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SCITAMA ECOLOGY LABORATORY LABORATORY (Exercises 1-4) Exercise 1: Data Sampling Techniques

Data sampling is used in order to capture, manipulate, analyze and display data This may be done manually (written) or electronically

1. Data Gathering 2. Statistical Analysis Parametric Comparison of the means and variances Normal Distribution Ex: Length, time, weight and temperature T-Test, Z-Test, Pearson and ANOVA

Non-Parametric RANKS Comparison of the Medians Ex: Number of males in the classroom

Expected value: sample size / (number of rows) (number of columns) Degrees of freedom: (rows -1) (columns -1)

Mann-Whitney, KruskalWallis, Wilcoxon, ChiSquare, Spearman

Chi-Square Test Non-Parametric Non-Parametric Test Two categorical variables from a single population Test for independence Test if there is a significant association between the two categories Test if there is a difference between the distributions of a categorical data Comparison of the observed and expected frequencies Are the two nominal values independent?

Formula

Sample Problem: 1. A reforested area consists of the tree species A, B, and C, and four species of endemic bird species 1,2,3, and 4. 4. The timber concession that owns the area is preparing to cut down the trees for use as wood pulp for paper manufacturing. As part of the deal with the WWF, the timber concession can only cut down one species of tree. To help them decide what species of tree to cut, the company hired an ornithologist who did a survey of each tree species, and what bird species was found utilizing each tree species. The results of the survey are listed as: Solution: 1. Make the null and alternate hypothesis HO: “the “the two variables are independent ”

SCITAMA

2. 3.

Chi-Square value is greater than the critical HA: “the two variables are not independent ” value, reject Ho. Get the expected value Compute for the Chi-square Kruskal-Wallis Test statistic Non-Parametric Test Degrees of Freedom Test for comparison Compare the Chi-square Only use the Kruskal-Wallis when statistic to the critical value comparing 3 or more conditions If the Chi-square statistic Bird 2 Bird 3 Bird 4 Total Kruskal-Wallis value

4. 5. 6.

Bird 1

Tree A

12

7

5

17

41

Tree B

14

6

22

9

51

Tree C

35

12

7

11

65

Total

61

25

34

37

157

critical value, reject Ho. Where H = Kruskal-Wallis value N = number of total scores k = sample size Ri = ranked total per sample ni = number of scores per sample

Computation for the expected value: = 61+25+34+37/ (3)(4) = 13 Chi-Square Statistic Bird 1

Bird 2 2

(12-13) / 13 = 0.077 2

(14-13) / 13 = 0.077 2

(35-13) / 13 = 37.23

Bird 3 2

(7-13) / 13 = 2.77

Bird 4 2

(5-13) / 13 = 4.92

2

2

(17-13) / 13 = 1.23

2

(6-13) / 13 = 3.77 2

(12-13) / 13 = 0.077

(22-13) / 13 = 6.23 2

(7-13) / 13 = 2.77

2

Mean rank

(9-13) / 13 =1.23

Level of confidence: 0.05 Critical X2= 12.592 (based on the table) Sum of all Chi-Square Statistic: 60.691

Bird 1

Bird 2

Bird 3

Bird 4

Tree A

12

7

5

17

Tree B

14

6

22

9

Tree C

35

12

7

11

2

(11-13) / 13 = 0.31

Computation for df: = (3-1) x (4-1) = 6

Answer: 60.691 > 12.592

Formula for df: Df= number of groups -1

Solution: 1. Get the mean rank per category 2. Get the Kruskal wallis value or the H value

SCITAMA 3. Compute for the degrees of he designated a single species Acanthurus olivaceus as the test freedom 4. Get the Critical value species, and 5. Compare the Critical value with established ten counting stations and the H value noted the number of A. olivaceus in 6. If the H value > critical value, each station and noted those in the reject Ho. data sheet. He did this for all areas and If H value < critical value, accept listed his data below. Ho. Mean Rank Sample Problem: 1. A marine biologist in charge of four marine reserves located on a small island noticed that one of the marine reserves (Area ‘A’) was twice the (‘B’,’C’ and ‘D’). other areas Considering that all other aspects of the marine reserves were equal except for size, the biologist wanted to find out if the size of the marine reserve had an effect on the overall number of A

