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...64 Thanks, Jeff Jeffrey. M. Miller, PhD Statistics & Data Anaysis Consultant CEO - AlphaPoint05, Inc <http://maps.yahoo.com/py/maps.py?Pyt=Tmap&addr=2792+SE+27th+Ave&csz=Gainesv ille%2C+FL+32641&country=us> 2792 SE 27th Ave Gainesville, FL 32641 <mailto:jeffmiller@alphapoint05.net> jeffmiller@alphapoint05.net <http://www.alphapoint05.net/> www.alphapoint05.net mobile: 352-359-6611 ************************************************************************ ********************************** This E-mail message is confidential, intended only for th...

Does anyone know how to obtain a plot of residuals by predicted values for a main-effects aov? I want to check that the residuals are distributed equally across treatment means. Thanks, Jeff [[alternative HTML version deleted]]

I thought more about this last night after my email about putting an end to "Read the Manual" replies. It seems that there are a handful of R Super-Programmers. If I were one, I would get tired of 100's of questions like "How can I get my Excel file into R!!!". Concurrently, R is experiencing an exponential increase in new users to the extent that there are now

Hi all, Does anyone know if an R function for the Mantel-Haenszel chi-square for a 2x2 table exists? I’ve also seen it called the randomization Q statistic. Note that I’m not looking for the Cochran-Mantel-Haenszel…I did see that out there as cmh.test. Thanks in advance, Jeff Internal Virus Database is out-of-date. Checked by AVG Free Edition. 11:27 AM [[alternative HTML version

Hi all, I hope someone can help me with this. Suppose I import a text file and one of the columns looks like this: New York New York England Spain Spain Orlando New York England France I want to add a variable that is based on the previous one US US Europe Europe Europe US US Europe Europe How do that? Also, I would like to be able to export the data as

Hi all, Suppose I have a column vector of 600 measurements taken in 1s intervals. What I want is a new vector with the averages for each min (so there would be 10 entries). Is there an efficient way to do this? I’ve been doing it with a ‘for’ loop but something tells me there is a simple command that is more efficient. Jeff Internal Virus Database is out-of-date. Checked by

Does anyone know of an existing R package or code to run a mixed Hurdle model? I found glmmADMB, but that seems to be ZIP. Any recommendations? Thanks, Jeff [[alternative HTML version deleted]]

Hi all, Cameron and Trivedi in their 1998 Regression Analysis of Count Data refer to NB1 and NB2 NB1 is the negative binomial model with variance = mu + (alpha * mu^1) yielding (1+alpha)*mu NB2 sets the power to 2; hence, variance = mu + (alpha*mu^2) I think that NB2 can be requested via negbin2<-glm(hhm~sex+age,family=quasi(var="mu^2",link="log")) Is

Has anyone successfully linked Mplus and R and have code on how to do it? It would be great to use R to simulate datasets and then ship to Mplus for SEM analysis. I know one researcher who does this quite regularly but I can’t get ahold of him right now. Sincerely, Jeff Miller **************************************** Jeffrey M. Miller, PhD HYPERLINK

I just a ran a fisher.test on a 9x5 table and received the following message Error in fisher.test(apaslg.t) : FEXACT error 30. Stack length exceeded in f3xact. This problem should not occur. I agree. The problem should not occur. I did some searching but only found other reports of the problem. Does anyone know of a fix? Jeff Internal Virus Database is out-of-date.

I posted this last night but I think I figured out the problem. I checked on the underlying equation for the Fisher Exact Test for tables greater than 2X2. It looks like I have an insane amount of factorials to be multiplied. Is that problem? Is it an overflow issue? Jeff Last night’s post: I just a ran a fisher.test on a 9x5 table and received the following message

Hi all, This should be easy, but I can't seem to figure it out. I have a table like this named newtable a1 a2 a3 a4 Cnts Score 1 1 0 0 4 3.28 1 0 1 1 2 2.63 I want the following: a1 a2 a3 a4 Cnts Score 1 1 0 0 4 3.28 1 1 0 0 4 3.28 1 1 0 0 4 3.28 1 1 0 0

Does anyone know if the discriminant coefficients in lda are standardized? Thanks, Jeff [[alternative HTML version deleted]]

This seems like it's easy but I can't seem to get it. My equation is (a^b)/c = d. I know that I can solve for b via log(d*c) / log(a) = b However, I don't really have d in the first place I have d/sum(d) values...no d's...just d/sum(d) values that all add to 1.0 So, for example, if my data is a=4,9,12,3 b=unknown c=12,18,13,20 d/sum(d)=43,45,23,18 I need an