jingjing
在为生存数据模型做试验设计时，需要计算样本量的大小，下面是一些基本的假设：
试验包含两个组: Placebo group and Vitamin D group, each group has the 6 months dose, the study time point is : month 3, month 6, month 9.
Set alpha=0.05, power=0.8;
1)For Placebo group, assuming the hazard rate is 0.36 at each study point;
2)For Vitamin D group, assuming the hazard rate is 0.15 at each study point;
3) for both group, assuming the rate of losting to follow-up is 0.030 at each study point,
我打算选用SAS 的PROCEDURE POWER，但是只找到当失访率（lost to follow up at each treatment interval)服从指数分布的时候的OPTION（见下面的CODE），那么当失访率服从其他分布或者是一个稳定的数值时，还能使用POWER来计算吗？ 另外，这是我初次使用POWER 程序，还请各位高手看一下下面的代码是否正确，谢谢。
proc power;
twosamplesurvival test=logrank
curve("Placebo") = (3 to 9 by 3): (0.64 0.4096 0.262144)
curve("Vitamin D") =(3 to 9 by 3): (0.85 0.7225 0.614125)
groupsurvival = "Placebo" | "Vitamin D"
curve("Placebo1")= 3: 0.030 /*assuming the loss curve is the exponential loss curve*/
curve("Vitamin D1")=3: 0.030
grouploss = "Placebo1" | "Vitamin D1"
accrualtime = 6
followuptime =3
alpha=0.05
sides=2
power = 0.8
npergroup = .;
run;
