Lecture 10 Sample Size

S.F.Kelsey/class2181/lecture 4-sample size 1

DESIGN OF CLINICAL TRIALSEPIDEMIOLOGY 2181

“SAMPLE SIZE DETERMINATION AND STATISTICAL POWER IN CLINICAL TRIALS”

October 30, 2008Lecture 4

SHERYL F. KELSEY, PhD

Department of Epidemiology


QUIZAssume 90% Power, a = 0.05 two-sided

(x) more with A(y) more with B(z) the same

1. Mortality 20% vs 10% 40% vs 30%

2. Mortality 20% vs 10% 20% vs 15%

3. Diastolic 80 vs 85 mmHg 90 vs 95 mmHg BP

4. Diastolic 80 vs 85 mmHg 80 vs 85 mmHg BP

A B




(St Dev 10) (St Dev 10)

(St Dev 10) (St Dev 8)

How manysubjects?


1. More with B

2. More with B

3. The same

4. More with A

ANSWERSVariance of the binomialbigger 50% smaller 0% 100%

Small difference need more subjects

Only standard deviation matters

Bigger standard deviation more subjects


SHALL WE COUNT THE LIVING OR THE DEAD?

40% vs 20% 20% 50% “reduction” in mortality lower mortality



60% vs 80% 20% 33% “improvement” in survival higher mortality

80% vs 90% 10% 2.5% “improvement” in survival higher mortality

90% vs 95% 05% 5.6% “improvement” in survival higher mortality

Absolute Relative


Even more confusing with continuous variables

Blood pressure (St Dev 10)

5.9% “reduction” 80 vs 85 mmHg

5.3% “reduction” 90 vs 95 mmHg


Intervention A has “increased” survival

No: Intervention A has longer, better, greater survival

“Increase” should be used for changes over time


PERCENTS Absolute difference

Relative difference

Percents as continuous measures

Proceed with caution


Legal null hypothesis: innocent until proven guilty

Scientific null hypothesis: no difference in response between treatment groups

Inno

cent

Gui

lty

Innocent Guilty

Truth

Dec

isio

n of

Ju

dge/

Jury

ok

ok

guilty goes freetype IIerror ()

hang theinnocenttype Ierror ()

Treatment Different

TreatmentSame

Truth

Sam

eD

iffer

ent

Obs

erve

d D

ata

ok

ok

miss goodtreatmenttype IIerror ()

promoteworthlessTx type Ierror ()


FUNDAMENTAL POINT

Clinical trials should have sufficient statistical power to detect differences between groups considered to be of clinical interest. Therefore, calculation of sample size with provision for adequate levels of significance and power is an essential part of planning.


THE RAW INGREDIENTS

What is your question, precisely?

What is your outcome, precisely?

Who will be measured?

Type 1 and type 2 error rates

Variability


PRIMARY COMPARISONS

Dichotomous Response Variables

The event rates in the intervention group (Pi) and the control group (Pc) are compared

Continuous Response Variables

The true but unknown mean value in the intervention group is compared with the true, but unknown

mean value in the control group.

Survival Data

A hazard rate is often compared for the two study groups or at least is used for sample size

estimation.


SAMPLE SIZE ISSUES

Choice of outcome – primary endpoint

Change from baseline only if correlation > .5 when in doubt don’t

The difference____________ to detect you want you believe is clinically meaningful you believe is biologically credible you can afford to


ASSESSMENTS OF EVENT RATE IN THE CONTROL AND INTERVENTION GROUPS

The estimate for the control group event rate is usually obtained from a previous study of similar people. Good data base desirable

The investigator must choose the difference in event rate based on preliminary evidence of the potential effectiveness of the intervention or be willing to specify some minimum difference.

Calculation of several sample sizes based on a range of estimates helps one to assess how sensitive the sample size is to these estimates.


