Statistical Inference, 2nd Edition by
George Casella
Complete Chapter Solutions Manual
are included (Ch 1 to 12)
** Immediate Download
** Swift Response
** All Chapters included
,Table of Contents are given below
1. Probability Theory.
2. Transformations and Expectations
3. Common Families of Distributions
4. Multiple Random Variables
5. Properties of a Random Sample
6. Principles of Data Reduction
7. Point Estimation
8. Hypothesis Testing
9. Interval Estimation
10. Asymptotic Evaluations
11. Analysis of Variance and Regression
12. Regression Models
, Chapter 1
Probability Theory
“If any little problem comes your way, I shall be happy, if I can, to give you a hint or two as
to its solution.”
Sherlock Holmes
The Adventure of the Three Students
1.1 a. Each sample point describes the result of the toss (H or T) for each of the four tosses. So,
for example THTT denotes T on 1st, H on 2nd, T on 3rd and T on 4th. There are 24 = 16
such sample points.
b. The number of damaged leaves is a nonnegative integer. So we might use S = {0, 1, 2, . . .}.
c. We might observe fractions of an hour. So we might use S = {t : t ≥ 0}, that is, the half
infinite interval [0, ∞).
d. Suppose we weigh the rats in ounces. The weight must be greater than zero so we might use
S = (0, ∞). If we know no 10-day-old rat weighs more than 100 oz., we could use S = (0, 100].
e. If n is the number of items in the shipment, then S = {0/n, 1/n, . . . , 1}.
1.2 For each of these equalities, you must show containment in both directions.
a. x ∈ A\B ⇔ x ∈ A and x ∈/ B ⇔ x ∈ A and x ∈/ A ∩ B ⇔ x ∈ A\(A ∩ B). Also, x ∈ A and
x∈/ B ⇔ x ∈ A and x ∈ B c ⇔ x ∈ A ∩ B c .
b. Suppose x ∈ B. Then either x ∈ A or x ∈ Ac . If x ∈ A, then x ∈ B ∩ A, and, hence
x ∈ (B ∩ A) ∪ (B ∩ Ac ). Thus B ⊂ (B ∩ A) ∪ (B ∩ Ac ). Now suppose x ∈ (B ∩ A) ∪ (B ∩ Ac ).
Then either x ∈ (B ∩ A) or x ∈ (B ∩ Ac ). If x ∈ (B ∩ A), then x ∈ B. If x ∈ (B ∩ Ac ),
then x ∈ B. Thus (B ∩ A) ∪ (B ∩ Ac ) ⊂ B. Since the containment goes both ways, we have
B = (B ∩ A) ∪ (B ∩ Ac ). (Note, a more straightforward argument for this part simply uses
the Distributive Law to state that (B ∩ A) ∪ (B ∩ Ac ) = B ∩ (A ∪ Ac ) = B ∩ S = B.)
c. Similar to part a).
d. From part b).
A ∪ B = A ∪ [(B ∩ A) ∪ (B ∩ Ac )] = A ∪ (B ∩ A) ∪ A ∪ (B ∩ Ac ) = A ∪ [A ∪ (B ∩ Ac )] =
A ∪ (B ∩ Ac ).
1.3 a. x ∈ A ∪ B ⇔ x ∈ A or x ∈ B ⇔ x ∈ B ∪ A
x ∈ A ∩ B ⇔ x ∈ A and x ∈ B ⇔ x ∈ B ∩ A.
b. x ∈ A ∪ (B ∪ C) ⇔ x ∈ A or x ∈ B ∪ C ⇔ x ∈ A ∪ B or x ∈ C ⇔ x ∈ (A ∪ B) ∪ C.
(It can similarly be shown that A ∪ (B ∪ C) = (A ∪ C) ∪ B.)
x ∈ A ∩ (B ∩ C) ⇔ x ∈ A and x ∈ B and x ∈ C ⇔ x ∈ (A ∩ B) ∩ C.
c. x ∈ (A ∪ B)c ⇔ x ∈ / B ⇔ x ∈ Ac and x ∈ B c ⇔ x ∈ Ac ∩ B c
/ A or x ∈
c
x ∈ (A ∩ B) ⇔ x ∈
/ A∩B ⇔ x∈ / B ⇔ x ∈ Ac or x ∈ B c ⇔ x ∈ Ac ∪ B c .
/ A and x ∈
1.4 a. “A or B or both” is A∪B. From Theorem 1.2.9b we have P (A∪B) = P (A)+P (B)−P (A∩B).
, 1-6 Solutions Manual for Statistical Inference
b. “A or B but not both” is (A ∩ B c ) ∪ (B ∩ Ac ). Thus we have
P ((A ∩ B c ) ∪ (B ∩ Ac )) = P (A ∩ B c ) + P (B ∩ Ac ) (disjoint union)
= [P (A) − P (A ∩ B)] + [P (B) − P (A ∩ B)] (Theorem1.2.9a)
= P (A) + P (B) − 2P (A ∩ B).
c. “At least one of A or B” is A ∪ B. So we get the same answer as in a).
d. “At most one of A or B” is (A ∩ B)c , and P ((A ∩ B)c ) = 1 − P (A ∩ B).
