1. Sample Size and Power Part II
PQHS 450
Sample Size and Power Part II

2. Outline Hypothesis testing (two independent samples)
Determine n to achieve a desired power of the test
a) Continuous outcome – comparing two means
b) Binary outcome - comparing two proportions
c) Time-to-event outcome – comparing two survival curves
d) Longitudinal studies – comparing rates of change (slopes)

3. Sample size determination as part of the design process
Logical steps in study design
Key questions relevant to ask as a first step
in sample size determination
What type of outcome(s)?
- continuous
- binary (yes/no)
- survival (time-to-event)
What kind of design? - parallel design or paired?
- clustered?
- repeated measures?
3. What kind of inference?
- Estimation with conf. Intervals
- Hypothesis testing
-Trying to show one group is better?
-Trying to show equivalence or
noninferiority?
Specific Aims
Hypotheses
Primary Outcome
Protocol Design
Analysis plan
Sample size

4. Two types of errors in Hypothesis Testing
H0 True
H0 False
Accept H0
1- 
Reject H0 
1-
 = Type I error (Pr reject H0 when H0 is true)
False positive
 = Type II error = (Pr accept H0 when H0 is false)
False negative
Power = (1- )
Sample size depends on following factors in general:
α, β, effect size, 1 or 2-sided test, test statistic and type of outcome

5. Survival (Time-to-event) Outcomes
Design of a typical trial with time-to-event outcome:
Calendar time
Begin
enrollment
End
Follow-up
Recruitment period
Follow-up period
Patient follow-up ranges from T-T0 to T

6. Analysis Method 1.0 Time (years) on study
We assume that data will be analyzed by comparing the survival
curves of the two groups using the logrank test
1.0
Proportion
Surviving
Group 1
Group 2
Time (years) on study

7. Time-to-event trials
Analysis method: Compare survival curves using a method like the logrank test
(compares distribution of times to event, not just proportions with the event)
Basic Sample Size formula – based on an exponential model, which assumes there is a constant risk (hazard) over time. (Works with logrank test /Cox model also)
Let: C = hazard rate in control group
I = hazard rate in intervention group
Under the exponential model, if  is the hazard, then
Prob(survive to time t) = Pr(no event by time t) = exp(- t)
Relation between median survival time TM and :  = -ln(.5)/TM = .693/ TM
For example: If TM = 5 years, then  = .693/5 = .1386
Also, event rate at 1 year = 1-exp(-.1386(1)) =
event rate at 3 years = 1-exp(-.1386(3)) = .340

8. Sample size formula – time to event outcome
Parameters to specify:
T = total duration of trial (years)
T0 = duration (years) of recruitment period (T0 < T)
C = hazard rate in control group
I = hazard rate in intervention group
 = type I error
Is trial 1-sided or 2-sided?
 = type II error (or power = 1- )
Required sample size n per group:
where
1-sided
2-sided

9. Example
Suppose we are planning a six-year study, where subjects with end-stage renal disease (ESRD) will be recruited evenly over the first 3 years, and then followed 3 more years after the last patient is enrolled. Subjects will be randomized to an intervention (strict BP control) or a Control (usual therapy). The event is time to occurrence of the composite endpoint ESRD/doubling of serum creatinine/death. Suppose the median time until the event in the controls is 4 years, and it important to be able to detect a 30% reduction in risk due to the intervention using a 2-sided test at alpha=0.05, with power of 90%.
From this, the parameters we need are:
T = 6 years
T0 = 3 years
 = (2-sided)
 = .10 (power = .90)
C = hazard rate in controls = .693/4 = .1733
I = hazard rate in intervention group = (1-.30) x = .1213

10. Example, continued Then we calculate: Plugging these into the formula:
K, =(Z.025+Z.10)2 = ( )2 = 10.51
Plugging these into the formula:
Therefore, we need 355 per group (a total of 710 patients)

11. Excel Program: phicalculator.xls
Program to calculate "Phi(lamba)" using equation on page 186 of Friedman, Furburg, DeMets, Reboussin (5th ed)
lambda= hazard rate
T0= length of recruitment period
T= total duration of study
Phi = Phi(lamba) on page 186 of Text (5th ed)
P(Event) = proportion of subjects in group who are expected to have an event during the study = (lambda^2)/Phi
lambda T0 T
Phi
P(Event)
0.1733 3 6
0.0560
0.5363
0.1213
0.0352
0.4174
Using alpha=0.05, 80% power:
Expected Number Total Events*
Formula A
Formula B
Formula C n=
267
per group
254.7
253.6
248.3
Using alpha=0.05, 90% power:
355
338.6
337.0
330.0
*Formulas for Expected Numbers of Events:
A. Based on Probabilities of Event calculated in above table:
B. Events=K(alpha,beta)*((average lambda**2)/(lambda1-lambda2)**2
c. Events=4*K(alpha,beta)/(log(hazard ratio))**2
INPUT

