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JOURNAL OF BIOPHARMACEUTICAL STATISTICS
Vol. 12, No. 4, pp. 471–483, 2002
ON SAMPLE SIZE CALCULATION BASED ON
ODDS RATIO IN CLINICAL TRIALS
Hansheng Wang,1,* Shein-Chung Chow,1 and Gang Li2
1
StatPlus, Inc., Heston Hall, Suite 206, 1790 Yardley-Langhorne Road,
Yardley, PA 19067
2
Department of Mathematics, Statistics and Computer Science,
University of Illinois at Chicago, 851 S. Morgan St,
Chicago, IL 60608
ABSTRACT
Sample size calculation formulas for testing equality, noninferiority, superiority, and equivalence based on odds ratio were derived under both parallel and
one-arm crossover designs. An example concerning the study of odds ratio
between a test compound (treatment) and a standard therapy (control) for
prevention of relapse in subjects with schizophrenia and schizoaffective
disorder is presented to illustrate the derived formulas for sample size
calculation for various hypotheses under both a parallel design and a crossover
design. Simulations were performed to assess the adequacy of the sample size
calculation formulas. Simulation results were given at the end of the paper.
Key Words:
Odds ratio; Therapeutic equivalence; Noninferiority; Superiority
INTRODUCTION
In clinical trials, it is often of interest to investigate the relative effect
(e.g., risk or benefit) of the treatments for the disease under study. The odds ratio
*Corresponding author. E-mail: [email protected]
471
DOI: 10.1081/BIP-120016231
Copyright q 2002 by Marcel Dekker, Inc.
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472
WANG, CHOW, AND LI
has been frequently used to assess the association between a binary exposure
variable and a binary disease outcome (see Refs. [2,4]). Let pT be the probability
of observing an outcome of interest for a patient treated by a test compound
(treatment) and pC for a patient treated by a standard therapy (control). For patients
receiving the treatment, the odds that a patient will have an outcome of interest
over that he/she will not have an outcome of interest is defined as
pT
:
1 2 pT
OT ¼
Similarly, the odds for a patient receiving the control is defined as
pC
:
1 2 pC
OC ¼
As a result, the odds ratio between the treatment and the control is defined as
OR ¼
OT pT ð1 2 pC Þ
¼
:
OC pC ð1 2 pT Þ
The odds ratio is always positive and has a range from 0 to 1. When OR ¼ 1; i.e.,
pT ¼ pC ; it implies that there is no difference between the treatment and control in
terms of the outcome of interest. When OR . 1; treatment is more likely to
produce the outcome of interest than control. Note that (1 2 OR) is usually
referred to as relative odds reduction in the literature. Intuitively, OR can be
estimated by
c ¼ p^ T ð1 2 p^ C Þ ;
OR
p^ C ð1 2 p^ T Þ
ð1Þ
where p̂T and p̂C are the maximum likelihood estimators of pT and pC, which are
respectively given by
p^ T ¼
xT
xC
and p^ C ¼ ;
nT
nC
where xT and xC are the respective numbers of outcomes of interest observed in the
treatment and control groups, nT and nC are the numbers of patients receiving the
treatment and the control, respectively. For a two-arm parallel trial, the asymptotic
c can be obtained as
variance of logðOR
ORÞ
c ¼
var½logðOR
ORÞ
1
1
þ
:
nT pT ð1 2 pT Þ nC pC ð1 2 pC Þ
Note that the above asymptotic variance can be estimated by simply replacing
pT and pC with their maximum likelihood estimators p̂T and p̂C (see Refs. [3,5]).
c under a crossover design is
However, the asymptotic variance of logðOR
ORÞ
unknown.
In clinical trials, commonly considered hypotheses include point hypotheses
for testing equality and interval hypotheses for testing equivalence/noninferiority
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SAMPLE SIZE CALCULATION BASED ON ODDS RATIO
473
and superiority. In what follows, sample size calculations based on odds ratio
between test and control will be derived under these hypotheses.
PARALLEL DESIGN
Formulas for sample size calculation for testing various hypotheses based on
odds ratio will be derived under a parallel design.
