Sample Size Estimation

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

• Why?

• n is large enough to provide a reliable answer
  to the
• question
• too small n ? a waste of time
• too many n ? A waste of money other
  resources
• May be unethical
• e.g., delayed
  beneficial therapy

• Study Objective
• - Hypothesis generating (Pilot study)
• ? No sample size estimation
• - Estimation of parameter, Hypothesis
  confirmation
• ? Sample size estimation

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n is usually determined by the primary objective
of the study
More than one primary outcomes 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.
Method of calculating n should be given in the
proposal, together with the assumptions made in
the calculation
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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.
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Pilot study
Example (Mar 23,1999, 1515) This study
for n20 eligible burn patients will generate
hypothesis about the predictive values of
various patient characteristics for predicting
number of days to return to work.
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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. N24 normal healthy
volunteers will be randomized to parallel groups
to study the effect of 4 antidepressant drugs
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Estimation study, Hypothesis confirmation study
Sample size determination 2 Objectives I.
Estimation of parameter(s) ? Precision (95 CI)

Specify ? error
- Estimate prevalence, sensitivity,
specificity - Estimate single mean, single
proportion - etc.
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II. Test H0 ? Statistical power (1- ?)

Specify ?, ? error
2.1 Single group - Test of single
proportion, mean - Test of Pearsons
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 gt 2 groups
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?, ? (Efficacy trial)
Truth
H0 true
H0 false
(AB) (A?B,
Difference) Decision Accept H0 No error
(1- ?) ? (from p-value) (p gt
?) Reject H0
? No error (1- ?)
(p ? ?)
Power
? Pr (incorrect conclusion of
difference ) False positive
(FP) ? Pr (incorrect conclusion of
equivalence) False negative (FN) 1
- ? Pr ( correct conclusion of
difference ) True positive (TP)
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Sample size for estimating parameter, testing
hypothesis
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nQuery Advisor
PASS
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PS (Power and Sample Size)
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STATCALC
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1. Estimation
1.1) Estimate single proportion ? 95 CI of ? p
d

• where ? Pr. of type I error 0.05
  (2-sided)
• z0.025 1.96
• p Estimated proportion of
• q 1-p
• d Margin of error in
  estimating ?

1.2) Estimate single mean ? 95 CI of ? x d
n z?/2 SD / d2
where SD Standard deviation of
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95 CI for ? p ? z?/2 SE(p)
p ? z?/2 ?p(1-p)/n
p ? d Thus, d z?/2
?p(1-p)
n d2 z?/22 p(1-p)
n
n z?/22 p(1-p)
d2
Derivation of formulas in (1.1), (1.2)
95 CI of ? x ? z?/2 SE(x)
x ? z?/2 SD / ?n
x ? d Thus, d z?/2 SD
?n
n z?/2 SD 2
d
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n z?/22 pq / d2
-----------(1)
a) If no idea about p, let p 0.5 ? biggest pq
0.25 At 2-sided ? 0.05 then
n 1.962 (0.5)(0.5) ? 1
-----------(2)
d2 d2
b) If finite popn (N is known), adjust n from
eqn. (1) n? n
------------(3)
1 n/N
z2pq N
------------(4)
d2N z2pq If n 1/d2 as in eqn. (2),
then n? N
------------(5)
(1d2N)
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Conclusion n? N
---------(5)

(1d2N) where N Population size
d Margin of error in estimating popn
proportion Formula 5 is appropriate
under 1) Objective To estimate a single
proportion 2) Assume p 0.5 to get the
biggest sample size 3) Assume population is
finite (known popn size N) 4) Assume sample
is selected by simple random sampling
(SRS)
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nQuery Advisor
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2) Test 2.1) Test of difference in 2
independent proportions
H0 ?1 - ?2 0 H1 ?1 - ?2 ? 0
where ? Probability of type I error
? Probability of type II error
p1 Proportion of in gr. 1
q1 1-p1 p2
Proportion of in gr. 2 q2
1-p2 p (p1p2)/2
q 1-p
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2.2) Test of difference in 2 independent means
H0 ?1 - ?2 0 H1 ?1 - ?2 ? 0
where ? Probability of type I error
? Probability of type II error
? Common standard deviation of
in group 1, 2
? Difference in mean between 2
groups ?1 - ?2
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2.3) Test of significance of 1 proportion
H0 ? p0 H1 ? p1
where p0 Proportion of under H0
p1 Proportion of under H1
2.4) Test of significance of 1 mean
H0 ? ?0 H1 ? ?1
where ? Standard deviation of
? ?1 - ?0
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2.5) Test of significance of 1 correlation
H0 ? ?0 H1 ? ?1
n (Z?/2 Z?) 2
F(Z0) - F(Z1)
where F(Z0) Fishers Z
transformation of ?0
0.5 ln (1?0)/(1-?0) F(Z1)
Fishers Z transformation of ?1
0.5 ln (1?1)/(1-?1)
ln Natural logarithm
3
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Examples
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Example 1 (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 echocardi
ogram. A sample of n 140 will provide
95 CI for true prevalence rate of PHT of 0.10
? 0.05.
1.962 (0.1)(0.9) / 0.052 139
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• 2-sided CI p d
• 2) 1-sided CI p d
• or 1-sided CI p - d

