similar to: a statistic question about chisq.test() (aprilsun)

Displaying 20 results from an estimated 4000 matches similar to: "a statistic question about chisq.test() (aprilsun)"

2003 Mar 26
3
a statistic question about chisq.test()
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
2003 Aug 21
0
The two chisq.test p values differ when the contingency table (PR#3896)
>>>>> dmurdoch writes: >> Date: Wed, 16 Jul 2003 01:27:25 +0200 (MET DST) >> From: shitao@ucla.edu >>> 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)}
2003 Apr 22
4
fisher exact vs. simulated chi-square
Dear All, I have a problem understanding the difference between the outcome of a fisher exact test and a chi-square test (with simulated p.value). For some sample data (see below), fisher reports p=.02337. The normal chi-square test complains about "approximation may be incorrect", because there is a column with cells with very small values. I therefore tried the chi-square with
2007 Mar 07
1
No fit statistics for some models using sem
Hi, New to both R and SEM, so this may be a very simple question. I am trying to run a very simple path analysis using the sem package. There are 2 exogenous (FARSCH, LOCUS10) and 2 endogenous (T_ATTENT, RMTEST) observed variables in the model. The idea is that T_ATTENT mediates the effect of FARSCH and LOCUS10 on RMTEST. The RAM specification I used is FARSCH -> T_ATTENT, y1x1, NA
2005 Jun 06
1
chisq.test and anova problems
we just started to use R and having some problems that no one in our school could solve. I hope someone here can help me out. the first problem is with the chisquare test. we want to exclude the missing values from the data. we used na.omit and made two new variables. now we want to use the chi square method but get the error x and y must have the same length. how do i use the chisquare method
2005 Aug 04
1
exact goodness-of-fit test
Hello, I have a question concerning the R-function chisq.test. For example, I have some count data which can be categorized as follows class1: 15 observations class2: 0 observations class3: 3 observations class4: 4 observations I would like to test the hypothesis whether the population probabilities are all equal (=> Test for discrete uniform distribution) If you have a small sample size
2013 Jul 09
0
probable bugs in stats::loglin calculation of pearson chisq
In running the following example of a loglinear model for the Titanic data, I was surprised to see NaN reported for the Pearson chisq > loglin(Titanic, margin=list(1:3, 4)) 2 iterations: deviation 2.273737e-13 $lrt [1] 671.9622 $pearson [1] NaN $df [1] 15 $margin $margin[[1]] [1] "Class" "Sex" "Age" $margin[[2]] [1] "Survived" Tracing it back,
2003 Jul 16
1
The two chisq.test p values differ when the contingency table is transposed! (PR#3486)
Full_Name: Tao Shi Version: 1.7.0 OS: Windows XP Professional Submission from: (NULL) (149.142.163.65) > x [,1] [,2] [1,] 149 151 [2,] 1 8 > c2x<-chisq.test(x, simulate.p.value=T, B=100000)$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
2008 Nov 16
3
chisq.test with simulate.p.value=TRUE (PR#13292)
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)) >
2009 Mar 09
2
path analysis (misspecification?)
hi, I have following data and code; cov <- c (1.670028 ,-1.197685 ,-2.931445,-1.197685,1.765646,3.883839,-2.931445,3.883839,12.050816) cov.matrix <- matrix(cov, 3, 3, dimnames=list(c("y1","x1","x2"), c("y1","x1","x2"))) path.model <- specify.model() x1 -> y1, x1-y1 x2 <-> x1, x2-x1 x2 <->
2012 Mar 09
1
Multiple Correspondence Analysis
You should send this to r-help@stat.math.ethz.ch. On 03/09/2012 09:21 AM, Andrea Sica wrote: > Hello everybody, I'm looking for someone who is able with MCA and > would like to gives some help. > > If what I'm doing is not wrong, according to the purpose I have, I > need to understand how to create a dependence matrix, where I can > analyze the > dependence between
2002 Dec 02
1
Monte Carlo chisq test
Dear all, I have a question about the chisq.test command. As an option one can chose the computation of p-values by Monte-Carlo simulation (simulate.p.value=T). Is there any documentation available how this calculations are done and how this simulation based test behaves in small samples? Thanks Klaus Abberger University of Konstanz, Germany [[alternate HTML version deleted]]
2012 Jun 26
5
chisq.test
Dear list! I would like to calculate "chisq.test" on simple data set with 70 observations, but the output is ''Warning message:'' Warning message: In chisq.test(tabele) : Chi-squared approximation may be incorrect Here is an example:         tabele <- matrix(c(11, 3, 3, 18, 3, 6, 5, 21), ncol = 4, byrow = TRUE)         dimnames(tabela) <- list(        
2003 Nov 07
2
chisq.test error: x and y must have at least 2 levels
Hi, I use a little script? to make a chi-square-test on 162 factors (it makes no difference if I take the numeric variant of the factors). At factor nr. 4 is stops with an error: [1] "v1= V7.KARTM v11= V7.KAR1M" Error in chisq.test(d1, d2) : x and y must have at least 2 levels But x and y /have/ two levels ("nein", "ja"): > fbhint.spss1$V7.KARTM [1] nein
2005 Oct 30
4
Yates' correction for continuity in chisq.test (PR#8265)
Full_Name: foo ba baz Version: R2.2.0 OS: Mac OS X (10.4) Submission from: (NULL) (219.66.32.183) chisq.test(matrix(c(9,10,9,11),2,2)) Chi-square value must be 0, and, P value must be 0 R does over correction when | a d - b c | < n / 2 &#65292;chi-sq must be 0
2010 Dec 29
1
Problem applying Chi-square in R and Cochran's Recommendations
Sir, I have a problem here while applying chisquare test to the following Data ( below the subject of this mail) ...when I wanted to test the significance using three different free statistical packages, here R, EpiInfo and OpenEpi. *Only OpenEpi accepts the test based on Cochran's Recommendations. * R says " chi squared approximation may be incorrect." Does it mean the same as
2007 Feb 13
1
lme4/lmer: P-Values from mcmc samples or chi2-tests?
Dear R users, I have now tried out several options of obtaining p-values for (quasi)poisson lmer models, including Markov-chain Monte Carlo sampling and single-term deletions with subsequent chi-square tests (although I am aware that the latter may be problematic). However, I encountered several problems that can be classified as (1) the quasipoisson lmer model does not give p-values when
2005 Sep 15
4
Rcommander and simple chisquare
In this years biostat teaching I will include Rcommander (it indeed simplifies syntax problems that makes students frequently miss the core statistical problems). But I could not find how to make a simple chisquare comparison between observed frequencies and expected frequencies (eg in genetics where you expect phenotypic frequencies corresponding to 3:1 in standard dominant/recessif
2007 Jun 27
1
how to use chi-square to test correlation question
Hi There, There are 300 boy students and 100 girl students in a class. One interesting question is whether boy is smarter than girl or not. first given the exam with a difficulty level 1, the number of the student who got A is below 31 for boy, 10 for girl. Then we increase the difficulty level of the exam to level 2, the number of the student who got A is below 32 for boy, 10 for girl. We
2003 Jul 15
0
Why two chisq.test p values differ when the contingency
Hi Tao: The P-values for 2x2 table are generated based on a random (discrete uniform distribution) sampling of all possible 2x2 tables, conditioning on the observed margin totals. If one of the cells is extremely small, as in your case, you get a big difference in P-values. Suppose, you changed the cell with value 1 to, say, 5 or 6, then the two P-values are nearly the same. However, I