Two habitats can contain the same number of species but have different diversity. If one species dominates almost every count in the first habitat, while the second has similar numbers of each species, a richness count alone misses the difference. Simpson's diversity measures combine information about species numbers and their relative abundance.
1. Check which Simpson formula you have
The word “Simpson” does not identify a single reporting convention. Always write the equation before interpreting a result. The table uses λ for concentration so that the symbol D does not hide the convention.
| Measure used here | Equation | Direction |
|---|---|---|
| Simpson concentration | λ = Σn(n − 1) ÷ N(N − 1) | Larger means greater concentration in a few species |
| Gini–Simpson diversity | 1 − λ | Larger means greater diversity |
| Reciprocal Simpson | 1 ÷ λ | Larger means greater diversity |
Here, n is the number of individuals of one species, N is the total across all species, and Σ means add the species contributions. You may also meet λ = Σp², where p = n/N. The two concentration equations are related but differ for finite samples. Use the equation supplied in your exam or analysis and compare results calculated by the same method.
2. Worked example: a balanced sample
Imagine counting four identifiable species in a teaching survey. The data below are invented, so they illustrate arithmetic rather than describe a real habitat.
| Species | n | n(n − 1) |
|---|---|---|
| A | 10 | 90 |
| B | 10 | 90 |
| C | 10 | 90 |
| D | 10 | 90 |
| Total | 40 | 360 |
N(N − 1) = 40 × 39 = 1560.
λ = 360 ÷ 1560 = 0.2308.
Gini–Simpson = 1 − 0.2308 = 0.7692.
Reciprocal Simpson = 1560 ÷ 360 = 4.3333.
The reciprocal calculated from this finite-sample concentration estimator can exceed the observed richness. Do not apply the “maximum equals richness” rule indiscriminately: that bound applies to the reciprocal of Σp². With p = 0.25 for each of these four species, Σp² = 0.25, so its reciprocal is exactly 4.
3. Compare an uneven sample
For Habitat B, the counts are 34, 2, 2 and 2. Richness is still four and N is still 40. However, Σn(n − 1) = 34 × 33 + 2 × 1 + 2 × 1 + 2 × 1 = 1128.
Using the same equation, λ = 1128/1560 = 0.7231 and 1 − λ = 0.2769. The balanced sample has the higher diversity value: 0.7692 versus 0.2769. The difference arises from evenness, not richness.
4. What the number means
For the n(n − 1) version, concentration describes the probability that two individuals drawn without replacement from the observed sample belong to the same species. Its complement describes different-species draws. That interpretation explains why dominance lowers the diversity complement.
One score does not describe every aspect of biodiversity. It does not show which species are present, whether rare species were missed, or whether a habitat contains threatened species. Keep the original abundance table alongside the index.
5. Design a fair comparison
- Use a consistent sampling method and comparable sampling effort at each site.
- Choose sampling positions using a stated random or systematic design rather than selecting especially varied patches.
- Keep the season and relevant conditions comparable, or record differences.
- Apply the same identification rules and treatment of uncertain specimens.
- Repeat sampling so that spatial variation can be considered.
For plants that grow as connected clumps, define the counting unit before sampling. Do not insert percentage-cover values into an individuals-without-replacement calculation as though they were counts of separate organisms. If using proportional cover, state that basis and use a method appropriate to those data.
6. Errors to avoid
Wrong direction: a larger concentration value indicates more dominance, while a larger diversity complement indicates more diversity. Wrong total: N is all individuals, not the number of species. Mixed equations: do not calculate one site with n(n − 1) and another with p². Early rounding: keep intermediate precision and round the final answer.
MCQs: answer before opening
1. Two samples each contain five species. What can differ? A) Only richness B) Evenness C) Their definition of a species must differ D) Nothing
B. Counts may be evenly distributed or dominated by one species.
2. If λ = 0.35, the diversity complement is: A) 0.35 B) 2.86 C) 0.65 D) 35
C. Calculate 1 − 0.35 = 0.65.
3. In Σn(n − 1), n refers to: A) Number of species B) Individuals of one species C) Number of habitats D) Quadrat area
B. Calculate a contribution for each species, then add the contributions.
4. Which comparison is valid? A) Compare scores without equations B) Use different equations without declaring it C) Use a stated shared equation and comparable sampling D) Select only diverse patches
C. Method and sampling differences can change the comparison.
Independent calculation challenge
A sample contains 8 individuals of species A and 2 of species B. Find N, Σn(n − 1), λ and 1 − λ using the without-replacement equation.
Reveal the worked answer
N = 10. The numerator is 8 × 7 + 2 × 1 = 58. The denominator is 10 × 9 = 90. λ = 58/90 = 0.6444; diversity complement = 0.3556. By contrast, Σp² = 0.8² + 0.2² = 0.68, showing why you must state the equation.
Frequently asked questions
Should I multiply the index by 100?
Usually it is reported as an index, not a percentage. Follow the question's requested format and identify your convention.
Can I calculate the n(n − 1) version when N = 1?
No. Its denominator is zero. You need more than one counted individual for that calculation.
Does a higher diversity score prove a habitat is healthier?
No. Interpretation depends on the habitat, sampling and ecological question. A score alone is not a complete conservation assessment.
Continue studying
Explore Plant Ecology II, browse biodiversity and conservation, and build revision practice with the FSc paper generator.
Sources
Formula conventions and teaching context checked against the Field Studies Council's Calculating biodiversity and Maths skills resources. Worked datasets, calculations and diagram are original.
