1. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 21 Lecture 2 Two-stage designs for normally distributed data 2.1 A typical fixed-sample design 2.2 A two-stage design 2.3 Example 2.4 Actual conduct of the two-stage test

2. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 22 2.1 A typical fixed-sample design Trial:Randomised, parallel group phase II or III study Treatments:E: Experimental C: Control where (E:C) is (1:1) Primary efficacyY, which is continuous and normally variable:distributed

3. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 23 Let m be the sample size per treatment and n = 2m the total sample size For the h th patient on treatment j, the response is Y hj Y hj ~ N(  j,  2 ), h = 1,.., m; j = E, C Suppose that the variance  2 is known Parameterise the advantage of E over C by  =  E   C

4. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 24 Objective To determine whether or not to PROCEED, from phase II to phase III, or from phase III to a licence application We require P( PROCEED ;  = 0) =  P( PROCEED ;  =  R ) = 1  for some specific value of  R > 0 PROCEED  reject H 0 :  = 0 in favour of the one-sided alternative H 1 :  > 0  is the one-sided type I error rate

5. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 25 Planned analysis PROCEED if Note that so that

6. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 26 Sample size We wish to satisfy the 2 equations: P( PROCEED ;  = 0) =  and P( PROCEED ;  =  R ) = 1   in 2 unknowns: k and n The first equation is: which is where z  is the 100  percentage point of N(0, 1)

7. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 27 The second equation is: which is so that and

8. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 28 As, it follows that The sample size  as ,  or  R  The sample size  as  For R:1 randomisation, E:C, replace 4 by (R + 1) 2 /R

9. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 29 Graphically: Z k n sample size PROCEED : E > C ABANDON E

10. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 210 Actual analysis When the trial is over, we have n E patients on E and n C on C where n E and n C each  m The estimate S of standard deviation may not be equal to the anticipated value  So, PROCEED if

11. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 211 If n is large, then t 1 ,n  2  z 1  If n E = n C = m = n/2 and S = , then so that PROCEED if t  t 1 ,n  2 is equivalent to: PROCEED if Using the t-test guarantees type I error, while power is achieved to a good approximation

12. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 212 Z u1u1 u2u2 n1n1 n2n2 n 1 2.2 A two-stage design PROCEED : E > C ABANDON E

13. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 213 Interim look after n 1 patients: stop if Z 1  ( 1, u 1 ) Final look after n 2 patients: final test statistic is Z 2 (2.1) Thus

14. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 214 Let denote the treatment means for the n + = (n 2 – n 1 ) new patients who respond after the interim analysis Then so that and

15. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 215 Thus

16. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 216 Hence so that

17. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 217 Design calculation We wish to satisfy the 2 equations: P( PROCEED ;  = 0) =  and P( PROCEED ;  =  R ) = 1   Now P( PROCEED ) which is

18. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 218 This is which is

19. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 219 where for In SAS probnorm(x) =  (x) probbnrm(x 1,x 2,  ) =  2 (x 1,x 2,  )

20. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 220 The equations: P( PROCEED ;  = 0) =  and P( PROCEED ;  =  R ) = 1   are thus (2.2) and (2.3)

21. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 221 Two equations Five unknowns: n 1, n 2, 1, u 1, u 2 Need up to three constraints If < 3 constraints, then need an optimality criterion – search for a solution Notice that equation (2.2) concerns 1, u 1 and u 2 only If the constraints are of the form 1 = cu 1, u 2 = ku 1 and n 1 = rn 2, for known c, k and r, then we can solve (2.2) first for u 1, and then (2.3) to obtain the sample sizes

22. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 222 2.3 Example Requirements:  = 0.025, 1  = 0.90,  R = 0.7 Assumption:  = 1 Constraints: 1 =  u 1 ; u 2 =  (n 1 /n 2 )u 1 ; n 1 /n 2 = 0.6 Solution:n 1 = 54, n 2 = 90 u 1 = 2.572, 1 =  2.572, u 2 = 1.9923  fixed sample size = 86

23. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 223 Properties 11 11 E(n*)P( PROCEED ) 00.00506 89.60.02499 0.350.000010.0992286.40.37676 0.700.000000.5000072.00.90991

24. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 224 2.4Actual conduct of the two-stage test Following Jennison & Turnbull (2000), after a suggestion by Pocock (1977) At each analysis, compute (2.4) which ~ t on (n i – 2) df under H 0 S i is the sample standard deviation at the i th analysis

25. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 225 Then find (2.5) where T denotes the t distribution function on (n i – 2) df Now, under H 0

26. MPS/MSc in StatisticsAdaptive & Bayesian - Lect 226 Thus, under H 0, If the intended sample sizes are realised,  is close to the true standard deviation and n 1 and n 2 are large, then and Option 1: Use in place of Z 1 and Z 2, as planned Option 2: Use in place of Z 1 as planned, then revise the design in the light of the actual values of n E1, n C1 and S 1 Option 3: Use the actual values of n E1, n C1 and S 1 to revise the design before the interim