Christmas Bird Count Diversity Index lab report

Restoration Ecology Lab

Shannon Diversity Index ​s H = ∑ - (P​i​ * ln P​i​) ​ i=1 where: H = the Shannon diversity index P​i ​= fraction of the entire population made​ ​up of species i S = numbers of species encountered ∑ = sum from species 1 to species S Note: The power to which the base e (e = 2.718281828.......) must be raised to obtain a number is called the​ natural logarithm​ (ln) of the number. To calculate the index:

1. Divide the number of individuals of species #1 (N​1​) you found in your sample by the total number of individuals of all species. For this exercise, we will use the Christmas Bird Count Data found in supplemental spreadsheet for field trip #3 to STA. This is P​i

2. Multiply the fraction by its natural log (P​1​ * ln P​1​) 3. Repeat this for all of the different species that you have. The last species is

species “s” 4. Sum all the - (P​i​ * ln P​i​) products to get the value of H

For example:

H = 0.223

High values of H would be representative of more diverse communities. A community with only one species would have an H value of 0 because P​i ​would equal 1 and be multiplied by ln P​i ​which would equal zero. If the species are evenly distributed then the H value would be high. So the H value allows us to know not only the number of species but how the abundance of the species is distributed among all the species in the community. For this lab report I want you to calculate the Shannon Diversity Index for TWO sites found in STA 5/6 [choose from: L1, Vitambi, Miami, West, East, 3B/3A, Deer, Gate, Manly, Blumberg]. I will want to see two tables (one for each site) like the above example. Please note that there are many blanks in the data for species that were not observed – I am expecting to see only observed species at each site in the table.

Birds N​i P​i ln P​i - (P​i​ * ln P​i​) Pigeon 96 .96 -.041 .039 Robin 1 .01 -4.61 .046 Starling 1 .01 -4.61 .046 Crow 1 .01 -4.61 .046 House sparrow

1 .01 -4.61 .046