Transcript Sample Size Estimation Dr. Tuan V. Nguyen Garvan Institute of Medical Research Sydney, Australia.

Sample Size Estimation
Dr. Tuan V. Nguyen
Garvan Institute of Medical Research
Sydney, Australia
The classical hypothesis testing
• Define a null hypothesis (H0) and a null hypothesis (H1)
• Collect data (D)
• Estimate p-value = P(D | H0)
• If p-value > a, accept H0; if p-value < a, reject H0
Diagnosis and statistical reasoning
Disease status
Present
Absent
Test result
+ve
True +ve
Significance Difference is
Present
Absent
False +ve
Test result
Reject Ho
True -ve
Accept Ho
(sensitivity)
-ve
False -ve
(Specificity)
(Ho not true)
(Ho is true)
No error
1-b
Type I err.
a
Type II err. No error
b
1-a
a : significance level
1-b : power
Study design issues
•
•
•
•
•
•
Setting
Participants: inclusion / exclusion criteria
Design: survey, factorial, etc
Measurements: outcome, covariates
Analysis
Sample size / power issues
Sample size issues
• How many judges / consumers?
– Practical and statistical issues
– Ethical issues
• Ethical issues
– Unnecessarily large number of judges may be
deemed unethical
– Too small a sample may also be unethical as the
study can’t show anything.
Practical difference vs statistical significance
Outcome
Improved
Group A Group B
Outcome
9
18
Improved
No improved
21
12
Total
30
% improved
30%
Group A Group B
6
12
No improved
14
8
30
Total
20
20
60%
% improved
30%
60%
Chi-square: 5.4; P < 0.05
“Statistically significant”
Chi-square: 3.3; P > 0.05
“Statistically insignificant”
Effect of sample size on preference proportion
Number
of judges
%
preferred
A (p)
Variance
of p
20
0.65
0.0114
30
0.65
40
50
Std Dev
of p
Z test for
p=0.5
P-value
0.1067
1.41
0.160
0.0076
0.0871
1.72
0.085
0.65
0.0057
0.0754
1.99
0.047
0.65
0.0046
0.0675
2.22
0.026
Effect of sample size: a simulation
True mean: 100
True SD: 15
True mean: 100
True SD: 35
Sample size
Est. M SD
Est. M SD
10
50
100
200
500
1000
2000
10000
100000
98.0
100.4
101.3
99.9
99.8
99.5
99.7
100.1
100.0
108.9
95.3
99.1
100.3
98.9
99.9
99.9
99.9
100.0
11.0
13.6
14.4
15.2
15.3
15.1
15.0
15.0
15.0
32.2
41.4
35.5
33.2
33.8
35.0
34.7
35.0
35.0
What are required for sample size estimation?
•
•
•
•
Parameter (or outcome) of major interest
Magnitude of difference in the parameter
Variability of the parameter
Bound of errors (type I and type II error
rates)
Parameter of interest
• Type of measurement of primary interest:
– Continuous or categorical outcome
• Examples:
– Proportion: proportion (or probability) of preference for
a product
– Hedotic scale: 0-10
– Nominal scale
Variability of the parameter of interest
• If the parameter is a continuous variable:
– What is the standard deviation (SD) ?
• If the parameter is a categorical variable:
– SD can be estimated from the proportion/probability.
Magnitude of difference of interest
• Distinction between practical and statistical relevance.
• Examples:
– Probability of preference: 85% vs 50%
– Tasting scores: difference between products by 1 SD.
The Normal distribution
0.95
0.95
Z2
-1.96
0
1.96
0.025
0
0.025
Prob.
0.80
0.90
0.95
0.99
Z1
0.84
1.28
1.64
2.33
Z2
1.28
1.64
1.96
2.81
Z1
1.64
0.05
The Normal deviates
Alpha
c
0.20
0.10
0.05
0.01
Za
(One-sided)
0.84
1.28
1.64
2.33
Za/2
(Two-sided)
1.28
1.64
1.96
2.81
Power
0.80
0.90
0.95
0.99
Z1-b
0.84
1.28
1.64
2.33
Study design and outcome
• Single population
• Two populations
• Continuous measurement
• Categorical outcome
• Correlation
Single group
Sample size for estimating a population mean
• How close to the true mean
• Confidence around the sample mean
• Type I error.
• N = (Za/2)2 s2 / d2
s: standard deviation
d: the accuracy of estimate (how
close to the true mean).
Za/2: A Normal deviate reflects the
type I error.
• Example: we want to estimate the
average weight in a population,
and we want the error of
estimation to be less than 2 kg of
the true mean, with a probability
of 95% (e.g., error rate of 5%).
• N = (1.96)2 s2 / 22
Effect of standard deviation
Sample size
96
138
188
246
311
384
400
350
300
Sample size
Std Dev (s)
10
12
14
16
18
20
450
250
200
150
100
50
0
0
5
10
15
Standard deviation
20
25
Sample size for estimating a population proportion
• How close to the true proportion
• Confidence around the sample
proportion.
• Type I error.
• N = (Za/2)2 p(1-p) / d2
p: proportion to be estimated.
d: the accuracy of estimate (how
close to the true proportion).
Za/2: A Normal deviate reflects the
type I error.
• Example: The proportion of
preference for product A is around
80%. We want to estimate the
preference p in a community
within 5% with 95% confidence
interval.
• N = (1.96)2 (0.8)(0.2) /
0.052 = 246 consumers.
Effect of accuracy
• N = (1.96)2 (0.3)(0.7) /
0.022 = 2017 subjects.
2500
2000
Sample size
• Example: The proportion of
preference in the general
population is around 30%. We
want to estimate the prevalence p
in a community within 2% with
95% confidence interval.
1500
1000
500
0
0
0.02
0.04
0.06
Standard deviation
0.08
0.1
Sample size for estimating a correlation coeffcient
• In observational studies which involve estimate a correlation (r) between two
variables of interest, say, X and Y, a typical hypothesis is of the form:
– Ho: r = 0
vs
H1: r not equal to 0.
• The test statistic is of the Fisher's z transformation, which can be written as:
1
1  r 
t  log e 
n-3

