AF-20-Module.pdf

Republic of the Philippines
SURIGAO DEL NORTE STATE UNIVERSITY
Mainit Campus
Magpayang, Mainit, Surigao del Norte, 8407
“For Nation’s Greater Heights”
AF 20
AGROFORESTRY STATISTICS AND MEASUREMENTS
module
Prepared by:
JULIE E. ASUBAR, RF

Introduction to
STATISTICS
For. Julie E. Asubar
Instructor

OUTLINE
1. Definitions
A. Statistics
I. Descriptive Statistics
II. Inferential Statistics
III. Variable
a. Categorical or Qualitative Variables
b. Numerical-Valued or Quantitative Variables
▪ Discrete
▪ Continuous
IV. Scales of Measurement
a. Nominal Scale
b. Ordinal Scale
c. Interval Scale
d. Ratio Scale
V. Population
a. Sample
VI. Parameters

Definitions: Statistics
▪ STATISTICS is a branch of applied mathematics which deals
with collection, organization, presentation, analysis and
interpretation of data.
▪ Statistician develop and apply appropriate methods in
collecting and analyzing data. They guide the design of a
research study then analyze the results. The interpretation of
the results is the basis of the statisticians in making inferences
about the population.

Definitions: Descriptive Statistics
▪ DESCRIPTIVE STATISTICS deals with the collection and
presentation of data and the collection of summarizing values
to describe its group characteristics.
▪ The most common summarizing values are the measures of
central tendency and variation.
• For instance, one can describe a class of 40 students by the mean score
in a given examination in Forest Economics. Suppose the mean score is
56 and the passing score is 45, then one can see that the general
majority of the students passed the test.
• If only 12 out of 60 students obtained scores above 45, then it means
that the exam is too difficult or the teaching is not effective so either the
teacher has to give another examination which is less difficult or re-
teach the topic.

Definitions: Inferential Statistics
▪ INFERENTIAL STATISTICS deals with the predictions and
inferences based on the analysis and interpretation of the
results of the information gathered by the statistician.
▪ Some of the common statistical tools of inferential statistics are
t-test, z-score, analysis of variance and Pearson r. These will be
discussed in the next topic.

Definitions: Variable
▪ VARIABLE is a numerical characteristics or attribute associated
with the population being studied.
▪ Types of Variable:
1. Categorical or Qualitative Variables – classified according to some
attributes or categories.
2. Numerical-Valued or Quantitative Variables – classified according to
numerical characteristics such as height, age, pulse rate, number of
children and speed. Numerical-valued variable are often grouped into
class intervals, and classified as discrete and continuous.

▪ Categorical or Qualitative Variables (Examples)
1. Gender, eye color, religion, blood type, civil status, year level, course,
profession and socio-economic status.
2. Categories may be ordered, which may or may not be assigned
specific numerical values such as:
a. Performance Rating (poor, fair, good, very good, and excellent)
b. I.Q. Score (low, average, high)
c. Satisfaction Level
1 – poor
2 – average
3 – good

▪ Numerical-Valued or Quantitative Variables (Examples)
1. Discrete – is a variable whose values are obtained by counting.
Example: number of children, number of ex-boyfriends/ex-girlfriends,
score in exams, students enrolled in statistics course, number of views,
number of mahogany tree in the campus, number ni sir
2. Continuous – is a variable whose values are obtained by measurement.
Example: temperature, distance, area, volume, density, age, height,
weight, etc. all of which cannot be put into a list because they can
have any value in some interval of real numbers.

Definitions: Scales of Measurement
▪ Measurement are usually obtained when gathering data by any method.
Examples of measurements are height in feet, weight in kilograms, age in
years, I.Q. scores, temperature in degree Celsius, incidence rate, mortality
rate, etc.
▪ In selecting the statistical tool to be used for drawing inferences on a
random sample, the type of measurement scale must be carefully chosen.
▪ Measurement are classified into four scales:
1. Nominal scale
2. Ordinal Scale
3. Interval Scale
4. Ratio Scale

▪ Nominal Scale – is a measurement scale that classifies elements into two or
more categories or classes, the numbers indicating that the elements are
different but not according to order or magnitude.
Course &
Year Level
Male Female LGBTQIA+ Total
BSAF 1 64 76 12 152
BSAF 2 57 72 7 136
BSAF 3 43 61 9 113
BSAF 4 34 50 18 102
Total 198 259 46 503

▪ Ordinal Scale – is a measurement scale that ranks individuals in terms of the
degree to which they posses a characteristics of interest.
Descriptive
Interpretation
Diversity (H’)
Very High > 3.50
High 3.00 – 3.49
Moderate 2.50 – 2.99
Low 2.00 – 2.49
Very Low < 1.99
Sites Shannon Diversity
San Pablo, Jabonga 4.12
Crossing, Kitcharao 3.66
Pungtod, Alegria 2.58
Motorpool, Tubod 3.49
Marayag, Mainit 3.98
Tolingon, Mainit 4.15
Overall 4.67
Lake Mainit species diversity as of 2021

▪ Interval Scale – is a measurement scale, in addition to ordering scores from
high to low or low to high, it also establishes a uniform unit in the scale so
that any equal distance between two scores is of equal magnitude.
▪ Example, aptitude scores from 80 – 90 are of equal difference as aptitude
scores from 90 – 100 (both being equal to 10).
▪ There is no absolute zero in this scale. For example, a place where the
temperature reading is 0° Celsius does not mean that there is no
temperature in that place.
Interval f
91 – 100 5
81 – 90 4
71 – 80 3
61 - 70 4
Interval f
0 – 2 4
3 – 5 3
6 – 8 4
9 – 11 5

▪ Ratio Scale – is a measurement scale, in addition to being an interval scale,
that also has an absolute zero in the scale.
Example: height, weight, volume, speed, rate of doing work, amount of
money deposited in a bank, etc.
Name
Height Weight
cm ft & in kg lbs.
Yeonjun 181 5”11” 65 143.3
Soobin 185 6”1” 69 152.1
Beomgyu 179 5”11” 55 121.3
Taehyun 177 5”10” 55 121.3
Huening Kai 183 6’ 61 134.5

Definitions: Population
▪ Population is defined as groups of people, animals, places, things or ideas
to which any conclusions based on characteristics of a sample will be
applied.
▪ Population means a group of organisms occupying a particular space,
living together and exchange genetic information. Every animal belongs to
a population, and each individual is influences by the size and
characteristics of and the social groupings within the population.
▪ Sample is a subgroup of the population that will represent the population as
a whole.

Definitions: Parameters
▪ Parameter is a numerical measure that describes a characteristics of a
population.
Example.
1. The population mean score of the BSAF 3rd year students in statistics exam
is 87.
2. The population average of farmers using organic approaches in farming in
the province of Surigao del Norte is 14.
▪ Statistics is a numerical measure that describes a characteristics of a
sample.
1. The sample mean score of the 30 BSAF 3rd year students in statistics exam is
85.
2. The sample average of farmers using organic approaches in farming in the
province of Surigao del Norte is 3.

Data Collection
▪ There is no formula for selecting the best method to be used in
gathering data. It depends on the researcher’s design of the
study, the type of data, the time available to complete the
study, and the financial capacity.
▪ Common methods used in data collection are:
1. Interview method
2. Questionnaire
3. Observation
4. Test
5. Experiment
6. Registration
7. Mechanical devices

Data Collection: Interview Method
▪ Direct Method – the researcher personally interviews the respondent.
Usually the interviewer calls for a meeting with the interviewee or visits him
or her at home. The method is appropriate to use if the needed
information is minimal and the number of respondents is few, less than 30
individuals.
▪ Indirect Method – the researcher uses the telephone (communication
gadgets) to interview he respondents. This method is quit expensive if there
are so many respondents. Also considered impolite and biased.

Data Collection: Questionnaire Method
▪ Questionnaire is a list of well planned questions written on paper which can
be either personally administered or mailed by the researcher to the
respondents using any of the following forms:
1. Guided-Response type
2. Recall type
3. Recognition type
4. Dichotomous type
5. Multiple choice type
6. Multiple response type
7. Free response type
8. Rating scale type

▪ Guided-Response type – the respondents is guided in making his/her reply
or answer.
1. Have you been studying?
Yes No
If your answer is yes, go to question 2. If your answer is no, go to question number 3.
2. Aptitude towards statistics: In answering the questions/statements below
refer to the following verbal equivalent of the used in the table.
1 – disagree
2 – slightly disagree
3 – moderately agree
4 – strongly agree
Statements 1 2 3 4
I love statistic as a subject
I’m always excited for my statistic classes
I will have a 1.0 grade in statistic

▪ Recall type
1. Name
2. Age
3. Gender
4. Civil status
5. Home address
6. Favorite subject
7. Scientific name of Narra

▪ Recognition type
1. Which of the following is a fruit?

▪ Dichotomous type – closed-ended questions, which only offers two possible
answers, and usually presented as Yes or No, True or False, Agree or
Disagree and Fair or Unfair.
1. Dinosaurs are extinct. True or False?
2. Am I single? Yes or No?
3. I deserve a 1.0 grade in statistics. Agree or Disagree?
4. You got 2.0 grade in statistics. Fair or Unfair?

▪ Multirole choice type
1. Which do you prefer?
a. Flower
b. Chocolate
c. Sex
2. Which of the following wildlife species previously known as monkey-
eating eagle?
a. Gorilla
b. Ape
c. Philippine Eagle
d. Vulture

▪ Multiple response type
1. Which of the following species are endemic to the country? Encircle
the letter of your answers.
a. Gmelina
b. Mahogany
c. Narra
d. Mangium
e. Yakal
f. Magkono
g. Sagimsim
h. Kamagong
i. Cacao
j. Agoho

▪ Rating scale type
1. How serious is your instructor in statistics?
very serious
serious
fairly serious
not serious
not a problem
2. Aptitude towards statistics: In answering the questions/statements
below refer to the following verbal equivalent of the used in the table.
1 – disagree
2 – slightly disagree
3 – moderately agree
4 – strongly agree
Statements 1 2 3 4
I love statistic as a subject
I’m always excited for my statistic classes
I will have a 1.0 grade in statistic

Data Collection: Observation Method
▪ Observation is commonly used in psychological and anthropological
studies. It is a method of obtaining data through seeing, hearing, tasting,
touching and smelling.
▪ Through observation, additional information maybe gathered which can
not be obtained using the other methods such as the questionnaire.
▪ The observer may participate in the activities of the group being studied
(participant observation) or he may just be a bystander (non-participant
observation).
▪ When observation is done in a laboratory as in the case of experimental
studies, the type of observation is called controlled observation.

Data Collection: Test Method
▪ Test method is widely used in psychological research and psychiatry.
Standard tests are used because of their validity, reliability and usability.
Example:
1. Aptitude test
2. IQ test
3. Achievement test
4. Entrance exam

Data Collection: Registration Method
▪ Examples:
1. Philippine Statistics Office
2. Land Transformation Office
3. DepEd
4. CHED
5. Supreme court
6. Population Commission
7. NBI
8. Other government agencies

Data Collection: Mechanical Devices
▪ Mechanical devices that can be used for social and educational research
in data gathering are the camera, projector, video tape, tape recorder,
etc.
▪ In chemical, biological and medical researches, the common methods
are x-ray machine, microscope, ultrasound, weighing scale, CT scan, etc.
▪ In astronomy, space and atmospheric researches, the telescope,
barometer computer, radar machines, camera and satellites are
commonly used.

Sampling Techniques
▪ Before the collection of data, it is necessary to determine the sample size if
the population is very large.
▪ The Slovin’s Formula:
n =
N
1 + Ne2
n – sample size
N – number of cases
e – margin of error

Sampling Techniques
▪ Example: A researcher wants to know the average income of the families
living in Barangay A which has 2,500 families. Calculate the sample size
that the researcher will need if a 5% margin of error is allowed.
n =
N
1 + Ne2
n =
2500
1 + 2500 (0.05)2
n = 344.8 or 345

Sampling Techniques
▪ The Raosoft sample calculator is basically a software that primarily
calculates or generates the sample size of a research or survey.
▪ To access the calculator, use the link below:
http://www.raosoft.com/samplesize.html?nosurvey
▪ For tutorial on how to use the calculator, click the link below:
https://www.youtube.com/watch?v=hQBK0N9dqF8

ST: Random Sampling
▪ In this method, all members of the population have equal chances of
being included in the study. This is applicable if the target population is not
classified into clusters, sections, levels, or classes. This method is easy to
apply, but not when the population is very large, say a thousand or more.
▪ Lottery Method is the most common and easiest method of random
sampling.
▪ The names of the respondents are written on small pieces of paper then
rolled and placed in a jar. The respondents who are included in the study
are those names written on the piece of paper picked at random from the
jar.

ST: Systematic Sampling
▪ Stratified Random Sampling is applied when the population is divided into
different strata or classes and each class must be represented in the study.
▪ Example: suppose a researcher wants to determine the average income
of the families in a barangay having 3,000 families that are distributed in 5
purok. Compute for the sample size n at 5% margin error.
n =
N
1 + Ne2
n =
3000
1 + 3000 (0.05)2
n = 353
Purok Population Percentage nk
1 800 27% 0.27 X 353 = 95
2 400 13% 0.13 X 353 = 46
3 500 17% 0.17 X 353 = 60
4 600 20% 0.20 X 353 = 71
5 700 23% 0.23 X 353 = 81
N 3000 100% 353

ST: Systematic Sampling
▪ Cluster Sampling is applicable when the geographical area where the
study is set to be conducted is too big and target population is too large.
In this technique, the selection of sample units is not by individual but by
groups called cluster. The area is divide into clusters then select a desired
number of clusters at random.
▪ Example: A forester wants to make a nationwide study on the diversity of
endemic plants in the country. He decided to take all 17 regions of the
country which can be considered as clusters. If 5 of the 17 regions / clusters
are the desired sample units, then the Forester can proceed to lottery
method procedures.

