Educational

Statistics
100
80
Chemistry

60
40
20
0
0 20 40 60 80 100
Mathematics

EPS 211
Prof. Y. K. A. Etsey,
Dept. of Educational
S Foundations
August 2012
 TABLE OF CONTENTS

Pages

UNIT 1 Introduction to Statistics 3

UNIT 2 Data Representation 9

UNIT 3 Measures of Central Tendency/Location 33

UNIT 4 Measures Of Variability 45

UNIT 5 Measures of Relative Position 60

UNIT 6 Measures of Relationships 65

UNIT 7 Simple Regression Analysis 78

UNIT 8 The Normal Distribution 84

Appendix 92

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 UNIT 1

INTRODUCTION TO STATISTICS

DEFINITIONS OF STATISTICS
There are three basic definitions of the term, Statistics.
1. Statistics (plural) is the body of numbers or data collected in any field.
For example industrial statistics – number of employees in an industry,
number of products, value of products; vital statistics – measurements of
bust, waist, hips; population statistics – number of people in a region, number
of people with secondary education
2. Statistics (singular) is the study of methods and procedures used in
collecting, organizing, analyzing, and interpreting a body of numbers for
information and decision making.
3. Statistics (plural; statistic - singular) are the values computed from a body
of numerical data. For example the “average” age of Level 200 students in
UCC, the proportion of EPS 211 students who are males.

From the definitions above, Educational Statistics can be defined as;

1. the body of numbers or data in the field of education. For example, school
statistics – number pupils in a school, number of teachers in a school, number
of textbooks in a school; number of teachers in a region, number of pupils in
a district.
2. the study of the methods and procedures used in collecting, organizing,
analyzing and interpreting a body of numbers related to education for
information and decision making.

The main concern of this course is the second definition.

WHY STUDY STATISTICS?

1. It helps teachers to use the appropriate statistics in describing the performance of
their classes to others. For example, using the mean and the median a teacher can
describe the performance as above average or below average. This is better than
saying the performance is poor or bad, good or very good because these terms are
relative.
2. It puts teachers in a position to better understand the information they receive
from test scores on students. For example, a frequency polygon or histogram
constructed from raw test scores can show the direction of performance. They

3
 would indicate whether the general performance of a class is low, average or
high.
3. It helps teachers to evaluate course grades and the differences in ability
represented by different grades. For students personal report cards, grades alone
do not provide enough information on a student‟s level of performance in a
subject. This information should be combined with the ranking in the subject.
4. It helps the teacher in the critical reading and understanding of professional
journals in education. Journals such as the Journal of Educational Management,
Journal of Educational Research, Journal of Educational Development and
Practice, Journal of Research and Development in Education often use statistics
in their analysis of results.
5. It is useful for research purposes. Statistics are used for data analysis in project
work and dissertations/thesis. Teachers would also use it in their own research in
the teaching profession.
6. It helps the teacher to understand information from standardized achievement
test manuals. The statistical information provided in the test manuals describes
the quality of the test and the interpretation of the test scores.

INTRODUCTORY CONCEPTS
Variables
A variable is any characteristic of an individual or object that can take on
different values. A value is an assigned number or label representing the attribute of
a given individual or object. For example, marital status as a variable can be broken
down into categories and given values as never married - 1, married - 2, divorced - 3
and widowed - 4. Number of children in a family as a variable can be given the
values 0, 1, 2, 3, 4 etc. Height can take on values such as 1.2 metres, 1.7 metres, 2.0
metres and 2.2 metres. Religious affiliation can be broken down to categories and
given values as: Christian – 1, Moslem – 2, Traditionalist – 3, Buddhist - 4.

Variables can be classified as ordered or unordered.

For ordered variables, the attributes differ in magnitude along a quantitative
dimension. For example, the number of pupils in a class can be 20, 25, 32, 40, 45 etc
where 25 is greater than 20 and 40 is less than 45.
For unordered variables, the attributes are classified into two or more mutually
exclusive categories that are qualitatively different. For example, gender is classified
into male and female. Undergraduate halls of residence on UCC campus are Oguaa,
Adehye, Atlantic, Casely Hayford, VALCO, Kwame Nkrumah.

4
 Variables can also be classified as discrete or continuous.
Discrete variables have values which in theory assume only certain distinct
values or whole numbers on a number line. These variables usually represent
counts of indivisible entities, for example, 8, 12, 20, 45, 100. For example the
number of goals scored in a soccer game or the number of students in a class.
Continuous variables have values which in theory assume any value on a number
line between two points. The values can differ by infinitesimal amounts, for
example, 10.5, 14.16, 42.001, 56.2222278. For example the height of a student
or the weight of a car is a continuous variable.

Descriptive and Inferential Statistics

Descriptive statistics provides summary data about a group. A single number
is often used to represent a group. For example, a mean age of 30 years can be
used to represent the ages of 60 Masters in Education students. A median age
of 21 years can be used to represent the ages of all Level 300 students in the
University of Cape Coast.

Inferential statistics uses data from a small group called a sample to make
statements or generalizations about a much larger group called a population.
For example, to know the mean age of first year university students (i.e.
population) in Ghana, a small group (i. e. sample) say 200, of first year students
can be used. Their mean age could be used as an estimate of the mean age of all
first year university students in Ghana.

Scales of Measurement
Depending upon the traits/attributes/characteristics and the way they are
measured, different kinds of data result representing different scales of
measurement. There are 4 types of measurement scales. These are Nominal,
Ordinal, Interval and Ratio.

Nominal Scales: A nominal scale classifies persons or objects into two or more
categories. Whatever the classification, a person can be in one and only one
category, and members of a given category have a common set of
characteristics. For identification purposes, categories are numbered. e.g.
Gender: Male - 1, Female - 2. All males have a common characteristic and all
females have a common characteristic which is different from males.

