Sample Size Estimation Chulaluk Komoltri DrPH (Bios) Faculty of Medicine Siriraj Hospital

Transcription

Sample 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 the
question
- too small n Æ a waste of time
- too many n Æ a waste of money & other resources
May be unethical
e.g., delayed beneficial therapy
placebo
• Study Objective:
- Hypothesis generating (Pilot study)
Æ No sample size estimation
- Hypothesis confirmation
Æ
Sample size estimation
- n is usually determined by the primary objective
of the study
- method of calculating n should be given in the
proposal, together with the assumptions made
in 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.
Pilot study (cont’d)
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…
Hypothesis confirmation study
Sample size determination: 2 Objectives
I. Estimation of parameter(s) Æ Precision (95% CI)
Specify α error
- Estimate prevalence, sensitivity, specificity
- Estimate single mean, single proportion
- etc.
II. Test H0
Æ Statistical power (1- β)
Specify α, β error
2.1 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
Commonly used formulas: Summary
1)
Estimation
1.1) Estimate single proportion Æ 95% CI of π = p ± d
(π = True pop’n proportion
p = Expected proportion e.g., prevalence
q = 1-p
d = Margin of error in estimating p)
2
zα/2 pq
n =
d2
1.2) Estimate single mean Æ 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
2)
Test
2.1) Test of difference in 2 independent proportions (p1, p2)
p1, p2 = Proportion of … in group 1 and 2
p
= (p1+p2)/2
n/group = [
zα/2 2pq + z β p1q1 + p 2 q 2
p1 − p 2
2.2) Test of difference in 2 independent means (Δ)
σ = Common SD of outcome var. in group 1, 2
Δ = Difference in mean b/t 2 groups
n/group = 2 [
(zα/2 + z β )σ
Δ
]
2
2.3) Test of difference in > 2 independent means
]2
2.4) Test of significance of 1 proportion
H0: π = π0
H1: π = π1
n =[
zα p 0 q 0 + z β p1q1
p 0 − p1
2.5) Test of significance of 1 mean
H0: μ = μ0
H1: μ = μ1
Δ = |μ1 - μ0|
σ = SD of outcome var.
n = [
(z α + z β )σ
Δ
]2
]2
2.6) Test of significance of 1 correlation
n
F(Z)
=
=
(Zα/2 + Zβ) 2 + 3
[F(Z0) + F(Z1)]
0.5 ln [(1+ρ)/(1-ρ)]
I. Estimation
1.1 Estimate Prevalence
Example: (Mar 18, 2000, #1688)
This is a cross-sectional study of the prevalence
of pulmonary hypertension (PHT) in patients aged
15-70 years with sickle cell disease.
The primary endpoint is PHT diagnosis based
on observed pulmonary pressure by droppler
echocardiogram.
A sample of n = 140 will provide 95% CI for true
prevalence rate of PHT of 0.10 ± 0.05.
95% CI for true prevalence (π) = 0.10 ± 0.05
(1- α)100%
″
= p ± d
= p ± zα/2√pq/n
d = zα/2√pq/n Æ solve for n
2
zα/2 pq
n =
2
d
where
p = estimated prevalence = 0.1
q = 1-p
= 0.9
d = allowable error in estimating prevalence
(margin of error)
= 0.05
α = probability of type I error
= 0.05 (2-sided), z 0.025 = 1.96
1.96 2 (0.1)(0.9)
n =
= 138.3 = 139
2
0.05
How big is d ?
1. Absolute d
2. Relative d: d ≤ 20% of prevalence(p)
p
d
95% CI
n
0.80
0.05*p = 0.04
0.05
0.10*p = 0.08
0.10
0.15*p = 0.12
0.15
0.20*p = 0.16
0.20
0.76, 0.84
0.75, 0.85
0.72, 0.88
0.70, 0.90
0.68, 0.92
0.65, 0.95
0.64, 0.96
0.60, 1.00
384
246
96
62
43
28
24
16
2
zα/2 pq
n =
2
d
pq
d (error)
α
n
I. Estimation (cont’d)
1.2 Estimate Sensitivity, Specificity
Example:
- 95% CI for Sensitivity = 85% ± 5% Æ nDi
=?
- 95% CI for Specificity = 90% ± 5% Æ nNon-Di = ?
Gold standard
+
Test +
a
b
c
d
a+c
b+d
Sensitivity = a / (a+c)
Specificity = d / (b+d)
Sensitivity
Specificity
p±d
85 ± 5
90 ± 5
95% CI
(80, 90)
(85, 95)
n
= 196
nDi
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 Asawavichianginda
Design:
Diagnostic study
Subjects:
Dizzy patients, aged 18-80 yrs, onset < 2 wks
Dizzy pts.