1.5 a. A ∩ B ∩ C = {a U.S. birth results in identical twins that are female}
1
b. P (A ∩ B ∩ C) = 90 × 13 × 12
1.6
p0 = (1 − u)(1 − w), p1 = u(1 − w) + w(1 − u), p2 = uw,
p0 = p2 ⇒ u+w =1
p1 = p2 ⇒ uw = 1/3.
These two equations imply u(1 − u) = 1/3, which has no solution in the real numbers. Thus,
the probability assignment is not legitimate.
1.7 a.
( 2
1 − hπr
A if i = 0
P (scoring i points) = πr2 (6−i)2 −(5−i)2 i
A 52 if i = 1, . . . , 5.
b.
P (scoring i points ∩ board is hit)
P (scoring i points|board is hit) =
P (board is hit)
πr2
P (board is hit) =
A
πr2 (6 − i)2 − (5 − i)2
P (scoring i points ∩ board is hit) = i = 1, . . . , 5.
A 52
Therefore,
(6 − i)2 − (5 − i)2
P (scoring i points|board is hit) = i = 1, . . . , 5
52
which is exactly the probability distribution of Example 1.2.7.
1.8 a. P (scoring exactly i points) = P (inside circle i) − P (inside circle i + 1). Circle i has radius
(6 − i)r/5, so
2 2 2
π(6 − i) r2 π ((6−(i + 1)))2 r2 (6 − i) −(5 − i)
P (sscoring exactly i points) = − = .
52 πr2 52 πr2 52
b. Expanding the squares in part a) we find P (scoring exactly i points) = 11−2i 25 , which is
decreasing in i.
c. Let P (i) = 11−2i
25 . Since i ≤ 5, P (i) ≥ 0 for all i. P (S) = P (hitting the dartboard) = 1 by
definition. Lastly, P (i ∪ j) = area of i ring + area of j ring = P (i) + P (j).
1.9 a. Suppose x ∈ (∪α Aα )c , by the definition of complement x ̸∈ ∪α Aα , that is x ̸∈ Aα for all
α ∈ Γ. Therefore x ∈ Acα for all α ∈ Γ. Thus x ∈ ∩α Acα and, by the definition of intersection
x ∈ Acα for all α ∈ Γ. By the definition of complement x ̸∈ Aα for all α ∈ Γ. Therefore
x ̸∈ ∪α Aα . Thus x ∈ (∪α Aα )c .
George Casella
Complete Chapter Solutions Manual
are included (Ch 1 to 12)
** Immediate Download
** Swift Response
** All Chapters included
,Table of Contents are given below
1. Probability Theory.
2. Transformations and Expectations
3. Common Families of Distributions
4. Multiple Random Variables
5. Properties of a Random Sample
6. Principles of Data Reduction
7. Point Estimation
8. Hypothesis Testing
9. Interval Estimation
10. Asymptotic Evaluations
11. Analysis of Variance and Regression
12. Regression Models
, Chapter 1
Probability Theory
“If any little problem comes your way, I shall be happy, if I can, to give you a hint or two as
to its solution.”
Sherlock Holmes
The Adventure of the Three Students
1.1 a. Each sample point describes the result of the toss (H or T) for each of the four tosses. So,
for example THTT denotes T on 1st, H on 2nd, T on 3rd and T on 4th. There are 24 = 16
such sample points.
b. The number of damaged leaves is a nonnegative integer. So we might use S = {0, 1, 2, . . .}.
c. We might observe fractions of an hour. So we might use S = {t : t ≥ 0}, that is, the half
infinite interval [0, ∞).
d. Suppose we weigh the rats in ounces. The weight must be greater than zero so we might use
S = (0, ∞). If we know no 10-day-old rat weighs more than 100 oz., we could use S = (0, 100].
e. If n is the number of items in the shipment, then S = {0/n, 1/n, . . . , 1}.
1.2 For each of these equalities, you must show containment in both directions.
a. x ∈ A\B ⇔ x ∈ A and x ∈/ B ⇔ x ∈ A and x ∈/ A ∩ B ⇔ x ∈ A\(A ∩ B). Also, x ∈ A and
x∈/ B ⇔ x ∈ A and x ∈ B c ⇔ x ∈ A ∩ B c .
b. Suppose x ∈ B. Then either x ∈ A or x ∈ Ac . If x ∈ A, then x ∈ B ∩ A, and, hence
x ∈ (B ∩ A) ∪ (B ∩ Ac ). Thus B ⊂ (B ∩ A) ∪ (B ∩ Ac ). Now suppose x ∈ (B ∩ A) ∪ (B ∩ Ac ).
