Why Estimate a Population Total?
A survey often aims to estimate an amount for an entire target population: the total number of people with a condition, the total volume of water used, or the total expenditure during a specified period. A sample sum alone usually does not answer that question. It adds the observed values for sampled units, but does not account for how many population units those sampled units represent.
A survey weight provides that representation. To estimate a population total, multiply each sampled unit’s observed value by its weight and add the products. The resulting estimate is a weighted total. Its interpretation depends on the target population, the meaning and scale of the weights, and the units of the variable being summed.
The weight is not the observed value and is not generally a probability. In many applications it is an expansion weight: a sampled unit with weight 250 represents approximately 250 units in the target population for this calculation. Final survey weights may also reflect design, nonresponse adjustments, or other construction steps. Use weights intended for the target population and analysis at hand; the formula does not by itself make an inappropriate weight appropriate.
Interpreting a Weighted Total
The units of \(\hat{T}_y\) follow from the units of \(y_i\) and the meaning of \(w_i\). If weights represent population units, multiplying a measurement per population unit by a weight and summing estimates the corresponding population-wide amount. For example, weights representing households combined with monthly electricity use in kilowatt-hours per household produce a total in kilowatt-hours per month. When \(y_i\) is a binary indicator, its weighted total estimates a count.
A weighted total is not the same as the sample sum \(\sum_{i\in s} y_i\). The sample sum treats each sampled observation as one contribution, whereas the weighted total scales each contribution according to its weight. Nor should the weighted total be divided by the sum of weights when the requested quantity is a total: that division changes the estimand to an average-type quantity.
Worked Example: Total Monthly Electricity Use
Suppose a hypothetical household survey has three responding households. Their final weights represent numbers of households in the target population. The observed monthly electricity use is 310, 480, and 260 kilowatt-hours, and the corresponding weights are 120, 95, and 140 households. Estimate total monthly electricity use for the target population.
Apply the weighted-total formula: $$ \hat{T}_y=\sum_{i\in s}w_i y_i. $$ Substitute the three weight-value pairs: $$ \hat{T}_y=(120)(310)+(95)(480)+(140)(260). $$ The contributions are \(37{,}200\), \(45{,}600\), and \(36{,}400\) kilowatt-hours per month. Therefore, $$ \hat{T}_y=37{,}200+45{,}600+36{,}400=119{,}200 $$ kilowatt-hours per month.
The estimate is a population total of monthly electricity use, not an estimate of use by one household. It is an estimate for the target population represented by these weights; it is not a claim that the true total is exactly 119,200 kilowatt-hours per month.
Binary Outcomes Give Estimated Counts
For a binary outcome, code each unit as \(y_i=1\) if the outcome is present and \(y_i=0\) otherwise. Then \(w_i y_i\) equals \(w_i\) for units with the outcome and zero for units without it. Consequently, the weighted total adds the weights of the sampled units with the outcome. When the weights represent population units, this estimates the number of target-population units with that outcome.
Worked Example: Estimated Number of Patients With a Condition
In a hypothetical survey of patients in a defined region, the outcome is whether a patient has a specified condition. The binary outcome and final patient weights for five sampled patients are shown below. The weights represent patients in the target population.
| Patient | Outcome \(y_i\) | Weight \(w_i\) |
|---|---|---|
| A | 1 | 80 |
| B | 0 | 120 |
| C | 1 | 60 |
| D | 1 | 140 |
| E | 0 | 100 |
The estimated population count is the weighted total: $$ \hat{T}_y=\sum_{i\in s}w_i y_i. $$ Substituting the observed outcomes and weights gives $$ \hat{T}_y=(80)(1)+(120)(0)+(60)(1)+(140)(1)+(100)(0) =80+0+60+140+0=280. $$ The estimated number of patients with the condition in the target population is 280 patients. The two sampled patients coded 0 contribute zero to this estimated count, although their weights remain part of the survey data for other possible summaries.
Why Inverse-Probability Weights Estimate Totals
A standard case is a probability sample from a finite population. Let \(\pi_i=P(i\in s)\) be the probability that population unit \(i\) is included in the sample. If every population unit has a known, positive inclusion probability and its value is observed when selected, the inverse-probability weight is \(w_i=1/\pi_i\). The resulting estimator is the Horvitz–Thompson estimator of the population total.
