Dear All,
I have a question regard to the calculation of the type I error as shown in the snapshot below, anyone could help me out ? In SAS I know there is procedures about alpha spending. But how to get the type i error if we don't adjust the alpha as shown in the hightlighted area.
Thanks .
Jack
Hello @Jack2012,
In C. Jennison, B.W. Turnbull (2000): Group Sequential Methods with Applications to Clinical Trials (which is frequently referred to in the PROC SEQDESIGN documentation), section "2.4 Pocock's test" I found a fairly simple (N(0,1)-distributed) test statistic Zk (formula (2.4), p. 25) based on normal distributions, which I simulated with the code below and reproduced the type I error probabilities for k=1, 2, 5 and 10 of your table (k = #interim analyses). These appear to be largely independent of all three parameters used: the sample size increment n (=m in the book) and the normal distribution parameters m (=mA=mB under H0), s (=s). (Actually, it should be possible to compute these probabilities from the multivariate normal distribution of the vector (Z1, Z2, ..., ZK), but I haven't tried to do so.)
%let n=20; /* arbitrary sample size increment per group */
%let m=0; /* arbitrary mean and */
%let s=1; /* arbitrary std of the common normal distr. of the trt. groups A, B under H0 */
/* Simulation of the absolute value c[k] of statistic Z_k, 1<=k<=10, from Jennison/Turnbull, p. 25 */
data test(drop=j);
call streaminit(27182818);
array c[10] _temporary_;
do i=1 to 100000;
call missing(a, b, of c[*]);
do k=1 to 10;
do j=1 to &n;
a+rand('normal',&m,&s);
b+rand('normal',&m,&s);
end;
c[k]=1/sqrt(2*&n*k*&s**2)*abs(a-b);
r=(max(of c[*])>probit(0.975)); /* indicator for rejection of H0 at any stage 1, ..., k */
output;
end;
end;
run;
/* Computation of relative frequencies of indicator variable r
(approximating the type I error probabilities) */
proc summary data=test nway;
class k;
var r;
output out=want(drop=_:) mean=p;
run;
proc print data=want noobs;
where k in (1, 2, 5, 10);
run;
Result:
k p 1 0.05054 2 0.08340 5 0.14176 10 0.19369
Jennison and Turnbull cite a 1969 paper reporting 0.142 for k=5.
Insufficient information. Type error results depend on specific test conducted, data and model. You have provided none of those elements.
Different procedures often use different options to display/control Type error information.
Hello @Jack2012,
In C. Jennison, B.W. Turnbull (2000): Group Sequential Methods with Applications to Clinical Trials (which is frequently referred to in the PROC SEQDESIGN documentation), section "2.4 Pocock's test" I found a fairly simple (N(0,1)-distributed) test statistic Zk (formula (2.4), p. 25) based on normal distributions, which I simulated with the code below and reproduced the type I error probabilities for k=1, 2, 5 and 10 of your table (k = #interim analyses). These appear to be largely independent of all three parameters used: the sample size increment n (=m in the book) and the normal distribution parameters m (=mA=mB under H0), s (=s). (Actually, it should be possible to compute these probabilities from the multivariate normal distribution of the vector (Z1, Z2, ..., ZK), but I haven't tried to do so.)
%let n=20; /* arbitrary sample size increment per group */
%let m=0; /* arbitrary mean and */
%let s=1; /* arbitrary std of the common normal distr. of the trt. groups A, B under H0 */
/* Simulation of the absolute value c[k] of statistic Z_k, 1<=k<=10, from Jennison/Turnbull, p. 25 */
data test(drop=j);
call streaminit(27182818);
array c[10] _temporary_;
do i=1 to 100000;
call missing(a, b, of c[*]);
do k=1 to 10;
do j=1 to &n;
a+rand('normal',&m,&s);
b+rand('normal',&m,&s);
end;
c[k]=1/sqrt(2*&n*k*&s**2)*abs(a-b);
r=(max(of c[*])>probit(0.975)); /* indicator for rejection of H0 at any stage 1, ..., k */
output;
end;
end;
run;
/* Computation of relative frequencies of indicator variable r
(approximating the type I error probabilities) */
proc summary data=test nway;
class k;
var r;
output out=want(drop=_:) mean=p;
run;
proc print data=want noobs;
where k in (1, 2, 5, 10);
run;
Result:
k p 1 0.05054 2 0.08340 5 0.14176 10 0.19369
Jennison and Turnbull cite a 1969 paper reporting 0.142 for k=5.
Million thanks for your help. This is what I am hunting for. Thank you very much.
Join us for SAS Innovate 2025, our biggest and most exciting global event of the year, in Orlando, FL, from May 6-9. Sign up by March 14 for just $795.
ANOVA, or Analysis Of Variance, is used to compare the averages or means of two or more populations to better understand how they differ. Watch this tutorial for more.
Find more tutorials on the SAS Users YouTube channel.