John DiNardo 125
First, let’s illustrate the problem. In experiment A, the question is: “How often
would we expect to see 9 black balls out of 12 balls under the null hypothesis?”:
P(̂μ≥
3
4
|H 0 ) ≡ P(X≥ 9 |H 0 )
=
∑^12
x= 9
(
12
x
)
μx( 1 −μ)^12 −x
=
(
12
9
)
1
2
9
( 1 −
1
2
)^3 +
(
12
10
)
1
2
10
( 1 −
1
2
)^2 +...
=
220 + 66 + 12 + 1
212
=
299
212
= 0.073.
In experiment B, the question is: “Under the null hypothesis, what is the prob-
ability of drawing 9 or more black balls before drawing a third red ball?” Letr= 3
be the pre-specified number of red balls to be drawn before the experiment is to be
stopped. Letxindex the number of black balls drawn, and letn=x+r.
This is a straightforward application of the negative binomial distribution
where:
P(X≥ 9 |H 0 ) =
∑∞
x= 9
(
r+x− 1
r− 1
)
μx( 1 −μ)r
=
∑∞
x= 9
(
x+ 2
2
)
μx( 1 −μ)r.
It is very helpful to observe in doing the calculation that:
∑∞
x=j
(
x+ 2
2
)
(
1
2
)x=
8 + 5 j+j^2
2 j
.
We can then write:
=
∑∞
x= 9
(
x+ 2
2
)
μx( 1 −μ)^3
=
(
1
2
) 3
8 + 5 ( 9 )+ 92
29
=
1
8
(
134
512
)
= 0.0327.