Species A | Species B | Species C |

11 | 21 | 31 |

12 | 21 | 37 |

13 | 22 | 40 |

15 | 18 | 44 |

16 | 18 | 43 |

12 | 25 | 42 |

15 | 27 | 40 |

17 | 29 | 42 |

16 | 20 | 30 |

15 | 26 | 38 |

Calculate the Shannon’s Diversity Index along with the Simpson’s Diversity Index for Species A, Species B and Species C respectively then state which species is more diverse and explain why it is more diverse.

## CALCULATIONS

**SPECIES A**

**A) Shannon – Weiner Index**

H^{1}=[NlnN-∑(n_{i }ln n_{i})]/ N

where N is the total number of species and n_{i} is the number of individuals in species.

For species A

N=11+12+13+15+16+12+15+17+16+15=142

For species A

Species A |

n_{i}=11 |

n_{i=}12 |

n_{i=}13 |

n_{i=}15 |

n_{i=}16 |

n_{i=}12 |

n_{i=}15 |

n_{i=}17 |

n_{i=}16 |

n_{i=}15 |

**B) SIMPSON’ DIVERSITY**

**Species A**

N is the total number of species in the sample and n_{i} is the total number of individual species .

From the above calculation N is equal to 142 and n_{i }is shown in the table below-

Species A |

n_{i}=11 |

n_{i=}12 |

n_{i=}13 |

n_{i=}15 |

n_{i=}16 |

n_{i=}12 |

n_{i=}15 |

n_{i=}17 |

n_{i=}16 |

n_{i=}15 |

**SPECIES B**

**a) Shannon – Weiner Index**

H^{1}=NlnN-∑(n_{i }ln n_{i})/ N

where N is the total number of species and n_{i} is the number of individuals in species.

For Species b N=21+21+22+18+18+25+27+29+20+26=227

Species B |

n_{i}=21 |

n_{i}=21 |

n_{i}=22 |

n_{i}=18 |

n_{i}=18 |

n_{i}=25 |

n_{i}=27 |

n_{i}=29 |

n_{i}=20 |

n_{i}=26 |

**SPECIES B**

**SIMPSON’ DIVERSITY**

b) N is the total number of species in the sample and n_{i} is the total number of individual species.

From the above calculation N is equal to 227 and n_{i }is shown in the table below-

Species B |

n_{i}=21 |

n_{i}=21 |

n_{i}=22 |

n_{i}=18 |

n_{i}=18 |

n_{i}=25 |

n_{i}=27 |

n_{i}=29 |

n_{i}=20 |

n_{i}=26 |

**SPECIES C**

**a) Shannon – Weiner Index**

H^{1}=NlnN-∑(n_{i }ln n_{i})/ N

where N is the total number of species and n_{i} is the number of individuals in species.

For species c N =31+37+40+44+43+42+40+42+30+38=387

SPECIES C |

n_{i}=31 |

n_{i = }37 |

n_{i =}40 |

n_{i }=44 |

n_{i }=43 |

n_{i =}42 |

n_{i }=40 |

n_{i =}42 |

n_{i =}30 |

n_{i =}38 |

**SPECIES C**

**b) SIMPSON’ DIVERSITY**

N is the total number of species in the sample and n_{i} is the total number of individual species.

I have found N to be 387 from the above the calculation.

SPECIES C |

n_{i}=31 |

n_{i = }37 |

n_{i =}40 |

n_{i }=44 |

n_{i }=43 |

n_{i =}42 |

n_{i }=40 |

n_{i =}42 |

n_{i =}30 |

n_{i =}38 |

The compliment to SIMPSONS D, the value is –

1-0.092=0.908

Species A is more diverse .

The value of Simpson’s D ranges from 0 to 1, with 0 representing infinite diversity and 1 representing no diversity, so the larger the value of D , the lower the diversity. For this reason, Simpson’s index is usually expressed as its inverse (1/D) or its compliment (1-D).

Using the inverse, the value of this index starts with 1 as the lowest possible figure. The higher the value of this inverse index the greater the diversity.

The higher value of the inverse index is species A which is 10.4932

Therefore the more diverse species is species A.