rank

B

rank

C

rank

D

rank

78

17

78

17

79

20.5

77

12.5

88

38.5

78

17

73

3

69

1

87

37

83

33.5

79

20.5

75

6.5

88

38.5

81

27.5

75

6.5

70

2

83

33.5

78

17

77

12.5

74

4

30.5 78 88

81 87

27.5 88 83

78 82

17 81 80

8380

33.5 89

78

27.5 78

81 83

81 27.5 78

80 81

81 23.5 82

8076

7623.5

79

23.5 73

82 79

75 30.5 77

78 78

80 17 78

7583

846.5

77

23.5 69

76 75

70 1074

83 83

80 33.5 75

7676

7510

89

40

76

10

84

36

75

6.5

TOTAL

309.5

TOTAL

217.5

TOTAL

190

TOTAL

106

A

rank

B

rank

C

rank

D

rank

Area 82 A Area B81 Area C 80 Area D80

fishes living within them. To test this,

Degrees of freedom: 4-1 = 3 Level of confidence = 0.05 Critical x2 value = 7.8147 H value = 16.34 Answer: 16.34 >7.8147 Reject Ho. ANOVA (Analysis of Variance Test) Parametric Test To test if there is any significant difference between the means of three or more independent (unrelated) groups Any group differences Two-way or One-way ANOVA Used for Comparisson

SCITAMA Sample Problem 1. A marine biologist in charge of four marine reserves located on a small island noticed that one of the marine reserves (Area ‘A’) was twice the size of the other areas (‘B’, ‘C’, and ‘D’). Considering that all other aspects of the marine reserves were equal except for size, the biologist wanted to find out if the size of the marine reserve had an effect on the overall size of fish species living within them. To test this, he designated a single fish species Acanthurus olivaceous as the test species, and collected 10 specimens of this fish in each of the four marine reserves. He measured each fish (in cm) and tabulated the Area A 78 Area B 78 Area C 79 Area D 77

88

87

88

83

82

81 80

80

89

78

83

81

78

81

81

76

76

73

79

75

77

78

80 78

83

84

69

75

70

74

83

80 75

76

75

data below.

Solution

82

Reject H0 if the computed F value > critical F value. =3; =36; level of confidence = 0.05 9.0024>2.87 Reject H0.

SCITAMA

How to solve ANOVA 1. Compute for the mean of each category 2. Compute for the grand mean: Mean of a+b+c+d/4 3. Compute for Xc 4. Compute for the Treatment of the Sum of the Squares 5. Compute for the Error for the Sum of the Squares 6. Compute for the Total Sum of the Squares 7. Make the Data Summary Table

nutrients, which undergo cycling between the aerobic zone (upper layer) and the anaerobic zone (lower layer).

Exercise 2: Winogradsky Column

Microbial communities are found in pond mud, and these organisms are capable of producing metabolic by-products that are required for the survival of other organisms within the environment. By using the Winogradsky column, these products can be studied, and the interdependent relationship of the microorganisms can be observed. This column can act as

a replica of the microbial environment. The Winogradsky column is a miniature, self-contained ecosystem which models ecological conditions in varying ways. It was invented by the Russian bacteriologist Sergei N. Winogradsky in 1880. The column is composed of a transparent cylindrical container which is filled with a few substrates (ex. Soil/mud) and marine or freshwater. The column is usually covered to prevent evaporation. Illumination (sunlight) is provided to promote the growth of microscopic organisms (ex. phototrophs). This composition will provide the information needed to study sulfur, nitrogen, carbon, phosphorus, and other

Gradients (light gradient, temperature gradient, nutrient, O2 and H2S concentration gradients) result in a complex interaction of microbes with their environment and with one another resulting in a series of community successions and, ultimately, stratification of microbial populations in the water column.

Key to potential observations:

SCITAMA a) Aerobic colors Green – eukaryal cyanobacteria

algae

or

Red/brown – cyanobacteria or thiobacilli Red/purple – purple non-sulfur Bacteria

White – sulfur oxidizing Bacteria b) Anaerobic colors

c)

Red/purple Bacteria

– purple

sulfur

Green – green sulfur Bacteria

Black – sulfate reducers

Gas

In the water column is probably O2 from oxygenic photosynthesis

In the aerobic zone is probably CO2 from respiration In the anaerobic probably CH4 methanogenesis

zone is from

Tracks in the upper layers of the sediment are formed by “worms” Small specks swimming in the water column are crustaceans, e.g. Daphnia & Cyclops

Interpretation of the Results:

Over time there is more oxygen at the top of a column than at the bottom, and this means that microbes that can tolerate or produce oxygen will be found at the top. Microbes that cannot tolerate free oxygen (called anaerobic bacteria) will be further down.