ST: Purposive Sampling
▪ The respondents of the study are chosen based on their knowledge of the
information required by the researcher.
▪ Example: Suppose a forester wants to study the ethnobotany of plants in
the watershed area of Lake Mainit and the target participants are the
senior citizens since they know more on ethnobotany. If there are 2,000
senior citizens and a margin of error allowed is 3%, then the sample size is
714. These 714 senior citizens will be chosen purposively based on the
preference of the researcher.

ST: Quota and Convenience Sampling
▪ Quota sampling is defined as a non-probability sampling method in which
researchers create a sample involving individuals that represent a
population. Commonly used in opinion polls.
▪ Convenience sampling is a type of non-probability sampling that involves
the sample being drawn from that part of the population that is close to
hand. This type of sampling is most useful for pilot testing.

Measures of Location
▪ Describes a dataset by citing a value within the range of the data that lies in a
specified location or position relative to the entire data set
1. Minimum
2. Maximum
3. Measures of Central Tendency
a. Mean
b. Median
c. Mode
4. Percentile
5. Quartile
6. Decile

Measures of Location
1. Minimum (Min) - the lowest value in the dataset.
2. Maximum (Max) - the highest value in the dataset.
3. Measures of Central Tendency - represent the value(s) where the data
observations tend to concentrate or cluster.
a. Mean (or arithmetic mean) – a single value defined as the sum of the data
values divided by the total number of data values.
b. Median - a single value at the middle of an array of data observations.
c. Mode - refers to the most frequent value in the dataset. When all the data
values have equal frequency, the mode doesn’t exist.
4. Percentile - divides an array into 100 equal parts, each part having one
percent of the distribution of the data values.
5. Decile - divide an array into ten equal parts.
6. Quartile - divide an array into four equal parts

*** Data
▪ Data are individual facts, statistics, or items of information, often
numeric, that are collected through observation. In a more technical
sense, data are a set of values of qualitative or quantitative variables
about one or more persons or objects

*** Types of Data
▪ Ungrouped data - when the data has not been placed in any
categories and no aggregation/summarization has taken placed on
the data then it is known as ungrouped data. Ungrouped data is also
known as raw data.
78 99 54 45 67
44 67 79 90 92
34 55 67 78 87
87 34 52 33 78
84 56 77 45 67
98 32 51 62 45

*** Types of Data
▪ Grouped data - when raw data have been grouped in different classes
then it is said to be grouped data.
Class Interval Tally Frequency (f)
91 – 100 III 3
81 – 90 IIII 4
71 – 80 IIIII 5
61 – 70 IIIII 5
51 – 60 IIIII 5
41 – 50 IIII 4
31 – 40 IIII 4
n = 30

Mean (Ungrouped Mean)
Example 1. The scores of 20 students in Wildlife
Management 20 items pre-test are 13, 17, 11, 9,
20, 12, 12, 15, 19, 12, 14, 4, 9, 11, 16, 17, 18, 19,
20, 11. Calculate the mean of the scores.
Solution:
Mean = 13 + 17 + 11 + 9 + 20 + 12 +
12 + 15 + 19 + 12 + 14 + 4 + 9
+ 11 + 16 + 17 + 18 + 19 + 20
+ 11
20
Mean = 279 / 20 = 13.95

Mean (Grouped Mean)
Example 1. The scores of 30 students in
Wildlife Management 100 items pre-test
are:
78 99 54 45 67
44 67 79 90 92
34 55 67 78 87
87 34 52 33 78
84 56 77 45 67
98 32 51 62 45
Calculate the mean of the scores.

Mean (Grouped Mean)
Steps 1. Prepare a frequency distribution
▪ Find the range of the scores.
▪ Decide on the number of groups or steps.
Maximum number of groups = 20
Minimum number of groups = 7
The ideal number of groups = 10 to 15
▪ Determine the class interval or step interval (ci)
Length of class interval (h) = R
ideal no.
▪ Decide where to start the classes.
▪ Group the scores based on the lowest limit then
tally the scores.
Steps 2. Prepare five (5) columns
▪ 1st column for steps or scores distribution
/ interval
▪ 2nd column for the tally
▪ 3rd column for the frequency (f)
▪ 4th column for the midpoint
▪ 5th column for the product of
frequency and midpoint

Mean (Grouped Mean)
Example 1. The scores of 30 students in
Wildlife Management 100 items pre-test
are:
78 99 54 45 67
44 67 79 90 92
34 55 67 78 87
87 34 52 33 78
84 56 77 45 67
98 32 51 62 45
Calculate the mean of the scores.
Highest number
Lowest number
Range (R) = Highest – Lowest
= 99 – 32
= 67
Length of class interval (h)
= R ÷ ideal no. of groups
= 67 ÷ 7
= 9.57 or 10
Note for selecting the number of groups:
▪ Maximum number of groups = 20
▪ Minimum number of groups = 7
▪ The ideal number of groups = 10 to 15

Mean (Grouped Mean)
Mean = Ʃfx ÷ n
= 1935 ÷ 30
= 64.5
Interpretation
Out of the 100 items pre-test in
wildlife management subject, the
30 students obtained an average
(mean) score of 64.5.
Class Interval Tally Frequency (f) Midpoint (x) fx
91 – 100 III 3 95.5 286.5
81 – 90 IIII 4 85.5 342.0
71 – 80 IIIII 5 75.5 377.5
61 – 70 IIIII 5 65.5 327.5
51 – 60 IIIII 5 55.5 277.5
41 – 50 IIII 4 45.5 182.0
31 – 40 IIII 4 35.5 142.0
n = 30 Ʃfx = 1935.0
78 99 54 45 67
44 67 79 90 92
34 55 67 78 87
87 34 52 33 78
84 56 77 45 67
98 32 51 62 45
Example 1. The scores of
30 students in Wildlife
Management 100 items
pre-test are:
Calculate the mean of
the scores.

Problem 1. Calculate the ungrouped Mean of the following data 3, 13, 11, 15, 5,
4, 2, 3, 2.
Problem 2. Calculate the grouped mean of the data in table. (Use 10
classes/groups with an interval of 9).
45 29 34 45 61 83 45 62 66 83
67 56 50 50 81 88 90 100 23 19
84 57 33 94 50 21 23 59 69 70
56 78 33 23 45 80 30 21 44 88
23 56 89 98 76 58 54 56 78 91

Median (Ungrouped Median)
Note: For an odd number of the dataset, the median is the middle value.
Example 1. The scores of 15 students in Wildlife Management 20 items pre-test are 13, 17, 11, 9,
20, 12, 19, 12, 14, 4, 9, 11, 16, 17, 18. What is the median of the scores?
First, arrange the scores in ascending order.
4, 9, 9, 11, 11, 12, 12, 13, 14, 16, 17, 17, 18, 19, 20
Then determine the middle score/value.
Interpretation:
Out of the 20 items pre-test in wildlife management subject, the 15 students obtained a
median score of 13.

Median (Ungrouped Median)
Note: For an even number of the dataset, the median is the average of the two middle
values.
Example 2. The scores of 20 students in Wildlife Management 20 items pre-test are 13, 17, 11, 9,
20, 12, 12, 15, 19, 12, 14, 4, 9, 11, 16, 17, 18, 19, 20, 11. What is the median of the scores.
First, arrange the scores in ascending order
4, 9, 9, 11, 11, 11, 12, 12, 12, 13, 14, 15, 16, 17, 17, 18, 19, 19, 20, 20
Identify the middle scores (13 & 14), then add the scores and divide by 2.
The Median = (13 + 14) ÷ 2 = 27 ÷ 2 = 13.5
Interpretation: Out of the 20 items pre-test in wildlife management subject, the 20 students
obtained a median score of 13.5

Median (Grouped Median)
(N/2) – F
Median = l + ------------ * h
f
l – lower limit of the median class
f – frequency of the median class
F – cumulative frequency of the class preceding the median class
N – total number of the observations
h – width of the median class
78 99 54 45 67 87 34 52 33 78
44 67 79 90 92 84 56 77 45 67
34 55 67 78 87 98 32 51 62 45
Example 1.
Note: The class whose cumulative
frequency is greater than the
value of N/2 is called the median
class

Median (Grouped Median)
Class
Interval
Tally
Frequency
(f)
cf
91 – 100 III 3 30
81 – 90 IIII 4 27
71 – 80 IIIII 5 23
61 – 70 IIIII 5 18
51 – 60 IIIII 5 13
41 – 50 IIII 4 8
31 – 40 IIII 4 4
n = 30
(N/2) – F
Median = l + ------------ * h
f
(30/2) – 13
= 61 + ------------ * 10
5
15 – 13
= 61 + ------------ * 10
5
2
= 61 + ------------ * 10
5
= 61 + 0.4 * 10
= 65
(f) frequency of the
median class
The median class
(F) cumulative
frequency of the class
preceding the
median class
(l) lower limit of the
median class

Problem 1. What is the ungrouped median of the following data 3, 13, 11, 15, 5,
4, 2, 3, 2?
Problem 2. Calculate the grouped median of the data in table. (Use 10
45 29 34 45 61 83 45 62 66 83
67 56 50 50 81 88 90 100 23 19
84 57 33 94 50 21 23 59 69 70
56 78 33 23 45 80 30 21 44 88
23 56 89 98 76 58 54 56 78 91

Mode (Types of Mode)
▪ Unimodal is a score distribution with one mode.
▪ Bimodal is a score distribution with two modes.
▪ Trimodal is a score distribution with three modes. It is also considered as multi-
modal which refers to more than two modes.

Mode (Ungrouped Mode)
Example 1. The scores of 20 students in Wildlife Management 20 items pre-test
are 13, 17, 11, 9, 20, 12, 12, 15, 19, 12, 14, 4, 9, 11, 16, 17, 18, 19, 20, 11. What is the
mode of the scores.
*Arrange the scores in ascending or descending order;
4, 9, 9, 11, 11, 11, 12, 12, 12, 13, 14, 15, 16, 17, 17, 18, 19, 19, 20, 20
*Find the observation which occurs the maximum number of times.
4, 9, 9, 11, 11, 11, 12, 12, 12, 13, 14, 15, 16, 17, 17, 18, 19, 19, 20, 20
Bimodal

Mode (Grouped Mode)
▪ Step 1. Find the maximum
class frequency.
▪ Step 2. Find the class
corresponding to this
frequency. It is called the
modal class.
▪ Step 3. Find the class size.
(upper limit – lower limit.)
▪ Step 4. Calculate mode
using the formula.
f1 – f0
Mode (Mo) = l + ---------------- * h
2 f1 – f0 – f2
l – lower limit of the modal class
h – size of the class interval
f1 – frequency of the modal class
f0 – frequency of the class preceding the modal class
f2 – frequency of the class succeeding the modal class
Note: The class with the highest frequency is the
modal class

Mode (Grouped Mode)
Example 1. The scores of 30 students in Wildlife Management 100 items pre-test
are:
64 99 54 45 67 87 34 52 33 78
44 67 79 90 92 84 56 77 45 67
34 55 67 78 87 98 32 51 62 45
Calculate the mode of the scores.