1. Dix-Hallpike test
2. Side-lying test
BPPV
No BPPV
Sample size: Based on 95% CI of true sensitivity (Sn) = 0.9 ± 0.1
2
zα/2 pq
n =
2
d
where
p
q
d
α
=
=
=
=
expected sensitivity = 0.9
1-p = 0.1
allowable error = 0.1
0.05 (2-sided), Z0.025 = 1.96
So, n = 34.56 = No. of patients with BPPV from Dix-Hallpike test
Since 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
1.3 Estimation of 1 Mean
การศึกษานี้มีวัตถุประสงคเพื่อประมาณคาเฉลี่ยของ subcarinal angle
ในคนไทยปกติ และจากการศึกษาของ ... ในคนปกติจํานวน 100 รายอายุ ... ป
พบวาคาเฉลี่ยของ subcarinal angle เทากับ 60.8 (SD=11.8)
ถากําหนดให 95% confidence interval (CI) ของคาเฉลี่ยของ
subcarinal angle ในประชากรไทย (μ) มีคาเทากับ 61 ± 2 (SD=13)
จะตองทําการศึกษาในคนไทยปกติจํานวน 163 คนดังรายละเอียดการคํานวณดังนี้
เมื่อ
ดังนั้น
n = [zα/2 SD / d]2
SD = Standard deviation ของ subcarinal angle
= 13
d = Margin of error ในการประมาณคาเฉลี่ย = 2
α = Probability of type I error (2-sided) = 0.05
z0.025 = 1.96
n = [1.96*13/2]2 = 162.31 = 163
II. Test
2.1 Test for Difference in 2 Independent Proportions
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 detect
an absolute difference in sustained viral response rate
of 11% (7% vs. 18%) at 2-sided α of 0.05.
Clinical significance vs. Statistical significance
an absolute difference in sustained viral response rate
of 11% (7% vs. 18%) at 2-sided α of 0.05.
Clinical (Practical) significance
Statistical significance
Hypotheses
or
or
where
H0
H1
H1
H1
:
:
:
:
π1 - π2
π1 - π2
π1 - π2
π1 - π2
=
≠
>
<
0
0 (2-sided)
0 (1-sided, upper tail)
0 (1-sided, lower tail)
π1 = True (population) response rate in group 1
π2 = True (population) response rate in group 2
1-sided, 2-sided test
n (2-sided test)
> n (1-sided test)
2-sided test is conservative Æ use more often
Decision to use either 1- or 2-sided test should be
made at the design stage, not after looking at the data
α, β (Efficacy trial)
Truth
H0 true
(A=B)
Decision
(from p-value)
Accept H0
Reject H0
No error (1- α)
α
H0 false
(A≠B, Difference)
β
No error (1- β)
Power
α
= Pr (incorrect conclusion of difference
= False positive (FP)
β
=
=
=
=
1-β
)
Pr (incorrect conclusion of equivalence)
False negative (FN)
Pr ( correct conclusion of difference )
True positive (TP)
Truth
H0 true
(Not guilty)
Decision Accept H0
(Not guilty)
Reject H0
(Guilty)
H0 false
(Guilty)
No error, 1-α
Type II error, β
Type I error, α
No error, 1-β,
Power
α = Probability of wrongly put innocent person into jail
β = 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,
Æ use small α (0.01, 0.02)
- To add to the body of the published knowledge,
type I error is less serious
Æ 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)
Calculation: n1 = n2 = n
Based on Chi-square test without continuity correction
Zα if 1-sided
n/group = [
where
zα/2 2pq + z β p1q1 + p 2 q 2
p1 − p 2
]2
p1 = response rate in group 1 = 0.07
= 0.93
q1 = 1 - p1
p2 = response rate in group 2 = 0.18
q2 = 1 – p2
= 0.82
p
q
= (p1 + p2) / 2
= 1–p
α = 0.05 (2-sided),
1- β = 0.80,
Æ n/group = 141
= 0.125
= 0.875
z0.025 = 1.96
z0.2
= 0.842
n / group
α = 0.05
2-sided 1-sided
p1
p2
Power
7
18
80
90
141
188
111
153
7
20
80
90
108
144
85
117
(p1 – p2)
Power
α
n
Based on Chi-square test with continuity correction
n′ =
n
4
⎤
⎡
4
⎥
⎢1 + 1 +
n p1 - p 2 ⎥⎦
⎢⎣
2
141 ⎡
4
=
⎢1 + 1 +
4 ⎢⎣
141 0.18 - 0.07
= 158.7 ~ 159
⎤
⎥
⎥⎦
2
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 arms
Subjects: 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
Randomization
Plastic wrap
No Plastic wrap
Body temp.
