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WinLD R (16)

WinLD https://www.biostat.wisc.edu/content/lan-demets-method-statistical-programs-clinical-trials WinLD.zip 2

2 1 α = 5% Type I error rate 1 5.0 % 2 9.8 % 3 14.3 % 5 22.6 % 10 40.1 % 3 Type I error rate = 1 (1α) k

Type I error 5% 1 α 5% α 1 α α 1 5.00 % 2 2.50 % 3 1.67 % 4 1.25 % 5 1.00 % 4

1. z-scoreb-value 2. Pocock O'brien-Fleming 3. Lan & DeMets α 4. p 5

2 X i i Y i i D i = Y i -X i N( δ, σ 2 ) σ 2 δ = 0 N Z N Z N 1 v N N i 1 D i S v N N, v N var( S N ) N var( D ) 1 n < N n Z { S S S } / v S / v ( S S ) / N n N n N 1n n+1n Z n B-value n N N n v N 6 var

2 t v n / v N var( S n ) / var( S N ) 0 1 = 0 = 1 = n/n = ()/() z-scores n /v 1/2 n Z() B-valueB() B ( t ) S n ( ), (1) (1) v N t Z t B Z S v N N 7 trial fractioninformation fraction

z-score B-value z-score B-value Z( ),, Z( ) B( ),, B( ) ^ θ = E[Z(1)] = δ / { var(δ) } 1/2... z-score E[Z()] = θ 1/2 cov[z( 1 ), Z( 2 )] = ( 1 / 2 ) 1/2 V[Z()] = 1 B-value E[B()] = θ cov[b( 1 ), B( 2 )] = 1 V[B()] = 8 B() = 1/2 Z()

B-value B-value θ B-value ( 1, B(1) ) = ( 1, Z(1) ) (, B() ) 0 1 9 B() N( θ, )

B-value ^ B() N( θ, ) θ = E[Z(1)] = δ/{ var(δ) } 1/2 B-value ^ ( 1, E[B(1) B(), θ=θ] ) (, B() ) slope = 0 ( 1, E[B(1) B(), θ=0] ) 0 1 10 Lan et.al.1988

1 200 100 1 1 2 100 z-score 1.7 = 100/200 = 0.5z-scoreZ(0.5) = 1.7 B-valueB(0.5) = (0.5) 1/2 Z(0.5) = 0.71.7 = 1.2 α = 0.005 2 Type I error 0.025 z-score c 1 c 2 Pr{ Z(0.5) > c 1 } = 0.005 Pr{ Z(0.5) > c 1 or Z(1.0) > c 2 } = 0.025 c 1 c 2 B-value a 1 a 2 Pr{ B(0.5) > a 1 } = 0.005 Pr{ B(0.5) > a 1 or B(1.0) > a 2 } = 0.025 a 1 a 2 11

22 2 200 100 100 0.3 0.4 ^ ^ = 100/200 = 0.5δ(0.5) = 0.3se{δ(0.5)} = 0.4 z-scorez(0.5) = 0.3/0.4 = 0.75 B-valueB(0.5) = (0.5) 1/2 Z(0.5) = 0.700.75 = 0.53 α = 0.005 Pr{ Z(0.5) > c 1 } = 0.005 c 1 c 1 = 2.576 0.75 < 2.576 12 seqnorm(0.005, lower=f) = 2.576

22 0.6 0.28 ^ ^ = 1δ(1.0) = 0.6se{δ(1.0)} = 0.28 ^ z-scorez(1) = 0.6/0.28 = 2.14θ = E[Z(1)] = 2.14 B-valueB(1) = (1) 1/2 Z(1) = 2.14 Type I error 0.025 Pr{ Z(0.5) > 2.576 or Z(1.0) > c 2 } = 0.025 c 2 c 2 = 2.003 2.14 > 2.003 13 ^ = var[δ(0.5)] -1 ^ / var[δ(1.0)] -1 ^ = { se[δ(0.5)] } -2 ^ / { se[δ(1.0)] } -2 = 0.5

