Chapter 14 Exercises in Sampling Theory Exercise 1 (Simple random...
Transcript of Chapter 14 Exercises in Sampling Theory Exercise 1 (Simple random...
SamplingTheory| Chapter 14 | Exercises in Sampling Theory | Shalabh, IIT Kanpur Page 1
Chapter 14 Exercises in Sampling Theory
Exercise 1 (Simple random sampling):
Let there be two correlated random variables and .X Y A sample of size n is drawn from a population by
simple random sampling without replacement. The observed paired sample is , , 1, 2,..., .i iX Y i n If the
sample totals are 1 1
ˆ ˆ, ,n n
tot i tot ii i
N NX X Y Y
n n
then find the covariance between ˆ andtotX totY .
Solution:
The covariance between andtot totX Y is
ˆ ˆ ˆ ˆ ˆ ˆ,
ˆ ˆ ˆ ˆ .
tot tot tot tot tot tot
tot tot tot tot
Cov X Y E X E X Y E Y
E X Y E X E Y
Note that
1
1
ˆ
ˆ
N
tot tot ii
N
tot tot ii
E X X X NX
E Y Y Y NY
where 1 1
1 1, ,
N N
i ii i
X X Y YN N
1
1 1
1
1.
( 1)
N
i i i ii
N N
i j i ii j j
E X Y X YN
E X Y X YN N
Also,
1 1 1 1 1
2
1 1 1
N N N N N
i j i i i ji j i i j j
N N N
i j i ii j j i
X Y X Y X Y
X Y N XY X Y
where 1 1
1 1, .
N N
i ii i
X X Y YN N
Then
SamplingTheory| Chapter 14 | Exercises in Sampling Theory | Shalabh, IIT Kanpur Page 2
22
1 1
22
1 1 1 1
22
1 1 1
22
ˆ ˆ,
( 1)
1
( 1)
1
n n
tot tot i ii i
N N N N
i j i ii j j i j j
N N N
i i i ii i j j
i i i
NCov X Y E X Y N XY
n
NE X Y X Y N XY
n
N n n nX Y X Y N XY
n N N N
N n n nX Y N XY X
n N N N
2
1 1
22 2
1
2
1
2
( 1) ( 1)1
( 1) 1
( ) 1
1
( )
1where .
1
N N
ii j
N
i ii
N
i ii
XY
XY i i
Y N XY
N n n n n nX Y N XY N XY
n N N N N N
N N nX X Y Y
Nn N
N N nS
Nn
S X X Y YN
Exercise 2 (Simple random sampling):
Under the simple random sampling without replacement, find xyE s where
1
1 1
1
1
1 1, .
n
xy i ii
n n
i ii i
s X x Y yn
x X y Yn n
Solution:
Consider
1
1
21 1 1
1 1 1 1
1 1 1
1
1
1
1
1
1
1 1
1
1 1 1.
1
n
xy i ii
n
i ii
n n n
i i i ii i i
n n n n
i i i i i ji i i j j
n n n
i i i ji i j j
s X x Y yn
X Y nxyn
nX Y x y
n n
X Y X Y X Yn n
nX Y X Y
n n n
SamplingTheory| Chapter 14 | Exercises in Sampling Theory | Shalabh, IIT Kanpur Page 3
1 1
1 1 1 1
Since
( 1).
( 1)
n N
i i i ii i
n n n n
i j i ji j j i j
nE X Y X Y
N
n nE X Y X Y
N N
Thus
1 1 1
1 1 1
2
1 1
1
11 1
1 1
1 1
1
1 1
1
1 1
( 1) 1
1
1
N N N
xy i i i ji i j
N N N
i i i ji i j j
N N
i i i ii i
N
i ii
i
n nn NE s X Y X Y
n n N nN N
X Y X YN N N
X Y N XY X YN N N
NX Y XY
N N N N
XN
1
1
1
1
.
n
ii
N
i ii
XY
Y NXY
X X Y YN
S
Exercise 3 (Simple random sampling):
Suppose in a population of size 1, and NN Y Y are two extreme values in the sense that 1Y denotes the
extremely low and NY denotes the extremely high values among the sample units 1 2, ,..., NY Y Y . Instead of
using the sample mean 1
1 n
ii
y yn
as an estimator of the population mean 1
1 N
ii
Y YN
in such a case,
the sample mean estimator is modified as follows
1
1
if sample contsins but not
if sample contsins but not
for all other samples
N
N
y k Y Y
y y k Y Y
y
where 0k is known constant. Determine if y is an unbiased estimator of population mean Y and find its
variance.
