Size of the Phase 3 Trial

We choose this representation for efficiency since the Phase 3 sample size is explicitly represented as n3 instead of indirectly as fN3.

With this representation for efficiency, ∂E/∂n3 will be greater than zero if

C n n p z n p z n n

z n p z n p

(1 ) 2

(1 ) 0

( )( ) ( ) ( ) ( )

+ + ⋅ ⋅Φ − δ + − ⋅α  × φ − δ ⋅ −δ  ⋅

− Φ − δ ⋅ ⋅Φ − δ + − ⋅α  >

(16.2)

or

C n n

z n n

p z n p z n n

z n

(1 ) 2 0

( )

( ) ( ) ( )

+ ⋅ φ − δ ⋅ −δ

 ⋅

+ ⋅Φ − δ + − ⋅α ⋅ φ − δ ⋅ −δ

 

−Φ − δ











 >

(16.3)

or finally

p z n p C n n z n n

z n n z n (1 )

/ ( / 2)

( / 2) 2 2

( ) ( )( ) ( ) ( )

⋅ Φ − δ + − ⋅α < + ⋅φ − δ ⋅ δ

φ − δ ⋅ −δ − Φ − δα α

since

z n n z n 2

03 3 33 3( ) ( )φ − δ ⋅ −δ  − Φ − δ <α α (16.5)

for δ < 0, which can be seen by graphing z x x z x 23 3

( ) ( )Φ − ⋅ −  − Φ −α α for x < 0. If the trial is powered at 80 percent with a type 1 error of 0.025, then

z n 0.2833( )φ − δ =α (16.6)

z n 0.8033( )Φ − δ =α (16.7)

− δ   =n2 1.403 (16.8)

So, the Phase 3 sample size should be increased if

p z n p C n n

(1 ) / 0.28 ( 1.4)

2( ) ( )⋅ Φ − δ + − ⋅α < + ⋅ ⋅ − ⋅ −

α (16.9)

= (C + n2)/n3 · 0.96 (16.10)

That is, if the Phase 2 trial is such that the chance of moving forward to Phase 3 is less than (C + n2)/n3 · 0.96, then the change in efficiency with respect to n3 is greater than zero, which means efficiency is improved by increasing the Phase 3 sample size. On the other hand, if the chance of moving forward to Phase 3 is greater than (C + n2)/n3 · 0.96, then the change in efficiency with respect to n3 is less than zero and one should conduct the 80 percent powered trial, which is the smallest trial one can conduct in Phase 3 and hope to gain regulatory approval. If we are developing a new molecule, we would expect

C to be about the same size as n3 and so the size of the Phase 3 trial should be increased beyond what it would be for an 80 percent powered trial.