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Sampling Solved Exercises 3 (Stratification)

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Sampling Solved Exercises 3 (Stratification)

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4
Stratification




4.1 Definition
Consider a population U split into H parts Uh , called ‘strata’, such that
<
H
Uh = U, and Uh ∩ Ui = ∅,
h=1

for all (h, i) with h = i. A design is called stratified if in each stratum Uh we
select a random sample Sh of fixed size, and that the sample selection in each
stratum is taken independently of the selection done in all other strata (see
Figure 4.1).


Fig. 4.1. Stratified design
U Uh U
1 H




S1 Sh SH




4.2 Estimation and variance
We furthermore assume throughout this chapter that the designs are simple
without replacement within each stratum. The population size Uh is denoted

,122 4 Stratification

Nh and the sample size Sh is denoted nh , where h = 1, ..., H. Since the
inclusion probability is πk = nh /Nh , for all k ∈ Uh , the Horvitz-Thompson
estimator of the total becomes
 yk 
H
Nh  
H
Yπ = = yk = Nh Y h ,
πk nh
k∈S h=1 k∈Sh h=1

where Y h is the unbiased mean estimator for stratum h:
1 
Y h = yk .
nh
k∈Sh

The variance of Yπ is

H 2
Nh − nh Syh
var(Yπ ) = Nh2 ,
Nh nh
h=1

where 
1
2
Syh = (yk − Y h )2 ,
Nh − 1
k∈Uh
and
1 
Yh = yk .
Nh
k∈Uh
The variance can be estimated by

H 2
Nh − nh syh
 Yπ ) =
var( Nh2 ,
Nh nh
h=1

where 
(yk − Y h )2 .
1
s2yh =
nh − 1
k∈Sh
The choice of the nh specifies different stratified designs:
• designs stratified with proportional allocation,
Nh
;
nh = n (4.1)
N
• designs stratified with optimal allocation to estimate a total (case of iden-
tical survey unit cost in all strata),
Nh Syh
nh = n 7 H . (4.2)
i=1 Ni Syi

Expressions (4.1) and (4.2) do not generally give an integer value for nh ; it
is therefore necessary to turn to a rounding procedure. Furthermore, Expres-
sion (4.2) sometimes leads to having nh > Nh . In this case, we take a census
in the strata where this problem exists, and we restart the calculation of nh
for the remaining strata.

, Exercise 4.1 123

EXERCISES
Exercise 4.1 Awkward stratification
Given a population U = {1, 2, 3, 4} and y1 = y2 = 0, y3 = 1, y4 = −1, the
values taken by the characteristic y.
1. Calculate the variance of the mean estimator for a simple random design
without replacement of size n = 2.
2. Calculate the variance of the mean estimator for a stratified random design
for which only one unit is selected per stratum and the strata are given
by U1 = {1, 2} and U2 = {3, 4}.

Solution
1. The mean of y is zero. Indeed,
1  1
Y = yk = (0 + 0 + 1 − 1) = 0.
N 4
k∈U

The population variance is
1  1
2
2
Sy2 = (yk − Y )2 = 0 + 02 + 12 + (−1)2 = .
N −1 4−1 3
k∈U

We thus have
 N − n S2 4−2 1 2
var Y =
y 1
= × × = .
N n 4 2 3 6
2. For the stratified design, we start by calculating the parameters within the
strata
1  1
Y1 = yk = (0 + 0) = 0,
N1 2
k∈U1

1  1
Y2 = yk = (1 − 1) = 0,
N2 2
k∈U2

1  1 2
2
Sy1 = (yk − Y 1 )2 = (0 + 02 ) = 0,
N1 − 1 1
k∈U1

and 
1 1
2

2
Sy2 = (yk − Y 2 )2 = 1 + (−1)2 = 2.
N2 − 1 1
k∈U2

The variance of the Horvitz-Thompson estimator is, with regards to a pro-
portional allocation,
 N −n 
2
4−2
var Y π =
Nh 2 1
S = (2 × 2 + 2 × 0) = .
nN N yh 2 × 42 4
h=1

, 124 4 Stratification

We therefore see that the variance for a stratified design is larger than for the
simple design, despite proportional allocation. This surprising result recalls
that stratification does not lead to a systematic improvement in accuracy; it
is due to the fact that, in this example, the inter-strata variance is zero and
that the population size is small.



Exercise 4.2 Strata according to income
Among the 7500 employees of a company, we wish to know the proportion P of
them that owns at least one vehicle. For each individual in the sampling frame,
we have the value of his income. We then decide to construct three strata in
the population: individuals with low income (stratum 1), with medium income
(stratum 2), and with high income (stratum 3). We denote:

Nh = the stratum size h,
nh = the sample size in stratum h (simple random sampling),
ph = the estimator of the proportion of individuals in stratum h
owning at least one vehicle.

The results are given in Table 4.1.

Table 4.1. Employees according to income: Exercise 4.2

h=1 h=2 h=3
Nh 3500 2000 2000
nh 500 300 200
ph 0.13 0.45 0.50

1. What estimator P of P do you propose? What can we say about its bias?
2. Calculate the accuracy of P, and give a 95% confidence interval for P .
3. Do you consider the stratification criteria to be adequate?

Solution

1. The Horvitz-Thompson estimator for the stratified design is given by


3
N h ph 1
P = = (3500 × 0.13 + 2000 × 0.45 + 2000 × 0.50)
N 7500
h=1
= 0.314.

This estimator is unbiased.

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