J. Chen, STAT551B
1
Solution of Homework 2
Q: 5.3.4
a)
y
P ( 1.64 < < y + 2.33 ) = P (1.64 < < 2.33)
y
n
n
n
= P (Z < 2.33) P (Z < 1.64)
= 0.9901 0.0505 = 93.96%
b)
P ( < < y + 2.58 ) = P (Z < 2.58) =
J. Chen, STAT551A 2013
1
Final Exam is Tuesday, December 17, 10.30am-12.30pm, In-class
Covering:
1. Discrete random variables and continuous random variables
2. Probability density function, cumulativ
J. Chen, STAT551B
1
Solution of Homework 1
Q: 5.2.1 The likelihood function is
8
ki (1 )1ki =
L() =
8
i=1
ki
8
i=1
(1 )8
ki ,
i=1
and log likelihood function is
8
l() = ln L() =
8
ki ln + (8
i=1
ki
J. Chen, STAT551B
1
Solution of Homework 5
Q: 7.3.7
a) F0.5,6,7 = 0.983
b) F0.001,15,5 = 0.132
c) F0.90,2,2 = 9.000
Q: 7.3.9
a) P (0.109 < F4,6 < x) = 0.95 P (F4,6 < x) = 0.975 x = 6.23
b) P (0.427 <
J. Chen, STAT551B
1
Solution of Homework 3
Q: 6.2.1
a) H0 : = 120 vs. H1 : < 120
Since Y = 114.2, n = 25, = 18, = 0.08,
By Normal table, we have Z0.08 = 1.41
Then Z =
Y 0
/ n
=
114120
18/5
= 1.66 < 1.
J. Chen, STAT551B
1
Solution of Homework 4
Q: 6.3.1
a) H0 : p = 0.4 vs. H1 : p > 0.4
Z=
p p0
=
p0 (1p0 )
n
24
52
0.4
0.40.6
52
= 0.9058 < Z0.05 = 1.645
Failed to reject H0 .
b)
p value = P (Z > Zobs
J. Chen, Handout 3, STAT551B
1
Probability and Mathematical Statistics
http:/www-rohan.sdsu.edu/ jchenyp/
Chapter 5. Estimation
1, Introduction
Let X1 , , Xn be random sample (iid) having normal distr
J. Chen, Handout 2, STAT551B
1
Probability and Mathematical Statistics
Introduction of Statistical Inference
The purposes of the course
1, Random Variable and Probability Distribution
Random variable
J. Chen, STAT551B, 2010
1
Solution of Homework 7
Q: 11.2.1
15
i=1
Since
15
i=1
xi = 249.8,
x2 = 4200.56,
i
15
i=1
yi = 1200.6,
15
i=1
xi yi = 20127.47,
The LS estimators are
2
15 Xi Yi nX Yi
15 20127