search for: numdf

...e ~ day + stereotypy, + random = ~ 1 | bear, data = learning, family = binomial) > fm2 <- glmmPQL(choice ~ day + envir + stereotypy, + random = ~ 1 | bear, data = learning, family = binomial) Individually, I get results from anova(): > anova(fm1) numDF denDF F-value p-value (Intercept) 1 2032 7.95709 0.0048 day 1 2032 213.98391 <.0001 stereotypy 1 2032 0.42810 0.5130 > > anova(fm2) numDF denDF F-value p-value (Intercept) 1 2031 5.70343 0.0170 day 1 2031 213.21673 <.0...

Hi I am working on a Nested one-way ANOVA. I don't know how to implement R code to test the significance of the random factor My R code so far can only test the fixed factor : anova(lme(PCB~Area,random=~1|Sites, data = PCBdata)) numDF denDF F-value p-value (Intercept) 1 12 1841.7845 <.0001 Area 1 4 4.9846 0.0894 Here is my data and my hand calculation. > PCBdata Area Sites PCB 1 A 1 18 2 A 1 16 3 A 1 16 4 A 2 19 5 A 2 20 6 A 2 19 7...

...ation in a nested design, OK? I need to know: the species are different in proportion? the size affect the species's proportion? existe interaction between size and species? I make the analysis. > m.lme <- lme(nsp/tot~size*specie,random=~1|size/specie) > anova(m.lme) numDF denDF F-value p-value (Intercept) 1 16 374.7121 <.0001 size 1 2 37.8683 0.0254 specie 1 2 18.2036 0.0508 size:specie 1 2 9.3203 0.0926 > This is the correct mean to make this analysis? or > m.lme <- lme(nsp/tot~size*specie,random=~...

...655 > #a friend told me that it is possible to do multiple comparisons for lme in a simplest way, i.e. : > anova(lm1,L=c("treatmentcontrol"=1,"treatmentAl200"=-1)) F-test for linear combination(s) treatmentAl200 treatmentcontrol -1 1 numDF denDF F-value p-value 1 1 12 2.538813 0.1371 > anova(lm1,L=c("treatmentcontrol"=1,"treatmentAl400"=-1)) F-test for linear combination(s) treatmentAl400 treatmentcontrol -1 1 numDF denDF F-value p-value 1 1 12 17.30181 0....

...ing Subjects within Experiments, but it is expected to have much slower RT (reaction time) in the second experiment, since the task is more complex, so it would not make much sense. That is why I kept analyses separated: (A) lme(RT ~ F2 + MI, random =~ 1 | Subject, data = exp1) ANOVA: numDF denDF F-value p-value (Intercept) 1 1379 243012.61 <.0001 F2 1 1379 47.55 <.0001 MI 1 1379 4.69 0.0305 Fixed effects: RT ~ F2 + MI Value Std.Error DF t-value p-value (Intercept) 6.430962 0.03843484 1379 167.32118 0.0000 F2...

..."terms" "residuals" [4] "coefficients" "aliased" "sigma" [7] "df" "r.squared" "adj.r.squared" [10] "fstatistic" "cov.unscaled" x$fstatistic value numdf dendf 72.04064 1.00000 31.00000 But can not find the p value of F statistics. Thanks Ted -- View this message in context: http://www.nabble.com/extract-the-p-value-of-F-statistics-from-the-lm-class-tp22891475p22891475.html Sent from the R help mailing list archive at Nabble.com.

...day + stereotypy, + random = ~ 1 | bear, data = learning, family = binomial) > fm2 <- glmmPQL(choice ~ day + envir + stereotypy, + random = ~ 1 | bear, data = learning, family = binomial) Individually, I get results from anova(): > anova(fm1) numDF denDF F-value p-value (Intercept) 1 2032 7.95709 0.0048 day 1 2032 213.98391 <.0001 stereotypy 1 2032 0.42810 0.5130 > > anova(fm2) numDF denDF F-value p-value (Intercept) 1 2031 5.70343 0.0170 day 1 2031 213.21673 <....

...Value Std.Error DF t-value p-value (Intercept) 24.937897 6.662475 11 3.743038 0.0032 kjday 0.108143 0.152540 7 0.708945 0.5013 treat3 -1.506605 0.485336 7 -3.104254 0.0172 #generating an anova table to get the F-distribution > anova(incub.lme2) numDF denDF F-value p-value (Intercept) 1 11 1176.6686 <.0001 kjday 1 7 5.7060 0.0483 treat 1 7 9.6364 0.0172 > -- View this message in context: http://www.nabble.com/F-distribution-from-lme%28%29--tf4729757.html#a13524346 Sent from the R help mailing...

...ackages lme4 and nlme, more specifically in the denominator degrees of freedom. I used data Orthodont for the two packages. The commands used are below. require(nlme) data(Orthodont) fm1<-lme(distance~age+ Sex, data=Orthodont,random=~1|Subject, method="REML") anova(fm1) numDF DenDF F-value p-value (Intercept) 1 80 4123.156 <.0001 age 1 80 114.838 <.0001 Sex 1 25 9.292 0.0054 The DenDF for each fixed effect is 80, 80 and 25. Using the package lme4: require(lme4) data(Orthodont) fm2<-lme(distance~age+...