Similarly, microbes that need light to make energy (via photosynthesis or a similar process) will need to live where they can get light in the column. Some green coloring should appear in the columns receiving light on the illuminated sides. This is mostly due to cyanobacteria and algae, which needs light. The column in the dark should remain dark brown. In the column that had egg yolk, areas of darker green, purple, and/or black coloring may have developed over time near the bottom— these colorings could be groups of certain anaerobic bacteria: green sulfur bacteria, purple sulfur bacteria, and sulfatereducing bacteria, respectively. Sulfate-reducing bacteria actually eat sulfur and make hydrogen sulfide gas, which is eaten by the green and purple sulfur bacteria. In the column that had newspaper, some areas of brown, orange, red or purple may be evident near the middle— these colorings could be groups of purple nonsulfur bacteria, which need a carbon source to thrive. In addition, worms, snails, shrimp or other small organisms in the water, but probably not many (if

SCITAMA any) in the bottle with the egg yolk, because hydrogen sulfide is toxic to most organisms.

declared fully operational on April 1995. It consists of 24 active satellites (21 GPS and 3 spare satellites) that circle the globe once every 12 hours in order to provide worldwide position, time and velocity information.

Exercise 3: Global Positioning System

The Global Positioning System (GPS) is the best example. The GPS was first designed in 1960 under of the United States (U.S.) Air Force for military purposes. The first satellites were launched into space in 1978 but the system was

The GPS is used to locate positions anywhere on earth with the use of the location coordinates. The information it calculated is transmitted via the ground stations on earth along with the satellites. It is composed of three segments such as the space, ground and user segment. The space segment currently consists of 28 satellites that orbit the earth on 6 different orbital planes, and orbit at the height of 20,180 km above the earth’s surface. The ground or control segment observe the satellite movement, compute orbital data, monitor satellite data, and further relay information such as satellite health, clock hours or orbital data of the satellites. Lastly, the user segment, known as the receiver, with four satellites determines the position, transit time and velocity (Zogg, 2001). The GPS receiver specifies the geographic position (longitude and latitude) within 100 meters from the device. It gets its information from three of the four satellites. The receiver may also have a screen that shows a map, and

SCITAMA pinpoints the position on the map.2. Line Transect The accuracy of the information3. Point-Quarter that the receiver transmits is dependent on the distance of theIndexes Used: unit from the ground stations and1. Simpson the satellites. Usually, readings are2. Shannon-Wiener within 10 to 16 feet of the actual3. Jaccard location. The Simpson index is used to measure the species siversity. The main index (D) indicates total diversity. The (d), which is 1-D is the index of diversity which indicates the dominance of the species. The ShannonWiener index of diversity indicates the diversity of pseudo-species per quadrat or sample set. This takes into account the number of species present or absent. Lastly, the Jaccard index is known as the similarity coefficient. This index indicates the similarity and diversity between two sample sets or quadrats. This uses the absence and presence of a pseudo-species.

Exercise 4. Terrestial Sampling Techniques Methods Used: 1. Quadrat

A.1

Simpson’s Index

SCITAMA ni = number of individuals (per species) Pi: number of individualsTOTAL number of individuals (all species) Simpson’s Index (D): Sum of all Pi 2 values Simpson’s Index of Diversity: 1 -D Simpson’s reciprocal index: 1/D

Formula and sample solution for Average Density A. 4.

A.2 Shannon-Wiener Index

ni = number of individuals (per species) Pi: number of individualsTOTAL number of individuals (all species) Ln(pi) Pi x ln (Pi) H (max) = ln (number of species)

A.5. Formula

and sample solution for Standard Error

A. 3 Jaccard Index

K1: Sum of all species in Quadrat 1 K2: Sum of all species in Quadrat 2 K12: Sum of the Square of all species in Quadrat 1 K22: Sum of the Square of all species in Quadrat 2 K1K2: Combination of all species in Quadrat 1 and 2 Jaccard

Index

=

K1K2/ K1+K2+

K1K2 1.c.

√ A.6 Formula

and sample used for Standard Deviation

Jaccard Index for Quadrat 1.