2
1 - r 
• Where n is the sample size and r is the observed correlation coefficient.
• It can be shown that t is normally distributed with mean 0 and unit variance,
and the sample size to detect a statistical significance of t can be derived as:
Z
a  Z1- b 
N
3
1
1

r


log


e
4 
1
r


2
Sample size for estimating r: example
• Example: According to the literature, the correlation between salt intake and
systolic blood pressure is around 0.3. A study is conducted to test the
correlation in a population, with the significance level of 1% and power of
90%. The sample size for such a study can be estimated as follows:
N
2.33  1.282
1
 1  0.3 
log

e
4 
 1 - 0.3 
 3  87
• A sample size of at least 87 subjects is required for the study.
Sample size for difference between two means
• Hypotheses:
Ho: m1 = m2 vs. Ha: m1 = m2 + d
• Let n1 and n2 be the sample sizes
for group 1 and 2, respectively; N
= n1 + n2 ; r = n1 / n2 ; s:
standard deviation of the variable
of interest.
• Then, the total sample size is
given by:
2

 2
r  1 Z  Z
 s
1- b 
 a
N
rd 2
Where Za and Z1-b are Normal deviates
• If we let Z = d / s be the “effect
size”, then:

r  1 Z  Z

a
1
b


N
rZ 2
2
• If n1 = n2 , power = 0.80, alpha
= 0.05, then (Za + Z1-b)2 = (1.96
+ 1.28)2 = 10.5, then the
equation is reduced to:
N
21
Z2
Two-group comparisons
Sample size for two means vs.“effect size”
Total sample size (N)
2400
2000
1600
1200
800
400
0
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
Effect size (d / s)
For a power of 80%, significance level of 5%
2
Sample size for difference between 2 proportions
• Hypotheses:
Ho: p1 = p2 vs. Ha: p1 = p2 + d .
• Let p1 and p2 be the sample proportions (e.g. estimates of p1 and p2) for
group 1 and group 2. Then, the sample size to test the hypothesis is:

Z
n
a
2 p1 - p   Z
p 1 - p  p 1 - p
 p - p 2
1- b
1
1
2
1
2
2
2
Where: n = sample size for each group ; p = (p1 + p2) / 2 ; Za and Z1-b are
Normal deviates
A better (more conservative) suggestion for sample size is:

n
4
n  1  1 

4
np -p 
a
1
2
2
Sample size for difference between 2 prevalence
• For most diseases, the prevalence in the general population is small (e.g. 1
per 1000 subjects). Therefore, a difference formulation is required.
• Let p1 and p2 be the prevalence for population 1 and population 2. Then, the
sample size to test the hypothesis is:
n
Z
0.00061arcsin
a
 Z1- b 2
p1 - arcsin p2 
2

Where: n = sample size for each group; Za and Z1-b are
Normal deviates.
Sample size for two proportions: example
• Example: The preference for product A is expected to be 70%, and for
product B 60%. A study is planned to show the difference at the significance
level of 1% and power of 90%.
• The sample size can be calculated as follows:
– p1 = 0.6; p2 = 0.7; p = (0.6 + 0.7)/2 = 0.65; Z0.01 = 2.81; Z1-0.9 = 1.28.
– The sample size required for each group should be:

