Power and Sample Size

Martin L. Lee, PhD
Overview of Lecture

• Basic parameters of power/sample size 

calculations

• Formulae for different settings

• Software
What we need to know: Study 

design

• Randomization: yes/no, type

• Matching

• Number of groups

• Measurement type (quantitative/qualitative/timeto-event-censored)

• Repeated measures

• Multivariate adjustment

• One vs. two-tailed
What we need to know: Study 

goals

• Estimation versus hypothesis testing

• Superiority versus equivalence/lack of 

inferiority

• Δ: minimum clinical important difference 

(superiority); equivalence definition
Statistical errors in hypothesis 

testing

 Significance Difference is

Present Absent

(Ho not true) (Ho is true)

Test result

Reject Ho No error Type I err.

1-β α

Accept Ho Type II err. No error

β 1-α
Choosing error rates

• α: “consumer” risk; would like to choose as 

small as possible, but impractical; 5% is a 

compromise

• β: “producer” risk; depends on true Δ

(inverse relationship) and sample size
The effect of Δ on β-error 
The effect of n on β-error
Information needed for sample size 

calculation

• Errors: α and β

• Δ

• Quantitative data: estimate of SD 

(literature or “rule of 4”)

• Effect size: Δ/SD or difference in 

proportions for qualitative data
Sample size for difference between two 

means

( )

2

2 2

12/ 1

∆





 





− + +

=

r

ZZr

N

σ

α β

 Hypotheses:

Ho: μ1 = μ2 vs. Ha: μ1 ≠ μ2

 Let n1 and n2 be the sample sizes 

for group 1 and 2, respectively; N

= n1 + n2 ; r = n1 / n2 ; σ: standard 

deviation of the variable of 

interest. 

 The sample size is adjusted 

slightly for the use of the t-test

 If we let η = Δ / σ be the 

“effect size”, then:

where Zα/2 and Z1-β are normal deviates

 If n1 = n2 , power = 0.80, 

alpha = 0.05, then (Zα/2 + Z1-

β)2 = (1.96 + 1.28)2 = 10.5, 

then the equation is reduced 

to:

2

21

η

N =
0

400

800

1200

1600

2000

2400

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

Effect size (d/ s)

Total sample size (N)

Sample size for 2 means versus effect 

size
An example: Oral contraceptive use

• Study of OC 

use

• p=0.46 

• Δ=5.4, which 

is large

• Power to 

detect Δ is 

0.13 

n Mean 

BP

SD

OC 

user 

8 132.8 15.3

NonOC 

user 

21 127.4 18.2
An example: OC use

• With the same Δ (about 5), average SD of 

about 16.7 and power of 80%, we need 

154 subjects

• Notice δ=0.3 standard deviations here, 

which is small-medium ES (Cohen, 1988)

Output from PASS 2008: QOL scale 

example

Power N1 N2 Ratio Alpha BetaMean1Mean2 S1 S2

0.90148 133 133 1 0.050.0985210.00 11.00 2.50 2.50

0.80365 100 100 1 0.050.1963510.00 11.00 2.50 2.50

0.90150 34 34 1 0.050.0985010.00 12.00 2.50 2.50

0.80749 26 26 1 0.050.1925110.00 12.00 2.50 2.50

0.90719 16 16 1 0.050.0928110.00 13.00 2.50 2.50

0.80208 12 12 1 0.050.1979210.00 13.00 2.50 2.50
Sample size for 2 means: hypothesis of 

equivalence

• H0: |μ1- μ2|≥Δ, where 

Δ is the definition of 

clinical 

bioequivalence

where δ=true difference

( )

2

)(

2 2

12/ 1

∆−





 





− + +

=

δ

σ

α β

r

ZZr

N
( ( ) ( ) ( ))

( )2

2 12 1 1

21

2/ 2111 2

pp

ppppZppZ

n

−

+− −+−

=

α −β

 Hypotheses:

Ho: p1 = p2 vs. Ha: p1 = p2 + d .

 Let p1 and p2 be the sample proportions (e.g. estimates of p1 and p2) 

for group 1 and group 2. Then, the sample size to test the hypothesis 

is: 

where: n = sample size for each group ; p = (p1 + p2) / 2 ; Za/2 and 

Z1-b are normal deviates

Sample size for difference between 2 

proportions
 Example: The rate of adverse events for drug A is expected to be 

70%, and for drug B 60%. A study is planned to show the difference 

at the significance level of 1% and power of 90%. 

 The sample size can be calculated as follows:

 p1 = 0.6; p2 = 0.7; p = (0.6 + 0.7)/2 = 0.65; Z0.01 = 2.58; Z1-0.9 = 

1.28. 

