Thank you very much JS Huang. It helps.
Thanks & Regards,
Arnab
Arnab Kumar Maity
Graduate Teaching Assistant
Division of Statistics
Northern Illinois University
DeKalb, IL 60115
Email: ma...@math.niu.edu
Ph: 779-777-3428
On Fri, Feb 13, 2015 at 5:03 PM, JS Huang [via R] <
ml-node+s789695n4703244...@n4.nabble.com> wrote:
> Hi,
>
> Here assume there are four elements in the ordinal set y and take a
> random sample of size 10 according to the cumulative distribution given, or
> probability distribution p below.
>
>
> > y <- c(levels = c("First", "Second", "Third", "Fourth"))
> > y
> levels1 levels2 levels3 levels4
> "First" "Second" "Third" "Fourth"
> > (x <- c(1/9, 1/4, 3/5, 1))
> [1] 0.1111111 0.2500000 0.6000000 1.0000000
> > p <- c(x[1], x[2]-x[1], x[3]-x[2], x[4]-x[3])
> > p
> [1] 0.1111111 0.1388889 0.3500000 0.4000000
> > sample(y, 10, replace=TRUE, prob=p)
> levels2 levels4 levels4 levels3 levels1 levels3 levels4 levels2
> levels4 levels3
> "Second" "Fourth" "Fourth" "Third" "First" "Third" "Fourth" "Second"
> "Fourth" "Third"
>
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