542-05-#1 STATISTICS 542 Introduction to Clinical Trials SAMPLE SIZE ISSUES Ref: Lachin, Controlled...

542-05-#1

STATISTICS 542STATISTICS 542Introduction to Clinical TrialsIntroduction to Clinical Trials

SAMPLE SIZE ISSUESSAMPLE SIZE ISSUES

Ref: Lachin, Controlled Clinical Trials 2:93-113, 1981.

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Sample Size IssuesSample Size Issues

• Fundamental Point

Trial must have sufficient statistical power to detect differences of clinical interest

• High proportion of published negative trials do not have adequate power

Freiman et al, NEJM (1978)

50/71 could miss a 50% benefit

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Example: How many subjects?Example: How many subjects?

• Compare new treatment (T) with a control (C)

• Previous data suggests Control Failure Rate (Pc) ~ 40%

• Investigator believes treatment can reduce Pc by 25%

i.e. PT = .30, PC = .40

• N = number of subjects/group?

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• Estimates only approximate– Uncertain assumptions– Over optimism about treatment– Healthy screening effect

• Need series of estimates– Try various assumptions– Must pick most reasonable

• Be conservative yet be reasonable

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Statistical ConsiderationsStatistical Considerations

Null Hypothesis (H0):No difference in the response exists between treatment and control groups

Alternative Hypothesis (Ha):A difference of a specified amount () exists between treatment and control

Significance Level (): Type I ErrorThe probability of rejecting H0 given that H0 is true

Power = (1 - ): ( = Type II Error)The probability of rejecting H0 given that H0 is not true

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Standard Normal DistributionStandard Normal Distribution

Ref: Brown & Hollander. Statistics: A Biomedical Introduction. John Wiley & Sons, 1977.

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Standard Normal TableStandard Normal Table


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Distribution of Sample Means (1)Distribution of Sample Means (1)


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Distribution of Test StatisticsDistribution of Test Statistics

• Many have a common form• Theta = population parameter (eg

difference in means)• Thetahat = sample estimate• Then

– Z = Thetahat – E(thetahat)/SE(thetahat)

• And then Z has a Normal (0,1) distribution

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• If statistic z is large enough (e.g. falls into red area of scale), we believe this result is too largeto have come from a distribution with mean O (i.e. Pc - Pt = 0)

• Thus we reject H0: Pc - Pt = 0, claiming that their exists 5% chance this result could have come from distribution with no difference

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Normal DistributionNormal Distribution


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Two Groups0

00

Tc

11 TC

OR )1,0(~/2

Nn

XXZ TC

or

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Test StatisticsTest Statistics

)~

(

)~

(~

~

v

estimate sample

parameter population

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Test of HypothesisTest of Hypothesis• Two sided vs. One sidede.g. H0: PT = PC H0: PT < PC

• Classic test z = critical value

If |z| > z If z > z

Reject H0 Reject H0

= .05 , z = 1.96 = .05, z = 1.645

where z = test statistic• Recommend

z be same value both cases (e.g. 1.96) two-sided one-sided = .05 or = .025 z = 1.96 1.96

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Typical Design Assumptions (1)Typical Design Assumptions (1)

1. = .05, .025, .01

2. Power = .80, .90

Should be at least .80 for design

3. = smallest difference hope to detect

e.g. = PC - PT

= .40 - .30

= .10 25% reduction!

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Typical Design Assumptions (2)Typical Design Assumptions (2)

Z 1 - Z 0.05 1.96 0.80 0.84 0.025 2.24 0.90 1.282 0.01 2.58 0.95 1.645

Two SidedPowerSignificance Level

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Sample Size ExerciseSample Size Exercise

• How many do I need?

• Next question, what’s the question?