Æ Hypothermia
Body temp.
Æ Hypothermia
Sample size estimation: Based on Test of 2 independent proportions
Our unit Æ hypothermia in low birth weight, premature infants = 55%
(p1 = 0.55)
Assume that plastic wrap would reduce hypothermia to 20%
(p2 = 0.2)
n/group = [
zα/2 2pq + z β p1q1 + p 2 q 2
p1 − p 2
]
2
ตัวอยาง
ในการศึกษาเพื่อหาปจจัยตางๆที่มีความสัมพันธกับการพยายามทํารายตนเอง
ผูวิจัยสนใจในผลของการมีประวัติการเจ็บปวยดวยโรคทางจิตตอการพยายามทํา
รายตนเอง
การศึกษาในอดีตพบวาผูปวยที่ไมเคยทํารายตนเอง (control) มีประวัติการ
เจ็บปวยดวยโรคทางจิต 4% และผูวิจัยคาดวาผูปวยที่เคยทํารายตนเองแตไม
เสียชีวิต (case) จะมีประวัติการเจ็บปวยดวยโรคทางจิตมากกวาคือ 10%
เนื่องจากโดยปกติจํานวนผูปวยที่ไมเคยทํารายตนเองมีมากกวาจํานวนผูปวยที่
เคยทํารายตนเอง จึงกําหนดให control มีจํานวนเปน 2 เทาของ case
เพื่อใหการศึกษานี้มี power 80% ในการพบวาความแตกตางของประวัติการ
เจ็บปวยดวยโรคทางจิต 10% vs. 4% มีนัยสําคัญทางสถิติที่ 2-sided type
I error = 0.05 จะตองใช 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 = 2
 case = 0.10
p1 = สัดสวนการมีประวัตกิ ารเจ็บปวยดวยโรคทางจิตในกลุม
q1 = 1- p1 = 0.90
 control = 0.04
p2 = สัดสวนการมีประวัตกิ ารเจ็บปวยดวยโรคทางจิตในกลุม
q2 = 1- p2 = 0.96
p = (p1 + rp2) / (r+1) = 0.06
q = 1- p = 0.94
α =
β =
ดังนั้น
โอกาสที่จะเกิด type I error = 0.05 (2-sided), z0.025 = 1.96
โอกาสที่จะเกิด type II error = 0.2, z0.2 = 0.842
ncase
=
[0.8062 + 0.3935]2 = 199.9
0.0072
II. Test (cont’d)
2.2 Test for Difference of 2 Independent Means
Example: (Aug 24, 1999, #1575)
For patients with idiopathic membranous
glomerulopathy, a phase II, randomized (1:1),
double-blind, placebo-controlled, multi-center
study of drug A will be conducted to determine efficacy.
The primary efficacy endpoint is the change from
baseline in proteinuria at Week 18.
a 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 group
will be recruited.
id
trt
1
2
0
0
n1
0
n1+1
n1+2
1
1
n1+n2
1
ln_pro0
ln_pro18
pro_d
trt: 0=placebo, 1=Drug A
ln_pro0 = loge of proteinurea at Wk 0 (baseline)
ln_pro18 = loge of proteinurea at Wk 18
pro_d
= change at Wk18 of loge of proteinurea from baseline
Hypotheses
or
or
H0
H1
H1
H1
:
:
:
:
μ1 - μ2
μ1 - μ2
μ1 - μ2
μ1 - μ2
=
≠
>
<
0
0 (2-sided)
0 (1-sided, upper tail)
0 (1-sided, lower tail)
where μ1 = true (population) mean in group 1
μ2 = true (population) mean in group 2
Zα if 1-sided
n/group = 2 [
(zα/2 + z β )σ
Δ
]
2
σ = Common standard deviation of change in
loge urine protein
= 1.30
(σ1 = σ2 = σ)
Δ = Difference in mean change between 2 groups
that is considered clinically important
= (-1.22) – (-2.00)
= 0.78
Δ / σ = Effect size (ES)
= effect of treatment in SD unit
α = 0.05 (2-sided), z0.025 = 1.96
z0.2
= 0.842
1 - β = 0.80,
where
Æ n / group =
Drop-out 20% Æ n / group =
44
44
= 55
(1- dropout)
Δ
σ
0.78
0.50
n/group = 2 [
Power
n / group
1.30
80
90
45
60
1.50
80
60
1.30
80
108
1.50
80
143
(zα/2 + z β )σ
Δ
]2
(mean1 – mean2)
σ
Power
α
n
Ex:
Title: Can knee immobilization after total knee replacement (TKA)
save blood from wound drainage
Investigator: Dr. Vajara Wilairatana
Design:
Randomized controlled trial
Subjects:
Pts. with hip disease that require TKA
Pts. with hip disease that require TKA
Randomization
Knee elevation 40°
Blood loss
A-P splint and
Knee elevation 40°
Blood loss
n/group = 2 [
(zα/2 + z β )σ
Δ
]
2
where
Δ = 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 and
local anesthesia
Investigator: Dr. Sahapol Anannamcharoen
Design:
RCT
Subjects:
Pts. with grade 3 or 4 hemorrhoidal disease
Pts. with hemorrhoidal disease
Randomization
Spinal anesthesia
Perianal nerve block
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
2.4 Test of Significance of 1 Proportion
Example: (Feb 29, 2000, #1649)
N = 100 subjects with nontuberculous mycobacteria
infection will be recruited for this multi-center study.