Ordinal Scales: An ordinal scale not only classifies subjects but also ranks them
in terms of the degree to which they possess a characteristic/attribute of interest.

5
 An ordinal scale puts subjects in order from highest to lowest, or from most to
least. With respect to height, 5 students can be ranked from 1 to 5, the subject
with rank 1 being the shortest. Though ordinal scales do indicate that some
subjects are higher or better than others, they do not indicate how much higher
or better. The intervals between the ranks are not equal.

Interval Scales: An interval scale has all the characteristics of both nominal and
ordinal scales and in addition has equal intervals. The zero point is arbitrary
and does not mean the absence of the characteristics/trait. Values can be added
and subtracted to and from each other but not multiplied or divided. Examples
include Celsius temperature, academic achievement.

Ratio Scales: A ratio scale has all the advantages of the types of scales and in
addition it has a meaningful true zero point. Height, weight and time are
examples. Values can be added, subtracted, multiplied and divided. Sixty
(60) minutes can be said to be 3 times as long as 20 minutes.

Arithmetic Comparisons

Scale Arithmetic Function

Addition Subtraction Multiplication Division
(+) (─) (X) ÷
Nominal
Ordinal
Interval √ √
Ratio √ √ √ √

Practice Exercises
1. Statistics is important for classroom teachers because it
A. enables them to write appropriate objectives.
B. helps them to construct good test items.
C. helps them to evaluate students‟ grades.
D. is useful for promotion and certification.

2. Which one of the following variables can be classified as unordered?

A. Enrolment in a course
B. Halls of residence
C. Height of a student
D. Scores in a quiz

6
3. A district director of education measures many variables on a sample of schools.
An example of a variable measured in an ordinal scale is the
A. enrolment of the classes in each school.
B. income in cedis of the teachers.
C. professional qualification of the teachers.
D. years of service for each teacher.

Use the information below to answer questions 4 and 5.

A study was conducted to see how well reading success in primary three
could be predicted from various kinds of information obtained in
kindergarten (reading readiness, age, gender, and socio-economic status).

4. Which one of the variables represents a nominal scale?

A. Age
B. Gender
C. Reading readiness
D. Reading success

5. Which one of the variables represents a ratio scale?

A. Age
B. Gender
C. Reading success
D. Socio-economic status

6. Which one of the following variables can be classified as continuous?

A. Age of students in a psychology class.
B. Colour of dresses students wear to class.
C. Number of photocopy machines on a college campus.
D. Percentage of female students in the University of Cape Coast.

7. Which one of the following variables can be classified as ordered?

A. Parents occupation
B. Region of country
C. Religious affiliation
D. Statistics achievement

7
8. The grades, A, B, C, D, E, F in a test were changed to 4, 3, 2, 1, 0 for statistical
purposes. What scale of measurement was used?
A. Interval
B. Nominal
C. Ordinal
D. Ratio

9. .An example of a variable that can best be classified as discrete is

A. Dancing ability of students.
B. Number of computers in a school.
C. Speed in a 200 metre race.
D. Study habits of students.

10. Which one of the following variables is an example of an interval scale?

A. Achievement scores on an intelligence test
B. Gender of students in a statistics class.
C. Number of cars in the University of Cape Coast.
D. Types of mobile phones students own.

11. Which one of the following variables, measured from Primary 6 pupils, has
an interval scale?
A. Dancing ability
B. Languages spoken
C. Region of birth
D. Religious affiliation

12. Which one of the following variables can be classified as unordered?

A. Ages of students in the EPS 211 class in UCC.
B. Names of houses of residence in a Senior High School.
C. Number of students in each class in a Senior High School.
D. Points obtained by houses in inter-house athletics competition.

8
 UNIT 2
DATA REPRESENTATION
Raw scores are often represented by graphics/pictures or tables. The
representation of data in these forms enables more information to be derived from the
scores. In educational statistics, more emphasis is placed on pictorial
representations.

Nsowah-Nuamah (2005) gave 4 purposes of pictorial representations of data.

1. To make it easier to hold in mind the mass of figures they represent.

2. To show more clearly than tables any relations which may exist between
figures.
3. To show degrees of difference between items.
4. To make a quick, lasting and accurate impression of relations, trends and
comparisons.

Pictorial representations of data should have the following properties.

1. They should be easier to understand than the figures they represent.
2. They should be simple and clear in their meaning.
3. They should carry all the information necessary to appreciate then fully.
4. They should be adequately labelled.
5. They should be titled.
6. They should have a key which may be included within or below the body of
the chart.
7. They should have footnotes or source notes, when appropriate, and these
should be given below the diagram.

Bar graph/Charts

Data that are from nominal scales are represented in graphic form with the
use of bar graphs. Bar graphs give a pictorial description of the data and emphasize
how groups compare with one another. They are used to compare the sizes of the
various parts. The height of the bars is the basis for the comparisons and not the area
of the bars.
Bar graphs are either column or horizontal. Column graphs are more popular
in education. Column bar graphs are simple, compound (multiple) or component.
Examples are shown below.

9
 Table 1
Performance in Inter-House Athletics at BASS
House Total Points
One 120
Two 100
Three 150
Four 60
Five 170

The figure below is a simple column bar graph showing performance in an Inter-
House Athletics competition at BASS.
Inter-House Athletics at BASS

180

160

140

120
Total Points

100

80

60

40

20

One Two Three Four Five

Houses

10
 Table 2
School Enrolment at Texas JSS
Form Male Female
1A 18 35
1B 30 25
1C 36 12
2A 22 30
2B 25 25
2C 40 18

The figure below is a compound column bar graph showing school enrolment at
Texas JSS by gender. School Enrolment at Texas JSS by Gender

45

40

35

30

25
Enrolment

Male
20

Female
15

10

0
1A 1B 1C 2A 2B 2C
Forms

Component Bar Chart

It is known as composite or stacked bar chart. It is used when a set of data combines
to form a total. The total is the length/height of the bar. It allows for visual
comparisons between different components ie how components contribute to the
total of the category.

