Equivalence and Noninferiority Tests for Quality, Manufacturing and Test Engineers

43

3

Variances (Standard Deviations)

and Coefficients of Variation

Test 3.1 Single Variance (One-Sided)

Parameters:

σ=

0

2

maximum tolerable variance

1 – β = power to reject the null if variance equals

σ

0

2

Hypotheses:

σ>σH :

0

2

0

2

σ≤σH :

1

2

0

2

Data:

S

n

XX

=samplestandarddeviation =

1

1

i

i

n

2

1

∑

()

−

−

=

Critical value(s):

Reject H

0

if

≤

σ

−

χ−

−β

S

n

n

1

(1

)

2

0

2

1

2

where

n(1) 100(1 )

1

2

χ−=−β

−β

percentile of a chi-squared distribution with n−

1 degrees of freedom.

Discussion:

The quantity

−

σ

nS

(1)

a

2

2

44 Equivalence and Noninferiority Tests

has a chi-squared distribution when

σ=σ

a

22

. Suppose that

kk

fo

r>1

a

22

0

2

σ= σ .

Then

−

σ

=

−

σ

nS

nS

k

(1)(1)

a

2

2

2

2

0

2

and

−

σ

≤χ −σ=σ













=

−

σ

≤

χ−













−β

−β

nS

n

nS

n

k

Pr

(1)

(1)| Pr

(1)

(1

)

a

a

2

2

1

222

2

0

2

1

2

2

.

For k = 1, this probability is 1 – β. As k gets large, the probability, which is the

power, decreases.

Example:

Suppose:

4.0

0

2

σ=

, n = 20, and S = 2.1. If 1 – β = 0.95, then

n(1)

30.144

1

2

χ−≈

−β

.

Since

=≤

σ

−

χ−

=≈

−β

S

n

n4.41

1

(1)

4

19

30.144 6.35

2

0

2

1

2

we reject H

0

. The power curve for this test is given in Figure3.1 (power as a

function of the multiplier, k).

Condence interval formulation:

σ=

−

χ−

β

nS

n

ˆ

(1)

(1

)

U

2

2

is an upper 100(1 − β) percent condence limit on σ. In the example,

nS

n

ˆ

(1)

(1)

(201)4.41

10.117

2.878 20 1 10.117

U

2

2

0.05

2

()

()

σ=

−

χ−

=

−

≈χ−≈

β

.

45Variances (Standard Deviations) and Coefficients of Variation

Computational considerations:

• SAS code

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

Programs & Output';

data calc;

set stuff.d20121107_test_3_1_example_data;

run;

proc means data = calc;

var X sig0 beta;

output out = onemean MEAN = xbar popsig betaprob

VAR=sampvar N = n1;

run;

data outcalc;

set onemean;

upplim = (popsig**2)*cinv(1-betaprob,n1-1)/(n1-1);

run;

1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0

Power

0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 21.91.81.71.61.51.41.31.2

k

FIGURE 3.1

Test 3.1, power curve for testing a single variance.

46 Equivalence and Noninferiority Tests

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

sampvar n1 upplim */

run;

The SAS System 13:53 Wednesday, November 7, 2012 2

The MEANS Procedure

Variable Label N Mean Std Dev Minimum Maximum

ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ

X X 30 0.1613638 0.7395784 -1.5758867 1.5200489

sig0 sig0 30 1.2000000 0 1.2000000 1.2000000

beta beta 30 0.0500000 0 0.0500000 0.0500000

ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ

The SAS System 13:53 Wednesday, November 7, 2012 3

Obs _TYPE_ _FREQ_ xbar popsig betaprob sampvar n1 upplim

1 0 30 0.161 1.200 0.05 0.54698 30 2.11317

• JMP Data Table and formulas (Figure3.2)

FIGURE 3.2

Test 3.1, JMP.

47Variances (Standard Deviations) and Coefficients of Variation

Test 3.2 Comparison of Two Variances (One-Sided)

Parameters:

σ

1

2

= variance of population 1

σ

2

2

= variance of population 2

k

0

2

= maximum tolerable ratio of the two variances

σ

σ

1

2

2

2

1 – β = power to reject the null if

σ

σ

= k

1

2

2

2

0

2

.

Hypotheses:

σ

σ

>

Hk

:

0

1

2

2

2

0

2

σ

σ

≤

Hk

:

1

1

2

2

2

0

2

Data:

S

1

= sample standard deviation, sampled from population 1

S

2

= sample standard deviation, sampled from population 2

n

1

= sample size from population 1

n

2

= sample size from population 2

In general, the formula employed here for sample standard deviation is

assumed to be

∑

()

=

−

−

=

S

n

XX

1

1

i

i

n

2

1

.

Critical value(s):

Reject H

0

if:

−

−

≤−

−

−β

nS

nSk

Fn n

(1)

(1)

1

(1

,1)

11

2

22

2

0

2

11 2

.