3.4  Small Population Example
3.4  Small Population ExampleExample 34: Wheat Production
(
Reference Section 6.4 of the text)unit (Farm) i 
1

2

3

Selection Prob, \(p_i\) 
0.3

0.2

0.5

Wheat produced 
11

6

25

N = 3 farms, n = 2 sample with replacement.
s

p(s)

y_{s}

1, 1

0.3(0.3) = 0.09

(11, 11)

2, 2

0.2(0.2) = 0.04

(6, 6)

3, 3

0.5(0.5) = 0.25

(25, 25)

1, 2

0.3(0.2) = 0.06

(11, 6)

2, 1

0.2(0.3) = 0.06

(6, 11)

1, 3

0.3(0.5) = 0.15

(11, 25)

3, 1

0.5(0.3) = 0.15

(25, 11)

2, 3

0.2(0.5) = 0.10

(6, 25)

3, 2

0.5(0.2) = 0.10

(25, 6)

Question: Compute the HansenHurwitz estimator.
Answer
When (1,1) is sampled, the HansenHurwitz estimator is:
\(\hat{\tau}_p=\dfrac{1}{2}\left(\dfrac{y_1}{p_1}+\dfrac{y_1}{p_1}\right)=\dfrac{1}{2}\left(\dfrac{11}{0.3}+\dfrac{11}{0.3}\right)=36.67\)
When (1,2) is sampled, the HansenHurwitz estimator is:
\(\hat{\tau}_p=\dfrac{1}{2}\left(\dfrac{y_1}{p_1}+\dfrac{y_2}{p_2}\right)=\dfrac{1}{2}\left(\dfrac{11}{0.3}+\dfrac{6}{0.2}\right)=33.33\)
Similarly, we can fill out the table and get the HansenHurwitz estimators as shown below:
s

p(s)

y_{s}

\(\hat{\tau}_p\)

1, 1

0.3(0.3) = 0.09

(11, 11)

36.67

2, 2

0.2(0.2) = 0.04

(6, 6)

30.00

3, 3

0.5(0.5) = 0.25

(25, 25)

50.00

1, 2

0.3(0.2) = 0.06

(11, 6)

33.33

2, 1

0.2(0.3) = 0.06

(6, 11)

33.33

1, 3

0.3(0.5) = 0.15

(11, 25)

43.33

3, 1

0.5(0.3) = 0.15

(25, 11)

43.33

2, 3

0.2(0.5) = 0.10

(6, 25)

40.00

3, 2

0.5(0.2) = 0.10

(25, 6)

40.00

Question: Compute the HorvitzThompson estimator.
Answer
\(\pi_1=0.09+0.06+0.06+0.15+0.15=0.51\)
\(\pi_2=0.04+0.06+0.06+0.10+0.10=0.36\)
\(\pi_3=0.25+0.15+0.15+0.10+0.10=0.75\)
When (1,1) is sampled, the HorvitzThompson estimator is:
\(\hat{\tau}_\pi=\left(\dfrac{11}{0.51}\right)=21.57\)
When (1,2) is sampled, the HorvitzThompson estimator is:
\(\hat{\tau}_\pi=\left(\dfrac{11}{0.51}+\dfrac{6}{0.36}\right)=38.24\)
Similarly, we can fill out the table and get the HorvitzThompson estimators as shown below:
s

p(s)

y_{s}

\(\hat{\tau}_p\) 
\(\hat{\tau}_\pi\)

1, 1

0.3(0.3) = 0.09

(11, 11)

36.67

21.57

2, 2

0.2(0.2) = 0.04

(6, 6)

30.00

16.67

3, 3

0.5(0.5) = 0.25

(25, 25)

50.00

33.33

1, 2

0.3(0.2) = 0.06

(11, 6)

33.33

38.24

2, 1

0.2(0.3) = 0.06

(6, 11)

33.33

38.24

1, 3

0.3(0.5) = 0.15

(11, 25)

43.33

54.90

3, 1

0.5(0.3) = 0.15

(25, 11)

43.33

54.90

2, 3

0.2(0.5) = 0.10

(6, 25)

40.00

50.00

3, 2

0.5(0.2) = 0.10

(25, 6)

40.00

50.00

Mean 
42

42


Variance 
34.67

146.46

\(\text{Mean of }\hat{\tau}_p = E(\hat{\tau}_p)=\sum{p(s) \cdot \hat{\tau}_p (s)}\)
\(= 0.09\times36.67+0.04\times30.00+0.25\times50.00+0.06\times33.33+0.06\times33.33\)
\(+0.15\times43.33+0.15\times43.33+0.10\times40.00+0.10\times40.00\)
\(= 42\)
From the table above we can see that both \(\hat{\tau}_p\) and \(\hat{\tau}_\pi\)are unbiased. This example is a small population example to illustrate conceptually the properties of these estimators. We can compute the variance for \(\hat{\tau}_p\)and the variance for \(\hat{\tau}_\pi\) directly from the definition of variance.
Since mean of \(\hat{\tau}_p\) is 42 and E(g) = \(\sum p(s)*g(s)\), we can compute the variance of \(\hat{\tau}_p\) as:
\(Var(\hat{\tau}_p) = E[\hat{\tau}_p\text{mean of }\hat{\tau}_p]^2\)
\(= 0.09\times(36.6742)^2+0.04\times(30.0042)^2+0.25\times(50.0042)^2+0.06\times(33.3342)^2\)
\( +0.06\times(33.3342)^2+0.15\times(43.3342)^2+0.15\times(43.3342)^2+0.10\times(40.0042)^2\) \( +0.10\times(40.0042)^2\)
\(=34.67\)
Similarly, we can compute that the variance of \(\hat{\tau}_\pi\)is 146.46.
Now, how do we compute the MSE of \(\hat{\tau}_p\)?
By definition, \( MSE (\hat{\tau}_p) = E(\hat{\tau}_p\tau)^2\), in this case, since \(\hat{\tau}_p\)is unbiased and \(E(\hat{\tau}_p) =\tau\) ,\( MSE (\hat{\tau}_p) \)is the same as \(Var(\hat{\tau}_p)\).
Remark 1. The above demonstration is just a teaching tool. In reality we will not know the population and will not come across small population problems like this other than in exams and homeworks. What we know are:
Unit 
1

2

3

Selection probability 
0.3

0.2

0.5

And, we draw a sample. If the sample we draw is (1,2) then \(\hat{\tau}_p\) = 33.33 and\(\hat{\tau}_\pi=38.24\) .
We will not be able to find the real population total nor the real variance of the estimator. However, we will be able to estimate them.
Remark 2. Now, should we use \(\hat{\tau}_p\)or should we use \(\hat{\tau}_\pi\)?
There are no clear answers. Both estimators ar acceptable when \(y_i\) and \(p_i\) are proportional.