Test for Equality
For testing equality of pT and pC between the treatment and control groups,
the hypotheses of interest are given by
H 0 : OR ¼ 1 vs: H a : OR – 1:
Testing the above hypotheses is equivalent to testing the following hypotheses
H 0 : logðORÞ ¼ 0 vs: H a : logðORÞ – 0:
Under the null hypothesis, the test statistic
21=2
1
1
c
þ
T ¼ logðOR
ORÞ
nT p^ T ð1 2 p^ T Þ nC p^ C ð1 2 p^ C Þ
ð2Þ
follows the standard normal distribution when nT and nC are sufficiently large.
Thus, we reject the null hypothesis H 0 : logðORÞ ¼ 0 if jTj . za=2 ; where za/2 is
the upper (a/2)th percentile of the standard normal distribution. Under the
alternative hypothesis H a : logðORÞ – 0; the power of the above test can be
approximated by
!
21=2
1
1
þ
2za=2 ;
F jlogðORÞj
nT pT ð1 2 pT Þ nC pC ð1 2 pC Þ
where F is the cumulative distribution of the standard normal random variable. As
a result, the sample size needed for achieving a desired power of ð1 2 b) can be
obtained by solving
21=2
1
1
þ
2za=2 ¼ zb :
jlogðORÞj
nT pT ð1 2 pT Þ nC pC ð1 2 pC Þ
Under the assumption that nT =nC ¼ k; we have
ðza=2 þ zb Þ2
1
1
þ
:
nC ¼
log2 ðORÞ kpT ð1 2 pT Þ pC ð1 2 pC Þ
ð3Þ
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WANG, CHOW, AND LI
Test for Noninferiority/Superiority
The problem of testing noninferiority and superiority can be unified by the
following hypotheses:
H 0 : OR # d 0 vs: H a : OR . d 0 ;
where d 0 is the noninferiority or superiority margin on the raw scale. If we let
d ¼ logðd 0 Þ; testing the above hypotheses is equivalent to testing the following
hypotheses
H 0 : logðORÞ # d vs: H a : logðORÞ . d;
where d is the noninferiority or superiority margin on the log-scale. When d . 0;
the rejection of the null hypothesis indicates superiority over the reference value.
When d , 0; the rejection of the null hypothesis implies noninferiority against the
reference value.
When logðORÞ ¼ d; the test statistic
21=2
1
1
c 2 dÞ
T ¼ ðlogðOR
ORÞ
þ
nT p^ T ð1 2 p^ T Þ nC p^ C ð1 2 p^ C Þ
follows the standard normal distribution when nT and nC are sufficiently large.
Thus, we reject the null hypothesis at the a level of significance if T . za : Under
the alternative hypothesis H a : logðORÞ . d; the power of the above test is
given by
!
21=2
1
1
2za :
F ðlogðORÞ 2 dÞ
þ
nT pT ð1 2 pT Þ nC pC ð1 2 pC Þ
As a result, the sample size needed for achieving a desired power of (1 2 b) can be
obtained by solving
21=2
1
1
þ
ðlogðORÞ 2 dÞ
2za ¼ zb :
nT pT ð1 2 pT Þ nC pC ð1 2 pC Þ
Under the assumption that nT =nC ¼ k; we have
ðza þ zb Þ2
1
1
þ
nC ¼
:
ðlogðORÞ 2 dÞ2 kpT ð1 2 pT Þ pC ð1 2 pC Þ
ð4Þ
Test for Equivalence
To establish equivalence, the following hypotheses are usually considered
H 0 : jORj $ d 0 vs: H a : jORj , d 0 :
