Session 9 Power and sample size

Transcription

1 Session 9 Power and sample size 9.1 Measure of the treatment difference 9.2 The power requirement 9.3 Application to a proportional odds analysis 9.4 Limitations and alternative approaches 9.5 Sample size reviews Session 9 1

2 9.1 Measure of the treatment difference Let θ measure the advantage of T over C θ > 0 T superior θ = 0 No difference θ < 0 T inferior Session 9 2

3 Binary p T : probability of success on T p C : probability of success on C p (1 p ) T C θ = log e p C(1 p T ) (log - odds ratio) Ordered categorical data (assuming proportional odds) Q kt : probability of being in C k or better on T Q kc : probability of being in C k or better on C log Q (1 Q ) kt kc θ = e Q kc(1 Q kt ) (log - odds ratio) Session 9 3

4 9.2 The power requirement The null hypothesis of no treatment difference should be rejected at significance level α (2-sided), with probability (1 β), for a given magnitude θ = θ R of treatment difference Prior to conducting the study, it is necessary to impose a model for the responses, in order to define the reference improvement Session 9 4

5 Information needed Test of null hypothesis based on the assumption Reject H 0 if ˆ 1 θ ~ N θ, w ˆ w c θ > where w will be a function of 1. sample size 2. unknown parameters Session 9 5

6 We need ( ˆ ) P θ w > c; θ = 0 = α i.e. ( ˆ ) P θ w > c; θ = 0 = α 2 (i) and i.e. ( ˆ ) R P θ w > c; θ = θ = 1 β ( ˆ ) R P θ w > c; θ = θ = 1 β (ii) as it is most unlikely that θ ˆ w < c when θ =θ R Session 9 6

7 Standard normal density If X ~ N (0, 1) then P(X > u γ ) =γ γ γ u 1 γ u 1 γ = u γ u γ by symmetry Now θ ˆ w ~ N( θ w,1) From (i) ( ˆ ) P θ w > c; θ = 0 = α 2 When θ = 0 θˆ w ~ N(0,1) So c = (iii) 2 u α Session 9 7

8 From (ii) (( ˆ ) ( ) ) R R R P θ θ w > c θ w ; θ = θ = 1 β and when θ =θ R, ( θˆ θ ) R w ~ N(0,1) c θ w = u = uβ (iv) So R 1 β Session 9 8

9 Eq (iii) Eq (iv) gives θ w = u + u R α 2 β i.e. w uα 2 + uβ = θr 2 Session 9 9

10 This formula has general validity Can use w = V (Fisher s information) (Whitehead, 1996) To obtain a sample size, w must be related to n - This is the most approximate part of the procedure Session 9 10

11 9.3 Application to a proportional odds analysis Assume proportional odds Denote the log-odds ratio, measuring the advantage of T over C, by θ Specify the difference sought (for which power is to be 1 β) as a value θ R > 0 of θ Session 9 11

12 From Session 3 m 3 ntncn nk V = 1 2 3( n + 1 ) k= 1 n Suppose that it is intended that n T n C, and anticipated that n k /n p, k = 1,, m. Then k m n V 1 p 12 k= 1 3 k so that n = ( ) 2 α 2 + β 12 u u θ m 2 3 R 1 pk j= 1 (9.1) Session 9 12

13 Example: Head injury trial Patients Head injury Treatments Experimental drug vs placebo Response Glasgow Outcome Scale at 3 months Anticipated responses in placebo arm Category Absolute Prob (p kc ) Cumulative Prob (Q kc ) Good recovery Moderate disability Severe disability Vegetative/ Dead Session 9 13

14 Example: Head injury trial Significance test 5% (two-sided) Power 0.9 Clinically relevant difference proportion in Good recovery and Moderate disability categories to move from 0.42 on placebo to 0.52 on experimental drug 0.520( ) θ R = loge = ( ) Session 9 14

15 Under the proportional odds model so that e = ( ) ( ) Q 1 Q θr kt kc Q 1 Q kc kt kt θr e Q QkT = 1 Q 1 Q kc kc that is Q kt = θr e QkC θr 1 Q + e Q ( ) kc kc for k = 1,2,3 Session 9 15

