search for: pchisq

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...

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] 0.4795...

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 instead of 0

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 >

...---- > > From: Spencer Graves [SMTP:spencer.graves at PDF.COM] > > Sent: Thursday, February 13, 2003 6:12 PM > > To: Jose Mar?a Fedriani Laffitte > > Cc: s-news at lists.biostat.wustl.edu > > Subject: Re: [S] Exact p-values > > > > > > Try ( 1-pchisq(29.8, df=1)): With S-Plus 6.1, I got 4.78992e-008. > > > > By the way, the distribtion functions in R have more > arguments. > > For example, pchisq(29.8, df=1, lower.tail=F) produces the same > > answer, and pchisq(29.8, df=1, lower.tail=F, log=T) produc...

Full_Name: Alan Bain Version: 2.4.0 OS: XP Submission from: (NULL) (155.140.122.227) Code for pnchisq contains following if (tSml) { if (x> f+ theta+ 5*sqrt(2*(f+2*theta) ))){ return 1.; /* better than 0 --- but definately FIXME */ } } This needs to check which tail has been requested; it is only correct if the default lower_tail=1 has been requested; for upper tail should return 0

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). I would assume that the former is more appropriate, so I was just curious if this is an intentional chang...

I have a quick question which is very simple but I seem to have a mental block! I'm using the pchisq function to specify a Chi Squared distribution with 9 df which I'm then going to use in the Kolmogorov-Smirnov Test to test some simulated values. so simply: pchisq(q, df=9) I know that q is the vector of quantiles but could anybody tell me what exactly this vector needs to contain? Many...

hello everybody i'm very new at using R so probably this is a very stupid question. I have a problem calculating a p-value. When i do this with excel i can use the method CHIDIST for 1.2654 with 1 freedom degree i get the answer 0.261 i just want to do the same thing in R but i can't find a method. can somebody help me friendly regards richard

...s at PDF.COM] > >> > Sent: Thursday, February 13, 2003 6:12 PM > >> > To: Jose Mar?a Fedriani Laffitte > >> > Cc: s-news at lists.biostat.wustl.edu > >> > Subject: Re: [S] Exact p-values > >> > > >> > > >> > Try ( 1-pchisq(29.8, df=1)): With S-Plus 6.1, I got 4.78992e-008. > >> > > >> > By the way, the distribtion functions in R have more > >>arguments. > >> > For example, pchisq(29.8, df=1, lower.tail=F) produces the same > >> > answer, and pchisq...

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?

...le). 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 based on the chi-square distribution by using: > pchisq(12, df= 6, lower.tail=F) [1] 0.0619688 > pchisq(5.1, df= 6, lower.tail=F) [1] 0.5310529 Is there a way to calculate the equivalent Monte Carlo p-values? The values were 0.02 and 0.138 respectively. The use of the approximate chi-square probabilities for Durbin's test are considered not...

Hi, Thanks, everyone for all the help! So, here is my calling function in C (called test.c): #include<stdio.h> #include<stdlib.h> #include<Rmath.h> int main(void) { printf("%f \n",pchisq(2.,7., 1, 0)); printf("%f \n",pnchisq(2.,7.,0., 1, 0)); return EXIT_SUCCESS; } I compile using: gcc test.c -I/usr/lib/R/include -L/usr/lib/R/lib -lm -lR However, running ./a.out gives me: 1.000000 0.040160 The first is wrong, but the second non-central is...

...       >  hosmerlem <- function (y, yhat, g = 10) { cutyhat <- cut(yhat, breaks = quantile(yhat, probs = seq(0, 1, 1/g)), include.lowest = T) obs <- xtabs(cbind(1 - y, y) ~ cutyhat) expect <- xtabs(cbind(1 - yhat, yhat) ~ cutyhat) chisq <- sum((obs - expect)^2/expect) P <- 1 - pchisq(chisq, g - 2) c("X^2" = chisq, Df = g - 2, "P(>Chi)" = P) }   > hosmerlem(no.NA$repcnd, fitted(mod.fit))  X^2                       Df                           P(>Chi) 7.8320107            8.0000000            0.4500497     > # Option 2 - Hosmer–le Cessie omnibus l...

Hi, How do we get the value of a chi square as we usually look up on the table on our text book? i.e. Chi-square(0.01, df=8), the text book table gives 20.090 > dchisq(0.01, df=8) [1] 1.036471e-08 > pchisq(0.01, df=8) [1] 2.593772e-11 > qchisq(0.01, df=8) [1] 1.646497 > nono of them give me 20.090 Thanks, cruz

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 testing LR chi2 from the null model, but what does pval1 says? Many thanks in advance, Dunia [[alternative HTML version de...

...t as a goodness of fit (as suggested by a kind “R-helper” previously). To test whether the predictor (svl, or body size) has significant effect on predicting whether or not a female snake is pregnant, I used the differences between null deviance and residual deviance using a code as following:   1-pchisq(mod.fit$null.deviance - mod.fit$deviance, mod.fit$df.null - mod.fit$df.residual)   Could anyone tell me whether I did the test properly? I did this test because I thought Wald test/z score listed in the output from "summary(mod.fit)" is not appropriate for a kind of data I have.  Does R h...

...ons recognize that df=Inf is equivalent to the standard normal distribution: > pt(1,df=Inf) [1] 0.8413447 > pnorm(1) [1] 0.8413447 On the other hand, pf() will accept Inf for df1, but returns the wrong result: > pf(1,df1=Inf,df2=1) [1] 1 whereas the correct limiting value is > pchisq(1,df=1,lower.tail=FALSE) [1] 0.3173105 pf() returns NaN when df2=Inf: > pf(1,df1=1,df2=Inf) [1] NaN Warning message: NaNs produced in: pf(q, df1, df2, lower.tail, log.p) although the correct value is available as > pchisq(1,df=1) [1] 0.6826895 Gordon > version _ platfor...

...he log-likelihoods of the glm model and of the glmer model as follows: lrt <- function (obj1, obj2){ L0 <- logLik(obj1) L1 <- logLik(obj2) L01 <- as.vector(- 2 * (L0 - L1)) df <- attr(L1, "df") - attr(L0, "df") list(L01 = L01, df = df, "p-value" = pchisq(L01, df, lower.tail = FALSE)) } gm0 <- glm(cbind(female,male)~date,family = binomial, data = liz3) gm1 <- glmer(cbind(female,male)~date+(1|dam),family=binomial,data= liz3) lrt(gm0, gm1) and I have compared the deviances as follows: (d0 <- deviance(gm0)) (d1 <- deviance(gm1)) (LR...

...late the G for each level of tran (likelihood ratios) G = vector() for (i in levels(tran)) { obsi = obscounts[tran==i] expi = expcounts[tran==i] G[i] = 2*sum((obsi*log(obsi/expi))) } dfs = rep(max(as.integer(levels(cov)))-1, max(as.integer(levels(cov)))) probs = pchisq(G, dfs, lower.tail=F) #and the lower tail probability of the G #get a weighted average for the pooled expectations weightexp = list() weights = vector() for (i in levels(tran)) { expi = expprop[tran==i] lengi = tleng[as.integer(i)] counti = sum(obscou...