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SAMPLE SIZE CALCULATION BASED ON ODDS RATIO
475
Similarly, testing the above hypotheses is equivalent to testing the following
hypotheses
H 0 : jlogðORÞj $ d vs: H a : jlogðORÞj , d;
where d ¼ logðd 0 Þ: The above hypotheses can be tested using the two one-sided
tests (TOST) procedure (see Ref. [1]). We then reject the null hypothesis at the a
level of significance if
21=2
1
1
c 2 dÞ
, 2za
ðlogðOR
ORÞ
þ
nT p^ T ð1 2 p^ T Þ nC p^ C ð1 2 p^ C Þ
and
c
ðlogðOR
ORÞ þ dÞ
21=2
1
1
. za :
þ
nT p^ T ð1 2 p^ T Þ nC p^ C ð1 2 p^ C Þ
Then the power of the above test can be approximated by
8 h
i21=2
>
1
1
>
d
þ
2z
if logðORÞ ¼ 0
2F
>
a 21
>
nT pT ð12pT Þ
nC pC ð12pC Þ
<
h
i21=2
>
1
1
>
>
F ðd 2 jlogðORÞjÞ nT pT ð12pT Þ þ nC pC ð12pC Þ
2za
if logðORÞ – 0:
>
:
Under the assumption that nT =nC ¼ k; the sample size is given by
8
h
i
ðza þzb=2 Þ2
1
1
>
>
¼
þ
n
2
C
<
kpT ð12pT Þ
pC ð12pC Þ
d
h
i
2
ðza þzb Þ
1
1
>
>
n
¼
þ
: C ðd2jlogðORÞjÞ2 kpT ð12pT Þ pC ð12pC Þ
if logðORÞ ¼ 0
if logðORÞ – 0:
ð5Þ
Note that very similar formulas were also obtained by Tu.[5]
CROSSOVER DESIGN
Consider a one-arm crossover design. For simplicity, we assume that there
are no period effects and carryover effects. Without loss of generality, we further
assume that every subject will first receive the control and then the treatment. Let
xij, j ¼ 1; . . . ; n be 1 if an outcome of interest is observed from the jth subject in the
ith period and 0 otherwise, then the
Pnumber of outcomes of interest observed in the
control group is given by xC ¼ nj¼1 x1j : And xT can be similarly defined. Then
the odds ratio between the treatment and control can be estimated according to
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476
WANG, CHOW, AND LI
Eq. (1). By Taylor’s expansion, we can obtain
pffiffiffi
pffiffiffi
1
1
c 2 logðORÞÞ ¼ n
ð p^ C 2 pC Þ 2
ð p^ T 2 pT Þ
nðlogðOR
ORÞ
pC ð1 2 pC Þ
pT ð1 2 pT Þ
n
1 X
x1j 2 pC
x2j 2 pT
þ op ð1Þ ¼ pffiffiffi
2
n j¼1 pC ð1 2 pC Þ pT ð1 2 pT Þ
þ op ð1Þ !d Nð0; sd2 Þ;
where
sd2 ¼ var
Let
dj ¼
x1j 2 pC
x2j 2 pT
2
:
pC ð1 2 pC Þ pT ð1 2 pT Þ
ð6Þ
x1j
x2j
2
:
p^ C ð1 2 p^ C Þ p^ T ð1 2 p^ T Þ
Then, sd2 can be estimated by sample variance of dj, j ¼ 1;...;n; which is denoted
by s^d2 :
Test for Equality
Similarly, the hypotheses of interest are given by
H 0 : OR ¼ 1 vs: H a : OR – 1:
Testing the above hypotheses is equivalent to testing the following hypotheses
H 0 : logðORÞ ¼ 0 vs: H a : logðORÞ – 0:
Under the null hypothesis, the test statistic
pffiffiffi
c
nlogðOR
ORÞ
T¼
s^d
follows the standard normal distribution when n is sufficiently large. Thus, we
reject the null hypothesis H 0 : OR ¼ 1 if jTj . za=2 : Under the alternative
hypothesis H a : OR – 1; the power of the above test can be approximated by