Two habitats can contain the same number of species but have different diversity. If one species dominates almost every count in the first habitat, while the second has similar numbers of each species, a richness count alone misses the difference. Simpson's diversity measures combine information about species numbers and their relative abundance.
1. Check which Simpson formula you have
The word “Simpson” does not identify a single reporting convention. Always write the equation before interpreting a result. The table uses λ for concentration so that the symbol D does not hide the convention.
| Measure used here | Equation | Direction |
|---|---|---|
| Simpson concentration | λ = Σn(n − 1) ÷ N(N − 1) | Larger means greater concentration in a few species |
| Gini–Simpson diversity | 1 − λ | Larger means greater diversity |
| Reciprocal Simpson | 1 ÷ λ | Larger means greater diversity |
Here, n is the number of individuals of one species, N is the total across all species, and Σ means add the species contributions. You may also meet λ = Σp², where p = n/N. The two concentration equations are related but differ for finite samples. Use the equation supplied in your exam or analysis and compare results calculated by the same method.
2. Worked example: a balanced sample
Imagine counting four identifiable species in a teaching survey. The data below are invented, so they illustrate arithmetic rather than describe a real habitat.
| Species | n | n(n − 1) |
|---|---|---|
| A | 10 | 90 |
| B | 10 | 90 |
| C | 10 | 90 |
| D | 10 | 90 |
| Total | 40 | 360 |
N(N − 1) = 40 × 39 = 1560.
λ = 360 ÷ 1560 = 0.2308.
Gini–Simpson = 1 − 0.2308 = 0.7692.
Reciprocal Simpson = 1560 ÷ 360 = 4.3333.
The reciprocal calculated from this finite-sample concentration estimator can exceed the observed richness. Do not apply the “maximum equals richness” rule indiscriminately: that bound applies to the reciprocal of Σp². With p = 0.25 for each of these four species, Σp² = 0.25, so its reciprocal is exactly 4.
3. Compare an uneven sample
For Habitat B, the counts are 34, 2, 2 and 2. Richness is still four and N is still 40. However, Σn(n − 1) = 34 × 33 + 2 × 1 + 2 × 1 + 2 × 1 = 1128.
Using the same equation, λ = 1128/1560 = 0.7231 and 1 − λ = 0.2769. The balanced sample has the higher diversity value: 0.7692 versus 0.2769. The difference arises from evenness, not richness.
4. What the number means
For the n(n − 1) version, concentration describes the probability that two individuals drawn without replacement from the observed sample belong to the same species. Its complement describes different-species draws. That interpretation explains why dominance lowers the diversity complement.
One score does not describe every aspect of biodiversity. It does not show which species are present, whether rare species were missed, or whether a habitat contains threatened species. Keep the original abundance table alongside the index.
5. Design a fair comparison
- Use a consistent sampling method and comparable sampling effort at each site.
- Choose sampling positions using a stated random or systematic design rather than selecting especially varied patches.
- Keep the season and relevant conditions comparable, or record differences.
- Apply the same identification rules and treatment of uncertain specimens.
- Repeat sampling so that spatial variation can be considered.
For plants that grow as connected clumps, define the counting unit before sampling. Do not insert percentage-cover values into an individuals-without-replacement calculation as though they were counts of separate organisms. If using proportional cover, state that basis and use a method appropriate to those data.
6. Errors to avoid
Wrong direction: a larger concentration value indicates more dominance, while a larger diversity complement indicates more diversity. Wrong total: N is all individuals, not the number of species. Mixed equations: do not calculate one site with n(n − 1) and another with p². Early rounding: keep intermediate precision and round the final answer.
MCQs: answer before opening
1. Two samples each contain five species. What can differ? A) Only richness B) Evenness C) Their definition of a species must differ D) Nothing
B. Counts may be evenly distributed or dominated by one species.
2. If λ = 0.35, the diversity complement is: A) 0.35 B) 2.86 C) 0.65 D) 35
C. Calculate 1 − 0.35 = 0.65.
3. In Σn(n − 1), n refers to: A) Number of species B) Individuals of one species C) Number of habitats D) Quadrat area
B. Calculate a contribution for each species, then add the contributions.
4. Which comparison is valid? A) Compare scores without equations B) Use different equations without declaring it C) Use a stated shared equation and comparable sampling D) Select only diverse patches
C. Method and sampling differences can change the comparison.
Independent calculation challenge
A sample contains 8 individuals of species A and 2 of species B. Find N, Σn(n − 1), λ and 1 − λ using the without-replacement equation.
Reveal the worked answer
N = 10. The numerator is 8 × 7 + 2 × 1 = 58. The denominator is 10 × 9 = 90. λ = 58/90 = 0.6444; diversity complement = 0.3556. By contrast, Σp² = 0.8² + 0.2² = 0.68, showing why you must state the equation.
Frequently asked questions
Should I multiply the index by 100?
Usually it is reported as an index, not a percentage. Follow the question's requested format and identify your convention.
Can I calculate the n(n − 1) version when N = 1?
No. Its denominator is zero. You need more than one counted individual for that calculation.
Does a higher diversity score prove a habitat is healthier?
No. Interpretation depends on the habitat, sampling and ecological question. A score alone is not a complete conservation assessment.
Continue studying
Explore Plant Ecology II, browse biodiversity and conservation, and build revision practice with the FSc paper generator.
Sources
Formula conventions and teaching context checked against the Field Studies Council's Calculating biodiversity and Maths skills resources. Worked datasets, calculations and diagram are original.
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