similar to: Box's M test

Dear List I am looking for a script that will calculate the Box M test to test the homogeneity of the variance/covariance matrix between two matrices. If anyone could send me the script I would appreciate it. I am aware of the scepticism about this test, where due to extreme sensitivity a p value of 0.01 is recommended. Despite this however Box's M test is the established

I have a large dataset organised in choice groups see below. Each choice represents a separate occasion with 1 product chosen out of the 6 offered.            +-------------------------------------------------------------------------------------------------+      | dn   obs   choice      acid   br                 date       cdate   situat~n   mth   year   set |     

Hopefully a quick question: is there a package/routine to perform Box's M (like proc discrim in SAS) or some other test for whether two multivariate samples (presumed multivariate normal) have the same v-c matrix? I've looked, but can't find it. If not, I can certainly code it up myself (and perhaps submit it for inclusion in the mvtnormtest package), but I wanted to check if it

On 27/10/2017 8:10 AM, Morkus via R-devel wrote: > Trying to get past a frustrating error to do a "simple" Box's M test using Java. > > The Box's M test says it will work with a data.frame. > > Here's the setup code: > > REXP myDf = REXP.createDataFrame(new RList( > new REXP[] > { > new REXPDouble(d1), > new REXPDouble(d2), > new

Hi all, Is there a box's M test for R? I have looked around, but have been unable to find it. Thanks Wade

Trying to get past a frustrating error to do a "simple" Box's M test using Java. The Box's M test says it will work with a data.frame. Here's the setup code: REXP myDf = REXP.createDataFrame(new RList( new REXP[] { new REXPDouble(d1), new REXPDouble(d2), new REXPDouble(d3), new REXPDouble(d4), new REXPInteger(d5) })); Here's the call: REXP boxMResult =

Just print the string you are asking to R to evaluate. It doesn't make any sense as an R expression. Fix that, and things will work. Duncan Murdoch On 27/10/2017 3:41 PM, Morkus via R-devel wrote: > It can't be this hard, right? I really need a shove in the right direction here. Been spinning wheels for three days. Cannot get past the errors. > > I'm doing something

Does it work if you supply the closing parenthesis on the call to boxM? The parser says the input is incomplete and a missing closing parenthesis would cause that error.. // create a string command with that variable name.String boxVariable = "boxM(boxMVariable [,-5], boxMVariable[,5]"; // try to execute the command... // FAILS with org.rosuda.REngine.Rserve.RserveException: eval

Hey Duncan, Since Java is the #1 language and R is extremely popular, I think the most telling thing is that nobody on the "R-devel" forum (where people do "programming with R") is doing R and Java like I'm doing: calling R from Java and passing data structures. So it appears I'm clearly pushing R somewhere it doesn't want to go. And, the boxM issue is more or

On 28/10/2017 6:26 AM, Morkus wrote: > I'm not sure what you mean. Could you please be more specific? You were trying to eval an expression that you constructed in Java. I was suggesting that before you eval it, you print it. > > If I print the string, I get: *boxM(boxMVariable[, -5], boxMVariable[, 5])* Right, that's what I was suggesting you do. Now you've fixed the

On 28/10/2017 7:12 AM, Morkus wrote: > Thanks Duncan. Awesome ideas! > > I think we're getting closer! > > I tried what you suggested and got a possibly better error... > . > . > . > rConnection.assign("boxMVariable", myDf); > > *String resultBV *= *"str(boxMVariable)"*; *// your suggestion.* > > *RESULTING ERROR:* > >

Dear list I have a response variable coded 0/1 i.e. a binary response. There are 20,000 individual responses that I would like to aggregate into numbers of each category (i.e. 0/1) by group called dn (350 different groups) and by month mth (there are several hundred responses per month. What is the simplest way to perform this operation in R?

While I don't know anything about Box's M test, I googled around and found a Matlab M-file that computes it. Below is my straight-forward translation of the code, without knowing Matlab or the formula (and done in a few minutes). I hope this demonstrates one of Prof. Ripley's point: If you really want to shoot yourself in the foot, you can probably program R to do that for you. [BTW:

Hello. Not an easy question at all, and it has little to do with software, alas! Veeeeeery loosely speaking: if the homogeneity hypothesis is rejected, then, depending on data availability, you may still be able to treat the data like a panel by: a) ignoring the results of the poolability test b) allowing the coefficients to vary. Of course, a) requires some courage while b) requires more

On 28/10/2017 8:59 AM, Morkus wrote: > Hey Duncan, > > Hard to debug? That's an understatement. Eyes bleeding.... > > In any case, I tried all your suggestions. To get "integer" for the > final column, I had to change the code to get integers instead of strings. The last column in iris is actually a factor. That's stored as an S3-classed integer vector

On 29/10/2017 7:26 AM, Morkus wrote: > Thanks Duncan. I can't tell you how helpful all your terrific replies > have been. > > I think the biggest surprise is that nobody appears to be using Java and > R together like I"m trying to do. I suppose it should be a surprise > since there are no books on the subject and almost no technical > documentation other than a

Breslow-Day test A statistical test for the homogeneity of odds ratios. Homogeneity In <javascript:void(0);> systematic reviews homogeneity refers to the degree to which the results of studies included in a review are similar. "Clinical homogeneity" means that, in studies included in a review, the participants, interventions and outcome measures are similar or comparable.

I am testing the homogeneity of variances via bartlett.test and fligner.test. Using the following example, how should I interpret the p-value in order to accept or reject the null hypothesis ? set.seed(5) x <- rnorm(20) bartlett.test(x, rep(1:5, each=4)) Bartlett test of homogeneity of variances data: x and rep(1:5, each = 4) Bartlett's K-squared = 1.7709, df = 4, p-value =

I'm not sure what you mean. Could you please be more specific? If I print the string, I get: boxM(boxMVariable[, -5], boxMVariable[, 5]) From this code: . . . // assign the data to a variable.rConnection.assign("boxMVariable", myDf); // create a string command with that variable name.String boxVariable = "boxM(boxMVariable[, -5], boxMVariable[, 5])";

Dear Group Members, Forgive me if I am a little bit out of subject. I am looking for a good way to test the homogeneity of two variance-covariance matrices using R, prior to a Hotelling T test. Youll probably tell me that it is better to use a robust version of T, but I have no precise idea of the statistical behaviour of my variables, because they are parameters from the harmonics of