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1 Sample size power calculation Sample Size Estimation AZTPIAIDS AIDSAZT AIDSPI AIDSRNA AZTPr (S A ) = π A, PIPr (S B ) = π B AIDS (sampling)(inference) π A, π B π A - π B = 0.20 PI 20 20AZT, PI HIV-RNA 20 n A = AZT S A = AZTn B = PI S B = PI π A = S A /n A, π B = S B /n B, AIDS =δ = π A - π B 1

2 H 0 : = 0 H A : 0 δh 0 reject H A accept δ π π π π δ Sample α/2 = α/2 = You Sample H 0 God Truth H A H 0 A=B (OK) Type II error (β error) H A Type I error A is not B (α error) truth A B A B Type I error(α error) A B Type II error(β error)fda α error β error 2

3 Truth Sample β -error Power = 1 - β -error power? If difference of means increase, If sample number increased β-error one side 2 AIDS AZT 20PI 30 HIV-RNA α = 0.05 (two sided), β = 0.20 ( power = 80%) α = 0.05 (two sided) δ = ( )/0.3 (1 0.3) (1 0.2) = n =[(Zα + Zβ)/d] 2 = [( )/0.1644] 2 = 290 AZT 290 PI

4 Power 90two sided test α = 0.05, 35% % 45% ( ) / (0.45 x x 0.65) = [( ) / 0.145] = 500 / 50% 60% ( ) / (0.60 x x 0.50) = [( ) / 0.143] = 513 / 4

5 α 5

6 6

7 7

8 success / failure AIDSAZT 24 HIV-RNA µ A PIµ B H 0 : µ A - µ B =0 H A :µ A - µ B 0 AIDS X A, X B µ A - µ B X A - X B X A - X B sample variance (Zα + Zβ) 2 δ 2 δ = µ 1 µ 2 8

9 SD (Zα + Zβ) 2 δ 2 δ = µ 1 µ SD [ 2 x S 2 (Zα + Zβ) 2 / 2 ] = [2 x 20 2 x ( ) 2 ]/10 2 = 63/arm [ 2 x S 2 (Zα + Zβ) 2 / 2 ] = [2 x 20 2 x ( ) 2 ]/6 2 = 175/arm 95(confidence interval) 95(confidence interval) SE Sample size 95CI (π B π A ) π π π π standard error (SE) phase II trial (1 arm) 95CI

10 one arm ππ When π = 0.8, n = 62 When π = 0.7, n = 81 When π = 0.6, n = 93 When π = 0.5, n = 96 When π = 0.4, n = 93 When π = 0.3, n = 81 When π = 0.2, n = 62 When π = 0.1, n = Clinical Equivalence Trials Bio-equivalence under the curve Cmax Clinical Equivalence Trials 10 Clinical Equivalence Trials one side AZT AIDS ddi AIDS D < D < 0.1 ddi AZT AZT OK (standard) 0.7 (new)

11 = πs πn 95% CI upper %CI 0.2 OK 95% CI upper 95CI 0.2 (πs πn)+ 1.65[πS (1 - πs)/n + πs (1 - πs)/n] πs = πn=0.7 sample size 1.65[πS (1 - πs)/n + πs (1 - πs)/n] = [2 x 0.7(1 0.7)/n]=0.2 n = 29 / arm 0.6 πs πn= [πS (1 - πs)/n + πs (1 - πs)/n] = [2 x 0.7(1 0.7)/n]=0.1 n = 115 / arm sample size 11

12 (πs πn)+ 1.65[πS (1 - πs)/n + πs (1 - πs)/n]=( ) [0.80 (1 0.80)/ (1 0.75)/100] = % p 1 p 2 95CI 10simple mastectomynon-equivalent α β δ δ α β δ equivalent 12

13 Hazard function Prob (T>t) = e λt H 0 : λ 1(t) = λ 2(t) H A : λ 1(t) = constant λ 2(t) 2 *hazard rate λ 1 /λ 2 * = ln(λ 1 /λ 2 )/2 d =[(Zα + Zβ)/*] 2 d: X 1 (λ 1 ) (λ 2 )α = 0.05, power = 80% * = ln(λ 1 /λ 2 )/2 = ln(1.5)/2 = d = [( ) / 0.287] 2 = Years of additional follow-up Years of accrual Accrual follow-up accrual 1 follow-up 2 3 Prob (T>t) = e λt

14 t = 11 Prob (T>1) = e λ1 = 0.5 ln (0.5) = λ 1 = 0.69 λ 1 /λ 2 = 1.5 = 0.69/λ 2 λ 2 = 0.46 Hazard function λ Accrual years = A, Follow-up years = F F A/2 A/2 + F accrual 2 2 2/2 + 2 = Prob (T>t) = e λt 3 T failure time Prob (T>3) = e λt = e 0.69 x 3 = Prob (T<3) = = Prob (T>3) = e λt = e 0.46 x 3 = Prob (T<3) = = sample size event %, 74.8% 14

15 (sample size) accrual AIDS randomized clinical trial AIDS AIDS (accrual)1 Type I error 5%, type II error 20% Prob (T>t) = e λt H 0 : λ 1(t) = λ 2(t) H A : λ 1(t) = constant λ 2(t) * = ln(λ 1 /λ 2 )/2 = ln(2.0/1.5)/2 = d = [( ) / 0.203] 2 = Prob (T>t) = e λt 1.5 Prob (T>1.5) = e λ x 1.5 = 0.5 ln (0.5) = λ 1 = 0.46 λ 1 /λ 2 = 2.0/1.5 = 1.33= 0.46/λ 2 λ 2 =