• Reason is that sample size depends on the outcome being measured, and the method of analysis to be used

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Simple Case - BinomialSimple Case - Binomial1. H0: PC = PT

2. Test Statistic (Normal Approx.)

3. Sample Size

Assume

• NT = NC = N

• HA: = PC - PT

)/1/1)(1(

ˆˆ

TC

TC

NNpp

ppZ

TC

TTCC

NN

PNPNp

ˆˆ

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Sample Size Formula (1)Sample Size Formula (1)Two ProportionsTwo Proportions

Simpler Case

• Z = constant associated with P {|Z|> Z } = two sided!

(e.g. = .05, Z =1.96)

• Z = constant associated with 1 - P {Z< Z} = 1-

(e.g. 1- = .90, Z =1.282)

• Solve for Z (1- ) or

2

2 )1()(2

ppZZ

N

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Sample Size Formula (2)Sample Size Formula (2)Two ProportionsTwo Proportions

• Z = constant associated with P {|Z|> Z } = two sided!

(e.g. = .05, Z =1.96)

• Z = constant associated with 1 - P {Z< Z} = 1- (e.g. 1- = .90, Z =1.282)

2

2])1()1()1(2[

TTCC PPPPZppZN

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Sample Size FormulaSample Size Formula

Power

• Solve for Z 1-

Difference Detected

• Solve for

Zqp

NZ

2

2

)(

N

qpZZ

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Simple Example (1)Simple Example (1)

• H0: PC = PT

• HA: PC = .40, PT = .30

= .40 - .30 = .10

• Assume

= .05 Z = 1.96 (Two sided)

1 - = .90 Z = 1.282

• p = (.40 + .30 )/2 = .35

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Simple Example (2)Simple Example (2)

Thus

a.

N = 4762N = 952

b.

2N = 956 N = 478

2

2

)3.4(.

])6)(.4(.)7)(.3(.282.1)65)(.35(.296.1[

N

478)3.4(.

)65)(.35(.)282.196.1(22

2

N

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Approximate* Total Sample Size for Comparing Various Approximate* Total Sample Size for Comparing Various Proportions in Two Groups with Significance Level (Proportions in Two Groups with Significance Level () )

of 0.05 and Power (1-of 0.05 and Power (1-) of 0.80 and 0.90) of 0.80 and 0.90True Proportions = 0.05

(one-sided)

= 0.05(two-sided)

pC(Control)

pI(Invervention)

1-0.90

1-0.80

1-0.90

1-0.80

0.60 0.50 850 610 1040 7800.40 210 160 260 2000.30 90 70 120 900.20 50 40 60 50

0.50 0.40 850 610 1040 7800.30 210 150 250 1900.25 130 90 160 1200.20 90 60 110 80

0.40 0.30 780 560 960 7200.25 330 240 410 3100.20 180 130 220 170

0.30 0.20 640 470 790 5900.15 270 190 330 2500.10 140 100 170 130

0.20 0.15 1980 1430 2430 18100.10 440 320 540 4000.05 170 120 200 150

0.10 0.05 950 690 1170 870

*Sample sizes are rounded up to the nearest 10

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Comparison of MeansComparison of Means

• Some outcome variables are continuous– Blood Pressure– Serum Chemistry– Pulmonary Function

• Hypothesis tested by comparison of mean values between groups, or comparison of mean changes

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Comparison of Two MeansComparison of Two Means

• H0: C = TC - T = 0

• HA: C - T =

• Test statistic for sample means ~ N ()

• Let N = NC = NT for design

• Power

)/1/1(2TC

TC

NN

XXZ

~N(0,1) for H0

2

2

2

22

)/(

)(2)(2

ZZZZN

ZNZ )/(2/

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ExampleExamplee.g. IQ = 15 = 0.3x15 = 4.5

• Set 2 = .05

= 0.10 1 - = 0.90

• HA: = 0.3 / = 0.3

• Sample Size

• N = 234

2N = 468

222

2

)3.0(

02.21

)3.0(

)51.10(2

)3.0(

)282.196.1(2

N

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Comparing Time to Event Comparing Time to Event DistributionsDistributions

• Primary efficacy endpoint is the time to an event

• Compare the survival distributions for the two groups

• Measure of treatment effect is the ratio of the hazard rates in the two groups = ratio of the medians