Solution:
SamplingTheory| Chapter 14 | Exercises in Sampling Theory | Shalabh, IIT Kanpur Page 4
Suppose the population units 1 2, ,..., NY Y Y are labeled as 11, 2,..., . So and NN Y Y now corresponds to labels 1
and N respectively.
Since the sample space corresponding to y has three possible situations, so it can be divided into three
disjoint subsets as follows:
1
2
All the units in thesample
All the units in the sample are such that unit label 1 is present and unit label is absent
All the units in the sample are such that unit labael is present and unit label 1 is absent
N
N
3 1 2.
When a sample of size n is drawn from a population of size ,N there are N
n
possible subsets. Similarly,
the number of possible subsets of sample drawn from the population
* 1
2are .
1
N
n
* 2
2are .
1
N
n
* 3
2 2are .
1 1
N N N
n n n
Under SRSWOR,
1 2 3
1 2 3
1
1
2 21
1 1
1 .
E y yN
n
y k y k yN
n
N Ny y y k k
N n n
n
y YN
n
So y is an unbiased estimator of the population mean.
SamplingTheory| Chapter 14 | Exercises in Sampling Theory | Shalabh, IIT Kanpur Page 5
The variance is
1 2 3
1 2 3
2 3
3
2
2 2
2 2 2
2
2 2
1
1
1
2 2 2
1
212 2
1
Var y y YN
n
y k Y y k Y y YN
n
y Y y Y y YN
n
NC C y Y y Y
n
Ny Y C C y Y y Y
N n
n
2
.
Observe that
*
2
1
1
N
n N nn
N N N
n
* 2
1
N
n
units in 1 containing the unit label 1, 3
2
N
n
of them have unit labels
2,3,..., 1j N and none of them contains the unit label .N
Thus
1 1 1
1
12
1
12
2 3 21
1 2 1
2 21 1.
1 12
N
jj
N
jj
y Y y Y
N N NY Y Y
n n nN
N NnY Y Y
n nn N
Similarly
2
1
2
2 21 1
1 12
N
N jj
N Nny Y Y Y Y
n nn N
.
SamplingTheory| Chapter 14 | Exercises in Sampling Theory | Shalabh, IIT Kanpur Page 6
Finally,
12
12
1
2
2
1
2 21 ( ) 1 1( ) 2 2
1 1( 1) 2
2 21 12
1 12
2 2
1
N
jj
N
N jj
yN
N Nn N n nV y Var y k k Y Y Y
N N nN N n N
n
N Nnk Y Y Y
N nn N
SN kY Y nk
N n N
.
Example 4 (Simple random sampling):
Let a sample of size 2 is drawn from a population of size 3 having units 1 2 3, andY Y Y . The following
estimator is used to estimate the population mean depending upon which sampling units are chosen
1 21 2
1 22 3
322 3
if sample is ,2 2
2if sample is ,
2 3
if sample is , .2 3
Y YY Y
Y Yy Y Y
YYY Y
Verify if y is an unbiased estimator of the population mean or not and find its variance.
Solution: The sample space 1 2 3, , .Y Y Y
3 31 2 1 2
1 2 3
21( )
3 2 2 2 3 2 3
2
1
3
Y YY Y Y YE y
Y Y Y
Y
which implies that y is an unbiased estimator of .Y Its variance is
2 2223 31 2 1 2
22 2 2 1 3 2 3 1 2 31 2
1 2 3
22 231 2 1
21( )
3 2 2 2 3 2 3
2
21 1 1 1 1 4 1 2
3 4 4 4 4 9 9 4 6 6 3
21
9 2 2 3
Y YY Y Y YVar y Y
YY Y Y Y Y YYYY Y Y
YY Y Y
22 3 .
2
YY Y
SamplingTheory| Chapter 14 | Exercises in Sampling Theory | Shalabh, IIT Kanpur Page 7
Exercise 5 (Simple random sampling):
The population coefficient of variation of a variable Y is defined as YY
SC
Y where
22
1
1
1
N
Y ii
S Y YN
,
1
1 N
ii
Y YN
is based on population size N . Let the sample mean 1
1 n
ii
y yn
is
based on sample size n by SRSWR which is usually employed to estimate .Y Assuming YC to be known,
improve the sample mean in the sense that the estimator has a minimum mean squared error. Find out the
relative efficiency of this estimator relative to sample mean.
