6 Equivalence and Noninferiority Tests
Pr{X ≤ X
U
| P
0
+ Δ
0
} ≈ 1 − β (usually 0.95).
That is, each side of the null hypothesis is treated as a one-sided test.
Theprobability formulas are
XXPn
n
k
PP
Pr |, ()1
L
k
nk
kX
n
00 00 00
L
∑
{}
()
≥−∆=
−∆ −+∆
−
=
and
XX Pn
n
k
PP
Pr |, ()1
U
k
nk
k
X
00 00
00
0
U
∑
{}
()
≤+∆=
+∆ −−∆
−
=
.
Example:
Suppose
P
0
= 0.50
Δ
0
= 0.025
n = 100
X = 59
1 − β = 0.95
Then
P
L
= P
0
– Δ
0
= 0.50 – 0.025 = 0.475
P
U
= P
0
+ Δ
0
= 0.50 + 0.025 = 0.525
X
L
is the value such that
XXP
n
k
Pr{|}(0.475)(1 0.475) 0.95
L
kn
k
kX
n
00
L
∑
≥−∆=
−≈
−
=
and X
U
is the value such that
XX P
n
k
Pr{|}(0.525)(1 0.525)
0.95
kn
k
k
X
U0 0
0
U
∑
≤+∆=
−≈
−
=
.
So X
L
= 40, and
n
k
(0.475)(1 0.475) 0.9460
knk
kX
n
L
∑
−≈
−
=
7Proportions and Binomial Random Variables
and X
U
= 60, so
n
k
(0.525)(1 0.525)
0.9460
knk
k
X
0
U
∑
−≈
−
=
.
Since 40 < 59 < 60, the null hypothesis is rejected.
Condence interval formulation:
Let
=P
X
n
ˆ
.
Then P
L
is the value such that
n
k
PP()(1 )
L
k
L
nk
kX
n
∑
−≈β
−
=
.
and P
U
is the value such that
n
k
PP()(1 )
U
k
U
nk
k
X
0
∑
−≈β
−
=
.
Then (P
L
, P
U
) is a 100(1 − 2β) percent condence interval for P. In the
example, with n = 100, X = 59, P
L
≈ 0.503, and P
U
≈ 0.673.
Computational considerations:
The same procedures used for Test 1.1 apply to Test 1.2.
Test 1.3 Difference of Two Proportions (Two-Sided)
Parameters:
p
1
= Pr{“success,” group or treatment 1}
p
2
= Pr{“success,” group or treatment 2}
n
1
= sample size, group or treatment 1
n
2
= sample size, group or treatment 2
Δ
L
= minimum allowable difference between p
1
and p
2
Δ
H
= maximum allowable difference between p
1
and p
2
8 Equivalence and Noninferiority Tests
Hypotheses:
H
0
: p
1
− p
2
< Δ
L
OR p
1
− p
2
> Δ
H
H
1
: Δ
L
≤ p
1
− p
2
≤ Δ
H
Data:
X
1
= number of “success,” group or treatment 1
X
2
= number of “success,” group or treatment 2
=p
X
n
ˆ
1
1
1
=p
X
n
ˆ
2
2
2
Critical value(s):
Reject H
0
if
−+ ≥∆
−β
ppzSE
ˆˆ
L121
and
−− ≤∆
−β
ppzSE
ˆˆ
U
121
, where:
=
−
+
−
SE
pp
n
pp
n
ˆ
(1
ˆ
)
ˆ
(1
ˆ
)
11
1
22
2
z
1 − β
= the upper 100(1 – β) percentile of a standard normal distribution.
Discussion:
The exact distribution of the difference between two proportions was
derived by Nadarajah and Kotz (2007). However, nding percentiles of this
distribution would be at best a difcult numerical approximation exercise.
Therefore, only the normal approximation formulas and methodology are
presented. The method provides the most accurate results when the popula-
tion proportions in question are “close” to 0.50 (Armitage, 1971).
Using the normal approximation, the power to reject the null hypothesis
is given by
{}
≥
∆−∆
−
−β
z
SE
zPr
L
1
or
{}
≤
∆−∆
+
−β
z
SE
zPr
U
1
where z ~ N(0, 1).
9Proportions and Binomial Random Variables
Example:
Suppose
n
1
= n
2
= 100
X
1
= 50
X
2
= 62
Δ
L
= −0.01
Δ
H
= +0.01
Then
== =p
X
n
ˆ
50
100
0.50
1
1
1
== =p
X
n
ˆ
62
100
0.62
2
2
2
=
−
+
−
=+≈SE
pp
n
pp
n
ˆ
(1
ˆ
)
ˆ
(1
ˆ
)
0.50(0.50)
100
0.62(0.38)
100
0.0697
11
1
22
2
1 − β = 0.95
z
1 − β
≈ 1.645.
The critical values are
p
1
− p
2
+ z
1−β 
SE ≈ −0.0054 ≥ −0.01 = Δ
L
*
and
p
1
− p
2
− z
1−β 
SE ≈ −0.2346 ≤ +0.01 = Δ
U
.
Therefore, the null hypothesis of nonequivalence is rejected. Figure 1.3
illustrates the power curve for this example. Note that only the curve for Δ
L
is presented.
Condence interval formulation:
The interval
()
−− −+
−β −β
ppzSEp pzSE
ˆˆ
,
ˆˆ
121121
is a 100(1 − 2β) percent condence interval for p
1
– p
2
.
*
Wellek (2003) has suggested to use odd’s ratio in lieu of differences between two proportions.
However, no text for equivalence based on odd’s ratios will be presented in this book.
10 Equivalence and Noninferiority Tests
Computational considerations:
Example: n = 30, β = 0.05, Δ
L
= 0.05, Δ
U
= 0.05
• SAS code
libname stuff 'H:\Personal Data\Equivalence & Noninferiority\
Programs & Output';
data calc;
set stuff.d20121026_test_1_3_example_data;
delL = -0.05;
delU = 0.05;
run;
proc freq data = calc;
tables group1/out = prop1;
tables group2/out = prop2;
run;
data prop_first;
set prop1;
count1 = COUNT;
1
0.9
0.8
0.7
0.6
0.5
0.4
Pr{Reject}
0.3
0.2
0.1
0
–0.275 –0.25 –0.225 –0.2 –0.175 –0.15
Delta = p1–p2
–0.125 –0.1 –0.075 –0.05 –0.025
0
FIGURE 1.3
Power curve for p
1
− p
2
≤ Δ
L
= −0.01.

Get Equivalence and Noninferiority Tests for Quality, Manufacturing and Test Engineers now with the O’Reilly learning platform.

O’Reilly members experience books, live events, courses curated by job role, and more from O’Reilly and nearly 200 top publishers.

Start your free trial