22 Z(1.0) Z(1.0) 2.003 Z(0.5) Z(0.5) 2.576 1 2.576 2 2 1 Pr{ Z(1.0) > 2.003 } Pr{ Z(0.5) 2.576, Z(1.0) > 2.003 } 14

2 15 "Statistical Monitoring of Clinical Trials" findroot()

1 = 0.5 Pr{ Z(0.5) > c 1 } = 0.005 c 1 2 z 1-0.005 = z 0.995 c 1 = 2.576 2 = 1.0 0.005 Pr{ Z(0.5) > 2.576 or Z(1.0) > c 2 } = 0.025 Pr{ Z(0.5) > 2.576 } + Pr{ Z(0.5) 2.576, Z(1.0) > c 2 } = 0.025 Pr{ Z(0.5)2.576, Z(1.0) > c 2 } = 0.02 c 2 2 c 2 = 2.003 1 2 α = 0.025 16

Pr{ Z(0.5)2.576, Z(1.0) > c 2 } = 0.02 c 2?? { Z(0.5)Z(1.0) } (0, 0 ) ( 1, 1 ) (0.5/1.0) 1/2 = 0.7071 c 2 Z(1.0) c 2 0.02 c 2 Z(0.5) 17 2.576

^ ^ δ( i ) c i se{δ( i )} - 2 100 0.3 0.4 c 1 = 2.576 0.3 2.5760.4 = ( -0.73, 1.33 ) 0.6 0.3 c 2 = 2.003 0.6 + 2.0030.3 = ( -, 1.20 ) Stagewise ordering 18 repeated confidence interval

2 2 2 = n/n = ()/() 2 2 19

32 2 2 200 50 100 2 1 3 α 0.025 c 1 c 2 c 3 Pr{ Z(0.25) > c 1 or Z(0.5) > c 2 or Z(1.0) > c 3 } = 0.025 c 1 c 2 c 3 50 100 Z() B-value B() 50 1 = 0.25Z(0.25) = 1.0B(0.25) = 0.25 1/2 1.0 = 0.5 100 2 = 0.5Z(0.5) = 1.7B(0.2) = 0.5 1/2 1.7 = 1.2 20

32 180 1 = 50/180 = 0.28 2 = 0.56 3 =1 Pr{ Z(0.28) > c 1 or Z(0.56) > c 2 or Z(1.0) > c 3 } = 0.025 c 1 c 2 200 200 180 (Z( ), Z( ), Z( )) 0 1 ( / ) 1/2 3 (50/200)(100/200) = (50/180)(100/180) = 50/100 Z( ) ( Z( ), Z( )) 21

32 180 1 = 50/180 = 0.28 2 = 0.56 3 =1 c 3 Pr{ Z(0.28) > c 1 or Z(0.56) > c 2 or Z(1.0) > c 3 } = 0.025 c 3 c 1 c 2 (Z( ), Z( ), Z( )) 0 1 ( / ) 1/2 3 50/200 50/180 100/200 100/180 ( Z( ), Z( ), Z( )) 22

32 50 100 200 1 α 0.005 Pr{ Z(0.25) > c 1 } = 0.005 c 1 c 1 = 2.58 2 α 0.01 Pr{ Z(0.25) > 2.58 or Z(0.5) > c 2 } = 0.01 c 2 c 2 = 2.49 3 α 0.025 Pr{ Z(0.25) > 2.58 or Z(0.5) > 2.49 or Z(1.0) > c 3 } = 0.025 c 3 c 3 = 2.09 180 Pr{ Z(0.28) > 2.58 or Z(0.56) >2.49or Z(1.0) >c 3 } = 0.025 c 3 c 3 = 2.08 23

32 24

32 25

1. z-scoreb-value 2. Pocock O'brien-Fleming 3. Lan & DeMets α 4. p 26

Pocock O'brien-Fleming (Z( ),, Z( )) 0 1 ( / ) 1/2 k Type I error0.025 N N/k Pocock z-score O'brien-Fleming B-value z-score 27