Solution: Consider a scalar multiple of y to estimate Y as
ky ky
where k is any scalar. The mean squared error of ky is
2
2
22 2
22 2 2
22 2 2
1
1 0
11
11 .
k
SRSWR
Yn
Y
M y E ky Y
E k y Y k Y
k Var y k Y
Nk S k Y
N
NY k C k
Nn
Now we use the principle of maxima/minima s follows to find the optimum value of k for which kM y is
minimum
2 2
2
2
1 2 2 1 .
Substituting 0
1 1 1
1 *, say.
11
k
Y
k
Y
Y
dM Y NY k C k
dk Nn
dM Y
dk
Nk C
Nn
k kN
CNn
The second order condition for maxima/minima is satisfied.
SamplingTheory| Chapter 14 | Exercises in Sampling Theory | Shalabh, IIT Kanpur Page 8
Thus the optimum estimator is
2
*1
1k
Yn
yy k y
NC
N
and its minimum mean squared error is
2 22 * 2 *
242
22 2
2 2
2 2
22
2
2 2
2
2
11
11
1 11 1
11
11
1
1
11
( ).
11
k Y
YY
Y Y
Y
Y
Y
Y
Y
SRSWR
Y
NM y Y k C k
Nn
NN CCNnNnY
N NC C
Nn Nn
NY C NNn C
NnNC
Nn
NY C
NnN
CNn
Var YN
CNn
12( ) 1
Relative efficiency 1 .( )
kY
M Y NC
Var y Nn
Exercise 6 (Sampling for proportions):
Suppose we want to estimate the proportion of men employees (P) in an organization having 1500 total
employees. In addition, suppose 3 out of 10 employees are men. Suppose a sample is drawn by simple
random sampling. Find the sample size to be selected so that the total length of confidence interval with
confidence level 0.05 is less than 0.02 for SRSWR and SRSWOR.
Solution:
First, consider that a sample of size n is drawn from a population of size N by SRSWR. The
100 1 % confidence interval for sample mean y or equivalently the sample proportion wrp is
given by
SamplingTheory| Chapter 14 | Exercises in Sampling Theory | Shalabh, IIT Kanpur Page 9
2 2
2 2
2 2
, ,
1 1or , .
1 1
y y
wr wr wr wrwr
s sy z y z
n n
p p p pp z p z
n n
The total length of confidence interval is
2
12 .
1wr wrp p
zn
It is given that
2
2
2
2
2
1 2 0.02
1
1or 0.0001
1
or 1 1 .
wr wr
wr wr
wr wr
p pz
n
p pz
n
n z p p
For 0.05, 2
2
31.96 and 0.3,
10wrz p we have
21.96 0.3 0.7
1 80690.0001
n
.
The sample size is greater than the given population size in this case but since SRSWR is used, so it is not
impossible.
Next, we consider the case when the sample is drawn by SRSWOR. The related 100(1 )% confidence
interval for y or equivalently the proportion worp is
2 2
2 2
2 2
, ,
or
(1 ) (1 ), .
1 1
y y
wor wor wor worwor wor
N n N ny z s y z s
Nn Nn
p p p pN n N np z p z
Nn n Nn n
The total length of the confidence interval is
2
(1 )2
1wor worp pN n
zNn n
.
SamplingTheory| Chapter 14 | Exercises in Sampling Theory | Shalabh, IIT Kanpur Page 10
It is given that
2
2
2
2
2
2
2
(1 ) 2 0.02
1
(1 )or 0.0001
1
0.0001 (1 )
or .(1 )
0.0001
wor wor
wor wor
wor wor
wor wor
p pN nz
Nn n
p pN nz
Nn n
z p p
np p
zn
2
2
2
3Since 0.3, 1.96, 1500, we have
10
0.0001 (1.96) 0.3(1 0.3)
0.3(1 0.3)0.0001 (1.96)
1500 1265.
worp z N
n
Exercise 7: (Varying Probability scheme):
Show that the Yates-Grundy estimator is non-negative under Midzuno sampling design.
Solution: A sufficient condition for the Yates-Grundy estimator to be non-negative is
0, , , 1, 2,..., .i j ij i j i j N
The expression of ,i j and ij are
1
1 1
1
1 1
1 1 2
1 2 1 2
ii
tot
jj
tot
i jij
tot
XN n n
N X N
XN n n
N X N
X XN n n n n
N N X N N
where 1
N
tot ii
X X
is the population total.
