similar to: p-value calculation

Dear all Just for fun, I have just downloaded the paper mentioned below and checked it with R-1.6.1. Everything is ok with exception of Table 2b, where I get always 1 instead of 0.5: > pbinom(1e15,2e15,0.5) [1] 1 Which value should be correct? Best regards Christian Stratowa ============================================== Christian Stratowa, PhD Boehringer Ingelheim Austria Dept NCE Lead

Hi list, I want to use the p-value from the survdiff function (package survival) to reuse within a function in a Kaplan-Meier plot. The p-value is somehow not a component of the value list ?! Thanks in advance -- A. Goralczyk G?ttingen, Ger.

How can I get the Log - Rank p value to be output? The chi square value can be output, so I was thinking if I can also have the degrees of freedom output I could generate the p value, but can't see how to find df either. > (survtest <- survdiff(Surv(time, cens) ~ group, data = surv,rho=0)) Call: survdiff(formula = Surv(time, cens) ~ group, data = surv, rho = 0) N Observed

Dear R users,I'm a new user of R. I'm trying to do a LM test an there is this type of error: Error in t(mX) %*% mX : requires numeric/complex matrix/vector arguments. To be clear I write down the code in which mY ( 126,1 ) mX (126,1) mZ(126,1) are matrix. LMTEST <- function(mY, mX, mZ)#mY, mX, mZ must be matrices!#returns the LM test statistic and the degree of freedom{iT =

if i want to plot the chi-square distribution with a different degree of freedom how can i plot it in the graph?Sometimes i plot the histogram and cut it in a lot of piece.It's distribution like a chi-square.So i want to plot the chi-square with a different degree of freedom to compare it . -- View this message in context:

HI R-users, I am trying to repeat an example from Rayner and Best "A contingency table approach to nonparametric testing (Chapter 7, Ice cream example). In their book they calculate Durbin's statistic, D1, a dispersion statistics, D2, and a residual. P-values for each statistic is calculated from a chi-square distribution and also Monte Carlo p-values. I have found similar p-values

The function 'pchisq' from the 'stats' library gives a wrong result if the argument equals exactly zero: # Upper tail of central 1-df chi^2 distribution > pchisq(1 , 1, ncp=0, lower.tail = F, log.p = FALSE) [1] 0.3173105 > pchisq(0.5 , 1, ncp=0, lower.tail = F, log.p = FALSE) [1] 0.4795001 > pchisq(0.01 , 1, ncp=0, lower.tail = F, log.p = FALSE) [1]

Dear all, I am fitting a GLM similar to library(MASS) anorex.1 <- glm(Treat~Postwt+Prewt,family = binomial, data = anorexia) I have found two ways of computing the p-value of the fitted model: pval1 <- 1-pchisq(anorex.1$deviance,anorex.1$df.residual) pval2 <- 1-pchisq(anorex.1$null.deviance - anorex.1$deviance, anorex.1$df.null - anorex.1$df.residual) pval2 is

Hello, I have to compute the pooled z-value and I would like to know which way is more appropriate b <- c( -0.205,1.040,0.087) s <- c(0.449,0.167,0.241) n <- c(310, 342, 348) z <- b/s Z <- sum(z)/sqrt(length(n)) P <- 2*(1-pnorm(abs(Z))) P w <- sqrt(n) Zw <- sum(w * z)/sqrt(sum(w^2)) Pw <- 1 - pchisq(Zw * Zw, 1) Pw Many thanks in advance, Cheba [[alternative HTML

Hi there, If I do an lm, I get p-vlues as p-value: < 2.2e-16 This is obtained from F =39540 with df1 = 1, df2 = 7025. Suppose am interested in exact value such as p-value = 1.6e-16 (note = and not <) How do I go about it? stephen

Hello, I am trying to fit gamma, negative exponential and inverse power functions to a dataset, and then test whether the fit of each curve is good. To do this I have been advised to calculate predicted values for bins of data (I have grouped a continuous range of distances into 1km bins), and then apply a chi-squared test. Example: > data <- data.frame(distance=c(1,2,3,4,5,6,7),

Dear R-community, I have a short question: How do I interpret the result of a likelihood ratio test correctly? I am fitting a parametric survival model (with aftreg {eha}) and the output tells me the overall p-value of my model is < 0.001. My simple question is: Does the result mean my model fits the data well OR does it mean my model DOES NOT fit the data well? Some side information how the

dear r experts: It seems my message got spam filtered, another try: i would appreciate advice on how to get the p-value from the object 'sr' created with the function survreg() as given below. vlad sr<-survreg(s~groups, dist="gaussian") Coefficients: (Intercept) groups -0.02138485 0.03868351 Scale= 0.01789372 Loglik(model)= 31.1 Loglik(intercept only)= 25.4

Sent this mail in rich text format before. Excuse me for this. ------------------------ Dear all, I'm using the lrm function from the package "Design", and I want to extract the p-values from the results of that function. Given an lrm object constructed as follows : fit <- lrm(Y~(X1+X2+X3+X4+X5+X6+X7)^2, data=dataset) I need the p-values for the coefficients printed by calling

Hi, This is more of a general question than a pure R one, but I hope that is OK. I want to combine one-tailed independent p-values using the weighted version of fisher's inverse chi-square method. The unweighted version is pretty straightforward to implement. If x is a vector with p-values, then I guess that this will do for the unweighted version: statistic <- -2*sum(log(x)) comb.p <-

Dear all, does anyone know how I can extract specific p-values for covariates from an aftreg object? After fitting a model with aftreg I can find all different variables by using str(), but there's no place where p-values are kept. The odd thing is that print() displays them correctly. EXAMPLE: > testdata start stop censor groupvar var1 var2 1 0 1 0

Hello, I want to use the quantile function so I read the doc but I don't understand with this > qchisq(seq(0.05,0.95,by=0.05),df=(length(don)-1)) [1] 62667.11 62795.62 62882.42 62951.47 63010.74 63064.00 63113.39 63160.27 63205.65 63250.33 63295.04 63340.48 63387.48 63437.03 63490.53 63550.14 63619.68 [18] 63707.24 63837.16 Can you help me please?

Hello I have a matlab MAT file that contains one single variable: a. The structure of a is as follows: a.river1.flow (flow values) a.river1.date_flow (date) a.river1.precip (precipitation values) a.river1.date_precip a.river2.flow a.river2.date_flow a.river2.precip a.river2.date_precip I have used readMat to load the variable a in R, however I have no idea how readMat translates a. I managed

Hi, I'm new to R and using it for Cox survival analysis. Thanks to this great forum I learned how to compute the HR with its confidence interval. My question would be: Is there any way to get the p-value for a hazard ratio in addition to the confidence interval? Thanks, Thierry -- Thierry Panje Visiting Student Researcher Department of Psychology Stanford

Dear R experts, I'm trying to call 'pchisq' from within a C subroutine. The following error is returned: ** NON-convergence in pgamma()'s pd_lower_cf() f= nan. This error message is not printed the first time I call 'pchisq' from the C subroutine, but the second time or the next time I call 'pchisq' from within R. My session output is shown below: