Sample Size EstimationSample Size Estimation

Chulaluk Komoltri DrPH (Bios)Faculty of Medicine Siriraj Hospital

Sample Size Estimation

• Why ?

- n is large enough to provide a reliable answer to thequestion

- too small n a waste of time

- too many n a waste of money & other resourcesMay be unethical e.g., delayed beneficial therapy

placebo

• Study Objective:- Hypothesis generating (Pilot study)

No sample size estimation

- Hypothesis confirmation Sample size estimation

- n is usually determined by the primary objectiveof the study

- method of calculating n should be given in theproposal, together with the assumptions madein the calculation

Pilot study

Example: (Mar 23,1999, #1515)

This study for n=20 eligible burn patients

will generate hypothesis about the predictive

values of various patient characteristics

for predicting number of days to return to

work.

Example: (Oct 27, 1998, #1465)

This is a pilot study providing preliminary

descriptive statistics that will be used to

design a larger, adequately powered study.

N=24 normal healthy volunteers will be

randomized to parallel groups to study the

effect of 4 antidepressant drugs…

Pilot study (cont’d)

Sample size determination: 2 Objectives

I. Estimation of parameter(s) Precision (95% CI)

Specify α error

- Estimate prevalence, sensitivity, specificity

- Estimate single mean, single proportion

- etc.

Hypothesis confirmation study

II. Test H0 Statistical power (1- β)

Specify α, β error2.1 Single group

- Test of single proportion, mean- Test of Pearson’s correlation- etc.

2.2 Two groups- Test difference of

- 2 independent proportions, means, survival curves- 2 dependent proportions, means

- Test equivalence of- 2 independent proportions, means

- etc.2.3 > 2 groups

2

2/2

dpq z n α=

1) Estimation1.1) Estimate single proportion 95% CI of π = p ± d

(π = True pop’n proportionp = Expected proportion e.g., prevalenceq = 1-pd = Margin of error in estimating p)

1.2) Estimate single mean 95% CI of μ =⎯x ± d(μ = True pop’n mean⎯x = Expected mean d = Margin of error in estimating mean)

n = [zα/2 SD / d]2

Commonly used formulas: Summary

2) Test

2.1) Test of difference in 2 independent proportions (p1, p2)p1, p2 = Proportion of … in group 1 and 2p = (p1+p2)/2

2

21

2211/2 ]pp

qp qpz pq2z[ n/group

−

++= βα

2/2 ])z (z

[ 2 n/groupΔ

+=

σβα

2.2) Test of difference in 2 independent means (Δ)σ = Common SD of outcome var. in group 1, 2Δ = Difference in mean b/t 2 groups

2.3) Test of difference in > 2 independent means

2

10

1100 ]pp

qpz qpz[ n

−

+= βα

2.4) Test of significance of 1 proportionH0: π = π0H1: π = π1

2.5) Test of significance of 1 meanH0: μ = μ0H1: μ = μ1Δ = |μ1 - μ0|σ = SD of outcome var.

2])z (z

[ nΔ

+=

σβα

n = (Zα/2 + Zβ) 2 + 3[F(Z0) + F(Z1)]

F(Z) = 0.5 ln [(1+ρ)/(1-ρ)]

2.6) Test of significance of 1 correlation

I. Estimation

Example: (Mar 18, 2000, #1688)

This is a cross-sectional study of the prevalenceof pulmonary hypertension (PHT) in patients aged15-70 years with sickle cell disease.

The primary endpoint is PHT diagnosis basedon observed pulmonary pressure by dropplerechocardiogram.

A sample of n = 140 will provide 95% CI for true prevalence rate of PHT of 0.10 ± 0.05.

1.1 Estimate Prevalence

95% CI for true prevalence (π) = 0.10 ± 0.05 (1- α)100% ″ = p ± d

= p ± zα/2√pq/nd = zα/2√pq/n solve for n

2

2/2

dpq z n α=

where p = estimated prevalence = 0.1q = 1 - p = 0.9d = allowable error in estimating prevalence

(margin of error) = 0.05α = probability of type I error

= 0.05 (2-sided), z 0.025 = 1.96

139 138.3 0.05(0.1)(0.9) 1.96 n 2

2

===

How big is d ? 1. Absolute d 2. Relative d: d ≤ 20% of prevalence(p)

p d 95% CI n0.80 0.05*p = 0.04 0.76, 0.84 384

0.05 0.75, 0.85 2460.10*p = 0.08 0.72, 0.88 96

0.10 0.70, 0.90 620.15*p = 0.12 0.68, 0.92 43

0.15 0.65, 0.95 280.20*p = 0.16 0.64, 0.96 24

0.20 0.60, 1.00 16

2

2/2

dpq z n α=

pqd (error) nα

Example:- 95% CI for Sensitivity = 85% ± 5% nDi = ?- 95% CI for Specificity = 90% ± 5% nNon-Di = ?

I. Estimation (cont’d)

1.2 Estimate Sensitivity, Specificity

Gold standard+ -

Test + a b- c d

a+c b+dSensitivity = a / (a+c) Specificity = d / (b+d)

p ± d 95% CI nSensitivity 85 ± 5 (80, 90) nDi = 196Specificity 90 ± 5 (85, 95) nNon-Di = 139

Ex.Title: Diagnosis of Benign Paraxysmal Positional Vertigo (BPPV)

by Side-lying test as an alternative to the Dix-Hallpike test

Investigator: Dr. Saowaros AsawavichiangindaDesign: Diagnostic studySubjects: Dizzy patients, aged 18-80 yrs, onset < 2 wks

Dizzy pts.

BPPV No BPPV

1. Dix-Hallpike test2. Side-lying test

Sample size: Based on 95% CI of true sensitivity (Sn) = 0.9 ± 0.1

where p = expected sensitivity = 0.9q = 1-p = 0.1d = allowable error = 0.1α = 0.05 (2-sided), Z0.025 = 1.96

So, n = 34.56 = No. of patients with BPPV from Dix-Hallpike testSince prevalence of BPPV among dizzy patients = 40%Thus, no. of dizzy patients = 34.56 = 86.4 = 87

0.4

Dix-Hallpike test (Gold std)+ (BPPV) - (No BPPV)

Side-lying test + Sn- 1 – Sn

35 52 87

2

2/2

dpq z n α=

การศึกษานี้มีวัตถุประสงคเพื่อประมาณคาเฉลี่ยของ subcarinal angleในคนไทยปกติ และจากการศึกษาของ ... ในคนปกติจํานวน 100 รายอายุ ... ป พบวาคาเฉลี่ยของ subcarinal angle เทากับ 60.8 (SD=11.8)ถากําหนดให 95% confidence interval (CI) ของคาเฉลี่ยของ subcarinal angle ในประชากรไทย (μ) มีคาเทากับ 61 ± 2 (SD=13) จะตองทําการศึกษาในคนไทยปกติจํานวน 163 คนดังรายละเอียดการคํานวณดังนี้

n = [zα/2 SD / d]2เมื่อ SD = Standard deviation ของ subcarinal angle

= 13d = Margin of error ในการประมาณคาเฉลี่ย = 2α = Probability of type I error (2-sided) = 0.05

z0.025 = 1.96 ดังนั้น n = [1.96*13/2]2 = 162.31 = 163

1.3 Estimation of 1 Mean

II. Test

Example: (May 25, 1999, #1549)

This is a randomized (1:1), double-blind,parallel-group, multi-center trial of drug A (dose1, 2)in chronic hepatitis C patients aged 18+ years.

The primary efficacy endpoint is sustained viral response rate after treatment.

N = 141 per group will provide 80% power to detectan absolute difference in sustained viral response rateof 11% (7% vs. 18%) at 2-sided α of 0.05.

2.1 Test for Difference in 2 Independent Proportions

N = 141 per group will provide 80% power to detectan absolute difference in sustained viral response rateof 11% (7% vs. 18%) at 2-sided α of 0.05.

Clinical (Practical) significance

Statistical significance

Clinical significance vs. Statistical significance

HypothesesH0 : π1 - π2 = 0H1 : π1 - π2 ≠ 0 (2-sided)

or H1 : π1 - π2 > 0 (1-sided, upper tail)or H1 : π1 - π2 < 0 (1-sided, lower tail)

where π1 = True (population) response rate in group 1 π2 = True (population) response rate in group 2

1-sided, 2-sided testn (2-sided test) > n (1-sided test)

2-sided test is conservative use more often

Decision to use either 1- or 2-sided test should bemade at the design stage, not after looking at the data

α, β (Efficacy trial)

TruthH0 true H0 false(A=B) (A≠B, Difference)

Decision Accept H0 No error (1- α) β(from p-value)

Reject H0 α No error (1- β)Power

α = Pr (incorrect conclusion of difference )= False positive (FP)

β = Pr (incorrect conclusion of equivalence)= False negative (FN)

1 - β = Pr ( correct conclusion of difference )= True positive (TP)

TruthH0 true H0 false

(Not guilty) (Guilty)

Decision Accept H0 No error, 1-α Type II error, β(Not guilty)

Reject H0 Type I error, α No error, 1-β, (Guilty) Power

α = Probability of wrongly put innocent person into jailβ = Probability of wrongly set the criminal free

1-β = Probability of correctly put criminal into jail

α is more important than β, so usually set β = 4 α

How big is α, β?

1. Type I error (α, test size, significance level)- To replace a standard drug with a new drug,

type I error is serious,use small α (0.01, 0.02)

- To add to the body of the published knowledge,type I error is less serious

use α = 0.05, 0.10

2. Type II error (β)- Power (1 - β)- Power is conventionally set at 80% - 90%- Typically, α is 4 times as serious as β

α = 0.05, β = 0.20 (power = 0.80)

2

21

2211/2 ]pp

qp qpz pq2z[ n/group

−

++= βα

where p1 = response rate in group 1 = 0.07q1 = 1 - p1 = 0.93p2 = response rate in group 2 = 0.18q2 = 1 – p2 = 0.82p = (p1 + p2) / 2 = 0.125q = 1 – p = 0.875

α = 0.05 (2-sided), z0.025 = 1.961- β = 0.80, z0.2 = 0.842

n/group = 141

Calculation: n1 = n2 = nBased on Chi-square test without continuity correction

Zα if 1-sided

α = 0.05p1 p2 Power 2-sided 1-sided

7 18 80 141 11190 188 153

7 20 80 108 8590 144 117

n / group

(p1 – p2) Power n α

Calculation: n1 = n2 = nBased on Chi-square test with continuity correction

159~ 158.7

0.07-0.181414 1 1

4141

p -pn 4 1 1

4n n

2

2

21

=⎥⎥⎦

⎤

⎢⎢⎣

⎡++=

⎥⎥⎦

⎤

⎢⎢⎣

⎡++=′

Ex:Title: Efficacy of polyethylene plastic wrap for the prevention of

hypothermia during the immediate postnatal period in low birth weight premature infants

Investigator: Dr. Santi Punnahitananda

Design: RCT, 2-parallel armsSubjects: Infants with ≤ 34 gestational wks, birth weight ≤ 1800 gms

Outcome: Infant’s body temperature taken on nursery admission

Infants, ≤ 34 gestational wks, BW ≤ 1800 gms

Plastic wrap No Plastic wrap

Body temp.Hypothermia

Body temp.Hypothermia

Randomization

Sample size estimation: Based on Test of 2 independent proportions

Our unit hypothermia in low birth weight, premature infants = 55%(p1 = 0.55)

Assume that plastic wrap would reduce hypothermia to 20%(p2 = 0.2)

2

21

2211/2 ]pp

qp qpz pq2z[ n/group

−

++= βα

ตัวอยาง

ในการศึกษาเพื่อหาปจจัยตางๆที่มีความสัมพันธกับการพยายามทํารายตนเอง ผูวิจัยสนใจในผลของการมีประวัติการเจ็บปวยดวยโรคทางจิตตอการพยายามทํา

รายตนเอง การศึกษาในอดีตพบวาผูปวยที่ไมเคยทํารายตนเอง (control) มีประวัติการ

เจ็บปวยดวยโรคทางจิต 4% และผูวิจัยคาดวาผูปวยที่เคยทํารายตนเองแตไม

เสียชีวิต (case) จะมีประวัติการเจ็บปวยดวยโรคทางจิตมากกวาคือ 10%

เนื่องจากโดยปกติจํานวนผูปวยที่ไมเคยทํารายตนเองมีมากกวาจํานวนผูปวยที่

เคยทํารายตนเอง จึงกําหนดให control มีจํานวนเปน 2 เทาของ case

เพื่อใหการศึกษานี้มี power 80% ในการพบวาความแตกตางของประวัติการเจ็บปวยดวยโรคทางจิต 10% vs. 4% มีนัยสําคัญทางสถิติที่ 2-sided type I error = 0.05 จะตองใช case 200 คนและ control 400 คน

n1 = ncase = [zα/2√(r+1)pq + zβ√r p1q1 + p2q2 ]2

r (p1 – p2)2

เมื่อ

r = n2/n1 = ncontrol / ncase = 2p1 = สัดสวนการมปีระวัตกิารเจ็บปวยดวยโรคทางจิตในกลุม case = 0.10q1 = 1- p1 = 0.90p2 = สัดสวนการมปีระวัตกิารเจ็บปวยดวยโรคทางจิตในกลุม control = 0.04q2 = 1- p2 = 0.96

p = (p1 + rp2) / (r+1) = 0.06q = 1- p = 0.94α = โอกาสที่จะเกิด type I error = 0.05 (2-sided), z0.025 = 1.96β = โอกาสที่จะเกิด type II error = 0.2, z0.2 = 0.842

ดังนั้น ncase = [0.8062 + 0.3935]2 = 199.9 0.0072

Example: (Aug 24, 1999, #1575)For patients with idiopathic membranous

glomerulopathy, a phase II, randomized (1:1),double-blind, placebo-controlled, multi-centerstudy of drug A will be conducted to determine efficacy.