Mode (Grouped Mode)
Class
Interval
Tally
Frequency
(f)
91 – 100 III 3
81 – 90 IIII 4
71 – 80 IIII 4
61 – 70 IIIII-I 6
51 – 60 IIIII 5
41 – 50 IIII 4
31 – 40 IIII 4
n = 30
f1 – f0
Mode (Mo) = l + ---------------- * h
2 f1 – f0 – f2
6 – 5
= 61 + ---------------- * 10
2 (6) – 5 – 4
1
= 61 + ---------------- * 10
3
= 61 + .33 * 10
= 64.3
Note: Interval is 10
f2 – frequency of the
class succeeding the
modal class
modal class
class preceding the
modal class
l – lower limit of the
modal class
The modal class

Problem 1. What is the ungrouped Mode of the following data 3, 13, 11, 15, 5, 4,
2, 3, 2?
Problem 2. Calculate the grouped mode of the data in table. (Use 10
45 29 34 45 61 83 45 62 66 83
67 56 50 50 81 88 90 100 23 19
84 57 33 94 50 21 23 59 69 70
56 78 33 23 45 80 30 21 44 88
23 56 89 98 76 58 54 56 78 91

Quartile, Decile & Percentile
Example 1. Calculate
Quartiles-2, Deciles-7,
Percentiles-20 from
the following data 3,
13, 11, 11, 5, 4, 2
* First arrange the
observations in
ascending order (2, 3,
4, 5, 11, 11, 13)

Quartile
Qi = i (n + 1) th
4
Q2 = 2 (7 + 1) th
4
= 16 th
4
= 4th observation
= 5
Percentiles-20 from
13, 11, 11, 5, 4, 2
* First arrange the
observations in
4, 5, 11, 11, 13)
2 , 3 , 4 , 5 , 1 1 , 1 1 , 1 3

Decile
Di = i (n + 1) th
10
D7 = 7 (7 + 1) th
10
= 56 th
10
= 5.6th observation
= 5th observation + 0.6 (6th – 5th)
= 11 + 0.6 (11 – 11)
= 11
Percentiles-20 from
13, 11, 11, 5, 4, 2
* First arrange the
observations in
4, 5, 11, 11, 13)
2 , 3 , 4 , 5 , 1 1 , 1 1 , 1 3
5th
observation
6th
observation
Separate decimals from
the whole number.
Example. 5.6 to 5 and 0.6

Percentile
Percentiles-20 from
13, 11, 11, 5, 4, 2
* First arrange the
observations in
4, 5, 11, 11, 13)
2 , 3 , 4 , 5 , 1 1 , 1 1 , 1 3
1st
observation
2nd
observation
Separate decimals from
the whole number.
Example. 1.6 to 1 and 0.6
Pi = i (n + 1) th
100
P20 = 20 (7 + 1) th
100
= 160 th
100
= 1.6th observation
= 1st observation + 0.6 (2nd – 1st)
= 2 + 0.6 (3 – 2)
= 2.6

Problem.
Following are the scores of the 3rd year BSAF students in the 50 items
statistics quiz. 45, 30, 36, 26, 16, 21 33, 40, 32, 14, 10, 29, 23, 39, 17, 11, 18, 34, 19,
24, 21, 35, 42, 37
Calculate the corresponding value of the following:
1. The 2nd quartile
2. The 5th decile
3. The 50th percentile

AF 20 - STATISTICS
& MEASUREMENTS
Julie E. Asubar, RF
MEASURE
MEASURE
VARIABILITY
VARIABILITY
of
Instructor

MEASURE OF VARIABILITY
Measures of variability or dispersion describe the spread of the data. This
include the range, interquartile range, standard deviation and variance.
A measure of variability is a summary statistic that represents the amount
of dispersion in a dataset. How spread out are the values?
While a measure of central tendency describes the typical value,
measures of variability define how far away the data points tend to fall from the
center.
We talk about variability in the context of a distribution of
values. A low dispersion indicates that the data points tend to be
clustered tightly around the center. High dispersion signifies that they
tend to fall further away.

MEASURES OF VARIATION
Measures of variation
give information on the
spread or variability of
the data value
In statistics, variability,
dispersion, and spread
are synonyms that denote
the width of the
distribution.

1. Range
▪ Range is the simplest
measure of variation .
▪ Range is the difference
between the largest and
the smallest values in a
data set
Formula:
Range = X Largest – X Smallest
78 99 54 45 67
44 67 79 90 92
34 55 67 78 87
87 34 52 33 78
84 56 77 45 67
98 32 51 62 45
Find the range of the data set.
Range = 99 – 32
= 67

1. Range
Disadvantages of Range
▪ Ignores the way in which data are distributed
▪ Sensitive to outliers

1.1 Range (Coefficient of Range)
Coefficient of Range = [L – S] ÷ [L + S]
Where:
L – is the largest value in the data set
S – is the smallest value in the data set
78 99 54 45 67
44 67 79 90 92
34 55 67 78 87
87 34 52 33 78
84 56 77 45 67
98 32 51 62 45
Find the range of the data set.
Coefficient of Range
= [99 – 32] ÷ [99 + 32]
= 67 ÷ 131
= 0.51
The coefficient of range is the ratio
of difference between the highest
and lowest value of frequency to
the sum of highest and lowest value
of frequency.

2. Interquartile Range
Interquartile Range = Q3 – Q1
Where:
Q3 – is the 3rd quartile
Q1 – is the 1st quartile
The interquartile range (IQR), also called the midspread, middle 50%, or H-spread, is
a measure of statistical dispersion, being equal to the difference between 75th and
25th percentiles, or between upper and lower quartiles.

With an Even Sample Size:
For the sample (n=10) the median of a 100 items pre-test is 71 (50% of the values are
above 71, and 50% are below). The quartiles can be determined in the same way we
determined the median, except we consider each half of the data set separately.
There are 5 values below the median (lower half), the middle value is 64 which is the first
quartile. There are 5 values above the median (upper half), the middle value is 77 which
is the third quartile. The interquartile range is 77 – 64 = 13; the interquartile range is the
range of the middle 50% of the data.

With an Odd Sample Size:
When the sample size is odd, the median and quartiles are determined in the same
way. Suppose in the previous example, the lowest value (62) were excluded, and the
sample size was n=9. The median and quartiles are indicated below.
When the sample size is 9, the median is the middle number 72. The quartiles are determined in
the same way looking at the lower and upper halves, respectively. There are 4 values in the
lower half, the first quartile is the mean of the 2 middle values in the lower half ((64+64)/2=64).
The same approach is used in the upper half to determine the third quartile ((77+81)/2=79).

3. Mean Absolute Deviation
The Mean Absolute Deviation (MAD) of a dataset is the average distance between
each data point and the mean. It gives us an idea about the variability in a dataset.
To find the mean absolute deviation, subtract the mean score from each raw score
then using the absolute values of the differences, get the sum of the results. The sum is called
the sum of the deviation from the mean. Next, divide the number by N, the total number of
cases.
In symbols
MAD = Ʃ [ x - x̄ ]
N
MAD = Ʃf [ x - x̄ ]
N
Ungrouped Data
Grouped Data

Example 1. The ages of the BSAF Instructors are 34, 35, 45, 56, 32, 25 and 40. Find the MAD of
the ages of BSAF Instructors.
x
34
35
45
56
32
25
40
Ʃx =
267
The raw data
Summation of all
raw data
The difference of the x
and the mean (x̄)
Multiply the difference of
the x and the mean (x̄)
by its sign
x - x̄
-4.14
-3.14
6.86
17.86
-6.14
-13.14
1.86
[ x - x̄ ]
4.14
3.14
6.86
17.86
6.14
13.14
1.86
Ʃ [ x - x̄ ] =
53.14
Mean ( x̄ ) = Ʃx ÷ N
= 267 ÷ 7
= 38.14
Summation of the
difference of the x and
the mean (x̄)
multiplied by its sign

Example 1. The ages of the BSAF Instructors are 34, 35, 45, 56, 32, 25 and 40. Find the MAD of
the ages of BSAF Instructors.
MAD = Ʃ [ x - x̄ ]
N
= 53.14
7
= 7.59
x x - x̄ [ x - x̄ ]
34 -4.14 4.14
35 -3.14 3.14
45 6.86 6.86
56 17.86 17.86
32 -6.14 6.14
25 -13.14 13.14
40 1.86 1.86
Ʃx =
267
Ʃ [ x - x̄ ] =
53.14
= 267 ÷ 7
= 38.14
Therefore, the mean
absolute deviation of
the BSAF Instructors
ages is 7.59

Example 2. Calculate the Mean Absolute Deviation (MAD) of a grouped data as shown in
the table.
Mean (x̄ ) = ƩfX ÷ N
= 10430 ÷ 115
= 90.7
Class
Interval
f
100-104 15
95-99 19
90-94 34
85-89 25
80-84 12
75-79 10
N =
115
X
102
97
92
87
82
77
fX
1530
1843
3128
2175
984
770
ƩfX =
10430
x - x̄
11.3
6.3
1.3
-3.7
-8.7
-13.7
[ x - x̄ ]
11.3
6.3
1.3
3.7
8.7
13.7
f [ x - x̄ ]
169.50
119.70
44.20
92.50
104.40
137.00
f [ x - x̄ ] =
667.30
given median
frequency
times median
median
minus mean
(median – mean)
frequency times
(median – mean)
number of observation Summation of frequency times median
Summation of frequency times
(median – mean) multiplied by its sign

Example 2. Calculate the Mean Absolute Deviation (MAD) of a grouped data as shown in
the table.
Therefore, the mean absolute
deviation of the grouped data is
at 5.8
Class
Interval
f X fX x - x̄ [ x - x̄ ] f [ x - x̄ ]
100-104 15 102 1530 11.3 11.3 169.50
95-99 19 97 1843 6.3 6.3 119.70
90-94 34 92 3128 1.3 1.3 44.20
85-89 25 87 2175 -3.7 3.7 92.50
80-84 12 82 984 -8.7 8.7 104.40
75-79 10 77 770 -13.7 13.7 137.00
N =
115
ƩfX =
10430
f [ x - x̄ ] =
667.30
Mean (x̄ ) = ƩfX ÷ N
= 10430 ÷ 115
= 90.7
MAD = Ʃf [ x - x̄ ]
N
= 667.30
115
= 5.8

4. Variance
The variance is a measure of variability. It is calculated by taking the average of
squared deviations from the mean.
Variance tells you the degree of spread in your data set. The more spread the data,
the larger the variance is in relation to the mean.
Why does variance matter?
Variance matters for two main reasons:
1. Parametric statistical tests are sensitive to variance.
2. Comparing the variance of samples helps you assess group differences.

4. Variance (Population Variance)
When you have collected data from every member of the population that you’re
interested in, you can get an exact value for population variance.
The population variance formula looks like this:
For Grouped Data For Ungrouped Data Where:
σ2 – population variance
Ʃ – sum of…
X – each value
ɥ - population mean
N – number of values in the
population
Ʃf (x - ɥ)2
σ2
N = -------------
N
Ʃ(x - ɥ)2
σ2
N = -------------
N

4. Variance (Sample Variance)
When you collect data from a sample, the sample variance is used to make
estimates or inferences about the population variance.
The sample variance formula looks like this:
For Grouped Data For Ungrouped Data Where:
s2 – sample variance
Ʃ – sum of…
X – each value
x̄ - population mean
N – number of values in the
population
NƩfx2
– (Ʃfx)2
s2
N-1 = -------------------
N (N – 1)
Ʃ(x - x
̄ )2
s2
N-1 = -------------
N – 1

4. Variance
Example 1. Find the population and sample variances of the following BSAF Instructor ages:
34, 35, 45, 56, 32, 25 and 40.
x
34
35
45
56
32
25
40
Ʃx =
267
The raw data
Summation of all
raw data
The difference of the x
and the mean (x̄)
x minus mean (x̄) raise to
the power of 2 or
multiplied by itself
x - x̄
-4.14
-3.14
6.86
17.86
-6.14
-13.14
1.86
(x - x̄)2
17.14
9.86
47.06
318.98
37.70
172.66
3.46
Ʃ(x - x̄)2 =
606.86
Summation of the difference
between x and the mean
(x̄) raise to the power of 2 or
multiplied by itself
= 267 ÷ 7
= 38.14

4. Variance
Example 1. Find the population and sample variances of the following BSAF Instructor ages:
34, 35, 45, 56, 32, 25 and 40
Mean ( ɥ or x̄ ) = Ʃx ÷ N
= 267 ÷ 7
= 38.14
x x - x̄ (x - x̄)2
34 -4.14 17.14
35 -3.14 9.86
45 6.86 47.06
56 17.86 318.98
32 -6.14 37.70
25 -13.14 172.66
40 1.86 3.46
Ʃx =
267
Ʃ(x - x̄)2 =
606.86
σ2N = Ʃ(x - ɥ)2
N
= 606.86
7
= 86.7
s2N-1 = Ʃ(x - x̄)2
N – 1
= 606.86
7 – 1
= 101.14
Ungrouped Data
Population Variance
Ungrouped Data
Sample Variance

4. Variance
Example 2. Calculate the population and sample variances of a grouped data as shown in
the table.
Class
Interval
f
100-104 15
95-99 19
90-94 34
85-89 25
80-84 12
75-79 10
N =
115
Mean (ɥ or x̄ ) = ƩfX ÷ N
= 10430 ÷ 115
= 90.7
given
number of observation
frequency
times median
median raise to
the power of 2
median
minus mean
the difference of
median minus mean
raise to the power of 2
frequency times the
difference of median
minus mean that was
raised to the power of 2
X
102
97
92
87
82
77
fX
1530
1843
3128
2175
984
770
ƩfX =
10430
x2
10404
9409
8464
7569
6724
5929
fX2
156060
178771
287776
189225
80688
59290
ƩfX2 =
951810
x - x̄
11.3
6.3
1.3
-3.7
-8.7
-13.7
(x - x̄)2
127.69
39.69
1.69
13.69
75.69
187.69
f(x - x̄)2
1915.35
754.11
57.46
342.25
908.28
1876.9
Ʃf(x - x̄)2 =
5854.35
The product of frequency
times the median that was
summation of frequency
times median
summation of the product of frequency times
the median that was raised to the power of 2
summation of the
frequency times the
difference of median
minus mean that was
median

4. Variance
the table.
Class
Interval
f X fX x2 fX2 x - x̄ (x - x̄)2 f(x - x̄)2
100-104 15 102 1530 10404 156060 11.3 127.69 1915.35
95-99 19 97 1843 9409 178771 6.3 39.69 754.11
90-94 34 92 3128 8464 287776 1.3 1.69 57.46
85-89 25 87 2175 7569 189225 -3.7 13.69 342.25
80-84 12 82 984 6724 80688 -8.7 75.69 908.28
75-79 10 77 770 5929 59290 -13.7 187.69 1876.9
N =
115
ƩfX =
10430
ƩfX2 =
951810
Ʃf(x - x̄)2 =
5854.35
= 10430 ÷ 115
= 90.7

4. Variance
σ2N = Ʃf (x - ɥ)2
N
= 5854.35
115
= 50.91
s2N-1 = NƩfx2 – (Ʃfx)2
N (N – 1)
= 115 (951,810) – (10,430)2
115 (115 – 1)
= 109,458,150 – 108,784,900
13,100
= 673,250
13,100
= 51.39
Grouped Data Population Variance Grouped Data Sample Variance

5. Standard Deviation
The standard deviation, σ for a population or s for a sample, is the square root of the
value of the variance.
The standard deviation is derived from variance and tells you, on average, how far
each value lies from the mean. It’s the square root of variance.
Both measures reflect variability in a distribution, but their units differ:
▪ Standard deviation is expressed in the same units as the original values (e.g., meters).
▪ Variance is expressed in much larger units (e.g., meters squared)
Since the units of variance are much larger than those of a typical
value of a data set, it’s harder to interpret the variance number intuitively.
That’s why standard deviation is often preferred as a main measure of
variability.
However, the variance is more informative about variability than the
standard deviation, and it’s used in making statistical inferences.

