search for: n11

...ion, we can calculate the probability mass of each possible table with same marginals as x. Ref.: Alan Agresti (1990). Categorical data analysis. New York: Wiley. Page 67. Hence, the result below suggests that the two-sided confidence interval has a confidence level of 97.5% as opposed to 95%. n11 <- 7:9 theta <- 0.2985103 choose(9,n11)*choose(15-9,13-n11)*theta^n11/ sum(choose(9,n11)*choose(15-9,13-n11)*theta^n11) #[1] 0.67344877 0.30154709 0.02500414 The 95% confidence interval with one-sided (greater) alternative appears to be correct. theta <- 0.4625314 choose(9,n11)*choose...

I'm wondering how odds ratio is computed. I thought that it is (n11/n12)/(n21/n22), but it is not what fisher.test() computes. Could somebody let me know? > n11=3 > n12=1 > n21=1 > n22=3 > > n1_=n11+n12 > n2_=n21+n22 > > n_1=n11+n21 > n_2=n12+n22 > > x=rbind(c(n11,n12),c(n21,n22)) > > threshold=dhyper(n11,n1_,n2_,n_1) >...

...Debut du programme for (n in N){ l1=0 # initialisation de la valeur permettant calculer le niveau de test ? 1% l2=0 # initialisation de la valeur permettant calculer le niveau de test ? 5% l3=0 # initialisation de la valeur permettant calculer le niveau de test ? 10% # Cr?ation d'une liste n11 qui contient les tailles des differents groupes n11=list() for (i in 1:q){ n11[[i]]=rep(as.integer(n/R[i]),R[i]) n11[[i]][R[i]]=n-((R[i]-1)*n11[[i]][1]) } # Cr?ation des listes P11 et P12 qui contient les probabilit?s et # les inverses des probabilites empiriques des differents groupes respectiv...

** I am using the following way to get p-values from fiser exact test. However, I do need to print for each pair the values "n00, n01, n10, n11". How can I print that as a table and not a matrix as below along with the p-value? Any help will be greatly appreciated fish <- function(y, x) {n00 = sum((1-x)*(1-y)); n01 = sum((1-x)*y); n10 = sum(x*(1-y)); n11 = sum(x*y); a = matrix(c(n00, n01, n10, n11), nrow = 2); pval = fisher.test(...

...quot;) 39 #Beginning of the program 40 for (n in N){ 41 l1=0 # initialization of the value to calculate the test level at 1%. 42 l2=0 # initialization of the value to calculate the test level at 5%. 43 l3=0 # initialization of the value to calculate the test level at 10%. 44 # Creation of an n11 list containing the sizes of the different groups 45 n11=list() 46 for (i in 1:q){ 47 n11[[i]]=rep(as.integer(n/R[i]),R[i]) 48 n11[[i]][R[i]]=n-((R[i]-1)*n11[[i]][1]) 49 } 50 # Creation of lists P11 and P12 which contain the probabilities and 51 # the inverses of the empirical probabilities of th...

...gt; 40 for (n in N){ > > > 41 l1=0 # initialization of the value to calculate the test level at 1%. > 42 l2=0 # initialization of the value to calculate the test level at 5%. > 43 l3=0 # initialization of the value to calculate the test level at 10%. > > 44 # Creation of an n11 list containing the sizes of the different groups > 45 n11=list() > 46 for (i in 1:q){ > 47 n11[[i]]=rep(as.integer(n/R[i]),R[i]) > 48 n11[[i]][R[i]]=n-((R[i]-1)*n11[[i]][1]) > 49 } > > > 50 # Creation of lists P11 and P12 which contain the probabilities and > 51 # the...

...> > > > 41 l1=0 # initialization of the value to calculate the test level at 1%. > > 42 l2=0 # initialization of the value to calculate the test level at 5%. > > 43 l3=0 # initialization of the value to calculate the test level at 10%. > > > > 44 # Creation of an n11 list containing the sizes of the different groups > > 45 n11=list() > > 46 for (i in 1:q){ > > 47 n11[[i]]=rep(as.integer(n/R[i]),R[i]) > > 48 n11[[i]][R[i]]=n-((R[i]-1)*n11[[i]][1]) > > 49 } > > > > > > 50 # Creation of lists P11 and P12 which contai...

Hi, I have a R output that looks as follow: Rad:0 Rad1:2 Rad3:3 I want to make a new matrix that looks like : sample size is 2400 Variable    n11  n12 Rad            0     2400-0=2400 Rad1          2       2400-2 Rad3  3      2400-3   Thanks a lot for your time and help:) Best,Farnoosh Sheikhi [[alternative HTML version deleted]]

...this in many ways: dat1<-read.table(text=" med1,med2,med3???? ?1,0,1?????? 0,1,1??? 2,0,0 ",sep=",",header=TRUE)?? #1st method library(reshape) dat2<-melt(dat1) dat3<-aggregate(dat2$value,by=list(dat2$variable),sum) ?colnames(dat3)<-c("name","sum(n11)") ?dat3 #? name sum(n11) #1 med1??????? 3 #2 med2??????? 1 #3 med3??????? 2 #2nd method res<-data.frame(colSums(dat1)) ?names(res)<-"sum(n11)" ?res #???? sum(n11) #med1??????? 3 #med2??????? 1 #med3??????? 2 #3rd method ?do.call(rbind,lapply(dat1,sum)) #???? [,1] #med1???...

