Equivalence and Noninferiority Tests for Quality, Manufacturing and Test Engineers

57

4

Exponential Rate Parameters

Test 4.1 Single Exponential Rate Parameter (One-Sided)

Parameters:

λ = exponential rate parameter (number of events per unit time)

λ

0

= maximum tolerable rate

1 – β = power to reject the null if λ = λ

0

Hypotheses:

H

0

: λ > λ

0

H

1

: λ ≤ λ

0

Data:

T averagetimebetween events=

T

1

=average rate (numberofeventsper unit time)λ=

Critical value(s):

If time to event, T, has an exponential distribution, then

Tn

T

ˆ

i

i

n

1

∑

γ= =

=

has a gamma distribution with parameters λ

0

and n. Often n is called the

“shape” parameter, and λ

0

is called the “scale” parameter. Thus, if

Gn

n

xe dx(

ˆ

|, )

()

n

nx

0

0

1

0

ˆ

0

∫

γλ=

λ

Γ

≥β

−

γ

−λ

58 Equivalence and Noninferiority Tests

then reject H

0

. That is, G(.) is the cumulative distribution function for a

gamma-distributed random variable with parameters λ

0

and n. Similarly,

nd the value, call it γ

c

, such that

Gn

n

xe dx(|,)

()

c

n

nx

0

0

1

0

c

0

∫

γλ=

λ

Γ

=β

−

γ

−λ

and reject H

0

if

ˆ

c

γ≥γ

.

One could also rely on the central limit theorem, and reject the null

hypothesis if

Tt

S

n

1

1

0

+≥

λ

−β

where S is the sample standard deviation of the T

i

.

Discussion:

The smaller the rate parameter, the longer the expected time to event. Thus,

demonstrating that the average time to event is at least 1/λ

0

is the same as

showing that the rate (number of events per xed unit of time) is no greater

than λ

0

.

In this formulation, we have implicitly assumed that the “events” of interest

are undesirable, such as a machine failure. Thus, high rates of occurrence

would be undesirable.

The probability of rejecting the null hypothesis under the alternative that

λ = λ

a

is

∫

−γ λ=−

λ

Γ

−

γ

−λ

Gn

n

xe

dx

1(|, )1

()

ca

a

n

nx

1

0

c

a

.

Example:

Let λ

0

= 0.75, or 1/ λ

0

≈ 1.33. So the hypotheses are

H

0

: λ > 0.75

H

1

: λ ≤ 0.75

Suppose:

T 1.29 (sampleaverage time to event)=

S = 1.08

59Exponential Rate Parameters

n = 20

β = 0.05

t

1 −

β ≈ 1.729

Tt

S

n

1.29 1.729

1.08

20

1.708

1

1.33

1

0

+≈+≈≥

λ

≈

−β

.

Therefore, we reject H

0

. Also, we have

ˆ

20 * 1.29 25.8γ≈ =

and

Gn

n

xe dx1(25.8|20, 0.75)1

()

0.5287

n

nx

0

0

1

0

25.8

0

∫

−=λ= =−

λ

Γ

≈>β

−−λ

.

Therefore, we would reject H

0

using the “exact” gamma critical value. Note

that for β = 0.05, with n = 20, the critical value, γ

c

, is 17.67. That is,

Gn

n

xe dx(17.67|20, 0.75)

()

0.05

n

nx

0

0

1

0

17.67

0

∫

=λ==

λ

Γ

≈=β

−−λ

.

Figure4.1 shows the power curve using the exact gamma critical value of

17.67 for this example.

Condence interval formulation:

Tt

S

n

1

+

−β

is a 100(1 − β) percent upper condence limit on 1/λ. For the example,

Tt

S

n

1.29 1.729

1.08

20

1.708

1

+≈

+≈

−β

is a 95 percent upper condence limit on 1/λ, the mean time to event.

Computational considerations:

Care should be taken when using built-in functions for computing

probabilities or quantiles from gamma distributions. In some cases, the

“scale” parameter may be expressed as a rate, for example, the expected

number of occurrences per unit time, and in some cases, the scale parameter

may be expressed as a mean time between events (R, JMP). In some cases,

60 Equivalence and Noninferiority Tests

such as SAS, the scale parameter is assumed to be 1, so that a change of

variables is required to compute probabilities or quantiles from distributions

where the scale parameter is not 1.

• SAS code

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

Programs & Output';

data calc;

set stuff.d20121108_test_4_1_example_data;

run;

proc means data = calc;

var T lamda0 beta;

output out = onemean MEAN = tbar lamda0val betaprob

SUM=sumt STD = sdt N = n1;

run;

data outcalc;

set onemean;

1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0

0.7 0.8 0.911.1 1.2 1.3 1.4 1.5 1.6 1.7

lamda_a

Power

FIGURE 4.1

Test 4.1, power curve for exponential time-to-event test—exact gamma method.

61Exponential Rate Parameters

crit_gam = gaminv(betaprob,n1)/lamda0val;

norm_stat = tbar + tinv(1-betaprob,n1-1)*sdt/sqrt(n1);

crit_norm = 1/lamda0val;

run;

proc print data = outcalc;/* has tbar lamda0val betaprob sumt

sdt n1 crit_gam norm_stat crit_norm */

run;

The MEANS Procedure

Variable Label N Mean Std Dev Minimum Maximum

ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ

T T 25 1.9768811 1.6434899 0.0482092 5.6483261

lamda0 lamda0 25 1.1000000 0 1.1000000 1.1000000

beta beta 25 0.0500000 0 0.0500000 0.0500000

ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ

The SAS System 08:52 Thursday, November 8, 2012 12

norm_ crit_

Obs _TYPE_ _FREQ_ tbar lamda0val betaprob sumt sdt n1 crit_gam stat norm

1 0 25 1.97688 1.1 0.05 49.4220 1.64349 25 15.8019 2.53924 0.90909

• JMP Data Table and formulas (Figure4.2)

FIGURE 4.2

Test 4.1, JMP screen.