11
School enrolment at Texas JSS

Constructing bar graphs/charts

1. Draw two axes, a vertical and horizontal. Label the vertical axis by the source of
the values/scores e.g. enrolment, points etc. Label the horizontal axis by the
names of the categories.
2. Divide vertical scale by points considering the lowest value and the highest
value. Choose appropriate scales such that the bars are not too tall or too short
and must start with zero.
3. Construct equally wide and equally spaced bars for each category with the height
of the bar being the value/score for the category on the horizontal axis, which has
the names of the categories as the label.
4. Where computer softwares such as Microsoft Excel and SPSS are not available, it
is recommended that graph sheets be used.
5. Shade/colour the bars to differentiate bars and components.

Strengths and Limitations

1. Bar charts are easy to draw

2. Values can be read easily from the vertical axis.
3. Comparisons can be made easily and the significance of the information is easily
grasped.
4. It cannot be effective with interval and ratio scales of measurement
5. Extreme values distort comparisons especially if some bars are too short and
some very tall.
6. For component bar graphs, too many subgroups make the graph crowded.
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 Uses

Teachers can use bar graphs in several ways. Enrolment by classes, courses and
subjects and inter-house competitions can be represented by bar graphs.

Pie Chart

Pie charts use nominal or categorical data. Pie charts are represented in the form of a
circle of 360 0 sliced into the shape of „pies‟. Each pie is cut from an angle at the
centre of the circle. The angle corresponds to the data for each category or group.
Pie charts give a pictorial view and the contributions of the parts that make a whole.
An example is shown below.

Table 3
Performance in Inter Hall House Athletics at BASS

House Total Points Degrees To calculate each degree:

One 120 72 For House One
Two 100 60 120
X 360 0 =72
Three 150 90 600
Four 60 36 For House Two

Five 170 102 100

X 360 0 =60
600
Total 600 360
Inter-House Athletics at BASS

One

Two

Three

Four

Five

13
 Constructing pie charts

1. Calculate the degree equivalents for the value of each category/group by dividing
the total point for each group by the overall total points and multiple the result by
3600.
For example, for House One above we have:
120 100
 360 0  72 0 and for House Two, we have  360 0  60 0
600 600
2. Use a pair of compass and protractor to draw the circle and the sectors based on
the degrees calculated.
3. Shade/Colour the sectors to differentiate one from the other.

Strengths and limitations

1. Individual parts of the whole are seen and can be compared.

2. It provides a visual impression of the proportion that each part contributes to the
overall total.
3. Angles are harder to compare.
4. They are not easy to draw especially where statistical softwares and computers
are not available.
5. Data that are in continuous form and of ratio and interval scales are not
appropriate.
6. The values of each component cannot be read from the chart but must be
provided.
7. It is not useful where there are many parts as these parts become too small.
8. It only gives a visual impression but not the details of the data.

Uses
Pie charts can be used by teachers and educational practitioners for examination
results by number of passes in various subjects, school enrolment by class, form
or subjects.

Line Graphs

Data that are related to time are best used for line graphs. Time could be days,
weeks, months and years. Line graphs show changes in the data over a period of
time. Data from interval and ratio scales are most appropriate. Line graphs
could be simple or compound. Simple line graphs give a pictorial description of
the data. Compound line graphs compare group data over a period of time.

14
 Examples are shown below.

Table 4
Attendance at monthly teachers‟ workshops

Month Total
January 120
February 85
March 100
April 150
May 90
June 85
July 100
August 60
September 90
October 75
November 100
December 150

The figure below is a simple line graph showing attendance at a monthly teachers‟
workshop.
Attendance at monthly teachers‟ workshops
160
140
Number of attendants

120
100
80
60
40
20

0
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
Month

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 Table 5
Attendance at monthly teachers‟ workshops

Attendance
Month Female Male
Jan 60 60
Feb 45 40
March 60 40
April 80 70
May 50 40
June 40 45
July 50 50
August 25 35
Sep 40 50
Oct 30 35
Nov 50 50
Dec 90 60

Compound line graph showing attendance at a monthly teachers‟ workshop by

gender.

Attendance at monthly teachers‟ workshops by gender

10
0
9
0
8
0
7
0
6
Attendance

0 Female
5 Male
0
4
0
3
0
2
0
1
0
0
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
Months

16
 Constructing line graphs

1. Draw two axes, a vertical and horizontal. Label the vertical axis by the source of
the values/scores e.g. attendance, enrolment, points etc. Label the horizontal axis
by the time period e.g. months, days, weeks etc.
2. Divide vertical scale by points considering the lowest value and the highest
value. Choose appropriate scales such that the graph is not too tall or too flat and
must start with zero.
3. Plot the value/quantity for each time period on the graph and join all the points
by a straight line.
4. Where computer softwares such as Microsoft Excel and SPSS are not available, it
is recommended that graph sheets be used.

Strengths and Limitations

1. Values can be read easily from the vertical axis.

2. Comparisons can be made easily and the significance of the information easily
grasped.
3. They are less appropriate for nominal scale data.
4. Graphs are distorted where there are extreme values.

Uses

Teachers and educational practitioners can use line graphs in several ways.
Examination results over a period of years in a subject, total school enrolment as
well as enrolment by subjects and courses for a period of time can be represented
by line graphs.

FREQUENCY DISTRIBUTIONS

Data normally comes in raw or ungrouped form as shown below for 40 students in a
Statistics class.

