Solutions Manual for
Statistical Inference, Second Edition
George Casella Roger L. Berger
University of Florida North Carolina State University
Damaris Santana
University of Florida
,0-2 Solutions Manual for Statistical Inference
“When I hear you give your reasons,” I remarked, “the thing always appears to me to be so
ridiculously simple that I could easily do it myself, though at each successive instance of your
reasoning I am baffled until you explain your process.”
Dr. Watson to Sherlock Holmes
A Scandal in Bohemia
0.1 Description
This solutions manual contains solutions for all odd numbered problems plus a large number of
solutions for even numbered problems. Of the 624 exercises in Statistical Inference, Second Edition,
this manual gives solutions for 484 (78%) of them. There is an obtuse pattern as to which solutions
were included in this manual. We assembled all of the solutions that we had from the first edition,
and filled in so that all odd-numbered problems were done. In the passage from the first to the
second edition, problems were shuffled with no attention paid to numbering (hence no attention
paid to minimize the new effort), but rather we tried to put the problems in logical order.
A major change from the first edition is the use of the computer, both symbolically through
Mathematicatm and numerically using R. Some solutions are given as code in either of these lan-
guages. Mathematicatm can be purchased from Wolfram Research, and R is a free download from
http://www.r-project.org/.
Here is a detailed listing of the solutions included.
Chapter Number of Exercises Number of Solutions Missing
1 55 51 26, 30, 36, 42
2 40 37 34, 38, 40
3 50 42 4, 6, 10, 20, 30, 32, 34, 36
4 65 52 8, 14, 22, 28, 36, 40
48, 50, 52, 56, 58, 60, 62
5 69 46 2, 4, 12, 14, 26, 28
all even problems from 36 − 68
6 43 35 8, 16, 26, 28, 34, 36, 38, 42
7 66 52 4, 14, 16, 28, 30, 32, 34,
36, 42, 54, 58, 60, 62, 64
8 58 51 36, 40, 46, 48, 52, 56, 58
9 58 41 2, 8, 10, 20, 22, 24, 26, 28, 30
32, 38, 40, 42, 44, 50, 54, 56
10 48 26 all even problems except 4 and 32
11 41 35 4, 20, 22, 24, 26, 40
12 31 16 all even problems
0.2 Acknowledgement
Many people contributed to the assembly of this solutions manual. We again thank all of those
who contributed solutions to the first edition – many problems have carried over into the second
edition. Moreover, throughout the years a number of people have been in constant touch with us,
contributing to both the presentations and solutions. We apologize in advance for those we forget to
mention, and we especially thank Jay Beder, Yong Sung Joo, Michael Perlman, Rob Strawderman,
and Tom Wehrly. Thank you all for your help.
And, as we said the first time around, although we have benefited greatly from the assistance and
,ACKNOWLEDGEMENT 0-3
comments of others in the assembly of this manual, we are responsible for its ultimate correctness.
To this end, we have tried our best but, as a wise man once said, “You pays your money and you
takes your chances.”
George Casella
Roger L. Berger
Damaris Santana
December, 2001
, 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).
