Hansen-Hurwitz estimator examples

Example 1

If we would be able to determine the selection probabilities \(p_i\) such that they were strictly proportional to the values \(y_i\) of the target variable.

Then, the true parametric total would be estimated perfectly by each single observation \(y_i\).

While this is certainly desirable, it is impossible because calculation of selection probabilities exactly proportional to the size of the target variable requires knowing the entire population. Then, there is no point in sampling. We, obviously, need to find an ancillary size variable of which we know that the target variable is as highly correlated as possible.

Example 2

Applying the Hansen Hurwitz estimator to the example population (Figure 1) with probabilities proportional to the plot area x, with replacement, we build a list as given in Figure 2. The parametric variances, with probabilities proportional to strip-plot area, for the estimated total and the estimated mean, respectively, are as follows:

\[var(\hat \tau) = \frac {1}{10} \sum_{i=1}^N p_i \left ( \frac {y_i}{p_i} - 212 \right )^2 = 118.97\,\] and

\[var(\bar y) = var (\hat \tau)/N^2 = 118.97/900 = 0.13129\]

Figure 1. Sample population
Number	y	x
1	2	50
2	3	50
3	6	100
4	5	100
5	6	125
6	8	130
7	6	130
8	7	140
9	8	140
10	6	130
11	7	140
12	7	150
13	9	160
14	8	170
15	10	180
16	9	200

Number	y	x
17	12	210
18	8	210
19	14	210
20	7	200
21	12	200
22	9	180
23	8	160
24	6	140
25	7	120
26	4	90
27	5	90
28	6	100
29	4	100
30	3	80
Mean	7.0667	13950
Pop. variance	7.1289	2087.25

Number	y	x	Selection probability (pi)	\(p_i * (\frac {y_i}{p_i} - 212)^2\)
1	2	50	0.011947	23.765352
2	3	50	0.011947	18.265352
3	6	100	0.023895	36.530705
[...]
29	4	100	0.023895	47.530705
30	3	80	0.019116	57.957064
Mean	7.0667	139.50
Pop. Variance	7.1289	2087.25
N	30
Sum	212	4185	1.000000	1189.685579
Figure 2. List sampling applied to the example population of 30 elements.