Presently, technology enables various businesses to operate globally. Even the business
with the smallest operations can have customers in various different zones. Various counties and
cultures sometimes have particular preferences; nevertheless, the good
Homework 1 Solutions
10.5
From class, we know that E[Fn (u)] = F (u) and E[Fn (v)] = F (v). Additionally, we have
#
" n
n
1X
1X
I(,u] (Xi )
I(,v] (Xj )
E[Fn (u)Fn (v)] = E
n i=1
n j=1
n
n
1 XX
=
E I(,u] (Xi )I(,v] (Xj )
2
n i=1 j=1
n
n
1 X
1 XX
=
P (Xi u
Homework3 solution
Courtesy of Crane & Fitzgerald with Modification
15.6) (a) A model for the data in this CRBD experiment would be y i j = + i + j + i j (i = 1, 2, 3;
j = 1, 2, , 7)
y
where i j is the typing score for the jth subject listening to the ith
Homework 2 Solutions
1.3-4
a. There are 4 5 2 = 40 possible experiments.
b. There are 2 2 2 = 8 possible experiments.
1.3-14
(n 1)!
n1
n1
(n 1)!
+
+
=
r!(n r 1)! (r 1)!(n 1 (r 1)!
r
r1
(n 1)!(n r) + r(n 1)!
=
r!(n r)!
n!
=
r!(n r)!
n
=
r
1.3-20
a. For
Homework 6 Solutions
6.5-1
a. p = 0.03738318
b. (0.02270901, 0.05205734)
c. (0.02524793, 0.0550219)
d. (0.02504435, 0.055451)
p
e. p + 1.645 p(1 p)/642 = 0.04969899.
6.5-3
a. p = 690/1000 = .69.
b. There are varying acceptable answers depending on which f
Homework 5 Solutions
6.3-12
For this data, xd = 0.07875 and sd = 0.2549222.
a. A point estimate of d is xd = 0.07875.
b. t.05;23 = 1.714, so the confidence interval is (0.01043932, ).
c. Since the lower bound is negative, there is not enough evidence to s
Stat. 6304 (Fall, 2011) Homework Assignment 3
1. (Problem 4-12) If X is a continuous random variable, argue that P (a < X b) = P (a < X b) =
P (a X < b) = P (a < X < b).
2. (Problem 4-28) Suppose f (x) = x/8 for 0 < x < 4.
a. Show that f (x) is a probabil
Homework 1 Solutions
5.1-4
Parts (c) and (d) will vary.
a.
0, x 0
x2 , 0 < x < 1
FX (x) =
1, x 1
b. We can simulate X by setting Y = FX (X) = X 2 X = Y .
We can then generate random U(0, 1) realizations and plug them in for
Y in the above to get realizat
Homework 3 Solutions
6.1-1
Let X1 , X2 , . . . , Xn be a random sample from the N (, 2 ), where
= cfw_ : < < .
Then the likelihood function is given by
P
(xi )2
2 n/2
L() = (2 )
exp
,
2 2
.
Then, the log-likelihood function is
P
2
`() = (n/2) ln(2 )
Homework 2 Solutions
5.6-1
Assuming n = 12 is large enough for the CLT to apply,
1
N 1/2,
.
X
12 12
Therefore
2/3) = P
P (1/2 X
1/2 1/2
2/3 1/2
Z
1/12
1/12
= P (0 Z 2) = 0.4472.
5.6-8
= E(X) = 24.43
a. E(X)
= V ar(X)/n = 2.20/30 = 0.0733
b. V ar(X)
c.
Stat. 6304 (Fall, 2011) HW 8 Solutions
10-55. Use Stat > Basic Statistics > 2 Variances for this problem.
a. H0 : men = women . Ha : men 6= women . = 0.02. Assuming that the individual values in
both groups are normal, F = 0.92 and the p-value = 0.843. Th
Stat. 6304 (Fall, 2011) HW 6 Solutions
9-99. This problem can be done in MINITAB, using STAT > Tables > Chi-Square goodness of fit (1
Var). Put the observed data in a column and the probabilities below (the probabilities X equals the
given values, except
Homework 5 Solutions
3.5-1
a. Here, X Gamma( = 10, = 3/2), so the p.d.f. is
f (x) =
b.
M (t) =
x9 e2x/3
(10)( 23 )10
1
1
=
,
(1 t)
(1 (3/2)t)10
t < 2/3
X = = 10(3/2) = 15
2
X
= 2 = 10(3/2)2 = 45/2
3.5-8
Here, X P ois( = 16).
a. W Gamma( = 7, = 1/16).
b. H
Homework 4 Solutions
2.5-1
a. This is Binomial
with n = 1 and p = 2/3. Thus, = 2/3, 2 = 2/9
q
and = 29
b. Thisq
is Binomial with n = 12 and p = 3/4. Thus, = 9, 2 = 9/4 and
= 94
2.5-5
This is an example of the Geometric distribution with p = 1/5.