8 1 2 3 28

Pocock k Pr{ 1Z ( i / k ) c ( k )} i α = 0.025 k=1c(k) = 1.96 k=2c(k) = 2.18 1 2 2.18 k=3c(k) = 2.29 1 3 2.29 k=4c(k) = 2.36 1 4 2.39 k=5c(k) = 2.41 1 5 2.41 29

Pocock k=3

Pocock 31 0.975 c

O'brien-Fleming Pr{ k i 1 B ( i / k ) a( k )} Pr{ k i 1 Z ( i / k ) c ( k )} Pr{ k i 1 Z ( i / k ) a( k ) / t 1 / 2 } α = 0.025k = 5 a(5) = 2.04 Z() = B()/ 1/2 c(1) = 2.04/(1/5) 1/2 = 4.56 c(2) = 2.04/(2/5) 1/2 = 3.23 c(3) = 2.04/(3/5) 1/2 = 2.63 c(4) = 2.04/(4/5) 1/2 = 2.28 c(5) = 2.04/(5/5) 1/2 = 2.04 32

O'brien-Fleming k=3k=5

O'brien-Fleming 34 0.975 c

Pocock O'brien-Fleming Pocock 5 O'brien-Fleming 5 35

Pocock O'brien-Fleming Pocock z-score O'brien-Fleming B-value z-score 36

1. z-scoreb-value 2. Pocock O'brien-Fleming 3. Lan & DeMets α 4. p 37

Lan & DeMets α 1 n 2 2n 3 3n Lan & DeMets α α() α(0) = 0 α(1) = 0.025 0.025 α() 38 } ) ( ) ( Pr{ ) ( ) ( ), 1, ( } ) ( Pr{ ) ( 1 1 1 1 j j i i j i j j i i j i j c t Z c t Z t t k j c t Z t

Lan & DeMets α Pocock α α p1 () = 0.025 log{ 1 - (e-1) } Pocock α α p2 () = 0.05 log{ 1 - (e-1) } O'brien-Fleming α α of1 () = 2 { 1 - Φ(z 0.0125 / 1/2 ) } = 2 { 1 - Φ(2.2414/ 1/2 ) } O'brien-Fleming α α of2 () = 4 { 1 - Φ(z 0.0125 / 1/2 ) } = 4 { 1 - Φ(2.2414/ 1/2 ) } Pocock O'brien-Fleming α 39 2.5% 5%

Lan & DeMets α 40

Lan & DeMets α Pocock z-score O'brien-Fleming B-value 41

4α = 0.2, 0.5, 0.8, 1.0 O'brien-Fleming α = 0.025 c 1 = 4.8769c 2 = 2.9626c 3 = 2.266c 4 = 2.0278 42

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4α 44

3 45

WinLD Bounds Interim analyses: 4 +[Enter] Info. times: User Input Test Bound.: One-Sided Overall Alpha: 0.025 Function: O'brien-Fleming Time Upper Bound

52 2 200 100 α 0.025 2 α O'brien-Fleming 100 1.5 = 100/200 = 0.5 1.5 2.96 α = 0.0015 2.96 α = 0.0015 3.0 1.5 < 3.0 47 qt(0.0015, df=200-2, lower=f)

52 48

2 2 200 2 1 α0.025 α α() = 0.025 1.5 1 2 50 100 = 0.250.5α() = 0.003125, 0.008838 c 1 = 2.73c 2 = 2.47 180 3 c 1 c 2 c 3 c 3 = 2.064 49

2 2 200 2 1 α0.025 α α() = 0.025 1.5 1 100 = 0.5α() = 0.008838 c 1 = 2.3723 250 = 0.4 = 0.5 α α i ' * ( t ) i { * ( t ) * ( t i )} * ( t i ) 51 α α i α i α α () α

2 50 100 = 0.250.5α() = 0.003125, 0.008838 250 = 0.20.4 = 0.250.5 α ' * ( t ) 0.008838 0.025 0.008838 0.025 0.025(0.4) 0.003364 0.021635t 1.5 150 = 0.6 α ' * (0.6) 0.01341985 = 0.4,0.6 c 2 = 2.4183 1.5 {0.025t 1.5 0.025(0.4) 1.5 } 52

