Power and Sample Size Calculation for

Log-rank Test under a Non-proportional

Hazards Model∗

Daowen Zhang

Department of Statistics

North Carolina State University

zhang@stat.ncsu.edu

http://www4.stat.ncsu.edu/∼dzhang2/

∗ Joint work with Hui Quan, Department of Biostatistics &

Programming, Sanofi-Aventis

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OUTLINE

1. Motivating example: Rimonabant trial on cardiovascular

risk

2. Review of the log-rank test statistic

3. Distributions of the log-rank test statistic

4. Detailed power calculation

5. Example and simulation Results

6. Summary

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1. Motivating example: Rimonabant trial

• Rimonabant trial: Assess the benefit of Rimonabant on reducingcardiovascular risk.

• Placebo-controlled

• Primary endpoint: time to cardiovascular event; event ratesexpected to be low in each group

• Log-rank test was proposed to assess the treatment effect.

• Power and sample size consideration should also be based on thelog-rank test

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• It is straightforward if treatment effect is characterized byλ1(t)

λ0(t)= eβ ,

λ1(t): hazard of cardiovascular event for treatment

λ0(t): hazard of cardiovascular event for placebo

• If β ≈ 0 and censoring process independent of treatment group,log-rank test statistic T has distribution (Schoenfeld, 1981

Biometrika)

Ta∼ N(β

√θ(1 − θ)D, 1),

θ: Allocation probability to the treatment

D: Expected total # of deaths (under Ha) from both groups

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• Can be used to calculate power and sample size if the treatmenteffect model (PH model) is reasonable.

• However ...

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• Other issues:1. censoring (information cannot be retrieved)

2. drop-out (information can be retrieved during the study)

• How to handle drop-out?1. treat it as censoring: assumption?

2. conduct ITT analysis: efficiency loss?

• Problem: how to calculate power and sample size for eachstrategy? which is better?

• Need to investigate the distribution of the log-rank test statisticfor our problem

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2. Review of the log-rank test statistic

• The (standard) log-rank test statistic

T =U√

v̂ar(U),

where

U =∑

x

{d1(x) − n1(x)

d(x)

n(x)

}

v̂ar(U) =∑

x

n1(x)n0(x)d(x){n(x) − d(x)}n2(x){n(x) − 1}

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• Under H0 : S1(t) = S0(t) ⇐⇒ H0 : λ1(t) = λ0(t),

Ta∼ N(0, 1)

So reject H0 if |T | ≥ zα/2.

• Under Ha : λ1(t) 6= λ0(t) (but λ1(t) ≈ λ0(t)) (Schoenfeld, 1981Biometrika)

Ta∼ N(φ, 1),

where

φ =

√n

∫ ∞0

log{λ1(t)/λ0(t)}π(t){1 − π(t)}V (t)dt[∫ ∞0

π(t){1 − π(t)}V (t)dt]1/2 ,

where V (t) describes process of observing deaths, π(t) −→ θ ifcensoring process is the same in both groups.

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• Special case: PH alternative

Ha :λ1(t)

λ0(t)= eβ (β ≈ 0),

then

Ta∼ N(β

√θ(1 − θ)D, 1),

• Can be used to calculate the power for PH alterative.

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3. Distribution of the log-rank test statistic

• It is reasonable to assume the alternative for our problem:

Ha :λ1(t)

λ0(t)=

1 t ∈ [0, t0)eβ (β ≈ 0) t ∈ [t0,∞)

λ1(t) = hazard of treated group

λ0(t) = hazard of untreated group

• Distributions of the log-rank test statistic under Ha for twostrategies?

1. Strategy 1: Treat drop-out as censoring

2. Strategy 2: Conduct ITT analysis

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Distribution for Strategy 1

• Direct use of the result of Schoenfeld, 1981 (Biometrika) =⇒

Ta∼ N(φ, 1),

φ ≈√

nβ∫ ∞

t0π(t){1 − π(t)}V (t)dt

[∫ ∞0

π(t){1 − π(t)}V (t)dt]1/2

≈ β√

θ(1 − θ) × D̃√D

,

D = total expected # of deaths from two groups in the study

D̃ = total expected # of deaths from two groups after t0.

• Power = P [Z > |φ| − zα/2].

• Concern: approximation good enough? better one?

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• The use of a series of double expectation theorem leads to

φ ≈√

θ(1 − θ) × (1 − e−β)D̃1 + (e

β − 1)D̃0√D

D̃1 = total # of deaths from treated group after t0D̃0 = total # of deaths from untreated group after t0

• Assumption: drop-out independent of the (unerlying) survivaltime had the patient not dropped out; the same in both groups.

