# Re: st: confidence limits of survival difference

 From Ricardo Ovaldia To statalist@hsphsun2.harvard.edu Subject Re: st: confidence limits of survival difference Date Tue, 25 Feb 2003 10:50:16 -0800 (PST)

```--- Giulio Rizzoli <giulio.rizzoli@unipd.it> wrote:
> Dear experts I have these medical data.
> that i have pooled from the literature. They concern
> survival of
> medical devices biologic or mechanic.
> at-risk are the unit at risk.
> Survival are the % in use at stated time.
>
> Time	at-risk-bio at-risk-mec	Survival-bio
> Survival-mec
> 0	477	391	100	100
> 1	362	286	86.5	86.9
> 5	249  186	73.6	73.5
> 10	94	84	62	60.2
> 15	29	22	46.7	47.8
>
> I would like to know if it is possible and how to
> calculate
> the 95% Confidence Interval of the Survival
> difference.

This is difficult because the number of censored
observations impact the accuracy of these estimates. A
naive estimate can be obtained from -prtest-. For
example, for t=15:

. prtesti 29      .467    22   .478

Two-sample test of proportion                      x:
Number of obs =       29
y: Number of obs =       22

Variable        Mean   Std. Err.      z    P>z
[95% Conf. Interval]

x        .467   .0926452                      .2854187
.6485813
y        .478   .1064971                      .2692695
.6867305

diff       -.011   .1411551
-.287659     .265659
under Ho:     .14114    -0.08   0.938

Ho: proportion(x) - proportion(y) = diff = 0

Ha: diff < 0            Ha: diff != 0             Ha:
diff > 0
z = -0.078               z = -0.078               z =
-0.078
P < z =  0.4689        P > z =  0.9379          P > z
=  0.5311

That is:

diff= -.011   95%C.I.= -0.287659, 0.265659

Best,
Ricardo.

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```