1 100 0.5 0.4 α 53

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α Pocock O'brien-Fleming α(0) = 0 α(1) = 0.025 α() = 0.025 α() = 0.025 1.5 Z 1, Z 2,... Type I error α 55

1. z-scoreb-value 2. Pocock O'brien-Fleming 3. Lan & DeMets α 4. p 56

p 9 z??? 57 1 2

p p p z p 2.5? 2.2 2.0 z 58 2.5 p 2.5 1 2

p =0.2z = 2.0 =0.5z = 2.5 (, z ) (, z ) z-score ( 2, z 2 ) ( 1, z 1 ) z-score ordering z-score ( 2, z 2 ) ( 1, z 1 ) z 2 z 1 B-value orderingb-value ( 2, B( 2 ) ) ( 1, B( 1 ) ) 2 1/2 z 2 1 1/2 z 1 MLE ordering Stagewise ordering z-score ( 2, z 2 ) ( 1, z 1 ) 2 1 or 2 = 1 z 2 z 1 59

p 6 3 2 (, z ) (0.2, 2.0) (0.5, 2.5) 2.2 z-score orderingpr{ (, Z ) (0.5,2.5) } p = Pr{ Z(0.2) 2.2 } + Pr{ Z(0.2) 2.2, Z(0.5) 2.5} + Pr{ Z(0.2) 2.2, Z(0.5) 2.2, Z(1.0) 2.5 } p B-value ordering Pr{ B() 0.5 1/2 2.5) } = Pr{ B() 1.8 } p = Pr{ B(0.2) 1.8 } + Pr{ B(0.2) 1.0, B(0.5) 1.8} + Pr{ B(0.2) 1.0, B(0.5) 1.6, B(1.0) 1.8 } p MLE ordering Stagewise ordering p 60 B() = 1/2 Z()1.0 = 0.2 1/2 2.21.6 = 0.5 1/2 2.2

Stagewise ordering p 6 3 2 j = 2 (, z ) (0.2, 2.0) (0.5, 2.5) 2.2 Stagewise ordering p = Pr{ Z(0.2) 2.2 } p Pr{ Z ( t ) c } Pr{ Z ( t ) j 1 i 1 i i j j + Pr{ Z(0.2) 2.2, Z(0.5) 2.5} = 0.018251.8% p z } 2.5 2.2 2.0 61 1 2

Stagewise ordering p 6 3 2 j = 2 (, z ) (0.2, 2.0) (0.5, 2.5) 2.2 Stagewise ordering p = Pr{ Z(0.2) 2.2 } p Pr{ Z ( t ) c } Pr{ Z ( t ) j 1 i 1 i i j j + Pr{ Z(0.2) 2.2, Z(0.5) 2.5} p z } 62

7p Probability Interim analyses: 5 +[Enter] Info. times: Equally Spaced Test Bound.: One-Sided Upper Bound Determine Bounds: User Input 4 1 = 0.2, 2 = 0.4, 3 = 0.6, 4 = 0.8, 5 = 1.0 Upper Bound 4.56, 3.23, 2.63, 2.28, 2.04 O'Brien Fleming 3 z-score 2.94 Stagewise Ordering p 0.001990.2% 63

X i i Y i i D i = Y i -X i N( δ, σ 2 ) σ 2 ^ ^ ^ ^ 95% ( δ L, δ U ) = ( δ - 1.96se(δ), δ + 1.96se(δ) ) ^ z-score z obs Z N( δ L /se(δ), 1 ) δ L Pr{ Z z obs } = α/2 = 0.025 δ L 64 Pr Z ˆ L se ( ˆ) z obs z PrZ PrZ se ( ˆ) L L se( ˆ) ˆ z z obs ˆ L se ( ˆ) L se ( ˆ) / 2 se ( ˆ) ˆ 1.96 se ( ˆ) 1 / 2 L 1 /2 α = 0.05