The following marks were obtained by a group of 40 students in a Statistics

examination.
76 88 93 75 70 93 73 62 69 75
71 80 52 76 66 54 73 80 79 89
83 62 53 79 69 56 81 75 71 72
52 65 49 80 67 59 88 87 91 82

17
The raw data alone does not give much information. In the example above, we can
best know the highest score (93) and the lowest score (49). A lot more information
can be obtained if the data are treated or put in other forms. One of the ways to
obtain information from data is to use frequency distributions.

A frequency distribution is any arrangement of data that shows the frequency of

occurrence of different values of the variable or the frequency of occurrence of
values falling within arbitrarily defined ranges of the variable. A frequency
distribution table is often used to show the frequency distribution which could either
be ungrouped or grouped.

Table 6
Ungrouped frequency distribution table

Score Tally Frequency

93 // 2
91 / 1
89 / 1
88 // 2
87 / 1
83 / 1
82 / 1
81 / 1
80 /// 3
79 // 2
76 // 2
75 /// 3
73 // 2
72 / 1
71 // 2
70 / 1
69 // 2
67 / 1
66 / 1
65 / 1
62 // 2
59 / 1
56 / 1
54 / 1
53 / 1
52 // 2
49 / 1
Total 40

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 Ungrouped frequency distributions are not very useful for further work. Zero
frequencies are often common and the tables are sometimes too tall.

For grouped frequency distributions, the individual scores are put into groups or
classes. The scores are most often put in groups/classes of 3, 5, 7, 9, and 10 as
group sizes. Column 2 provides the mid-points, column 3 the tallies, and
column 4 the frequencies.

Table 7
Grouped frequency distribution of Statistics students‟ performance

Class Class Mark Tally Frequency

91-95 93 /// 3
86-90 88 //// 4
81-85 83 /// 3
76-80 78 ///// // 7
71-75 73 ///// /// 8
66-70 68 ///// 5
61-65 63 /// 3
56-60 58 // 2
51-55 53 //// 4
46-50 48 / 1
Total 40

Features

1. Class. A group of scores.

2. Class interval. The range within which a group of scores lie. It has a number at
the beginning and at the end. E.g. 90-95.
3. Unequal class interval. These result where there are differences in the range of
the intervals. E.g. 91 – 95, 91 - 100
4. Open-ended classes. These are classes with a value at the beginning or the end.
e.g. 90 and above, 45 and below, Below 46, Above 90.
5. Class limits. The end points of a class interval. The smaller number is the lower
limit and the bigger number is the upper limit.
6. Class boundaries. The exact or real limits of a class interval. The lower class
boundaries are obtained by subtracting 0.5 from the lower class limit. The upper
class boundaries are obtained by adding 0.5 to the upper class limits. A class
interval with limits of 91 – 95 produces class boundaries of 90.5 - 95.5
7. Class size/class width. The number of distinct/discrete scores within a class
interval. They are obtained by finding the difference between successive lower
class limits or upper class limits in cases of equal class intervals. They can also
19
 be obtained by finding the difference between successive class marks in cases of
equal class intervals or between class boundaries for each interval.
8. Class mark: The midpoint for each class interval.
9. Frequency: The number of distinct scores from the given data that can be found
in a class interval.
10. Cumulative frequency. The successive sum of the frequencies starting from the
frequency of the bottom class.
11. Cumulative percentage frequency. The successive sum of the percentage
frequencies starting from the percentage frequency of the bottom class. It is also
obtained by expressing each cumulative frequency as a percentage.
12. Relative frequency. It is obtained by dividing each frequency by the total
frequency.
13. Cumulative relative frequency. The successive sum of the relative frequencies
starting from the frequency of the bottom class.

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Table 8
An expanded frequency distribution table

Class Class Tally Frequency Cumulative Cumulative Relative Cumulative

Mark Frequency Percentage Frequency Relative
Frequency Frequency
91-95 93 /// 3 40 100 0.075 1.0
86-90 88 //// 4 37 92.5 0.100 0.925
81-85 83 /// 3 33 82.5 0.075 0.825
76-80 78 ///// 7 30 75.0 0.175 0.750
//
71-75 73 ///// 8 23 57.5 0.200 0.575
///
66-70 68 ///// 5 15 37.5 0.125 0.375
61-65 63 /// 3 10 25.0 0.075 0.250
56-60 58 // 2 7 17.5 0.050 0.175
51-55 53 //// 4 5 12.5 0.100 0.125
46-50 48 / 1 1 2.5 0.025 0.025
Total 40 1.000

Constructing a grouped frequency distribution table

1. Draw a table with four columns with the headings – Class, Class mark, Tally,
Frequency.
2. Determine the range i.e., the difference between the highest score and the lowest
score. For example, from the raw scores of the 40 students in the Statistics
examination, the highest score is 93 and the lowest is 49. The range becomes 93
– 49 = 44.
3. Arbitrarily decide on a class size. Popular sizes are 3, 5, 7, 10. Odd-numbered
class sizes make computations easier. In Education, the most popular sizes are 5
and 10.
4. Determine the approximate number of classes by dividing the range by the class
size. E.g. Suppose a size of 5 is taken. The approximate number of classes would
44
be  8.8 which is rounded to 9 classes. Generally, number of classes is
5
between 5 and 20.
Alternatively, arbitrarily decide on the number of classes. This is normally between
10 and 20 for large data sets. Determine the approximate class width by dividing the
range by the number of classes. Suppose the number of classes taken is 10. The
44
approximate class size becomes,  4.4 which is rounded to 5.
10
21
5. Identify the highest value/score and write it down.
6. To obtain the topmost class, decide on whether to start with the lower or the
upper limit of the class interval. Determine the closest value to the highest value
identified in Step 5 that is a multiple of the class size. Choose one of the values
as a limit and use the class size to determine the other limit. For example, if a
class size is 5, and the highest score is 93, then the closest values are 90 and 95.
A possible lower limit is 90 and a possible upper limit is 95. Selecting the lower
limit, the topmost class becomes, 90-94 or selecting the upper limit the topmost
class becomes, 91-95.
7. Complete the first column with the rest of the classes using equal class sizes, then
the second column with the class marks.
8. Tally the scores in the third column which is the tally column. Take the scores
one by one and place slashes (/) or tally marks in the respective classes. Where
the tallies are five, bind them into one unit to facilitate counting.
9. Count the number of slash or tally marks and write them in the frequency
column. Add the frequencies and put the total at the bottom of the frequency
column.

