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...ome numbers, it is different. The following example, shows what happens: set.seed(12233) n <- 5 a <- runif(n,1,5) e <- runif(n, 5*(1:n),10*(1:n)) s <- runif(1, 1.2, 4) p <- runif(5, 3, 10) x <- c(e[-5], (sum(e*p)-sum(e[-5]*p[-5]))/p[5]) u <- a^(1/s)*x^((s-1)/s) dyn.load("sumF.so") u[1] <- u[1]+.0001 ### If we do not add .0001, all differences are 0 s1 <- sum(u) s2 <- .Fortran("sumf", as.double(u), as.integer(n), sf1=double(1), sf2=double(1))[3:4] s3 <- .C("sumc", as.double(u), as.integer(n), sC=double(1))[[3]] s1-s2...

...t; set.seed(12233) >> n <- 5 >> a <- runif(n,1,5) >> e <- runif(n, 5*(1:n),10*(1:n)) >> s <- runif(1, 1.2, 4) >> p <- runif(5, 3, 10) >> x <- c(e[-5], (sum(e*p)-sum(e[-5]*p[-5]))/p[5]) >> u <- a^(1/s)*x^((s-1)/s) >> dyn.load("sumF.so") >> >> u[1] <- u[1]+.0001 ### If we do not add .0001, all differences are 0 >> s1 <- sum(u) >> s2 <- .Fortran("sumf", as.double(u), as.integer(n), sf1=double(1), >> sf2=double(1))[3:4] >> s3 <- .C("sumc", as....

...ple, shows what happens: > > set.seed(12233) > n <- 5 > a <- runif(n,1,5) > e <- runif(n, 5*(1:n),10*(1:n)) > s <- runif(1, 1.2, 4) > p <- runif(5, 3, 10) > x <- c(e[-5], (sum(e*p)-sum(e[-5]*p[-5]))/p[5]) > u <- a^(1/s)*x^((s-1)/s) > dyn.load("sumF.so") > > u[1] <- u[1]+.0001 ### If we do not add .0001, all differences are 0 > s1 <- sum(u) > s2 <- .Fortran("sumf", as.double(u), as.integer(n), sf1=double(1), > sf2=double(1))[3:4] > s3 <- .C("sumc", as.double(u), as.integer(n)...

Hi all Forgive me, I'm a total R newbie, and this seems to be a straightforward simple bootstrap problem, but after a whole day of trying to figure out how to do it I'm ready to give up. Part of the problem is that every example and every help page seems to be about doing something more far more complex. I'm got a table with 40 columns and 750 rows. I sum all the values across the