Stat 804: Practice Exam
1. Consider the following nite, stratied population of size N = 4.
Stratum
Farm
Acreage xk
Corn Acreage yk
1
1
24
12
2
36
24
3
48
36
4
72
60
tx = 180
ty = 132
known
unknown
2
(a) SI of size 3.
Sample Measurements
Prob[S = s]
HT est
Stat 804
Chapter 9
9. Two Phase Sampling
Use of Auxiliary Information x
Design (pps, ST)
Estimation (Ratio, Regression)
When the auxiliary information is not available, we may
use the two phase sampling design.
Sample selection
Phase 1 : Using a design
Stat 804
Chapter 6
6. The Regression Estimator
Auxiliary Information
xk = (x1k , , xJk ) .
Assume known for every k U .
Difference Estimator
Suppose
yk =
J
j =1
Aj xjk = xk A
.
=yk ,
where A = (A1, , AJ ) . Then
ty =
U yk
=
ty,dif f =
V
ty,dif f =
0
Sta
Stat 804
Chapter 5
5. More Complex Estimation Problem
Parameter: Function of Totals
= f (t1, , tq ) ,
where
tj =
yjk ,
kU
j : j th variable of study, k : kth element.
Example)
1. Mean
2. Variance
3. Regression Coefcient
0
Stat 804
Chapter 5
The effect
Stat 804
Chapter 4
4. Cluster Sampling
Example: Want to estimate the total volume of apples in
orchard
Does not have a list of apples
Do have a list of trees
= Draw a sample of trees. For each selected tree, measure apples.
: Apple sizes on same tree a
Stat 804
Chapter 3
3.6. Probability Proportional-to-Size Sampling
Without Replacement (ps sampling)
Recall
1
V cfw_ =
t
2
yk
y
kl
l
k l
U
2
Let x1, , xN be the size variable that is correlated with
y . Dene the sampling design such that
k xk ,
(k = cxk
Stat 804
Chapter 3
Read Ch. 3.1. and 3.2.
3.3. Simple Random Sampling
3.3.1. Simple Random Sampling w/o Replacement
Result 3.3.1: Done
Result 3.3.2: The Horvitz-Thompson estimator for the population mean is
t
U = = ys .
y
N
The variance is
1f 2
SyU ,
n
a
Stat 804
Chapter 1 & 2
Example : Township with four farms.
Label
1
2
3
4
total
Acreage(xi)
4
6
6
20
36
Corn Acreage(yi)
1
3
5
15
24
The total acreage (xi) are known, but yi are unknown.
Interest in ty = total corn acres.
Take a random sample of size 2
5.2. Bernoulli sampling with k = = 0.3.
ty =
R=
tz
1
V R=2
z
t
S
1
=2
tz
s yk
s zk
= 1.099 .
kl yk Rzk yl Rzl
kl
k
l
(1 )
yk Rzk
2
S
2
= 0.0066
5.3.
2
BR R
1
1
nN
2
C VZU
1
1
2
C VZU 0.01
nN
1
0.01
1
2 +N
n
CVZU
1
0.01
2 +N
CVZU
n
Variable
u zk
2
u zk
M
Stat 804: HW2
Solution
3.3.
Note that ns = 71, = 0.5 and N = 124. Thus, an approximate 95% condence interval for the
proportion of countries belonging to Asia is
t
(1.96)
N
3.4.
21 1
N2
1
1
= 0.3387096774 0.1024375143.
The coefcient of variation of the
Stat 804: HW1
2.3.
Solution
U = U1 U2 U3 and N (1, 000) = N1 (600) + N2 (300) + N3 (100)
(a) k = 0.1 k U1 , k = 0.2 k U2 , k = 0.8 k U3 . The sample size is ns = n1 + n2 + n3 ,
where n1 b(600, 0.1), n2 b(300, 0.2) and n3 b(100, 0.8). Thus
E (ns ) = E (n1
STAT804 Exam 1: March. 9. 2006
Name: Solution
You may use a calculator. Show your work. Use your own paper if necessary.
1. Consider the following nite, clustered population of size N = 4.
Cluster, i
Cluster size, xi
k
yk
1
12
1
8
2
10
2
6
3
11
3
4
4
2
t