Points to note in constructing Frequency Distributions

1. In Education, the highest class intervals/classes are at the top so this convention
must be followed in constructing the frequency distribution table.
2. Use mutually exclusive classes. Make sure that an observation falls into one and
only class. Classes must not overlap at the class limits. For example, 70 – 80
and 80 – 90 contain overlapping class limits of 80.
3. There should be no class with a zero frequency. If this occurs, it is recommended
that the class size is changed. Preferably increase class size.
4. Open-ended classes should be avoided. These classes have only the lower limit
if it is the class at the top, or the upper limit if it is the class at the bottom. For
example, 51 and above, 20 and below.
5. Aim at classes with equal sizes or width. This facilitates interpretation of the
information from the frequency distribution.
6. The number of classes should not be too small (i.e. not less than 5) and not too
large (i.e. not more than 20). Where the number of classes is less than 5, class
size should be reduced but when the number of classes is more than 20, the class
size should be increased.

22
 Practice Exercise

Given the following scores of 50 students in a Statistics class, and using a class
width of 5, construct a grouped frequency distribution table. Also obtain the
cumulative percentage frequencies, and cumulative relative frequencies.

32 38 25 40 47 22 48 45 20 35
16 18 10 6 8 11 33 30 28 27
42 35 30 34 31 21 25 12 20 25
43 33 36 39 42 17 19 22 26 10
33 38 32 22 26 42 37 35 40 46

GRAPHIC REPRESENTATIONS OF
FREQUENCY DISTRIBUTIONS

Histogram

Histograms use data from ratio or interval scale and depend on frequency
distributions. It uses the classes and the frequencies from the frequency distribution
table. An example is shown below.

F 40
r
e 30
q
20

10

0 5 10 15 20 25 30 35 40 45 50 55 60
Classes

23
To construct a histogram
1. Draw two axes, a vertical and horizontal. Label the vertical axis by frequency
and the horizontal axis scores/classes.
2. Select an appropriate scale on the vertical axis considering the highest/largest
value. When using a graph sheet, the scale should be such that the bars are not
too tall nor too short.
3. Use class midpoints/marks or class boundaries or class limits to label the points
on the horizontal axis.
4. Draw bars of equal width representing the classes from a frequency distribution
table with corresponding heights as the frequencies.

Importance
1. It gives a pictorial description of the raw data, providing information about the
nature of the data.
2. It gives the direction of performance in terms of academic performance (i.e.
skewness).

F 40 F 40
r r
e 30 e 30
q q
20 20

10 10

0
5 10 15 20 25 30 0 5 10 15 20 25 30
Classes Classes
Skewed to the right Skewed to the left
Group performance tends to be low Group performance tends to be high

3. It provides an estimate of the most typical score. This is the intersection of the
two diagonals of the tallest bar.

Frequency Polygon
Frequency polygon uses data from ratio or interval scales and depends on
frequency distributions. It uses the classes and the frequencies from the frequency
distribution table. An example is shown below.

24
 F
r
e
q

Classes

To construct a frequency polygon

1. Draw two axes, a vertical and horizontal. Label the vertical axis by frequency
and the horizontal axis scores/classes.
2. Select an appropriate scale on the vertical axis considering the highest/largest
value. When using a graph sheet, the scale should be such that the polygon is not
too pointed or too short.
3. Use class midpoints/marks or class boundaries or class limits to label the points
on the horizontal axis.
4. Plot at the midpoint of each class or the midpoint of the histogram the relevant
heights as the frequencies. Join the midpoints with a straight line.
5. Where the line has not touched the horizontal axis, extend the line one class in
that direction so that the polygon touches the horizontal axis.

Importance
1. It gives a pictorial description of the raw data, providing information about the
nature of the data.
2. It provides an estimate of the most typical score. This is the point on the
horizontal axis where the highest point of the polygon is located.

Most typical score

3. It is used to compare the performance of groups. E.g. Performance in a class test

for Forms 1 and 2 can be shown as follows.

25
 Form 1

Form 2

10 20

The diagram shows that Form 2 class, which is more to the right, performs better.
The most typical scores, where the highest point of the polygon is located can be
used to confirm the comparisons. Where the total frequencies are not the same, use
relative frequencies in place of the actual frequencies to draw the polygon.

4. It gives the direction of performance (skewness). Consider three classes, A, B,

C.

A B C

Positive skewness Normal Negative skewness

Skewed to the right Skewed to the left
Tends to score low marks Tends to score high marks

Cumulative Percentage Frequency Polygon (Ogive)

Ogives are drawn from frequency distribution tables. Data from ratio or
interval scales are most appropriate.