The primary efficacy endpoint is the change frombaseline in proteinuria at Week 18.

N = 45 per group will provide 80% power to detecta difference in mean change in loge of urine protein of–1.22 for placebo and –2.00 for an active drug,assuming SD = 1.30, 2-sided α of 0.05.

A drop-out rate of 20% is expected, so N = 55 per groupwill be recruited.

II. Test (cont’d) 2.2 Test for Difference of 2 Independent Means

id trt ln_pro0 ln_pro18 pro_d 1 0 2 0

n1 0n1+1 1n1+2 1

n1+n2 1

trt: 0=placebo, 1=Drug Aln_pro0 = loge of proteinurea at Wk 0 (baseline)ln_pro18 = loge of proteinurea at Wk 18pro_d = change at Wk18 of loge of proteinurea from baseline

Hypotheses

H0 : μ1 - μ2 = 0H1 : μ1 - μ2 ≠ 0 (2-sided)

or H1 : μ1 - μ2 > 0 (1-sided, upper tail)or H1 : μ1 - μ2 < 0 (1-sided, lower tail)

where μ1 = true (population) mean in group 1 μ2 = true (population) mean in group 2

2/2 ])z (z

[ 2 n/groupΔ

+=

σβα

where σ = Common standard deviation of change inloge urine protein = 1.30(σ1 = σ2 = σ)

Δ = Difference in mean change between 2 groupsthat is considered clinically important

= (-1.22) – (-2.00) = 0.78Δ / σ = Effect size (ES)

= effect of treatment in SD unit α = 0.05 (2-sided), z0.025 = 1.96

1 - β = 0.80, z0.2 = 0.842

Calculation: n1 = n2 = n Zα if 1-sided

n / group = 44Drop-out 20% n / group = 44 = 55

(1- dropout)

Δ σ Power n / group

0.78 1.30 80 4590 60

1.50 80 60

0.50 1.30 80 108

1.50 80 143

(mean1 – mean2) σ nPowerα

2/2 ])z (z

[ 2 n/groupΔ

+=

σβα

Ex:Title: Can knee immobilization after total knee replacement (TKA)

save blood from wound drainage

Investigator: Dr. Vajara WilairatanaDesign: Randomized controlled trialSubjects: Pts. with hip disease that require TKA

Pts. with hip disease that require TKA

Knee elevation 40° A-P splint andKnee elevation 40°

Randomization

Blood loss Blood loss

2/2 ])z (z

[ 2 n/groupΔ

+=

σβα

whereΔ = Difference in mean postoperative blood loss

between 2 groups σ = SD of postoperative blood loss

Kim YH et al.Knee splint in 69 knees, mean wound drainage = 436 ml, SD = 210 ml

Ishii et al.30 non-splint knees, mean blood loss = 600 ml, SD = 293

Ex:Title: Early postoperative pain and urinary retention after closed

hemorrhoidectomy: Comparison between spinal andlocal anesthesia

Investigator: Dr. Sahapol AnannamcharoenDesign: RCTSubjects: Pts. with grade 3 or 4 hemorrhoidal disease

Pts. with hemorrhoidal disease

Spinal anesthesia Perianal nerve block

Randomization

Visual analogue scale (VAS) pain score (0-10)

Sample size if parametric test (2-sample t-test) is used

Sample size if Non-parametric (Mann-Whitney) test is used(VAS pain score is usually positively skewed !!)

II. Test

Example: (Feb 29, 2000, #1649)

N = 100 subjects with nontuberculous mycobacteriainfection will be recruited for this multi-center study.

The primary objective is to test if the frequency ofcystic fibrosis transmembrane conductance regulator(CFTR) gene mutation is 4%. If more CF carriers arefound at a statistically significant number, then this would suggest that CFTR alleles may be important inpredisposing to this disease.

N = 96 will provide 90% power to test H0 : π = π0 = 0.04,against 1-sided H1 : π = π1 = 0.115, using α = 0.05.