pffiffiffi
njlogðORÞj
2 za=2 :
F
sd
As a result, the sample size needed for achieving a desired power of (1 2 b) can be
obtained by solving
pffiffiffi
njlogðORÞj
2 za=2 ¼ zb :
sd
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SAMPLE SIZE CALCULATION BASED ON ODDS RATIO
477
This leads to
n¼
ðza=2 þ zb Þ2 sd2
:
log2 ðORÞ
ð7Þ
Test for Noninferiority/Superiority
As indicated earlier, the problem of testing noninferiority and superiority can
be unified by the following hypotheses:
H 0 : logðORÞ # d vs: H a : logðORÞ . d;
where d ¼ logðd 0 Þ is the noninferiority or superiority margin on log-scale. When
logðORÞ ¼ d; the test statistic
pffiffiffi
c 2 dÞ
nðlogðOR
ORÞ
T¼
s^d
follows the standard normal distribution when n is sufficiently large. Thus, we
reject the null hypothesis at the a level of significance if T . za : Under the
alternative hypothesis H a : logðORÞ . d; the power of the above test is
approximated by
pffiffiffi
nðlogðORÞ 2 dÞ
2 za :
F
sd
As a result, the sample size needed for achieving a desired power of 1 2 b can be
obtained by solving
pffiffiffi
nðlogðORÞ 2 dÞ
2 za ¼ zb :
sd
Thus, we have
nC ¼
ðza þ zb Þ2 sd2
:
ðlogðORÞ 2 dÞ2
ð8Þ
Test for Equivalence
To establish equivalence, the following hypotheses are considered
H 0 : jlogðORÞj $ d vs: H a : jlogðORÞj , d:
We then reject the null hypothesis at the a level of significance if
pffiffiffi
c 2 dÞ
nðlogðOR
ORÞ
, 2za
s^d
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478
WANG, CHOW, AND LI
and
pffiffiffi
c þ dÞ
nðlogðOR
ORÞ
s^d
. za :
The power of the above test is approximated by
8 pffiffi
nd
>
21
if logðORÞ ¼ 0
2F
2
z
>
a
<
sd
pffiffi
nðd2jlogðORÞjÞ
>
2 za
if logðORÞ – 0:
>
:F
sd
Then the sample size is given by
8
ðza þzb=2 Þ2 s 2d
>
if logðORÞ ¼ 0
<n ¼
d2
2
2
ðza þzb Þ s d
>
: n ¼ ðd2jlogðORÞjÞ
2
ð9Þ
if logðORÞ – 0:
AN EXAMPLE
Parallel Design
Suppose a sponsor is interested in conducting a clinical trial to study the
relative risk between a test compound (treatment) and a standard therapy (control)
for prevention of relapse in subjects with schizophrenia and schizoaffective
disorders. Based on the results from a previous study with 365 subjects (i.e., 177
subjects received the test compound and 188 received the standard therapy), about
25% (45/177) and 40% (75/188) of subjects receiving the test compound and the
standard therapy experienced relapse after the treatment, respectively. Subjects
who experienced the first relapse may withdraw from the study or stay in the trial.
The sponsor is interested in studying the odds ratio of the test compound as
compared to the standard therapy for prevention of experiencing the first relapse.
In addition, it is also of interest to examine the odds ratio for prevention of
experiencing the second relapse.
