Assignment 8 Solutions
1. The hypotheses are H0: 100 and H1: < 100. The sample size is 12, so there are 11
degrees of freedom. The significance level is .01, so reject H0 if t0 < t.01 = 2.718.
The test statistic is
88.92100
t 0=
=2.505
15.32/ 12
so we d
STAT
4710
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Exam
III
A
25 Aprit 2013
lJniversity of Missoufi
Department of Statistics
Introduction to Mathem atcal Statistics
Stat 4710 /771,0
Spring 201
Examination
,
;
III A
April 26,201,3
Vrx
Student Name
Student Number
Joke:
Wq F,t $,^a1u
,/
STAT
47rcfi7rc
Examination
IA
l5h
March 2013
UNTVERSITY of MISSOURI
Department of Statistics
Introduction to Mathematical Statistics
Stat 471.0 /7710
Spring 201 j.
Examination 1 A
15'h Ma rch 201,3
Time:
%
50 min
Insffuctions
1
2
'
Attempt all question
2 5 2
30. 25 =3,268,760; 5 5 5 = 0 =775,200
10 3 103 3 7
no; something that occurs by chance with probability .1667 is not unusually rare
32. (a) 26510=118,813,760
' 5
(b) [3]5=50
1
(0) E6
33. 2'6; 216 1 , since only one possibility consists of no por
69. Theorem 3.8.1 consists of3 parts.
1. Prove mx (t) = ekwl).
°°  x no k (x
m.<r)=E[e'X]=zeue*k (k'e)
F0 x! x.
(ke)2 (ke')3
=e" 1+ke+
+
2! 3!
I l l
= e k "e v1a the Maclaurm serles for e
_ k(e')
e
2. Prove E[X]=k.
de (t)
dt = ekeI) (ke)
[=0
=
WB5~1. A large insurance company serves a number of customers who have purchased
both a homeowner's policy and an automobile policy from the agency. For each type of policy, a
deductible amount must be specied. For an automobile policy, the choices are $1
(1) no, [1 = 3 lies in the condence interval
Review Exercises
1 9+1 '
5W) lJf(x)dx=I(1+9)x9dx=(l+9);+l
regardless of the value of 6?
1+6
_ :1,
6+1
0
I=l
a 9, x9' 1+6
bEX= 10 dx=16 +dx=16 =
U [1;x(+)x (+)0Ix (+)6+109+2
1+6
(0) Let X=m :>(6+2)X
WBSl. In a pollution study, a random sample of 8 air specimens contained an average of
2.26 microg/cum of suspended matter with a standard deviation of .56 microg/cum. Assume
that suspended matter amounts are normally distributed.
(a) Find a 99% cond
Assignment 7 Solutions
1. a. The book showed that the mle for for a normal distribution is the sample mean
X .
From Theorem 7.3.4, the sample mean is normally distributed with mean and variance
2 16
= =4 /3 . From Theorem 7.1.1, we know that it is unbias
STAT 4710/7710
Assignment 9 Solutions
1. (20 points) A business journal traveler publisher plans to survey a sample
of the subscribers to estimate the proportion p with annual household incomes
over $100,000.
a. How many subscribers must be surveyed to ob
Question
t
A
Final Exam
STAT 4710fl710
May 2013
Y) has ioiot density fi.rnction
Suppose that the tandom vectot (X,
_)
fxv(x,y)
=
r,
<x<
W
0
:'19,
L.
zerootherwise.
0.y.nJ
t.
o z K3
0>3r
',3.*
Ytcfw_
b.
Show that the marginal distribution
p*
cnl

c
o