Population Standard Deviation (σ) Sample Standard Deviation (s)
σ = σ2
N s = s2
N-1
Where:
σ – population standard deviation
σ2
N – population variance
Where:
s – population standard deviation
s2
N-1 – sample variance

Example 1. Find the standard deviation of the following BSAF Instructor ages: 34, 35, 45, 56,
32, 25 and 40.
Mean ( ɥ or x̄ ) = Ʃx ÷ N
= 267 ÷ 7
= 38.14
x x - x̄ (x - x̄)2
34 -4.14 17.14
35 -3.14 9.86
45 6.86 47.06
56 17.86 318.98
32 -6.14 37.70
25 -13.14 172.66
40 1.86 3.46
Ʃx =
267
Ʃ(x - x̄)2 =
606.86
σ2N = Ʃ(x - ɥ)2
N
= 606.86
7
= 86.7
s2N-1 = Ʃ(x - x̄)2
N – 1
= 606.86
7 – 1
= 101.14
Ungrouped Data
Population Variance
Ungrouped Data
Sample Variance

Example 1. Find the standard deviation of the following BSAF Instructor ages: 34, 35, 45, 56,
32, 25 and 40.
σ2N = Ʃ(x - ɥ)2
N
= 606.86
7
= 86.7
s2N-1 = Ʃ(x - x̄)2
N – 1
= 606.86
7 – 1
= 101.14
Ungrouped Data
Population Variance
Sample Standard
Deviation
Population Standard
Deviation
Ungrouped Data
Sample Variance
σ = σ2
N
= 86.7
= 9.31
s = s2
N-1
= 101.14
= 10.06

the table.
Class
Interval
f X fX x2 fX2 x - x̄ (x - x̄)2 f(x - x̄)2
100-104 15 102 1530 10404 156060 11.3 127.69 1915.35
95-99 19 97 1843 9409 178771 6.3 39.69 754.11
90-94 34 92 3128 8464 287776 1.3 1.69 57.46
85-89 25 87 2175 7569 189225 -3.7 13.69 342.25
80-84 12 82 984 6724 80688 -8.7 75.69 908.28
75-79 10 77 770 5929 59290 -13.7 187.69 1876.9
N =
115
ƩfX =
10430
ƩfX2 =
951810
Ʃf(x - x̄)2 =
5854.35
= 10430 ÷ 115
= 90.7

σ2N = Ʃf (x - ɥ)2
N
= 5854.35
115
= 50.91
N (N – 1)
= 115 (951,810) – (10,430)2
115 (115 – 1)
= 109,458,150 – 108,784,900
13,100
= 673,250
13,100
= 51.39
Grouped Data
Population Variance
Grouped Data
Sample Variance
Population Standard
Deviation
σ = σ2
N
= 50.91
= 7.14
Sample Standard
Deviation
s = s2
N-1
= 51.39
= 7.17

6. Coefficient of Variation
CV = ( s ÷ x̄ ) x 100%
Where:
s – standard deviation
x̄ – mean
Note: if the standard deviation and mean have the same units, their units
are canceled out and the CV has no unit
The coefficient of variation (CV) is a measure of relative variability. It is the ratio
of the standard deviation to the mean (average).
The CV is particularly useful when you want to compare results from two
different surveys or tests that have different measures or values.

CV (male) = ( s ÷ x̄ ) x 100%
= (10 ÷ 162) x 100%
= 6.17%
Example 1. Supposed two groups of students are to be compared in terms of height.
Group Mean height Standard Deviation CV
Male 162 cm 10 cm
Female 148 cm 4 cm
CV (female) = ( s ÷ x̄ ) x 100%
= (4 ÷ 148) x 100%
= 2.70%
Comparing the relative variations in height
of the male and female students, it can be
seen that the male students have higher
coefficient of variation in height than the
female students. Thus, male students height
are more varied.

CV (height) = ( s ÷ x̄ ) x 100%
= (12 ÷ 168) x 100%
= 7.14%
Example 2. Compare the variability of the height and the weight of the students given
the following data
Mean height Standard Deviation CV
Height in cm 168 cm 12 cm
Weight in lb 200 lb 20 lb
CV (weight) = ( s ÷ x̄ ) x 100%
= (20 ÷ 200) x 100%
= 10.00%
From the results, it can be seen that the
weight of the students is more varied than
the height.

Problem 1. Ungrouped data. The scores of BSAF 3 students in AF 20 midterm
examination are listed as follows.
50 25 37 40 28
35 50 55 32 30
45 70 75 55 40
Calculate the:
a. Range
b. Coefficient of Range
c. Interquartile Range
d. Mean Absolute Deviation (MAD)
e. Variance (Population and Sample)
f. Standard Deviation (Population and Sample)
g. Coefficient of Variation (CV)

Problem 2. Grouped data. The scores of BSAF 3 students in AF 20 midterm examination
are presented in the table.
Calculate the:
a. Range
b. Coefficient of Range
c. Interquartile Rage
d. Mean Absolute Deviation (MAD)
e. Variance (Population and Sample)
f. Standard Deviation (Population and Sample)
g. Coefficient of Variation (CV)
Class Interval
(scores)
Frequency
(f)
91-95 4
86-90 11
81-85 14
76-80 8
71-75 35
66-70 12
61-65 15

DISTRIBUTION
NORMAL
Instructor

NORMAL DISTRIBUTION
The distribution of some human abilities and characteristics such as mental
ability tends to follow a certain specific shape called the normal distribution. When the
distribution is normal, most of the observations (about 68%) tend to converge at the
middle and the rest are distributed to the left and the right ends of the distribution.
The normal curve is bell-shaped. The mean, median and mode values are
equal and coincide in one point when the graph is drawn.

THE STANDARD NORMAL DISTRIBUTION
The normal curve is the graph of the equation
Where:
z = z-score
σ = population standard deviation
e ≈ 2.718 (≈ means approximately equal to
𝝅 ≈ 3.14

1. z-score
A distribution which is not normal can be normalized by changing all scores in
the distribution into z-scores. The graph using z-scores as points is a normal curve. The
total area under the curve is 1 (obtained by applying the integral calculus in equation
1).
At the vertex of the normal curve lies the mean, median and mode values.
Since it is bell-shaped, it means the right side and left side of the curve are symmetric
with respect to the vertical axis. The area under the curve to the right is 0.5 and to the
left is also 0.5. At the vertex also lies the z-score 0, and all z-scores to the right side are
positive and to the left are the negative z-scores.
x = score
μ = mean
σ = population standard deviation
Formula:
z = x – μ
σ

1. z-score
Example 1. Convert the following scores to z-scores where μ = 75 and σ = 5
A. x = 75
z = x – μ
σ
z = 75 – 75
5
z = 0
a. 75
b. 80
c. 58
B. x = 80
z = x – μ
σ
z = 80 – 75
5
z = 1
C. x = 58
z = x – μ
σ
z = 58 – 75
5
z = -3.4
Note: many statistical problems can be solved by applying the z-scores and
areas under the normal curve.

1. z-score
Example 2. The following are the scores in AF 20 quiz of 27 students:
12 10 9 10 12 15 15 16 15
20 22 23 10 12 10 14 16 17
18 20 20 21 10 12 23 10 10
Solve following:
a. Convert scores 10 and 20 to z-scores.
b. What percent of the class obtained scores higher than 20?
c. How many students obtained a score less than 20?
d. How many students scored between 10 and 20?

1. z-score
a. Convert scores 10 and 20 to z-
scores.
Score z-scores
9 -1.37
10 -1.09
12 -0.64
14 -0.20
15 0.02
16 0.24
17 0.47
18 0.69
20 1.13
21 1.36
22 1.58
23 1.80
Given:
μ = 14.9
σ = 4.5
The first step is to calculate the
mean and the population standard
deviation of the scores.
Then use the z-score formula.

1. z-score
From the Table of Areas Under the Normal Curve, the area under the curve
when z is 1.13 is A = 0.3708 or 37.08%. This represent the total number of students who
scored 20 and below. The percentage of students who scored more than 20 (shaded
portion of the bell curve below) is:
0.50 – 0.3708 = 0.1292 or equivalent to 12.92% of the whole class.
Score z-scores
20 1.13

1. z-score
0.50 – 0.3708 = 0.1292 or equivalent
to 12.92% of the whole class.
Score z-scores
Area Under the
Normal Curve
20 1.13 0.3708 or 37.08%.

1. z-score
c. How many students obtained a score less than 20?
The number of students who
scored less than 20 is 0.3708 + 0.5 =
0.8708 or 87.08% of all the 27 students in
the class which is 23.51. Hence, there are
about 24 students who scored less than
20 in the test.
The desired area is the region
lying to the left of z = 1.13

1. z-score
d. How many students scored between 10 and 20?
The total area under the curve
for the interval (-1.09, 1.31) is 0.7329 or
73.29% of the 27 students or 19.79 or 20
students.
z-score of 20 = 1.13, Area = 0.3708
z-score of 10 = -1.09, Area = 0.3621
The desired area lies between z
= 1.09 and z = 1.13.
-1.09< z >1.13

2. Skewness
Skewness refers to the symmetry or asymmetry of the frequency distribution
and its measure can be obtained by using Pearson’s median skewness coefficient
formula.
Sk = 3 (x̄ – Md )
s
Where:
x̄ - mean
Md - median
s – standard deviation

2. Skewness
If the values of the mean
and median are equal, the
distribution is normal and the graph
is a bell curve. If the observations
are concentrated at the left side of
the vertical axis and has fewer
observation at the right side, it is
called a positively skewed
distribution and if the observations
are concentrated at the right side ,
we have what we call a negatively
skewed distribution.
In a positively skewed distribution, the mean is higher than the
median, while in a negatively skewed distribution, the mean is lower than
the median.

2. Skewness
Problem 1. Calculate the degree of
skewness of a distribution if the mean is
45, median is 40 and the standard
deviation is 5.
Sk = 3 (x̄ – Md )
s
= 3 (45 – 40)
5
= 3(5)
5
= 3
Problem 2. Calculate the degree of
skewness of a distribution if the mean is
90, median is 99 and a variance of is 81.
Hence, the distribution is positively skewed.
Sk = 3 (x̄ – Md )
s
= 3 (90 – 99)
9
= 3(-9)
9
= -3
Hence, the distribution is negatively skewed.

2. Skewness
Problem 3. Find the degree of skewness of the following
data.
Scores f
90-94 1
85-89 4
80-84 3
75-79 8
70-74 20
65-69 15
60-64 7
55-59 1
50-54 1
N =
60
X
92
87
82
77
72
67
62
57
52
fX
92
348
246
616
1440
1005
434
57
52
ƩfX =
4290
<CF
60
59
55
52
44
24
9
2
1
Class
Boundary
89.5-94.5
84.5-89.5
79.5-84.5
74.5-79.5
69.5-74.5
64.5-69.5
59.5-64.5
54.5-59.5
49.5-54.5
X2
8464
7569
6724
5929
5184
4489
3844
3249
2704
fX2
8464
30276
20172
47432
103680
67335
26908
3249
2704
ƩfX2 =
310220
Scores & frequency
Median
Median raise to
the power of 2
Frequency
times median
Cumulative
frequency
Lower limit minus 0.5
and upper limit plus 0.5
Frequency times the median that
was raised to the power of 2
Number of
distribution
Summation of
frequency times median
Summation of f frequency times the
median that was raised to the power of 2

2. Skewness
Problem 3. Find the degree of skewness of the following data.
Scores f X fX <CF
Class
Boundary
X2
fX2
90-94 1 92 92 60 89.5-94.5 8464 8464
85-89 4 87 348 59 84.5-89.5 7569 30276
80-84 3 82 246 55 79.5-84.5 6724 20172
75-79 8 77 616 52 74.5-79.5 5929 47432
70-74 20 72 1440 44 69.5-74.5 5184 103680
65-69 15 67 1005 24 64.5-69.5 4489 67335
60-64 7 62 434 9 59.5-64.5 3844 26908
55-59 1 57 57 2 54.5-59.5 3249 3249
50-54 1 52 52 1 49.5-54.5 2704 2704
N =
60
ƩfX =
4290
ƩfX2 =
310220
A. Solve the MEAN
Mean = ƩfX ÷ N
= 4290 ÷ 60
= 71.5

2. Skewness
B. Solve the MEDIAN
(N/2) – F
Median = l + ------------ * h
f
(60/2) – 24
= 70 + ------------ * 5
20
30 – 24
= 70 + ------------ * 5
20
6
= 70 + ------------ * 5
20
= 70 + 0.3 * 5
= 71.5
C. Solve the VARIANCE then STANDARD DEVIATION
N (N – 1)
= 60 (310,220) – (4,290)2
60 (60 – 1)
= 18,613,200 – 18,404,100
3,540
= 209,100
3,540
= 59.07
s = s2
N-1
= 59.07
= 7.69

2. Skewness
e. Solve the DEGREE OF SKEWNESS
Mean – 71.5
Median – 71.5
Standard Deviation – 7.69
Sk = 3 (x̄ – Md )
s
= 3 (71.5 – 71.5)
7.69
= 3(0)
7.69
= 0
Hence, the distribution is NORMAL.

