Equivalence and Noninferiority Tests for Quality, Manufacturing and Test Engineers

101

8

Sample-Based Criteria

In the previous seven chapters, critical values were always based on some

hypotheses stated a priori and on desired risk levels for rejecting the null

hypothesis. Unfortunately, sometimes a naïve approach is taken in choosing

critical values for a test, without regard to any hypotheses, sample size, or

associated risk levels. In this chapter, several such cases are discussed. Most

of these cases are formulated as acceptance sampling problems as opposed

to equivalence or noninferiority tests, per se.

Test 8.1 Single Proportion

Data:

X = number of successes out of n Bernoulli trials

=P

X

n

ˆ

Critical value(s):

Pass the test if

Pp

ˆ

c

≥

.

Discussion:

Given that the population proportion of successes, P, is unknown, and

there is no hypothesis concerning its minimum acceptable value or maxi-

mum unacceptable value, it is not possible to specify the risk of failing (or

passing). As a consequence, it is not possible to determine what sample size,

n, would be required. If the sample size is xed to some arbitrary value, then

a risk curve for that sample size could be constructed. In other words, the

critical number of successes would be:

X

c

= round (np

c

, 0)

where round(x, 0) means round x to the nearest integer.

For P = 0 to 1, the probability of passing the test is

∑

{}

()

≥=













−

=

−

XXPn

n

k

PP

Pr |, () 1

c

kX

n

k

nk

c

.

102 Equivalence and Noninferiority Tests

However, since no particular minimum acceptable probability of success

was specied, it is not possible to determine whether the sample size, n, is

sufcient, too small, or even too large.

Example:

Suppose n = 100, and the criterion for passing the test is that

P

ˆ

0.95≥

, so

that X

c

= round (100 * 0.95, 0) = 95. If the population’s probability of a “success”

is P = 0.95, then the probability of passing the criterion is:

∑

{}

()

≥=













≈

=

−

X

k

Pr 95|0.95,100

100

(0.95) 0.05 0.616

k

k

k

95

100

100

or about a 61.6 percent chance. Conversely, suppose that the sample size is

changed to 50, but the sample criterion remains the same, that

≥P

ˆ

0.95

in

order to pass. Then, under the same population condition, that is, P = 0.95,

the probability of passing is now:

∑

{}

()

≥=













≈

=

−

X

k

Pr 48|0.95,48

50

(0.95) 0.05 0.540

k

k

k

48

50

50

or about a 54.0 percent chance of passing, even though the sample criterion is

still 95 percent of n. Table8.1 shows the probabilities of “passing” the test for

n = 200, 100, and 50, under the conditions that P = 0.95 and P = 0.97.

Test 8.2 Single Mean

Data:

X sample mean=

n = sample size

TABLE8.1

Probabilities of Passing the Sample-

Based Criterion Test for Proportions

N P X

c

= 95% of N Pr{X ≥ X

c

}

200 0.95 190 58.31%

100 0.95 95 61.60%

50 0.95 48 54.05%

200 0.97 190 95.99%

100 0.97 95 91.92%

50 0.97 48 81.08%

103Sample-Based Criteria

Critical value(s):

Pass the test if:

Xx

c

≤

.

Discussion:

The central limit theorem dictates that the sample statistic,

X

, tends toward

having a normal distribution, that is,

XN

n

~,

2

µ

σ













.

Clearly, the probability of passing the test depends on the parameters μ, σ,

and n. Without specifying μ and σ, it is not possible to assess the risk curve for

such a test. Of course, as in the case of the single proportion, xing the accep-

tance criterion/critical value regardless of sample size results in risks that

potentially change for each instance of the test, if the sample size is not xed.

Example:

Suppose that the pass criterion is that:

Xx 100

c

≤= .

Suppose further that μ = 95, σ = 14, and n = 30. Then

Xn

ed

xPr 100|95, 14,30

30

14 2

0.9748

x 95

14/30

100

1

2

2

∫

{}

≤µ=σ===

π

≈

−

−













−∞

.

If the sample size were changed to n = 10, without changing the critical

value, the associated probability of passing would be:

XnedxPr 100|95, 14,10

10

14 2

0.8706

x 95

14/10

100

1

2

2

∫

{}

≤µ=σ===

π

≈

−

−













−∞

.

Test 8.3 Relative Difference between Two Means

Data:

X

a

=

sample mean of antecedent system

X

d

=

sample mean of descendent system