Re: Calculate Standard Deviation with R

On May 28, 1:28 pm, Bruce Weaver <bwea...@xxxxxxxxxxxx> wrote:
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R User wrote:
Hello,

I begin with R and I am searching a way to calculate a ST of a
weighed serie :

xi(10,20,30)
ni(1,5,10)

I found the function VAR which uses a vector of one dimension but
dot not care about coefficients.

You could either write the function yourself, or use cov.wt:

x=rnorm(10)
wt=1:10
cov.wt(data.frame(x), wt)

I found this using help.search("weighted variance").
Ok thank you,

If I understand you affect a coefficient of 1 on the value 10 ?
But how do I do if I have several observations ?
I tried this :
x=rnorm(496,497,498,499,500,501,502,503,504))
Erreur : erreur de syntaxe dans
"x=rnorm(496,497,498,499,500,501,502,503,504))"

You have 2 closed brackets. And after that, rnorm() will still be
screwed: check its help page (?rnorm): I was only generating x and wt
to give an example of R working.
ok thank you,
could you confirm what I means : this not a miracle function which gives
me the standard deviation ?
No, it gives you the weighted variance.
Do I need to compute all intermediaries steps ?

Just give it a vector of values, and a vector of weights, and it'll give
you the weighted covariance matrix (actually, it gives you a couple of
other things as well). Then take the square root of that: if you're
giving it the

How would you to calculate the standard deviation of a weighted serie
like this :
xi(5,6,8,9,10)
ni(1,5,3,2,6)

If ni are the weights, then this will do it:
xi=c(5,6,8,9,10)
ni=c(1,5,3,2,6)

sqrt(cov.wt(data.frame(xi), ni)$cov)
Thank you but it does not work. I tried with this serie
xi=c(496,497,498,499,500,501,502,503,504)
ni=c(1,3,6,8,13,9,5,3,2)
sqrt(cov.wt(data.frame(xi), ni)$cov)
xi
xi 1.967988

I found 1.967 as the standard deviation and if I computer it with excel I
find : 4.25

496 1
497 3
498 6
499 8
500 13
501 9
502 5
503 3
504 2

Average 500
Variance 18,1111111
STD 4,25571511

I made a mistake, here is the result :

xi-X (xi-X)^2*ni
496 1 -4 16
497 3 -3 27
498 6 -2 24
499 8 -1 8
500 13 0 0
501 9 1 9
502 5 2 20
503 3 3 27
504 2 4 32
TOTAL 163
Average 500 3,26
Variance 3,26
STD 1,80554701

it is even different from the R solution

This is how I would do it in SPSS.

data list list / x n (2f5.0).
begin data.
496 1
497 3
498 6
499 8
500 13
501 9
502 5
503 3
504 2
end data.
weight by n.
descriptives x /stat = min max mean stddev var.

Descriptive Statistics
|---------------|--|-------|-------|------|--------------|--------|
| |N |Minimum|Maximum|Mean |Std. Deviation|Variance|
|---------------|--|-------|-------|------|--------------|--------|
|x |50|496 |504 |500.06|1.823 |3.323 |
|---------------|--|-------|-------|------|--------------|--------|
|Valid N |50| | | | | |
|(listwise) | | | | | | |
|---------------|--|-------|-------|------|--------------|--------|

The SD and variance are computed with n-1 in the denominator. With
division by N, the variance would be approximately 3.323 * 49/50 = 3.165.

--
Bruce Weaver
bwea...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx/wv/bwhomedir

I used pure brute force in R :
cnt<- c(496, rep(497,3), rep(498,6), rep(499,8), rep(500,13),
rep(501,9 ), rep(502,5), rep(503,3), rep(504,2))

and get the same results.
sd(cnt)
[1] 1.822871
sqrt(3.165)
[1] 1.779045
var(cnt)
[1] 3.322857


.

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