3. Kurtosis
The degree of peakedness or flatness of a curve is called kurtosis denoted by ku.
This also known as percentile coefficient of kurtosis and the formula is given by:
Ku = QD ÷ PR
Where:
QD – quartile deviation
PR – percentile range
QD = (Q3 – Q1) ÷ 2
PR = P90 – P10
Where:
Q3 – 3rd quartile
Q1 – 1st quartile
Where:
P90 – 90th percentile

3. Kurtosis
When the value of Ku is:
a. Equal to 0.263, the curve
is a normal curve or
mesokurtic
b. Greater than 0.263, the
distribution is platykurtic
of flat
c. Lesser than 0.263, the
distribution is leptokurtic
or thin

3. Kurtosis
Scores f
90-94 1
85-89 4
80-84 3
75-79 8
70-74 20
65-69 15
60-64 7
55-59 1
50-54 1
N =
60
X
92
87
82
77
72
67
62
57
52
fX
92
348
246
616
1440
1005
434
57
52
ƩfX =
4290
<CF
60
59
55
52
44
24
9
2
1
Class
Boundary
89.5-94.5
84.5-89.5
79.5-84.5
74.5-79.5
69.5-74.5
64.5-69.5
59.5-64.5
54.5-59.5
49.5-54.5
Scores & frequency
Median
Frequency
times median
Cumulative
frequency
Lower limit minus 0.5
and upper limit plus 0.5
Number of
distribution
Summation of
frequency times median
Example 1. Calculate the percentile coefficient of kurtosis of the data below.

3. Kurtosis
Scores f X fX <CF
Class
Boundary
90-94 1 92 92 60 89.5-94.5
85-89 4 87 348 59 84.5-89.5
80-84 3 82 246 55 79.5-84.5
75-79 8 77 616 52 74.5-79.5
70-74 20 72 1440 44 69.5-74.5
65-69 15 67 1005 24 64.5-69.5
60-64 7 62 434 9 59.5-64.5
55-59 1 57 57 2 54.5-59.5
50-54 1 52 52 1 49.5-54.5
N =
60
ƩfX =
4290
First, find the QD
1N ÷ 4 = 60 ÷ 4 = 15
Q1 = LQ1 + [(1N/4 – <cfb) ÷ fQ1] h
In the CF look for value
greater than N/4, then
your lower limit lies their
Q1 = LQ1 + [(1N/4 – <cfb) ÷ fQ1] h
Q1 = 64.5 + [(15 – 9) ÷ 15] 5
= 64.5 + (0.4) 5
= 64.5 + 2 = 66.5
Where:
LQ1 - lower limit of 1st quartile
<cfb – CF preceding N/4
fQ1 – frequency 1st quartile
h – class interval

3. Kurtosis
Scores f X fX <CF
Class
Boundary
90-94 1 92 92 60 89.5-94.5
85-89 4 87 348 59 84.5-89.5
80-84 3 82 246 55 79.5-84.5
75-79 8 77 616 52 74.5-79.5
70-74 20 72 1440 44 69.5-74.5
65-69 15 67 1005 24 64.5-69.5
60-64 7 62 434 9 59.5-64.5
55-59 1 57 57 2 54.5-59.5
50-54 1 52 52 1 49.5-54.5
N =
60
ƩfX =
4290
First, find the QD
3(N) ÷ 4 = 3(60) ÷ 4 = 45
Q3 = LQ3 + [(3N/4 – <cfb) ÷ fQ3] h
greater than 3N/4, then
Q3 = LQ3 + [(3N/4 – <cfb) ÷ fQ3] h
Q3 = 74.5 + [(45 – 44) ÷ 8] 5
= 74.5 + (0.125) 5
= 74.5 + 0.625 = 75.125
Where:
LQ3 - lower limit of 3rd quartile
<cfb – CF preceding 3N/4
fQ3 – frequency 3rd quartile

3. Kurtosis
Scores f X fX <CF
Class
Boundary
90-94 1 92 92 60 89.5-94.5
85-89 4 87 348 59 84.5-89.5
80-84 3 82 246 55 79.5-84.5
75-79 8 77 616 52 74.5-79.5
70-74 20 72 1440 44 69.5-74.5
65-69 15 67 1005 24 64.5-69.5
60-64 7 62 434 9 59.5-64.5
55-59 1 57 57 2 54.5-59.5
50-54 1 52 52 1 49.5-54.5
N =
60
ƩfX =
4290
First, find the QD
QD = (Q3 – Q1) ÷ 2
Where:
Q3 – 3rd quartile
Q1 – 1st quartile
QD = (75.125 – 66.5) ÷ 2
= 4.3125

3. Kurtosis
Scores f X fX <CF
Class
Boundary
90-94 1 92 92 60 89.5-94.5
85-89 4 87 348 59 84.5-89.5
80-84 3 82 246 55 79.5-84.5
75-79 8 77 616 52 74.5-79.5
70-74 20 72 1440 44 69.5-74.5
65-69 15 67 1005 24 64.5-69.5
60-64 7 62 434 9 59.5-64.5
55-59 1 57 57 2 54.5-59.5
50-54 1 52 52 1 49.5-54.5
N =
60
ƩfX =
4290
Second, find the PR
10(N) ÷ 100 = 10(60) ÷ 100 = 6
P10 = LP10 + [(10N/100 – <cfb) ÷ fP10] h
P10 = LP10 + [(10N/100 – <cfb) ÷ fP10] h
P10 = 59.5 + [(6 – 2) ÷ 7] 5
= 59.5 + (0.57) 5
= 59.5 + 2.85 = 62.35
Where:
LP10 - lower limit of 10th percentile
fP10 – frequency 10th percentile

3. Kurtosis
Scores f X fX <CF
Class
Boundary
90-94 1 92 92 60 89.5-94.5
85-89 4 87 348 59 84.5-89.5
80-84 3 82 246 55 79.5-84.5
75-79 8 77 616 52 74.5-79.5
70-74 20 72 1440 44 69.5-74.5
65-69 15 67 1005 24 64.5-69.5
60-64 7 62 434 9 59.5-64.5
55-59 1 57 57 2 54.5-59.5
50-54 1 52 52 1 49.5-54.5
N =
60
ƩfX =
4290
Second, find the PR
90(N) ÷ 100 = 90(60) ÷ 100 = 54
P90 = LP90 + [(90N/100 – <cfb) ÷ fP90] h
P90 = LP90 + [(90N/100 – <cfb) ÷ fP90] h
P90 = 79.5 + [(54 – 52) ÷ 3] 5
= 79.5 + (0.66) 5
= 79.5 + 3.33 = 82.83
Where:
LP90 - lower limit of 90th percentile
fP10 – frequency 90th percentile

3. Kurtosis
Scores f X fX <CF
Class
Boundary
90-94 1 92 92 60 89.5-94.5
85-89 4 87 348 59 84.5-89.5
80-84 3 82 246 55 79.5-84.5
75-79 8 77 616 52 74.5-79.5
70-74 20 72 1440 44 69.5-74.5
65-69 15 67 1005 24 64.5-69.5
60-64 7 62 434 9 59.5-64.5
55-59 1 57 57 2 54.5-59.5
50-54 1 52 52 1 49.5-54.5
N =
60
ƩfX =
4290
Second, find the PR
PR = 82.83 – 62.35
= 20.48
PR = P90 – P10
Where:

3. Kurtosis
Finally, when the QD the PR are calculated, you can now calculate the kurtosis.
Ku = QD ÷ PR
Where:
Ku = 4.3125 ÷ 20.48
= 0.211
Since the coefficient is lower than 0.263, it means that the distribution is leptokurtic

Problem 1. Convert the following scores to z-scores:
a. x = 76
b. x = 40
c. x = 50
Mean = 50 Standard Deviation = 10
Problem 2. In a multiple choice examination in Wildlife & Biodiversity, students John
and Ford received scores of 82 and 65, respectively. If these scores are equivalent
to z-scores of 1.6 and -0.2, respectively, what are the mean and standard
deviation of the scores of all the students who took the examination in Wildlife &
Biodiversity?

Problem 3. Determine the degree of skewness and kurtosis of the following sets of
data.
a. Anxiety level of students during exams
Level Frequency
15-19 32
10-14 40
5-9 34
0-4 25
Where:
0-4 – low anxiety level
5-9 – slight anxiety level
10-14 – average anxiety level
15-19 high anxiety level

HYPOTHESIS
TESTING
Instructor

Introduction
▪ Testing the significance of difference between two means,
between two standard deviation, two proportions, or two
percentages, is an important area of inferential statistics.
▪ Comparison between two or more variables often arises in
research or in experiments and to make valid conclusions
regarding the result of the study, one has to apply an
appropriate statistics.

Hypothesis
▪ Hypothesis is a conjecture or a statement which aims to explain
certain phenomena in the real world.
▪ Statistical or not, many hypotheses are products of man’s
curiosity. To seek answers to his questions, he tries to find the
present evidences then test the validity of his hypothesis using
statistical tools and analysis.
▪ In statistical analysis, assumptions are given in the form of a null
hypothesis the truth of which will be either accepted or
rejected within certain critical interval.

The Null Hypothesis
▪ Null Hypothesis, denoted by H0, is a statement which states that
there is no significant relationship or no significant difference
between two or more variables, or one variable does not
affect another variable.
▪ In statistical research, hypothesis should be written in null from.

The Null Hypothesis
▪ For example, suppose we want to know whether method A is
more effective than method B in teaching high school
mathematics
▪ The null hypothesis for this study is “there is no significant
difference between the effectiveness of Method A and Method
B.”

The Alternative Hypothesis
▪ alternative hypothesis is a position that states something is
happening, a new theory is preferred instead of an old one
(null hypothesis).
▪ In statistics, alternative hypothesis is often denoted as Ha or H1.
Hypotheses are formulated to compare in a statistical
hypothesis test.

The Alternative Hypothesis
▪ For example, suppose we want to know whether method A is
more effective than method B in teaching high school
mathematics
▪ The null hypothesis for this study is “there is a significant
difference between the effectiveness of Method A and Method
B.”

Significance Level
▪ To test the null hypothesis, one must set the level of significance.
▪ The level of significance is the probability of making Type I error
and is denoted by the symbol α.
▪ Type I error is the probability of accepting the alternative
hypothesis (H1), when in fact the null hypothesis (H0) is true.

Significance Level
▪ The probability of accepting the null hypothesis (H0) when in
fact it is false is called the Type II error.
▪ The Type II error is denoted by the symbol β.
▪ The most commonly used level of significance is 5% and 1%.

Significance Level & Confidence Level
▪ The confidence level defines the distance for how close the
confidence limits are to sample mean.
▪ The significance level defines the distance the sample mean
must be from the null hypothesis to be considered statistically
significant.
▪ Both the significance level and the confidence level define a
distance from a limit to a mean.

▪ The confidence level is equivalent to 1 – the alpha level. So, if
your significance level is 0.05, the corresponding confidence
level is 95%.
❑ If the P value is less than your significance (alpha) level, the
hypothesis test is statistically significant.
❑ If the confidence interval does not contain the null
hypothesis value, the results are statistically significant.
❑ If the P value is less than alpha, the confidence interval will
not contain the null hypothesis value.

Testing Hypothesis
▪ The following are the steps in testing the truth of a hypothesis:
1. Specify the Null Hypothesis
2. Specify the Alternative Hypothesis
3. Set the Significance Level (a)
4. Calculate the Test Statistic and Corresponding P-Value
5. Drawing a Conclusion

Testing Hypothesis (Examples)
1. Specify the Null Hypothesis
▪ There is a low diversity of plants in the Lake Mainit
watershed area.
▪ There is no significant difference on the initial growth
performance of Narra branch cuttings as applied with
different levels of IBA hormone.
▪ There is no relationship between the demographic profile
and socio-economic status of the famers towards their
knowledge in agroforestry farming.

2. Specify the Alternative Hypothesis
▪ There is a high diversity of plants in the Lake Mainit
watershed area.
▪ There is a significant difference on the initial growth
performance of Narra branch cuttings as applied with
different levels of IBA hormone.
▪ There is a relationship between the demographic profile and
socio-economic status of the famers towards their
knowledge in agroforestry farming.

3. Set the Significance Level
▪ The significance level (denoted by the Greek letter alpha—
α) is generally set at 0.05. This means that there is a 5%
chance that you will accept your alternative hypothesis
when your null hypothesis is actually true. The smaller the
significance level, the greater the burden of proof needed
to reject the null hypothesis, or in other words, to support the
alternative hypothesis.

