ALL CHAPTERS COVERED
,Chapter 2
Section 2.1
Exercise 2.1 Equations (2.1) and (2.2) imply the system
k = Φ−1 (1 − α)
δ
k − = Φ−1 (β) .
p
σ 2/n
Therefore,
δ
p = Φ−1 (1 − α) − Φ−1 (β) .
σ 2/n
Expressing n , arrive at (2.3),
¢2 ³ −1 ´2
−1
¡
n = 2 σ/δ Φ (1 − α) − Φ (β) .
Exercise 2.2 In this exercise, H0 : µA = µB is tested against H1 : µA 6=
µB , α = 0.05, β = 0.15, δ = 7, and σ = 16. Denote by x̄A and x̄B the
sample mean responses for group A and group B, respectively. Under H0 ,
the distribution of x̄A − x̄B is N 0, 2σ 2 /n , and under a specific alternative
¡ ¢
H1 : µA − µB = δ , the distribution is N δ, 2σ 2 /n . The acceptance region
¡ ¢
for the test is
n x̄A − x̄B o n p p o
−k < p < k = − k σ 2/n < x̄A − x̄B < k σ 2/n ,
σ 2/n
where the critical value k > 0. The equations for α and β are of the form
à !
x̄A − x̄B ¯ x̄ − x̄
¯ A B
1 − α = P −k < p < k¯ p ∼ N (0, 1)
σ 2/n σ 2/n
2
, ¡¯ ¯ ¢
= P ¯ Z ¯ < k , where Z ∼ N (0, 1) ,
and
³ p p ¯ ´
β = P − k σ 2/n < x̄A − x̄B < k σ 2/n ¯ x̄A − x̄B ∼ N (δ, 2σ 2 /n)
¯
ï ¯ !
¯ δ ¯
= P ¯Z + p ¯ < k , where Z ∼ N (0, 1) .
¯ ¯
¯ σ 2/n ¯
From here,
k = Φ−1 1 − α/2 ,
¡ ¢
and
³ δ ´ ³ δ ´
β = Φ k − p −Φ −k − p .
σ 2/n σ 2/n
For α = 0.05 and β = 0.15, δ = 7, and σ = 16, the numerical solution is
k = 1.96 and n ≥ 93.82 . Thus, in practice, n = 94, which corresponds to
β = 0.1493 .
Exercise 2.3 Take X ∼ P oisson(λ) , and assume that H0 : λ ≥ λ0 is
tested against H1 : λ < λ0 for some λ0 . To compute the likelihood ratio
max λx e−λ /x!
λ ≥ λ0
Λ(x) =
max λx e−λ /x!
λ>0
consider the case x ≥ λ0 . The maximum likelihood estimator of λ is x,
therefore,
xx e−x /x!
Λ(x) = = 1.
xx e−x /x!
Consider the case x < λ0 . Since when λ ≥ x the function λx e−λ /x! is
strictly decreasing, the maximum for λ ≥ λ0 > x is achieved at λ = λ0 .
3
, Hence, the likelihood ratio is
λx0 e−λ0 /x!
Λ(x) = = (λ0 /x)x e−(λ0 − x) .
xx e−x /x!
© ª
The acceptance region for the likelihood ratio is given by x : Λ(x) > c
for some constant c. From the graph of Λ(x) below, this region is equivalent
to {x : x > x0 } for some x0 > 0. Since the distribution of X is discrete, x0
can be assumed integer.
Λ(x)
✻
1
c
e−λ0 ◦
✲
0 x0 λ0 x
(b) By definition, the probability of type I error equals
x0
X λi e−λ
α = max P(X ≤ x0 ) = max .
λ ≥ λ0 λ ≥ λ0
i=0
i!
Consider the function
x0
X λ i e−λ
g(λ) = .
i=0
i!
Taking the derivative of g(λ) , get
x0 x0
′
X i λ i − 1 e−λ X λ i e−λ
g (λ) = −
i=1
i! i=0
i!
xX
0 −1 x0
λ i e−λ X λ i e−λ λ x0 e−λ
= − = − < 0.
i=0
i! i=0
i! x0 !