Test for Equality
Assuming the relapse rates in the test group and the control group are 25%
and 40%, respectively. This yields a relative risk of
OR ¼
0:40ð1 2 0:25Þ
¼ 2:
ð1 2 0:40Þ0:25
According to Eq. (3) and n ¼ nT ¼ nC ðk ¼ 1Þ; the sample size per treatment
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SAMPLE SIZE CALCULATION BASED ON ODDS RATIO
479
group needed in order to achieve an 80% ðb ¼ 0:2Þ power at 5% ða ¼ 0:05Þ level
of significance is given by
ðz0:05=2 þ z0:2 Þ2
1
1
n¼
þ
0:4ð1 2 0:4Þ 0:25ð1 2 0:25Þ
log2 ð2Þ
¼
ð1:96 þ 0:84Þ2
ð4:17 þ 5:33Þ ¼ 156:4 < 157:
0:692
Test for Superiority
Suppose that a 20% difference in odds ratio (in log scale) is considered of
clinical importance. Hence, the superiority margin is chosen to be d ¼ 0:2:
According to Eq. (4), the sample size per treatment group required for achieving
an 80% power ðb ¼ 0:2Þ is given by
ðz0:05 þ z0:2 Þ2
1
1
þ
n¼
ðlogð2Þ 2 0:2Þ2 0:4ð1 2 0:4Þ 0:25ð1 2 0:25Þ
¼
ð1:64 þ 0:84Þ2
ð4:17 þ 5:33Þ ¼ 243:4 < 244:
0:492
Test for Equivalence
Assuming that (i) the true relapse rate of the test compound is approximately
the same as that of the standard therapy (i.e., logðORÞ ¼ 0) and (ii) the equivalence
limit of the odds ratio (in log scale) is 50% ðd ¼ 0:50Þ: According to Eq. (5) the
sample size needed to achieve an 80% ðb ¼ 0:2Þ power for establishment of
equivalence is given by
ðz0:05 þ z0:2=2 Þ2
1
1
þ
n¼
0:25ð1 2 0:25Þ 0:25ð1 2 0:25Þ
0:52
¼
ð1:64 þ 1:28Þ2
ð5:33 þ 5:33Þ ¼ 363:6 < 364:
0:25
Crossover Design
Now suppose that the sponsor consider to adopt a one-arm crossover design
for this trial. In other words, each subject will first receive the standard therapy
(control) then receive the test compound (treatment). Using the same example, it is
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480
WANG, CHOW, AND LI
assumed that the relapse rate for patients in the standard therapy is 25%. At the
next dosing period, the relapse rate increases to 40% after receiving the test
compound. Also, assuming that sd in Eq. (6) is 2.5. The sample size calculation for
various hypotheses can be carried out as follows.
Test for Equality
The relative risk between the standard therapy and the test compound is
given by
OR ¼
0:40ð1 2 0:25Þ
¼ 2:
ð1 2 0:40Þ0:25
According to Eq. (7), the sample size needed in order to achieve an 80% ðb ¼ 0:2Þ
power at the 5% ða ¼ 0:05Þ level of significance is given by
n¼
ðz0:05=2 þ z0:2 Þ2 sd2 ð1:96 þ 0:84Þ2 2:52
¼
¼ 102:9 < 103:
log2 ð2Þ
0:692
Test for Superiority
Suppose that a 20% difference in odds ratio (in log scale) is considered of
clinical importance. Hence, the superiority margin is chosen to be d ¼ 0:2:
According to Eq. (8), the sample size needed in order to achieve a power of 80%
ðb ¼ 0:2Þ is given by
n¼
ðz0:05 þ z0:2 Þ2 2:52 ð1:64 þ 0:84Þ2 2:52
¼
¼ 160:1 < 161:
0:492
ðlogð2Þ 2 0:2Þ2
Test for Equivalence
Assume that (i) there is no difference in the true relapse rate between the
standard therapy and the test compound ðlogðORÞ ¼ 0Þ and (ii) the equivalence
limit of the odds ratio (in log scale) is 50%. Accordig to Eq. (9), the sample size
needed in order to achieve a power of 80% ðb ¼ 0:2Þ is given by
n¼
ðz0:05 þ z0:2=2 Þ2 2:52 ð1:64 þ 1:28Þ2 2:52
¼
¼ 213:2 < 214:
0:52
0:25
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SAMPLE SIZE CALCULATION BASED ON ODDS RATIO
481
REMARKS
For testing the following hypotheses under a parallel design
H 0 : logðORÞ ¼ 0 vs: H a : logðORÞ – 0;
an alternative test can be constructed as follows
21=2
1
1
1
*
c
T ¼ logðOR
ORÞ
þ
;
nT nC p^ ð1 2 p^ Þ
ð10Þ
where
p^ ¼
nT p^ T þ nC p^ C
:
nT þ nC
Under the null hypothesis H 0 : logðORÞ ¼ 0; T* is asymptotically distributed as
the standard normal variable. Hence, we reject the null hypothesis at the a level of
significance if jT* j . za=2 : It should be noted that formula (10) is very similar to
c is different. In Eq.