• LetD1 = total expected # of deaths from treated group

D0 = total expected # of deaths from untreated group

D∗1 = total expected # of deaths from treated group before t0D∗0 = total expected # of deaths from placebo group before t0

D = D0 + D1, D̃1 = D1 − D∗1, D̃0 = D0 − D∗0

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Distribution for Strategy 2

• Lakatos (1988, Biometrics) derived an approx. dist. of thelog-rank test under any Ha : λ

∗1(t) 6= λ∗0(t) (λ∗1(t) ≈ λ∗0(t)).

• λ∗0(t) = hazard of the group randomized to placeboλ∗1(t) = hazard of the group randomized to treatment

• Partition patient time [0, L = A + F ) = ∪[ti, ti+1) with equalwidth ∆.

0

-

ti ti+1 F L = A + F

A = accrual period, F = follow-up time, L = study length.

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• Under Ha : λ∗1(t) 6= λ∗0(t) (λ∗1(t) ≈ λ∗0(t)):

Ta∼ N(φ, 1),

φ ≈∑

Di

{ξipi

1+ξipi− pi1+pi

}

{∑Di

pi(1+pi)2

}1/2

1. Di = {n1(ti)λ∗1(ti) + n0(ti)λ∗0(ti)}∆= total expected # of deaths in [ti, ti+1)

2. ξi = λ∗1(ti)/λ

∗0(ti)

3. pi = n1(ti)/n0(ti)

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4. n0(ti), n1(ti), number of patients at risk, can be calculated

iteratively:

nk(ti+1) =

nk(ti){1 − λ∗k(ti)∆} ti < Fnk(ti)

{1 − λ∗k(ti)∆ − ∆L−ti

}ti ≥ F

Assume constant accrual rate in [0, A].

• Need to know the hazard function for each (randomized) group.

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• Assume λ0(t) = λ0 =⇒ λ∗0(t) = λ0• Assume drop-out process (has no effect on untreated group)

Z ∼ exp(τ)

• Then it is reasonable to assume λ1(t|Z) as1. Case 1, Z ≤ t0: λ1(t|Z) = λ02. Case 2, Z > t0:

λ1(t|Z) =

λ0 t ∈ [0, t0)λ1 t ∈ [t0, Z)λ̃1 t ∈ [Z,∞)

where λ̃1 ∈ [λ1, λ0]; e.g.,

λ̃1 = wλ1 + (1 − w)λ0.

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• The survival function for group randomized to treatment:

S∗1(t) = E{I(T ≥ t)}= E[E{I(T ≥ t)|Z}]= E{S1(t|Z)}.

• Case 1, Z < t0:

S1(t|Z) = e−λ0t

• Case 2: Z ≥ t0:

S1(t|Z) = e−Λ1(t|Z) =

e−λ0t t ∈ [0, t0)e−λ0t0−λ1(t−t0) t ∈ [t0, Z)e−λ0t0−λ1(Z−t0)−λ̃1(t−Z) t ∈ [Z,∞)

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• Can calculate S∗1(t) and f∗1 (t) and hence

λ∗1(t) =f∗1 (t)

S∗1(t).

• Then can calculate the nc φ in N(φ, 1) for the log-rank test.

• For better numerical accuracy, ∆ needs to be small, say, 1/1000,if unit = year.

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4. Detailed power calculation for strategy 1

• Some assumptions:1. Other than drop-out, end-of-study is the only other censoring

(can be relaxed)

2. [0, A) is the accrual period, a = accrual rate (can be a(t))

3. F = follow-up period, L = A + F = total study length

4. F ≥ t0.5. λ0(t) = λ0.

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• Consider [t, t + dt) in [0, A):

0

-

t t + dt A L

• Average # of patients entering into study in [t, t + dt):

θadt treatment group

(1 − θ)adt placebo group(1)

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• The probability that a patient entering at t is observed to die inthe study (i.e., dies before L) is

P [T ≤ min(L − t, Z)]

• The probability that a patient entering at t is observed to diebefore t0 is

P [T ≤ min(t0, Z)]

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• For placebo group:

P [T ≤ min(L − t, Z)] = E[E{I[T ≤ min(L − t, Z)]|Z}]

The inner expectation can be shown to be

E{I[T ≤ min(L − t, Z)]|Z} =

1 − e−λ0(L−t) Z ≥ L − t1 − e−λ0Z Z < L − t

=⇒

P [T ≤ min(L − t, Z)] = λ0λ0 + τ

− λ0λ0 + τ

e−(λ0+τ)(L−t)

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• The total expected # of deaths in the study for placebo group:

D0 =

∫ A

0

a(1 − θ)P [T ≤ min(L − t, Z)]dt

=a(1 − θ)λ0

λ0 + τ

[A − e

−(λ0+τ)L

λ0 + τ{e(λ0+τ)A − 1}

].