2.4 Test of Significance of 1 Proportion

2

10

1100 ]pp

qpz qpz[ n

−

+= βα

where p0 = 0.04q0 = 1 – p0 = 0.96

p1 = 0.115q1 = 1 – p1 = 0.885

α = 0.05 (1-sided), z0.05 = 1.6451- β = 0.90, z0.1 = 1.282

n = 96

Calculation

H0 : π = π0 (π0 = 0.04 )H1 : π > π0 (π1 = 0.115)

Hypotheses

II. Test (cont’d)

Example:The average weight of men over 55 years of age

with newly diagnosed heart disease was 90 kg.

However, it is suspected that the average weight is now somewhat lower.

How large a sample would be necessary to test, at 5% level of significance with a power of 90%,whether the average weight is unchanged versusthe alternative that it has decreased from 90 to 85 kgwith an estimated SD of 20 kg?

2.5 Test of Significance of 1 Mean

HypothesesH0 : μ = μ0 (μ0 = 90)H1 : μ < μ0 (μ1 = 85)

Calculation 2]

)z (z[ n

Δ

+=

σβα

where σ = estimated SD = 20Δ = | μ1 - μ0| = 5

α = 0.05 (1-sided), z0.05 = 1.6451- β = 0.90, z0.1 = 1.282

n = 137.08 = 138

II. Test (cont’d)

2.6 Test of Significance of 1 Correlation CoefficientH0: ρ = ρ0H1: ρ = ρ1

จากวัตถุประสงคหลกัของการวิจัยเพื่อหาความสัมพนัธระหวาง

คาที่ไดจากการวัดพงัผดืในตบัดวยวิธี Transient Elastographyกับคาที่ไดจากการวินิจฉยัภาวะพังผืดดวยวิธีเจาะชิ้นเนือ้ตับ การคํานวณขนาดตัวอยางจึงเปนการคํานวณเพื่อทดสอบคา Correlation coefficient โดยมีสมมติฐานทางสถิติดงันี้

H0 : ρ0 = 0 H1 : ρ1 = 0.3 (Ref #1)

“การศึกษาความสัมพนัธระหวางการวัดพงัผดืในตบัดวยวิธีTransient Elastography กับการวินิจฉัยดวยวิธีเจาะชิ้นเนือ้ตับในผูปวยโรคเรื้อนกวาง (psoriasis)ที่ไดรับการรักษาดวยยาเมโธเทร็กเซท (methotrexate; MTX) ในคนไทย”

n = (Zα/2 + Zβ) 2 + 3[F(Z0) + F(Z1)]

α = Probability of type I error = 0.05 (2-sided) Z0.025 = 1.96

β = Probability of type II error = 0.11-β = Power = 0.90

Z0.1 = 1.282

F(Z) = Fisher’s Z transformation = 0.5 ln [(1+ρ)/(1-ρ)]

Under H0: ρ=0 F(Z0) = 0.5 x ln [(1+0)/(1-0)] = 0Under H1: ρ=0.3 F(Z1) = 0.5 x ln [(1+0.3)/(1-0.3)] = 0.31

Thus,n = [ (1.96+1.282)/(0-0.31) ]2 + 3

= 112.4 = 113

More than one primary outcome

If one of these endpoints is regarded as more importantthan others, then calculate n for that primary endpoint.

If several outcomes are regarded as equally important,then calculate n for each outcome in turn, and select the largest n as the sample size required toanswer all the questions of interest.

Caution:

Calculation of sample size needs a number of

assumptions and ‘guesstimates’,

so such calculation only provides a guide to the

number of subjects required.

Sample size estimation:

1. Formulas

2. Published tables, nomograms

3. Softwares e.g., - nQuery Advisor- PS (Power and Sample Size Program) - etc.

ReferencesBlackwelder WC. Proving the Null Hypothesis in Clinical

Trials. Controlled Clinical Trials 1982; 3: 345-353.

Breslow NE, Day NE. Statistical Methods in CancerResearch Vol. II – The Design and Analysis of Cohort Studies. Oxford : Oxford University Press; 1987.

Chow SC, Liu JP. Design and Analysis of Clinical Trials.Concept and Methodologies. New York: John Wiley & Sons,Inc. 1998.

Fleiss JL. Statistical Methods for Rates and Proportions. New York : John Wiley & Sons; 1981.

Karlberg J, Tsang K. Introduction to Clinical Trials: ClinicalTrials Research Methodology, Statistical Methods in ClinicalTrials, The ICH GCP Guidelines. Hong Kong: The ClinicalTrials Centre. 1998.

Lachin JM. Introduction to Sample Size Determinationand Power Analysis for Clinical Trials. Controlled Clinical Trials 1981; 2: 93-113.

Lemeshow S, Hosmer DW, Klar J, Lwanga SK.Adequacy of Sample Size in Health Studies. New York :John Wiley & Sons; 1990.