2015 Duke-Industry Statistics Symposium Sample Size Determination for a Three-arm Equivalence Trial of Poisson and Negative Binomial Data

Victoria Chang Senior Statistician Biometrics and Data Management October 23rd, 2015

Outline

• Equivalence Assessment: The Gold Standard Design

• Hypotheses based on Regression Model

• Numerical Results

• Summary

Equivalence Assessment: The Gold Standard Design

In a three-arm clinical trial, an equivalence assessment may be carried out by demonstrating the following two superiority tests and one equivalence test

Superiority Test

and

Equivalence Test

or

or

**

13

*

103

2:

:

RTa

RT

H

H*

2 RT

(at α = 2.5%)

(at α = 5%)

0:

0:

1

01

PTa

PT

H

H

H02 :mR -mP £ 0

Ha2 :mR -mP > 0

103 :

R

THmTmR

³d2

213 :

R

TaH

3

or

Equivalence Test for Discrete Data Poisson and Negative Binomial Distributed Data

• Example: Frequency of adverse events, asthma, migraines

• Assume lower incidence rate is more favorable

• Poisson : Mean=Variance

Negative Binomial: overdispersion parameter

4

Regression Model

Let 𝑌𝑖𝑗~𝑃𝑜𝑖𝑠 𝜇𝑗𝑡𝑖𝑗 and 𝑍𝑖𝑗~𝑁𝐵 𝜇𝑗𝑡𝑖𝑗 , 𝑘 ,

where 𝑖 = 1,2…𝑛𝑗 , 𝑗 = 0,1,2

Link Function:

log(𝜇𝑗𝑡𝑖𝑗) = log 𝑡𝑖𝑗 + 𝛽0 + 𝛽1𝑥𝑖1 + 𝛽2 𝑥𝑖2

𝑥𝑖1 and 𝑥𝑖2 are indicator functions

Then 𝜇0 = 𝑒𝛽0

𝜇1 = 𝑒𝛽0+𝛽1

𝜇2 = 𝑒𝛽0+𝛽2

5

Hypotheses

• Superiority Test Equivalence Test

𝐻01: 𝜇2

𝜇0≥ 1 𝐻02:

𝜇2

𝜇1≤ 𝛿1

′ or 𝜇2

𝜇1≥ 𝛿2

′

𝐻𝑎1: 𝜇2

𝜇0< 1 𝐻𝑎2: 𝛿1

′<𝜇2

𝜇1< 𝛿2

′

𝐻01: 𝛽2 ≥ 0 𝐻02: 𝛽2 − 𝛽1 ≤ 𝛿1 𝐻03: 𝛽2 − 𝛽1 ≥ 𝛿2

𝐻𝑎1: 𝛽2 < 0 𝐻𝑎2: 𝛽2 − 𝛽1 > 𝛿1 𝐻𝑎3: 𝛽2 − 𝛽1 < 𝛿2

where 𝛿𝑖 = log𝛿𝑖′ 𝑖 = 1,2

• Assume smaller incidence rate is more favorable

6

𝛼 = 2.5% 𝛼 = 5%

Test Statistics and Decision Rule

• Test Statistics:

𝑇1 =𝛽2

𝑉01, 𝑇2 =

𝛽2 −𝛽1 − 𝛿1

𝑉02 , 𝑇3 =

𝛽2 −𝛽1 − 𝛿2

𝑉03

where 𝑉0𝑖 is the variance under 𝐻0𝑖 , 𝑖 = 1,2,3

• Decision Rule:

Claim equivalence of the test and the reference treatments

by rejecting all null hypotheses if

𝑇1 < −𝑍0.025, 𝑇2 < −𝑍0.05 and 𝑇3 > 𝑍0.05

7

Power Function

• Power (𝜇0, 𝜇1, 𝜇2, 𝛿1, 𝛿2)

≈ 𝑃 𝑇1 < −𝑍0.025, 𝑇2 < −𝑍0.05 , 𝑇3 > 𝑍0.05 | 𝜇0, 𝜇1, 𝜇2, 𝛿1, 𝛿2

≈ 𝑃𝜇0,𝜇1,𝜇2,𝛿1,𝛿2 𝑍1 <−𝑍0.025 𝑉01 − 𝛽2

𝑉𝑎1,𝑍0.05 𝑉02 − 𝛽2 − 𝛽1 + 𝛿1

𝑉𝑎2< 𝑍2 <

−𝑍0.05 𝑉03 − 𝛽2 − 𝛽1 + 𝛿2

𝑉𝑎2

• 𝑍1 =𝛽2 −𝛽2

𝑉𝑎1, 𝑍2 =

𝛽2 −𝛽1 −(𝛽2−𝛽1)

