similar to: Singular model.matrix of nested designs

I read somewhere that t.test is equivalent to a hypothesis testing for one way ANOVA. But I'm wondering how they are equivalent. In the following code, the p-value by t.test() is not the same from the value in the last command. Could somebody let me know where I am wrong? > set.seed(0) > N1=10 > N2=10 > x=rnorm(N1) > y=rnorm(N2) > t.test(x,y) Welch Two Sample t-test data:

Hi, I don't understand what the meaning of the following lines returned by model.matrix(). Can somebody help me understand it? What can they be used for? attr(,"assign") [1] 0 1 2 2 attr(,"contrasts") attr(,"contrasts")$A [1] "contr.treatment" attr(,"contrasts")$B [1] "contr.treatment" Regards, Peng > a=2 > b=3 > n=4

Hi, I run the following commands. 'A' has 3 levels and 'B' has 4 levels. Should there be totally 3+4 = 7 coefficients (A1, A2, A3, B1, B2, B3, B4)? > a=3 > b=4 > n=1000 > A = rep(sapply(1:a,function(x){rep(x,n)}),b) > B = as.vector(sapply(sapply(1:b, function(x){rep(x,n)}), function(x){rep(x,a)})) > Y = A + B + rnorm(a*b*n) > > fr =

Hi there! I´m a noob when it comes to R and I´m using it to run statisc analysis. With the code for ARIMA below I´m getting this error: Error in solve.default(res$hessian * n.used) :Lapack routine dgesv: system is exactly singular The code is: > s.ts <- ts(x[,7], start = 2004, fre=12) > get.best.arima <- function (x.ts, maxord=c(1,1,1,1,1,1)) + { + best.aic <- 1e8 + n <-

I have the following code. I'm wondering why summary() doesn't show F value and Pr? Rscript multi_factor.R > a=3 > b=4 > c=5 > d=6 > e=7 > > A=1:a > B=1:b > C=1:c > D=1:d > E=1:e > > X=matrix(nr=a*b*c*d*e,nc=5) > colnames(X)=LETTERS[1:5] > > for(i_a in 1:a-1) { + for(i_b in 1:b-1) { + for(i_c in 1:c-1) { + for(i_d in 1:d-1) { +

Hi I have a problem with this error, I have searched the archives and found previous discussion about this, can I cannot understand how the explanations apply to what I am trying to do. I am trying to do Log_rank Survival analysis, I have included tables and str command, is it a factor/integer problem? If so how do I correct this, as all my attempt to recode the data have failed. >

Hi, Excuse me for asking this silly question. But I really couldn't understand why cov() and ccov() don't work for my calculation of covariance matrix. a <- matrix(1:8, 2, 4) a [,1] [,2] [,3] [,4] [1,] 1 3 5 7 [2,] 2 4 6 8 > ccov(a) Error in solve.default(cov, ...) : Lapack routine dgesv: system is exactly singular I also tried colume bind, but it

Dear all, The background is that I'm trying to fix this bug in the geometry package: https://r-forge.r-project.org/tracker/index.php?func=detail&aid=1993&group_id=1149&atid=4552 Boiled down, the problem is that there exists at least one matrix X for which det(X) != 0 and for which solve(X) fails giving the error "system is computationally singular: reciprocal condition

When trying to fit a generalized linear mixed model using glmmPQL: > fit0 <- glmmPQL(ifelse(response=="A",1,0)~gender,data=set1, random=~1|subject,family=binomial) iteration 1 Error in solve.default(pdMatrix(a, fact = TRUE)) : Lapack routine dgesv: system is exactly singular Could this be occuring because the paired responses for each subject are always the same? If

Hi, I don't quite understand what are the return values of aov. I know that it has 'coefficients'. But I need to know what all the other return values are. Can somebody let me know how to figure them? Value: An object of class 'c("aov", "lm")' or for multiple responses of class 'c("maov", "aov", "mlm",

Hi, I'm trying to use the solve() function in R to invert a matrix. I get the following error, "Lapack routine dgesv: system is exactly singular" However, My matrix doesn't appear to be singular. [,1] [,2] [,3] [,4] [1,] 0.99252358 0.93715047 0.7540535 0.4579895 [2,] 0.01607797 0.09616267 0.2452471 0.3088614 [3,] 0.09772828 0.58451468 1.4907090

Dear R-experts, I am trying to fit my model but I couldn't because of this error. here is the error "Error in solve.default(dial(m) -A) : Lapack routine dgesv: system is exactly singular" thank you all. sonam

Dear R-help, I'm very sorry to ask 2 questions in a week. I am using the package 'crr' and it does exactly what I need it to when I use the dataset a. However, when I use dataset b I get the following error message: Error in drop(.Call("La_dgesv", a, as.matrix(b), tol, PACKAGE = "base")) : system is computationally singular: reciprocal condition number =

Dear R group, I have to solve a hessian matrix 40*40, called M, in order to obtain the standart deviations of estimators. When I use the function solve(M), I have the following error message: "Error in solve.default(M) : Lapack routine dgesv: system is exactly singular" Do you know an alternative approach which could succeed? I have found some information about the function

I have the R code at the end. The last command gives me "1 observation deleted due to missingness". I don't understand what this error message. Could somebody help me understand it and how to fix the problem? > summary(afit) Df Sum Sq Mean Sq F value Pr(>F) A 2 0.328 0.16382 0.1899 0.82727 B 3 2.882 0.96057 1.1136 0.34644 C

Hello all I'm trying to learn how to fit a nested model in R. I found a toy example on internet where a dataset that have?3 areas and 4 sites within these areas. When I use Minitab to fit a nested model to this data, this is the ANOVA table that I got: Nested ANOVA: y versus areas, sites Analysis of Variance for y Source DF SS MS F P areas 2 4.5000 2.2500

I have a simple system of linear equations to solve for X, aX=b: > a [,1] [,2] [,3] [,4] [1,] 1 2 1 1 [2,] 3 0 0 4 [3,] 1 -4 -2 -2 [4,] 0 0 0 0 > b [,1] [1,] 0 [2,] 2 [3,] 2 [4,] 0 (This is ex Ch1, 2.2 of Artin, Algebra). So, 3 eqs in 4 unknowns. One can easily use row-reductions to find a homogeneous solution(b=0) of: X_1

Hello all- I have two time series, Index1stdiff and Comps1stdiff. I regressed the first on the second and R returned the summary stats I expected. Then I looked at and plotted the residuals. I then wanted to assess autocorrelation characteristics and tried to run a Durbin-Watson using: library(lmtest) dwtest(formula=Index1stdiff~Comps1stdiff,alternative=c("greater")) I am

Hi, I'm using e1071 package to do fuzzy cluster analysis. My dataset (ra) has 5237 observations and 2 variables - depth and velocity. I used fuzzy cmeans to create 6 fuzzy classes. >ra.flcust6<-cmeans(ra,6,iter.max=100,verbose=F,dist="euclidean",method="cmea ns",m=1.7,rate.par=NULL,weights=1) I would like to calculate the value of all the fuzzy validity

We have a problem inverting a matrix which has the following eigenvalues: > eigen(tcross, only.values=TRUE) $values [1] 7.917775e+20 2.130980e+16 7.961620e+13 8.241041e+12 2.258325e+12 [6] 3.869428e+11 6.791041e+10 2.485352e+09 9.863098e+08 9.819373e+05 [11] 3.263408e+05 2.929853e+05 2.920419e+05 2.714355e+05 8.733435e+04 [16] 8.127136e+04 6.543883e+04 5.335074e+04