• The total expected # of deaths for placebo group before t0:

D∗0 =

∫ A

0

a(1 − θ)P [T ≤ min(t0, Z)]dt

=aA(1 − θ)λ0

λ0 + τ{1 − e−(λ0+τ)t0}.

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• For treatment group:

P [T ≤ min(L − t, Z)] =(

τ

λ0 + τ− τ

λ1 + τ

)e−(λ0+τ)t0 +

λ0λ0 + τ

− λ1λ1 + τ

e−(λ0−λ1)t0−(λ1+τ)(L−t).

• The total expected # of deaths in the study for treatment group:

D1 = aθ

[KA − λ1

(λ1 + τ)2e−(λ0−λ1)t0−(λ1+τ)L

{e(λ1+τ)A − 1

}].

• The total expected # of deaths for treatment group before t0:

D∗1 =

∫ A

0

aθP [T ≤ min(t0, Z)]dt

=aAθλ0λ0 + τ

{1 − e−(λ0+τ)t0}.

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5. Example and simulation results

• Expect new treatment takes effect after 1 year =⇒ t0 = 1

• Rate to have cardiovascular risk 0.03 per year (λ0 = 0.03)

• Expect 25% reduction when new treatments takes its full effect(λ1 = 0.0225).

• Accrual rate a = 1000 patients/month

• Study length (L = 50) months

• Expect 10% (per year) drop-out rate

• Significance level α = 0.05; targeted power = 0.9

• How long should the accrual period (A) be? And sample size?

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1.0 1.2 1.4 1.6 1.8 2.0

0.85

0.90

0.95

Accrual period in years

Powe

r

Solution 1: 1.313 years

solution 2: 1.385 years

accrual rate: 12000 patients/yearstudy length: 4.17 years

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8000 9000 10000 11000 12000 13000 14000

0.75

0.80

0.85

0.90

0.95

Accrual rate (number of patients per year)

Powe

r

accrual period: 1.42 yearsstudy length: 4.17 years

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3.6 3.8 4.0 4.2 4.4

0.75

0.80

0.85

0.90

0.95

Study length in years

Powe

r

accrual rate: 12000 patients/yearaccrual period: 1.42 years

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0.0 0.2 0.4 0.6 0.8 1.0

0.86

0.88

0.90

0.92

Weight (w)

Powe

r

accrual rate: 12000 patients/yearaccrual period: 1.42 yearsstudy length: 4.17 years

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1.0 1.2 1.4 1.6 1.8 2.0

0.70

0.75

0.80

0.85

0.90

0.95

Accrual period in years

Powe

r

accrual rate: 12000 patients/yearstudy length: 4.17 years

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8000 9000 10000 11000 12000 13000 14000

0.70

0.75

0.80

0.85

0.90

0.95

Accrual Rate (number of patients per year)

Powe

r

accrual period: 17 monthsstudy length: 50 months

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3.6 3.8 4.0 4.2 4.4

0.70

0.75

0.80

0.85

0.90

0.95

Study Length in Year

Powe

r

accrual rate: 12000 patients/yearaccrual period: 17 months

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software: S-plus function logrankpower(

alpha=0.05, signifance level of the log-rank test

lambda0=, hazard for placebo

lambda1=, hazard for treatment

t0=0, t0 used in the formula

wt=0.5, weight for residual treatment effect

tau=0, drop-out rate

acrate=, accrual rate

acperiod=, accrual period

slength=, study lenght (slength-acperiod>t0)

theta=0.5, allocation prob

nsub=1000, number of sub-intervals for ITT analysis

itt=F flag for ITT analysis)

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6. Discussion

• Delayed treatment effect + drop-outs present challenge tostatisticians

• Proposed two strategies:1. Treat drop-outs as censored observations

(a) Assumption: drop-out process independent of (underlying

true) time to event

(b) Drop-out processes almost the same in both groups.

(c) Calculation straightforward

(d) Don’t need to specify the hazard for untreated group

2. Conduct ITT analysis:

(a) May be what regulatory agencies want

(b) May have enough power only if residual treatment effect is

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relatively large (70% in our example)

(c) Can be computationally intensive (small ∆)

(d) Have to specify the hazard for untreated group

• Derived formula easy to use; confirmed by simulation to havegood statistical properties

• Can include other censoring

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