search for: chisquar

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 alleles). Any idea where this feature might be hidden? Or could it be added to Rcommander? Thanks, Christian. ps: in cas...

...<-> FARSCH, x1x1, NA LOCUS10 <-> LOCUS10, x2x2, NA T_ATTENT <-> T_ATTENT, y1y1, NA RMTEST10 <-> RMTEST10, y2y2, NA LOCUS10 <-> FARSCH, x2x1, NA This model runs, but using the summary function does not return the usual model fit statistics, only the following: Model Chisquare = 0 Df = 0 Pr(>Chisq) = NA Chisquare (null model) = 8526.8 Df = 6 Goodness-of-fit index = 1 BIC = 0 If I omit the last line from the RAM specification(i.e., delete "LOCUS10 <-> FARSCH, x2x1, NA"), I DO get all the usual statistics: Model Chisquare = 1303.7...

I sort of asked this before, but perhaps not explicitly enough. In my ctest collection, there are several chisquare-based tests. For some of them, it may be useful to also return information on expected (and observed) counts. The question is, how should this be done. Of course, there is no problem adding the corresponding components to the list returned by the functions. However, as these are of class `htes...

...5 46 5 34 3 3 1 1 Aski_chart 12 1 1 2 4 5 2 4 29 2 15 4 25 5 2 34 3 1 Aski_deli 2 1 1 2 <NA> 45 12 4 6 2 5 46 5 3 3 4 1 1.......... The variables are factors and I would like to perform a ChiSquare Test, comparing for example X1(first Table) to X1(secondTable)(Null Hypothesis: Variables are equal) Depending on the variables are the factor levels 1 up to 9. But as you immediately see, there is often more than one answer per species in the variables (Example 129). My question is now: how...

From: Martin Maechler <maechler@stat.math.ethz.ch> To: Kurt.Hornik@ci.tuwien.ac.at CC: R-devel@stat.math.ethz.ch Subject: Re: R-alpha: class for chisquare tests; Thoughts on print & summary Kurt, I think we hopefully are coming to an agreement that 1) your ctest collection should make its way into 'R core' ((and you are the one who can make it happen now ..)) By all means - throw the present versions away and add yours. We may...

Hello, My data is discrete, taking values between around -5 and +5. I cannot give bounds for the values. So I consider it as numerical data and not categorical data. The histogram has a 'normal' shape, so I test for normality via a chisquare statistic (by calculating the expected values by hand). When I use the sample mean and variance, the normality hypothesis has to be rejected. But when I test for sample mean + a small epsilon, I get very high p-values. I am not sure if this right shift is a good idea. Any suggestions? Thanks,...

...ons 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 and therefore a sparse (1xr)-table, then assumptions for chisquare-goodness-of-fit test are violated (the numbers expected are less than 5 in more than 75% of the entries.) ####### R-Program: Chisquare-Test :######### mydata <- c(15,0,3,4) chisq.test(mydata, correct=TRUE, rescale.p = TRUE, simulate.p.value = TRUE, B = 2000) As you cannot ignore the small...

Hi, I would like to know if there is a program written in R to get the CDF (cumulative distribution function) of a bivariate non-central chi-square distribution. Hope someone will reply. Thank you, Rossita M Yunus yunus@usq.edu.au This email (including any attached files) is confidentia...{{dropped:19}}

Hi to all I found an formular of an ** ***p-Value Calculator for the Chi-Square test* *http://www.danielsoper.com/statcalc/calc11.aspx* *with the formula* *http://www.danielsoper.com/statkb/topic11.aspx* *what's the gamma function of this formula in r?* *df=5* *ch2=25.50878* *the following code does not give the result <0.001 for the values above * *p=

...0.0000 0.0000 0.0426 0.1702 0.2128 0.1596 0.1809 0.0957 0.0745 0.0106 0.0106 0.0000 0.0000 0.0000 a1,a2,a3 ...... a17 is the frequency of 17 alleles , the sum is 1. I want to test the significance of the distribution of 17 alleles between two populations. How can I do? I want to use chisquare, is is right for these data ? can anyone help me ? Thanks!! luan Yellow Sea Fisheries Research Institute , Chinese Academy of Fishery Sciences , Qingdao , 266071 __________________________________________________ 佈伵伝仮伱佲伔佈G佊伿佅佷仯伃佒佇伖侜伒佢佉伝伨侙佄佫伬伂伝侙佊伿伡侢伾仹伻伵伋伂伌侒佊伿佅佷

...according to the purpose I have, I > need to understand how to create a dependence matrix, where I can > analyze the > dependence between all my variables. > Till now this is what I was able to do: > > /p <- length(spain)/ #this is the number of the variables (91) > > /chisquare <- matrix(spain, nrow=(p-1), ncol=p)/ #it creates a > squared-matrix with all the variables (if I'm not already wrong) > > /for(i in (1:(p-1))){/ > /chisquare[i, (1:(p-1))] <- chisq.test(spain[,i], spain[, i+1])$statistic/ > /chisquare[i, p] <- chisq.test(spain[,i], sp...

