similar to: calculate correlation effect size using contrast analysis for an omnibus Chi-square test statistic

Does anybody know how to negate a string in a grep command, i.e., what I need is to return only strings that do NOT contain a second string anywhere in the entire string. for txt <- c("boo","goo","doob","foo","boofoo") I need a grep command that returns strings with "oo" except when "b" is present anywhere. I know that

Consider an analysis of covariance involving age and cohort. The goal is to assess whether the influence of cohort depends upon the age. The simplest case involves data as follows value Age Cohort x1 ????? 1?????? 3 x2?????? 1?????? 4 x3?????? 1?????? 5 x4 ????? 2 ????? 3 x5 ????? 2 ????? 4 x6 ????? 2 ????? 5 etc. Age is a factor. The numeric response variable is value and Cohort is a

I'm using lmer to fit mixed-effect logistic regression models. This is for a small data set. First, I fit a constant: Generalized linear mixed model fit using Laplace Formula: propm ~ (1 | study) Data: inducedSR71507.dat Family: binomial(logit link) AIC BIC logLik deviance 183.7 189.4 -89.84 179.7 Random effects: Groups Name Variance Std.Dev. study (Intercept)

I have a question about stepAIC and extractAIC and why they can produce different answers. Here's a stepAIC result (slightly edited - I removed the warning about noninteger #successes): stepAIC(glm(formula = (Morbid_70_79/Present_70_79) ~ 1 + Cohort + Cohort2, family = binomial, data = ghs_70_79, subset = ghs_70_full),direction = c("backward")) Start: AIC=3151.41

If one estimates a model using multinom, is it possible to perform the omnibus LR test ( the analogue to omnibus F in linear models ) using the output from multinom ? The residual deviance is there but I was hoping I could somehow pull out the deviance based on just using an intercept ? Sample code is below from the CAR book but I wasn't sure how to do it based on that example. Thanks

R 2.9.1 Windows XP I am fitting a random effects ANOVA with two factors Group which has two levels and Time which has three levels: fita<-lme(Post~Time+factor(Group)+factor(Group)*Time, random=~1|SS,data=blah$alldata) I want to get the omnibus significance tests for each factor and the interaction. I believe I can get the omnibus test for the interaction by running the model:

I''ve noticed that the copy of Mongrel installed by the camping-omnibus gem doesn''t work with Ruby 1.8.6. Or to be more specific, cgi_multipart_eof_fix (which Mongrel is dependent upon) doesn''t work: > $ sudo gem install mongrel --source http://code.whytheluckystiff.net > > ERROR: Error installing mongrel: > cgi_multipart_eof_fix requires Ruby version

gem install camping-omnibus --source code.whytheluckystiff.net Considering the number of optional dependencies now in Camping, I''ve put together an eigengem which will load all of the recommended libraries that most devoted campers will want: ActiveRecord, Mongrel, SQLite3, RedCloth and the acts_as_versioned gem. (Those last two are used in the examples.) Stuff like mosquito and

This is very possibly not a question on R. I was under the impression that the argument that gives rise to Fisher's LSD method in ANOVA works in other situations with three-way comparisons too, given that formal logic works the same ("if the omnibus test rejects, only two of the three groups may be equal, and therefore only one hypothesis can be rejected falsely"). However, when I

Hello, I've been running some HLMs using the lme function quite happily; it does what I want and I'm pretty sure I understand it. The issue is that I'm currently trying to estimate a model with a 14-level "nusiance" factor as an independent variable...which makes the output quite ugly. All I'm really interested in is the question of whether these factor as a whole

Daniel, When Group is entered as a factor, and the factor has two levels, the ANOVA table gives a p value for each level of the factor. What I am looking for is the omnibus p value for the factor, i.e. the test that the factor (with all its levels) improves the prediction of the outcome. You are correct that normally one could rely on the fact that the model

Try another console like ALT-F2. See if you have a black screen login prompt. I opened a bug with RedHat for a system that when I went from 6.7 to 6.8 booted but would not start the X server. Problem was in the MACH64 driver. It is missing an external symbol that the X server requires. -----Original Message----- From: CentOS [mailto:centos-bounces at centos.org] On Behalf Of Leroy Tennison

I figured out that the system boots OK and it is possible to login remotely. However the monitor does not start up. Do you know how to fix it? On Mon, Mar 20, 2017 at 3:15 PM, Rosenthal, Shoshana < srosenthal at cfa.harvard.edu> wrote: > You are right it is a Dell PowerEdge. However as I said it stops in the > middle of booting with no message. > > > > On Mon, Mar 20,

I am currently using Samba on several unix servers, and have numerous users accessing them from NT workstation machines. I have my samba security set equal to DOMAIN. And I have successfully added the unix servers to the domain and created the domain accounts. The problem I have is not a constant problem, but it is a major inconvenience. Sometimes, when a user attempts to access the samba share,

I am also using CentOS, but to get a bug fixed, you have to go upstream to RedHat. I think John Pierce is right, I had to change my run level to get the alternate consoles to work. I still have the machine at run level 3 hoping that a bug fix lets me go back to run level 5. Other machines work fine, just the 2 with the MACH64 video on the motherboard will no longer run X. Bob S.

hi all - is there a package or library that contains a function for partitioning the chi-square statistic of an I X J contingency table into its respective independent parts? i looked around for this, but i didn't find anything. perhaps there's another name for this sort of analysis? i know it as "g-squared". thanks, chris. [[alternative HTML version deleted]]

Dear R helpers, >From the lrm( ) model used for binary logistic regression, we used the L.R. model value (or the G2 value, likelihood-ratio chi-squared statistic) to evaluate the goodness-of-fit of the models. The model with the lowest G2 value consequently, has the best performance and the highest accuracy. However our model includes rsc() functions to account for non-linearity. We

Dear R list, From the lrm() binary logistic model we derived the G2 value or the likelihood-ratio chi-squared statistic given as L.R. model, in the output of the lrm(). How can this value be penalized for non-linearity (we used splines in the lrm function)? lrm.iRVI <- lrm(arson ~ rcs(iRVI,5), penalty=list(simple=10,nonlinear=100,nonlinear.interaction=4)) This didn’t work

On Wed, Jan 11, 2017 at 8:13 PM, Mehdi Amini <mehdi.amini at apple.com> wrote: > > Can you elaborate why? I’m curious. > The con of proposal c was that many passes would need to learn about many region intrinsics. With tokens, you only need to teach all passes about tokens, which they should already know about because WinEH and other things use them. With tokens, we can add as

Dear R-ers, I am using getGEO to download expression data from the Gene Expression Omnibus. With default settings, when a file is downloaded and parsed, lots of dotted lines are printed in the terminal, like this: .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... ..........