st: Problem with centile and normal confidence limits

 From James Shaw <[email protected]> To [email protected] Subject st: Problem with centile and normal confidence limits Date Wed, 5 Sep 2007 15:33:29 -0700 (PDT)

```Dear Statalist:

I am having trouble replicating the normal confidence
limits produced by -centile-.  The manual states that
the standard error (Sq) of a given quantile, Cq, may
be computed under the assumption of normality using
the equation on p. 202 in the Stata Reference Manual
(Release 9, A-G).  However, I have to multiply Sq by
sqrt(n) to get the same endpoints as Stata.  My sample
code and output are provided below.

I am not sure why I cannot replicate Stata's results
using the equation presented in the manual.  Given
that I can approximate the limits produced by
-centile, normal- using the bootstrap, I suspect that
the notation in the manual may be incorrect.  I do not
and am therefore unable to verify whether or not this
is the case.

Regards,

Jim

. drawnorm y
. qui: summ y
. scalar sm1= r(mean)
. scalar sd1 = r(sd)
. scalar sn = r(N)

. qui: centile y, normal
. scalar smd2 = r(c_1)
. scalar stul = r(ub_1)
. scalar stll = r(lb_1)

. /* from Stata */
. scalar li smd2 stul stll
smd2 =  .02788301
stul =  .10364515
stll = -.04787914

. scalar sq =
sqrt(50*(100-50))/(100*sn*normalden(smd2,sm1,sd1))

. /* using formula given in manual */
. scalar ul11 = smd2 + sq*invnormal(.975)
. scalar ll11 = smd2 - sq*invnormal(.975)
. scalar li smd2 ul11 ll11
smd2 =  .02788301
ul11 =  .03027881
ll11 =   .0254872

. /* multiplied by sqrt(n) */
. scalar ul12 = smd2 + sq*invnormal(.975)*sqrt(sn)
. scalar ll12 = smd2 - sq*invnormal(.975)*sqrt(sn)
. scalar li smd2 ul12 ll12
smd2 =  .02788301
ul12 =  .10364515
ll12 = -.04787914

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