4. Calculate the Test Statistic (depending on the topic/research)
▪ For diversity researches, the Fernando Biodiversity
Scale will be used as basis.
Descriptive Interpretation Diversity (H’)
Very High > 3.5
High 3.0 – 3.49
Moderate 2.5 – 2.99
Low 2.0 – 2.49
Very Low < 1.99

▪ For experimental researches, the following are commonly
used:
❑ T-test – a t-test is a statistical test that compares the
means of two samples.
❑ ANOVA – analysis of variance is used for three or more
groups of data, to gain information about the
relationship between the dependent and independent
variables.

▪ For socio-economic researches, correlation statistic
is commonly used.
Correlation
Coefficient
Verbal Interpretation
± 0.81 to ± 1.00
Very high correlation, very dependable
relationship
± 0.61 to ± 0.80 High correlation, marked relationship
± 0.41 to ± 0.60 Moderate correlation, substantial relationship
± 0.21 to ± 0.40 Slight correlation, definite but small relationship
± 0.00 to ± 0.20 Slight correlation, almost negligible relationship

5. Drawing a Conclusion
▪ Decide what hypothesis to be accepted and to be rejected
based on the results of the study / analysis.
▪ If the null hypothesis is true, then reject the alternative
hypothesis and accept the null hypothesis
▪ If the null hypothesis is false, then accept the alternative
hypothesis and reject the null hypothesis

FOREST BIOMETRY
▪ “Forest Bio” would literally mean all living things within the forest
while “metry” means measurement.
▪ Forest biometry is a branch of science which deals with the
application of mathematical and statistical principles in the
determination, estimation, and prediction of forest volume and
growth including the forest products derived there from (Del
Castillo, 1988).

FOREST MENSURATION
▪ Forest mensuration is one of the most fundamental disciplines
within forest and related sciences. It deals with the
measurement of trees and stands and the analysis of the
resultant information.”
▪ Forest mensuration is the art and science of providing the
quantitative information about trees and forest stands
necessary for forest management, planning and research.”
▪ The traditional (20th century) and apparently outdated term
for forest biometrics was forest mensuration. The general term
“mensuration” is a word used by scientists to mean activities
related to measuring.

FOREST MENSURATION vs BIOMETRY
▪ Biometrics encompasses the quantification of biological and
physical characteristics of trees and associated vegetation,
insects, disease, wildlife, topography, soils, and climate,
individually and collectively. These characteristics include all
quantifiable attributes within forestry, both temporal and
spatial. (James D. Arney, 2006, FBRI)
▪ Mensuration deals with the measurement of lengths of lines,
areas of surfaces, and volumes of solids. In forestry, it
encompasses the determination of dimensions, form, weight,
growth, volume, health and age of trees, individually or
collectively.

Topic Outline
▪ Basic Mathematics
▪ Tree Diameter Measurements
▪ Tree Height Measurements

BASIC
BASIC
MATHEMATICS
MATHEMATICS

BASIC MATHEMATICS (Length)
▪ Length is the distance from one end to the other end of an
object. For example, the length of the line segment AB is the
distance along the line from A to B.
A B

▪ Metric System – In metric system the unit of measurement of
length is meter (m). Smaller units are millimeter (mm),
centimeter (cm), micrometer (µm) and nanometer (nm). The
larger unit is kilometer (km).
▪ Relations among the various units are given below.
1 km = 1000 m
1 m = 100 cm
1 cm = 10 mm
1 mm = 1000 micrometer
1 micrometer = 1000 nanometer

▪ British System – In British system the units of measurement of
length are inch, foot, yard and mile.
▪ Conversion factors of one unit to another are given below.
1 foot = 12 inches
1 yard = 3 feet
1 mile = 1760 yards = 5280 feet

▪ Conversion factors between Metric and British Units
1 inch = 2.54 cm; 1 foot = 30.48 cm; I yard = 91.44 cm =
0.9144 meter
1 mile = 1609.34 m = 1.6093 km
1 cm = 0.3937 inch; 1 meter = 39.370 inch = 3.281 feet =
1.0936 yard
1 km = 1093.61 yard = 0.621 mile

BASIC MATHEMATICS (Area)
▪ Area is the extent of a two
dimensional surface or
shape. In other words,
area of a plane figure
means the number of
square units the figure
covers.
▪ The square units could be
square inch (in2), square
cm (cm2), square ft (ft2),
square m (m2) etc.

▪ Metric System – Commonly used metric system of units for area
measurement are square cm (cm2), square meter (m2), square
km (km2), hectare (ha) etc.
▪ Relations among these units are given below.
1 m2 = 10,000 cm2
1 ha = 10,000 m2
1 km2 = 100 ha = 1,000,000 m2

▪ British System - Commonly used British system of unite for area
measurement are square inch (in2), square feet (ft2), square
yard (yard2), square mile (mile2), acre etc.
1 ft2 = 144 inch2
1 yard2 = 9 ft2 = 1296 inch2
1 acre = 4840 yard2
1 mile2 = 640 acre = 3097600 yard2

▪ Conversion factors (approximate) between Metric and British
Units
1 inch2 = 6.45 cm2
1 ft2 = 929.03 cm2 = 0.093 m2
1 yard2 = 0.836 m2
1 mile2 = 2.590 km2
1 acre = 4046.24 m2
1 cm2 = 0.155 inch2
1 m2 = 1.196 yard2
1 km2 = 0.386 mile2
1 ha = 2.47 acre

BASIC MATHEMATICS (Volume)
▪ Volume is the space occupied by a material or object. In other
words, volume is the number of unit cubes that fit within the
solid figure of the object.

▪ Metric System – Commonly used units are cubic mm (mm3),
cubic cm or cc (cm3), cubic meter (m3), liter etc.
1 cm3 or 1 cc = 10 mm x 10mm x 10 mm = 1000mm3
1 liter = 1000 cc
1 m3 = 1m x 1m x 1m = 100 cm x 100 cm x 100 cm =
10,00,000 cc.

▪ British System – Commonly used units are cubic inch (inch3),
cubic feet (ft3), cubic yard (yard3) etc.
▪ Relations are as follows.
1ft3 = 1 foot x 1 foot x 1 foot
1ft3 = 12 inches x 12 inches x 12 inches = 1728 inch3
1 yard3 = 1 yard x 1 yard x 1 yard
1 yard3 = 3ft x 3 ft x 3 ft = 27ft3

system
1 cc = 0.061 inch3
1 m3 = 35.318 ft3 = 1.308 yard3
1 inch3 = 16.387 cc
1 ft3 = 0.0283 m3
1 yard3 = 0.764 m3

BASIC MATHEMATICS (Mass / Weight)
▪ Mass represents the amount of matter in an object. Weight, in
scientific terms, represents the amount of gravitational force
acting on an object, that is, the force with which the object is
attracted by earth towards its center.
▪ While mass of an object is constant, its weight varies from place
to place. However, weight of an object is proportional to its
mass, and in everyday usage, mass and weight are used
interchangeably. In forest measurement weight normally would
mean mass, that is, the amount of matter

▪ Metric System – Commonly used units are gram (gm), kilogram
(kg) , quintal and metric ton.
▪ The relations among the units are as follows.
1 kg = 1000 gm
1 quintal = 100 kg
1 metric ton = 1000 kg

▪ British System – Commonly used units are pound (lb.), imperial
ton or long ton etc.
1 Imperial ton = 2240 lbs.
1 short ton = 2000 lbs.

system
1 lb. = 453.59 grams = 0.45359 kg
1 imperial ton = 2240 lbs. = 1016.0469 kg = 1.016 metric ton
1 kg = 2.2046 lb.
1 metric tone = 1000 kg = 2204.6 lbs. = 0.984 imperial ton

BASIC MATHEMATICS (Density)
▪ Density of an object or substance is defined as mass per unit
volume. It is a measure of how much matter in unit volume.
▪ Loosely speaking density of a substance gives an indication of
how heavy or light is the substance.
▪ For example, density of water is 1 gm/cc. It would mean that 1
cc of water weighs 1 gram. In, comparison, iron, known to be
heavier than water, has density 7.87 gm/cc. That is, one cc of
iron weighs 7.87 grams. We may say iron is 7.87 times heavier
than water.

▪ Metric system – Commonly used units are gram per cubic cm
(gm/cc) and kg per cubic meter (kg/m3).
1 kg/m3 = 1000gm/1000000cm3 = 0.001 gm/cc,
1 gm/cc = 1000 kg/m3
▪ British system – Commonly used unit is:
Pounds per cubic feet (lbs./ft3)

system
1 lb./ft3 = 453.59 gm/28316.8 cc =0 .0160 gm/cc
1 gm/cc = 62.4280 lb./ft3
1 kg/m3 = 0.001 gm/cc = 0.062428 lb./ft3

TREE DIAMETER
▪ Tree diameter is a length from the outside of the bole through
the center to the opposite side of it.
Longitudinal / Side View
Cross-sectional View
diameter
diameter

TREE DIAMETER
▪ On standing trees, a reference diameter is normally required.
This is known as the diameter at breast height (DBH) where
breast height is defined at 1.37 meters (4 feet 6 inches) above
ground level.
▪ Other diameter measurements can be done on diameter
above buttress (DAB), stump diameter, top diameter, and
intermediate diameter.
▪ Diameter may be measured outside bark (DOB) and inside
bark (DIB) and the difference is twice the bark thickness.

TREE DIAMETER
▪ Average Bark Thickness (ABT) = (BT1 + BT2) ÷ 2
BT1
Where:
BT1 – bark thickness 1
BT2 – bark thickness 2
BT2

TREE DIAMETER
▪ Diameter Inside Bark (DIB) = DOB – 2ABT
DIB Where:
DOB – diameter outside bark
ABT – average bark thickness

TREE DIAMETER
▪ Diameter Outside Bark (DOB) = DIB + 2ABT
DOB Where:
DIB – diameter inside bark
ABT – average bark thickness

TREE DIAMETER
Example 1. What is the DOB of the log with a ABT of 1cm and a
DIB of 23cm
a. 21cm
b. 24cm
c. 25cm
d. 26cm
Solution:
DOB = 23cm + 2(1cm)
= 23cm + 2cm
= 25cm

TREE DIAMETER
Example 2. What is the ABT of the log with a DOB of 25cm and a
DIB of 23cm
a. 44cm
b. 24cm
c. 2cm
d. 1cm
Solution:
ABT = (25cm – 23cm) ÷ 2
= 2cm ÷ 2
= 1cm

TREE DIAMETER (DBH)
▪ The standard position for diameter measurement at standing
tree is at breast height. It is defined at 1.30 meter above
ground in most countries, but there are some countries where
diameter at breast height is measured at different heights.
Why DBH is preferred?
At breast height the instrument is easily handled
(convenience and ease). Also on most trees the influence of
buttress on the stem form is already much reduced at breast
height.

TREE DIAMETER (DBH)
HOW TO MEASURE DBH?
▪ Trees on slope: measure dbh at the standard height above the
floor/ground on the uphill side of the tree.
▪ Leaning tree: measure parallel to the lean on the lower side of
the lean
▪ Buttress tree: if the buttress height is more than one meter then
measure dbh from the point where buttress ends, otherwise
measure normally.

TREE DIAMETER (DBH)
▪ Abnormalities at breast height (swellings, knots, crooks, etc):
Measure the dbh above or below the abnormalities and
indicate the height at which diameter is measured. Sometimes
measurement is done at equal distance above and below
breast height and then dbh is estimated by taking the mean of
the two readings.
▪ Bifurcation: If a tree bifurcates above breast height then
measure dbh as usual. But, if a tree bifurcates below breast
height then measure dbh on each stem separately.

TREE DIAMETER (Dendrometers)
DENDROMETERS – instruments used in the measurement of tree
diameters
1. Caliper
2. Diameter tape
3. Biltmore stick
4. Meter stick
5. Tape Measure or Measuring Tape

▪ Caliper - Is the most efficient to measure dbh directly
whenever there is direct access to the tree. It can be made of
wood, metal or aluminum. It has two arms one fixed and a
graduated bar/beam on which the second arm slides.
▪ To measure with a caliper, hold it firmly and horizontally as well
as perpendicular to stem axis at the same time. Usually two
readings are taken perpendicular to each other rat breast
height and then the average value will be recorded.

▪ Diameter Tape - a diameter tape has an inch scale and a
diameter scale printed on a steel tape. The diameter scale side
is determined by the formula, circumference divided by pi or
3.1416.
▪ You wrap the tape level around a tree's trunk at 4.5 feet dbh
and read the diameter side of the tape for the tree diameter
determination.

CALIPERS VS DIAMETER TAPE
▪ Tapes are easy to carry than calipers (especially in dense
forest)
▪ Measuring with caliper is faster than with tape.
▪ Bigger trees can be measured with tapes easily (calipers have
an upper bound) tapes can be extended by joining them
▪ Tapes are good to maintain consistency in measuring diameter
regular
▪ The most accurate tool used in making a diameter
measurement is a tree caliper and is used most often in
exacting tree studies.

▪ Biltmore Stick - this "cruiser's stick" is a scaled "ruler" that is held
at arm's length (25 inches from the eye) and horizontal to the
tree's dbh. The left end of the stick is aligned with the outer tree
edge and the reading is taken where the opposite edge
intersects the stick.
▪ This is the least accurate method of the three and should be
used only for rough estimates.

▪ Meter Stick – usually divided with lines for each millimeter (1000
per meter) and numerical markings per centimeter (100 per
meter), with numbers either in centi- or millimeter.
▪ Tape Measure or Measuring Tape – is a flexible ruler used to
measure size or distance. It consists of a ribbon of cloth, plastic,
fibre glass, or metal strip with linear-measurement markings.

