Topics to be covered: Topics to be covered:

1. 1.

Parametric Tests

Parametric Tests - anything you can measure - anything you can measure

--can be between 2 pointscan be between 2 points

--comparison of the meanscomparison of the means e.g.

e.g.length, time, weight, temperaturelength, time, weight, temperature Non-Parametric Tests

Non-Parametric Tests- cannot find a value in between- cannot find a value in between - deals with

- deals withranksranks

- comparison of the medians - comparison of the medians e.g.

e.g.number of males in the number of males in the classroomclassroom Comparison Test

Comparison Test -- used to know whether 2 or moreused to know whether 2 or more groups are the

groups are the which is

which is Correlation Test

Correlation Test --used to know the relationship betweenused to know the relationship between 2 or sometimes 3 groups whether they 2 or sometimes 3 groups whether they are

are and and

NOTICE

NOTICEthat your value is either betweenthat your value is either between1 and 0 1 and 0 _{ or}_{ or}0 and 10 and 1

_{ VALUE}_{ VALUE}_{: the closer your value to zero, the}_{: the closer your value to zero, the}

weaker the relationship weaker the relationship

_{SIGN}_{SIGN}_{: positive sign means it is directly related;}_{: positive sign means it is directly related;}

negative means it is inversely related negative means it is inversely related

0.90-1.00

0.90-1.00 very very weak weak correlationcorrelation 0.70-0.89

0.70-0.89 weak weak correlationcorrelation 0.40-0.69

0.40-0.69 modest modest correlationcorrelation 0.20-0.39

0.20-0.39 strong strong correlationcorrelation 0.00-0.19

0.00-0.19 very very strong strong correlationcorrelation Association Test

Association Test -- used to know which group or groupsused to know which group or groups show an

show an Unmatched

Unmatched - -uses 2 different populationuses 2 different population Matched

Matched--uses the same populationuses the same population

Mann-Whitney

Mann-Whitney

-- nonparametric, comparison, unmatchednonparametric, comparison, unmatched Example

Example:: Problem Set D Problem Set D

A herpetologist studying the effect of a deadly fungal A herpetologist studying the effect of a deadly fungal disease on frogs wanted to find out if the altitude of the disease on frogs wanted to find out if the altitude of the frog’s habitat makes a difference in the prevalence of the frog’s habitat makes a difference in the prevalence of the disease among resident animals. She delineated two disease among resident animals. She delineated two study sites (A and B) found on different altitudinal areas study sites (A and B) found on different altitudinal areas (A = 20 masl, B =

(A = 20 masl, B = 350 masl), and set up eight 350 masl), and set up eight traps in eachtraps in each of the sites (total of 16 traps). She left the

of the sites (total of 16 traps). She left the traps in the sitestraps in the sites for a few days, and went back to collect the captured for a few days, and went back to collect the captured frogs and count how many tested positive for the fungal frogs and count how many tested positive for the fungal disease in each trap, Upon her return, she

disease in each trap, Upon her return, she found out thatfound out that one trap in site B was missing, so the

one trap in site B was missing, so the data for this trap wasdata for this trap was not counted. Tabulating her results, she arrived at the not counted. Tabulating her results, she arrived at the following values: following values: 8 8 12 15 21 12 15 21 25 44 44 25 44 44 6060 2 2 4 4 5 5 9 12 9 12 17 17 1919 Hypotheses: Hypotheses: H

H00: There is no significant difference between the two samples.: There is no significant difference between the two samples.

H

H11: There is a significant difference between the two samples.: There is a significant difference between the two samples.

1.

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

average rank of those items.

average rank of those items.

1

1

computed lower U value = 8.5

computed lower U value = 8.5

8.5<10 8.5<10 Fail to reject H Fail to reject H00..

