similar to: chisq.test

Hi I am running chisq as below and getting a warning. Can anyone tell me the significance or the warning? > chisq.test(c(10 ,4 ,2 ,6 ,5 ,3 ,4 ,4 ,6 ,3 ,2 ,2 ,2 ,4 ,7 ,10 ,0 ,6 ,19 ,3 ,2 ,7 ,2 ,2 ,2 ,1 ,32 ,2 ,3 ,10 ,1 ,3 ,9 ,4 ,10 ,2 ,2 ,4 ,5 ,7 ,6 ,3 ,7 ,4 ,3 ,3 ,7 ,1 ,4 ,2 ,2 ,3 ,3 ,5 ,5 ,4 ), p =c(0.01704142 ,0.017988166 ,0.018224852 ,0.017751479 ,0.017988166 ,0.018224852 ,0.017278107

I want to make a glm and then use predict. I have a fairly small sample (4000 cases) and I want to train on 90% and test on 10% but I want to do it in slices so I test on every 10th case and train on the others. Is there some simple way to get these elements? Stephen -- 21/10/2005 [[alternative HTML version deleted]]

Hi Say I have some data, two columns in a table being a binary outcome plus a predictor and I want to plot a graph that shows the percentage positives of the binary outcome within bands of the predictor, e.g. Outcome predictor 0 1 1 2 1 2 0 3 0 3 0

Hi I am trying to do this: chisq.test(c(11, 13, 12, 18, 21, 43, 15, 12, 9, 10, 5, 28, 22, 11, 15, 11, 18, 28, 16, 8, 15, 19, 44, 18, 11, 23, 15, 23, 2, 5, 4, 14, 3, 22, 9, 0, 6, 19, 15, 32, 3, 16, 14, 10, 24, 16, 24, 31, 29, 28, 16, 26, 11, 11, 4, 17, 16, 13, 20, 26, 16, 19, 34, 19, 17, 14, 22, 25, 17, 12, 23, 14, 19, 30, 18, 10, 23, 21, 17, 16, 10, 14, 6, 17, 17, 10, 21, 25, 20, 4, 11, 4,

Hi I am using R for logistic regression and finding it very useful. However, I wondered if anyone could point me to any course or notes on this subject using R. All help most welcome. Stephen -- Internal Virus Database is out-of-date. Checked by AVG Anti-Virus. [[alternative HTML version deleted]]

Hi I can get all my features by doing this: > logistic.model = glm(similarity ~ ., family=binomial, data = cData[3001:3800,]) I can get the product of all my features by this: logistic.model = glm(similarity ~ . ^ 2, family=binomial, data = cData[3001:3800,]) I don't seem to be able to get polys by doing this: logistic.model = glm(similarity ~ poly(.,2), family=binomial, data

I am trying to do some verification across a large dataset, cuData, that has 23 columns. Column 23 (similarity) is the outcome 0 or 1 and the other columns are the features. I do this: verificationglm.model <- glm(formula = similarity ~ ., family=binomial, data=cuData[1:1000,]) and produce the model: > summary(verificationglm.model) Call: glm(formula = similarity ~ ., family =

Hi, I have a data set with 999 observations, for each of them I have data on four variables: site, colony, gender (quite a few NA values), and cohort. This is how the data set looks like: > str(dispersal) 'data.frame': 999 obs. of 4 variables: $ site : Factor w/ 2 levels "1","2": 1 1 1 1 1 1 1 1 2 2 ... $ gender: Factor w/ 2 levels "0","1":

Hi I am trying to do a large glm and running into this message. Error: cannot allocate vector of size 3725426 Kb In addition: Warning message: Reached total allocation of 494Mb: see help(memory.size) Am I simply out of memory (I only have .5 gig)? Is there something I can do? Stephen [[alternative HTML version deleted]]

Is there some decision tree method available with R, like ripper, that ends up producing a list of the rules and can be used for prediction? Stephen Choularton 02 9999 2226 0413 545 182 -- 5:40 PM [[alternative HTML version deleted]]

Hi I am just starting to look at R and trading in futures, stock, etc Can anyone point me to useful background material? Stephen Choularton 02 9999 2226 0413 545 182 -- 11:39 PM [[alternative HTML version deleted]]

Hi I am looking for a couple of pointers using glm (family = binary). 1. I want to add all the products of my predictive features as additional features (and I have 23 of them). Is there some easy way to add them? 2. I want to drop each feature in turn and get the most significant, then drop two and get the next most significant, etc. Is there some function that allows me to do this?

I am working with a largish dataset of 25k lines and I am now tying to use predict. pred = predict(cuDataGlmModel, length + meanPitch + minimumPitch + maximumPitch + meanF1 + meanF2 + meanF3 + meanF4 + meanF5 + ratioF1ToF2 + rationF3ToF1 + jitter + shimmer + percentUnvoicedFrames + numberOfVoiceBreaks + percentOfVoiceBreaks + meanIntensity + minimumIntensity + maximumIntensity +

Hi Is there some function R that multiplies each coefficient by the standard deviation of the corresponding variable and produces a ranking? Stephen -- No virus found in this outgoing message. Checked by AVG Anti-Virus. [[alternative HTML version deleted]]

Hi I am trying to use randomForest for classification. I am using this code: > set.seed(71) > rf.model <- randomForest(similarity ~ ., data=set1[1:100,], importance=TRUE, proximity=TRUE) Warning message: The response has five or fewer unique values. Are you sure you want to do regression? in: randomForest.default(m, y, ...) > rf.model Call: randomForest(x = similarity ~ .,

Hello, What is the best way to get ranks for a vector of values, limit the range of rank values and create equal count in each group? I call this uniform ranking...uniform count/number in each group. Here is an example using three groups: Say I have values: x = c(3, 2, -3, 1, 0, 5, 10, 30, -1, 4) names(x) = letters[1:10] > x a b c d e f g h i j 3 2 -3 1 0 5 10 30 -1 4 I

Dear All I am doing the following: > x <- yacas("3/2") > for (i in 2:400) + x <- yacas(paste(x,"*",x)) > x expression(Inf^1.260864167e+117/Inf^6.304320836e+116) > Eval(x) [1] NaN No luck this way. However, I am successful with > y <- yacas("(3/2)^400") > y expression(7.05507910865533e+190/2.58224987808691e+120) > Eval(y) [1]

Full_Name: Tao Shi Version: 1.7.0 OS: Windows XP Professional Submission from: (NULL) (149.142.163.65) > x [,1] [,2] [1,] 149 151 [2,] 1 8 > c2x<-chisq.test(x, simulate.p.value=T, B=100000)$p.value > for(i in (1:20)){c2x<-c(c2x,chisq.test(x, simulate.p.value=T,B=100000)$p.value)} > c2tx<-chisq.test(t(x), simulate.p.value=T, B=100000)$p.value > for(i in

I'm using R1.7.0 runing with Win XP. Thanks, ...Tao ???????????????????????????????????????????????????????? >x [,1] [,2] [1,] 149 151 [2,] 1 8 >t(x) [,1] [,2] [1,] 149 1 [2,] 151 8 >chisq.test(x, simulate.p.value=T, B=100000) Pearson's Chi-squared test with simulated p-value (based on 1e+05 replicates) data: x X-squared = 5.2001, df =

Dear all, I would like to do a goodness-of-fit test on my data to see if they follow a mixture of 2 poisson distributions. I have small numbers for observed values. Most of them <5. The chisq.test gives warning message: Chi-squared approximation may be incorrect in: chisq.test(x , p = prob). However, the option sim=TURE would suppress the warning message. Does that mean with the option