Preliminary Exam Reviewer ECOLAB
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UST Ecology Laboratory...
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PRELIMINARY EXAMINATIONS Reviewer page 1
Topics to be covered: 1. T-test 2-Group
c i r t e m a r a P
Z-test
Comparison 3,4,5-Group Correlation
Pearson
Association
Chi-Square
Analysis of Variance (ANOVA)
Mann-Whitney
c i r t e m a r a P n o N
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 , if not which is ? which is ? Correlation Test - used to know the relationship between 2 or sometimes 3 groups whether they are and ? NOTICE that your value is either between 1 and 0 or 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 Unmatched - uses 2 different population Unmatched Matched-- uses the same population Matched
© 2015 Ellement
Mann-Whitney - nonparametric, comparison, unmatched Example:: Problem Set D Example 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
12 15 21 25 44 44 60 4 5 9 12 17 19
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 10 11 12 13.5 13.5 15
rank
rank
4 6.5 8 11 12 13.5 13.5 15
1 2 3 5 6.5 9 10
=8
Total of ranks of
=7
36.5
+ − −
2. Use the following formulae to solve for
where
:
= number of observations in first column = number of observations in second column = sum of the ranks in the second column
87 772 11 −36.5 . 87− . 47.5
and
3. Reject H0 if the computed lower U value > critical U value. =8; =7; level of confidence = 0.05 critical U value = 10 computed lower U value = 8.5 8.5 critical F value.
=3; =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 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
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
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.
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. 1. Rank the data. Data items that have equal values are given the
average rank of those items.
rank
rank
rank
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
rank
12
201
2412
144
40401
12
283
3396
144
80089
78
17
78
17
79
20.5
77
12.5
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
83
33.5
78
17
77
12.5
74
4
82
30.5
81
27.5
78
17
83
33.5
1654
16367.4
881.99
329626
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
6.5 106
2. Complete the ANOVA Table:
12 1 = − 3 1 −1
where
= Kruskal-Wallis value = number of total scores = sample size = ranked total per sample = number of scores per sample
12 1309.105 217.105 11090 10610} {4040 .4 −1 − 3 1 3
3. Reject H0 if the computed H value > critical X2 value.
=3; level of confidence = 0.05 critical X2 value = 7.8147 computed H value = 16.34 16.34>7.8147 Reject H0.
© 2015 Ellement
91.9
−∑∑ √ [∑ −∑ ][∑ − ∑] ∑ 1016367.4 − 91.91654
2. Using the formula below, get the value of r.
][10329626 −1654] 91. [ 10881. 9 9 − 9 √ 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
0.8058
Since , the relationship between the horn length and mass shows a strong positive correlation.
PRELIMINARY EXAMINATIONS Reviewer 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: 12 14 35
7 6 12
5 22 7
17 9 11
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.
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.
157 ∑ ∑ ∑ ∑ 12 13.08 ~13
3. Using the formula below, get the value of
12 14 35 61
7 6 12 25
5 22 7 34
and .
17 9 11 37
−
12 14 35
− 0.0769 0.0769 37.230
7 6 12
− 2.7692 3.7692 0.0769
3− 14− 1 − 1 −1
5 22 7
− 4.9230 6.2307 2.7692
60.6923
6
4. Reject H0 if the computed X2 value > critical X2 value.
=6; level of confidence = 0.05 critical X2 value = 12.592 computed X2 value =60.6923 60.6923>.592 Reject H0.
© 2015 Ellement
17 9 11
41 51 92 157
− 1.2307 1.2307 0.3076
PRELIMINARY EXAMINATIONS Reviewer page 5
2.
3.
- 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
a. Quadrat - applies to a square sample unit or plot - may be a single sa mple 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 d ensity 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 page 6
Abundance/Species Richness ( ) - count of number of species occurring within the community
Relative Abundance/Species Evenness
where
Whittaker plot/Rank-Abundance Curve - species ranking based on relative abundance, ranked from most to least abundant ) x-axis and relative abundance (y axis. axis) expressed on a - a 2D chart with relative abundance on the Y-axis and the abundance rank on the X-axis on Species Richness
log
- 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.
Simpson’s Index of Diversity1 1− Simpson’ s Diversity Index
Density
where
= total number of individuals of species
Relative Density
Diversity&Dominance Simpson’s Index
where
= Simpson’s index = number of individuals of species = 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.
′ ln
Shannon-Weiner’s Index
where
© 2015 Ellement
values of index,
ln
*Relative Dominance
= abundance of species = number of individuals of species = total number of individuals of all species
Rank-Abundance
When only one species is present, the value of H is 0. When all species are present in equal numbers, the maximum
= proportion of individuals found in species = number of individuals in species = total number of individuals of all species
, where = total number of species
- absolute dominance of species i divided by the sum of dominance for all species - usually done with trees
**Rank Dominance Frequency
# ℎ #
Relative Frequency
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