Questions and answers · Matched example · 3 readings per condition.
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Plan the comparison
Which comparison addresses the question?
- Compare raw plant counts from different-sized quadrats without recording their areas.
- Compare plant counts only at the most densely vegetated positions in each moisture zone.
- Compare plant counts at randomly selected positions across measured moisture zones.
Which set of records allows plant density to be compared with moisture?
- Plant count and moisture at each location, with no quadrat area.
- Plant count, quadrat area and moisture at each location.
- Plant presence or absence, quadrat area and moisture at each location.
Which sampling approach is useful?
- Use a pre-defined random selection of locations within each zone and consistent quadrat area.
- Count the same quadrat three times and treat the counts as three independent locations.
- Move each randomly chosen quadrat until it contains the target plant.
Calculate from evidence
Separate locations are selected within each defined zone without targeting dense patches.
Density = mean quadrat count ÷ 0.50 m². Estimated count in the comparable 200 m² zone = density × 200 m². Give numerical answers to about three significant figures.
Calculate density in the 30% moisture zone, in plants m⁻².
Estimate the number of plants in its comparable 200 m² zone.
Evaluate the comparison
Compare the representative-location and dense-patch examples. Explain why more biased quadrats do not solve the sampling problem, limit the inference from the moisture association and state when the 200 m² extrapolation is defensible.
Use the alternate example below alongside the data above.
Answers and reasoning
Which comparison addresses the question?
Compare plant counts at randomly selected positions across measured moisture zones. The question concerns a measured environmental gradient and plant density.
Which set of records allows plant density to be compared with moisture?
Plant count, quadrat area and moisture at each location. Counts and areas permit density calculation; moisture defines the environmental comparison.
Which sampling approach is useful?
Use a pre-defined random selection of locations within each zone and consistent quadrat area. Random selection within defined zones reduces deliberate patch selection; consistent area supports comparison.
Calculate density in the 30% moisture zone, in plants m⁻².
Mean = (6 + 7 + 8) ÷ 3 = 7. Density = 7 ÷ 0.5 = 14 plants m⁻².
Estimate the number of plants in its comparable 200 m² zone.
Conditional estimate = 14 × 200 = 2800 plants in the stated comparable zone.
Evaluation example
Choosing dense patches raises the example counts and systematically over-represents high-density places. More locations selected by the same biased rule can produce a precise but unrepresentative mean. Pre-defined random locations within each zone and consistent areas improve the comparison. The observed moisture association does not isolate a causal effect because other environmental factors may differ. Extrapolation to 200 m² is conditional on those samples representing that comparable zone, not every part of the habitat.
- I identify deliberate dense-patch selection and distinguish bias from random variation.
- I propose representative selection and avoid treating association as proof of causation.
- I limit extrapolation to the stated comparable zone rather than the whole habitat.
Original authored examples, not real experimental predictions, official marks or a laboratory protocol.