To Plan with continuous endpoints: Clinical difference worth detecting

1–Power Probability of obtaining a significant result if is true difference

Significance level, must specify one or two-tailed test

(Z Z )2 Multiplier which depends on level of significance and Power 1-

n Sample size for each of two groups

For continuous measures:

Standard deviation

2

222)(

ZZ

n

With a little algebra

Z=1.96 for =.05, two-sided

n

ZZ 222)(

Zn

z 2

(solve for power)

(Solve for difference)


For two proportions P1 vs P2, = P1 - P2 With a little algebra

Z = 1.64 for .05, one-sided

Z = 1.96 for .05, two-sided 221

2

arcsinarcsin2

)(

PP

ZZ

221

2

arcsinarcsin2 PP

ZZn

ZPPnZ 21 arcsinarcsin2

2

1

2

2 arcsin2

)(sin

P

n

ZZP

22

1 2

)(arcsinsin

n

ZZP

212

22112 11

PP

PPPPZZn


TABLE(Z + Z)2

Needed to determine the size of each sample

(Z22.32 1.645 1.28)

Desired Two-Tailed Tests One-Tailed Tests Power Level Level

Z P 0.01 0.05 0.10 0.01 0.05 0.10

Two groups of unequal size: Calculate the harmonic mean

}2/11

/{1

21

nn

n

This n is what is needed for 2 groups of equal size. Note that equal sized groups are the most efficient, that is the harmonic mean is less than the arithmetic mean.References: Snedecor and Cochran, 7th Edition Statistical Methods, 1980, pp 102-1- 5, 120, 130.

Fleiss, JL. Statistical Methods for Rates and Proportions, 1981, Chapter 3 & Tables. Schlesselman, JJ. Case Control Studies, 1981, Chapter 6 & Tables.

(Z 2.576 1.96 1.645)

0.84 0.80 11.7 7.9 6.2 10.0 6.2 4.5

1.28 0.90 14.9 10.5 8.6 13.0 8.6 6.6

1.645 0.95 17.8 13.0 10.8 15.8 10.8 8.6


Example: Compare .10 vs .05

= .05 one sidedPower 80%

arcsin

arcsin

2255.05.

3218.10.

221

2

)arcsin(arcsin2

)(

PP

ZZn

334

)0963(.2

2.62

n

So total study: 334 x 2 = 668

.10 vs .05 with 200 patients in each group

2002x

Power = 61%

with 100 patients Z = .28 39% power

50 patients Z = .68 25% power

nZ 2 | 21 arcsin PP | .0963| - 1.64 = .286| — Z arcsin


FURTHER SAMPLE SIZE CONSIDERATIONS

outcome is survival time discount for noncompliance/dropout projections subgroups more than 2 treatment groups unequal group sizes - less efficient - can get more information on new

treatment computer packages (PASS, SAS, MINITAB) ASSUMPTIONS ABOUT EVENT RATES PARAMOUNT

sensitivity analysis


ONE-SIDED VERSUS TWO-SIDED TESTS

I Drug A side effects/expensiveDrug B no side effects/cheap

A more efficaciousA&B the sameB more efficacious

II X Nutrition Intervention Strategy-Group sessionsY Nutrition Intervention Strategy-Individual program

X reduce sodium intake moreX&Y the sameY reduce sodium intake more


SAMPLE SIZE FOR TESTING “EQUIVALENCY”

OF INTERVENTIONS

The problem in designing positive control studies is that there is no statistical method to demonstrate complete equivalence.

Computing a sample size assuming no difference results in an infinite sample size.

One approach is to specify a value for difference in response such that interventions

with differences that are less than this might be considered equally effective or equivalent.