formula (2) except that the estimate of the variance of logðOR
ORÞ
c
(2), varðlogðOR
ORÞÞ is estimated by the maximum likelihood estimate (MLE)
without any constraints, while in Eq. (10) the same quantity is estimated by the
MLE under the null hypothesis that logðORÞ ¼ 0: We will refer to Eq. (2) as the
unconditional method and Eq. (10) as the conditional method. Under the null
hypothesis, both methods have an asymptotic size of a. In practice, it is a dilemma
regarding which method (i.e., unconditional or conditional) should be used
because one is not necessarily more powerful than the other. However, the
conditional approach for testing noninferiority/superiority and equivalence is
Table 1. The Simulation Result with Nominal Significance Level 0.05 of the Test for Equality
Under the Parallel Design
pT
0.20
0.20
0.20
0.25
0.25
0.25
0.30
0.30
0.30
0.35
0.35
0.35
0.40
pC
nT
nC
Power
pT
pC
nT
nC
Power
0.30
0.35
0.40
0.35
0.40
0.45
0.40
0.45
0.50
0.45
0.50
0.55
0.50
426
200
118
482
224
130
526
242
140
560
254
146
580
213
100
59
241
112
65
263
121
70
280
127
73
290
0.793
0.799
0.796
0.794
0.800
0.800
0.789
0.797
0.810
0.793
0.801
0.810
0.800
0.40
0.40
0.45
0.45
0.45
0.50
0.50
0.50
0.55
0.55
0.55
0.60
0.60
0.55
0.60
0.55
0.60
0.65
0.60
0.65
0.70
0.65
0.70
0.75
0.70
0.75
262
150
590
264
150
588
262
148
576
254
144
550
242
131
75
295
132
75
294
131
74
288
127
72
275
121
0.808
0.826
0.800
0.810
0.819
0.798
0.814
0.821
0.804
0.818
0.829
0.809
0.818
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482
WANG, CHOW, AND LI
Table 2. The Simulation Result with Nominal Significance Level 0.05 of the Test for Equivalence
Under the Parallel Design
pT
0.30
0.30
0.30
0.30
0.35
0.35
0.35
0.35
0.40
0.40
0.40
0.40
0.45
0.45
0.45
pC
nT
nC
Power
pT
pC
nT
nC
Power
0.30
0.35
0.40
0.45
0.35
0.40
0.45
0.50
0.40
0.45
0.50
0.55
0.45
0.50
0.55
122
140
260
636
112
128
228
528
108
120
212
488
104
116
210
61
70
130
318
56
64
114
264
54
60
106
244
52
58
105
0.785
0.780
0.800
0.798
0.784
0.785
0.802
0.796
0.783
0.773
0.800
0.786
0.794
0.774
0.800
0.45
0.50
0.50
0.50
0.50
0.55
0.55
0.55
0.55
0.60
0.60
0.60
0.65
0.65
0.70
0.60
0.50
0.55
0.60
0.65
0.55
0.60
0.65
0.70
0.60
0.65
0.70
0.65
0.70
0.70
494
102
116
216
544
104
120
234
672
108
130
272
112
144
122
247
51
58
108
272
52
60
117
336
54
65
136
56
72
61
0.798
0.769
0.768
0.784
0.785
0.790
0.766
0.787
0.784
0.785
0.772
0.780
0.779
0.773
0.781
difficult to carry out due to the fact that the MLE under the null hypothesis is
almost impossible to track.