• Must also consider the length of follow-up

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Assuming Exponential Assuming Exponential Survival DistributionsSurvival Distributions

H : =

10 11

2

2

H :

1a 11

2

2

If P (T > t) = e w here = in group 1,- t1

,

= in group 2, le t2

= = m ed m ed w here m ed1 2 2 1 i / / ln (. ) / 5 i

• Then define the effect size by

• Standard difference

•

ln ( ) 2

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Time to Failure (1)Time to Failure (1)• Use a parametric model for sample size

• Common model - exponential– S(t) = e-t = hazard rate– H0: I = C

– Estimate N

George & Desu (1974)

• Assumes all patients followed to an event

(no censoring)

• Assumes all patients immediately entered

NZ Z

c I

2 2

2

( )

[ln ( / )]

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Assuming Exponential Assuming Exponential Survival DistributionsSurvival Distributions

• Simple case

• The statistical test is powered by the total number of events observed at the time of the analysis, d.

d = 4(Z + Z )

[ln( )]

2

2

=

C

I

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Converting Number of Events Converting Number of Events (D) to Required Sample Size (2N)(D) to Required Sample Size (2N)

• d = 2N x P(event) 2N = d/P(event)• P(event) is a function of the length of total follow-

up at time of analysis and the average hazard rate• Let AR = accrual rate (patients per year)

A = period of uniform accrual (2N = AR x A)F = period of follow-up after accrual completeA/2 + F = average total follow-up at planned

analysis = average hazard rate

• Then P(event) = 1 – P(no event) =

1 e - (A / 2+ F)

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Time to Failure (2)Time to Failure (2)• In many clinical trials

1. Not all patients are followed to an event

(i.e. censoring)

2. Patients are recruited over some period of time

(i.e. staggered entry)

• More General Model (Lachin, 1981)

where g() is defined as follows

2

2

)(

)}()({)(

IC

IC ggzzN

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1. Instant Recruitment Study Censored At Time T

2. Continuous Recruiting (O,T) & Censored at T

3. Recruitment (O, T0) & Study Censored at T (T > T0)

Teg

1)(

2

)1()(

3

TeT

Tg

0

)(

2

0

1

)(

Tee

gTTT

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ExampleAssume = .05 (2-sided) & 1 - = .90

C = .3 and I = .2T = 5 years follow-upT0 = 3

0. No Censoring, Instant Recruiting

N = 128

1. Censoring at T, Instant Recruiting

N = 188

2. Censoring at T, Continual Recruitment

N = 310

3. Censoring at T, Recruitment to T0

N = 233

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Sample Size Adjustment Sample Size Adjustment for Non-Compliance (1)for Non-Compliance (1)

• References:1. Shork & Remington (1967) Journal of Chronic Disease

2. Halperin et al (1968) Journal of Chronic Disease

3. Wu, Fisher & DeMets (1988) Controlled Clinical Trials

• Problem

Some patients may not adhere to treatment protocol

• Impact

Dilute whatever true treatment effect exists

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Sample Size Adjustment Sample Size Adjustment for Non-Compliance (2)for Non-Compliance (2)

• Fundamental PrincipleAnalyze All Subjects Randomized

• Called Intent-to-Treat (ITT) Principle– Noncompliance will dilute treatment effect

• A SolutionAdjust sample size to compensate for dilution effect (reduced power)

• Definitions of Noncompliance– Dropout: Patient in treatment group stops taking

therapy– Dropin: Patient in control group starts taking

experimental therapy

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Comparing Two Proportions– Assumes event rates will be altered by

non‑compliance– Define

PT* = adjusted treatment group rate

PC* = adjusted control group rate

If PT < PC,

0

PT PC

PT * PC *

1.0

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Simple Model - Compute unadjusted N– Assume no dropins– Assume dropout proportion R– Thus PC* = PC

