similar to: Turning off continuation prompt?

I am trying to perform goodness of fit test using R. I am using this website http://wiener.math.csi.cuny.edu/Statistics/R/simpleR/stat013.html for help. However, I am unable to carry out the test successfully. My code follows. It is taken from the website just mentioned. freq=c(22,21,22,27,22,36) # frequencies obtained after rolling the dice 150 times. prob=c(1,1,1,1,1,1)/6 # specify expected

Hello all, Let say I have 2-way contingency table: Tab <- matrix(c(8, 10, 12, 6), nr = 2) and the Chi-squared test could not reject the independence: > chisq.test(Tab) Pearson's Chi-squared test with Yates' continuity correction data: Tab X-squared = 1.0125, df = 1, p-value = 0.3143 However I want to get all possible contingency tables under this independence

Dear UseRs, I'm running a chi-squared test where the expected matrix is the same as the observed, after rounding. R reports a X-squared of zero with a p value of one. I can justify this because any other result will deviate at least as much from the expected because what we observe is the expected, after rounding. But the formula for X-squared, sum (O-E)^2/E gives a positive value. What

Dear R list, I have one very elementary question regrading correlation between two variables. x = c(44,46,46,47,45,43,45,44) y = c(44,43,41,41,46,48,44,43) > cov(x, y) [1] -2.428571 However, if I try to calculate the covariance using the formula as covariance = sum((x-mean(x))*(y-mean(y)))/8       # no of of paired obs. = 8 or     covariance = sum(x*y)/8-(mean(x)*mean(y)) gives

Greetings! For some reason I am not managing to work out how to do this (in principle) simple task! As a result of applying strsplit() to a vector of character strings, I have a long list L (N elements), where each element is a vector of two character strings, like: L[1] = c("A1","B1") L[2] = c("A2","B2") L[3] = c("A3","B3")

many languages have shorthands for that operation like: variable += 1 or ++variable is there something like that in R ? -- View this message in context: http://r.789695.n4.nabble.com/increase-or-decrease-variable-by-1-tp3076390p3076390.html Sent from the R help mailing list archive at Nabble.com.

Dear Sir or Madam: Dose anyone know the R function which corresponds to "fcdf", a "F cumulative distribution function" of Matlab? http://esra.univ-paris1.fr/matlab5/toolbox/stats/fcdf.html Please guide me how to get this function if it is available. ..... or code it ab initio if no function in R language Thanks for kind reply further. Hsih-Te

Hi all, why factorial(150) shows the error out of range in 'gammafn'? I have to calculate the number of subset formed by 150 samples taking 10 at a time. How is this possible? best [[alternative HTML version deleted]]

Dear All,   I have a questions I would like to ask about and wonder if you have any thoughts to make it work in R.   1. I work in the field of medicine where physiologic variables are often simulated, and they can not have negative values. Most often the assumption is made to simulate this parameters with a normal distribution but in the "log-domain" to avoid from negative values to be

Greetings All. The function edit.data() allows a convenient spreadsheet-like view of a dataframe with too many rows/columns to fit on the screen (especially when there are many columns). Very useful when scanning through a dataset (row & column are conveniently identified by the labels at the side and above). However, there seens to be no option to set it "read-only" on start-up,

Greetings All. I want to do the following simple thing. I have defined a function med3x3() such that, given vectors X,Y, med3x3(X,Y) returns a 3x3 table where: Row 1: X > median(X) Row 2: X = median(X) Row 3: X < median(X) Col 1: Y < median(Y) Col 2: Y = median(Y) Col 3: Y > median(Y) (with intersections of these conditions for the individual cells). I can easily define fixed

Dear All, I assume this is an R-devel issue, apologies if I missed something obvious. I have a dataframe where the row names are country codes, based on ISO 3166, something like this: ------------ "v1" "v2" "UK" 1 2 "NA" 2 3 ------------ It happens that "NA" is the country code for "Namibia", and that creates problems on

Greetings all! A recent news item got me thinking that a problem stated therein could provide a teasing little exercise in R programming. http://www.bbc.co.uk/news/uk-england-cambridgeshire-17680326 Cambridge University hosts first European 'maths Olympiad' for girls The first European girls-only "mathematical Olympiad" competition is being hosted by Cambridge

Hello, I have done a scatterplot and now would like to add its regression line but it does not show. Below, the code I have used. lm3 <- lm(data$S_pH_KCl2.5_BCx~data$B_OleicoPF_BCx_per) plot(data$S_pH_KCl2.5_BCx, data$B_OleicoPF_BCx_per) abline(lm3) I have been able to do the complete operation using the software STATISTICA but it would be great to do it with R. If you require more details

A quick question: Is there anywhere a listing of the RGB components of the named colours listed by colors()? For example, where would I find the RGB for "orange1" or "salmon"? When I look at an EPS file from R where I have used these colours, it seems that for: "salmon": 0.9804 0.5020 0.4471 rgb "orange1": 1 0.6471 0 rgb However, this is a tedious way

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 =

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

Hi list, i've got a question about the chisq.test function. in the use of the "given probabilities" method (p= ...), normally there should be typed in probabilities in the range of 0 to 1 with the absolute sum of 1.0 (r-help) But it is possible to use probabilities > than 1. or the sum <1.! without any warning message Ok, now the question, what does r calcutate in these

Hi, Does anybody encounter the same problem when we overlap histogram and density     that the density line seem to shift to the right a little bit?           If you do have the same problem, what should we do to correct that?           Thank you.           par(mar=c(4,4,2,1.2),oma=c(0,0,0,0))     hist(datobs,prob=TRUE, main ="Volume of a catchment from four    

Greetings I want to remove numbers from a string of characters that identify sites so that I can merge two data frames. For example, a site in one frame is called "001a Frozen Niagara Entrance" whereas the same site in the other data frame is called "Frozen Niagara Entrance". It seems to me the easiest thing to do would be to remove the numbers from the first data