r - calc mean/std/ci from column -


is there package calculate each specific n number, mean/std/ci. in example starting data:

> n = c(0,0,0,0,0,0,0,2,2,2,2,5,5,5,5,8,8,8,8) > s = c(43,23,65,43,12,54,43,12,2,43,62,25,55,75,95,28,48,68,18) > df = data.frame(n, s) > df    n  s 1  0 43 2  0 23 3  0 65 4  0 43 5  0 12 6  0 54 7  0 43 8  2 12 9  2  2 10 2 43 11 2 62 12 5 25 13 5 55 14 5 75 15 5 95 16 8 28 17 8 48 18 8 68 19 8 18 

resulting as:

data n mean std ci 0 40   ..  .. 2 30   ..  .. 5 63   ..  .. 8 41   ..  .. 

you can use dplyr package.

here's code snippet. note, i'm assuming want build confidence interval using standard normal approximation @ 95% level can make whatever choice like.

n = c(0,0,0,0,0,0,0,2,2,2,2,5,5,5,5,8,8,8,8) s = c(43,23,65,43,12,54,43,12,2,43,62,25,55,75,95,28,48,68,18) df = data.frame(n, s)    df %>%   group_by(n) %>%   summarise(mean = mean(s),             std = sqrt(var(s)),             lower = mean(s) - qnorm(.975)*std/sqrt(n()),             upper = mean(s) + qnorm(.975)*std/sqrt(n()))  source: local data frame [4 x 5]    n     mean      std     lower    upper 1 0 40.42857 17.88721 27.177782 53.67936 2 2 29.75000 27.69326  2.611104 56.88890 3 5 62.50000 29.86079 33.236965 91.76303 4 8 40.50000 22.17356 18.770313 62.22969 

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