Now
SamplingTheory| Chapter 14 | Exercises in Sampling Theory | Shalabh, IIT Kanpur Page 11
2
2
1 1 21 1
1 1 1 2 1 2
1 1 1 1 1 2
1 1 1 2 1 1 2
1
j i jii j ij
tot tot tot
i j i j
tot tot
X X XN n n n nXN n n N n n
N n X N N n X N N N X N N
X X X XN n nN n n n n
N X N X N N N N N
N n
N
2
2 22
( ) ( 1 ( ) ( 11 .
1 2 1 2
i j i j
tot tot
X X X XN n n N n n
X XN N N N
Each of the terms on right hand side is nonnegative and so that Yates-Grundy estimator is always
nonnegative.
Exercise 8 (Stratified sampling):
Suppose that the population of size N can be expressed as N nk where n and k are integers. The
population is divided into n strata where the thh stratum contains units that are labeled as
1 , 1,2,..., , 1, 2,..., .hG h k j j k h n
Suppose only one unit is randomly selected from every stratum and a sample of size n is obtained. The
values of units in the population are modelled as , 1, 2,...,iY i i N where and are some
constants. Find the variance of population total.
Solution:
The estimate of population total in case of stratified sampling is
1
ˆ .L
hst h
h h
NY y
n
where 1
.L
h hjj
y y
Compared to the notations in stratified sampling, we have
Number of strata L n
thh stratum size hN k .
sample size from thh stratum 1.hn
Then
SamplingTheory| Chapter 14 | Exercises in Sampling Theory | Shalabh, IIT Kanpur Page 12
1
1
11
ˆ ˆ 11
.
n
st tot hh
n
hh
kY Y y j
k y
Since the variance of stY is
22
1
ˆ ,L
h h hst h
h h h
N N nVar Y S
N n
so we have
22
01 1
2
1 1
1 1ˆ.1 1
.
n k
hj hh j
n k
hj hh j
k kVar Y Y Y
k k
k Y Y
When population values are modeled by the relation , 1, 2,..., ,iY i i N then
1
1
22 2
1 1
2 222
2 2
1
1
11
( 1)1
2
1
2
11
4
1 2 1 1 1
6 4 2
1.
12
hj
k
h hjj
k
j
hj h
k k
hj hj j
Y h k j
Y Yk
h k jk
k kh k
kY Y j
kY Y j k j
k k k k k k k
k k
Thus
2 2 2 1ˆ .
12tot
nk kVar Y
SamplingTheory| Chapter 14 | Exercises in Sampling Theory | Shalabh, IIT Kanpur Page 13
Exercise 9 (Stratified sampling):
Suppose there are two strata of sizes 1 2andN N such that 1 2 ,N N N (population size). Let
1 21 2and
N N
N N . Assume the population variances of first and second strata are 2 2
1 2andS S and they
are equal, i.e., 1 2.S S Consider a cost function 1 1 2 2C C n C n where 1 2andC C are the cost of collecting
an observation from first and second strata respectively. Assuming 1 2andN N to be large, show that
1 1 2 22
1 1 2 1
( )
( )prop st
opt st
Var y C C
Var y C C
where andprop st opt stVar y Var y are the variance of stratum mean under proportional and optimum
allocations in stratified sampling.
Solution
We have
2 2
1
22 2
1
Ki i
st i ii i i
Ki
i ii i
N nVar y S
N n
Sn
when iN is large where K is the number of strata. If cost function, in general, is
1
K
i ii
C C n
then under proportional allocation
1 1
1
1
or
or
or
or .
i i
i i
K K
i i i ii i
K
i ii
K
i ii
n N
n dN
C n d C N
C d C N
Cd
C N
Thus
1
ii K
i ii
CNn
C N
SamplingTheory| Chapter 14 | Exercises in Sampling Theory | Shalabh, IIT Kanpur Page 14
is the required sample size.
In the set up of given framework, we have 1 22, ,K S S S so
1 1 2 2
, 1, 2,.ii
CNn i
C N C N
Substituting in in the expression for the variance
2 2 2
1 1 2 22
1 2
2 221 2
21 2
1 1 2 2 1 1 2 2
21 1 2 2 1 22
21 1 2 2
1 2
21 1 2 2
1
1
(Using )
.
prop st
N S N SVar y
N n n
N NS
N CN CNC N C N C N C N
C N C N N NS
C N
C N C N SN N N
C N
C CS
C
The sample size under optimum allocation for the fixed cost is
1
, 1, 2,.
i i
iL K
i i ii
N S
Cn i
N S C
The variance of sty under optimum allocation when iN is large is
SamplingTheory| Chapter 14 | Exercises in Sampling Theory | Shalabh, IIT Kanpur Page 15
2 22
21
2 221 2
21 2
2 22 2
1 1 2 121 1
2
1 1 2 2
1 2
2 221 1 1 2 2 1 1 1 1 2 2 2
1 221 2
1
assuming
i iopt st
i i
i i i ii i
N SVar y
N n
N NS
N n n
N N S C N N S CS
N N S C N S C
C C
N N C N C C N N C N C CSS S
N N C N C
21 1 2 2 1 1 2 2
2
2 2
1 1 2 2 .