𝑉𝑎2

• (𝑍1, 𝑍2)~𝑁00, 1𝜌𝜌1

with 𝜌 =𝑉𝑎𝑟 𝛽2 −𝐶𝑜𝑣(𝛽1 ,𝛽2 )

𝑉𝑎𝑟 𝛽2 ∙𝑉𝑎𝑟(𝛽2 −𝛽1 )

• 𝑉01, 𝑉02 , 𝑉03 are the variances of 𝛽2 and 𝛽2 − 𝛽1 under null

• 𝑉𝑎1, 𝑉𝑎2 are the variances of 𝛽2 and 𝛽2 − 𝛽1 under alternative space

8

Variance Derivation for Poisson Model

• 𝑙 𝜃; 𝑌 = −𝜇𝑗𝑡𝑖𝑗 + 𝑦𝑖𝑗 ∙ log 𝜇𝑗𝑡𝑖𝑗 − 𝑙𝑜𝑔𝑦𝑖𝑗!𝑛𝑗𝑖=1

2𝑗=0

• Assume the exposure time of the subjects in each arm are the same and denoted as 𝑡0, 𝑡1 and 𝑡2

• 𝐼−1 =1

𝑝

1 −1 −1

−1𝑝+𝑞

𝑞1

−1 1𝑝+𝑟

𝑟

where 𝑝 = 𝑛0𝑡0𝑒𝛽0, 𝑞 = 𝑛1𝑡1𝑒

𝛽0+𝛽1 and 𝑟 = 𝑛2𝑡2𝑒𝛽0+𝛽2

• 𝑉𝑎𝑟 𝛽1 = 𝑝−1 + 𝑞−1 =1

𝑛0𝑡0𝜇0+

1

𝑛1𝑡1𝜇1

• 𝑉𝑎𝑟 𝛽2 = 𝑝−1 + 𝑟−1 =1

𝑛0𝑡0𝜇0+

1

𝑛2𝑡2𝜇2

• 𝑉𝑎𝑟 𝛽2 − 𝛽1 = 𝑞−1 + 𝑟−1 =1

𝑛1𝑡1𝜇1+

1

𝑛2𝑡2𝜇2

9

Variances under Null

• Derive 𝑉01 under 𝐻01: 𝛽2 = 0 𝜇0 = 𝜇2

RMLE: 𝜇0 = 𝜇2 =𝜇0 𝑛0𝑡0+𝜇2 𝑛2𝑡2

𝑛0𝑡0+𝑛2𝑡2 , 𝜇1 = 𝜇1

𝑉01= 𝑉𝑎𝑟(𝛽2 )|𝐻01 =1

𝑛0𝑡0𝜇0 +

1

𝑛2𝑡2𝜇2

• Derive 𝑉02 , 𝑉03under 𝐻02 , 𝐻03: 𝛽2 − 𝛽1 = 𝛿𝑖 𝜇2 = 𝑒𝛿𝑖𝜇1

Apply Lagrange multiplier

RMLE: 𝜇1 =𝜇1 𝑛1𝑡1+𝜇2 𝑛2𝑡2

𝑒𝛿𝑖𝑛2𝑡2+𝑛1𝑡1, 𝜇2 = 𝑒

𝛿𝑖𝜇1

𝑉0𝑖= 𝑉𝑎𝑟(𝛽2 − 𝛽1 )|𝐻0𝑖 =1

𝑛1𝑡1𝜇1 +

1

𝑛2𝑡2𝜇2 where 𝑖 = 1,2

10

Sample Size Example for Poisson Data

𝝁𝒕 𝝁𝟎 𝝁𝟏 𝝁𝟐/𝝁𝟏 (𝒂, 𝒃) 𝒏𝟎 Total

3 3 2.4 0.9 (1,1) 132 396

(1,2) 95 380

(2,2) 66 330

1 (1,1) 55 165

(1,2) 44 176

(2,2) 34 170

1/0.9 (1,1) 148 444

(1,2) 131 528

(2,2) 105 525

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• Power = 80%

• 𝛿1, 𝛿2 = 0.8,1.25

• 𝑛1 = 𝑎 ∙ 𝑛0 and 𝑛2 = 𝑏 ∙ 𝑛0

Sample Size Example for Poisson Data

𝝁𝒕 𝝁𝟎 𝝁𝟏 𝝁𝟐/𝝁𝟏 (𝒂, 𝒃) 𝒏𝟎 Total

3 3 2.4 0.9 (1,1) 132 396

(test better)

