QUESTION 120 (5 marks)
Two teams independently sample the same uniform meadow at random positions. Team S uses ten 0.25 m² quadrats and counts 20 plants in total. Team L uses ten 1.00 m² quadrats and counts 80 plants in total. The meadow area is 600 m².
a) Calculate plant density and estimated meadow abundance for each team. [3 marks]
b) A student says Team L found a fourfold higher density. Explain why that conclusion is incorrect. [2 marks]
Practice marking scheme
Answer
Both density estimates are 8 plants m⁻²; both abundance estimates are 4800 plants. The larger total count reflects the larger area sampled.
Working
S samples 10×0.25 = 2.5 m², giving 20/2.5 = 8 plants m⁻². L samples 10 m², giving 80/10 = 8 plants m⁻². Each estimates 8×600 = 4800 plants. Counts must be standardised by sampled area before comparing density.
Marking criteria
- Calculates S density 8 plants m⁻². [1 mark]
- Calculates L density 8 plants m⁻². [1 mark]
- Obtains 4800 plants for both meadow estimates. [1 mark]
- Identifies the fourfold difference in area sampled. [1 mark]
- Explains that equal densities can give different raw totals. [1 mark]
Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
View the QCAA syllabusCompare your working with the guide above.