16 accrual 3 1 3/2 + 1 = Prob (T>t) = e λt 2.5 T failure time Prob (T>2.5) = e λt = e 0.46 x 2.5 = Prob (T<2.5) = = Prob (T>2.5) = e λt = e 0.35 x 2.5 = Prob (T<2.5) = = sample size event %, 58.3% (sample size)

17 STATA sample size randomized placebo controlled clinical trial randomization 4 68 pilot study placebo 498 ± 20.2 sec, 485 ± 19.5 sec 0.7 change method α = 0.05 (two sided), 90% power STATA command. sampsi , sd1(20.2) sd2(19.5) method(change) pre(1) post(3) r1(.7) Estimated sample size for two samples with repeated measures Assumptions: alpha = (two-sided) power = m1 = 498 m2 = 485 sd1 = 20.2 sd2 = 19.5 n2/n1 = 1.00 number of follow-up measurements = 3 correlation between follow-up measurements = number of baseline measurements = 1 correlation between baseline & follow-up = Method: CHANGE relative efficiency = adjustment to sd = adjusted sd1 = adjusted sd2 = Estimated required sample sizes: n1 = 20 n2 = 20 placebo 20 17

18 Clinical trials with repeated measures sampsi , sd1(20.2) sd2(19.5) method(change) pre(1) post(3) r1(.7) n1(1 5) n2(15) Estimated power for two samples with repeated measures Assumptions: alpha = (two-sided) m1 = 498 m2 = 485 sd1 = 20.2 sd2 = 19.5 sample size n1 = 15 n2 = 15 n2/n1 = 1.00 number of follow-up measurements = 3 correlation between follow-up measurements = number of baseline measurements = 1 correlation between baseline & follow-up = Method: CHANGE relative efficiency = adjustment to sd = adjusted sd1 = adjusted sd2 = Estimated power: power = placebo 20. sampsi , sd1(20.2) sd2(19.5) method(change) pre(1) post(3) r1(.7) n1(20) n2(15) Estimated power for two samples with repeated measures Assumptions: alpha = (two-sided) m1 = 498 m2 = 485 sd1 = 20.2 sd2 = 19.5 sample size n1 = 20 n2 = 15 18

19 n2/n1 = 0.75 number of follow-up measurements = 3 correlation between follow-up measurements = number of baseline measurements = 1 correlation between baseline & follow-up = Method: CHANGE relative efficiency = adjustment to sd = adjusted sd1 = adjusted sd2 = Estimated power: power =

20 Two-sample test of equality of proportions Yes/no type sample size H 0 reject α = 0.05, power 0.80 sample. sampsi , power(0.8) Estimated sample size for two-sample comparison of proportions Test Ho: p1 = p2, where p1 is the proportion in population 1 and p2 is the proportion in population 2 Assumptions: alpha = (two-sided) power = p1 = p2 = n2/n1 = 1.00 Estimated required sample sizes: n1 = 222 n2 = phase I trial 90. sampsi , power(0.9) Estimated sample size for two-sample comparison of proportions Test Ho: p1 = p2, where p1 is the proportion in population 1 and p2 is the proportion in population 2 Assumptions: alpha = (two-sided) power = p1 = p2 = n2/n1 = 1.00 Estimated required sample sizes: n1 = 287 n2 =

21 300 placebo 150. sampsi , n1(300) r(0.5) Estimated power for two-sample comparison of proportions Test Ho: p1 = p2, where p1 is the proportion in population 1 and p2 is the proportion in population 2 Assumptions: alpha = (two-sided) p1 = p2 = sample size n1 = 300 n2 = 150 n2/n1 = 0.50 Estimated power: power = placebo 2:1 80. sampsi , power(0.8) r(0.5) Estimated sample size for two-sample comparison of proportions Test Ho: p1 = p2, where p1 is the proportion in population 1 and p2 is the proportion in population 2 Assumptions: alpha = (two-sided) power = p1 = p2 = n2/n1 = 0.50 Estimated required sample sizes: n1 = 349 n2 =

22 One sample test of proportion golden standard 75 a = 0.05, 80. sampsi , power(0.8) onesample Estimated sample size for one-sample comparison of proportion to hypothesized value Test Ho: p = , where p is the proportion in the population Assumptions: alpha = (two-sided) power = alternative p = Estimated required sample size: n = historical comparison randomized clinical trial. sampsi , power(0.8) onesample Estimated sample size for one-sample comparison of proportion to hypothesized value Test Ho: p = , where p is the proportion in the population Assumptions: alpha = (two-sided) power = alternative p = Estimated required sample size: n = 29 22

23 Two sample test of equality of means endpoint 105 mmhg SD 10 mmhg 98 SD 10 placebo 2:1 80α = sampsi , p(0.8) r(2) sd1(10) sd2(10) Estimated sample size for two-sample comparison of means Test Ho: m1 = m2, where m1 is the mean in population 1 and m2 is the mean in population 2 Assumptions: alpha = (two-sided) power = m1 = 105 m2 = 98 sd1 = 10 sd2 = 10 n2/n1 = 2.00 Estimated required sample sizes: n1 = 25 n2 = 50 23

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