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2-sided 95 CI for ? p d d ? level
of precision p 0.10, d 0.15 ?
95 CI 0.10 0.15
-0.05, 0.25
???
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• How big is d ?
• Absolute d
• 2. Relative d d ? 20 of prevalence(p)

p d 95
CI n 0.80 0.05p 0.04
0.76, 0.84 384
0.05 0.75, 0.85 246
0.10p 0.08 0.72, 0.88 96
0.10 0.70, 0.90
62 0.15p 0.12 0.68,
0.92 43
0.15 0.65, 0.95 28
0.20p 0.16 0.64, 0.96 24
0.20 0.60, 1.00
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Example 2 ????????????????????????????????????
??????????????????????????? ???????????? 95
confidence interval (CI) ????????????? ???????????
?????????????????? 20 ? 3 ???? 17 - 23
???????????????????????????????????? 683
?????????????????????????? n z?/22
p(1-p) / d2 ????? p ????????????????????
?????????????????????????????? 0.2 d
???????????????????????????????????? 0.03 ?
?????????????? type I error
0.05(2-sided) z0.025
1.96 ??????? n 1.962
(0.2)(0.8)/0.032 682.95 683
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Example 3 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 lt 2 wks
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Sample size Based on 95 CI of true
sensitivity (Se) 0.9 0.1
where p expected sensitivity 0.9
q 1-p 0.1 d
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 Se
- 1 Se
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52 87
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Example 4
?????????????????????????????????????????????
??? subcarinal angle ???????????
????????????????? ... ????????????? 100 ???????
... ?? ????????????????? subcarinal angle
??????? 60.8 (SD11.8) ??????????? 95
confidence interval (CI) ???????????????
subcarinal angle ???????????? (?) ????????????
61 ? 2 (SD13) ????????????????????????????????
163 ????????????????????????????? n
z?/2 SD / d2 ????? SD Standard
deviation ??? subcarinal angle
13 d Margin of error
???????????????????? 2 ?
Probability of type I error (2-sided) 0.05
z0.025 1.96 ??????? n
1.9613/22 162.31 163
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Example 5 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 Infants body temperature taken on
nursery admission
Infants, ? 34 gestational wks, BW ? 1800 gms
Randomization
Plastic wrap
No Plastic wrap
Body temp. ? Hypothermia
Body temp. ? Hypothermia
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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)
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Example 6 Title Relationship between
microalbuminuria (MAU) and diabetic
retinopathy (DR) in type 2 diabetes Investigator
Dr. Attasit Srisubat Design
Cross-sectional study Subjects Type 2
diabetic pts., DM duration ? 5 yrs