Hi everyone, I'm using the function 'HWE.exact' of 'genetics' package to compute p-values of the HWE test. My data set consists of ~600 subjects (cases and controls) typed at ~ 10K SNP markers; the test is applied separately to cases and controls. The genotypes are stored in a list of 'genotype' objects, all.geno, and p-values are calculated inside the loop over all

I evaluated the Bayes factor in the k=2 allele case with a "triangular" prior under the null as in the example in the help file: HWETriangBF2(nvec=c(88,10,2)) [1] 0.4580336 When I swap the n11 entry and n22 entry of nvec, I received totally different Bayes factor: > > HWETriangBF2(nvec=c(2,10,88)) [1] 5.710153 > In my understanding, defining the genotype frequency as n11 or n22 are arbitrary. So I was expecting the same value of Bayes factor. This is the case for conjugate...

...ondering if I missed anything or if that's a problem with R: I want to have many x axis tickmarks and labels such that it makes sense to put them on multiple "lines". The newline character seems to be ignored though for axis labels: plot(1:33, axes=F) axis(1, 9, 9) axis(1, 11, "\n11") axis(1) # whereas title("one\ntwo") works as I expect it to work. In addition there is a little hook on top of the tickmark (the upper end of the vertical dash) that seems to be unintended. Best regards, Andreas Krause PS. R version 1.4.1 on Red Hat Linux. PPS. One way...

Without trying to understand your code in detail let me just assume you are trying to create a matrix, ret, whose i,j-th entry is some function, f, of row i of X and row j of X. In that case this should do it: apply(X,1,function(x)apply(X,1,function(y)f(x,y))) Date: Wed, 30 Jun 2004 15:28:47 -0300 (ART) From: <daniel at sintesys.com.ar> To: <r-help at stat.math.ethz.ch>

...I need a function that combines all the values from the dataset in order to produce the xi, xi*xj, xi*xj*xk values. The variables can take values in different ranges so I need that the function controls the scale. After that I would apply the regression to the obtained dataset. Example X1 X2 n11 n12 n21 n22 n31 n32 -> X1 X2 X1*X2 n11 n12 n11*n12 n21 n22 ... n31 n32 ... Is there any function like this in which, in addition of the above mentioned, I can specify the degree of combination? Thanks for your help, Miguel

Hello, I have a data with following format:      Predictors          n11     n12     n21     n22     Odds.Ratio    log.ratio    se.log.odds. 1 ProcOR respirato     2 ProcVaric vein      3 DiagCardiac anom   4 DiagAllergy        5  DiagOth skin dx    6    DiagGastritis          I want to plot odds ratio by command: forestplot in rmeta library, but I get the following err...

I have been trying to determine the size of the R user base, and was asked to share my findings with this mailing list. Although I still don't have any definite estimate of this number, I do have some interesting and indicative information: 1. It appears that there are about 100,000 S-PLUS users. Rationale: According to Insightful's 2002 Annual Report, over 100,000 people use

.../G" "G/G" "G/G" "G/G" "G/G" [73] "G/G" "G/G" "G/A" "G/G" "G/G" "G/G" "G/G" "G/G" Alleles: G A > HWE.exact(g1) Exact Test for Hardy-Weinberg Equilibrium data: g1 N11 = 71, N12 = 9, N22 = 0, N1 = 151, N2 = 9, p-value = 1 > control.data <- c("G/G","G/G","G/G","G/G","G/G","G/G","G/G","G/G","G/G","G/G","G/G","G/G","G/G","G/G...

Hello ... Are there any known problems or even gotchas to look out for when using a gzfile connection in read.csv/read.table in Windows? In the package PROcess, available at www.bioconductor.org/repository/devel/package/html/PROcess.html there are two files in the PROcess/inst/Test directory which are of the extension *.csv.gz. With both files, if I open up a gzfile connection, say: vv <-

Rhelpers: I'm curious what the appropriate analysis to use for testing the hypothesis that two histograms are statistically different from one another? Thanks! --j

...e function in Hmisc package. But it seems that it is meant for two groups such as reference/control versus cases. This is the sample of data in each of the two groups: Treatment_stat Death MonthsToEvent 1 0 60.0 2 1 55.3 0 1 36.7 0 1 39.9 number of cases in treatment strata : group 1: n12=110; n11=25;n10=15 (N2=150) group 2: n22=43;n21=8;n20=9 (N2=60) Coxph output indicates significant different survival in group 1 data but not in group 2. How can one be sure that the lack of significance of survival difference is true in group 2 and it is not because of fewer number of cases in group 1 ?...