...ot;,"x2"))) path.model <- specify.model() x1 -> y1, x1-y1 x2 <-> x1, x2-x1 x2 <-> x2, x2-x2 x1 <-> x1, x1-x1 y1 <-> y1, y1-y1 x2 -> y1, x2-y1 summary(sem(path.model, cov.matrix, N = 422)) and I get following results; Model Chisquare = 12.524 Df = 1 Pr(>Chisq) = 0.00040179 Chisquare (null model) = 812.69 Df = 3 Goodness-of-fit index = 0.98083 Adjusted goodness-of-fit index = 0.885 RMSEA index = 0.16545 90% CI: (0.09231, 0.25264) Bentler-Bonnett NFI = 0.98459 Tucker-Lewis NNFI = 0.9573 Bentler CF...

Dear R experts, I was wondering if there are any R functions that give the tail area of a sum of chisquare distributions of the type: a_1 X_1 + a_2 X_2 where a_1 and a_2 are constants and X_1 and X_2 are independent chi-square variables with different degrees of freedom. Thanks, Klaus -- "Feel free" - 5 GB Mailbox, 50 FreeSMS/Monat ...

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 were i exclude the missing values ? and can i use this method if...

...oi)*reg*inv, data) varlm = lm(I(residuals(olslm)^2)~log(aloi)*reg*inv, data) glslm = lm(log(rr)~log(aloi)*reg*inv, data, weights=1/fitted(varlm)) Testing both olslm and glslm with both ncvTest and bptest gives: > ncvTest(olslm) Non-constant Variance Score Test Variance formula: ~ fitted.values Chisquare = 46.88206 Df = 1 p = 7.538963e-12 > ncvTest(glslm) Non-constant Variance Score Test Variance formula: ~ fitted.values Chisquare = 0.001466426 Df = 1 p = 0.9694533 > bptest(olslm) studentized Breusch-Pagan test data: olslm BP = 213.1477, df = 7, p-value < 2.2e-16...

Hi, I am trying to test for pairwise associations between genotypes ( Rows=individuals, Columns =genes, data are up to 4 genotypes per gene, some with 2,3 or 4) where each chisquare comparison is different depending on the genes tested. The test is the observed multilocus (across columns for each individual) genotypes vs the expectation, which is the product of the individual frequency for each genotype times the total number of individuals. Simple test. I have set up a loo...

...erform confirmatory factor-analysis model using polychoric correlations why I do not get an estimated confidence interval for the RMSEA. My experience with these type models is that I would obtain a confidence interval estimate. I did not get any warning messages with the output. RESULTS: Model Chisquare = 1374 Df = 185 Pr(>Chisq) = 0 Chisquare (null model) = 12284 Df = 210 Goodness-of-fit index = 0.903 Adjusted goodness-of-fit index = 0.88 RMSEA index = 0.0711 90% CI: (NA, NA) Bentler-Bonnett NFI = 0.888 Tucker-Lewis NNFI = 0.888 Bentler CFI = 0.902 SRMR = 0.0682 BIC...

Hello. I am getting results from sem that are not correct (that's assuming that the results from my AMOS 4.0 software are correct). sem does not vary some of the parameters substantially from their starting values, and the final estimates of those parameters as well as the model chisquare value are incorrect. I've attached some code that replicates the problem. The parameters in question are the residual variances of the endogenous variables B and C. A couple of things: the coefficents for the one-headed paths are on a much different scale than those for the two-headed paths...

Hello, I estimated a model using SEM package in R, which was fit to a raw moment matrix, and includes an intercept term. The only goodness of fit statistics that are output are Model Chisquare, AIC, AICc, BIC, CAIC, and normalized residuals. How can I get the other goodness of fit statistics, like adjusted goodness of fit, RMSEA, and R-squared? And how can I get the final value of the log-likelihood of the model? Thanks, Maya [[alternative HTML version deleted]]

...d fitted.values vs residuals and somewhat intuitively believe, it isn't really increasing... 2. further I ran the following tests bptest (studentized and non-studentized), gqtest, ncv.test with the following results: ncv: Non-constant Variance Score Test Variance formula: ~ fitted.values Chisquare = 13.87429 Df = 1 p = 0.00194580 Goldfeld-Quandt test data: reg GQ = 1.7092, df1 = 327, df2 = 327, p-value = 7.93e-07 studentized Breusch-Pagan test data: reg BP = 15.8291, df = 23, p-value = 0.92 Breusch-Pagan test data: reg BP = 377.5604, df = 23, p-value < 2.8e-18 bptest and...