similar to: lme() F-values disagree with aov()

I don't have Kirk but with your data in file "Kirk.txt" in the working directory, here is how I would do it. --- > Kirk <- read.table("Kirk.txt") > names(Kirk) [1] "subject" "A" "B" "C" "response" > for(j in 1:4) Kirk[,j] <- factor(Kirk[,j]) ### Check this! > fm <- aov(response ~

##I want to remove the rows where the row sums are zero and this is as far as I have gotten ffg <- (structure(list(CD = c(0, 0, 0, 0, 3.125, 0, 0, 0, 0, 1.6, 3.125, 0, 0, 6.25, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3.125, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1.6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1.6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3.125, 0, 0, 0, 0, 0, 0, 0, 0,

Hi all Consider the classic data below from Darwin on the heights of 15 pairs of zea mays (corn) plants either cross-fertilized or self-fertilized, where the goal is to see if it makes a difference. > head(ZeaMays) pair pot cross self diff 1 1 1 23.500 17.375 6.125 2 2 1 12.000 20.375 -8.375 3 3 1 21.000 20.000 1.000 4 4 2 22.000 20.000 2.000 5 5 2 19.125

station month bas 190 5 0.000 190 7 1.563 190 10 0.000 190 11 0.000 202 4 18.750 202 5 18.750 202 7 6.250 202 10 4.800 202 11 3.125 198 4 18.750 198 5 31.250 198 7 3.125 198 10 3.200 198 11 12.500 205 4 0.000 205 5 0.000 205 7 0.000

I have a SUSE 9.2 box running 3.0.3-testing-rc3, and dom0 and domU disagree on the cpu frequency. dom0 has two CPUs and domU one vcpu. dom0 says: model name : Intel(R) Pentium(R) III CPU family 1400MHz stepping : 1 cpu MHz : 1558.703 bogomips : 2800.85 domU says: model name : Intel(R) Pentium(R) III CPU family 1400MHz stepping : 1 cpu MHz

Full_Name: Paul Bailey Version: 2.3.1 OS: OS X (10.4.6) Submission from: (NULL) (128.135.133.123) according to cat's help file, "'cat' converts numeric/complex vectors in the same way as 'print' (and not in the same way as 'as.character' which is used by the S equivalent), so 'options' '"digits"' and '"scipen"' are

The tone of your email makes it look like you think the exact output of cat and print in these instances is irrelevant and not worth thinking about and that the documentation is `close enough,' or up to the (perhaps implied) R documentation standard. This is a reasonable stance. If you hold it, please just circular file my bug report as irrelevant and let's not waste any more time on it.

I have a situation where the parameter estimates from lrm identify a binary predictor variable ("X") as clearly non-significant (p>0.3), but the ANOVA of that same model gives X a chi^2-df rank of > 200, and adjudicates X and one interaction of X and a continuous measure as highly significant. The N is massive and X has two categories, each with > 100,000 observations.

1 4 499492 150392 4496 4763380 0 0 192 552 1227 1094 2 0 75 24 0 0 5 499492 150656 4500 4763528 0 0 0 160 465 263 1 0 68 30 0 0 5 499492 150468 4500 4763532 0 0 0 0 177 93 1 0 69 31 0 1 5 499492 151020 4500 4763540 0 0 0 0 160 132 0 0 69 31 0 1 5 499492 151268 4500 4763540 0 0 0 0 304 143 1 0 69

Dear all, I have the following output generated by linear regression. Since there is only one regression intercept and one slope for one set of data, what is the meaning of std. error for intercept and that of slope? Thanks in advance. Sincerely, Minghua > data(thuesen) > attach(thuesen) > lm(short.velocity~blood.glucose) Call: lm(formula = short.velocity ~ blood.glucose)

Hi, I have a problem where Dovecot and its LDA seem to disagree on what character encoding should be used in Maildir names. I use Dovecot and Postfix on a Debian server. Mail is delivered by Dovecot LDA (called from .forward) and sorted according to a Sieve script into a number of Maildir folders in my home directory. Dovecot serves the mail over IMAP and I read it in Kmail. I originally

I am sure I am opening myself up to looking stupid, but I have two samples with medians of 613.5 and 189 (difference in location of 424 compared to the difference suggested from the wilcoxon of 291.5) > wilcox.test(pipwtCount,pipwdCount, conf.int=TRUE, na.rm=TRUE) Wilcoxon rank sum test data: pipwtCount and pipwdCount W = 822, p-value = 0.01227 alternative hypothesis: true location

I've noticed that glm and lrm give extremely different results if you attempt to fit a saturated model to a dataset with zero cells. Consider, for instance the data from, Agresti's Death Penalty example [0]. The crosstab table is: , , PENALTY = NO VIC DEF BLACK WHITE BLACK 97 52 WHITE 9 132 , , PENALTY = YES VIC DEF BLACK WHITE BLACK 6 11

I've posted the following to R-help on May 15. It has reproducible R code for real data -- and a real (academic, i.e unpaid) consultion background. I'd be glad for some insight here, mainly not for myself. In the mean time, we've learned that it is to be expected for anova(*, "marginal") to be contrast dependent, but still are glad for advice if you have experience. Thank

I have been using lme from nlme to do a 3-way anova with all the effects treated as random. I was wondering if someone could direct me to an example of how to do this using lmer from lme4. I have 3 main effects, tim, trt, ctr, and all the interaction effects tim*trt*ctr. The response variable is ge. Here is my lme code: dat <-

Hi R list, I imported values from Excel, there is a column with numbers like 45%, 65%, 12%. I want to find its mean. What should I use? strisplit() split() parse() Data from dput(), structure(c(78L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("", "-0.15%", "-0.34%", "-1.3%", "-10.77%", "-100.00%", "-11.45%",

Dear R-help group, How can I define a lme with 3 factors(a,b,c), where c is nested in b, and a is crossed with b/c? I think that: lme(response ~ ..., data = Data, random = pdBlocked(list(pdIdent(~ a - 1), pdIdent(~ b - 1)))) is one part of the answer and: lme(response~..., data=Data, random=~1|b/c) is the other part of the answer but how can I combine them?? Could anybody please help

Hello all: I am interested in computing what the multilevel modeling literature calls a multiple membership model. More specifically, I am working with a data set involving clients and providers. The clients are the lower-level units who are nested within providers (higher-level). However, this is not nesting in the usual sense, as clients can belong to multple providers, which I understand

#this is my stab at - I am sure that I am missing something. If this doesn't make sense then please ask for more details. #This may show my low level of programing knowledge hester. <- c(1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4) value <- rnorm(16) x <- data.frame(value, hester.) z <- (if(x[,"hester."]==1){ x[,"value"]*6.250} else if(x[,"hester."]==2){

I''ve run into an odd one on one of my new RHEL 6 boxes. Puppet and Facter seem to disagree about the value of a fact. Puppet version: 2.7.9 Facter version: 1.6.4 Module: puppetlabs-apache apache::params falls through its if structure: if $::osfamily == ''redhat'' or $::operatingsystem == ''amazon'' { ... } elsif $::osfamily ==