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
Frank E Harrell Jr Professor and Chair School of Medicine
Department of Biostatistics Vanderbilt University