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ComparisonComparison2-Group 2-Group T-test T-test Z-test Z-test 3,4,5-Group

3,4,5-Group _{ Variance (ANOVA)}_{ Variance (ANOVA)}Analysis ofAnalysis of

Correlation

Correlation PearsonPearson

Association

Association Chi-SquareChi-Square

N N o o n n - - P P a a r r a a m m e e t t r r i

i c c _{Comparison}_{Comparison}

2-Group 2-Group Mann-Whitney Mann-Whitney Wilcoxon Wilcoxon Statistics

Statistics and and ata ata ManagementManagement

same or equal

same or equal, if not, if not greater

greater? which is? which issmallersmaller?? directly or inversely related

directly or inversely related byby how much

how much??

affinity to a set of conditions affinity to a set of conditions

SITE A

SITE A _{n= 8}_{n= 8}

SITE B

SITE B _{n= 7}_{n= 7}

SITE A

SITE A rankrank rank rank

2 2 44 11 3 3 22 4 4 33 5 5 55 6.5 6.5 6.56.5 6.5 6.5 99 8 8 1010 9 9 10 10

##

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==77 11 11 36.536.5 12 12 13.5 13.5 13.5 13.5 15 15 2.2. Use the following formulae to solve forUse the following formulae to solve for

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where

where

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= number of observations in = number of observations in first columnfirst column###

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= number of observations in second column= number of observations in second column###

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= sum of the ranks in the second column= sum of the ranks in the second column###

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### 87

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3. 3. SITE B SITE B 4 4 5 12 5 12 6.56.5 8 15 8 15 88 9 21 9 21 1111 12 25 12 25 1212 1212 12 44 12 44 13.513.5 1717 15 44 15 44 13.513.5 1919 17 60 17 60 1515##

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==88 Total of ranks of Total of ranks of 19 19 21 21 SITE BSITE B 25 25 44 44 44 44 60 60### 87− 47.5

### 87− 47.5

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Reject HReject H00if the computed lower U value > critical U value.if the computed lower U value > critical U value.

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==88;;###

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==77;;level of confidence = 0.05level of confidence = 0.05critical U value = 10

critical U value = 10

3,4,5-Group

3,4,5-Group Kruskal-WallisKruskal-Wallis

C

Coorrrreellaattiioonn SSppeeaarrmmaann

Association

Analysis of Variance One-Way (1-Way ANOVA) - parametric, comparison, 3,4,5-group Example: Problem Set A

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 data below. 78 88 87 88 83 82 81 80 80 89 78 78 83 81 78 81 81 82 76 76 79 73 79 75 77 78 80 78 83 84 77 69 75 70 74 83 80 75 76 75 Hypotheses:

H0: There is no significant difference among the means.

H1: At least one of the means is different from the others.

1. Find the size

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, mean (###

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, and the grand mean (### ̿

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78 31.36 78 1.96 79 0.16 77 2.56 88 19.36 78 1.96 73 31.36 69 40.96 87 11.56 83 12.96 79 0.16 75 0.16 88 19.36 81 2.56 75 12.96 70 29.16 83 0.36_{78}1.96

_{77}2.56

_{74}1.96 82 2.56 81 2.56 78 0.36 83 57.76 81 6.76 81 2.56 80 1.96 80 21.16 80 12.96 82 6.76 78 0.36 75 0.16 80 12.96 76 11.56 83 19.36 76 0.36 89 29.16 76 11.56 84 29.16 75 0.16

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10###

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10###

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83.6 146.4 79.4 56.4 78.6 98.4 75.4 154.4###

9.25### ∑ ̅

_{ }

2. Complete the ANOVA Table:

Sources

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Treatments### .

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2.87 Errors### .

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Total### .

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where###

= Treatment Sum of Squares###

= Error Sum of Squares###

= Total Sum of Squares###

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= sample size### ̅

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= mean### ̿

= grand mean###

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= mean of the### (

_{ }

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where##

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= number of populations##

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= number of observations##

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3. Reject H0if the computed F value > critical F value.

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=3;###

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=36;level of confidence = 0.05 critical F value = 2.87 computed F value = 9.0024 9.0024>2.87 Reject H0. AREA A AREA B AREA C AREA D###

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Kruskal-Wallis

- nonparametric, comparison, 3,4,5-group Example: Problem Set B

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.

