Bio 203l Ecology Lab Per

PRELIMINARY EXAMINATIONS Reviewer Ecology LAB page 1

Topics to be covered: 1. Statistics and Data Management T-test

Parametric

2-Group Z-test

Comparison 3,4,5-Group Correlation

Pearson

Association

Chi-Square

Analysis of Variance (ANOVA)

Mann-Whitney

Non-Parametric

2-Group Wilcoxon

Comparison 3,4,5-Group Correlation

Spearman

Association

Chi-Square

Kruskal-Wallis

Parametric Tests - anything you can measure - can be between 2 points - comparison of the means e.g. length, time, weight, temperature Non-Parametric Tests - cannot find a value in between - deals with ranks - comparison of the medians e.g. number of males in the classroom Comparison Test - used to know whether 2 or more groups are the same or equal, if not which is greater? which is smaller? Correlation Test - used to know the relationship between 2 or sometimes 3 groups whether they are directly or inversely related and by how much? NOTICE that your value is either between 1 and 0 or 0 and 1

 VALUE: the closer your value to zero, the weaker the relationship  SIGN: positive sign means it is directly related; negative means it is inversely related 0.90-1.00 very weak correlation 0.70-0.89 weak correlation 0.40-0.69 modest correlation 0.20-0.39 strong correlation 0.00-0.19 very strong correlation

Association Test - used to know which group or groups show an affinity to a set of conditions Unmatched - uses 2 different population Matched - uses the same population

© 2015 Ellement

Mann-Whitney - nonparametric, comparison, unmatched Example: Problem Set D A herpetologist studying the effect of a deadly fungal disease on frogs wanted to find out if the altitude of the frog’s habitat makes a difference in the prevalence of the disease among resident animals. She delineated two study sites (A and B) found on different altitudinal areas (A = 20 masl, B = 350 masl), and set up eight traps in each of the sites (total of 16 traps). She left the traps in the sites for a few days, and went back to collect the captured frogs and count how many tested positive for the fungal disease in each trap, Upon her return, she found out that one trap in site B was missing, so the data for this trap was not counted. Tabulating her results, she arrived at the following values: 8 2

SITE A SITE B

12 15 21 25 44 44 60 n= 8 4 5 9 12 17 19 n= 7

Hypotheses: H0: There is no significant difference between the two samples. H1: There is a significant difference between the two samples. 1. Rank the data. Data items that have equal values are given the

average rank of those items. 1 2 3 4 5 6.5 6.5 8 9

2 4 5 8 9 12 12 15 17

SITE A

rank

SITE B

rank

8 12 15 21 25 44 44 60

4 6.5 8 11 12 13.5 13.5 15

2 4 5 9 12 17 19

1 2 3 5 6.5 9 10

10 11 12 13.5 13.5 15

19 21 25 44 44 60

𝑛1 =8

𝑛2 = 7

Total of ranks of SITE B

36.5

2. Use the following formulae to solve for 𝑈1 and 𝑈2 :

𝑛 (𝑛2 +1)

𝑈1 = 𝑛1 𝑛2 + 2 𝑈2 = 𝑛1 𝑛2 − 𝑈1

2

− 𝑅2

where 𝑛1 = number of observations in first column 𝑛2 = number of observations in second column 𝑅2 = sum of the ranks in the second column 7(7 + 1) 𝑈1 = (8)(7) + − 36.5 2 𝑼𝟏 = 𝟒𝟕. 𝟓 𝑈2 = (8)(7) − 47.5 𝑼𝟐 = 𝟖. 𝟓 3. Reject H0 if the computed lower U value > critical U value.

𝑛1 =8; 𝑛2 =7; level of confidence = 0.05 critical U value = 10 computed lower U value = 8.5 8.5 critical F value.

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

PRELIMINARY EXAMINATIONS Reviewer Ecology LAB page 3

Kruskal-Wallis - nonparametric, comparison, 3,4,5-group

Pearson Product-Moment Coefficient - parametric, correlation

Example: Problem Set B

Example: Problem Set J

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 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 number of fishes living within them. To test this he designated a single species Acantharus olivaceous as the test species, and established ten counting stations and noted the number of A. olivaceous in each station and noted those in the data sheet. He did this for all areas and listed his data below.