TREE DIAMETER (Upper Stem Diameter)
▪ Upper stem diameter of a tree is measured for instance to
describe the shape of a stem (derive taper curve).
▪ It is measured at various heights. Upper stem diameters are
most easily observed on felled trees; however there are
situations in which upper stem diameter need to be measured
on a standing tree.
▪ Upper stem diameter is measured either at a fixed point (X
meter) or at relative height (X% of height).

▪ Methods Used in Measuring Upper Stem Diameter of a Standing
Tree
❑ Using Finn caliper (Finnish parabolic caliper)
❑ Optical caliper (parallel beams)
❑ Measuring upper diameter with angle measurement
technique
❑ Measuring upper stem diameter with mirror relaskop

▪ Using Finn caliper (Finnish parabolic caliper)
❑ Used to measure diameter up to 7m
❑ Difficult to carry mostly beyond this height

▪ Optical caliper (parallel beams)
❑ Needs determination of height before or after diameter
measurement
❑ Read diameter when the two images of the trunk coincides
❑ It is independent of distance measurement

▪ Measuring upper diameter with angle measurement technique

alpha
Distance (e)

Beta (β)
β/2
Distance (e3)
Example:
β = 5 degrees
e3 = 5m
d = 2tan(β/2)e3
= 2tan(5/2)5m
= 2tan(2.5)5m
= 0.44m or 44cm

▪ Measuring upper stem diameter
with mirror relaskop
❑ Measurement of DBH
❑ Aim at DBH and determine the
number of relaskop units
❑ Calculate to how many cm
one unit corresponds
❑ Aim at upper diameter and
determine the number of
relaskop units
❑ Calculate upper diameter

▪ Measuring upper stem diameter with mirror relaskop

TREE DIAMETER (Diameter Classes)
▪ Tree diameters can be measured nearest to 0.1 inch. However,
it is more convenient and customary to group measurements
into diameter classes. When 2 inch diameter or 4 cm diameter
classes are used, group or class limits may be taken as follows:
2 inch Diameter Class 4cm Diameter Class
Diameter Limits Mean Diameter Diameter Limits Mean Diameter
7 - <9 8in 18 - <22 20cm
9 - <11 10in 22 - <26 24cm
11 - <13 12in 26 - <30 28cm
13 - <15 14in 30 - <34 32cm

TREE DIAMETER (Diameter Classes)
▪ Thus trees having diameter from 7 inch to less than 9 inch may
be taken as belonging to 8 inch diameter class, trees having
diameter from 9 inch to less than 11 inch will belong to 10 inch
diameter class and so on. Similarly, trees having diameter from
18 cm to less than 22 cm may be considered as belonging to
20 cm diameter class, trees having diameter from 22 cm to less
than 26 cm will belong to 24 cm diameter class.
▪ The convenience of grouping actual diameters into diameter
classes is that all trees of a diameter class may be considered
as having the same diameter corresponding to the said class
for the purpose of enumeration and calculation.

TREE HEIGHT
WHY TREE HEIGHT IS NEEDED TO BE MEASURED?
▪ Height is a tree variable that is used to estimate or determine
the volume of a tree. It (dominant height) also helps to deal
with the issues of site classification.
TREE HEIGHT VS TREE LENGTH
▪ Tree height is defined to be the perpendicular distance
between the ground level and the top of the tree.
While, Tree length is the distance between the stem
foot and the top along the stem

VS
Red – Tree Height
Black – Tree Length

TREE HEIGHT
TYPES OF TREE HEIGHT
▪ Total height - the distance between the ground and top of the
tree.
▪ Bole height - the distance between the ground and the crown
point.
▪ Merchantable height - the distance between the ground and
the terminal position of the last useable portion of the tree
stem.

TREE HEIGHT
TYPES OF TREE HEIGHT
▪ Stump height - the distance between the ground and the
position where a tree is cut.
▪ Merchantable length - is the distance between the top of the
stump and the terminal position of the last useable portion of
the tree stem.
▪ Dominant height - is the average height of 100 thickest trees
per hectare.

TREE HEIGHT (Hypsometers)
HYPSOMETERS – instruments used in the measurement of tree
heights
1. Clinometer
2. Relaskop
3. Laser range finder / electronic hypsometer
4. Merritt hypsometer on Biltmore stick

TREE HEIGHT (Hypsometers)
HYPSOMETERS – other instruments:
1. Abney level
2. Blume Leiss
3. Bottle opener dendrometer
4. Haga
5. Spiegel Relaskop
6. Suunto clinometer
7. Telerelaskop
8. Vertex
9. Criterion laser
10.LEM-300

Haga
Blume Leiss
Bottle Opener
Dendrometer
Abney Level

TREE HEIGHT
METHODS OF TREE HEIGHT MEASUREMENT
1. Direct method - it involves climbing or using height
measuring rods. It is rarely used and only for small trees.
2. Indirect method
a. Using geometric principle
b. Using trigonometric principle

TREE HEIGHT (Geometric Principle)
▪ A christen hypsometer or ruler of a certain length (30cm for
example) and a pole of constant length/height used to
estimate/measure tree height.
▪ Technique:
1. Place a pole of known length at upright position against the tree to be
measured.
2. Hold ruler (of known length) vertically and parallel to the tree to be
measured.
3. Find the sighting position by moving back and forth and/or right and
left so that the top of the ruler exactly aligned with the tip of the tree
and the bottom of the ruler with the base of the tree.
4. Take ruler reading in line with the top of the pole. Then apply the
following formula.

▪ Christen, Merritt or JAL altimeter use geometric principle, which is based
on the equation: A’C’/AC = A’B’/AB, where AB corresponds to the tree
height. Before mentioned instruments that apply this principle use fixed
distances of A’B’, A’C’ and AC, where A’B’ and A’C’ are given on the
instrument and AC is set by some reference fixed at the tree (see the
formula on the main photo).
▪ With the Christen altimeter, the visual image of the tree or part of the tree
to be measured must be fit exactly between the upper and lower ends of
the scale. The height or length of a tree, a stem, or a stem section is then
determined on the basis of a fixed reference length on the stem.
Instruments such as Christen altimeter are relatively simple in construction,
only one reading is necessary, and the measurement is not affected by
the inclination of the terrain.

Tree height (BC) = Known ruler length (bc) X Known length of pole (BD)
Ruler reading on the pole (bd)
▪ Advantages:
+ no distance measurement is required
+ height reading is not influenced by slope
▪ Drawbacks:
- In dense forest it is difficult to find suitable point of
observation
- Only with a steady hand can serious misreading be
avoided

TREE HEIGHT (Trigonometric Principle)
Formulas in Height Measurements
▪ HD = cosθ * SD
▪ H = tanθ * HD
▪ H = (%/100) * HD
▪ θ = (%/100) * tan-1
▪ % = tanθ (100)
Where:
HD – horizontal distance
SD – slope distance
θ – degree (angle)
% - percent

TREE ON LEVEL GROUND
TH
H2
H1
HD
Where:
H1 – height 1
H2 – height 2
A1 – angle 1
A2 – angle 2
TH – Total Height
A2
A1
eye level

TH
H2
H1
20m
15°
5°
eye level
Example 1. Calculate the total height

H1 = tanθ * HD
= tan5° * 20m
= 0.0875 * 20m
= 1.75m
H2 = tanθ * HD
= tan15° * 20m
= 0.2679 * 20m
= 5.358m
TH = H1 + H2
= 1.75m + 5.358m
= 7.108m
TH = (tan5° + tan15°) * HD
= (0.0875 + 0.2679) * 20m
= 0.3554 * 20m
= 7.108m

TREE ON SLOPING GROUND
TH
H2
H1
HD
Where:
SD – slope distance
H1 – height 1
H2 – height 2
A1 – angle 1
A2 – angle 2
TH – Total Height
A2
A1 eye level
SD

TH
H2
H1
HD
27°
5° eye level
25m

HD = cosθ * SD
= cos5° * 25m
= 0.9962 * 25m
= 24.905m
H1 = tanθ * HD
= tan5° * 24.905m
= 0.0875 * 24.905m
= 2.1792m
H2 = tanθ * HD
= tan27° * 24.905m
= 0.5095 * 24.905m
= 12.6891m
Note: H2 and TH are the same
for example 2

TH
H2
H1
31m
31°
7° eye level

H1 = tanθ * HD
= tan31° * 31m
= 0.6009 * 31m
= 18.6279m
H2 = tanθ * HD
= tan7° * 31m
= 0.1228 * 31m
= 3.8068m
TH = H1 + H2
= 18.6279m + 3.8068m
= 22.4347m
TH = (tan31° + tan7°) * HD
= (0.6009 + 0.1228) * 31m
= 0.7237 * 31m
= 22.4347m

LEANING TREE
HD
Where:
AC – leaning distance
BC – tree height
AB – tree length
A1 – angle 1
A1
A
B
C

LEANING TREE
18m
33°
A
B
C
Example 4. Calculate the tree length (AB)
2.5m

LEANING TREE
BC = (HD + AC) * tanθ
= (18m + 2.5m) * tan33°
= 20.5m * 0.6494
= 13.3127m
Pythagorean Theorem
AB = √AC2 + BC2
= √2.52 + 13.31272
= √6.25 + 177.228
= √183.478
= 13.5454m

LEANING TREE
28m
33°
A
B
C
3m
13°
D
0.75m
BD
CD
BC

LEANING TREE
CD = (HD + AD) * tanθ
= (28m + 0.75m) * tan13°
= 28.75m * 0.2309
= 6.6384m
Pythagorean Theorem
AB = √AC2 + BC2
= √3.752 + 26.76982
= √14.0625 + 716.6222
= √730.6847
= 27.0312m
BD = (HD + DC) * tanθ
= (28m + 3m) * tan33°
= 31m * 0.6494
= 20.1314m
BC = CD + BD
BC = 6.6384m + 20.1314m
BC = 26.7698m
AC = AD + CD
AC = 0.75m + 3m
AC = 3.75m

1. Covert the following:
a. 1 mile to micrometer
b. 247 acre to square meter
c. 100 cubic meter to cubic centimeter
2. Determine the diameter of the following in cm:
a. A Binuang tree with a radius or 0.53m
b. A Tindalo with a BT1 of .75cm, BT2 of 1.25cm, and a DIB
of 0.2m. Calculate the DOB.
c. A circumference of 39.37in

3. Calculate TH and MH (merchantable height)
TH
H1
MH
H2
H3
HD = 59.04ft
4°
7.5°
20°

TREE VOLUME
Volume – is a measure of solid content or capacity, usually
expressed in cubic units. It is the most widely used measure of
wood quantity.
▪ In a tree measurement, the most important of a tree to
consider in terms of usable wood, is the stem.
▪ A log is solid which is assumed to have a circular cross-
section at any point on its axis.
▪ The cubic volume of a cylinder which is equal to its basal
area (BA) multiplied by its length or height is used as the
basis for log volume.
▪ The basal area is found from its diameter (BA = 0.7854*D2)

TREE VOLUME (Formulas for Round Timber)
1. Direct Measurement
a. Xylometric process
2. Indirect Measurement
a. Huber formula
b. Smalian formula
c. Newton’s formula
d. Government Scaling formula (DENR formula)

▪ The fluid displacement method, also called xylometry,
accurately measures gross biological volume. Essentially, the
tree stem is cut into manageable sections and immersed in a
bath. The amount of water displaced equals the volume of the
section.
▪ This method does not require any assumptions to be made
about tree shape and thus has no theoretical bias. However,
xylometry is expensive and rarely used outside research.
XYLOMETRIC PROCESS

XYLOMETRIC PROCESS
Example 1.
Drum full of water Wood submerge in
a drum full of water
Remaining water after removing
the wood submerged in the drum

XYLOMETRIC PROCESS
Example 1.
1. Calculate the volume of
water present in the drum
before submerging the
wood.
2. What is the volume of
remaining water after
submerging the wood?
3. What is the volume of the
wood?
Given:
Drum length – 1.2m
Drum diameter – 70cm
Length of remaining water after
submerging the wood – 80cm

XYLOMETRIC PROCESS
Example 1.
Step 1. Calculate the Volume of Water
or Drum
Volume of the cylinder = 𝜋r2h
VC = 3.1416 (0.35m)2 x 1.2m
VC = 3.1416 (0.1225m2) x 1.2m
VC = 0.4618m3
Step 2. Calculate the Volume of
Remaining Water in the Drum
Volume of the cylinder = 𝜋r2h
VRW = 3.1416 (0.35m)2 x 0.8m
VRW = 3.1416 (0.1225m2) x 0.8m
VRW = 0.3079m3
Step 3. Calculate the Volume of Wood or Displaced Water
VDW = Volume of water or drum – Volume of remaining water
VDW = VC – VRW
VDW = 0.4618m3 – 0.3079m3
VDW = 0.1539m3

▪ The cubic volume is calculated from the measurement of the
length of the log and a single diameter measurement on the
outside of the bark (dob) from the middle of the log (Freese
1973, Grosenbaugh 1948). Huber's formula for estimating the
cubic volume of a log is as follows:
HUBER’S FORMULA
V – Volume in cubic
V = 0.7854 (Dm)2 x L Dm – Diameter at the middle section of the log
L – Length of the log