(1,2) 95 380

(2,2) 66 330

1 (1,1) 55 165

(1,2) 44 176

(2,2) 34 170

1/0.9 (1,1) 148 444

(reference better)

(1,2) 131 528

(2,2) 105 525

12

• Power = 80%

• 𝛿1, 𝛿2 = 0.8,1.25

• 𝑛1 = 𝑎 ∙ 𝑛0 and 𝑛2 = 𝑏 ∙ 𝑛0

Power vs. sample size for the placebo arm of Poisson and negative binomial data for 𝑻 = 𝟑 , 𝝁𝟏 = 𝟏. 𝟖, 𝝁𝟐 = 𝟐, 𝒂 = 𝒃 = 𝟐, and dispersion parameter k =1.

13

0 200 400 600 800 1000

0.0

0.2

0.4

0.6

0.8

1.0

mut=3_Mu1=1.8_Mu2=2_a=2_b=2

n0

po

we

r

Mu0=3Mu0=2.2

Mu0=2.1

Poisson

𝜇0 = 3

𝜇0 = 2.2

𝜇0 = 2.1

𝑛0

0 200 400 600 800 1000

0.0

0.2

0.4

0.6

0.8

1.0

mut=3_Mu1=1.8_Mu2=2_a=2_b=2

n0

po

we

rMu0=3

Mu0=2.2Mu0=2.1

Negative Binomial

𝜇0 = 3

𝜇0 = 2.2 𝜇0 = 2.1

𝑛0

Power vs. placebo incidence rate of Poisson and negative binomial data for 𝑻 = 𝟏 , 𝝁𝟏 = 𝝁𝟐= 𝟐, 𝒂 = 𝒃 = 𝟐 and dispersion parameter k =1.

14

2.5 3.0 3.5 4.0

0.0

0.2

0.4

0.6

0.8

1.0

Poisson_mut=1_Mu1=2_Mu2=2_a=2_b=2

mu0

po

we

r

n.0=100n.0=200

Poisson

𝑛0 = 100 𝑛0 = 200

𝜇0

2.5 3.0 3.5 4.0

0.0

0.2

0.4

0.6

0.8

1.0

NB_Mut=1_Mu1=2_Mu2=2_a=2_b=2_k=1

Mu0

po

we

r

n.0=100

n.0=200

Negative Binomial

𝑛0 = 100 𝑛0 = 200

𝜇0

1.8 1.9 2.0 2.1 2.2

0.2

0.4

0.6

0.8

1.0

Poisson_Mut=1,Mu0=3,Mu1=2,n0=200

mu2

po

we

r

(a,b)=(3,3)

(a,b)=(1,5)(a,b)=(5,1)

Poisson

𝜇2/𝜇1

𝜇2

0.9 0.95 1 1.05 1.1

Power vs. incidence rate of test treatment of both Poisson and negative binomial data for 𝑻 = 𝟏 , 𝝁𝟎 = 𝟑 , 𝝁𝟏= 𝟐, 𝒏𝟎 = 𝟐𝟎𝟎 and k =1.

15

1.8 1.9 2.0 2.1 2.2

0.2

0.4

0.6

0.8

1.0

NB_mut=1_Mu0=3_Mu1=2_N0=200

Mu2

po

we

r

(a,b)=(3,3)

(a,b)=(1,5)(a,b)=(5,1)

Negative Binomial

𝜇2/𝜇1

𝜇2

0.9 0.95 1 1.05 1.1

Power vs. k for different assigned parameter values with required sample size calculated based on the Poisson Model.

16

Misuse of a Poisson Model

Power vs. k for different assigned parameter values. Each power curve is generated under the condition of total sample size = 500.