S
N C N C N C N CS
N C C
SC C
C
From the expression of and ,prop st opt stVar y Var y we find
1 1 2 22
1 1 2 1
( ).
( )prop st
opt st
Var y C C
Var y C C
Exercises 10 (Stratified sampling):
Suppose there are two strata and a sample of sizes 1 2andn n are drawn from these strata. Let a be the
actual ratio of 1 2andn n , i.e., 1
2
.a
n
n Let opt be the ratio of 1 2andn n under optimum allocation, i.e.,
1( )
2( )
optopt
opt
n
n . Let anda optV V be the variances of stratum mean under actual and optimum allocations,
respectively. Show that irrespective of the values 1 2 1 1 2, , , andN N S S S , the ratio opt
a
V
V is never less than
2
4
(1 )
when 1 2andN N are large and .a
opt
Solution:
When 1 2andN N are large, then
SamplingTheory| Chapter 14 | Exercises in Sampling Theory | Shalabh, IIT Kanpur Page 16
2 2 2 2
1 1 2 2
1 2st a
N S N SVar y V
n n
and variance under optimum allocation is
2
1 1 2 2
1.opt st optVar y V N S N S
n
Thus
2
1 1 2 2
2 2 2 21 1 2 2
1 2
2
2 2
1 12 22 2
2 21 2 1 1
1
11
.1
opt
a
N S N SV nN S N SV
n n
N Sn N S
N Sn n N S
Since under optimum allocation
1 11
1 1 2 2
2 22
1 1 2 2
1 1 1
2 2 2
1
2
1 2 2
2 1 1
.
opt
a
a
opt
N Sn n
N S N S
N Sn n
N S N S
n N S
n N S
n
n
n N S
n N S
Thus
2
2
12
2 22
1 1 2
2
1 2
21 2
11
1
1
.
opt
a
nV n n
nVn n n
n nn
n n
SamplingTheory| Chapter 14 | Exercises in Sampling Theory | Shalabh, IIT Kanpur Page 17
In this expression, replace
n by 1 2n n and
2 2
1 2 1 2 1 2by 4 ,n n n n n n
then we have
2
1 2 1 22 2
1 2 1 2
4.
1
opt
a
V n n n n
V n n n n
Using the result that
whenever 0,z
zz
over the expression opt
a
V
V, we notice that since 2
1 2 1 24 1n n n n is always true, so we have that
2
4.
(1 )opt
a
V
V
Exercise 11 (Stratified sampling):
Suppose there are two strata and equal size of samples are drawn from both the strata i.e., 1 2n n . Another
option to draw the samples is optimum allocation. Let ande optV V denotes the variances under equal
allocation 1 2n n and optimum allocation respectively then assuming 1 2andN N are large, show that
2
1
2
1where ,
1opte opt
opt opt
nV V
V n
1 2andopt optn n are the sample sizes drawn from first and second strata.
Solution:
The variance of sty is
2
2
1
22
21
1
Ki i i
st ii i i
Ki
ii i
N N nVar y S
N N n
NS
N n
assuming iN to be large.
SamplingTheory| Chapter 14 | Exercises in Sampling Theory | Shalabh, IIT Kanpur Page 18
When 1 2 ,2
nn n say, then the variance of sty under equal allocation is
2 2 2 21 1 2 22
2e stV Var y N S N S
nN .
Under optimum allocation,
1 11
1 1 2 2
1 22
1 1 2 2
N Sn n
N S N S
N Sn n
N S N S
and
2 2 2 21 1 2 2
21 2
2 2 2 21 1 2 2
21 1 2 2
1 1 2 2 1 1 2 2
2
1 1 2 22
1
1
1.
opt stopt opt
N S N SV Var y
N n n
N S N S
N nN S nN S
N S N S N S N S
N S N S
N n
Since 1 1
2 2
, soN S
N S
2 2 22 22
22 22 22
2 222 2 2 2 2
2 2
2 222 2
2
2
21
21
2 11 1
11
1.