N(δ,1) z obs δ z α/2 ^ N(δ L /se(δ),1) Pr{Zz obs } = α/2 α/2 ^ δ L /se(δ) z obs 65 95%

X i Y i D i = Y i -X i N( δ, σ 2 ) σ 2 ^ ^ ^ 95% ( δ L, δ U ) = ( δ - 1.96se(δ), δ + 1.96se(δ) ) ^ z-score z obs Z N(δ U /se(δ), 1 ) δ U Pr{ Z z obs } = 0.025 δ U ^ δ L /se(δ) z obs ^ δ U /se(δ) 66 95%

δ L δ U ^ ^ δ = 3se(δ) = 1z obs = 3 [ -7, 7 ] δ L δ L = -7 Z N( -7, 1 ) Pr{ Z 3 } = 1-Φ(10) = 0.000... δ L = 0 Z N( 0, 1 ) Pr{ Z 3 } = 1-Φ(3) = 0.001 δ L = 3.5 Z N( 3.5, 1 ) Pr{ Z 3 } = 1-Φ(-0.5) = 0.308 δ L = 1.7 Z N( 1.7, 1 ) Pr{ Z 3 } = 1-Φ(1.3) = 0.096 δ L = 0.8 Z N( 0.8, 1 ) Pr{ Z 3 } = 1-Φ(2.2) = 0.014 δ L = 1.0 Z N( 1.0, 1 ) Pr{ Z 3 } = 1-Φ(2) = 0.023 δ L = 1.04 Z N( 1.04, 1 ) Pr{ Z 3 } = 1-Φ(1.96) = 0.025 δ L = 1.04 δ U = 4.96 δ L δ U ( = grid search) 67 δ L = 3-1.961 = 1.04

Stagewise ordering Pr{ (, Z ) ( obs,z obs ) } = 0.025 δ L Pr{ (, Z ) ( obs,z obs ) } = 0.975 δ U Pr{ (, Z ) ( obs,z obs ) } = 0.5 δ mid ^ θ = δ / se(δ) Pr{ (, Z ) ( obs,z obs ) } = 0.025 θ L Pr{ (, Z ) ( obs,z obs ) } = 0.975 θ U Pr{ (, Z ) ( obs,z obs ) } = 0.5 θ mid 68

Stagewise ordering Pr{ (, Z ) ( obs,z obs ) } = 0.025 θ L Pr{ (, Z ) ( obs,z obs ) } = 0.975 θ U Pr{ (, Z ) ( obs,z obs ) } = 0.5 θ mid z 2.2 2.5 Pr{ Z(0.2) 2.5 } = 0.025 ^ θ L = δ L /se(δ) Z(0.2) H 0 θ = θ L N( 0.2 1/2 θ L, 1 ) 1 ( = 0.2 ) 2 ( = 0.5 ) 69

Stagewise ordering Pr{ (, Z ) ( obs,z obs ) } = 0.025 θ L Pr{ (, Z ) ( obs,z obs ) } = 0.975 θ U Pr{ (, Z ) ( obs,z obs ) } = 0.5 θ mid z 2.2 2.0 1 ( = 0.2 ) 2.5 2 ( = 0.5 ) Pr{ Z(0.2) 2.5 } + Pr{ Z(0.2) 2.2, Z(0.5) 2.5 } ^ = 0.025 θ L = δ L /se(δ) ( Z(0.2), Z(0.5) ) H 0 θ = θ L (0.2 1/2 θ L, 0.5 1/2 θ L ) 1 (0.2/0.5) 1/2 = 0.63 70

WinLD Stagewise ordering Pr{ (, Z ) ( obs,z obs ) } = 0.025 θ L Pr{ (, Z ) ( obs,z obs ) } = 0.975 θ U Pr{ (, Z ) ( obs,z obs ) } = 0.5 θ mid ^ ^ θ = δ / se(δ) se(δ) 71