...controls for the two other lines so I have set the following contrasts for lines: [,1] [,2] [,3] 18 1 0 1 25 -1 0 1 l 0 1 -1 s 0 -1 -1 If I do the following: mod1<-lme(area ~ line * temp, random = ~1|replicate/temp, mydata) anova(mod1) I get: numDF denDF F-value p-value (Intercept) 1 336 41817.83 <.0001 line 3 8 14.38 0.0014 temp 1 8 338.21 <.0001 line:temp 3 8 0.62 0.6211 I have a significant effect of selection line. Eyeballing the interction.plot, it is clear the the line...

Hi, I am using the lme package to fit mixed effects models to a set of data. I am having a difficult time understanding the *meaning* of the numDF (degrees of freedom in the numerator), denDF (DF in the denomenator), as well as the Intercept term in the output. For example: I have a groupedData object called 'Soil', and am fitting an lme model as follows: ## fit a simple model # errors partitioned among replicates fit1 <- lme...

...icted scores (no error) proc mixed data = dataSet; class treat group; model y = treat*time ; random intercept / subject=group(treat); PARMS (2.1) (1.2) / NOITER; contrast 'slopes' treat*time 1 -1 0,treat*time 1 0 -1; ods output contrasts=c; run; data dataSet; set c; alpha=0.05; ncparm=numdf*fvalue; fc=finv(1-alpha,numdf,dendf,0); power=1-probf(fc,numdf,dendf,ncparm); run; proc print;run; cheers, Wilfried Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm

Hi all I tried to reproduce an example with lme and used the Orthodont dataset. library(nlme) fm2a.1 <- lme(distance ~ age + Sex, data = Orthodont, random = ~ 1 | Subject) anova(fm2a.1) > numDF denDF F-value p-value > (Intercept) 1 80 4123.156 <.0001 > age 1 80 114.838 <.0001 > Sex 1 25 9.292 0.0054 or alternatively (to get the same result) fm2a.2 <- lme(distance ~ age + Sex, data = Orthodont, random = list(Subject = ~ 1)) a...

...7.63 20.576 0.0002855 *** Residuals 10 426.53 42.65 --- Signif. codes: 0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1 Error: Within Df Sum Sq Mean Sq F value Pr(>F) Residuals 36 33.287 0.925 > anova(lme(fixed=score~Machine,random=~1|Worker/Machine,data=Machines)) numDF denDF F-value p-value (Intercept) 1 36 773.5709 <.0001 Machine 2 10 20.5762 3e-04 No problem here: the results are essentially the same, which is expected. Now I turn to an ANCOVA with a random grouping factor. > data(Orthodont) > OrthoFem <- Orthodont[Orthod...

...0.554146 0.43 The usual conclusion would be that the two models are equivalent and to keep the null model for parsimony (!). However, an anova shows that the variable 'log(1e-04 + transat)' is significantly different from 0 in model 2 (lmmedt9) > anova(lmmedt9) numDF denDF F-value p-value (Intercept) 1 20 289.43109 <.0001 log(1e-04 + transat) 1 20 31.18446 <.0001 Has anyone an opinion about what looks like a paradox here ? Patrick

...siduals 40 881875 22047 I have tried the following lme function to specify that Site is random: > lme1 <- lme(sp~Location, random=~1|Site, data=mavric) > lme2 <- lme(sp~Location, random=~1|Location/Site, data=mavric) > anova(lme1) numDF denDF F-value p-value (Intercept) 1 40 3.418077 0.0719 Location 4 5 1.152505 0.4294 This gives me the correct F-value for Location from MSLocation/MSLocation:Transect, but the p-value doesn't seem to be correct (by my calculations in Microsoft Excel it should be 0.345...

...- I start by assuming that the appropriate denominator degrees lies between n - p and and n - q, where n=number of observations, p=number of fixed effects (rank of model matrix X), and q=rank of Z:X. - I then conclude that good estimates of P values on the F ratios lie between 1 - pf(F.ratio, numDF, n-p) and 1 - pf(F.ratio, numDF, n-q). - I further surmise that the latter of these (1 - pf(F.ratio, numDF, n-q)) is the more conservative estimate. When I use these criteria and compare my "ANOVA" table to the results of analysis of Helmert contrasts using MCMC sample with highest...

...24 0.19 0.848 bstime:typeSnow Cap 0.9 0.4 24 2.25 0.034 However in Milliken & Johnson all df are 23. Values (estimates) are almost identical, but there are some small differences in SE and t. Using anova(LME.1) I obtain numDF denDF F-value p-value type 6 0 18.19 NaN bstime:type 6 24 4.04 0.0061 but in the book it is: numDF denDF F-value p-value type...

...says: When only one fitted model object is present, a data frame with the sums of squares, numerator degrees of freedom, denominator degrees of freedom, F-values, and P-values The output of fm1 <- lme(distance ~ age, data = Orthodont) # random is ~ age anova(fm1) gives columns numDF denDF F-value p-value -- i.e. the sums of squares aren't there! (For fairly good reasons; lme doesn't actually compute them internally, and it might not always be straightforward to compute them, for more complex models. They would mostly be useful for comparison with simpler, method-of...

...e to use. I also have some missing values as NA as part of the dataset. I have also set Subj, Region, and Call as factors using as.factor. I have run the model which I am sure is wrong: model1<-anova(lme(egr ~ Trt* Region, random=~1|Subj/Region,na.action=na.omit)) The output I am getting is: numDF denDF F-value p-value Min. : 1.00 Min. : 32 Min. : 2.891 Min. :0.000e+00 1st Qu.: 1.75 1st Qu.:197 1st Qu.: 12.120 1st Qu.:0.000e+00 Median : 6.50 Median :252 Median : 20.275 Median :1.146e-05 Mean : 9.00 Mean :197 Mean : 87.353...