26
Plot the graph using the upper class boundaries of each class against the cumulative
percentage frequencies.

C 100
U 80
M 70
% 60
50
F 40
R 30
E 20
Q 10

0 10 20 30 40 50 60 70 80
CLASSES

To construct an ogive,
1. Obtain cumulative percentage frequencies.
2. Plot the cumulative percentage frequencies in each class on the vertical scale.
Choose appropriate scales, on a graph sheet, such that the ogive is not distorted.
3. Label the horizontal axis as scores or classes.
4. Plot at the upper class boundary of each class the relevant values of the
cumulative frequency. Join the points with a straight line.
5. Extend the line one class to the left so that the polygon touches the horizontal
axis.

Importance
1. It is used for comparisons of distributions of performance especially for
distributions where the class/group sizes are not the same. Generally, the graph
that moves more to the right has better performance. The median score obtained
at the cumulative frequency of 50 is also used.

27
 Given the following performances in a test, draw two ogives. Which school
performed better?

School A School B
Classes Frequency Cum. % Freq Frequency Cum. % Freq
91 - 100 1 100 7 100
81 – 90 2 99 17 95.3
71 – 80 11 97 30 84
61 – 70 24 86 25 64
51 – 60 20 62 15 47.3
41 – 50 16 42 11 37.3
31 – 40 12 26 19 30
21 – 30 8 14 14 17.3
11 - 20 4 6 6 8
1 - 10 2 2 6 4
Total 100 150

2. It is used to determine percentiles and percentile ranks. Later in the course, you
will learn how to obtain the percentiles and percentile ranks.

Box and Whisker Plot

It is used to compare distributions by noting the 10th percentile (P10), first quartile
(Q1), median (Q2), third quartile (Q3), and 90th percentile (P90).

A box and whisker plot is drawn below. Later in the course, you will learn how to
obtain the percentiles and quartiles.

Q1 Q2 Q3
P10 P90

An example.

Assume that the following values were obtained for two classes, Form 1A and Form
1B in a class test in Mathematics.

P10 Q2 P90
Form 1A 10 40 73
Form 1B 28 56 91

28
The information is presented below by two box and whisker plots.
Form 1A
Q1 Q2 Q3
P10 P90

Form 1B
Q1 Q2 Q3
P10 P90

0 25 50 75 100
10 28 40 56 73 91
It can be observed that P10 , Q1, Q2, Q3, and P90 values are greater in Form 1B than
in Form 1A. This means that performance is better in Form 1B than in Form 1A.
Also note that the graph for Form 1B has moved more to the right towards higher
values than that of Form 1A.

29
 Practice Exercises
1. You have data on the long vacation earnings of a sample of 1,000 University of
Cape Coast students. What kind of graph is most appropriate to use to describe
the distribution of their earnings?
A. Bar chart.
B. Box and Whisker.
C. Histogram.
D. Pie chart.

2. You are writing an article for the SRC newspaper about the cost of attending a
university. You want to make a graph to compare costs at your institution and
three similar institutions. The most appropriate choice of a graph would be a
A. Bar chart.
B. Frequency polygon.
C. Histogram.
D. Pie chart.

3. The Registrar of the University of Cape Coast wants a pictorial representation of

the degree classifications of the 2004 graduated class. Which of the following
graphs would be the most appropriate for him to use?
A. Frequency polygon
B. Histogram
C. Line graph
D. Pie Chart

Study the frequency distribution below and answer questions 4 – 5.

Distribution of examination scores for Level 100 students

Classes Frequency
91 - 100 10
81 – 90 8
71 – 80 15
61 – 70 10
51 – 60 4
41 – 50 2
31 - 40 1
Total 50

30
4. What is the relative frequency for the class, 61 – 70?
A. 0.50
B. 0.20
C. 10.0
D. 17.0

5. The class mark for the class, 81 – 90, is

A. 79.5
B. 80.5
C. 85.5
D. 90.5

6. Histograms are most useful for representing data when the scale of measurement
is
I. Interval II. Nominal III. Ordinal IV. Ratio

A. I only.
B. IV only.
C. I and IV.
D. I, III, IV.

Study the frequency distribution below and answer questions 7 – 10.

Distribution of examination scores for Statistics students

Classes Frequency
46 - 50 12
41 - 45 14
36 - 40 12
31 – 35 6
26 – 30 5
21 – 25 1
Total 50

31
7. The relative frequency for the class, 36-40, is
A. 0.024
B. 0.24
C. 0.48
D. 0.52

8. What is the cumulative frequency for the class, 36 – 40?

A. 12
B. 22
C. 24
D. 38

9.. The lower class boundary for the class 41 – 45, is

A. 40.0
B. 40.5
C. 45.0
D. 45.5
10. The cumulative relative frequency for the class, 36-40, is
A. 0.24
B. 0.48
C. 24.0
D. 48.0

32
 UNIT 3

MEASURES OF CENTRAL TENDENCY/LOCATION

These measures are also called Averages. They provide single values which are
used to summarise a set of observations/data. The three main measures are the
Mean, Median and Mode.
1. They are used as single scores to describe data.
2. They help to know the level of performance by comparing with a given standard
of performance. Performance may be above average or below average where the
average is a standard such as the mean or median..
3. They give the direction of student performance.
Where Mean >Median, the distribution is skewed to the right (positive skewness)
showing that performance tends to be low.
Where Mean < Median, the distribution is skewed to the left (negative skewness)
showing that performance tends to be high.

Illustration

Mode Mean Mean Mode

Median Median
Positive skewness Negative skewness

33
 THE MEAN ( X )

There are three types. These are Arithmetic, Geometric and Harmonic. In Education,
the Arithmetic mean is the most useful.

The Arithmetic Mean. It is the sum of the observations divided by the total number
of observations.
i.e. Add the values and divide by the number of observations.
15
i.e. 4 + 2 + 3 + 1 + 5 = 15 Mean = 3
5

Methods
The Arithmetic Mean ( X ) can be obtained from both the ungrouped and grouped
data. It can also be easily obtained from Microsoft Excel.

