[R] CI

[Frank E Harrell Jr]

Liaw, Andy wrote:
> Here's one way:
> 
> R> x <- c(6,11,5,14,30,11,17,3,9,3,8,8)
> R> confint(lm(x~1), level=.9)
>                  5 %    95 %
> (Intercept) 6.546834 14.2865
> 
> Andy 

Or without assuming normality,

library(Hmisc)
smean.cl.boot(x, conf.int=.9, B=10000)
      Mean     Lower     Upper
10.416667  7.333333 14.083333

This took 0.33 seconds using code that is optimized for bootstrapping 
the mean to get percentile confidence intervals.  The lack of symmetry 
in the bootstrap CL correctly reflects the asymmetry in the data.

Or:

library(boot)
w <- boot(x, function(a,b)mean(a[b]), R=10000) # took 1 sec.
boot.ci(w, conf=.9)  # 2.6 sec.
BOOTSTRAP CONFIDENCE INTERVAL CALCULATIONS
Based on 10000 bootstrap replicates

CALL :
boot.ci(boot.out = w, conf = 0.9)

Intervals :
Level      Normal              Basic
90%   ( 6.99, 13.79 )   ( 6.75, 13.50 )

Level     Percentile            BCa
90%   ( 7.33, 14.08 )   ( 7.67, 14.92 )
Calculations and Intervals on Original Scale

I choose 10000 reps to get virtually the same CI on multiple runs.
Frank

> 
> From: Ethan Johnsons
>> I have a quick question, please.
>>
>> Does R have function to compute i.e. a 90% confidence 
>> interval for the mean for these numbers?
>>
>>> mean (6,11,5,14,30,11,17,3,9,3,8,8)
>> [1] 6
>>
>> I thought pt or qt would give me the interval, but it seems not.
>>
>> thx much.
>>
>> ej

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