K1

1+1+1+2+1+1+2+5 = 14

K2

2+1+4+2+2=11

K1

1+1+1+4+1+1+25=38

K2

4+1+16+4+4=29

K1UK2

(1x1)+(2x4)+(1x2)+(2x2)=15 0.375

Jaccard Index 15/14+11+15

A.7. Formula

and sample solution for Spatial Dispersion

SCITAMA ̅

A.8. Formula

and sample solution for

Density

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Data sampling is used in order to capture, manipulate, analyze and display data This may be done manually (written) or electronically

1. Data Gathering 2. Statistical Analysis Parametric Comparison of the means and variances Normal Distribution Ex: Length, time, weight and temperature T-Test, Z-Test, Pearson and ANOVA

Non-Parametric RANKS Comparison of the Medians Ex: Number of males in the classroom

Expected value: sample size / (number of rows) (number of columns) Degrees of freedom: (rows -1) (columns -1)

Mann-Whitney, KruskalWallis, Wilcoxon, ChiSquare, Spearman

Chi-Square Test Non-Parametric Non-Parametric Test Two categorical variables from a single population Test for independence Test if there is a significant association between the two categories Test if there is a difference between the distributions of a categorical data Comparison of the observed and expected frequencies Are the two nominal values independent?

Formula

Sample Problem: 1. A reforested area consists of the tree species A, B, and C, and four species of endemic bird species 1,2,3, and 4. 4. The timber concession that owns the area is preparing to cut down the trees for use as wood pulp for paper manufacturing. As part of the deal with the WWF, the timber concession can only cut down one species of tree. To help them decide what species of tree to cut, the company hired an ornithologist who did a survey of each tree species, and what bird species was found utilizing each tree species. The results of the survey are listed as: Solution: 1. Make the null and alternate hypothesis HO: “the “the two variables are independent ”

SCITAMA

2. 3.

Chi-Square value is greater than the critical HA: “the two variables are not independent ” value, reject Ho. Get the expected value Compute for the Chi-square Kruskal-Wallis Test statistic Non-Parametric Test Degrees of Freedom Test for comparison Compare the Chi-square Only use the Kruskal-Wallis when statistic to the critical value comparing 3 or more conditions If the Chi-square statistic Bird 2 Bird 3 Bird 4 Total Kruskal-Wallis value

4. 5. 6.

Bird 1

Tree A

12

7

5

17

41

Tree B

14

6

22

9

51

Tree C

35

12

7

11

65

Total

61

25

34

37

157

critical value, reject Ho. Where H = Kruskal-Wallis value N = number of total scores k = sample size Ri = ranked total per sample ni = number of scores per sample

Computation for the expected value: = 61+25+34+37/ (3)(4) = 13 Chi-Square Statistic Bird 1

Bird 2 2

(12-13) / 13 = 0.077 2

(14-13) / 13 = 0.077 2

(35-13) / 13 = 37.23

Bird 3 2

(7-13) / 13 = 2.77

Bird 4 2

(5-13) / 13 = 4.92

2

2

(17-13) / 13 = 1.23

2

(6-13) / 13 = 3.77 2

(12-13) / 13 = 0.077

(22-13) / 13 = 6.23 2

(7-13) / 13 = 2.77

2

Mean rank

(9-13) / 13 =1.23

Level of confidence: 0.05 Critical X2= 12.592 (based on the table) Sum of all Chi-Square Statistic: 60.691

Bird 1

Bird 2

Bird 3

Bird 4

Tree A

12

7

5

17

Tree B

14

6

22

9

Tree C

35

12

7

11

2

(11-13) / 13 = 0.31

Computation for df: = (3-1) x (4-1) = 6

Answer: 60.691 > 12.592

Formula for df: Df= number of groups -1

Solution: 1. Get the mean rank per category 2. Get the Kruskal wallis value or the H value

SCITAMA 3. Compute for the degrees of he designated a single species Acanthurus olivaceus as the test freedom 4. Get the Critical value species, and 5. Compare the Critical value with established ten counting stations and the H value noted the number of A. olivaceus in 6. If the H value > critical value, each station and noted those in the reject Ho. data sheet. He did this for all areas and If H value < critical value, accept listed his data below. Ho. Mean Rank Sample Problem: 1. A marine biologist in charge of four marine reserves located on a small island noticed that one of the marine reserves (Area ‘A’) was twice the (‘B’,’C’ and ‘D’). other areas Considering that all other aspects of the marine reserves were equal except for size, the biologist wanted to find out if the size of the marine reserve had an effect on the overall number of A

