On May 26, 2009, at 4:37 , Frank E Harrell Jr wrote:
Manuel Morales wrote:
On Mon, 2009-05-25 at 06:22 -0500, Frank E Harrell Jr wrote:
Jarle Bjørgeengen wrote:
On May 24, 2009, at 4:42 , Frank E Harrell Jr wrote:
Jarle Bjørgeengen wrote:
On May 24, 2009, at 3:34 , Frank E Harrell Jr wrote:
Jarle Bjørgeengen wrote:
Great,
thanks Manuel.
Just for curiosity, any particular reason you chose standard
error , and not confidence interval as the default (the
naming of the plotting functions associates closer to the
confidence interval .... ) error indication .
- Jarle Bjørgeengen
On May 24, 2009, at 3:02 , Manuel Morales wrote:
You define your own function for the confidence intervals.
The function
needs to return the two values representing the upper and
lower CI
values. So:
qt.fun <- function(x) qt(p=.975,df=length(x)-1)*sd(x)/
sqrt(length(x))
my.ci <- function(x) c(mean(x)-qt.fun(x), mean(x)+qt.fun(x))
Minor improvement: mean(x) + qt.fun(x)*c(-1,1) but in general
confidence limits should be asymmetric (a la bootstrap).
Thanks,
if the date is normally distributed , symmetric confidence
interval should be ok , right ?
Yes; I do see a normal distribution about once every 10 years.
Is it not true that the students-T (qt(... and so on) confidence
intervals is quite robust against non-normality too ?
A teacher told me that, the students-T symmetric confidence
intervals will give a adequate picture of the variability of the
data in this particular case.
Incorrect. Try running some simulations on highly skewed data.
You will find situations where the confidence coverage is not very
close of the stated level (e.g., 0.95) and more situations where
the overall coverage is 0.95 because one tail area is near 0 and
the other is near 0.05.
The larger the sample size, the more skewness has to be present to
cause this problem.
OK - I'm convinced. It turns out that the first change I made to
sciplot
was to allow for asymmetric error bars. Is there an easy way (i.e.,
existing package) to bootstrap confidence intervals in R. If so, I'll
try to incorporate this as an option in sciplot.
library(Hmisc)
?smean.cl.boot
H(arrel)misc :-)
Thanks for valuable input Frank.
This seems to work fine. (slightly more time consuming , but what do
we have CPU power for )
library(Hmisc)
library(sciplot)
my.ci <- function(x) c(smean.cl.boot(x)[2],smean.cl.boot(x)[3])
lineplot
.CI
(V1
,V2
,data
=
d
,col
=
c
(4
),err
.col
=
c
(1
),err
.width
=
0.02
,legend=FALSE,xlab="Timeofday",ylab="IOPS",ci.fun=my.ci,cex=0.5,lwd=0.7)
Have I understood you correct in that this is a more accurate way of
visualizing variability in any dataset , than the students T
confidence intervals, because it does not assume normality ?
Can you explain the meaning of B, and how to find a sensible value (if
not the default is sufficient) ?
Best regards
Jarle Bjørgeengen
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