1
1
Proportions and Binomial Random Variables
Test 1.1 Single Binomial Proportion (One-Sided;
Probability of “Success”)
Parameters:
P
0
= probability of “success”
n = number of Bernoulli trials
Hypotheses:
H
0
: P < P
0
H
1
: P ≥ P
0
Data:
X = number of successes out of n trials
Critical value(s):
X
c
= minimum number of successes needed to REJECT H
0
.
Discussion:
The experimenter chooses X
c
so that:
Pr{X ≥ X
c
|P
0
} ≈ 1 − β (usually 0.95)
where:
XXPn
n
k
PP
Pr |, () 1
c
kX
n
k
nk
c
∑
{}
()
≥=
−
=
−
.
The power curve is generated by plotting Pr{X ≥ X
c
|P
a
} on the vertical
axisagainst P
a
on the horizontal. For values of P
a
< P
0
, the power will be less
than 1 − β.
2 Equivalence and Noninferiority Tests
Example:
It is desired that at least 95 percent of the units of some machine part
manufactured satisfy the criteria for some quality inspection. Then P
0
= 0.95.
To test the hypotheses H
0
: P < 0.95 versus H
1
: P ≥ 0.95, it was decided to use
a sample of n = 100 units. Let 1 – β = 0.95. The critical number of “successes”
was chose to be X
c
= 92, since:
k
100
(0.95) 0.05 0.9369 1
0.95
k
k
nk
92
100
∑
()
≈≤−β=
=
−
.
Thus, if X = 92 or higher, the null hypothesis would be rejected. Do not be
confused by the fact that, in this example, P
0
= 0.95 and 1 − β = 0.95. Figure1.1
shows a power curve for this test.
Condence interval formulation:
The lower condence limit, P
L
, is the value such that, given X successes
observed out of n trials:
n
k
PP
n
k
PP11 1
L
k
L
nk
kX
n
L
k
L
nk
k
X
0
1
∑∑
() ()
−=−
−≈β
−
=
−
=
−
.
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0.8 0.825 0.85 0.875 0.9
Pa
Power
0.925 0.95 0.97
51
FIGURE 1.1
Power curve for H
0
: P < 0.95 vs. H
1
: P > = 0.95.
3Proportions and Binomial Random Variables
In the example, if X = 93 out of n = 100, with 1 − β = 0.95, the value P
L
=0.8726
yields:
p
k
PP
k
PP
100
11
100
1 0.0502
L
k
L
k
k
L
k
L
k
k
100
93
100
100
0
92
∑∑
() ()
=
−=−
−≈
−
=
−
=
so 1 – p ≈ 0.9498 ≤ 1 − β = 0.95, so the lower condence limit for P is P
L
=
0.8726.
Computational considerations:
• SAS code
libname stuff 'H:\Personal Data\Equivalence & Noninferiority\
Programs & Output';
data calc;
set stuff.d20121026_test_1_1_example_data;
xc = 21;
power = 1 - probbnml(0.8,30,xc-1);/* Pr{Binomial > = xc} */
run;
proc freq data = calc;
tables outcome;
run;
proc print data = calc;/*dataset calc has columns n muL muA
sigma delta nc power */
run;
The SAS System 08:19 Friday, October 26, 2012 3
The FREQ Procedure
outcome
Cumulative Cumulative
outcome Frequency Percent Frequency Percent
ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ
0 4 13.33 4 13.33
1 26 86.67 30 100.00
The SAS System 08:19 Friday, October 26, 2012 4
Obs outcome xc power
1 1 21 0.93891
2 1 21 0.93891
4 Equivalence and Noninferiority Tests
3 0 21 0.93891
4 1 21 0.93891
5 1 21 0.93891
6 1 21 0.93891
7 1 21 0.93891
8 1 21 0.93891
9 1 21 0.93891
10 1 21 0.93891
11 1 21 0.93891
12 1 21 0.93891
13 1 21 0.93891
14 1 21 0.93891
15 1 21 0.93891
16 1 21 0.93891
17 1 21 0.93891
18 1 21 0.93891
19 1 21 0.93891
20 1 21 0.93891
21 1 21 0.93891
22 1 21 0.93891
23 1 21 0.93891
24 0 21 0.93891
25 0 21 0.93891
26 1 21 0.93891
27 1 21 0.93891
28 0 21 0.93891
29 1 21 0.93891
30 1 21 0.93891
• JMP data table and formulas:
Example: n = 30; P
0
= 0.80, β = 0.05 (Figure1.2).
Test 1.2 Single Binomial Proportion (Two-Sided)
Parameters:
P
0
= probability of “success”
Δ
0
= allowable difference
n = number of Bernoulli trials
Hypotheses:
H
0
: P < P
0
− Δ
0
OR P > P
0
+ Δ
0
H
1
: P
0
− Δ
0
≤ P ≤ P
0
+ Δ
0
5Proportions and Binomial Random Variables
Data:
X = number of successes out of n trials
Critical value(s):
X
L
= minimum number of successes needed to REJECT H
0
.
X
U
= maximum number of successes needed to REJECT H
0
.
If X < X
L
or X > X
U
, then H
0
is rejected in favor of H
1
.
Discussion:
As mentioned earlier, in the case of two-sided tests, the two one-sided test
(TOST) philosophy will be adopted. Thus, X
L
is chosen so that:
Pr{X ≥ X
L
| P
0
– Δ
0
} ≈ 1 − β (usually 0.95)
and
FIGURE 1.2
Test 1.1: JMP screen.

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