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
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