1. Ungrouped data

Given the following scores, 15, 12, 10, 10, 9, 20, 14, 11, 13, 16, to obtain the mean,
all the scores are added and divided by the total number of observations. The mean

is represented by the symbol, X

15  12  10  10  9  20  14  11  13  16 130
X    13
10 10

Generally, the letter, X, is used with a subscript to differentiate the numbers as

follows.
E.g. 15, 12, 10, 10 , 9, 20, 14, 11, 13, 16
X1, X2, X3 , X4, X5, X6, X7, X8, X9 , X10

The formula used for ungrouped data is: X 

 X , OR X   X where the
n N
Greek letter, ∑ (sigma), shows summation and n or N is the total number of
observations.

2. Grouped data
Two methods can be used. These are the long method and the coding method. The
methods are used with frequency distributions.

Long method: X 
 fx OR X   fx where f is the frequency and x, the class
n N
marks.

34
Example using the long method
Scores Midpoint Freq
X f fx
46 – 50 48 4 192
41 – 45 43 6 258
36 – 40 38 10 380
31 – 35 33 12 396
26 – 30 28 8 224
21 – 25 23 7 161
16 – 20 18 3 54
Total 50 1665

Long method X 
 fx  1665  33.3
n 50

Coding method: X  AM 
 fd i , which is used for distributions with equal
n
class intervals. AM, is the assumed mean, f, is the frequency, d is the code for each
class, n is the total frequency and i, the class size.
To use the coding method, class intervals must have the same size. The class
in the middle or the class with the highest frequency is chosen for the code of 0.
Classes above the zero coded class are given positive codes and those below are
given negative codes in steps of 1.

Example using the coding method

Scores Midpoint Freq code
X f d fd
46 – 50 48 4 3 12
41 – 45 43 6 2 12
36 – 40 38 10 1 10
31 – 35 33 12 0 0
26 – 30 28 8 -1 -8
21 – 25 23 7 -2 -14
16 – 20 18 3 -3 -9
Total 50 3

 fd i 35  33  15  33.3
Coding method X  AM   33 
n 50 50

35
OPTIONAL
Using Microsoft Excel

1. Open Excel
2. Type in data to be used in one column, if data is not yet entered.
3. Click an empty cell where you want the result to be and type in Mean.
4. Click the empty cell directly below where you typed Mean.
5. Click white space to the right of the fx symbol.
6. Type in =AVERAGEA(cell number where data begins from:cell number
where data ends at). E.g. =AVERAGEA(B2:B32). This means that data
begins at cell B2 and ends at cell B32.
7. Press Enter. (The mean is given in the empty cell clicked.

An example is shown below.

36
Properties of the Mean

1. The mean is influenced by every score or value that makes it up. If a score is
changed, the values of the mean changes.
3, 4, 2, 4, 7 Mean = 4
3, 4, 7, 4, 7 Mean = 5. The change of the score 2 to 7 has changed the
mean to 5.

2. The mean is very sensitive to extreme scores (outliers).

4, 2, 3, 6, 5 Mean = 4
4, 2, 23, 6, 5 Mean = 8. All the scores are below 7 and the presence of 23,
an outlier has moved the mean from 4 to 8.

3. The mean is a function of the sum (or aggregate or total) of the scores.

X 
X
N
NX   X This implies that the number of observations
multiplied by the mean gives the sum of the scores.

Of the three measures it is the only one that is a function of the sum of the scores.
It is also possible to calculate the mean for a combined group if only the means and
number of scores (N) are available.

e.g. Mr Mensah‟s class Mean =5 N = 20

Ms Addo‟s class Mean = 6 N = 30

(5 X 20)  (6 X 30) 280

Mean for the total group: X    5.6
50 50

4. If the mean is subtracted from each individual score and the differences are
summed, the result is 0.
4 – 4 =0
2 – 4 = -2
3 – 4 = -1
6–4=2
5–4=1
The distance of the score from the mean is known as the deviation.

5. If the same value is added to or subtracted from every number in a set of scores,
the mean goes up or goes down by the value of the number.
For example, given 8 2 10 4 X  6.
37
Now add 2 to each score: 10 4 12 6 X  8 ie 6 + 2

6. If each score is multiplied or divided by the same value, the mean increases or
decreases by the same value.
For example, given 8 2 10 4, X 6.
Now multiple each score by 3. 24 6 30 12 X  18 ie 6 × 3

Uses of the mean

1. It is useful when the actual magnitude of the scores is needed to get an average.
E.g. total sales for a new product, selecting a student to represent a whole class in
a competition.
2. It is useful for further statistical work e.g. standard deviation, correlation
coefficient.
3. It is useful when the scores are symmetrically distributed (i.e. normal).
4. It provides a direction of performance, compared with other measures of location
especially the median. Where Mean >Median, the distribution is skewed to the
right (positive skewness) showing that performance tends to be low and where
Mean < Median, the distribution is skewed to the left (negative skewness)
showing that performance tends to be high.
5. It serves as a standard of performance with which individual scores are
compared. For example, for normally distributed scores, where the mean is 56,
an individual score of 80 can be said to be far above average/. Also performance
can be described as just above average or far below average or just below
average.

THE MEDIAN (Mdn)

It is a score such that approximately one-half (50%) of the scores are above it and
one-half (50%) are below it when the scores are arranged sequentially.
E.g. Given the scores, 8, 4, 9, 1, 3, the Median, after sequentially arranging the
scores as 1, 3, 4, 8, 9 is 4.

(n  1)
For odd set of numbers, median occupies the th position.
2
For even set of numbers, find the mean of the two middle numbers or the number at
(n  1)
the th position.
2

The median can be obtained from both ungrouped and grouped data and also from
Microsoft Excel.

