Estimating Dandelions on the School Oval
A student wants to estimate how many dandelion plants are on the school oval, which is about 80 m by 50 m. She throws a 1 m × 1 m quadrat ten times and counts 3, 0, 5, 12, 1, 0, 4, 2, 0 and 25 dandelions. Two of the throws landed close to the fence, where dandelions are dense; one landed on the edge of the cricket pitch, where the grass is mown short. She calculates the mean of her ten counts, multiplies by 4000, and reports about 20 800 dandelions.
1. Check her arithmetic and her method. Is the figure of 20 800 defensible? Give at least two reasons for your judgement.
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Arithmetic: the counts sum to 3 + 0 + 5 + 12 + 1 + 0 + 4 + 2 + 0 + 25 = 52. Mean = 52 ÷ 10 = 5.2 dandelions per m². The oval is 80 × 50 = 4000 m², so 5.2 × 4000 = 20 800. The calculation is correct.
Method - problems:
- Throwing is not random. She controls where the quadrat lands, so it is biased - she may unconsciously favour open, easy-to-reach ground, or the throw may cluster near where she stands. The two fence-side throws and the 25-count suggest clustered placement.
- The mean is dominated by one sample. Nine of the ten counts are 12 or below; one count of 25 drags the mean up. A single quadrat has disproportionate influence, which signals too few samples for a clustered plant.
- The area is not uniform. The mown cricket-pitch edge has far fewer dandelions, yet it is treated as representative of all 4000 m². Dandelions are patchy, so a uniform mean over a mixed area is the wrong model.
So 20 800 is arithmetically right but statistically shaky; the true number could easily be half that.
2. The count of 25 came from one quadrat near the fence. Recalculate the estimate without it and explain what the change shows.
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Remaining counts: 3, 0, 5, 12, 1, 0, 4, 2, 0 → sum = 27 over 9 quadrats, giving a mean of 3.0 per m².
3.0 × 4000 = about 12 000 dandelions.
Removing a single quadrat out of ten changes the estimate from about 20 800 to about 12 000 - a drop of roughly 40%. An estimate that swings that far on one sample is not reliable. It shows that:
- The population is clustered (most quadrats had 0–5, one had 25), so the variance is high and ten samples is too few.
- A mean is only a good estimator when the samples truly represent the whole area. Here the dense fence zone is over-represented relative to its share of the oval.
3. Design a better sampling plan for the oval and explain how it would fix the problems identified.
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Plan:
- Divide the oval into a grid (say 10 m × 10 m cells) and use a random number generator to choose quadrat positions, rather than throwing. This removes personal bias and makes the sample random.
- Take many more samples - 30 to 50 quadrats, not 10 - so a few unusually dense or bare spots cannot dominate the mean.
- Stratify by zone. Map the oval into zones that look different (fence line, open playing area, cricket-pitch edge) and measure each zone's area. Sample each zone separately, then combine: total = (mean of zone A × area of A) + (mean of zone B × area of B) + …
- Record a fixed, defined unit - individual dandelion rosettes per m² - and have a second person recount a few quadrats to check consistency.
How it fixes the problems: random placement removes selection bias; more samples shrink the influence of any single count; stratification stops a sparse zone being described by a dense-zone average, and vice versa. The result is an estimate with a much smaller and more honest margin of error than 'about 20 800'.