The Jackson’s chameleon is a very popular animal among reptile keepers owing to the horns possessed by the males. The larger the horns, the more expensive the price. An exotic animal breeder wanted to find out if the length of the horns of males are related to the mass (weight) of the animal rather than size (length). He collected data from his captive stock males and got the following data:

78 78 79 77

AREA A AREA B AREA C AREA D

88 78 73 69

87 83 79 75

88 81 75 70

83 78 77 74

82 81 78 83

81 81 80 80

80 82 78 75

80 76 83 76

89 76 84 75

Hypotheses: H0: There is no significant difference in the distribution of fishes from four marine reserves. H1: There is a significant difference in the distribution of fishes from four marine reserves.

HORN LENGTH MASS (g)

6.6 6.9 7.3 8.2 8.3 11 12 12 9.4 10.2 86 92 71 74 185 185 201 283 255 222

Hypotheses: H0: There is no correlation between the 2 groups. H1: There is either a positive or negative correlation between the 2 groups. 1. Compute for the xy, x2, and y2. Calculate their summation. horn length (cm) mass (g) xy x2 y2

1. Rank the data. Data items that have equal values are given the

average rank of those items.

6.6

86

567.6

43.56

7396

6.9

92

634.8

47.61

8464

7.3

71

518.3

53.29

5041

8.2

74

606.8

67.24

5476

8.3

185

1535.5

68.89

34225

11

185

2035

121

34225

A

rank

B

rank

C

rank

D

rank

12

201

2412

144

40401

78

17

78

17

79

20.5

77

12.5

12

283

3396

144

80089

88

38.5

78

17

73

3

69

1

9.4

255

2397

88.36

65025

87

37

83

33.5

79

20.5

75

6.5

10.2

222

2264.4

104.04

49284

88

38.5

81

27.5

75

6.5

70

2

𝟐

83

33.5

78

17

77

12.5

74

4

82

30.5

81

27.5

78

17

83

33.5

81

27.5

81

27.5

80

23.5

80

23.5

80

23.5

82

30.5

78

17

75

6.5

80 89 TOTAL

23.5

76 76 TOTAL

10

83 84 TOTAL

33.5

76 75 TOTAL

10

40 309.5

10 217.5

36 190

12 𝑅𝑖 2 𝐻 = (∑ ) − 3(𝑁 + 1) 𝑁(𝑁 + 1) 𝑛𝑖 𝑖=1

𝑑𝑓 = 𝑘 − 1 where 𝐻 𝑁 𝑘 𝑅𝑖 𝑛𝑖

= Kruskal-Wallis value = number of total scores = sample size = ranked total per sample = number of scores per sample 12 309.52 217.52 1902 1062 𝐻 = {( )( + + + )} 40(40 + 1) 10 10 10 10 − 3(𝑁 + 1) 𝐻 = 𝟏𝟔. 𝟑𝟒 𝑑𝑓 = 4 − 1 𝑑𝑓 = 3

3. Reject H0 if the computed H value > critical X2 value.

𝑑𝑓 =3; level of confidence = 0.05 critical X2 value = 7.8147 computed H value = 16.34 16.34>7.8147 Reject H0.

© 2015 Ellement

∑ 𝒙𝑦

∑𝒙

∑ 𝑦2

91.9

1654

16367.4

881.99

329626

𝑟 =

6.5

𝑘

∑𝑦

2. Using the formula below, get the value of r.

106

2. Complete the ANOVA Table:

∑𝑥

𝑟 =

𝑁 ∑ 𝑥𝑦 − ∑ 𝑥 ∑ 𝑦 √[𝑁 ∑ 𝑥 2 − (∑ 𝑥)2 ][𝑁 ∑ 𝑦 2 − (∑ 𝑦)2 ] 10(16367.4) − (91.9)(1654)

√[10(881.99) − (91.9)2 ][10(329626) − (1654)2 ] 𝑟 = 0.8058 3. Based from the following, determine the correlation between

the two groups.

