Chapter 6 SAMPLE SIZE ISSUES

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Chapter 6 SAMPLE SIZE ISSUES

• Ref Lachin, Controlled Clinical Trials 293-113,
  1981.

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Sample 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?

• 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 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 Error The 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 Distribution
Ref Brown Hollander. Statistics A Biomedical
Introduction. John Wiley Sons, 1977.
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Standard Normal Table
Ref Brown Hollander. Statistics A Biomedical
Introduction. John Wiley Sons, 1977.
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Distribution of Sample Means (1)
Ref Brown Hollander. Statistics A Biomedical
Introduction. John Wiley Sons, 1977.
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Distribution of Sample Means (2)
Ref Brown Hollander. Statistics A Biomedical
Introduction. John Wiley Sons, 1977.
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Distribution of Sample Means (3)
Ref Brown Hollander. Statistics A Biomedical
Introduction. John Wiley Sons, 1977.
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Distribution of Sample Means (4)
Ref Brown Hollander. Statistics A Biomedical
Introduction. John Wiley Sons, 1977.
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Distribution of Test Statistics

• Many have a common form
• population parameter (eg
  difference in means)
• sample estimate
• Then
• Z E( )/SE( )
• 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
  large
• to have come from a distribution with mean O
  (i.e. Pc - Pt 0)
• Thus we reject H0 Pc - Pt 0, claiming that
  there exists 5 chance this result could have
• come from distribution with no difference

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Normal Distribution
Ref Brown Hollander. Statistics A Biomedical
Introduction. John Wiley Sons, 1977.
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Two Groups
OR
or
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Test Statistics
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Test of Hypothesis

• Two sided vs. One sided
• e.g. H0 PT PC H0 PT lt PC
• Classic test za critical value
• If z gt z? If z gt 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)

• 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)
Two Sided
Power
Significance Level
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Sample Size Exercise

• How many do I need?
• Next question, whats 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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One Sample Test for Mean

• H0 ? ?0 vs. HA ? ?0 ?

za constant associated with a P Zgt za
a two sided
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One Sample Test for Mean

• Under the alternative hypothesis that ?
• ?0 ?, the test statistic
• follows a standard normal variable.

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One Sample Test for Mean
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Simple Case - Binomial

• 1. H0 PC PT
• 2. Test Statistic (Normal Approx.)
• 3. Sample Size
• Assume
• NT NC N
• HA? PC - PT

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Sample Size Formula (1)Two Proportions

• Simpler Case
• Za constant associated with a
• P Zgt Za a two sided!
• (e.g. a .05, Za 1.96)
• Zb constant associated with 1 - b
• P Zlt Zb 1- b
• (e.g. 1- b .90, Zb 1.282)
• Solve for Zb (? 1- b) or D

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Sample Size Formula (2)Two Proportions

• Za constant associated with a
• P Zgt Za a two sided!
• (e.g. a .05, Za 1.96)
• Zb constant associated with 1 - b
• P Zlt Zb 1- b
• (e.g. 1- b .90, Zb 1.282)

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Sample Size Formula

• Power
• Solve for Zb ? 1- b
• Difference Detected
• Solve for D

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Simple Example (1)

• H0 PC PT
• HA PC .40, PT .30
• ? .40 - .30 .10
• Assume
• a .05 Za 1.96 (Two sided)
• 1 - b .90 Zb 1.282
• p (.40 .30 )/2 .35

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Simple Example (2)

• Thus
• a.
• N 476
• 2N 952
• b.
• N 478
• 2N 956

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Approximate Total Sample Size for Comparing
Various Proportions in Two Groups with
Significance Level (a) of 0.05 and Power (1-b) of
0.80 and 0.90
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Comparison 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 Means

• H0 ?C ?T ? ?C - ?T 0
• HA ?C - ?T ?
• Test statistic for sample means N (?, ?)
• Let N NC NT for design
• Power

N(0,1) for H0
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Example

• e.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

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Comparing Time to Event Distributions

• 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 Survival Distributions

• Then define the effect size by
• Standard difference

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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

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Assuming Exponential Survival Distributions

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

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Converting Number of Events (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 complete
• A/2 F average total follow-up at planned
  analysis
• average hazard rate
• Then P(event) 1 P(no event)

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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

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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
  gt T0)

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• Example
• Assume ? .05 (2-sided) 1 - ? .90
• ?C .3 and ?I .2
• T 5 years follow-up
• T0 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 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 for Non-Compliance (2)

• Fundamental Principle
• Analyze All Subjects Randomized
• Called Intent-to-Treat (ITT) Principle
• Noncompliance will dilute treatment effect
• A Solution
• Adjust 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 lt PC,

1.0
0
PC
PT
PC
PT
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Adjusted Sample Size

• Simple Model -
• Compute unadjusted N
• Assume no dropins
• Assume dropout proportion R
• Thus PC PC
• PT (1-R) PT R PC
• Then adjust N
• Example
• R 1/(1-R)2 Increase
• .1 1.23 23
• .25 1.78 78

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Sample Size Adjustment for Non-Compliance

• Dropouts dropins (R0, RI)
• Example
• R0 R1 1/(1- R0- R1)2 Increase
• .1 .1 1.56 56
• .25 .25 4.0 4 times

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Multiple 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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Sample 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 Programwww.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