Diabetic retinopathy
DR No
DR Microalbuminurea With p1
1-p1 n1
Without p2 1-p2
n2
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where p1 Proportion of DR in type 2 DM w/
microalbuminuria 0.43 q1 1 - p1 p2
Proportion of DR in type 2 DM w/o
microalbuminuria 0.28 q2 1 - p2 p (p1
rp2) / (r1) q 1 - p r n2/n1 2
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Example 7
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
A-P splint and Knee elevation 40
Blood loss
Blood loss
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where ? Difference in mean
postoperative blood loss
between 2 groups ? ? 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
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Example 8
Title Whether long-term AED therapy has
significant adverse effect on bone
mineral density in Thai pre-menopausal
female epileptic pts. compared with healthy
controls Investigator Dr. Rungsan
Chaisewikul Design Cross-sectional
study Subjects - Pre-menopausal female
epileptic pts. receiving
long-term antiepileptic drugs (AEDs)
- Pre-menopausal healthy females,
having the same age range and
living in the same house or
neighborhood as pts.
Epileptic pts., Healthy controls Bone mineral
density (BMD)
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BMD
Mean ? SD Bone site Pre-menopausal
Epileptic pts.
healthy females receiving AEDs Distal
radius 0.69 ? 0.06 Femoral neck
1.00 ? 0.13 Lumbar spine 1.15 ? 0.13
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Example 9
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
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Sample size if parametric test (2-sample t-test)
is used
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Sample size if Non-parametric (Mann-Whitney) test
is used (VAS pain score is usually
positively skewed)
1
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2
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Example 10 (Feb 29, 2000, 1649) N 100
subjects with nontuberculous mycobacteria infectio
n 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 ? ?0
0.04, against 1-sided H1 ? ?1 0.115, using
? 0.05.
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Hypotheses
H0 ? ?0 (?0 0.04 ) H1 ? gt ?0 (?1
0.115)
Calculation
where p0 0.04
q0 1 p0 0.96
p1 0.115
q1 1 p1 0.885 ?
0.05 (1-sided), z0.05 1.645
1- ? 0.90, z0.1
1.282 ? n 96
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Example 11
?????????????????????????????????????????????????
??? Transient Elastography ???????????????????????
?????????????????????????????????????
(psoriasis)???????????????????????????????????
(methotrexate MTX) ???????
????????????????????????????????????????????
?????????????????????????????????????????????????
Transient Elastography ??????????????????????????
??????????????????????????????????
????????????????????????????????????????????????
Correlation coefficient ??????????????????????????
? H0 ?0 0 H1 ?1 0.3
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Examples How to write a sample size section
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The study was conducted in Mushin, a high-density
inner city area of Lagos in Nigeria with a
population of 11,689 school entrants in 76
public primary schools. EPI Info (6.04)
statistical package was used to compute the
initial sample size of 256 based on the
following formula Sample size (n) k /
1(k/population) where, k z2 p(1-p)
/ d2 z Z-score
corresponding to 95 CI (i.e. 1.96),
p available local prevalence rate in the
target population
(which was 2.8 obtained from
a comparable local study),
d margin of error allowed (2) By
building in an attrition rate of 20 the sample
size was increased to 306 as a baseline.
Olusanya BO, Okolo AA, Adeosun AA. Predictors of
hearing loss in school entrants in a developing
country. J Postgrad Med 200450173-179.
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To achieve a 15-mm clinical difference in VAS
score between control and intervention groups,
with a slightly overestimated SD of 26, an ?
risk for 0.05, and power (1-?) of 0.80, the
number of participants needed in each group was
54. With an estimated 10 of patients lost to
follow-up, we sought to include 120 patients
but recruited only 112.
Rannou F, Dimet J, Boutron I, Baron G, Fayad F,
Mace Y et al. Splint for base-of-thumb
osteoarthritis. Ann Intern Med 2009150661-669.
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Sample size calculation was based on a
hypothesized 0.8 difference in hemoglobin A1c
levels in the motivational enhancement therapy
plus cognitive behavior therapy group (or
motivational enhancement therapy alone group)
compared with usual diabetes care. We assumed
that the SD of the changes was approximately
1.65. At a power of 90, a type I error of
0.05 (2-tailed), a randomization ratio of 111,
and a 20 withdrawal rate, we estimated a sample
size of 339 participants (n113 in each group).
Ismail K, Thomas SM, Maissi E, Chalder T, Schmidt
U, Bartlett J et al. Motivational enhancement
therapy with and without cognitive behavior
therapy to treat type 1 diabetes. Ann Intern Med
2008149708-719.
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Sample size was estimated by analyzing previous
data from studies comparing verbal rating-scale
pain scores between patients receiving a COX-2
selective inhibitor and those receiving placebo
after spinal fusion surgery. With 90 power, a
mean difference of 2.9, a standard deviation of
1.0, and ? 0.05, a power analysis of ANOVA
testing on four independent means would require
16 patients per group. To be conservative, we
planned to initially enroll 20 patients per
group. At the end of the study, if there was a
drop-out in any group due to a protocol
violation, new patients would be enrolled
and randomly assigned to treatment groups to
increase the total to 20 per group.
Reuben SS, Buvanendran A, Kroin JS, Raghunathan
K. The analgesic efficacy of celecoxib,
pregabalin, and their combination for spinal
fusion surgery. Anesth Analg 20061031271-7.
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We designed the study as a noninferiority
study. A difference of 0.5 has been recognized
as the minimum clinically important difference
to distinguish treatments on the dyspnea
subscale of the CRQ. By using this value in the
sample size calculation for a noninferiority
study, with an ? level of 0.025, 1-?, 0.90, and
SD of 1.1 (slightly greater than previous
similar studies), the required sample size was
204 (102 patients per group). On the basis of
our previous study in a similar patient sample,
we anticipated 15 attrition, so we planned to
randomly assign 240 patients.
Ismail K, Thomas SM, Maissi E, Chalder T, Schmidt
U, Bartlett J et al. Effects of home-based
pulmonary rehabilitation in patients with chronic
obstructive pulmonary disease. Ann Intern Med
2008149869-878.
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Methodology, Statistical Methods in Clinical
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Clinical Trials Centre. 1998.
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Lachin JM. Introduction to Sample Size
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