similar to: Minimum-chi-square estimation

Hi R People: Sorry about the off topic question. Does anyone know the reference for "-2 Log Lambda is approx dist. Chi square", please? It may be Bartlett, but I'm not sure.... thanks in advance! Sincerely, Laura Holt mailto: holtlaura at gmail.com

Hi. I have made 100 experiments of an M/M/1 queue, and for each one I have calculated both, mean and variance of the queue size. Now, a professor has told me that variance is usually chi-squared distributed. Is there a way in R that I can find the parameter that best fits a chi-square to the variance data? I know there's fitdistr()m but this function doesn't handle chi-square. I believe

Hi, Can somebody explain what is the hex string following the path to the RSA key used for authentication? For an example, see last line in the debug snippet shown below. debug1: SSH2_MSG_SERVICE_REQUEST sent debug2: service_accept: ssh-userauth debug1: SSH2_MSG_SERVICE_ACCEPT received debug2: key: /Users/blah/.ssh/id_rsa (0x6fef81c1d114) How is that value generated? Is it something I can

Hi Dragos, The issue is that not all packet loss maybe congestion related. Often, reducing bitrate seems to have no impact on improving packet loss. Thanks, Manpreet. On Tue, Dec 16, 2014 at 2:09 AM, Dragos Oancea <droancea at yahoo.com> wrote: > > Hi > > You can start a VOIP call with 50 kbps bitrate and reduce the bitrate if > there is packet loss. You know if

if i want to plot the chi-square distribution with a different degree of freedom how can i plot it in the graph?Sometimes i plot the histogram and cut it in a lot of piece.It's distribution like a chi-square.So i want to plot the chi-square with a different degree of freedom to compare it . -- View this message in context:

I want to calculate chi square test of goodness of fit to test, Sample coming from Poisson distribution. please copy this script in R & run the script The R script is as follows ########################## start ######################################### No_of_Frouds<- c(4,1,6,9,9,10,2,4,8,2,3,0,1,2,3,1,3,4,5,4,4,4,9,5,4,3,11,8,12,3,10,0,7) N <- length(No_of_Frouds) # Estimation of

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

Hi! This is going to be a real newbie question, but I can't figure it out. I'm trying to plot densities of various functions of chi-square. A simple chi-square plot I can do with dchisq(). But e.g. chi.sq/degrees of freedom I only know how to do using density(rchisq()/df). For example: plot(1, type="n", xlab="", ylab="", xlim=c(0,2), ylim=c(0,7)) for (i

Hi, How do we get the value of a chi square as we usually look up on the table on our text book? i.e. Chi-square(0.01, df=8), the text book table gives 20.090 > dchisq(0.01, df=8) [1] 1.036471e-08 > pchisq(0.01, df=8) [1] 2.593772e-11 > qchisq(0.01, df=8) [1] 1.646497 > nono of them give me 20.090 Thanks, cruz

I am running lrm() with a single factor. I then run anova() on the fitted model to obtain a p-value associated with having that factor in the model. I am noticing that the "Model L.R." in the lrm results is almost the same as the "Chi-Square" in the anova results, but not quite; the latter value is always slightly smaller. anova() calculates the p-value based on

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 ...

Dear all, I met a problem while doing SEM by sem package. I got a negative chi-square of null model. Because the theoretical value of chi-square cannot be negative, I checked the source code of sem.R in sem package and I found the Chi-square of null model was computed by the following expression: result$chisqNull <- (N - 1) * (sum(diag(S %*% diag(1/diag(S)))) + log(prod(diag(S)))) I think

Dear help list. I would like to plot the chi-square distribution for n=1 through 10, on one plot. How do I go about doing that? I know how to plot one chi-square (using curve() for example), but plotting 10 on one plot, that is too difficult for me. I greatly appreciate your help. -- fb isabellesup at fastmail.fm --

R 2.12 windows 7 I am summary data that I would like to make into a 2x2 table representing counts positive vs. negative counts: 28/289 20/276 My table should look something like the following: group1 group2 Positive 28 20 Negative 289 276 How can a (1) create the 2x2 table (2) run a chi square test on the table? I have tried the following code, but I

Hi, I am new to R and I need to perform multiple chi-square tests. I manage to perform one at a time, but is there a specific command to do multiple tests? For example, I have a table that looks like this: Marker Treatment Genotype1 Genotype2 Genotype3 1 A 23 57 32 1 B 43 59 12 ... ... n A ## ## ## n

Dear R : Sorry for the off topic question, but does anyone know the reference for the -2 Ln Lambda following a Chi Square distribution, please? Possibly one of Bartlett's? Thanks in advance! Sincerely, Laura Holt mailto: lauraholt_983 at hotmail.com

Timothy, I believe you are mistaken. Fisher's exact test give the correct answer even in the face of small expected values for the cell counts. Pearson's Chi-square approximates Fisher's exact test and can give the wrong answer when expected cell counts are low. Chi-square was developed because it is computationally "simple". Fisher's exact test, particularly with tables

Is Pearson's Chi-Square test for contingency tables asymptotically unbiased for large tables (large degrees of freedom) regardless of the expected values in each cell? The rule of thumb is that Pearson's Chi-square should not be used when large numbers of cells have expected values < 5. However, I compared the results on 4x4 contingency tables for R's chisq.test using chi-square

I was asked by my boss to do an analysis on a large data set, and I am trying to convince him to let me use R rather than SPSS. I think Sweave could make my life much much easier. To get me a little closer to this goal, I ran my analysis through R and SPSS and compared the resulting values. In all but one case, they were the same. Given the matrix [,1] [,2] [1,] 110 358 [2,] 71 312 [3,]

I tried to run a GLM model using S-Plus and R. The same model and data were applied to the both packages. I got different results of P(>|Chi|) for Anova, between S-PLus and R. Other estimates values are the same. P(>|Chi|) for S-Plus: 0.99565 0.99682 0.04871 0.84597 1.00000 0.99999 1.00000 1.00000 1.00000 0.95834 P(>|Chi|) for R: 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000