similar to: chisq warning

Hi I am trying to use this function. Can anyone show me how I would input the following example? Chi-Squared = (40-30)^2 + (20-30)^2 + (30-30)^2 30 30 30 = 3.333 + 3.333 + 0 = 6.666 (p value = 0.036) I want to be able to use different denominators so can you show me how I can do it to accommodate these rather than assuming they are all the

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,

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]]

Full_Name: Reginaldo Constantino Version: 2.8.0 OS: Ubuntu Hardy (32 bit, kernel 2.6.24) Submission from: (NULL) (189.61.88.2) For many tables, chisq.test with simulate.p.value=TRUE gives a p value that is obviously incorrect and inversely proportional to the number of replicates: > data(HairEyeColor) > x <- margin.table(HairEyeColor, c(1, 2)) >

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 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 Stephen -- Internal Virus Database is out-of-date.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, In the chisq.test(), if the expected frequency for some categories is <5, there will be a warning message which says Warning message: Chi-squared approximation may be incorrect in: chisq.test(x, p = probs) I am wondering whether there are some methods to get rid of this mistake... Seems the ?chisq.test() doesn''t provide more options to solve this problem. Or, the only choice is

Hello everybody, I'm seeing some strange behavior on R 1.8.1 on Intel/Linux compiled with gcc 3.2.2. The p-value calculated from the chisq.test function is incorrect for some input values: > chisq.test(matrix(c(0, 1, 1, 12555), 2, 2), simulate.p.value=TRUE) Pearson's Chi-squared test with simulated p-value (based on 2000 replicates) data: matrix(c(0, 1, 1,

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 =

for example, an expression such as chisq(df=1,ncp=0) ? thanks -- View this message in context: http://www.nabble.com/how-to-calculate-chisq-value-in-R-tp15338943p15338943.html Sent from the R help mailing list archive at Nabble.com.

> On 28 Dec 2017, at 13:08 , Kurt Hornik <Kurt.Hornik at wu.ac.at> wrote: > >>>>>> Jan Motl writes: > >> The chisq.test on line 57 contains following code: >> STATISTIC <- sum(sort((x - E)^2/E, decreasing = TRUE)) > > The preceding 2 lines seem relevant: > > ## Sorting before summing may look strange, but seems to be >

Hello, I received the following warning when running chi-square; n Is there a way to catch the 'error' code of 'warning' after run chisq.test(x)? n What does this error mean? Thank you for your help. [[alternative HTML version deleted]]

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 am getting "Chi-squared approximation may be incorrect in: chisq.test(x)" with the data bleow. Frequency distribution of number of male offspring in families of size 5. Number of Male Offspring N 0 518 1 2245 2 4621 3 4753 4 2476 5

I am very new to R. I have some data from a CVS stored in vdata with 4 columns labeled: X08, Y08, X09, Y09. I have created two new "columns" like so: Z08 <- (vdata$X08-vdata$Y08) Z09 <- (vdata$X09-vdata$Y09) I would like to use chisq.test for each "row" and output the p-value for each in a stored variable. I don't know how to do it. Can you help? so far I have

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

Hello all, I would like to know what the difference is between chisq.test and fisher.test when using the Monte Carlo method with simulate.p.value=TRUE? Thank you -- View this message in context: http://r.789695.n4.nabble.com/Difference-in-Monte-Carlo-calculation-between-chisq-test-and-fisher-test-tp2322494p2322494.html Sent from the R help mailing list archive at Nabble.com.

Where does the value 2.2e-16 come from in p-values for chisq tests such as those reported below? > Anova(cm.mod2) Analysis of Deviance Table (Type II tests) Response: Freq LR Chisq Df Pr(>Chisq) B 11026.2 1 < 2.2e-16 *** W 7037.5 1 < 2.2e-16 *** Age 886.6 8 < 2.2e-16 *** B:W 3025.2 1 < 2.2e-16 *** B:Age 1130.4 8 < 2.2e-16 *** W:Age 332.9 8 < 2.2e-16 *** --- Signif.