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, 2009/02/01 17:03:

chapter 15, table 15.4 95% CI for C statistic - in the example Program for testicular cancer analyses: t821 the line ”## NRI and IDI calculations with Harrell's functions improveProb(x1=as.numeric(NoLDH)-1, x2=as.numeric(YesLDH)-1, y=full$y)” did't work. how can I calculate the 95% CI for c statistic with a tree model? thank you

, 2009/03/30 18:05:

Calculations for the 95%CI of c are based on the rcorr.cens function in Harrell's Design library. The function gives the SE of Somer's D statistic; divide by 2, multiply by 1.96 for 95%CI.

, 2011/03/09 12:58:

chapter 15, table 15.4 how can i calculate 95%CI for C statistic in internal valdation. R code t821 internal validation, the line “cat(cstat[1], ”[”, cstat[1]-1.96/2*cstat[3], ” - ”, cstat[1]+1.96/2*cstat[3],”]”)” did not work. 'can not find the cstat object'. can you tell me how can i complete the R code for 95%CI for C statistic in internal valdation. cstat ← rcorr.cens(?, ?)

, 2011/12/23 09:51:

The issue of estimating a 95% CI after internal validation by bootstrapping is thorny. I discussed this with Frank Harrell several times. One simple approach is use the SE of the original c statistic for the optimism-corrected c statistic. Specific on the programming: The R program needs the object 'cstat' it seems; see help for the rcorr.cens function.

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