16 For θ R = and anticipated Q kc values Category Cumulative Prob (Q kc ) Cumulative Prob (Q kt ) Good recovery Moderate disability Severe disability Vegetative/ Dead Absolute Prob (p kc ) Absolute Prob (p kt ) Average Absolute Prob ( ) p k Session 9 16

17 Is proportional odds assumption sensible? Session 9 17

18 Is proportional odds assumption sensible? Session 9 18

19 α = β=0.9 uα 2 = u = β p k= 1 3 k = = Hence n = ( + ) = 863 That is the total sample size: 432 patients on each treatment Session 9 19

20 9.4 Limitations and alternative approaches Method is accurate if θ R < 1, and should be avoided if θ R > 2 - in the example θ R = 1 n = 140 θ R = 2 n = 35 To overcome use an exact method (Hilton and Mehta, 1993) bootstrap simulate Session 9 20

21 Kolassa (1995) improves on equation (9.1), using a Cornish-Fisher approximation to the null distribution in place of the normal approximation The method is implemented in the software nquery Advisor Session 9 21

22 nquery Advisor: main menu Session 9 22

23 Entry of category probabilities for the two groups Session 9 23

24 Calculation of power Power is 0.89 for 432 patients per group - as found from equation (9.1) Power is 0.90 for 436 patients per group Session 9 24

25 Lesaffre et al. (1993) present an alternative method based on simulation Hilton (1996) evaluates the robustness of formula (9.1) Julious and Campell (1996) examine (9.1) in the special case of binary data Julious and Campell (1998) present formulae for the calculation of sample size for paired or matched ordered categorical data Session 9 25

26 9.5 Sample size reviews Equation (9.1) is valid provided that: proportional odds hold p 's are anticipated correctly k The latter can be checked at a sample size review Session 9 26

27 Idea 1. Guess p 1,..., pm 2. Calculate n from equation (9.1) : denote value by n 0 3. Take cn 0 observations, c (0, 1) 1 ( e.g. c = 2 ) 4. Estimate p 1,..., pm from blinded data 5. Use estimates to recalculate n, denote value by n 1 6. Collect the remaining data needed to achieve this sample size Session 9 27

28 Notes Final sample size must be cn 0 Can limit to values (n 0, 2n 0 ), for example Gould (1992, 1995) investigated the binary case, showed that type I error unaffected Session 9 28

29 Example in head injury (Bolland et al., 1998) Patients: suffering from severe head injury Treatments: eliprodil vs placebo Outcome: Glasgow Outcome Scale (GOS) six months after randomisation - ordinal Proportion in each category GR MD SD/V/D Placebo Eliprodil Improvement to detect: GR + MD from 0.47 to 0.62 Session 9 29

30 Q (1 Q ) θ R = = = kt kc loge for k 1, 2 Q kc(1 Q kt ) n 12 Uα 2 + Uβ = θ 3 1 p k R j 2 (9.1) α = 0.05, 1 β = 0.9, p1 = 0.222, p2 = 0.323, p3 = giving n = 394 Total sample size rounded up to n 0 = 400 Session 9 30

31 Planned sample size review after responses from 100 patients timing at just beyond 9 months into trial after about 180 patients recruited (assuming entry rate of about 30/month) assessment of the need to adjust sample size for stratification new sample size to be used, n 400 if n 1 + n n = n 1 + n 2 if 400 < n 1 + n 2 < if n 1 + n Session 9 31

32 Actual sample size review responses from 93 patients 2 years into the trial stratification for Glasgow Coma Score at day 0 (4-5 vs 6-8) n 12 Uα 2 + Uβ = 2 θ S 1 p h= 1 3 h kh R j 2 where S h is the proportion of patients in stratum h Session 9 32

33 Proportion in each category GCS at Proportion GR MD SD/V/D day 0 of patients n 1 + n 2 = 444, rounded to 450 recommendation of modest increase from 400 to 450 accepted by the Trial Steering Committee Session 9 33