SIMULATION
A total of four simulation studies were carried out to evaluate the finite
sample performance of the derived sample formulae. Since both tests for equality
Table 3. The Simulation Result with Nominal Significance Level 0.05 of the Test for Equality
Under the Crossover Design
pT
0.20
0.20
0.20
0.25
0.25
0.25
0.30
0.30
0.30
0.35
0.35
0.35
0.40
pC
sd
n
Power
pT
pC
sd
n
Power
0.30
0.35
0.40
0.35
0.40
0.45
0.40
0.45
0.50
0.45
0.50
0.55
0.50
3.004753
2.971249
2.983305
2.724457
2.817950
2.799687
2.658232
2.675998
2.714261
2.614447
2.619666
2.648626
2.541418
244
118
73
254
130
77
285
135
81
307
141
82
309
0.804
0.829
0.830
0.786
0.815
0.820
0.792
0.813
0.800
0.822
0.776
0.810
0.817
0.40
0.40
0.45
0.45
0.45
0.50
0.50
0.50
0.55
0.55
0.55
0.60
0.60
0.55
0.60
0.55
0.60
0.65
0.60
0.65
0.70
0.65
0.70
0.75
0.70
0.75
2.599919
2.648516
2.533354
2.575073
2.636148
2.571822
2.628982
2.706164
2.588022
2.679054
2.827535
2.692324
2.801336
145
84
313
142
82
316
142
81
301
135
78
292
129
0.802
0.818
0.792
0.813
0.806
0.825
0.817
0.810
0.815
0.804
0.844
0.808
0.794
MARCEL DEKKER, INC. • 270 MADISON AVENUE • NEW YORK, NY 10016
©2002 Marcel Dekker, Inc. All rights reserved. This material may not be used or reproduced in any form without the express written permission of Marcel Dekker, Inc.
SAMPLE SIZE CALCULATION BASED ON ODDS RATIO
483
Table 4. The Simulation Result with Nominal Significance Level 0.05 of the Test for Equivalence
Under the Crossover Design
pT
0.30
0.35
0.35
0.40
0.40
0.45
0.45
0.50
0.50
pC
sd
n
Power
pT
pC
sd
n
Power
0.20
0.20
0.25
0.25
0.30
0.30
0.35
0.35
0.40
2.959132
2.963281
2.820795
2.835270
2.666669
2.709451
2.579485
2.603712
2.544874
255
1003
182
528
142
364
122
289
114
0.779
0.814
0.784
0.796
0.797
0.802
0.773
0.786
0.796
0.55
0.55
0.60
0.60
0.65
0.65
0.70
0.70
0.75
0.40
0.45
0.45
0.50
0.50
0.55
0.55
0.60
0.60
2.599655
2.556995
2.596150
2.529833
2.618966
2.615441
2.659312
2.680022
2.794186
270
113
269
112
293
126
351
143
513
0.797
0.782
0.782
0.796
0.781
0.787
0.778
0.779
0.778
and superiority/noninferiority are similar to each other. The simulation studies
were only performed for testing equality and equivalence under both parallel and
crossover design. All simulations were performed based on 1000 iterations. For
crossover design, correlated binary random variables ½e:g:; ðZ T ; Z R Þ were
generated according to the following procedure:
1. Generate independent uniform (0,1) random variables denoted by X0,
XT, XR.
2. Let Yi, i ¼ T; R be a random variable with probability 0.5 be X0 and 0.5
be Xi.
3. For a given pi, i ¼ T; R, Zi is defined to be 1 if Y i , pi ; and 0 otherwise.
It can be noted that Zi is a binary response with success probability pi. Since, there
is 0.25 probability Y T ¼ Y R ¼ X 0 ; ZT and ZR is correlated in someway.
The simulation results are summarized in Tables 1 – 4. From those tables it
can be seen clearly that our sample size formula works satisfactorily with usually
no more than 3% deviation from the target value.
REFERENCES
1.
2.
3.
4.
5.
Chow, S.C.; Liu, J.P. Design and Analysis of Clinical Trials; John Wiley & Sons: New
York, 1998.
Deeks, J. Swots Corner: What Is an Odds Ratio? Bandolllier 1996, 25, 6 –9.
Haldane, J.B.S. The Estimation and Significance of Logarithm of a Ratio of
Frequencies. Ann. Hum. Genet. 1955, 20, 309 – 311.
Lachin, J.M. Biostatistical Methods: The Assessment of Relative Risk; John Wiley &
Sons: New York, 2000.
Tu, D. On the Use of the Ratio or Odds Ratio of Cure Rates in Therapeutic
Equivalence Clinical Trials with Binary Endpoints. J. Biopharm. Stat. 1998, 8,
263 – 282.