John DiNardo 125First, 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 )=∑^12x= 9(
12
x)
μx( 1 −μ)^12 −x=(
12
9)
1
29
( 1 −
1
2)^3 +(
12
10)
1
210
( 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.