 VALUE: the closer your value to zero, the weaker the relationship  SIGN: positive sign means it is directly related; negative means it is inversely related Since 𝑟 = 0.8058, the relationship between the horn length and mass shows a strong positive correlation.

PRELIMINARY EXAMINATIONS Reviewer Ecology LAB page 4

Chi-Square - non/parametric, association Example: Problem Set I A reforested area consists of three tree species A, B, and C, and four species of endemic bird species 1, 2, 3, and 4. The timber concession that owns the area is preparing to cut down 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: BIRD 1

BIRD 2

BIRD 3

BIRD 4

12 14 35

7 6 12

5 22 7

17 9 11

TREE A TREE B TREE C

Hypotheses: H0: The number of bird inhabitants does not depend on the species of the trees. H1: The number of bird inhabitants depend on the species of the trees. 1. Get the total of the rows and columns.

TREE A TREE B TREE C TOTAL

BIRD 1

BIRD 2

BIRD 3

BIRD 4

TOTAL

12 14 35 61

7 6 12 25

5 22 7 34

17 9 11 37

41 51 92 157

2. In an ideal world, it is expected to have equal distribution of the

birds. To get the expected value, divide the grand total with the number of cells.

𝐸 =

∑𝑤 + ∑𝑥 +∑𝑦 + ∑𝑧 𝑁

157 12 𝐸 = 13.08 ~13 3. Using the formula below, get the value of 𝑋 2 and . 𝐸 =

TREE A TREE B TREE C TOTAL

BIRD 1

BIRD 2

BIRD 3

BIRD 4

TOTAL

12 14 35 61

7 6 12 25

5 22 7 34

17 9 11 37

41 51 92 157

(𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 − 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑)2 𝑋2 = ∑ [ ] 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 1 A B C

12 14 35

(𝑶 − 𝑬)2 𝑬 0.0769 0.0769 37.230

2

7 6 12

(𝑶 − 𝑬)2 𝑬 2.7692 3.7692 0.0769

3

5 22 7

(𝑶 − 𝑬)2 𝑬 4.9230 6.2307 2.7692

𝑋 2 =60.6923

𝑑𝑓 = (𝑟 − 1)(𝑐 − 1) 𝑑𝑓 = (3 − 1)(4 − 1) 𝑑𝑓 = 6 4. Reject H0 if the computed X2 value > critical X2 value. 𝑑𝑓 =6; level of confidence = 0.05 critical X2 value = 12.592 computed X2 value =60.6923 60.6923>.592 Reject H0.

© 2015 Ellement

4

17 9 11

(𝑶 − 𝑬)2 𝑬 1.2307 1.2307 0.3076

PRELIMINARY EXAMINATIONS Reviewer Ecology LAB page 5

2. Global Positioning System - relies on a constellation of 24 NAVSTAR satellites launched and maintained by the U.S. Department of Defense - uses at least 5 satellites/space vehicles (SVs)  Satellites are used to transmit the signal by letting these signal bounce on them since sound and light travel in a straight line.  SVs orbit at an altitude of about 21,000 km  SVs keep time using an atomic clock that loses or gains one second every 30,000 years  Unlike other devices like communication gadgets, GPS doesn’t need an importance signal.  It only needs the signal to be bounced back to the recipient.  1st Live Telecast via satellite: 1964 Summer Olympics in Tokyo REMINDERS in using the GPS:

 Use it in an open area.  Move slowly, do not run.  Do not cover the transmitter.

Other information GPS can give you:

     

Direction Distance Depth Elevation Speed Temperature

3. Terrestrial Sampling Techniques a. Quadrat

- applies to a square sample unit or plot - may be a single sample unit or be divided into subplots  the richer the flora, the larger or more numerous the quadrats must be

b. Transect Line

- across section of an area - used to relate changes in vegetation within it to changes in the environment

c. Point-Quarter

- most useful in sampling communities in which individuals are widely spaced or in which the dominant plants are large shrubs or trees

The classic distance method is the point quarter method which was developed by the first land surveyors in the U.S.A. in the nineteenth century. The four trees nearest to the corner of each section of land (1 sq. mile) were recorded in the first land surveys and they form a valuable data base on the composition of the forests in the eastern U.S. before much land had been converted to agriculture. The point quarter technique has been a commonly used distance method in forestry. It was first used in plant ecology by Cottam et al. (1953) and Cottam and Curtis (1956). Figure 5.10 illustrates the technique. A series of random points is selected often along a transect line with the constraint that points should not be so close that the same individual is measured at two successive points. The area around each random point is divided into four 90° quadrants and the distance to the nearest tree is measured in each of the four quadrants. Thus, 4 point-to-organism distances are generated at each random point, and this method is similar to measuring the distances from a random point to the 1st, 2nd, 3rd and 4th nearest neighbors.