rank

B

rank

C

rank

D

rank

78

17

78

17

79

20.5

77

12.5

88

38.5

78

17

73

3

69

1

87

37

83

33.5

79

20.5

75

6.5

88

38.5

81

27.5

75

6.5

70

2

83

33.5

78

17

77

12.5

74

4

30.5 78 88

81 87

27.5 88 83

78 82

17 81 80

8380

33.5 89

78

27.5 78

81 83

81 27.5 78

80 81

81 23.5 82

8076

7623.5

79

23.5 73

82 79

75 30.5 77

78 78

80 17 78

7583

846.5

77

23.5 69

76 75

70 1074

83 83

80 33.5 75

7676

7510

89

40

76

10

84

36

75

6.5

TOTAL

309.5

TOTAL

217.5

TOTAL

190

TOTAL

106

A

rank

B

rank

C

rank

D

rank

Area 82 A Area B81 Area C 80 Area D80

fishes living within them. To test this,

Degrees of freedom: 4-1 = 3 Level of confidence = 0.05 Critical x2 value = 7.8147 H value = 16.34 Answer: 16.34 >7.8147 Reject Ho. ANOVA (Analysis of Variance Test) Parametric Test To test if there is any significant difference between the means of three or more independent (unrelated) groups Any group differences Two-way or One-way ANOVA Used for Comparisson

SCITAMA Sample Problem 1. A marine biologist in charge of four marine reserves located on a small island noticed that one of the marine reserves (Area ‘A’) was twice the size of the other areas (‘B’, ‘C’, and ‘D’). Considering that all other aspects of the marine reserves were equal except for size, the biologist wanted to find out if the size of the marine reserve had an effect on the overall size of fish species living within them. To test this, he designated a single fish species Acanthurus olivaceous as the test species, and collected 10 specimens of this fish in each of the four marine reserves. He measured each fish (in cm) and tabulated the Area A 78 Area B 78 Area C 79 Area D 77

88

87

88

83

82

81 80

80

89

78

83

81

78

81

81

76

76

73

79

75

77

78

80 78

83

84

69

75

70

74

83

80 75

76

75

data below.

Solution

82

Reject H0 if the computed F value > critical F value. =3; =36; level of confidence = 0.05 9.0024>2.87 Reject H0.

SCITAMA

How to solve ANOVA 1. Compute for the mean of each category 2. Compute for the grand mean: Mean of a+b+c+d/4 3. Compute for Xc 4. Compute for the Treatment of the Sum of the Squares 5. Compute for the Error for the Sum of the Squares 6. Compute for the Total Sum of the Squares 7. Make the Data Summary Table

nutrients, which undergo cycling between the aerobic zone (upper layer) and the anaerobic zone (lower layer).

Exercise 2: Winogradsky Column

Microbial communities are found in pond mud, and these organisms are capable of producing metabolic by-products that are required for the survival of other organisms within the environment. By using the Winogradsky column, these products can be studied, and the interdependent relationship of the microorganisms can be observed. This column can act as

a replica of the microbial environment. The Winogradsky column is a miniature, self-contained ecosystem which models ecological conditions in varying ways. It was invented by the Russian bacteriologist Sergei N. Winogradsky in 1880. The column is composed of a transparent cylindrical container which is filled with a few substrates (ex. Soil/mud) and marine or freshwater. The column is usually covered to prevent evaporation. Illumination (sunlight) is provided to promote the growth of microscopic organisms (ex. phototrophs). This composition will provide the information needed to study sulfur, nitrogen, carbon, phosphorus, and other

Gradients (light gradient, temperature gradient, nutrient, O2 and H2S concentration gradients) result in a complex interaction of microbes with their environment and with one another resulting in a series of community successions and, ultimately, stratification of microbial populations in the water column.

Key to potential observations:

SCITAMA a) Aerobic colors Green – eukaryal cyanobacteria

algae

or

Red/brown – cyanobacteria or thiobacilli Red/purple – purple non-sulfur Bacteria

White – sulfur oxidizing Bacteria b) Anaerobic colors

c)

Red/purple Bacteria

– purple

sulfur

Green – green sulfur Bacteria

Black – sulfate reducers

Gas

In the water column is probably O2 from oxygenic photosynthesis

In the aerobic zone is probably CO2 from respiration In the anaerobic probably CH4 methanogenesis

zone is from

Tracks in the upper layers of the sediment are formed by “worms” Small specks swimming in the water column are crustaceans, e.g. Daphnia & Cyclops

Interpretation of the Results:

Over time there is more oxygen at the top of a column than at the bottom, and this means that microbes that can tolerate or produce oxygen will be found at the top. Microbes that cannot tolerate free oxygen (called anaerobic bacteria) will be further down.