2.81
n
2
2  0.65  0.35  1.28 0.6  0.4  0.7  0.3 
 759
2
0.6 - 0.7 
• Adjusted / conservative sample size is:

759 
4
n 
1  1 
  836
4 
759 0.6 - 0.7 
2
a
Sample size for two proportions vs. effect size
Difference from p1 by:
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
424
625
759
825
825
759
625
424
131
173
198
206
198
173
131
73
67
82
89
89
82
67
45
.
41
47
50
47
41
31
.
.
28
30
30
28
22
.
.
.
19
20
19
17
.
.
.
.
14
14
13
.
.
.
.
.
10
9
8
.
.
.
.
.
P1
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Note: these values are “unadjusted” sample sizes
Sample size for estimating an odds ratio
• In case-control study the data are usually summarized by an odds ratio (OR),
rather then difference between two proportions.
• If p1 and p2 are the proportions of cases and controls, respectively, exposed
to a risk factor, then:
p1 1 - p2 
OR 
p2 1 - p1 
• If we know the proportion of exposure in the general population (p), the total
sample size N for estimating an OR is:
1  r 2 Za  Z1- b 2
N
2
r ln OR  p1 - p 
• Where r = n1 / n2 is the ratio of sample sizes for group 1 and group2; p
is the prevalence of exposure in the controls; and OR is the hypothetical
odds ratio. If n1 = n2 (so that r = 1) then the fomula is reduced to:
4Za  Z1- b 
2
N
ln OR 2 p1 - p 
Sample size for an odds ratio: example
• Example: The prevalence of vertebral fracture in a population is 25%. It is
interested to estimate the effect of smoking on the fracture, with an odds
ratio of 2, at the significance level of 5% (one-sided test) and power of 80%.
• The total sample size for the study can be estimated by:
41.64  0.85
N
 275
2
ln 2  0.25  0.75
2
Sample size for 2 correlation coefficients
• In detecting a relevant difference between two correlation coefficients r1 and
r2 obtained from two independent samples of sizes n1 and n2, respectively,
we need to firstly transform these coefficients into z value as follows:
 1  r1 

z1  0.5 log e 
 1 - r1 
 1  r2 

z2  0.5 log e 
 1 - r2 
• The total sample size N required to detect the difference between two
correlation coefficients r1 and r2, with a significance level of a and power 1b, can be estimated by:
2
4Za  Z1- b 
N
 z1 - z2 2
Where Za and Z1-b are Normal deviates
Sample size for two r’s: example
• The sample size required to detect the difference between r1 = 0.8 and r2 =
0.4 with the significance level of 5% (two-tailed) and power of 80% can be
solved as follows:
– z1 = 0.5 ln ((1+0.4) / (1-0.4)) = 0.424
– z1 = 0.5 ln ((1+0.8) / (1-0.8)) = 1.098
41.96  1.28
N
2  92
0.424 - 1.098
2
• 46 subjects is needed in each group.
Some comments
•
•
•
•
•
The formulae presented are theoretical.
They are all based on the assumption of Normal distribution.
The estimator [of sample size] has its own variability.
The calculated sample size is only an approximation.
Non-response must be allowed for in the calculation.
Computer programs
• Software program for sample size and power evaluation
– PS (Power and Sample size), from Vanderbilt Medical Center. This can be
obtained from me by sending email to ([email protected]). Free.
• On-line calculator:
– http://ebook.stat.ucla.edu/calculators/powercalc/
• References:
–
–
–
–
–
–
Florey CD. Sample size for beginners. BMJ 1993 May 1;306(6886):1181-4
Day SJ, Graham DF. Sample size and power for comparing two or more treatment groups in clinical trials. BMJ
1989 Sep 9;299(6700):663-5.
Miller DK, Homan SM. Graphical aid for determining power of clinical trials involving two groups. BMJ 1988
Sep 10;297(6649):672-6
Campbell MJ, Julious SA, Altman DG. Estimating sample sizes for binary, ordered categorical, and continuous
outcomes in two group comparisons. BMJ 1995 Oct 28;311(7013):1145-8.
Sahai H, Khurshid A. Formulae and tables for the determination of sample sizes and power in clinical trials for
testing differences in proportions for the two-sample design: a review. Stat Med 1996 Jan 15;15(1):1-21.
Kieser M, Hauschke D. Approximate sample sizes for testing hypotheses about the ratio and difference of two
means. J Biopharm Stat 1999 Nov;9(4):641-50.