Figure 5.10 Point-quarter method of density estimation. The area around each random point is subdivided into four 90° quadrants and the nearest organism to the random point is located in each quadrant. Thus four point-to-organism distances (blue arrows) are obtained at each random point. This method is commonly used on forest trees. Trees illustrate individual organisms

© 2015 Ellement

PRELIMINARY EXAMINATIONS Reviewer Ecology LAB page 6

Abundance/Species Richness (𝑆)

- count of number of species occurring within the community

Relative Abundance/Species Evenness

𝑅𝐷𝑖 = where

𝑛𝑖 𝑁

𝑅𝐷𝑖 = abundance of species 𝑖 𝑛𝑖 = number of individuals of species 𝑖 𝑁 = total number of individuals of all species

Rank-Abundance Whittaker plot/Rank-Abundance Curve

- species ranking based on relative abundance, ranked from most to least abundant ) x-axis and relative abundance (yaxis) expressed on a log10 axis. - a 2D chart with relative abundance on the Y-axis and the abundance rank on the X-axis on Species Richness - reflected by the greater length of the curve

on Species Evenness

- equitable distribution of individuals among species - indicated by the more gradual slope of the curve

e.g.

Density

𝐷𝑖 = where

𝐴𝑖 𝑎𝑟𝑒𝑎

𝐴𝑖 = total number of individuals of species 𝑖

Relative Density

𝑅𝐷𝑖 =

𝐷𝑖 𝑡𝑜𝑡𝑎𝑙 𝑝𝑙𝑎𝑛𝑡 𝑑𝑖𝑣𝑒𝑟𝑠𝑖𝑡𝑦

Diversity&Dominance Simpson’s Index

𝑛𝑖 2 𝐷 = ∑( ) 𝑁 Simpson’s Index of Diversity = 1 − 𝐷 1 Simpson’s Diversity Index = 𝐷 where

𝐷 = Simpson’s index 𝑛𝑖 = number of individuals of species 𝑖 𝑁 = total number of individuals of all species  The greater the value of D, the lower the diversity  The greater the Simpson’s Index of Diversity, the greater the diversity  A D value of 1 represents complete dominance meaning only one species is present in the community.

Shannon-Weiner’s Index

𝑛𝑖 𝑁 𝐻′ = ∑(𝑝𝑖 )(ln 𝑝𝑖 ) 𝑝𝑖 =

where

© 2015 Ellement

𝑝𝑖 = proportion of individuals found in species 𝑖 𝑛𝑖 = number of individuals in species 𝑖 𝑁 = total number of individuals of all species

 When only one species is present, the value of H is 0.  When all species are present in equal numbers, the maximum values of index, 𝐻𝑚𝑎𝑥 = ln 𝑆, where 𝑆 = total number of species

*Relative Dominance

- absolute dominance of species i divided by the sum of dominance for all species - usually done with trees

**Rank Dominance Frequency

𝐹𝑖 =

# 𝑜𝑓 𝑞𝑢𝑎𝑑𝑟𝑎𝑡𝑠 𝑤𝑖𝑡ℎ 𝑠𝑝𝑒𝑐𝑖𝑒𝑠 𝑖 𝑡𝑜𝑡𝑎𝑙 # 𝑜𝑓 𝑞𝑢𝑎𝑑𝑟𝑎𝑡𝑠 𝑠𝑎𝑚𝑝𝑙𝑒𝑑

Relative Frequency

𝑅𝐹𝑖 =

𝐹𝑖 𝑡𝑜𝑡𝑎𝑙 𝑝𝑙𝑎𝑛𝑡 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