Choose aggregation weights that match the environmental quantity

Frame the environmental question

A reported average hides a choice of weights. The mean concentration experienced at a randomly chosen time differs from the mean concentration attached to a randomly chosen litre of delivered water when flow changes. Neither mean is universally correct; each answers a different question. Define the sampling unit and target quantity before selecting an averaging rule.

Build the measurement or system model

The original day has two equal-duration intervals. The first delivers 100 m³ and the second 900 m³. Their concentrations are constant within intervals by assumption. Equal time weights therefore differ from volume weights of 0.1 and 0.9. Multiplying a time-average concentration by total volume can misestimate delivered mass if concentration and flow covary.

Work at the right scale

Compute interval mass before combining it. Concentration in mg/L multiplied by volume in m³ is divided by 1000 to give kilograms. Summing those masses and dividing by total volume gives the volume-weighted concentration. The numerical equality of a daily mass and weighted concentration in this particular 1000 m³ packet is a scaling coincidence, not permission to discard units.

Challenge the inference

Aggregation can also change comparisons across sites or periods. A site with lower concentrations in matched flow conditions can appear different after its sample mixes shift toward a high-concentration condition. Stratification, matched intervals and transparent weighting help distinguish within-condition change from composition change. An average alone does not preserve the joint variation of flow and concentration.

Design a defensible comparison

In a field plan, justify temporal resolution and sampling representativeness. A few convenient daytime grabs can miss storm pulses or diurnal cycles; an automated composite is not automatically valid if its weights do not match the question. This lesson analyzes supplied virtual intervals and does not request sampling in hazardous water.

Worked original packet

With concentrations 5 then 1 mg/L, time mean 3 differs from volume-weighted 1.4 mg/L. Interval masses 0.5 and 0.9 kg sum 1.4 kg/day. Using 3 mg/L with the full volume gives 3 kg/day, an overestimate 1.6 kg/day.

Investigation

Swap concentrations 5/1 to 1/5 while retaining interval volumes. Compare time means, volume means and mass; explain why the unchanged time mean conceals a changed delivery.

Changed-case transfer

For concentrations 2 then 4 mg/L, find the two means and daily mass. Explain why neither mean alone estimates an individual organism’s exposure.

Feedback

Time mean 3 mg/L; volume mean 3.8 mg/L; mass 3.8 kg/day. Organism location, duration, uptake and movement remain unknown.

Sources and evidence boundary

Original instructional prose, values and exercises. Factual context checked 30 October 2026: primary teaching or agency source. These are synthetic computational comparisons, not measured field data, verified mastery or authorization for physical intervention. Retain your prediction, assumptions and source of each observation.