12. * Comparison of Survival Curves; proc power;
* Using SAS Proc Power;
* Comparison of Survival Curves;
proc power;
twosamplesurvival test=logrank
accrualtime=3
followuptime=3
gsurvexphazards=(.1733,.1213)
alpha=0.05
power=0.90
npergroup=.;
run; T0
T-T0

13. Output: Log-Rank Test for Two Survival Curves Fixed Scenario Elements
Method Lakatos normal approximation
Form of Survival Curve Exponential
Form of Survival Curve Exponential
Accrual Time
Follow-up Time
Alpha
Group 1 Survival Exponential Hazard
Group 2 Survival Exponential Hazard
Nominal Power
Number of Sides
Number of Time Sub-Intervals
Group 1 Loss Exponential Hazard
Group 2 Loss Exponential Hazard
Computed N Per Group
Actual N Per
Power Group

14. * Comparison of Survival Curves; * Specifying hazard in control
* Using SAS Proc Power;
* Comparison of Survival Curves;
* Specifying hazard in control
and hazard ratio;
proc power;
twosamplesurvival test=logrank
accrualtime=3
followuptime=3
refsurvexphazard=.1733
hr=0.70
alpha=0.05
power=0.90
npergroup=.;
run; T0
T-T0

15. Output:
Log-Rank Test for Two Survival Curves
Fixed Scenario Elements
Method Lakatos normal approximation
Form of Survival Curve Exponential
Form of Survival Curve Exponential
Accrual Time
Follow-up Time
Alpha
Reference Survival Exponential Hazard
Hazard Ratio
Nominal Power
Number of Sides
Number of Time Sub-Intervals
Group 1 Loss Exponential Hazard
Group 2 Loss Exponential Hazard
Computed N Per Group
Actual N Per
Power Group

16. Note: Sample size can be rewritten in terms of required number of events (deaths)
P = expected proportion in the group
who will have an event occur
during the trial
=2/()
where
This can be rewritten as:
, or
Thus, can calculate total events based on λ’s, α,. Then determine T and To
along with number recruited to achieve the required number of events

17. Required number of events (deaths)
In the literature there are two general formulae for calculating the
total required number of deaths (D) given I, C, , and : 1)
George and Desu (1974) – exponential survival curves
Schoenfeld (1981, Biometrika) – logrank test
Schoenfeld (1983, Biometrics) – Cox model 2)
(as on previous slide)
Freedman (1982 Stat in Med)
Results from the two are usually close

18. * Using SAS Proc Power;
* Comparison of Survival Curves;
* Incorporating 5% yearly LTFU;
* Exponential hazard for loss=-ln(.95)=.0513;
proc power;
twosamplesurvival test=logrank
accrualtime=3
followuptime=3
refsurvexphazard=.1733
hr=0.70
alpha=0.05
power=0.90
grouplosshazards=(.0513,.0513)
npergroup=.;
run;

19. Output:
Log-Rank Test for Two Survival Curves
Fixed Scenario Elements
Method Lakatos normal approximation
Form of Survival Curve Exponential
Form of Survival Curve Exponential
Accrual Time
Follow-up Time
Alpha
Reference Survival Exponential Hazard
Hazard Ratio
Group 1 Loss Exponential Hazard
Group 2 Loss Exponential Hazard
Nominal Power
Number of Sides
Number of Time Sub-Intervals
Computed N Per Group
Actual N Per
Power Group
Up from 343 when no
loss to follow-up is
assumed

20. Example – Effect of changing design parameters:
Scenarios: Required
n/group
Original scenario
Duration=T=6, Recruitment = T0=
Keep the same recruitment period (T0=3) but extend the
follow-up by 1 year, (T=7 year study, T-T0=4)
c) Keep it as a 6-year study, but recruit patients more quickly 328
by adding more centers (sites). T0=2 year recruitment period,
with T-T0=4
d) Keep T and T0 as before. Change eligibility criteria to enroll 295
more severely ill patients, so that median time to event for
controls is 3 years instead of 4 years (increases the event rates)
C = .2310, I = .1617

21. What if we could enroll all the patients at the very beginning?
Here use T0=0.01, T=6. In this scenario all patients are followed 6 yrs.
Using the survival analysis approach: n=292 / group
This can be compared to the approach where we simply calculate
the 6-year survival rates as binomial proportions. Here, the expected
proportions surviving to the end of the study are
controls: exp(-.1733x6) = .354
intervent: exp(-.1213x6)=.483
Using the formula for comparing 2 proportions, the
required n is 306/group ( only slightly larger)