48 Equivalence and Noninferiority Tests
Discussion:
If
σ= σk
1
2
0
2
2
2
then
−σ
−σ
nSk
nS
(1)
(1)
11
2
0
2
2
2
22
2
2
2
has an F-distribution with n
1
− 1 and n
2
− 1 degrees of freedom in the numer-
ator and denominator, respectively. Thus,
−
−
−−
nS
nS
kF
nn
(1)
(1)
~(1,
1)
11
2
22
2
0
2
12
.
Therefore,
kFnn Fn n
Fn n
k
Fn n
Pr (1,1)(1, 1)
Pr (1,1)
1
(1
,1
)
1
0
2
12 11 2
12
0
2
11 2
{}
−−≤−−
=−−≤
−−
=−β
−β
−β
so
−
−
≤−−
=−β
−β
nS
nSk
Fn nPr
(1)
(1)
1
(1,1
)1
11
2
22
2
0
2
11 2
.
If in fact
kk
k,
aa
1
22
2
2
0
σ=
σ>
then, as
k
a
2
gets larger than
k
0
2
, the power decreases. That is, the power is
givenby:
≤−
−
−β
Pr F
k
k
Fn n(1
,1
a
0
2
2
11 2
.
Example:
Suppose:
k
0
2
= 1.0
49Variances (Standard Deviations) and Coefficients of Variation
S
1
= 5.0
S
2
= 6.0
n
1
= n
2
= 10
and 1 – β = 0.95, so
−
−
=≈
nS
nS
(1)
(1)
9*5.0
9*6.0
0.694
11
2
22
2
2
2
−−≈
−β
k
Fn n
1
(1,1)
3.1789
0
2
11 2
.
Since
−
−
≤−−≈
−β
nS
nSk
Fn n
(1)
(1)
1
(1,1) 3.1789
11
2
22
2
0
2
11 2
we reject H
0
.
The power curve for this example is given as a function of k
a
≥ k
0
= 1.0 in
Figure3.3.
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0.75 1 1.25 1.5 1.7522.25 2.5 2.75 3 3.25 3.5 3.75 4
Power
ka
FIGURE 3.3
Test 3.2, power curve for comparison of two variances.
50 Equivalence and Noninferiority Tests
Condence interval formulation:
=
−
−
−−
−β
k
nS
nS
Fn n
ˆ
(1)
(1)
(1
,1
)
U
22
2
11
2
11 2
is an upper 100(1 − β) percent condence limit for
=
σ
σ
k
1
2
.
The example data yield:
k
nS
nS
Fn n
ˆ
(1)
(1)
(1,1)
(101)36
(101)25
(3.1789) 4.578
U
22
2
11
2
11 2
=
−
−
−−=
−
−
≈
−β
.
Note that as n
1
and n
2
get large, F
1−β
(n
1
− 1, n
2
− 1) will get smaller.
Computational considerations:
• SAS code
libname stuff 'H:\Personal Data\Equivalence & Noninferiority\
Programs & Output';
data calc;
set stuff.d20121107_test_3_2_example_data;
run;
proc means data = calc;
var X1 X2 k0;
output out = onemean MEAN = xbar1 xbar2 k0val VAR = v1 v2
N= n1 n2;
run;
data outcalc;
set onemean;
beta = 0.05;
var_ratio = ((n1-1)*v1)/((n2-1)*v2);
crit_val = finv(1-beta,n1-1,n2-1)/k0val**2;
run;
proc print data = outcalc;/* has vars xbar1 xbar2 k0val v1 v2
n1 n2 beta var_ratio crit_val */
run;
51Variances (Standard Deviations) and Coefficients of Variation
The SAS System 13:53 Wednesday, November 7, 2012 7
The MEANS Procedure
Variable Label N Mean Std Dev Minimum Maximum
ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ
X1 X1 35 0.1920832 1.8406642 -3.8176650 4.3223639
X2 X2 35 0.1100396 2.0376728 -4.5550806 4.7211140
k0 k0 35 1.0500000 0 1.0500000 1.0500000
ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ
The SAS System 13:53 Wednesday, November 7, 2012 8
var_
Obs _TYPE_ _FREQ_ xbar1 xbar2 k0val v1 v2 n1 n2 beta ratio crit_val
1 0 35 0.192 0.110 1.05 3.38804 4.15211 35 35 0.05 0.81598 1.60732
• JMP Data Table and formulas (Figure3.4)
FIGURE 3.4
Test 3.2, JMP screen.
52 Equivalence and Noninferiority Tests
Test 3.3 Single Coefficient of Variation (One-Sided)
Parameters:
=
σ
µ
c
= coefcient of variation (cv)
c
0
= upper tolerable bound on c.
1 – β = power to reject the null if c equals c
0
.
Hypotheses:
H
0
: c > c
0
H
1
: c ≤ c
0
Data:
c
ˆ
ˆ
ˆ
sample
cv
=
σ
µ
= .
n = sample size.
Critical value(s):
Reject H
0
if
≤
′
β−
c
n
Tn
n
c
ˆ
,1;
0
.
′
β−
Tn
n
c
,1;
0
is the 100(β) percentile of a noncentral t-distribution with n – 1 degrees of
freedom and noncentrality parameter
n
c
.
0
Discussion:
The critical value and subsequent power calculations are all based on an
approximation to the distribution of the sample
=
σ
µ
cCV,
ˆ
ˆ
ˆ
.

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