T = Innovative Therapy

S = Standard Therapy

“Superiority”H0: death rate T = death rate S

Halt:death rate T < death rate S

Equivalence H0: death rate T death rate S +

Halt:death rate T < death rate S +

In general equivalence studies require more patients


Patients: Acute MI

Treatment: Double bolus vs accelerated Alteplace

Outcome: 30 day mortality

COBALT Equivalence

Death rate within 0.4%

GUSTO III SuperiorityDouble bolus reduce mortality by 20%

WARE AND ANTMAN EDITORIAL


MORTALITY RESULTS

COBALT GUSTO III

N 7169 15059

Double bolus 7.98% 7.47%

Accelerated 7.53% 7.24%

Difference 0.45% 0.23%

95% CI Approx. (-.85%, 1.66%) (-.66%, 1.10%)

Action reject equivalence accept null not significantly different from zero


DESIGN OF CLINICAL TRIALSEPIDEMIOLOGY 2181

RANDOMIZATION IN CLINICAL TRIALS

SHERYL F. KELSEY, PH.D


WHY RANDOMIZE? Best way to assure comparability

In the long run balance of factors known unknown

Statistical hypothesis test based on random assignment

Selection is impartial: “dice not trying to prove a point” - must convince others of validity of comparison


RANDOMIZATIONFIXED ALLOCATION: Assigns with pre-specified probability (not necessarily, though usually, equal)

ADAPTIVE: Changes probabilities during study Baseline adaptive:

• on basis of number per group• on basis of variables

Responsive adaptive:• depends on prior outcome• assumes:

• rapid response• stable population source


STEPS IN THE RANDOMIZATION OF A PATIENT

Check eligibility

Informed consent

Formal identification

RANDOMIZE

Confirmation of patient entry


HOW RANDOM TREATMENT ASSIGNMENTS ARE MADE

Model: Slips in a hat or flipping a coin

Masked drugs numbered and given in order: pharmacy, drug manufacturer

Envelopes

Telephone to central unit Microcomputer at the site

Central computer – internet access


HOW TO DO THE SCHEME

Simple randomization

Biased coin, urn models

Example:

Start with 2 balls, one black and one white

Draw-replace and add one of opposite color

Prevents imbalance with high probability early on

Random permuted block

Balance at the end of block

Could predict with unmasked trial


BLOCKS OF SIZE 4

4

24

2 2

4 3 2 1

2 1 2 16

!

! !

* * *

* * *1) 1100

2) 1010

3) 1001

4) 0110

5) 0101

6) 0011


HOW TO USE BLOCKS WHEN TREATMENT IS NOT MASKED

Choose the block sizes at random, too

Example: 2 treatment, equal allocation

Block sizes 4, 6, and 8

Balance in each block


SHOULD YOU STRATIFY?

Clinical sites - generally yes

Prognostic variables - generally no

Size

Practical considerations

Often governed by custom rather than statistical justification

Stratified ANALYSIS is generally preferable


MINIMIZATION

Advantages: Balance several prognostic factors Balance marginal treatment totals Good for small trials (<100 patients) Computer makes this fairly easily

Disadvantages: Can’t prepare treatment assignment Scheme in advance Need up-to-date record Not really random - could predict but can introduce

random element by using say 3/4, 1/4


TABLE 5.7. - TREATMENT ASSIGNMENTS BY THE FOUR PATIENT FACTORS FOR 80 PATIENTS IN AN ADVANCED BREAST CANCER TRIAL

Factor Level No. on each Next treatment patient A B

Performance status Ambulatory 30 31 Non-ambulatory 10 9

Age <50 18 17 50 22 23

Disease-free interval <2 years 31 32 2 years 9 8

Dominant metastatic lesion Visceral 19 21 Osseous 8 7 Soft tissue 13 12

Pocock S. Clinical Trials: A Practical Approach. John Wiley & Sons, Chichester, England, 1991, p. 85.

Thus, for A this sum = 30 + 18 + 9 + 19 = 76while for B this sum = 31 + 17 + 8 + 21 = 77


INTERNAL VALIDITY

compare treatmentsExternal Validity/ Generalizability

extrapolate to other patients

Not realistic to find a random sample of patients for recruitment (at the very least they have to consent)

More important to establish efficacy of treatment before deciding if it can be broadly applied

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Lecture 10 Sample Size

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