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GCRC Research-Skills Workshop

October 26, 2007

William D. DupontDepartment of Biostatistics

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How to do Power & Sample Size Calculations Part 2

Survival Analysis

Follow-up time

Fate at exit

Statistic: Log-rank test

Power: Schoenfeld & Richter, Biometrics 1982

Time

Sam

ple

Siz

e

0

Accrual time A Additional

follow-up F

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00P

roba

bilit

y o

f Hem

orrh

age-

Fre

e S

urvi

val

0 10 20 30 40 50Months of Follow-up

Homozygous e3/e3At least one e2 or e4 allele

Hemorage-free survival in patients with previous lobular introcerebral hemmorrhage subdivided by apolipoprotein E genotype (O’Donnell et al. 2000).

Example: Hemorrhage-free survival and genotype Control group = patients with an e2 or e4 allele

Pr[Type I error] = 0.05Power = 0.8Relative Risk (control/experiment) R = 2Medial survival time of controls m1 = ?

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00P

roba

bilit

y o

f Hem

orrh

age-

Fre

e S

urvi

val



Hemorage-free survival in patients with previous lobular intracerebral hemmorrhage subdivided by apolipoprotein E genotype (O’Donnell et al. 2000).

Example: Hemorrhage-free survival and genotype Control group = patients with an e2 or e4 allele

Pr[Type I error] = 0.05Power = 0.8Relative Risk (control/experiment) R = 2Medial survival time of controls m1 = 38Accrual A = 12 monthsAdditional Follow-up F = 24 months

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

Pro

babi

lity

of H

emor

rhag

e-F

ree

Sur

viva

l



What if we wanted to study a treatment of Homozygous e3/e3 patients?

t = some time

p = probability of survival at time t

mediant =log(1/ 2)

log( )t

p

If survival has an exponential distribution

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

Pro

babi

lity

of H

emor

rhag

e-F

ree

Sur

viva

l



t = some time

p = probability of survival at time t

mediant =log(1/ 2)

log( )t

p

If survival has an exponential distribution

=log(1/ 2)40

171log(.85)

=

0

.1

.2

.3

.4

.5

.6

.7

.8

.9

1

Pro

babi

lity

of s

urvi

val

0 1 2 3 4 5 6 7 8 9 10Years since recruitment

Median Survival

Experimental treatment

Control treatment

Relative Risk (Hazard Ratio) for controls relative to experimental subjects

Median survival for experimental subjects

Median survival for controls=

= = 263

For t tests power calculations for increased or decreased response relative to control response are symmetric.

i.e. The power to detect is the same as the power to detect

c c

This is not true for Survival Analysis.

The power to detect a two-fold increase in hazard does not equal the power to detect a 50% decrease in hazard.

0

.1

.2

.3

.4

.5

.6

.7

.8

.9

1

Pro

babi

lity

of s

urvi

val

0 1 2 3Years since recruitment

Power to detect treatment 1 vs. control greater than treatment 1 vs.control

Treatment 1

Treatment 2

Control

Relative Risk

Control vs. Treatment 1 2Control vs. Treatment 2 0.5

It is important to read the parameter definitions carefully.

Power Calculations for Linear Regression

20

25

30

35

40

45

5 10 15 20 25 30Estriol (mg/24 hr)

Birt

hwei

ght (

g/10

0)

Rosner Table 11.1Am J Obs Gyn 1963;85:1-9

sx = 4.75

20

25

30

35

40

45

5 10 15 20 25 30Estriol (mg/24 hr)

Birt

hwei

ght

(g/1

00)


y = standard deviation of y variable

x = standard deviation of x variable

20

25

30

35

40

45

5 10 15 20 25 30Estriol (mg/24 hr)

Birt

hwei

ght (

g/10

0)


Estimated by s = root mean squared error (MSE)