PT* = (1-R) PT + R PC

– Then adjust N

– ExampleR 1/(1-R)2 % Increase

.1 1.23 23% .25 1.78 78%

2)1(*

R

NN

Adjusted Sample SizeAdjusted Sample Size

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Sample Size Adjustment Sample Size Adjustment for Non-Compliancefor Non-Compliance

Dropouts & dropins (R0, RI)

– ExampleR0 R1 1/(1- R0- R1)2 %

Increase

.1 .1 1.56 56%

.25 .25 4.0 4 times%

20 )1(

*IRR

NN

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• More Complex ModelRef: Wu, Fisher, DeMets (1980)

• Further Assumptions– Length of follow-up divided into intervals– Hazard rate may vary– Dropout rate may vary– Dropin rate may vary– Lag in time for treatment to be fully effective

Sample Size AdjustmentsSample Size Adjustments

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• Used complex model

• Assumptions

1. = .05 (Two sided) 1 - = .902. 3 year follow-up

3. PC = .18 (Control Rate)

4. PT = .13 Treatment assumed

28% reduction5. Dropout

26% (12%, 8%, 6%)6. Dropin

21% (7%, 7%, 7%)

Example: Beta-Blocker Example: Beta-Blocker Heart Attack Trial (BHAT) (1)Heart Attack Trial (BHAT) (1)

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Unadjusted Adjusted

PC = .18 PC* = .175

PT = .13 PT* = .14

28% reduction 20% reduction

N = 1100 N* = 2000

2N = 2200 2N* = 4000

Example: Beta-Blocker Example: Beta-Blocker Heart Attack Trial (BHAT) (2)Heart Attack Trial (BHAT) (2)

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““Equivalency” or Non-Inferiority Equivalency” or Non-Inferiority TrialsTrials

• Compare new therapy with standard

• Wish to show new "as good as"

• Rationale may be cost, toxicity, profit

• Examples– Intermittent Positive Pressure Breathing Trial

Expensive IPPB vs. Cheaper Treatment– Nocturnal Oxygen Therapy Trial (NOTT)

12 Hours Oxygen vs. 24 Hours

• Problem

Can't show H0: = 0

• A SolutionSpecify minimum difference = min

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Sample Size Formula Sample Size Formula Two ProportionsTwo Proportions

Simpler Case

• Z = constant associated with

• Z = constant associated with 1 -

• Solve for Z (1- ) or

2

2 )1()(2

ppZZ

N

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Difference in EventsTest Drug – Standard Drug

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Multiple Response VariablesMultiple Response Variables

• Many trials measure several outcomes

(e.g. MILIS, NOTT)

• Must force investigator to rank them for importance

• Do sample size on a few outcomes (2-3)

• If estimates agree, OK

If not, must seek compromise

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Mid Stream AdjustmentsMid Stream Adjustments

• Murphy's Law applies to sample size

• May find event rate assumptions way off from early results, power of study very inadequate

• Problem– Quit?– Continue for almost certain doom?– Adjust sample size?– Extend followup?

• Early Decision

Best to decide early, not look at treatment comparisons

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Adaptive DesignsAdaptive Designs

• One class allows re-estimating the sample size once the trial is underway– Chung et al– Chen, Lan & DeMets

• Methods have been criticized for allowing bias (eg Mehta & Tsiatis)

• Thus, methods still not widely used– AHEFT Trial one example

• Will be discussed later in data monitoring lecture

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Sample Size SummarySample Size Summary

• Ethically, the size of the study must be large enough to achieve the stated goals with reasonable probability (power)

• Sample size estimates are only approximate due to uncertainty in assumptions

• Need to be conservative but realistic

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Demo of Sample Size ProgramDemo of Sample Size Programwww.biostat.wisc.edu/www.biostat.wisc.edu/

• Program covers comparison of proportions, means, & time to failure

• Can vary control group rates or responses, alpha & power, hypothesized differences

• Program develops sample size table and a power curve for a particular sample size

542-05-#1 STATISTICS 542 Introduction to Clinical Trials SAMPLE SIZE ISSUES Ref: Lachin, Controlled...

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