0.0 0.5 1.0 1.5

0.0

0.2

0.4

0.6

0.8

1.0

mut=3_Mu1=2.4_n0=100

k

po

we

r

Mu1/Mu2=1

Mu1/Mu2=0.9Mu1/Mu2=1/0.9

𝑇 = 3, 𝜇1= 2.4, 𝑛0= 100, 𝑎, 𝑏 = 2,2

𝜇2/𝜇1 = 1

𝜇2/𝜇1 = 0.9 𝜇2/𝜇1 = 1/0.9

0.0 0.5 1.0 1.5

0.0

0.2

0.4

0.6

0.8

1.0

mut=3_Mu1=2.4

k

po

we

r

Mu1/Mu2=1, (a,b)=(2,2)

Mu1/Mu2=1, (a,b)=(1,2)Mu1/Mu2=0.9, (a,b)=(2,2)

Mu1/Mu2=0.9, (a,b)=(1,2)Mu1/Mu2=1/0.9, (a,b)=(2,2)Mu1/Mu2=1/0.9, (a,b)=(1,2)

𝑇 = 3, 𝜇1= 2.4

𝜇2/𝜇1 = 1, (a,b)=(2,2)

𝜇2/𝜇1 = 1, (a,b)=(1,2)

𝜇2/𝜇1 = 0.9, (a,b)=(2,2)

𝜇2/𝜇1 = 0.9, (a,b)=(1,2)

𝜇2/𝜇1 = 1/0.9, (a,b)=(2,2)

𝜇2/𝜇1 = 1/0.9, (a,b)=(1,2)

𝑘 → 0 ∶ 𝑃𝑜𝑖𝑠𝑠𝑜𝑛

Summary – Poisson and Negative Binomial Data

• Power is affected by 𝜇0 when it is very close to 𝜇2, otherwise the power is dominated by the equivalence test.

• When sample size for the test arm and the reference arm are equal, it gives the highest power.

• Assign more subjects to the better treatment gives a higher power for Poisson Model.

• A wrong model may cause a very low power and incorrect interpretation.

17

Reference

• Pigeot, I., Schafer, J., Rohmel, J., & Hauschke, D. (2003). Assessing non-inferiority of a new treatment in a three-arm clinical trial including a placebo. Stat Med, 22(6), 883-899.

• Rohmel, J., & Pigeot, I. (2010). A comparison of multiple testing procedures for the gold standard non-inferiority trial. J Biopharm Stat, 20(5), 911-926.

• Esinhart, J. D., & Chinchilli, V. M. (1994). Extension to the use of tolerance intervals for the assessment of individual bioequivalence. J Biopharm Stat, 4(1), 39-52.

• EMEA (2001). Note for Guidance on the Investigation of Bioavailability and Bioequivalence, The European Medicines Agency Evaluation of Medicines for Human Use. EMEA/EWP/QWP/1401/98, London, United Kingdom.

• FDA (2001). Guidance on Statistical Approaches to Establishing Bioequivalence, Center for Drug Evaluation and Research, the US Food and Drug Administration, Rockville, Maryland, USA.

• FDA (2003a). Guidance on Bioavailability and Bioequivalence Studies for Orally Administrated Drug Products – General Considerations, Center for Drug Evaluation and Research, the US Food and Drug Administration, Rockville, Maryland, USA.

• FDA (2003b). Guidance on Bioavailability and Bioequivalence Studies for Nasal Aerosols and Nasal Sprays for Local Action, Center for Drug Evaluation and Research, the US Food and Drug Administration, Rockville, Maryland, USA.

• Friede, T., Mitchell, C., & Müller‐Velten, G. (2007). Blinded Sample Size Reestimation in Non‐Inferiority Trials with Binary Endpoints. Biometrical Journal, 49(6), 903-916.

• Generic drug (2013). from http://en.wikipedia.org/wiki/Generic_drug

• Hsieh, T-C., Chow, S.-C., Liu, J-P., Hsiao, C-F., Chi, E. (2010). Statistical test for evaluation of biosimilarity in variability of follow-on biologics. J. Biopharm. Stat. 20(1): 75-89.

• Hauschke, D., Kieser, M., Diletti, E., and Burke, M. (1999). Sample Size Determination for Proving Equivalence Based on the Ratio of Two Means for Normally Distributed Data. Stat Med ,18: 93–105.

• Hilton, J. F. (2006). Designs of superiority and noninferiority trials for binary responses are noninterchangeable. Biometrical journal, 48(6), 934-947.

Thank you!

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