search for: 2.2e

Dear R users, I am using plm packege in R to build my model, but from the result I can't quite figure out what it is... Can anyone tell me why? Am I missing something? R Results: *> ar1<-plm(formula=ADOP~lag(ADOP,1)+PE+WOR, + data=well, effect="time",model="within") > summary(ar1)* Oneway (time) effect Within Model Call: plm(formula = ADOP ~ lag(ADOP, 1) + PE

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.

##I am trying to test for fixed factor main effects in an unbalanced mixed effects model but when I fit the reduced model for "mic" factor effects, the extra degrees of freedom are being allocated to a nested term rather than the residuals. The model has inc, mic and spp are independent variables and vial nested within spp. inc and spp are already coded as factors since they were

In an effort to select the most appropriate number of clusters in a mixture analysis I am comparing the expected and actual membership of individuals in various clusters using the Fisher?s exact test. I aim for the model with the lowest possible p-value, but I frequently get p-values below 2.2e-16 and therefore does not get exact p-values with standard Fisher?s exact tests in R. Does anybody know

Hello, I'm trying to test my data for normality. I enter the data (95ish species counts) run >ks.test (data,pnorm) and get a p- value <2.2e-16 But this seems to be the p-value no matter what the data I enter. (I have multiple datasets and am testing them all for normality). [Actually, I just entered a vector of 1's and the p-value changed.] When I use the >Shapiro.test command,

I'm running an ANOVA on some data for respiration in a forest. I am having a problem with my degrees of freedom. For one of my variables I get one fewer degrees of freedom than I should. I have 12 plots and I therefore expected 11 degrees of freedom, but instead I got 10. Any ideas? I have some code and output below: > class(Combined.Plot) [1] "character" >

Hi, I'm still trying to figure out that GLM procedure in SAS. Let's start with the simple example: PROC GLM; MODEL col1 col3 col5 col7 col9 col11 col13 col15 col17 col19 col21 col23 =/nouni; repeated roi 6, ord 2/nom mean; TITLE 'ABDERUS lat ACC 300-500'; That's the same setup that I had in my last email. I have three factors: facSubj,facCond and facRoi. I had this pretty

>>>>> Martin Maechler >>>>> on Tue, 8 Sep 2020 10:40:24 +0200 writes: >>>>> Hugh Parsonage >>>>> on Tue, 8 Sep 2020 18:08:11 +1000 writes: >> I can only reproduce on Windows, but reliably (both 4.0.0 and 4.0.2): >> $> R --vanilla >> x <- c(0L, -2e9:2e9) >> # > Segmentation

I calculate an anova test in the following way: expdata<-read.table("/home/dorien/UA/meta-music/optimuse/optimuse1-build-desktop/results/results_processedCP", header=TRUE)

Hi, For further calculations I need to extract indivdual Variances of different random effects from a fitted model. I found out how to extract the correlations (VarCorr(m1)@reSumry$group1) but I was not able to find a way to extract the other components individually. To extract the Residuals I tried: (ranef(m1)@ stdErr) which unfortunately did not work. Thank you very much for your help!

Dear all How does print.htest display the p-value in scientific notation? > (x <- cor.test(iris[[1]], iris[[3]])) Pearson's product-moment correlation data: iris[[1]] and iris[[3]] t = 21.65, df = 148, p-value < 2.2e-16 alternative hypothesis: true correlation is not equal to 0 95 percent confidence interval: 0.8270 0.9055 sample estimates: cor 0.8718 Above the p-value comes

Dear all, thanks for your feedback so far. With the help of a colleague I think I found the solution to my problem: > pt(10,100,lower=FALSE) [1] 4.950844e-17 IS *NOT* EQUAL TO > 1-pt(10,100,lower=TRUE) [1] 0 This means that R is capable of providing p-values < 2.2e-16, however, if the value is used in a substraction or addition then the default value of the machine epsilon

On Tue, 8 Sep 2020, Hugh Parsonage wrote: > Thanks Martin. On further testing, it seems that the segmentation > fault can only occur when the amount of obtainable memory is > sufficiently high. On my machine (admittedly with other processes > running): > > $ R --vanilla --max-mem-size=30G -e "x <- c(0L, -2e9:2e9)" > Segmentation fault > > $ R --vanilla

>>>>> luke-tierney >>>>> on Tue, 8 Sep 2020 09:42:43 -0500 (CDT) writes: > On Tue, 8 Sep 2020, Martin Maechler wrote: >>>>>>> Martin Maechler >>>>>>> on Tue, 8 Sep 2020 10:40:24 +0200 writes: >> >>>>>>> Hugh Parsonage >>>>>>> on Tue, 8 Sep 2020

Hi, I 'm a novice user of R statistics and my hands-on experience with it is minimal. I want to create a table for my MBA course assignment that looks like the ones that SPSS and MS Excel produces ,the data that the table has to include are the following : > table(agec) agec 1 2 3 749 160 32 > x=table(agec) > x agec 1 2 3 749 160 32 > > prop.table(x) agec

Hi,  I have a problem with the output of my anova. > Tabelle2 <- data.frame(sb_ek_ber, JE, ABWHALT_C, ALLEINF, Alter_Ältester_inklVN_C, Alter_Jüngster_C_inkl_AlterNutz, ALTERKAU_C, Wuerfel, BEGKUNDE_C, Geoscore_C, GESCHL_VN, +  NUTZART, NUTZKREIS, NICHTK_C_korrigiert, TARIFDAT_C, TARIFGRP_C, WOHNEIGEN, WOHN_C, ZWEITFAHRZ, ZAHLWEISE, LASTSCHR)  str(Tabelle2) 'data.frame':   1237838

I have a stupid question on how to get the real p-values for wilcox.test and correlation. the minmun can be reached is 2.2E-16 using the R version 2.6.2. I do not think it is the R version causing this but other issues. Any help is highly appreciated. Charles -- View this message in context: http://www.nabble.com/minmun-p-value-for-wilcox.test-and-correlation-tp23637568p23637568.html Sent

Hello, I am hoping someone can clarify why I might obtain a quite different value in R & SPSS for a McNemar test I ran. Firstly, here is the R syntax & output R OUTPUT > mctest <- as.table(matrix(c(128,29,331,430), + ncol =2, dimnames = list(group=c("preMHT","postMHT"), + assault=c("yes","no")))) > mctest assault

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

Sometimes the p.value returned by t.test() is the same that I calculate using pt() and sometimes it's not. I don't understand the difference. I'm sure there is a simple explanation but I haven't been able to find it, even after looking at the code for t.test.default. I apologize if this is a basic and obvious question. For example: > data(sleep) >