= standard deviation of the regression errors

( ) ( )22ˆi iy y n= - -å

r = 0

r = -1

0 < r < 1

r = 1

c) r

( )/xy x yr =s s s is estimated by ( )/xy x yr s s s= {1.2}

Correlation coefficient

5

10

15

20

25

30

35

2 4 6 8 10 12Independent Variable x

Re

spo

nse

Var

iabl

e y

1 3 5 7 9 11

Treatment

A0.9

B0.6

Both treatments haveidentical values ofx = 6, y = 20x = 2, y = 5

Measures of association between continuous variables

Correlation vs. linear regression

5

10

15

20

25

30

35


Re

spo

nse

Var

iabl

e y

1 3 5 7 9 11

Treatment

A2.250.92.18

B1.50.64.0

Both treatments haveidentical values ofx = 6, y = 20x = 2, y = 5

x j

xj

yjj th Regression Error

Treatment Dose Level

RegressionLine x

1 Unitof

Dosage

0

Pat

ient

R

espo

nse

•

20

25

30

35

40

45

5 10 15 20 25 30Estriol (mg/24 hr)

Birt

hwei

ght (

g/10

0)


c) Slope parameter estimate

a.k.a. ) is estimated by b = r sy

/sx

r = 0

r = -1

0 < r < 1

r = 1

is estimated by b = r

sy /sx

20

25

30

35

40

45

5 10 15 20 25 30Estriol (mg/24 hr)

Birt

hwei

ght (

g/10

0)


Studies can be either experimental or observational

Normally distributed unless investigator

chooses level

Treatment level

Chosen by investigator

Estimating directly is often difficult

21/ 1x

2 2 2y x

=

=

If we can estimate

and x

Warning:

If the anticipated value of in your experiment is different from that found in the literature then your value of will also be different.

x

or and x y

5

10

15

20

25

30

35


Re

spo

nse

Var

iabl

e y

1 3 5 7 9 11

Relationship between BMI and exercise time

n = 100 women willing to follow a diet exercise program for six months

Interquartile range (IQR) of exercise time = 15 minutes (pilot data) = (IQR / 2) / z0.25 = 7.5 / 0.675 = 11.11x

= 4.0 kg/m2 = standard deviation of BMI for women obtained by Kuskowska-Wolk et al. Int J Obes 1992;16:1-9

y

Would like to detect a true drop in BMI of = -0.0667 kg/m2 per minute of exercise

(1/2 hour of exercise per day induces a drop of 2 kg/m2 over 6 months)

a

When

0.675

0.250.25

-0.675

InterquartileRange

= (IQR / 2) / z0.25xIn general

Impaired Antibody Response to Pneumococcal Vaccineafter treatment for Hodgkin’s Disease

Siber et al. N Engl J Med 1978;299:442-448.

n = 17 patients treated with subtotal radiation. vaccinated 8 to 51 months later

A linear regression of log antibody response against time from radiation to vaccination gave

ˆ 0.01 0.11

0.40

P

r

Suppose we wanted to determine the sample size for a new study with

1 0.90

0.05

patients randomized to vaccination at 10, 30, or 50 weeks

Vit

al C

ap

aci

ty (

liters

)

Age (years)

Exposed > 10 years Unexposed

20 30 40 50 60

2

4

6

Composing Slopes of Two Linear Regressin Lines

Armitage and Berry (1994) gave age and pulmonary vital capacity for28 cadmium industry workers with > 10 years of exposure44 workers with no exposure

112x

29.19x

1̂ 0.0306

2̂ 0.0465

(unexposed)

(exposed)

pooled error variance

2 2 21 1 2 2 1 22 2 /( 4) 0.395

0.574

s s n s n n n

s

2 1ˆ ˆ 0.0159 ( 0.26)P

How many workers do we need to detect with

ratio of unexposed workers m = 44/28 = 1.57?

1 0.8, 0.05

Need 167 exposed workers and167 x 1.57 = 262 unexposed workers

112x

29.19x

1̂ 0.0306

2̂ 0.0465

(unexposed)

(exposed)

pooled error variance 2 0.395

0.574

s

s

2 1ˆ ˆ 0.0159 ( 0.26)P

How many workers do we need to detect with

ratio of unexposed workers m = 44/28 = 1.57?

2 1 0.0159 1 0.8, 0.05

Need 167 exposed workers and167 x 1.57 = 262 unexposed workers

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