Equivalence and Noninferiority Tests for Quality, Manufacturing and Test Engineers

53Variances (Standard Deviations) and Coefficients of Variation

Kang et al. (2007) used the approximation that

′

−













c

n

Tn

n

c

ˆ

~

1,

where

′

−













Tn

n

c

1,

is a noncentral t random variable with n – 1 degrees of freedom and

noncentrality parameter

n

c

.

Using T′ to represent the appropriate noncentral t variable, the power to reject

the null hypothesis is:

≤

′

β−









































=

′

≤

′

β−









































=

′

≥

′

β−

































c

n

Tn

n

c

n

T

n

Tn

n

c

TT n

n

c

Pr

ˆ

,1,

Pr

,1,

Pr ,1

,.

Example:

Suppose:

c

0

= 0.06

H

0

: c > 0.06

H

1

: c ≤ 0.06

c

ˆ

0.065=

n = 64

54 Equivalence and Noninferiority Tests

1 – β = 0.95

Then

=≈

n

c

64

0.06

133.33

0

Tn

n

c

,1; 116.43

0

′

β−













≈

n

Tn

n

c

,1;

8

116.43

0.0678

0

′

β−













≈≈

.

Since

=<c

ˆ

0.065 0.0687

, the null hypothesis is rejected. Figure3.5 shows the

power curve for this example.

1

0.9

0.8

0.7

0.6

0.5

Power

0.4

0.3

0.2

0.1

0

5.75 6 6.25 6.5 6.75 7 7.25 7.5 7.7588.25 8.5 8.75 99.25 9.5 9.75

%CV

FIGURE 3.5

Test 3.3, power curve for test of single CV.

55Variances (Standard Deviations) and Coefficients of Variation

Condence interval formulation:

′

β−













n

Tn

n

c

,1;

ˆ

is a 100 percent (1 − β) upper condence limit on the coefcient of varia-

tion, c.

The example data yield:

′

β−













=

′













≈≈

n

Tn

n

c

T,1;

ˆ

64

0.05,63;

64

0.065

8

107.46

0.0744.

Computational considerations:

• SAS code

libname stuff 'H:\Personal Data\Equivalence & Noninferiority\

Programs & Output';

data calc;

set stuff.d20121107_test_3_3_example_data;

run;

proc means data = calc;

var X c0 beta;

output out = onemean MEAN = xbar c0val betaprob STD = sdx

N= n1;

run;

data outcalc;

set onemean;

samp_cv = sdx/xbar;

crit_val = sqrt(n1)/tinv(betaprob,n1-1,sqrt(n1)/c0val);

run;

proc print data = outcalc;/* has vars xbar c0val betaprob sdx

n1 sampe_cv crit_val */

run;

56 Equivalence and Noninferiority Tests

The MEANS Procedure

Variable Label N Mean Std Dev Minimum Maximum

ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ

X X 25 9.6829027 1.0045857 7.4892100 12.0170228

c0 c0 25 0.1000000 0 0.1000000 0.1000000

beta beta 25 0.0500000 0 0.0500000 0.0500000

ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ

The SAS System 13:53 Wednesday, November 7, 2012 10

Obs _TYPE_ _FREQ_ xbar c0val betaprob sdx n1 samp_cv crit_val

1 0 25 9.68 0.10 0.05 1.00 25 0.10375 0.12351

• JMP Data Table and formulas (Figure3.6)

FIGURE 3.6

Test 3.3, JMP screen.