sciplot question

Jarle Bj?rgeengen wrote:

On May 24, 2009, at 4:42 , Frank E Harrell Jr wrote:

Jarle Bj?rgeengen wrote:

On May 24, 2009, at 3:34 , Frank E Harrell Jr wrote:

Jarle Bj?rgeengen 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 Bj?rgeengen
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.

Frank

Best rgds
Jarle Bj?rgeengen