P (X = 4
Stat. 6304 (Fall, 2011) HW 3 Solutions
1. P (a X b) = P (a < X b) + P (X = a) = P (a < X b) + 0 = P (a < X b). P (a < X b) =
P (a < X < b) + P (X = b)P (a < X < b) + 0 = P (a < X < b). P (a X b) = P (a X < b) + P (X =
b) = P (a X < b) + 0 = P (a X < b).
2
Stat. 6304 (Fall, 2011) HW 2 Solutions
3.3 The possible values are cfw_0, 1, 2, ., 99999.
3.4 The possible values are cfw_0, 1, 2, ., 10.
3.24 The probability mass function is
0.3
p(x) = 0.5
0.2
0
if x=$10,000,000
if x=$5,000,000
if x = $1, 000, 000
othe
Stat. 6304 (Fall, 2011) HW 1 Solutions
6.1 No, the sample mean does not have to correspond to a data point. For example, suppose the data set
is cfw_0, 1. The sample mean is 0.5, which is not part of the data set.
6.2 No. Using the Jordan data set from cl
Associate Professor J. Kerr
Final Exam
STAT 6501
Fall 2015
Instructions: You have 110 minutes to complete the exam. Please write all solutions in
your greenbook starting each new problem on a new page, including all pertinent work for
full credit. Partial
Homework 2 Solutions
8.6
a. Since f (x) =
n
x
px (1 p)nx , the log-likelihood function is
(p) = ln
n
+ x ln(p) + (n x) ln(1 p).
x
The rst derivative w.r.t. p is
d (p)
x nx
=
dp
p
1p
1
x
which, when set to zero, yields p 1 = n 1 p = n .
x
The second deri
Associate Professor J. Kerr
Midterm II
STAT 6501
Fall 2015
Instructions: You have 110 minutes to complete the exam. Please write all solutions in
your bluebook starting each new problem on a new page, including all pertinent work for
full credit. Partial
Homework 3 Solutions
1. Find the m.l.e. of based on a random sample X1 , X2 , . . . , Xn from
each of the following p.d.f.s.
(a)
f (x|) = x1 ,
0 < x < 1, 0 <
n
() = n ln() + ( 1)
ln(xi )
i=1
d ()
n
= +
d
and
ln(xi )
i=1
d2 ()
n
= 2
d2
So
n
ln(xi )
=
and
Associate Professor J. Kerr
Midterm I
STAT 6501
Fall 2015
Instructions: You have 110 minutes to complete the exam. Please write all solutions in
your bluebook starting each new problem on a new page, including all pertinent work for
full credit. Partial c
Homework 5 Solutions
8.6
b. The Cramr-Rao inequality is
e
V (T )
1
n I()
[here n = 1]
where (from homework 2)
I(p) = E( (p, X) = E
X
nX
2
p
(1 p)2
=
n
E(X) n E(X)
+
=
.
2
2
p
(1 p)
p(1 p)
Therefore,
p(1 p)
.
n
From homework 2, the m.l.e. of p is p = X s
Homework 4 Solutions
8.62
We have f (x|) = ex ; x > 0, > 0 with x = 5.1, n = 20. The prior is
1 e .
h() =
()
Therefore, the posterior is
f (|x1 , . . . , xn ) f (x1 , . . . , xn |)h() n+ e(+
xi )
.
We see that this is proportional to the gamma p.d.f. with
Associate Professor J. Kerr
Take Home Midterm
STAT 6502
Spring 2014
Instructions: A hard copy and digital copy of your solutions are due by the start of the
in-class portion of the midterm. There is to be no collaboration of any kind on this exam.
You mus
What are differences and describe the three ways that a researcher can investigate for differences?
Every marketer relies on market segmentation, which holds that within a product market, there are
different types of customers who have different needs and
Why is Japan so heavily affected by the global economic crisis?. (2013). Retrieved from
http:/www.voxeu.org/article/why-has-japan-been-so-hard-hit-global-crisis?
quicktabs_tabbed_recent_articles_block=1
What Companies Benefit (Or Not) From A Weaker Yen?.
The Kindle also has several functions that customers can find helpful while using it. They
can bookmark pages, highlight selections of the text or even type notes. These features allow the
Kindle to potentially replace hardcopy textbooks in the future, so
From my point of view, it is imperative for Sony to manufacture more products in the United States
and Europe because it will not only increase their sales but will also increase the revenue while
decreasing costs. In order to minimize the risks and contr