STAT 409
Fall 2011
Examples for 09/19/2011
The sample proportion:
p= x
n
where x is the number of elements in the sample found to belong
to the category of interest (the number of "successes"), and n is
the sample size.
E( P ) = p,
p (1 p ) ,
n
Var( P ) =
STAT 409
1.
Examples for 09/09/2011
Fall 2011
Suppose the lifetime of a particular brand of light bulbs is normally distributed with
standard deviation of = 75 hours and unknown mean.
a)
What is the probability that in a random sample of 49 bulbs, the ave
STAT 409
1.
Practice Problems 4
Let X 1 , X 2 , , X n be a random sample from the distribution with the
probability density function
f X ( x ; ) = ( + 1 ) (1 x ) ,
a)
n
Y = ln ( 1 X i
i =1
Recall:
)
0 < x < 1,
> 1.
has Gamma ( = n, usual =
1
)
+1
distri
Discussion session 1
September 5, 2017
1. Let X1 , X2 , . . . , Xn be a random sample from:
N (, 2 )
is the MLE for .
,where is unknown and is known. Show that
=X
2. Let X1 , X2 , . . . , Xn be a random sample from:
N (, 2 )
,where is known and is unkno
Discussion session 5
October 3, 2017
1. Province Ranch is an insurance company that provides homeowners policies to tenants and nontenants in Neverland. A random sample of 16 policies for tenants yielded the sample mean basic
premium of 250 and the sample
Examples for 5.5, 7.1
X 1 , X 2 , , X n be i.i.d. N , 2
Let
.
Let
X=
S
2
X i
n
=
X 1 + X 2 + . + X n
( sample mean )
n
(X i X )
=
2
( sample variance )
n 1
Then
X and S 2 are independent;
2
X has N ,
n
X
distribution;
has N ( 0 , 1 ) distribution;
n
(
STAT 409
Fall 2011
Examples for 08/29/2011
In general, if X 1 , X 2 , , X n is a random sample of size n from a continuous distribution with
cumulative distribution function F ( x ) and probability density function f ( x ), then
F max X ( x ) = P ( max X
STAT 409
5.
Let X 1 , X 2 , , X n be a random sample of size n from a uniform
distribution on the interval ( 0 , ).
1
f ( x; ) =
0
E( X ) =
a)
otherwise
Var ( X ) =
2
x<0
0
x
F( x; ) =
1
0< x<
2
0< x<
x>
12
~
Obtain the method of moments estimator of ,
STAT 409
p.m.f. or p.d.f.
1.
Fall 2011
Examples for 08/24/2011
f ( x ; ),
.
Suppose = cfw_ 1, 2, 3 and the p.m.f.
parameter space.
f ( x ; ) is
= 1:
f ( 1 ; 1 ) = 0.6,
f ( 2 ; 1 ) = 0.1,
f ( 3 ; 1 ) = 0.1,
f ( 4 ; 1 ) = 0.2.
= 2:
f ( 1 ; 2 ) = 0.2,
f
STAT 409
1.
Examples for 11/02/2011 (1)
Fall 2011
The developers of a math proficiency exam to be used at Anytown State University
believe that 60% of all incoming freshmen will be able to pass the exam. In a random
sample of 200 incoming freshmen, 105 pa
STAT 409
Fall 2011
Examples for 10/14/2011 (2)
H 0 true
H 0 is NOT true
Do Not Reject H 0
Type II Error
Reject H 0
Type I Error
= significance level = P ( Type I Error ) = P ( Reject H 0 | H 0 is true )
= P ( Type II Error ) = P ( Do Not Reject H 0 | H
Left tailed test
H 0 : p = p0
vs.
H1 : p < p0
If H 0 is TRUE :
Use p 0 .
Reject H 0
Do NOT Reject H 0
Type I Error
Correct decision
a
0
a+1
n
Rejection Rule for a Left tailed test:
Find a such that P( Y a ) = CDF @ a .
( using Binomial ( n , p 0 ) tables
STAT 409
Examples for 10/07/2011 (1)
Comparing Two Population Proportions
Fall 2011
H0 : p1 = p2
Test Statistic:
p1 p 2
z=
1.
p (1 p ) 1 n + 1 n
1
2
where
,
Y +Y n p + n p
p= 1 2 = 1 1 2 2
n1 + n 2
n1 + n 2
In a comparative study of two new drugs, A and
STAT 409
1.
Fall 2011
Examples for 09/16/2011
A machine makes -inch ball bearings. In a random sample of 41 bearings, the
sample standard deviation of the diameters of the bearings was 0.02 inch. Assume
that the diameters of the bearings are approximately
STAT 409
Fall 2011
Examples for 09/14/2011
Two independent samples:
X1, X2, , Xn
Y1, Y2, , Yn
1
from population 1
2
from population 2
mean 1 , std. dev. 1
mean 2 , std. dev. 2
If n 1 and n 2 are large, or population 1 and population 2 are approximately no
STAT 409
1.
Fall 2011
Examples for 09/12/2011
A manufacturer of TV sets wants to find the average selling price of a
particular model. A random sample of 25 different stores gives the mean
selling price as $342 with a sample standard deviation of $14. Ass
STAT 409
1.
Fall 2011
Examples for 09/02/2011
Let X 1 , X 2 , , X n be a random sample of size n from a Poisson distribution
with mean . That is,
k e
f ( k; ) = P ( X 1 = k ) =
,
k = 0, 1, 2, 3, .
k!
a)
Use Factorization Theorem to find Y = u ( X 1 , X 2
1 4.
If the random variable Y denotes an individuals income, Paretos law claims
that P ( Y y ) = k
It follows that
y , where k is the entire populations minimum income.
1
fY( y ) = k
+1
y k;
,
y
> 0.
The income information has been collected on a rand