similar to: how to calculate chisq value in R

Full_Name: Jerry W. Lewis Version: 2.6.1 OS: Windows XP Professional Submission from: (NULL) (24.147.191.250) pchisq(0,0,ncp=lambda) returns 0 instead of exp(-lambda/2) pchisq(x,0,ncp=lambda) returns NaN instead of exp(-lambda/2)*(1 + SUM_{r=0}^infty ((lambda/2)^r / r!) pchisq(x, df + 2r)) qchisq(.7,0,ncp=1) returns 1.712252 instead of 0.701297103 qchisq(exp(-1/2),0,ncp=1) returns 1.238938

Dear List, If any of observed and/or expected data has less than 5 frequencies, then chisq.test (Pearson's Chi-squared Test for Count Data from package:stats) gives warning messages. For example, x<-c(10, 14, 10, 11, 11, 7, 8, 4, 1, 4, 4, 2, 1, 1, 2, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1) y<-c(9.13112391745095, 13.1626482033341, 12.6623267638188, 11.0130706413029, 9.16415925139016,

I have dowloaded the Source Code of R,and I want to know the process of chi-sqared test,but how can I found it? [[alternative HTML version deleted]]

Dear colleague, I not sure this R code is correctly ? I would to show the number of Sample Size at Sample Size Axis that line draw from Power Axis (80%) from R code. How I show this and select the most appropriate of this power (.79955687 - 80983575). Thank for your help and answer. Best Regards, Nikom Thanomsieng, Email: nikom at kku.ac.th .... #Power analysis: Sample size for

Full_Name: Richard Mott Version: 1.9.0 OS: Windows XP Submission from: (NULL) (81.178.233.208) Shouldn't these give the same answer? > pchisq(67.60644,df=1,lower.tail=F,ncp=0) [1] 3.219647e-15 > pchisq(67.60644,df=1,lower.tail=F) [1] 1.996145e-16 >

Dear List, recently tried to reproduce the results of some custom model selection function after updating R, which unfortunately failed. However, I ultimately found the issue to be that testing with pchisq() in drop1() seems to have changed. In the below example, earlier versions (e.g. R 2.4.1) produce a missing P-value for the variable x, while newer versions (e.g. R 2.7.1) produce 0 (2.2e-16).

Where does the value 2.2e-16 come from in p-values for chisq tests such as those reported below? > Anova(cm.mod2) Analysis of Deviance Table (Type II tests) Response: Freq LR Chisq Df Pr(>Chisq) B 11026.2 1 < 2.2e-16 *** W 7037.5 1 < 2.2e-16 *** Age 886.6 8 < 2.2e-16 *** B:W 3025.2 1 < 2.2e-16 *** B:Age 1130.4 8 < 2.2e-16 *** W:Age 332.9 8 < 2.2e-16 *** --- Signif.

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?

Hi, I want to determine the confidence interval on the sum of two sigma's. Is there an easy way to do this in R? I guess I have to use some sort of chisquare convolution algorithm??? Thanx, Roy -- The information contained in this communication and any atta...{{dropped}}

Hello all, Here's my query: Running R 1.6.2 on FreeBSD 5.0, and on WinXP, and I find that the following hangs the process: dchisq(alpha=0.01, df=1, ncp=295) but it does work for ncp < 294.92. Is this general? Best wishes to all, Andrew Andrew Robinson Ph: 208 885 7115 Department of Forest Resources Fa: 208 885 6226 University of Idaho E : andrewr at uidaho.edu PO

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 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

Dear List, How do I extract the approximate Wald test for the frailty (in the following example 17.89 value)? What about the P-values, other Chisq, DF, se(coef) and se2? How can they be extracted? ######################################################> kfitm1 Call: coxph(formula = Surv(time, status) ~ age + sex + disease + frailty(id, dist = "gauss"), data = kidney)

I compiled the 1.9.1 src.rpm with the standard gnu tools and it works. I tried compiling the 1.9.1 src.rpm with the Intel 8 C and FORTRAN compilers and it bombs out during the testing phase: comparing 'd-p-q-r-tests.Rout' to './d-p-q-r-tests.Rout.save' ...267c267 < df = 0.5[1] "Mean relative difference: 5.001647e-10" --- > df = 0.5[1] TRUE

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),

Full_Name: Robert King Version: 1.5.0 OS: windows Submission from: (NULL) (134.148.4.19) Also occurs in 1.6.0 on linux anova.glm(fitted.object,test="Chisq") is giving strange answers in this situation > resptime sex task time 1 m s 210 2 m s 300 3 m s 420 4 f s 250 5 f s 310 6 f s 390 7 m c 310 8 m c 400 9 m c 600

I am getting "Chi-squared approximation may be incorrect in: chisq.test(x)" with the data bleow. Frequency distribution of number of male offspring in families of size 5. Number of Male Offspring N 0 518 1 2245 2 4621 3 4753 4 2476 5

Full_Name: Viktor Witkovsky Version: 2.9.2 OS: Windows XP Submission from: (NULL) (78.98.89.227) Hello, I have found strange behavior of the function qchisq (the non-central qchisq is based on inversion of pchisq, which is further based on pgamma). The function gives wrong results without any warning. For example: qchisq(1e-12,1,8.94^2,lower.tail=FALSE) gives 255.1840972465858 (notice that

Dear R-Users, How can I use chisq.test() as a goodness of fit test? Reading man-page I?ve some doubts that kind of test is available with this statement. Am I wrong? X2=sum((O-E)^2)/E) O=empirical frequencies E=expected freq. calculated with the model (such as normal distribution) See: http://www.itl.nist.gov/div898/handbook/eda/section3/eda35f.htm for X2 used as a goodness of fit test. Any

Hi Folks, I note that, while the "chisq" functions dchisq(x, df, ncp=0, log = FALSE) pchisq(q, df, ncp=0, lower.tail = TRUE, log.p = FALSE) qchisq(p, df, ncp=0, lower.tail = TRUE, log.p = FALSE) rchisq(n, df, ncp=0) all have a slot for the non-centrality parameter "ncp", of the functions for the t and F distributions: dt(x, df, log = FALSE)