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Sample Size Calculations for Survival Analysis

Mark A. Weaver, PhDFamily Health International

Office of AIDS Research, NIHICSSC, FHI

Goa, India, September 2009

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Design Considerations

• Will study enroll a fixed number of patients and follow each for some specified period of time?

• Or, will study continue until a sufficient number of events have been observed?

• Will any interim analyses be performed?

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Components in Sample Size Calculations1. Null hypothesis to be tested2. Test statistic3. Assumed effect size4. Size of the test (significance level, α)5. Desired power6. Sample size (in terms of number of events)7. Probability of an event during study8. Expected rate of loss9. Sample size (in terms of number of patients)10.Adjustments for interim analyses

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Null Hypothesis to be Tested• Recall for log-rank tests

H0: S1(t) = S2(t), for all t↔

H0: h1(t) = h2(t), for all t↔

H0: h1(t) / h2(t) = 1, for all t↔

H0: HR = 1, where HR = h1(t) / h2(t) for all t assuming proportional hazards

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Test Statistic

• Can use either 1. log-rank test (or some variation thereof) or 2. HR estimated from Cox model

• to test the null hypothesis of no difference

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Effect Size

• Assume proportional hazards• Hazard ratio: HR = h1(t) / h2(t)• HR = 1 implies no difference between treatments• HR > 1 implies “survival” is longer on treatment 2• HR < 1 implies “survival” is longer on treatment 1

• Want to base sample size calculations on having sufficient power to detect some minimally clinically important effect, which may have no relation to the actual treatment effect– e.g., maybe a 30% reduction in incidence (HR = 0.7)

would be clinically meaningful

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Significance Level

• Magic number: α = 0.05 (two-sided)– other values can be used, but should be justified– e.g., could use α = 0.001 if very strong evidence of a

treatment effect is desired– e.g., could use α = 0.1 if type I error presents little risk

3.290.0011.960.05

1.6450.1

zα/2α

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Power

• Power: typically desire power of at least 80%, 90%, or 95%– Recall that for means and proportions, power is a

function of sample size– However, for survival data, power is entirely driven by

number of events

95%90%80%

Power

1.6450.051.2820.100.8420.20

zββ

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Required Number of Events

• zα/2 and zβ are standard normal percentiles– See tables previous 2 slides

• π1 and π2 are the proportions to be allocated to groups 1 and 2– for equal allocation π1 = π2 = 1/2

( )( )221

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HRlogevents

ππβα zz +

=

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Example: Required Number of Events

• Suppose α = 0.05 and β = 0.10 (power = 90%)– zα/2 = 1.96, zβ = 1.282

• Equal allocation (π1 * π2 = ¼)

* Always round calculations up. Why?– Note: have not yet discussed background event rate

( )( )221

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HRlogevents

ππβα zz +

=

880.503310.708450.80

Events Required *HR

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Components in Sample Size Calculations

1. Null hypothesis to be tested2. Test statistic3. Assumed effect size4. Size of the test (significance level, α)5. Desired power6. Sample size (in terms of number of events)7. Probability of an event during study8. Expected rate of loss9. Sample size (in terms of number of patients)10.Adjustments for interim analyses

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How Many People?

• Suppose we designed study to have 90% power to detect a hazard ratio of 0.5 (e.g., a 50% reduction in event rate)

• We have already determined that this would require observing 88 events

• But, how many people should we plan to enroll to get that number of events?

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A Simplification

• Will consider only case in which each patient is to be followed for some specified period of time, T

• More general case: – patients recruited during accrual period of length a– after recruitment ends, additional follow-up of length f– patients remain in study from time recruited until end of

study– first patient in is followed for a + f– last patient in is followed for f– for calculations in general case, see e.g., Collett (1994),

Modelling Survival Data in Medical Research

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Calculating Number of People

• Need to consider probability of event during study• Once we have this probability, total number of

people can then essentially be calculated as

– ignoring loss for now…

{ }eventPrevents

=n

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Probability of an Event

• Recall that π1 and π2 are the proportions to be allocated to groups 1 and 2– for equal allocation π1 = π2 = ½

• How do we get values for S1(T) and S2(T)?

{ } ( ))()(1eventPr 2211 TSTS ππ +−=

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Assume Exponential Survival Times• For exponential failure times, IR = λt

– e.g., if IR = 10 events / 100 p-y = 0.1 events / p-y, then λt = 0.1 at t = 1 year

• Recall that S(t) = exp{ -λt }• We can use an assumed IR to calculate an

assumed failure probability for SS calculations– e.g., if λ*1 = 0.1, then S(1) = exp{ -0.1 } = 0.905– Thus, we assume that 1 – 90.5% = 9.5% of

participants will have an event within 1 year

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Example 1• One-year study (T = 1), equal allocation• Suppose have previous estimate of IR for control

group of 10 / 100 p-y, or .1 / p-y• For power calculations, assume HR = 0.5• S1(1) = exp(-0.1) = 0.905• S2(1) = exp(-0.1 * 0.5) = 0.951• Pr{ event } = 1 – (0.905 + 0.951)/2 = 0.072• n = events / Pr{ event } = 88 / 0.072 = 1223

– 612 / group

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Example 2• 6 month study (T = 0.5), equal allocation• As before, estimate of IR for control group is

.1 events / p-y (or .05 events / person-half-year)• For power calculations, assume HR = 0.5• S1(0.5) = exp( -0.05 ) = 0.951• S2(0.5) = exp(-0.05 * 0.5) = 0.975• Pr{ event } = 1 – (0.951 + 0.975)/2 = 0.037• n = events / Pr{ event } = 88 / 0.037 = 2380

– 1190 / group

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What About Loss?

• Approximate method for handling lossnadj = n / (1 – loss)

• Example, assume at most 10% loss– From Example 1, nadj = 612 / 0.90 = 680 per group– From Example 2, nadj = 1190 / 0.90 = 1323 per group

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Adjusting for Interim Analyses• Conducting multiple interim analyses entails a

slight loss of power• Must inflate sample size to account for this

– Sample size multipliers for 2-sided α = 0.05

1.211.231.031.0351.181.201.021.0241.151.171.021.0231.101.111.011.012

Power = 90%

Power = 80%

Power = 90%

Power = 80%

PocockO’Brien-FlemingNumber of planned analyses

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Example of Interim Analysis Adjustment• Suppose plan for 4 analyses (3 interim plus final)

– Decide to use a Pocock boundary• Number of events required = 88 * 1.18 = 104• Number of people required = 104 / 0.072 = 1444• Adjusting for 10% LTF = 1444 / 0.9 = 1604

– 802 per group

• With O-F boundary, verify that would require about 695 per group

• Compare these to 680 per group without interim analyses

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Questions?

Don’t worry, you’ll get to try one on your own…

but you won’t have to do it “by hand.”