Can I get to an approximation of the population with knowledge of the expansion factor?

Question

Say, I have a set of variables for different villages and for each village I have an expansion factor so I would get a sample representative of each village -- the expansion factor does not change with the observations within the village. Could I get an approximation of the population, in each village, by just multiplying the number of observations in the sample by the expansion factor? Would I incur in a great error? If so, is there any other way to get this approximation?

Answer 1

Introduction

The "expansion factor" is also known as the probability weight associated with an observation. In your situation, the factors are constant within village, so we will call that constant factor $W_{h*}$. Suppose the number of people selected in the village is $n_h$. Your basic question: is the quantity $T_h = n_h\times W_{h*}$ an estimate of $N_h$? The answer is "Yes" only if there was no sampling of villages, in other words, if the study villages constitute the entire population of interest. Moreover, the estimate is unbiased. If villages were sampled, the answer is "No".

I'll cover the case where villages are sampled first, because this is the most likely scenario in practice. Then I'll specialize to the case where villages are not sampled (are effectively sampling strata). All this follows from the multi-stage theory of weights. contained in Kish's standard text (1965).

Reference

Leslie Kish (2005) Survey Sampling, Wiley.

Case 1: Villages are sampled

Here is a possible multi-stage design that would have given rise to the village sample. I'm going to eliminate consideration of strata and higher level stages and assume that primary sampling unit is the village and that there is a single stratum. Then there are three likely stages of sampling: 1. Village 2. Households 3. People. \ A design of two stages (1. Villages; 2. People) is impossible under the terms of the question, since the second stage would require knowing the number of people in the village, which, by assumption, is unknown.

Let $f_h$ be the probability that village $h$ is selected into the sample.

The sampling could be simple random sampling without replacement, but might (and, if villages differ in size should ) be sampling with probability proportional to size (PPS) sampling.

Let $f_{j|h}$ be the conditional probability that in village $h$, household $j$ is selected

In small villages, all households could be located on a map;and common practice is to list the households, draw a random start and take a systematic sample, e.g. 1 in $k$.

Let $f_{i|jh}$ be the conditional probability that within household $j$, of village $h$, individual $i$ is selected.

Then:

The total probability that individual $i$ in household $j$ of village $j$ is selected into the household is the product of the conditional probabilities

$$ f_{hji} = f_h \times f_{j|h} \times f_{i|jh} \quad\quad\quad (1) $$

the last two terms in equation (1).

The final weight associated with individual $i$ in household $j$ of village $j$ is

$$ W_{hji} = 1/f_{hji} = (1/f_h)(1/f_{j|h})(1/f_{i|jh}) \quad\quad\quad (2) $$

If there is only a single constant "expansion factor" for each village, this must be it. In other words, $$ W_{hji} = W_{h*} \quad \quad \quad (3) $$

The intuitive understanding of a weight variable $W$ for a given person is that it is the number people in the entire population represented by that person. So, an estimate of the total population count is:

$$ \hat{N}_{h} = \sum_{hji}W_{hji} = \sum_{h}W_{h*} \quad \quad \quad (4) $$

Suppose for illustration that the population has 100 villages and the sample has 20 villages. If villages are selected by a simple random sample, then every each sampled village has probability $f_h = 20/100 = 1/5$. In other words, every selected village "represents" five villages, including itself.

If the constant expansion factor in the village is $W_{hji}= 10$, then each person represents 10 people in the population of 100 villages, not 10 people in his or her own village.Thus if there are 45 people in the sample, the sum of the expansion factor for the village is:

Thus:

$$ T_h = 45 \times 10 = 450 $$

This is not an estimate of the population total for that village alone, but for the five villages "represented" by that village.

Case 2. Estimating the number in a given village: village selection probability known

Given that village $h$ is selected, the conditional probability of selecting person $i$ in family $j$ is:

$$ f_{ij|h} = f_{j|h} \times f_{i|jh} \quad \quad \quad (5) $$

If we consider just village $h$ as the "population", then the weight associated with person {ij}:

$$ W_{ij|h}= 1/f_{ij|h} = W_{h*}\times f(_h) \quad \quad \quad (6) $$

If you know $f_h$, then you can sum the $W_{ij|h}$, to estimate the number of people in the village. Note that knowledge of $f_h$ is extra information not specified in the original question.

In our example, $W_{ij|h} = 10/5 = 2$, and the estimated number of people in the village is $2 \times 45$ = 90.

Case 3. Villages are not sampled

When there is no sampling of villages, then $f_h=1$. In this case, multiplying the expansion factor times the sample size will estimate the population total of the village.

Are these "good" estimates of population and village totals?

The expansion estimate will give an unbiased estimate of the relevant population totals in the statistical sense. However the estimate can be bad if (Case 1) villages differ greatly in size, but the village selection probability does not take this into account; or (Cases 2 & 3), households differ greatly in size, but the household selection probabilities do not take this into account. To take a simple example, consider a population of 10 villages, where 9 are of similar size (population 200), but one has population 1,000). Then the total population size is 2800, with the larger constituting 35.7% of the population. A simple random sample of size $n = 3$ is likely to miss the large village. Therefore most samples would give an estimated population size 10*200 = 2,000, an error of 40%.