Similarly, microbes that need light to make energy (via photosynthesis or a similar process) will need to live where they can get light in the column. Some green coloring should appear in the columns receiving light on the illuminated sides. This is mostly due to cyanobacteria and algae, which needs light. The column in the dark should remain dark brown. In the column that had egg yolk, areas of darker green, purple, and/or black coloring may have developed over time near the bottom— these colorings could be groups of certain anaerobic bacteria: green sulfur bacteria, purple sulfur bacteria, and sulfatereducing bacteria, respectively. Sulfate-reducing bacteria actually eat sulfur and make hydrogen sulfide gas, which is eaten by the green and purple sulfur bacteria. In the column that had newspaper, some areas of brown, orange, red or purple may be evident near the middle— these colorings could be groups of purple nonsulfur bacteria, which need a carbon source to thrive. In addition, worms, snails, shrimp or other small organisms in the water, but probably not many (if

SCITAMA any) in the bottle with the egg yolk, because hydrogen sulfide is toxic to most organisms.

declared fully operational on April 1995. It consists of 24 active satellites (21 GPS and 3 spare satellites) that circle the globe once every 12 hours in order to provide worldwide position, time and velocity information.

Exercise 3: Global Positioning System

The Global Positioning System (GPS) is the best example. The GPS was first designed in 1960 under of the United States (U.S.) Air Force for military purposes. The first satellites were launched into space in 1978 but the system was

The GPS is used to locate positions anywhere on earth with the use of the location coordinates. The information it calculated is transmitted via the ground stations on earth along with the satellites. It is composed of three segments such as the space, ground and user segment. The space segment currently consists of 28 satellites that orbit the earth on 6 different orbital planes, and orbit at the height of 20,180 km above the earth’s surface. The ground or control segment observe the satellite movement, compute orbital data, monitor satellite data, and further relay information such as satellite health, clock hours or orbital data of the satellites. Lastly, the user segment, known as the receiver, with four satellites determines the position, transit time and velocity (Zogg, 2001). The GPS receiver specifies the geographic position (longitude and latitude) within 100 meters from the device. It gets its information from three of the four satellites. The receiver may also have a screen that shows a map, and

SCITAMA pinpoints the position on the map.2. Line Transect The accuracy of the information3. Point-Quarter that the receiver transmits is dependent on the distance of theIndexes Used: unit from the ground stations and1. Simpson the satellites. Usually, readings are2. Shannon-Wiener within 10 to 16 feet of the actual3. Jaccard location. The Simpson index is used to measure the species siversity. The main index (D) indicates total diversity. The (d), which is 1-D is the index of diversity which indicates the dominance of the species. The ShannonWiener index of diversity indicates the diversity of pseudo-species per quadrat or sample set. This takes into account the number of species present or absent. Lastly, the Jaccard index is known as the similarity coefficient. This index indicates the similarity and diversity between two sample sets or quadrats. This uses the absence and presence of a pseudo-species.

Exercise 4. Terrestial Sampling Techniques Methods Used: 1. Quadrat

A.1

Simpson’s Index

SCITAMA ni = number of individuals (per species) Pi: number of individualsTOTAL number of individuals (all species) Simpson’s Index (D): Sum of all Pi 2 values Simpson’s Index of Diversity: 1 -D Simpson’s reciprocal index: 1/D

Formula and sample solution for Average Density A. 4.

A.2 Shannon-Wiener Index

ni = number of individuals (per species) Pi: number of individualsTOTAL number of individuals (all species) Ln(pi) Pi x ln (Pi) H (max) = ln (number of species)

A.5. Formula

and sample solution for Standard Error

A. 3 Jaccard Index

K1: Sum of all species in Quadrat 1 K2: Sum of all species in Quadrat 2 K12: Sum of the Square of all species in Quadrat 1 K22: Sum of the Square of all species in Quadrat 2 K1K2: Combination of all species in Quadrat 1 and 2 Jaccard

Index

=

K1K2/ K1+K2+

K1K2 1.c.

√ A.6 Formula

and sample used for Standard Deviation

Jaccard Index for Quadrat 1.

K1

1+1+1+2+1+1+2+5 = 14

K2

2+1+4+2+2=11

K1

1+1+1+4+1+1+25=38

K2

4+1+16+4+4=29

K1UK2

(1x1)+(2x4)+(1x2)+(2x2)=15 0.375

Jaccard Index 15/14+11+15

A.7. Formula

and sample solution for Spatial Dispersion

SCITAMA ̅

A.8. Formula

and sample solution for

Density

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