8 repeated confidence interval 1 = 0.35, 2 = 0.65, 3 = 1.02 3.5521, 2.5581, 1.9893 2 z = 2.8 = 0.083 Pr{ (, Z ) ( obs,z obs ) } = 0.025 θ L Pr{ Z(0.35) 3.5521 } + Pr{ Z(0.35) 3.5521, Z(0.65) 2.8} = 0.025 θ L ( Z(0.35), Z(0.65) ) 0.35 1/2 θ L 0.65 1/2 θ L 1 (0.35/0.65) 1/2 = 0.73 θ L grid-search θ L = 1.0 (0.35 1/2, 0.65 1/2 ) = (0.6, 0.8) 1 0.73 Pr{ Z(0.35)3.5521 } + Pr{ Z(0.35)3.5521, Z(0.65)2.8 } 0.9750.5 95% [1.03430.083, 5.90160.083] = (0.085, 0.489) 3.95540.083 = 0.328 72

8 73

8WinLD CI Determine Bounds: Spending Function User Input Standardized Statistics: 2.8 Confidence interval: 0.95 Overall Alpha: 0.05 Function: O'brien-Fleming Time () 1 = 0.35, 2 = 0.65, 3 = 1.02 O'brien-Fleming 3.6128, 2.5503,... 2 z = 2.8 = 0.083 95% [1.03430.083, 5.90160.083] = (0.085, 0.489) 74

9 repeated confidence interval CI Overall Alpha: 0.05 Function: O'brien-Fleming Determine Bounds: Spending Function User Input Standardized Statistics: 0.405 Confidence interval: 0.95 Time 1 = 0.36, 2 = 0.65, 3 = 1.0 O'brien-Fleming 3.5521, 2.5581, 1.9893 3 z = 0.405 = 0.046 95% [-1.55500.046, 2.36560.046] = (-0.071, 0.108) 75

10Median Unviased Estimator Drift Determine Bounds: User Input Power: 0.5 4 () () 1 = 0.15, 2 = 0.25, 3 = 0.4, 4 = 0.7, 5 = 1.03 5.67, 4.33, 3.36, 2.44, 2.00 3 z = 3.4785 = 0.06 5.49820.06 = 0.330 76

Stagewise ordering Stagewise ordering p p α 1 p Stagewise ordering p p α θ L > 0 θ U < 0 1 repeated confidence interval Stagewise ordering Stagewise ordering 77 mid-point estimator

1. 2. z-scoreb-value 3. Pocock O'brien-Fleming 4. Lan & DeMets α 5. p 78

Probability 1 = 0.17, 2 = 0.33, 3 = 0.50, 4 = 0.67, 5 = 0.83, 6 = 1.0 5.029, 3.556, 2.903, 2.514, 2.249, 2.053 1 = 59.035 9.9 θ = δ/(2σ 2 /18) 1/2 = 3.333 δ = 2, σ = 1.8 89% 79

B() = b B(1) > z α/2 θ z-score E[ B(1) - B() ] = θ(1 - ) V[ B(1) - B() ] = 1 - E θ [ B(1) B()=b ] = b + θ(1 - ) V θ [ B(1) B()=b ] = 1 - CP ( t ) z 1 E [ B (1) B ( t ) / 2 1 t θ b] θ θ = 0 θ ^ = B()/ 80

= 0.75B(0.75) = 0.5 slope = E[B()]=3.84E[B(1) B(0.75)=0.5]=1.46 V[B(1) B(0.75)=0.5]=0.25CP 3.84 = 1 - Φ{ (1.96-1.46)/0.25 1/2 } = 1 - Φ(1) = 0.16 ( α=0.05 ) slope = 0 CP 0 = 1 - Φ{ (1.96-0)/0.25 1/2 } 0 ( 1, E[B(1) B(), θ=3.84] ) (, B() ) slope = 0 ( 1, E[B(1) B(), θ=0] ) 0 1 81 Lan et.al.1988

Statistical Monitoring of Clinical Trials Michael A. Proschan et. al.springer The B-Value: A Tool for Monitoring Data K.K.Grodon Lan et.al.biometrics1988 Multiple Comparisons Using RFrank Bretz et. al.crc press The R Tips 2 R 82