To verify the result, define \(I_i=1\) if unit \(i\) is included and \(I_i=0\) otherwise. Then the sample sum can be written over all population units as \(\hat{T}_{HT}=\sum_{i\in U} I_i y_i/\pi_i\). Since \(P(I_i=1)=\pi_i\), the expected inclusion indicator is \(E(I_i)=\pi_i\). By linearity of expectation,
This argument does not require inclusion indicators for different units to be independent; it uses their individual inclusion probabilities and linearity of expectation. The positivity condition matters: if a population unit has zero probability of selection, its contribution cannot be recovered by an inverse-probability weight. The result also does not say that every realized sample produces the exact population total. It says that the estimator’s average over repeated samples under the design equals that total.
Not every practical final weight is exactly \(1/\pi_i\). Adjusted or calibrated weights can still be used in the same weighted-sum calculation, but the proof above establishes design-unbiasedness specifically for inverse inclusion-probability weights under the stated design conditions. Do not infer that a different weight construction has the same property without additional justification.
Worked Example: Inverse-Probability Estimate of Crop Output
Suppose a hypothetical finite population consists of farms, and the goal is to estimate total crop output in kilograms. A probability sample selects three farms with the inclusion probabilities shown below. Assume each selected farm’s output is observed, and all farms in the target population have positive inclusion probabilities.
| Farm | Inclusion probability \(\pi_i\) | Observed output \(y_i\) (kg) |
|---|---|---|
| A | 0.10 | 1,000 |
| B | 0.20 | 600 |
| C | 0.25 | 800 |
The inverse-probability weights are \(1/0.10=10\), \(1/0.20=5\), and \(1/0.25=4\). The estimated total is $$ \hat{T}_{HT} =\frac{1{,}000}{0.10}+\frac{600}{0.20}+\frac{800}{0.25} =10{,}000+3{,}000+3{,}200 =16{,}200\text{ kg}. $$ Thus, the estimated total crop output is 16,200 kilograms. Under the stated probability design and positivity conditions, this estimator is design-unbiased for the finite-population total. The value 16,200 kilograms is nevertheless an estimate from this particular sample, not a guaranteed exact total.
Checking a Total Before Reporting It
A short calculation can be numerically correct yet target the wrong quantity. Before reporting a weighted total, check the variable definition, the target population, the weight scale, and the units. A weight intended for people should not silently be treated as a weight for households if the unit of analysis is a household. Likewise, a total over a subgroup should include only eligible sampled units in that subgroup, using weights appropriate to the target represented by the estimate.
State the population and what is being totaled, including any time period or eligibility condition.
For each included sampled unit, pair its observed \(y_i\) with its intended weight \(w_i\).
Calculate \(w_i y_i\) for each included unit and add the contributions.
Give the estimated total, its units, and the target population it represents.
Common Mistake: Dividing a Total by the Sum of Weights
A frequent error is to calculate the weighted sum and then divide it by \(\sum_{i\in s}w_i\), even when the question asks for a population total. For the electricity example, this would divide 119,200 kilowatt-hours per month by 355 represented households, producing about 335.8 kilowatt-hours per household per month. That calculation is not a total; it has units per household and answers an average-type question. For the requested total, retain the weighted sum: 119,200 kilowatt-hours per month.
A second error is to call a weighted total an observed population count. A weighted sum estimates a population quantity; it does not turn sampled observations into a census. State it as an estimate and retain the units. For a binary indicator the units are people, households, or whichever population units the weight represents; for a continuous measurement, the units combine the measurement with the population total being represented.
Check Your Understanding
Use the definitions and calculations above to answer these questions.
- A sample unit has a monthly water-use value of 24 cubic metres and a weight of 75 households. What is its weighted contribution, and what units does that contribution have?
- For a binary outcome, explain why a sampled unit coded 0 contributes nothing to the estimated count while a unit coded 1 contributes its weight.
- A sample has weighted contributions of 900, 1,200, and 700 dollars per year. What is the estimated total, and why is it not a sample average?
- State the positivity condition needed for the inverse-probability estimator and explain why a unit with zero inclusion probability causes a problem.
- Why does design-unbiasedness not mean that one realized weighted estimate must equal the true population total?