 The sample size required for each group should be:

Sample size for 2 proportions: an example

( )

( )

675 2 7.06.0

2 3.07.04.06.028.135.065.0258.2

≈

−

×+×+××

n =
Sample size for two proportions vs. effect 

size

 Difference from p1 by: 

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

P1 

0.1 424 131 67 41 28 19 14 10

0.2 625 173 82 47 30 20 14 9 

0.3 759 198 89 50 30 19 13 8 

0.4 825 206 89 47 28 17 . . 

0.5 825 198 82 41 22 . . . 

0.6 759 173 67 31 . . . . 

0.7 625 131 45 . . . . . 

0.8 424 73 . . . . . .

Note how sample size increases near the middle of the range of the 

baseline proportion

Output from PASS 2008: Remission rate 

example

SampleSampleProp|H1Prop 

Size Size Diff Diff 

TrtmntControlH0 H1TargetActual

Power N1 N2 P1 P2 D0 D1AlphaAlpha Beta

0.9002 266 266 0.2 0.1 0 0.1 0.05 0.0998

0.8001 199 199 0.2 0.1 0 0.1 0.05 0.1999

0.9023 79 79 0.3 0.1 0 0.2 0.050.05160.0977

0.8051 59 59 0.3 0.1 0 0.2 0.050.05070.1949

0.9010 40 40 0.4 0.1 0 0.3 0.050.05870.0990

0.8126 31 31 0.4 0.1 0 0.3 0.050.05520.1874
Sample size for difference between 2 

proportions: Equivalence

( ( ) ( ) ( ))

( ) 2 ( )

2 12 1 1

21

2/ 2111 2

pp

ppppZppZ

n

−−

+− −+−

= −

δ

α β

where, δ is the true difference between the proportions
Sample size for two proportions when 

there are multiple controls per case

( ( ) ( ))

( )2

2 )1()/)1(( 1/1

21

2/ 111 2 2

pp

ppcppZppccZ

n

−

+−+ −+−

=

α −β

where c = number of cases per control 

and p = (cp1 +p2)/(1+c) 

n = number of cases and c x n = number of controls
Effect of adding more controls on power

Note that α=0.05, power=0.80 and an odds ratio of 2.0
Sample size for estimating an odds ratio

( ) ( )

( ) ( ) ppORr

ZZr

N −

++

= −

1ln

1

2

2

1

2

α β

 In case-control study the data are usually summarized by an odds 

ratio (OR), rather then difference between two proportions.

 If p1 and p2 are the proportions of cases and controls, respectively, 

exposed to a risk factor, then: 

( )

( ) 2 1

1 2

1

1

pp

pp OR −

− =

 If we know the proportion of exposure in the general population (p),

the total sample size N for estimating an OR is:

 Where r = n1 / n2 is the ratio of sample sizes for group 1 and 

group2; p is the prevalence of exposure in the controls; and OR 

is the hypothetical odds ratio. If n1 = n2 (so that r = 1) then the 

fomula is reduced to: ( )

( ) ( ) ppOR

ZZ

N −

+

= −

1ln

4

2

2

α 1 β
 Example: The prevalence of vertebral fracture in a population is 25%. 

It is interested to estimate the effect of smoking on the fracture, with 

an odds ratio of 2, at the significance level of 5% (one-sided test) and 

power of 80%.

 The total sample size for the study can be estimated by:

( )

( ) 275

ln 75.025.02

85.064.14

2

2

=

××

+ N =

Sample size for an odds ratio: example
Sample size for rare events

• Goal of study is to estimate the probability 

of a rare event

• Example: frequency of HIV after infusion 

of a blood product

• Hopefully, zero events

• What is the 95% upper bound for the 

probability, θ ? 
Sample size for rare events

• θ≤3/n (“Rule of 3’s”)

• Defines n for a desired upper bound

Example: How many patients do we need to 

study in order to ensure with 95% 

confidence that the rate of HIV 

transmission is less than 1%?

Answer: 300 by rule of 3’s
Software 

• PASS 2008

• nQuery

• Internet calculators


Sample size for estimating the mean and 

proportion

 How close to the true mean

 Confidence around the sample 

mean

 n = (Zα/2)2 s2 / d2

s: standard deviation

d: the accuracy of estimate 

(how close to the true mean).

Zα/2: normal deviate reflects the 

confidence level.

 How close to the true proportion

 Confidence around the sample 

proportion.

 n = (Zα/2)2 p(1-p) / d2

p: proportion to be estimated.

d: the accuracy of estimate 

(how close to the true 

proportion).

Zα/2: normal deviate reflects the 

confidence level.
Examples of estimating the mean and 

proportion

 Example: we want to estimate 

the average weight in a 

population, and we want the 

error of estimation to be less 

than 2 kg of the true mean, 

with 95% confidence and we 

suppose σ=20

 n = (1.96)2 202 / 22 = 384 

 Example: The expected 

prevalence of a disease is 

around 20%. We want to 

estimate the true prevalence in 

a community within 5% with a 

95% confidence interval.

 n = (1.96)2 (0.2)(0.8) / 

0.052 = 246 individuals.