UNIVERSITY OF REGINA
DEPARTMENT OF MATHEMATICS & STATISTICS
Statistics 160 001, 002
Final Examination
201410
Time : 3 hours
NAME :
Instructor: Dr. Y. Zhao (001)
STUDENT NO. :
Dr. R. McIntosh (002)
SECTION :
Show all your work on the pages of this examinat
Stat 351 Fall 2008
Assignment #2 Solutions
Problem 2: We verify that Q(A) is a probability by checking the three conditions.
Since P () = 0, we conclude
Q() = P (|B) =
P ( B)
P ()
=
=0
P (B)
P (B)
since cfw_ B = . Similarly,
Q() = P (|B) =
P (B)
P ( B)
=
Stat 351 Fall 2008
Assignment #3 Solutions
Problem #3, page 27: Suppose that T t(n) so that the density of T is given by
( n+1 )
x2
fT (x) = 2 n 1 +
n
n ( 2 )
(n+1)/2
, < x < .
Let Y = T 2 . If y 0, then the distribution function of Y is given by
FY (y) =
Stat 351 Fall 2008
Solutions to Assignment #4
Problem #3, page 55: Suppose that X + Y = 2. By denition of conditional density,
fX,X+Y (x, 2)
.
fX+Y (2)
fX|X+Y =2 (x) =
We now nd the joint density fX,X+Y (x, 2). Let U = X and V = X + Y so that X = U and
Y
Stat 351 Fall 2008
Assignment #1 Solutions
2. (a) If X Unif[1, 3], then FX (x) =
x1
2
for 1 x 3, and if Y N (0, 1), then
y
1
2
eu /2 du
2
FY (y) =
for < y < . Since X and Y are independent, we conclude that
FX,Y (x, y) = FX (x) FY (y) =
x1
2
y
1
2
eu /2
Stat 351 Fall 2008
Solutions to Assignment #5
1. (a) We begin by calculating E(Y1 ). That is,
E(Y1 ) = 1 P (Y = 1) + (1) P (Y = 1) = p (1 p) = 2p 1.
We now notice that Sn+1 = Sn + Yn+1 . Therefore,
E(Sn+1 |X1 , . . . , Xn ) = E(Sn + Yn+1 |X1 , . . . , Xn
Stat 351 Fall 2008
Assignment #9 Solutions
1. (a) By Denition I, we see that X1 X2 is normally distributed with mean
E(X1 X2 ) = E(X1 ) E(X2 ) = 0
and variance
var(X1 X2 ) = var(X1 ) + 2 var(X2 ) 2 cov(X1 , X2 ) = 1 + 2 22 = 1 2 .
That is, X1 X2 = Y where
Stat 351 Fall 2008
Assignment #8 Solutions
1. Recall from Stat 251 that if X N (0, 1), then X 2 2 (1). Furthermore, recall that if Z1 , . . . , Zn
are independent with Zj 2 (pj ), then
n
Zj 2 (p1 + + pn ).
j=1
(That is, the sum of independent chi-squared
Stat 351 Fall 2008
Assignment #6 Solutions
1. We nd
P (X < Y < Z) =
f (x, y, z)dx dy dz.
cfw_x<y<z
Since f (x, y, z) =
sion are
0
z
0
y
0
x
for 0 < x, y, z < , the six equivalent iterated integrals for this expres
exyz dx dy dz =
0
z
=
0
exyz
y
exyz dy dz
Stat 351 Fall 2007
Assignment #7 Solutions
1. If X1 , X2 are independent N (0, 1) random variables, then by Denition I, Y1 = X1 + 3X2 2 is
normal with mean E(Y1 ) = E(X1 ) + 3E(X2 ) 2 = 2 and variance var(Y1 ) = var(X1 + 3X2 2) =
var(X1 ) + 9 var(X2 ) + 6
Statistics 252 Winter 2006 Midterm #3 Solutions
1. (a) By denition, the signicance level is the probability of a Type I error; that is, the proba2
bility under H0 that H0 is rejected. Hence, since 4S2 2 ,
4
4S 2
4 1.945
>
1
1
= PH0 (reject H0 ) = P (S 2
Statistics 451 (Fall 2013)
Basic Set Theory
Let N denote the set of natural numbers, namely N = cfw_1, 2, 3, . . ..
Let Z denote the set of integers, namely Z = cfw_. . . , 3, 2, 1, 0, 1, 2, 3, . . ..
The non-negative integers are cfw_0, 1, 2, 3, . . . =
Statistics 451 (Fall 2013)
Some More Basic Set Theory
Suppose that f : Rn Rm is a function.The range of f is
cfw_y Rm : f (x) = y for some x Rn .
