MPS/MSc in StatisticsAdaptive & Bayesian - Lect 21 Lecture 2 Two-stage designs for normally...

MPS/MSc in Statistics Adaptive & Bayesian - Lect 2 1

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


2.1 A typical fixed-sample design

Trial: Randomised, parallel groupphase II or III study

Treatments: E: ExperimentalC: Controlwhere (E:C) is (1:1)

Primary efficacy Y, which is continuous and normally variable: distributed


Let m be the sample size per treatment and n = 2m the total sample size

For the hth patient on treatment j, the response is Yhj

Yhj ~ 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


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 H0: = 0 in favour of the one-sided alternative H1: > 0

is the one-sided type I error rate


Planned analysis

PROCEED if

Note that

so that

E CnZ y y k

2

2 2

E C 2

n n 2 2Z ~ N ,2 4 n n

Z n 2 ~ N 0,1


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)

P Z k 0

11 k k z


The second equation is:

which is

so that

and

RP Z k 1

R R RP Z n 2 k n 2 1

R1 k n 2 1

R 1k n 2 z


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

21 12

R

z zn 4

1k z


Graphically:

Zk

nsample size

PROCEED: E > C

ABANDON E


Actual analysis

When the trial is over, we have nE patients on E and nC on C

where nE and nC each m

The estimate S of standard deviation may not be equal to the anticipated value

So, PROCEED if

E C1 ,n 21 1

E C

y yt tS n n


If n is large, then t1,n2 z1

If nE = nC = m = n/2 and S = , then

so that PROCEED if t t1,n2 is equivalent to: PROCEED if

Using the t-test guarantees type I error, while power is achieved to a good approximation

E CE C1 1

E C

y y nt y y2S n n

1Z t z k


Z

u1 u2

n1 n2 n

1

2.2 A two-stage design

PROCEED: E > C

ABANDON E


Interim look after n1 patients: stop if Z1 (1, u1)

Final look after n2 patients: final test statistic is Z2

(2.1)

Thus

ii Ei Ci

nZ y y , i 1,2

2

2 2

i ii E C 2

i i

i

n n 2 2Z ~ N ,2 4 n n

n N ,1 , i 1,2

2


Let denote the treatment means for the n+ = (n2 – n1) new patients who respond after the interim analysis

Then

so that

and

E Cy and y

2 j2 1 j1 jn y n y n y , j E,C

2 E2 C2 1 E1 C1 E Cn y y n y y n y y

2 2 1 1 E C2 n Z 2 n Z n y y


Thus

1 1 2 2 1 1 1 1 E C

2 21 1 1 E C 1

1 1 1 1 E C

2 21 1 1 E C 1

cov 2 n Z ,2 n Z cov 2 n Z ,2 n Z n y y

E 4 n Z 2 n n y y Z

E 2 n Z E 2 n Z n y y

E 4 n Z E 2 n n y y Z

2

1 1 1 1 E C

1 1

21

E 2 n Z E 2 n Z E n y y

var 2 n Z

4 n


Hence

so that

1 2 1 1 2 221 2

21 1

221 2

1cov Z ,Z cov 2 n Z ,2 n Z4 n n

4 n n n4 n n

1 1 1 2

2 2 1 2

Z n 2 1 n n0~ N ,

0Z n 2 n n 1


Design calculation

We wish to satisfy the 2 equations:


Now P(PROCEED)

which is

1 1 1 1 1 2 2P Z u or Z ,u and Z u

1 1 1 1 2 2

1 1 2 2

P Z u P Z and Z u

P Z u and Z u


This is

which is

1 1 1 1

1 1 1 1

2 2 2 2

1 1 1 1

2 2 2 2

P Z n 2 u n 2

P Z n 2 n 2

and Z n 2 u n 2

P Z n 2 u n 2

and Z n 2 u n 2

1 1 2 1 1 2 2

2 1 1 2 2

u n 2 n 2 , u n 2 ,

u n 2 , u n 2 ,


where

for

In SAS

probnorm(x) = (x)probbnrm(x1,x2,) = 2(x1,x2, )

2 1 2 1 1 2 2x ,x , P X x ,X x

11 2

2

X 0 1~ N , ; n n

X 0 1


The equations:


are thus

(2.2)and

(2.3)

1 R 1 2 1 R 1 2 R 2

2 1 R 1 2 R 2

u n 2 n 2 , u n 2 ,

u n 2 , u n 2 , 1

1 2 1 2 2 1 2u , u , u , u ,


Two equations

Five unknowns: n1, n2, 1, u1, u2

Need up to three constraints

If < 3 constraints, then need an optimality criterion – search for a solution

Notice that equation (2.2) concerns 1, u1 and u2 only

If the constraints are of the form 1 = cu1, u2 = ku1 and n1 = rn2, for known c, k and r, then we can solve (2.2) first for u1, and then (2.3) to obtain the sample sizes


2.3 Example

Requirements: = 0.025, 1 = 0.90, R = 0.7

Assumption: = 1

Constraints: 1 = u1; u2 = (n1/n2)u1; n1/n2 = 0.6

Solution: n1 = 54, n2 = 90u1 = 2.572, 1 = 2.572, u2 = 1.9923

fixed sample size = 86


Properties

1 1 E(n*) P(PROCEED)

0 0.00506 0.00506 89.6 0.02499

0.35 0.00001 0.09922 86.4 0.37676

0.70 0.00000 0.50000 72.0 0.90991

1 1 1 1 1P Z n 2

1 1 1 1 1P Z u 1 u n 2

2 2 1 1 1E n n n n


2.4 Actual 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 (ni – 2) df under H0

Si is the sample standard deviation at the ith analysis

Ei Cii 1 1

i Ei Ci

y yt , i 1,2S n n


Then find

(2.5)

where T denotes the t distribution function on (ni – 2) df

Now, under H0

i

1i n 2 iZ T t , i 1,2

i

i

i

i i

1i n 2 i

n 2 i

1i n 2

1n 2 n 2

P Z z P T t z

P T t z

P t T z

T T z

z


Thus, under H0,

If the intended sample sizes are realised, is close to the true standard deviation and n1 and n2 are large, then

and

Option 1: Use in place of Z1 and Z2, as planned

Option 2: Use in place of Z1 as planned, then revise the

design in the light of the actual values of nE1, nC1 and S1

Option 3: Use the actual values of nE1, nC1 and S1 to revise

the design before the interim

1 2Z ~ N 0,1 and Z ~ N 0,1

i iZ Z , i 1,2

1 2 1 2cov Z ,Z n n

1 2Z and Z

1Z

MPS/MSc in StatisticsAdaptive & Bayesian - Lect 21 Lecture 2 Two-stage designs for normally...

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