16 Equivalence and Noninferiority Tests
Power calculations are similar to those for the Δ paradigm, with some
important differences.
Note that under the condition that μ
1
= (1 − p
a
)μ
2,
p
a
≠ p
0
EX pX pppp1(1)(1 )
aa
102202
02
() ()
−−
=− µ− −µ=−µ
As a result, the noncentrality parameter for the test statistic is
pp n
p11
a
02
0
2
()
()
−µ
σ+−
Thus, in order to compute the probability of rejecting the null, estimates
of both σ and μ
2
are required. Figure 2.7 shows the power curve for the
example, assuming μ
2
= 100 and σ = 2.8. In this example, Pr{Reject H
0
|p
a
=
0.0835} ≈ 0.051.
Condence interval formulation:
The expression
Power
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0.04 0.05 0.06 0.07
pa
0.08 0.09
FIGURE 2.7
Test 2.3, power curve for equivalence of two means—percent paradigm.
17Means
XpXtSE1
1021
()
−− +
−β
is a one-sided 100(1 − β) percent upper condence limit for μ
1
– (1 – p
0
)μ
2
.
From the example, the 95 percent upper condence limit for μ
1
– (1 – p
0
)μ
2
is
()
−− += +≈
−β
XpXtSE1 93.5 –(0.95)*100.0 (1.701)*0.997
0.197
1021
.
Computational considerations:
• SAS code
libname stuff 'H:\Personal Data\Equivalence & Noninferiority\
Programs & Output';
data calc;
set stuff.d20121104_test_2_3_prop;
run;
proc means data = calc;
var X1 X2 p0 beta;
output out = onemean MEAN = xbar1 xbar2 p0val betaprob STD =
sd1 sd2 N = n1 n2;
run;
data outcalc;
set onemean;
se = sqrt(sd1**2/n1 + ((1-p0val)**2)*sd2**2/n2);
w1 = ((sd1**2/n1)/(sd1**2/n1 + sd2**2/n2))**2/(n1-1);
w2 = ((sd2**2/n1)/(sd1**2/n1 + sd2**2/n2))**2/(n2-1);
dfe = 1/(w1 + w2);
lowlim = xbar1 - (1-p0val)*xbar2 + tinv(1-betaprob,dfe)*se;
run;
proc print data = outcalc;/* has vars xbar1 xbar2 p0val
betaprob n1 n2 se lowlim */
run;
The MEANS Procedure
Variable Label N Mean Std Dev Minimum Maximum
ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ
X1 X1 25 8.8799300 1.0305582 7.3220509 11.6006410
X2 X2 25 9.6829027 1.0045857 7.4892100 12.0170228
p0 p0 25 0.1000000 0 0.1000000 0.1000000
beta beta 25 0.0500000 0 0.0500000 0.0500000
ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ
18 Equivalence and Noninferiority Tests
The SAS System 06:27 Sunday, November 4, 2012 2
Obs _TYPE_ _FREQ_ xbar1 xbar2 p0val betaprob sd1
1 0 25 8.88 9.68 0.10 0.05 1.03
Obs sd2 n1 n2 se w1 w2 dfe lowlim
1 1.00 25 25 0.27419 0.010955 .009891787 47.9688 0.62520
• JMP Data Table and formulas (Figures2.8, 2.9, and 2.10)
Test 2.4 Single Mean (Two-Sided)
Parameters:
μ = population mean
σ = population standard deviation
FIGURE 2.8
Test 2.3, JMP screen 1.
19Means
FIGURE 2.9
Test 2.3, JMP screen 2.
FIGURE 2.10
Test 2.3, JMP screen 3.
20 Equivalence and Noninferiority Tests
μ
L
= lower acceptable limit for μ (L for “low”)
μ
H
= upper acceptable limit for μ (H for “high”)
n = sample size
1 – β = power to reject the null if μ = μ
L
or μ = μ
H
Hypotheses:
H
0
: μ < μ
L
OR μ > μ
H
H
1
: μ
L
≤ μ ≤ μ
H
Data:
=X sample mean
S = sample standard deviation
n = sample size
Critical value(s):
If
Xt
S
n
L1
+≥
µ
−β
and
Xt
S
n
H1
−≤
µ
−β
where t
1 − β
= 100*(1 − β) percentile of a central t-distribution with n − 1 degrees
of freedom, then reject H
0
.
Discussion:
This is an implementation of the two one-sided test, or TOST, philosophy
of schuirmann (1987). That is, we do not split β in half, but apply all of this
risk to each side of the hypothetical interval, (μ
L
,μ
H
). The reasoning for using
TOST is twofold:
1. The “OR” in the null hypothesis statement is an exclusive “or.” That
is, μ cannot be both less than μ
L
and greater than μ
H
.
2. Inasmuch as failing to reject the null is not the desired state, reduc-
ing β by splitting it in half would be a less conservative criterion
than not splitting the risk.
As a result, the power calculations are identical to those of the single mean
(one-sided) test.
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