search for: n_2

...best to enter the information but here goes... I'm trying to transfer the following integral into R to solve for ln(gamma_1), on the left, for multiple instances of gamma_i and variable N_i. gamma_i is, for example, (0, 0.03012048, 0.05000000, 0.19200000, 0.44000000, 0.62566845) N_i (N_1 or N_2) is between 0 and 1 so that N_1+N_2=1, so if N_1=(0,.166,.180,.250,.325,.374), then N_2=(1.000, 0.834, 0.820, 0.750, 0.675, 0.626) a_i (a_1 or a_2) So, for gamma_i (in this case gamma_2), N_i (N_2), and a_i (a_2) first the following a_i = ln(gamma_i)/(1-N_i)^2 then, ln(gamma_1) = -a_2*N_1*N*2...

...s into a list, where each element would represent one day, something like list[[1]] "price at time 1, day 1" "price at time 2, day 1" ... "price at time n_1, day 1" list[[2]] "price at time 1, day 2" "price at time 2, day 2" ... "price at time n_2, day 2" and so on. As the "n_1", "n_2", etc, suggest, each day have its own number of observations (this reflects the fact that the market may open and close at varying daytimes). I have guessed that a list would be a better way to store my data, since a matrix would be f...

...; > next: > ret void > > [grosser at greina0 build]$ bin/dominators > ../test/Analysis/RegionInfo/test.bc > > New Dominator Tree: > [0] %virtual_entry IO{0, 7}, R{nullptr}, P{nullptr} > [1] %entry_n_1 IO{1, 6}, R{virtual_entry}, P{virtual_entry} > [2] %n_2 IO{2, 3}, R{entry_n_1}, P{entry_n_1} > [2] %n_3 IO{4, 5}, R{entry_n_1}, P{entry_n_1} > > New Dominator Tree: > [0] %virtual_entry IO{0, 7}, R{nullptr}, P{nullptr} > [1] %entry_n_1 IO{1, 6}, R{virtual_entry}, P{virtual_entry} > [2] %n_2 IO{2, 3}, R{entry_n_1}, P{...

Dear R users, how can I test, whether two correlations differ significantly. (I want to prove, that variables are correlated differently, depending on the group a person is in.) Greetings from Freiburg im Breisgau (Germany), Timo Stolz

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) > probability=dhyper(0:n_1,n1_,n2_,n_1) > sum(probability[probability<=threshold]) [1] 0.4857143 > (n11/n12)/(n21/n22) [1] 9 > fisher.test(x) Fisher's Exact Test for Count Data...

Dear all, Could any one please tell me the exact formula R uses to calculate the sample size for one-sample and two-sample t-tests? Thanks, Caimiao [[alternative HTML version deleted]]

Greetings, I have the following? Problem: Given k (=10) discrete independent random variables X_i with n_i (= 5 to 20) values each,compute quantiles of the distribution of the sum X = X_1+...+X_k. Here X has n=n_1 x n_2 ... n_k distinct values which is too large to list them all together with their probabilities. I tried several approaches: (A) Convolution: each X_j is approximated with Y_j=X_j+Z, where Z is an N(0,sigma) variable with small sigma. Then Y_j is a probability mixture of the normal variables N(x...

...Dsii) / NN ss <- sqrt(NN * (DsD - DD^2) / (NN-1)) ### descriptive statistics print(" \n------------- Data ---------------------------------------\n") print(paste("N_1 = ",Ni ," : Mean_1 ? s_1 = ", Di," ? ",si, sep="")) print(paste("N_2 = ",Nii," : Mean_2 ? s_2 = ", Dii," ? ",sii, sep="")) print(paste("N = ", NN ," : Mean_comb ? s_comb = ", DD," ? ",ss, sep="")) print(" ") print(paste("s_L = ",sL,", s_H = ",sH,"...