Note that the range of f is a subset of Rm . If A Rn , then the image of A under f is the
subset of the range
Stat 351 Fall 2009
Assignment #1 Solutions
Problem 1:
(a) If X Unif[1, 3], then FX (x) =
x1
2
for 1 x 3, and if Y N (0, 1), then
y
FY (y) =
1
2
eu /2 du
2
for < y < . Since X and Y are independent, we conclude that
FX,Y (x, y) = FX (x) FY (y) =
x1
2
y
1
Stat 351 Fall 2009
Assignment #5 Solutions
1. If X1 , X2 are independent N (0, 1) random variables, then by Denition I, Y1 = X1 + 3X2 2 is
normal with mean E(Y1 ) = E(X1 ) + 3E(X2 ) 2 = 2 and variance var(Y1 ) = var(X1 + 3X2 2) =
var(X1 ) + 9 var(X2 ) + 6
Statistics 351 Fall 2006 Midterm #2 Solutions
1. (a) Recall that a square matrix is strictly positive denite if and only if the determinants
of all of its upper block diagonal matrices are strictly positive. Since
2 2
2 3
=
we see that det( 1 ) = det(2) =
Stat 351 Fall 2009
Assignment #6 Solutions
1. Recall from Stat 251 that if X N (0, 1), then X 2 2 (1). Furthermore, recall that if Z1 , . . . , Zn
are independent with Zj 2 (pj ), then
n
Zj 2 (p1 + + pn ).
j=1
(That is, the sum of independent chi-squared
Stat 351 Fall 2009
Chapter 2 Solutions
Problem #2. Suppose that X + Y = 2. By denition of conditional density,
fX,X+Y (x, 2)
.
fX+Y (2)
fX|X+Y =2 (x) =
We now nd the joint density fX,X+Y (x, 2). Let U = X and V = X + Y so that X = U and
Y = V U . The Jaco
Stat 351 Fall 2009
Chapter 5 Solutions
Problem #2. Let X = (X, Y ) with
0
,
0
XN
1
1
,
and consider the change of variables to polar coordinates (R, ) . The inverse of this transformation
is given by
x = r cos and y = r sin
for 0 < 2, r > 0 so that the
Stat 351 Fall 2009
Chapter 4 Solutions
Problem #3. If 0 y 1/2, then
1y
1y
2 dz = 2(1 2y).
fX(1) ,X(2) (y, z) dz =
fY (y) =
y
y
On the other hand, if 1/2 y 1, then
y
y
fY (y) =
1y
2 dz = 2(2y 1).
fX(1) ,X(2) (y, z) dz =
1y
Problem #5. Since E[F (X(n) ) F (
Stat 351 Fall 2009
Chapter 1 Solutions
Problem #2. Suppose that X C(m, a) so that the density of X is given by
fX (x) =
1
a
2
, < x < .
a + (x m)2
Let Y = 1/X. If X = 0, then for < y < the distribution function of Y is
1/y
FY (y) = P (Y y) = P (1/X y) =
Stat 257: Solutions to Assignment #3
(3.1) Let X denote the assessed yields, and let Y denote the actual yields. Our goal is to estimate
YT . In addition to the data clearly given in the problem, note that we also know the following:
N = 280, n = 25, and
Stat 257: (Selected) Solutions to Assignment #5
(5.2) We begin with the observation that
M
L
M
M
(Yij Y )2 + 2
i=1 j=1
L
L
(Yij Y )(Yik Y ) =
i=1 j<k
(Yij Y )(Yik Y ).
()
i=1 j=1 k=1
The two expressions on the left side of () are easy to deal with. From e
Stat 257: Solutions to Assignment #2
(2.1) In order to calculate y, we must pool the averages from each of the two simple random samples
remembering to weight by the appropriate sample sizes. That is,
n1
n2
300
500
y +
y =
2.98 +
3.42 = 3.255.
n1 + n2 1
Stat 257: Solutions to Assignment #4
(4.1) From the data given in the problem, we have N = 235 public libraries which are grouped
into three strata: strata 1 consists of N1 = 74 small libraries, strata 2 consists of N2 = 133 medium
libraries, and strata 3
Stat 252.01 Winter 2007
Assignment #6 Solutions
1. Textbook
(9.73) If Y1 , . . . , Yn are i.i.d. exponential(), then each Yi has common density
1
fY (y|) = ey/ ,
y > 0.
Therefore, the likelihood function is
n
n
fY (yi |) =
L() =
i=1
i=1
1
1 yi /
e
= n exp
Stat 252 Winter 2007
Assignment #5 Solutions
1. Textbook
(8.40) (a) By denition, if 0 < y < , then
y
FY (y) =
y
fY (u) du =
0
2( u)
(2u u2 )
du =
2
2
y
=
0
(2y y 2 )
2y y 2
=
2.
2
That is,
FY (y) =
0,
2y
y 0,
y2
,
2
0 < y < 1,
y .
1,
(b) If U = Y /, then
Stat 252 Winter 2007
Assignment #4 Solutions
2. Textbook
(8.30) If = Y , then E() = E(Y ) = so that Y is an unbiased estimator of . Since the
is
standard error of
= Var(Y ) =
,
n
a natural guess for the estimated standard error is
=
.
n
(8.36) (a) If