HUBER’S FORMULA
Example 2.
36cm 34cm 32cm
Given:
Big End Diameter (D) – 36cm
Middle Diameter (Dm) – 34cm
Small End Diameter (d) – 32cm
Log Length (L) – 10m
10m

HUBER’S FORMULA
Example 2.
Calculate the Volume of the Wood Using the Huber Formula
V = 0.7854 (Dm)2 x L
V = 0.7854 (0.34m)2 x 10m
V = 0.7854 (0.1156m2) x 10m
V = 0.9079m3

▪ A cubic volume formula used in log scaling, expressed as cubic
volume = [(D + d)/2] L, where D = the cross-sectional area at
the large end of the log, d = the cross-sectional area at the
small end of the log, and L = log length.
SMALIAN’S FORMULA
V = 0.7854 (D)2 + 0.7854 (d)2
2
* L
Big End Diameter (D)
Small End Diameter (d)
Log Length (L)

Example 3.
36cm 34cm 32cm
Given:
Big End Diameter – 36cm
Middle Diameter – 34cm
Small End Diameter – 32cm
Log Length – 10m
10m
SMALIAN’S FORMULA

Calculate the Volume of the Wood Using the Smalian Formula
V = 0.7854 (D)2 + 0.7854 (d)2
2
V = 0.7854 (0.36m)2 + 0.7854 (0.32m)2
2
V = 0.7854 (0.1296m2) + 0.7854 (0.1024m2)
2
V = 0.1018m2 + 0.0804m2
2
Example 3.
SMALIAN’S FORMULA
V = 0.1822m2
2
V = 0.0911m2 * 10m
V = 0.911m3
* L
* 10m
* 10m
* 10m
* 10m

▪ Newton's formula is the standard against which other cubic
formula's are compared. The top (small), middle and base
(big) diameters are used with the log length for the
determination of cubic volume.
NEWTON’S FORMULA
V = 0.7854 (D)2 + 0.7854 [4 (Dm)2] + 0.7854 (d)2
6
* L
Where:
Big End Diameter (D)
Middle Diameter (Dm)
Small End Diameter (d)
Log Length (L)

Example 4.
36cm 34cm 32cm
Given:
Log Length – 10m
10m
NEWTON’S FORMULA

Calculate the Volume of the Wood Using the Newton Formula
V = 0.7854 (D)2 + 0.7854 [4 (Dm)2] + 0.7854 (d)2
6
V = 0.7854 (0.36m)2 + 0.7854 [4 (0.34m)2] + 0.7854 (0.32m)2
6
V = 0.7854 (0.1296m2) + 0.7854 [4 (0.1156m2)] + 0.7854 (0.1024m2)
6
V = 0.7854 (0.1296m2) + 0.7854 (0.4624m2) + 0.7854 (0.1024m2)
6
V = 0.1018m2 + 0.3632m2 + 0.0804m2
6
Example 4.
V = 0.5454m2
6
V = 0.0909m2 * 10m
V = 0.909m3
* L
* 10m
* 10m
* 10m
* 10m
NEWTON’S FORMULA
* 10m

▪ The Department of Environment and Natural Resources (DENR)
formula of the Government Scaling Formula use the big end
diameter multiplied to length of the log and 0.7854
GOVERNMENT SCALING FORMULA
V – Volume in cubic
V = 0.7854 (D)2 x L D – Big end diameter
L – Length of the log

Example 5.
36cm 34cm 32cm
Given:
Log Length – 10m
10m

Example 5.
Calculate the Volume of the Wood Using the Government
Scaling Formula
V = 0.7854 (D)2 x L
V = 0.7854 (0.36m)2 x 10m
V = 0.7854 (0.1296m2) x 10m
V = 1.0179m3

Example 6.
Calculate the volume of the standing tree in cubic feet with an
estimated lower stem diameter of 33in, a middle stem diameter of
29in and 24in upper stem diameter, and an estimated
merchantable height of 100ft.
A. Huber formula
B. Government scaling formula
C. Smalian formula
D. Newton’s formula

Example 6.
B. Gov’t Scaling Formula
V = 0.7854 (D)2 x L
V = 0.7854 (2.75ft)2 x 100ft
V = 0.7854 (7.5625ft2) x 100ft
V = 593.9588ft3
A. Huber Formula
V = 0.7854 (Dm)2 x L
V = 0.7854 (2.4167ft)2 x 100ft
V = 0.7854 (5.8404ft2) x 100ft
V = 458.705ft3

C. Smalian Formula
V = 0.7854 (D)2 + 0.7854 (d)2
2
V = 0.7854 (2.75ft)2 + 0.7854 (2ft)2
2
V = 0.7854 (7.5625ft2) + 0.7854 (4ft2)
2
V = 5.9396ft2 + 3.1416ft2
2
V = 9.0812ft2
2
V = 4.5406ft2 * 100ft
V = 454.06ft3
* L
* 100ft
* 100ft
* 100ft
* 100ft
Example 6. cont.

D. Newton’s Formula
V = 0.7854 (D)2 + 0.7854 [4 (Dm)2] + 0.7854 (d)2
6
V = 0.7854 (2.75ft)2 + 0.7854 [4 (2.4167ft)2] + 0.7854 (2ft)2
6
V = 0.7854 (7.5625ft2) + 0.7854 [4 (5.8404ft2)] + 0.7854 (4ft2)
6
V = 0.7854 (7.5625ft2) + 0.7854 (23.3616ft2) + 0.7854 (4ft2)
6
V = 5.9396ft2 + 18.3482ft2 + 3.1416ft2
6
V = 27.4294ft2
6
V = 4.5716ft2 * 100ft
V = 457.16ft3
* L
* 100ft
Example 6. cont.
* 100ft
* 100ft
* 100ft
* 100ft

REGIONAL
VOLUME
EQUATION
REGIONAL
VOLUME
EQUATION

TREE VOLUME (Regional Volume Equation)
REGION
SPECIES GROUP
DIPTEROCARP NON-DIPTEROCARP
Northern Luzon
(Region 1, 2 and 3)
V = 0.00005203 (D2 * H) V = 0.00005109 (D2 * H)
Southern Luzon
(Region 4 & 5 except Palawan)
V = 0.00005171 (D2 * H) V = 0.00005204 (D2 * H)
Western Visayas
(Region 6, 7 and Palawan)
V = 0.00004649 (D2 * H) V = 0.00004874 (D2 * H)
Eastern Visayas
(Region 8 and Bohol)
V = 0.00005231 (D2 * H) V = 0.00005109 (D2 * H)

REGION
SPECIES GROUP
DIPTEROCARP NON-DIPTEROCARP
Eastern Mindanao
(Region 11 and 13)
(Portion of Agusan, East of
Agusan River, Davao & Surigao)
V = 0.00005087 (D2 * H) V = 0.00004961 (D2 * H)
Central Mindanao
(Region 10, 12 and ARMM)
(Bukidnon, Cotabato, Lanao,
Misamis, Portion of Agusan, West
of Agusan River)
V = 0.00005019 (D2 * H) V = 0.00005039 (D2 * H)
Western Mindanao
(Region 9)
(Basilan, Sulu, Zamboanga)
V = 0.00004668 (D2 * H) V = 0.00004840 (D2 * H)

Example 1.
Given:
Location: Isabela (Region 2)
Species: Binuang (Non-Dipterocarp)
DBH (cm): 41
MH (m): 10
V = 0.00005109 (D2 * L)
V = 0.00005109 (41)2 * 10
V = 0.00005109 (1681) * 10
V = 0.00005109 (16810)
V = 0.8588m3
V = 0.5109 (D2 * L)
V = 0.5109 (0.41m)2 * 10m
V = 0.5109 (0.1681m2) * 10m
V = 0.5109 (1.681m3)
V = 0.8588m3
or

Example 2.
Given:
Location: Catanduanes (Region 5)
Species: Apitong (Dipterocarp)
DBH (cm): 37
MH (m): 9
V = 0.00005171 (D2 * L)
V = 0.00005171 (37)2 * 9
V = 0.00005171 (1369) * 9
V = 0.00005171 (12321)
V = 0.6371m3
V = 0.5171 (D2 * L)
V = 0.5171 (0.37m)2 * 9m
V = 0.5171 (0.1369m2) * 9m
V = 0.5171 (1.2321m3)
V = 0.6371m3
or

Example 3.
Given:
Location: Cebu (Region 7)
Species: Bagtikan
DBH (cm): 25
MH (m): 7
V = 0.00004649 (D2 * L)
V = 0.00004649 (25)2 * 7
V = 0.00004649 (625) * 7
V = 0.00004649 (4375)
V = 0.2034m3
V = 0.4649 (D2 * L)
V = 0.4649 (0.25m)2 * 7m
V = 0.4649 (0.0625m2) * 7m
V = 0.4649 (0.4375m3)
V = 0.2034m3
or

Example 4.
Given:
Location: Surigao del Norte
Species: Toog
DBH (in): 30.7087
MH (m): 17
V = 0.00004961 (D2 * L)
V = 0.00004961 (78)2 * 17
V = 0.00004961 (6084) * 17
V = 0.00004961 (103428)
V = 5.1311m3
V = 0.4961 (D2 * L)
V = 0.4861 (0.78m)2 * 17m
V = 0.4961 (0.6084m2) * 17m
V = 0.4961 (10.3428m3)
V = 5.1311m3
or
First, covert inches to cm
30.7086in x 2.54cm = 78cm
1in

TREE VOLUME (Volume of Geometric Solid)
Conoid
Paraboloid
Neiloid
BA x L BA – Basal Area
VN = --------------- L – length of log basal portion
2
VP = --------------- L – length of log middle portion
3
VC = --------------- L – length of log upper portion
4

3m
3m
3m
32cm
34cm
36cm
Calculate the following:
a. Neiloid volume
b. Paraboloid volume
c. Conoid volume
d. Total volume
Example 1.

3m
3m
3m
32cm
34cm
36cm
A. Neiloid Volume
VN = (BA * L) ÷ 2
VN = [0.7854 (0.36m)2 (3m)] ÷ 2
VN = [0.7854 (0.1296m2) (3m)] ÷ 2
VN = 0.3054m3 ÷ 2
VN = 0.1527m3
Example 1. cont.

3m
3m
3m
32cm
34cm
36cm
B. Paraboloid Volume
VP = (BA * L) ÷ 3
VP = [0.7854 (0.34m)2 (3m)] ÷ 3
VP = [0.7854 (0.1156m2) (3m)] ÷ 3
VP = 0.2724m3 ÷ 3
VP = 0.0908m3
Example 1. cont.

3m
3m
3m
32cm
34cm
36cm
C. Conoid Volume
VC = (BA * L) ÷ 4
VC = [0.7854 (0.32m)2 (3m)] ÷ 4
VC = [0.7854 (0.1024m2) (3m)] ÷ 4
VC = 0.2413m3 ÷ 4
VC = 0.0603m3
Example 1. cont.

3m
3m
3m
32cm
34cm
36cm
D. Total Volume
V = VN + VP + VC
V = 0.1527m3 + 0.0908m3 + 0.0603m3
V = 0.3038m3
Example 1. cont.

Example 2.
27in diameter
and 6ft length
26in diameter
and 5ft length
24in diameter
and 4ft length
Calculate the total volume of the log in cubic feet.

Example 2. cont.
Neiloid Volume
VN = (BA * L) ÷ 2
VN = [0.7854 (2.25ft)2 (6ft)] ÷ 2
VN = [0.7854 (5.0625ft2) (6ft)] ÷ 2
VN = 23.8565ft3 ÷ 2
VN = 11.9283ft3
Paraboloid Volume
VP = (BA * L) ÷ 3
VP = [0.7854 (2.1667ft)2 (5ft)] ÷ 3
VP = [0.7854 (4.6946ft2) (5ft)] ÷ 3
VP = 18.4357ft3 ÷ 3
VP = 6.1452ft3

Example 2. cont.
Conoid Volume
VC = (BA * L) ÷ 4
VC = [0.7854 (2ft)2 (4ft)] ÷ 4
VC = [0.7854 (4ft2) (4ft)] ÷ 4
VC = 12.5664ft3 ÷ 4
VC = 3.1416ft3
D. Total Volume
V = VN + VP + VC
V = 11.9283ft3 + 6.1452ft3 + 3.1416ft3
V = 21.2151ft3

Problem 1. Calculate the volume of the following using the
regional volume equations in cubic meter:
Location Species DBH
Merchantable
Height
a. Dinagat Islands Magkono 74cm 2000cm
b. Surigao del Sur Kamagong 0.5m 11m
c. Surigao del Norte Mayapis 24in 252ft
d. Agusan del Sur Narig 2ft 21ft

Problem 2. Calculate the total volume of the following using the
volume of geometric solid formulas in cubic meter:
Log # Neiloid Paraboloid Conoid
1 D – 76.2cm
L – 2.1342m
D – 63.5cm
L – 3.0488m
D – 53.34cm
L – 2.1342m
2 D – 30in
L – 7ft
D – 25in
L – 10ft
D – 21in
L – 7ft

Problem 3. Calculate the volume of the following standing trees
using the indirect volume measurement formulas (Huber, Smalian,
Newton and Government Scaling Formula in cubic meter.
Log # D Dm d MH
1 33cm 12.2047in 0.9515ft 775cm
2 1.0827ft 31cm 11.4174in 305.1181in
3 12.9922in 1.0171ft 29cm 25.4265ft

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