ) > 2.5 ] Note that in most cases it is worth going the extra step of making your

Dear list, I am slightly confused about how I can do the following in R. I want to perform robust multiple regression. I?ve used the Boot function in CAR package to find confidence intervals and standard errors. Inadition to these, I want to find the robust estimates for the F test and r-square. Finally, I would like to know the significance levels of bootstrap results. Below I explain

Dear list members, I want to save residuals above or less than a certain value to an R object. I have performed a multiple linear regression, and now I want to find out which cases have a residual of above + 2.5 and ? 2.5. Below I provide the R commands I have used. Reg<-lm(a~b+c+d+e+f) # perform multiple regression with a as the dependent variable. Residuals<-residuals(reg) # store

Hi all, I want to make a contingency table in R. I want to tabulate two variables, one as the independent and second as the dependent variable. The IV has two categories, namely, birth complications, and no birth complications. The frequency of birth complication category is fifty, and the frequency of no birth complication category is 34. The categories and frequencies of DV follows.

Dear Faiz, Bootstrapping R^2 using Boot() is straightforward: Simply write a function that returns R^2, possibly in a vector with the regression coefficients, and use it as the f argument to Boot(). That will get you, e.g., bootstrapped confidence intervals for R^2. (Why you want that is another question.) See the example in ?Boot that shows how to bootstrap the estimated error variance (without

Hi all, I would like to read the results of the commands type in the terminal window in Microsoft Word. As a blind user my options are somewhat limited and are time consuming if I want to see the results of the commands that I have type earlier. for example if my first two commands were x<-c(1,2,3,4,5) mean(x) and I have typed ten more commands after the first two commands it is not easy

?text says "'adj' allows _adj_ustment of the text with respect to '(x,y)'. Values of 0, 0.5, and 1 specify left/bottom, middle and right/top, respectively." But it looks like 0, 1 specify top, bottom respectively in the y direction. plot(1:4) text(2,2, "adj=c(0,0)", adj=c(0,0)) text(2,2, "adj=c(0,1)", adj=c(0,1), col=2) #the red

Dear list, I have the following how-to-do it in R, questions. Suppose I have ten independent variables, and one dependent variable. I want to find the Pearson correlation of all the IVs with the DV, but not the correlation between the IVs. What I know so far, about R, that I have to type the cor () function ten times, each time requesting for a correlation between one IV and the DV. I was

Hi, I am wondering if anyone knows how to perform an F-test on the change in R square between hierarchical models in R? SPSS provides this information and a researcher that I am working with is interested in getting this information. Alternatively, if someone knows how I can calculate the test statistic (SPSS calls it F-change?) and dfs that would be helpful as well. The output and the test I am

Hi For first question, maybe I am completely wrong but cor(swiss[,-1], swiss[,1]) should give you what you want in one step. Second question Without an example it is hard to say but maybe aggregate is the way forward. > aggregate(iris[,1:4], list(iris$Species), function (x) c(mean=mean(x), sd=sd(x))) Group.1 Sepal.Length.mean Sepal.Length.sd Sepal.Width.mean Sepal.Width.sd 1

Hi, I saw you discussed about the meaning of the R squared in a nls regression. Do you have a source or a quotation please? kind regards, Pierre Grison Tel: 06 01 79 74 22 Mail: pgrison@hotmail.fr [[alternative HTML version deleted]]

Hi, Sorry for the naive question, but what exactly does the 'Adjusted R-squared' coefficient in the summary of linear model adjust for? Sample code: > x <- rnorm(15) > y <- rnorm(15) > lmr <- lm(y~x) > summary(lmr) Call: lm(formula = y ~ x) Residuals: Min 1Q Median 3Q Max -1.7828 -0.7379 -0.4485 0.7563 2.1570 Coefficients:

Dear R experts, I was wondering if there are any R functions that give the tail area of a sum of chisquare distributions of the type: a_1 X_1 + a_2 X_2 where a_1 and a_2 are constants and X_1 and X_2 are independent chi-square variables with different degrees of freedom. Thanks, Klaus -- "Feel free" - 5 GB Mailbox, 50 FreeSMS/Monat ...

Hi all, I am trying to use type message with sink, like this: sink("all.Rout", type="message") 1+3 sink() readLines(con = "all.Rout") So to get the following output: > 1+3 [1] 4 Obviously this doesn't work. I tried some variations (based on the explanations in the help) but am missing something on how to make it work. Any suggestions? (p.s: I need

Dear Sir or Madam, I am a student at MSc Probability and Finance at Paris 6 University/ Ecole Polytechnique. I am using R and I can't find an answer to the following question. I will be very thankful if you can answer it. I have two vectors rendements_CAC40 and rendements_AlcatelLucent. I use the lm function as follows, and then the sumarry function: regression=lm(rendements_CAC40 ~

We have extracted parameters from physiological measurements by fitting SSlogis-like curves with nlsList and nlme. We presented residuals plot in a paper, but a referee argues that these cannot be included (too technical), and r-square values should be given instead to compare the goodness of fit with those of other authors. I remember that 30 years ago in my stat 101, I learned that r-square is

Greetings R Users, I have a system of equations for which I would like to output all the R-squares. Assume there are four equations in my system, the only way I found to output all the R-squares is by calling them out one by one as this: summary(fitSUR$eq[[1]])$r.squared summary(fitSUR$eq[[2]])$r.squared summary(fitSUR$eq[[3]])$r.squared summary(fitSUR$eq[[4]])$r.squared But isn't there a

Hello: I want to use values from the output of linear models done using permuted data to construct a random distribution. The problem I am having is the extraction of a value, say the p-value or the regression coefficient, from the summary of a linear model. When summarizing a linear model I get this: Call: lm(formula = fitness ~ mm) Residuals: Min 1Q Median 3Q Max

Dear R-list, I apologize for my many emails but I think I know how to desctribe my problem differently and more clearly. My question is how to compare linear models with intercept and those without intercept using maximizing adjusted R^2 strategy. Now I do it like the following: > library(leaps) > n=20 > x=matrix(rnorm(n*3),ncol=3) > b=c(1,2,0) > intercept=1 >