4
,Chapter 2
Section 2.1
Exercise 2.1 Equations (2.1) and (2.2) imply the system
k = Φ−1 (1 − α)
δ
k − = Φ−1 (β) .
p
σ 2/n
Therefore,
δ
p = Φ−1 (1 − α) − Φ−1 (β) .
σ 2/n
Expressing n , arrive at (2.3),
¢2 ³ −1 ´2
−1
¡
n = 2 σ/δ Φ (1 − α) − Φ (β) .
Exercise 2.2 In this exercise, H0 : µA = µB is tested against H1 : µA 6=
µB , α = 0.05, β = 0.15, δ = 7, and σ = 16. Denote by x̄A and x̄B the
sample mean responses for group A and group B, respectively. Under H0 ,
the distribution of x̄A − x̄B is N 0, 2σ 2 /n , and under a specific alternative
¡ ¢
H1 : µA − µB = δ , the distribution is N δ, 2σ 2 /n . The acceptance region
¡ ¢
for the test is
n x̄A − x̄B o n p p o
−k < p < k = − k σ 2/n < x̄A − x̄B < k σ 2/n ,
σ 2/n
where the critical value k > 0. The equations for α and β are of the form
à !
x̄A − x̄B ¯ x̄ − x̄
¯ A B
1 − α = P −k < p < k¯ p ∼ N (0, 1)
σ 2/n σ 2/n
2
, ¡¯ ¯ ¢
= P ¯ Z ¯ < k , where Z ∼ N (0, 1) ,
and
³ p p ¯ ´
β = P − k σ 2/n < x̄A − x̄B < k σ 2/n ¯ x̄A − x̄B ∼ N (δ, 2σ 2 /n)
¯
ï ¯ !
¯ δ ¯
= P ¯Z + p ¯ < k , where Z ∼ N (0, 1) .
¯ ¯
¯ σ 2/n ¯
From here,
k = Φ−1 1 − α/2 ,
¡ ¢
and
³ δ ´ ³ δ ´
β = Φ k − p −Φ −k − p .
σ 2/n σ 2/n
For α = 0.05 and β = 0.15, δ = 7, and σ = 16, the numerical solution is
k = 1.96 and n ≥ 93.82 . Thus, in practice, n = 94, which corresponds to
β = 0.1493 .
Exercise 2.3 Take X ∼ P oisson(λ) , and assume that H0 : λ ≥ λ0 is
tested against H1 : λ < λ0 for some λ0 . To compute the likelihood ratio
max λx e−λ /x!
λ ≥ λ0
Λ(x) =
max λx e−λ /x!
λ>0
consider the case x ≥ λ0 . The maximum likelihood estimator of λ is x,
therefore,
xx e−x /x!
Λ(x) = = 1.
xx e−x /x!
Consider the case x < λ0 . Since when λ ≥ x the function λx e−λ /x! is
strictly decreasing, the maximum for λ ≥ λ0 > x is achieved at λ = λ0 .
3
, Hence, the likelihood ratio is
λx0 e−λ0 /x!
Λ(x) = = (λ0 /x)x e−(λ0 − x) .
xx e−x /x!
© ª
The acceptance region for the likelihood ratio is given by x : Λ(x) > c
for some constant c. From the graph of Λ(x) below, this region is equivalent
to {x : x > x0 } for some x0 > 0. Since the distribution of X is discrete, x0
can be assumed integer.
Λ(x)
✻
1
c
e−λ0 ◦
✲
0 x0 λ0 x
(b) By definition, the probability of type I error equals
x0
X λi e−λ
α = max P(X ≤ x0 ) = max .
λ ≥ λ0 λ ≥ λ0
i=0
i!
Consider the function
x0
X λ i e−λ
g(λ) = .
i=0
i!
Taking the derivative of g(λ) , get
x0 x0
′
X i λ i − 1 e−λ X λ i e−λ
g (λ) = −
i=1
i! i=0
i!
xX
0 −1 x0
λ i e−λ X λ i e−λ λ x0 e−λ
= − = − < 0.
i=0
i! i=0
i! x0 !
4
