NRI 95% CI using Mark Lunt NRI package.

[Gunnar Austad]

Hello,

I use the NRI package from Mark Lunt (net from http://personalpages.manchester.ac.uk/staff/mark.lunt).

I am calculating the categorical NRI for the new risk factor "plaqueburden". I set the risk thresholds at 5%, 10% and 15%.

I use the following syntax:
nri priorCVD plaqueburden sex age smk lpa diabetes CVDinfamily, at(0.05 0.10 0.15)

Results are as follows:
"In subjects with priorCVD == 1, 18.64% had increased probabilities
In subjects with priorCVD == 1, 15.25% had decreased probabilities

In subjects with priorCVD == 0, 6.07% had increased probabilities
In subjects with priorCVD == 0, 10.86% had decreased probabilities

Net Reclassification Improvement for plaqueburden = 0.0818
Standard Error = 0.0542
z = 1.5109
P-value (one-sided) = 0.0654
P-value (two-sided) = 0.1308"

However, I have trouble obtaining the 95% CI. Would very much appreciate if anyone could provide the correct syntax.

[George Ford]

https://www.statalist.org/forums/forum/general-stata-discussion/general/1608133-net-reclassification-index-nri-and-integrated-discrimination-index-idi-computing-confidence-intervals

[George Ford]

di "CI = (" %5.3f 0.0818-1.96*0.0542 ", " %5.3f 0.0818+1.96*0.0542 ")"

[George Ford]

di "CI = (" %5.3f r(nri)-1.96*r(se) ", " %5.3f r(nri)+1.96*r(se) ")"

or maybe

di "CI = (" %5.3f r(nri)-r(z)*r(se) ", " %5.3f r(nri)+r(z)*r(se) ")"

since the help feils says r(z) is the "Z-statistic for testing null hypothesis NRI = 0"