The primary objective is to test if the frequency of
cystic fibrosis transmembrane conductance regulator
(CFTR) gene mutation is 4%. If more CF carriers are
found at a statistically significant number, then this
would suggest that CFTR alleles may be important in
predisposing 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.
Hypotheses
H0 : π = π0 (π0 = 0.04 )
H1 : π > π0 (π1 = 0.115)
Calculation
n =[
where
zα p 0 q 0 + z β p1q1
p 0 − p1
p0 =
q0 = 1 – p0
0.04
= 0.96
p1 =
q1 = 1 – p1
0.115
= 0.885
]
2
α = 0.05 (1-sided), z0.05 = 1.645
1- β = 0.90,
z0.1 = 1.282
Æ n = 96
II. Test (cont’d)
2.5 Test of Significance of 1 Mean
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 versus
the alternative that it has decreased from 90 to 85 kg
with an estimated SD of 20 kg?
Hypotheses
H0 : μ = μ0 (μ0 = 90)
H1 : μ < μ0 (μ1 = 85)
Calculation
n = [
where
(z α + z β )σ
Δ
σ = estimated SD
Δ = | μ1 - μ0|
]
2
= 20
= 5
α = 0.05 (1-sided), z0.05 = 1.645
1- β = 0.90,
z0.1 = 1.282
Æ n = 137.08 = 138
II. Test (cont’d)
2.6 Test of Significance of 1 Correlation Coefficient
H0: ρ = ρ0
H1: ρ = ρ1
“การศึกษาความสัมพันธระหวางการวัดพังผืดในตับดวยวิธี
Transient Elastography กับการวินิจฉัยดวยวิธีเจาะชิ้นเนือ
้ ตับใน
ผูปวยโรคเรื้อนกวาง (psoriasis)ที่ไดรับการรักษาดวยยาเมโธเทร็กเซท
(methotrexate; MTX) ในคนไทย”
จากวัตถุประสงคหลักของการวิจัยเพื่อหาความสัมพันธระหวาง
คาที่ไดจากการวัดพังผืดในตับดวยวิธี Transient Elastography
กับคาที่ไดจากการวินิจฉัยภาวะพังผืดดวยวิธีเจาะชิ้นเนือ้ ตับ การ
คํานวณขนาดตัวอยางจึงเปนการคํานวณเพื่อทดสอบคา Correlation
coefficient โดยมีสมมติฐานทางสถิติดงั นี้
H0 : ρ 0 = 0
H1 : ρ1 = 0.3 (Ref #1)
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 = 0.1
1-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)]
Under 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
= 0
More than one primary outcome
If one of these endpoints is regarded as more important
than 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 to
answer 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.
References
Blackwelder WC. Proving the Null Hypothesis in Clinical
Trials. Controlled Clinical Trials 1982; 3: 345-353.
Breslow NE, Day NE. Statistical Methods in Cancer
Research 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: Clinical
Trials Research Methodology, Statistical Methods in Clinical
Trials, The ICH GCP Guidelines. Hong Kong: The Clinical
Trials Centre. 1998.
Lachin JM. Introduction to Sample Size Determination
and 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.

Sample Size Estimation Chulaluk Komoltri DrPH (Bios) Faculty of Medicine Siriraj Hospital

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