78 88 87 88 83 82 81 80 80 89 78 78 83 81 78 81 81 82 76 76 79 73 79 75 77 78 80 78 83 84 77 69 75 70 74 83 80 75 76 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.

1. Rank the data. Data items that have equal values are given the average rank of those items.

rank rank rank 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 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 23.5 76 10 83 33.5 76 10 89 40 76 10 84 36 75 6.5

TOTAL 309.5 TOTAL 217.5 TOTAL 190 TOTAL 106

2. Complete the ANOVA Table:

## 12

_{ 1}

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### =

## − 3 1

## −1

where

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= Kruskal-Wallis value###

= number of total scores###

= sample size###

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= ranked total per sample###

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= number of scores per sample### { 12

_{4040 1309.5}

_{10 217.5}

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_{10 190}

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_{10 106}

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3. Reject H0if the computed H value > critical X2 value.

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=3;level of confidence = 0.05 critical X2_{ value = 7.8147}computed H value = 16.34 16.34>7.8147 Reject H0. AREA A AREA B AREA C AREA D

Pearson Product-Moment Coefficient - parametric, correlation

Example: Problem Set J

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:

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 y}2_{. Calculate their summation.}

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 12 201 2412 144 40401 12 283 3396 144 80089 9.4 255 2397 88.36 65025 10.2 222 2264.4 104.04 49284

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_{91.9}

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881.99###

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3296262. Using the formula below, get the value of r.

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### 1016367.4 − 91.91654

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_{][10329626 −1654}

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### 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

### 0.8058

, the relationship between the horn length and mass shows a strong positive correlation.HORN LENGTH MASS g)

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 7 5 17

14 6 22 9

35 12 7 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 7 5 17 41

14 6 22 9 51

35 12 7 11 92

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.

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### 13.08 ~13

3. Using the formula below, get the value of

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and .12 7 5 17 41 14 6 22 9 51 35 12 7 11 92 61 25 34 37 157

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12 0.0769_{7}2.7692

_{5}4.9230

_{17}1.2307 35 37.230 12 0.0769 7 2.7692 11 0.3076

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60.6923## 1 −1

_{1}

4.
computed X2_{ value =}60.6923 60.6923>.592 Reject H0.

BIRD 1 BIRD 2 BIRD 3 BIRD TREE A

TREE B TREE C

BIRD 1 BIRD 2 BIRD 3 BIRD 4 TOTAL TREE A

TREE B TREE C

TOTAL _{61 } _{25 } _{34 } _{37 } _{157}

BIRD 1 BIRD 2 BIRD 3 BIRD 4 TOTAL TREE A TREE B TREE C TOTAL 14 0.0769 6 3.7692 22 6.2307 9 1.2307

###

## −

_{ 3− 14 −}

### 6

Reject H0if the computed X2 value > critical X2 value.

###

=6;level of confidence = 0.05 critical X2_{ value = 12.592}

2.

- 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.

_{1}st_{ 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.
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 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

Abundance/Species Richness (

##

)- count of number of species occurring within the community Relative Abundance/Species Evenness

##

###

##

_{}

###

where

###

###

= abundance of species###

###

###

= number of individuals of species###

###

= total number of individuals of all species Rank-AbundanceWhittaker plot/Rank-Abundance Curve

- species ranking based on relative abundance, ranked from most to least abundant ) x-axis and relative abundance (y-axis) expressed on a

### log

###

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

##

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##

_{}

###

where

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###

= total number of individuals of species###

Relative Density##

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##

_{ }

##

###

Diversity&Dominance Simpson’s Index##

###

##

###

## Simpson’s Index of Diversity 1−

## Simpson’s Diversity Index 1

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. Shannon-Weiner’s Index

##

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##

_{}

###

## ′

###

## ln

###

##

where

###

###

= proportion of individuals found in species###

###

###

= number of individuals in species###

###

= 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,

###

###

### ln

, where###

= 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

##

###

## # ℎ

_{ # }

Relative Frequency