similar to: exact values for p-values - more information.

Hi there, If I do an lm, I get p-vlues as p-value: < 2.2e-16 Suppose am interested in exact value such as p-value = 1.6e-16 (note = and not <) How do I go about it? stephen

inputfille snpid indid genotype gvariable probeid gene geneexpression rs1040480 CHB_NA18524 C/T 2 GI_19743926-I PTPRT 5.850586 rs1040480 CHB_NA18526 C/C 1 GI_19743926-I PTPRT 6.028641 rs1040480 CHB_NA18529 C/C 3 GI_19743926-I PTPRT 5.944392 rs1040481 CHB_NA18532 C/C 1 GI_19743926-I PTPRT 5.938578 rs1040481 CHB_NA18537 C/C 2 GI_19743926-I PTPRT 5.874439 rs1040481 CHB_NA18540 C/C 3 GI_19743926-I

This is just a suggestion/wish that it would be nice for the F-distribution functions to recognize limiting cases for infinite degrees of freedom, as the t-distribution functions already do. The t-distribution functions 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

i tried to use df(x,df1,df2) but the values arent the same when i looked it up in a F Distribution table.. How to get the same F Distribution values in R as in the f table?

I would suggest the method of Sokal and Rholf (1995) S. 498, using the F test. Below I repeat the analysis by Spencer Graves: Spencer: > df1 <- data.frame(x=1:3, y=1:3+rnorm(3)) > df2 <- data.frame(x=1:3, y=1:3+rnorm(3)) > fit1 <- lm(y~x, df1) > s1 <- summary(fit1)$coefficients > fit2 <- lm(y~x, df2) > s2 <- summary(fit2)$coefficients > db <-

Hi, I make a upgrade to R 2.1.0 and in some analysis I give an error: anova(model,test="F") Analysis of Deviance Table Model: binomial, link: logit Response: landing/total Terms added sequentially (first to last) Df Deviance Resid. Df Resid. Dev F Pr(>F) NULL 16 105.079 trat 1 93.149 15 11.930 93.15 Warning message:

Hi, yet another anova.mlm problem - it doesn't seem to end. This time, I have a setup with a few within-subject factors and a between-subject factor (SGROUP). Consider the most simple case with only one within-factor (apo): > mlmfit0 <- lm(data.n ~ 0 + SGROUP) > mlmfit1 <- lm(data.n ~ 1 + SGROUP) > anova(mlmfit1,mlmfit0,test="Spherical",M=~hemi,X=~1) Analysis of

Dear all, I found the following behaviour > rf(5,Inf,Inf) [1] 1 1 1 1 1 but > qf(0.1,Inf,Inf) [1] NaN Warning messages: 1: In qf(0.1, Inf, Inf) : value out of range in 'lgamma' 2: In qf(p, df1, df2, lower.tail, log.p) : NaNs produced incidentally, > pf(1.00000000000001,Inf,Inf) [1] 1 > pf(1.0000000000000001,Inf,Inf) [1] 0.5 Is this the expected behaviour? Thanks

Estimados Escribo para consultar sobre el uso de modelos mixtos anidados. Los datos que estoy analizando provienen de censos de malezas en cuatro tipos de paisajes de la región pampeana, en los que seleccioné al azar igual número de lotes agrícolas cultivados con tres cultivos (maíz, soja y trigo-soja). En cada lote censé el número de especies de malezas en tres posiciones: el alambrado, el borde

Hi everyone. I have a problem that I have been unable to determine either the best way to proceed and why the methods I'm trying to use sometimes fail. I'm using the pf() function in an optimization function to find a noncentrality parameter that leads to a specific value at a specified quantile. My goal is to have a general function that returns the noncentrality parameter that

Full_Name: Long Qu Version: 2.3.1 OS: Windows XP Submission from: (NULL) (64.113.93.235) The QQ-plot of two versions of simulating noncentral F-distributed random numbers has quite different scales: > qqplot(rf(1000,2,15,3),qf(runif(1000),2,15,3)) The rf() function reads: > rf function (n, df1, df2, ncp = 0) { if (ncp == 0) .Internal(rf(n, df1, df2)) else rchisq(n, df1,

I want to calculate the probability that a group will include a particular point using the squared Mahalanobis distance to the centroid. I understand that the squared Mahalanobis distance is distributed as chi-squared but that for a small number of random samples from a multivariate normal population the Hotellings T2 (T squared) distribution should be used. I cannot find a function for

Dear addresses, I need perform a batch of 10 000 simulations for each of 4 options considered. (The idea is to obtain the parameter estimates in a heteroskedastic linear regression model - with additive or mixed heteroskedasticity - via the Kenward-Roger small-sample adjusted covariance matrix of disturbances). For this purpose I wrote an R program which would capture all possible options (true

Hello r helpers! Below is the whole coding for my programme. Before proceed more further, let me explain for you. First of all, I need to compute trimmed mean. Till that step is ok. Then I need to compute ssdw which is sum of square deviation. If I do equal trimming at both tail of distribution that I chose, I will use the first ssd formulae which is "a". But if I am doing unequal

OK, what is the trick to extracting the overall p value from an lm object? It shows up in the summary(lm(model)) output but I can't seem to extract it: > test2 = apply(aa, 1, function(x) summary(lm(x[,1] ~ 0 + x[,3] + x[,6]))) > test2[[1]] Call: lm(formula = x[, 1] ~ 0 + x[, 3] + x[, 6]) [omitted summary output] F-statistic: 40.94 on 2 and 7 DF, p-value: 0.0001371 It does not seem

Hi there, I am stuck trying to solve what should be a fairly easy problem. I have a data frame that essentially consists of (ID, time as seqMonth, variable, value) and i want to find the regression coefficient of value vs time for each combination of ID and Variable. I have tried several approaches and none of them seems to work as i expected. For example, i have tried:

Dear R users I have run an regression and want to extract the p value of the F statistics, but I can find a way to do that. x<-summary(lm(log(RV2)~log(IV.m),data=b)) Call: lm(formula = log(RV2) ~ log(IV.m), data = b[[11]]) Residuals: Min 1Q Median 3Q Max -0.26511 -0.09718 -0.01326 0.11095 0.29777 Coefficients: Estimate Std. Error t value Pr(>|t|)

hello R-friends, I'm looking for a quantile function for the noncentral f-distribution in the area of equivalence hypotheses testing. Can somebody help me? Many thanks ----------------------------------------------------------------- Dipl. Inform. J. Hedderich Institut f?r Medizinische Informatik Phone : 0431 / 5973182 und Statistik im Klinikum an der CAU

Hi R-users I'm trying to do multivariate analysis of variance of a experiment with 3 treatments, 2 variables and 5 replicates. The procedure adopted in SAS is as follow, but I'm having difficulty in to implement the contrasts for comparison of all treatments in R. I have already read manuals and other materials about manova in R, but nothing about specific contrasts were found in them,

Hello I have 2 data frames DF1 and DF2 where DF2 is a subset of DF1: DF1= data.frame(V1=1:6, V2= letters[1:6]) DF2= data.frame(V1=1:3, V2= letters[1:3]) How do I create a new data frame of the difference between DF1 and DF2 newDF=data.frame(V1=4:6, V2= letters[4:6]) In my real data, the rows are not in order as in the example I provided. Thanks much Joseph [[alternative HTML version