Then either x ∈ (B ∩ A) or x ∈ (B ∩ Ac ). If x ∈ (B ∩ A), then x ∈ B. If x ∈ (B ∩ Ac ),
then x ∈ B. Thus (B ∩ A) ∪ (B ∩ Ac ) ⊂ B. Since the containment goes both ways, we have
B = (B ∩ A) ∪ (B ∩ Ac ). (Note, a more straightforward argument for this part simply uses
the Distributive Law to state that (B ∩ A) ∪ (B ∩ Ac ) = B ∩ (A ∪ Ac ) = B ∩ S = B.)
c. Similar to part a).
d. From part b).
A ∪ B = A ∪ [(B ∩ A) ∪ (B ∩ Ac )] = A ∪ (B ∩ A) ∪ A ∪ (B ∩ Ac ) = A ∪ [A ∪ (B ∩ Ac )] =
A ∪ (B ∩ Ac ).
1.3 a. x ∈ A ∪ B ⇔ x ∈ A or x ∈ B ⇔ x ∈ B ∪ A
x ∈ A ∩ B ⇔ x ∈ A and x ∈ B ⇔ x ∈ B ∩ A.
b. x ∈ A ∪ (B ∪ C) ⇔ x ∈ A or x ∈ B ∪ C ⇔ x ∈ A ∪ B or x ∈ C ⇔ x ∈ (A ∪ B) ∪ C.
(It can similarly be shown that A ∪ (B ∪ C) = (A ∪ C) ∪ B.)
x ∈ A ∩ (B ∩ C) ⇔ x ∈ A and x ∈ B and x ∈ C ⇔ x ∈ (A ∩ B) ∩ C.
c. x ∈ (A ∪ B)c ⇔ x ∈ / B ⇔ x ∈ Ac and x ∈ B c ⇔ x ∈ Ac ∩ B c
/ A or x ∈
c
x ∈ (A ∩ B) ⇔ x ∈
/ A∩B ⇔ x∈ / B ⇔ x ∈ Ac or x ∈ B c ⇔ x ∈ Ac ∪ B c .
/ A and x ∈
1.4 a. “A or B or both” is A∪B. From Theorem 1.2.9b we have P (A∪B) = P (A)+P (B)−P (A∩B).
, 1-6 Solutions Manual for Statistical Inference
b. “A or B but not both” is (A ∩ B c ) ∪ (B ∩ Ac ). Thus we have
P ((A ∩ B c ) ∪ (B ∩ Ac )) = P (A ∩ B c ) + P (B ∩ Ac ) (disjoint union)
= [P (A) − P (A ∩ B)] + [P (B) − P (A ∩ B)] (Theorem1.2.9a)
= P (A) + P (B) − 2P (A ∩ B).
c. “At least one of A or B” is A ∪ B. So we get the same answer as in a).
d. “At most one of A or B” is (A ∩ B)c , and P ((A ∩ B)c ) = 1 − P (A ∩ B).
1.5 a. A ∩ B ∩ C = {a U.S. birth results in identical twins that are female}
1
b. P (A ∩ B ∩ C) = 90 × 13 × 12
1.6
p0 = (1 − u)(1 − w), p1 = u(1 − w) + w(1 − u), p2 = uw,
p0 = p2 ⇒ u+w =1
p1 = p2 ⇒ uw = 1/3.
These two equations imply u(1 − u) = 1/3, which has no solution in the real numbers. Thus,
the probability assignment is not legitimate.
1.7 a.
( 2
1 − hπr
A if i = 0
P (scoring i points) = πr2 (6−i)2 −(5−i)2 i
A 52 if i = 1, . . . , 5.
b.
P (scoring i points ∩ board is hit)
P (scoring i points|board is hit) =
P (board is hit)
πr2
P (board is hit) =
A
πr2 (6 − i)2 − (5 − i)2
P (scoring i points ∩ board is hit) = i = 1, . . . , 5.
A 52
Therefore,
(6 − i)2 − (5 − i)2
P (scoring i points|board is hit) = i = 1, . . . , 5
52
which is exactly the probability distribution of Example 1.2.7.
1.8 a. P (scoring exactly i points) = P (inside circle i) − P (inside circle i + 1). Circle i has radius
(6 − i)r/5, so
2 2 2
π(6 − i) r2 π ((6−(i + 1)))2 r2 (6 − i) −(5 − i)
P (sscoring exactly i points) = − = .
52 πr2 52 πr2 52
b. Expanding the squares in part a) we find P (scoring exactly i points) = 11−2i 25 , which is
decreasing in i.
c. Let P (i) = 11−2i
25 . Since i ≤ 5, P (i) ≥ 0 for all i. P (S) = P (hitting the dartboard) = 1 by
definition. Lastly, P (i ∪ j) = area of i ring + area of j ring = P (i) + P (j).
1.9 a. Suppose x ∈ (∪α Aα )c , by the definition of complement x ̸∈ ∪α Aα , that is x ̸∈ Aα for all
α ∈ Γ. Therefore x ∈ Acα for all α ∈ Γ. Thus x ∈ ∩α Acα and, by the definition of intersection
x ∈ Acα for all α ∈ Γ. By the definition of complement x ̸∈ Aα for all α ∈ Γ. Therefore
x ̸∈ ∪α Aα . Thus x ∈ (∪α Aα )c .