1
e
opt
e opt
opt
V N SnN
V N SnN
N SN SV V n n n
N SVn n
SamplingTheory| Chapter 14 | Exercises in Sampling Theory | Shalabh, IIT Kanpur Page 19
Exercise 12 (Ratio method of estimation)
Consider a generalized form of ratio estimator ˆR
yY X
x
for estimating the population mean Y where
is some scalar. Derive the approximate bias and mean squared error of ˆ .RY For what value of , the
mean squared error is minimum. Derive the minimum mean squared error. Note that for ˆ1, RY reduces
to usual ratio estimator.
Solution:
Define
0 0
1 1
0 1
2 2 2 20 1 0 1
2 22 2
2 2
1
1
0, 0
, ,
, , ,
X Y X Y
X YX Y
x Xx X
Xy Y
y YY
E E
f f fE C E C E C C
n n n
S SN nf C C
N X Y
is the population correlation coefficient between and .X Y
Write
1
0
1 0
21 0 0 0
21 0 0 1 0
21 0 0 1 0
ˆ
1
1
1 1
1 (1 ) 1 ... (assuming 1)
2
1 1 ....
2
1ˆ (upto second order of approximati2
R
R
yY X
xY
XX
Y
Y
Y
Y Y Y
on).
The bias ˆRY is given as
SamplingTheory| Chapter 14 | Exercises in Sampling Theory | Shalabh, IIT Kanpur Page 20
21 0 0 1 0
2
ˆ ˆ
1
2
10 0
2
1.
2
R R
X Y X
XX Y
Bias Y E Y Y
YE
f fY C C C
n n
YfCC C
n
The mean squared error of ˆRY is given as
2
2 2 21 0 0 1
2 2 2 2
ˆ ˆ
2 (upto second order of approximation)
2 .
R R
Y X X Y
MSE Y E Y Y
E
fY C C C C
n
.
To obtain the value of for which the ˆRMSE Y is minimum, we use the principle of maxima/minima as
follows:
2
min
ˆ
0 2 2 0
or , say.
R
X X Y
Y
X
dMSE YC C C
dC
C
The second-order condition for maxima/minima is satisfied.
The minimum value of mean squared error is obtained by substituting min in the expression for
ˆRMSE Y as
22 2 2 2
2
2 2
ˆ 2
1 .
Y YR Y X X Y
X X
Y
C CfMin MSE Y Y C C C C
n C C
fC
n
SamplingTheory| Chapter 14 | Exercises in Sampling Theory | Shalabh, IIT Kanpur Page 21
Exercise 13 (Varying probability scheme):
Suppose there are 5 units in a population with values
1 2 3 4 5
8 8 21, 1, , ,
3 3 3y y y y y .
The probabilities of selecting a sample of size 2 with different units as
1 2 3 4 3 5 4 5
1 1 1 1, , , , , , ,
2 6 6 6p y y p y y p y y p y y .
Calculate the first and second-order inclusion probabilities and find the probability distribution of the -
estimator of the total.
Suppose the sample total HTY is used which is based on Horvitz-Thompson estimator is used to estimate
the square root of population total totY . Find the probability distribution of HTY and find if it is
unbiased for totY or not. Calculate the variance of HTY .
Solution:
The inclusion probabilities are
1 2 3 4 5 12 34 35 45
1 1 1 1 1 1 1 1 1, , , , , , , , ,
2 2 3 3 3 2 6 6 6 0ij for all other pairs of
( , ).i j
The Horvitz Thompson estimator of population total is
1 1 14 with probability
1/ 2 1/ 2 2ˆ
8 / 3 8 / 3 1 1 1 116 with probability .
1/ 3 1/ 3 6 6 6 2
Y
Since the sample size is 2, the estimator of variance is
2
ˆ j i j iji
i j ij
yyVar Y
.
Thus for the
sample 1 2
1 ˆ( , ), , 0.2sy y p Var Y
sample 3 4
1 ˆ( , ), , 0.6sy y p Var Y
SamplingTheory| Chapter 14 | Exercises in Sampling Theory | Shalabh, IIT Kanpur Page 22
sample 3 5
1 ˆ( , ), , 0.6sy y p Var Y
sample 4 5
1 ˆ( , ), , 0.6sy y p Var Y
Next
12 with probability
21
4 with probability .2
Y
Thus
1 12 4 3 10 .
2 2 totE Y Y
So Y is a biased estimator of totY and it underestimates it.
The variance of Y is
2
ˆ ˆ
10 9 1.
Var Y E Y E Y