38
To find the median from ungrouped data
1. Arrange all observations in order of size from smallest to largest or vice versa.
2. If the number of observations, n, is odd, the median is the number at the centre or
(n  1)
the number at the th position.
2
3. If the number of observations, n, is even, the median is the mean of the two
centre observations.

Examples

1. For odd set of numbers

Given a set of observations as: 8 11 26 7 12 9 6 20 14
Note that there are 9 observations, which is odd.
1. Rearrange the scores in a sequential order: 6 7 8 9 11 12 14 20 26
(n  1) (9  1) 10
2. Find th position i.e. th  th  5th position
2 2 2
3. The score at the 5th position is 11.

2. For even set of numbers

Given a set of numbers as: 48 52 36 54 62 71 69 45 58 32
1. Rearrange the scores in a sequential order: 32 36 45 48 50 54 58 62 69 71
(n  1) (10  1) 11 1
2. Find th position ie th  th  5 th position. This means that
2 2 2 2
the median lies half-way between the 5th and 6th positions.
3. The score at the 5th position is 50 and at the 6th position is 54. Half-way
(50  54) 104
between 50 and 54 is   52 . The median is therefore 52.
2 2

To find the median from grouped data

Example
Classes Midpoint Freq Cum Freq
X f cf
46 – 50 48 4 50
41 – 45 43 6 46
36 – 40 38 10 40
31 – 35 33 12 30
26 – 30 28 8 18
21 – 25 23 7 10
16 – 20 18 3 3
Total 50

39
Step 1. Identify the median class. It is the class that will contain the middle score.
N
Find the value of , where N is the total score. This is the position of the middle
2
score. Checking from the cumulative frequency column, find the number equal to
the position or the smallest
N 50
number that is greater than the position. From the table above,   25 ,
2 2
therefore the number is 30. The class that this number belongs to is the median class.
From the table above, the median class is 31 – 35.
Step 2. Use the formula below to obtain the Median.

N 
 2  cf 
Mdn = L  i where
1
 f mdn 
 
L1 is the lower class boundary of the median class
N is the total frequency
cf is the cumulative frequency of the class just below the median class
i is the class size/width
fmdn is the frequency of the median class

Substituting the values in the table in the formula above, we have:

Mdn =
 50 
 2  18   25  18  7
30.5   5  30.5    5  30.5   5  30.5  0.585  30.5  2.9  33.4
 12   12  12 
 

OPTIONAL

To find the median from Microsoft Excel

1. Open Excel
2. Type in data to be used in one column, if data is not yet entered.
3. Click an empty cell where you want the result to be and type in Median.
4. Click the empty cell directly below where you typed Median.
5. Click white space to the right of the fx symbol.

40
 6. Type in =MEDIAN(cell number where data begins from:cell number where
data ends
at). E.g. =MEDIAN(B2:B32). This means that data begins at cell B2 and
ends at cell B32.
7. Press Enter. (The median is given in the empty cell clicked.

An example is shown below.

Features of the median

1. It is not influenced by extreme scores. For example, the median for the following
numbers, 2, 3, 4, 5, 6 is 4. If 6 changes to 23 as an extreme score, the median
remains 4.
2. It does not use all the scores in a distribution but uses only one value.
3. It has limited use for further statistical work.
4. It can be used when there is incomplete data at the beginning or end of the
distribution.
5. It is mostly appropriate for data from interval and ratio scales.

41
6. Where there are very few observations, the median is not representative of the
data.
7. Where the data set is large, it is tedious to arrange the data in an array for
ungrouped data computation of the median.

Uses of the median

1. It is used as the most appropriate measure of location when there is reason to
believe that the distribution is skewed.
2. It is used as the most appropriate measure of location when there are extreme
scores to affect the mean. E.g. Typical income in a company of senior and junior
staff.
3. It is useful when the exact midpoint of the distribution is wanted.
4. It provides a standard of performance for comparison with individual scores when
the score distribution is skewed. For example, if the median score is 60 and an
individual student obtains 55, performance can be said to be below
average/median. Also performance can be described as just above average or
far below average or just below average.
5. It can be compared with the mean to determine the direction of student
performance.
Where Median < Mean, the distribution is skewed to the right (positive
skewness) showing that performance tends to be low and where Median > Mean,
the distribution is skewed to the left (negative skewness) showing that
performance tends to be high.

THE MODE
It is the number that occurs most frequently in a distribution.
Given the following scores, 1, 2, 4, 6, 4, 6, 7, 2, 4 the number that occurs most
frequently is 4. This is the Mode. This number appears 3 times.
Given the following scores,11, 22, 14, 26, 34, 6, 27, 12, 40 no number occurs most
frequently. There is therefore no mode.

1. The main advantage is that it is the only measure that is useful for nominal scale.
2. It is used when there is the need for a rough estimate of the measure of location.
3. It is used when there is the need to know the most frequently occurring value e.g.
dress styles.
4. It is not useful for further statistical work because the distribution can be bi-modal
or trimodal or no mode at all.

42
 Practice Exercises
Use the histogram below to answer questions 1 and 2

40
F
30
R
20
E
10
Q
0 10 20 30 40 50 60 70 80
Classes

1. The overall shape of this distribution is

A. Roughly symmetric.
B. Single-peaked.
C. Skewed to the left.
D. Skewed to the right.

2. Because of the shape of the distribution, the mean is

A. About the same as the median.
B. Approximately 50.0.
C. Larger than the median.
D. Smaller than the median.

3. The mean score obtained by 10 students in a quiz in statistics out of a total of 20

was 11.5. It was observed that a student who obtained 2 really had 20. How
would the discovery affect the mean?
A. More information is needed.
B. New mean is greater than old mean.
C. Old mean is greater than the new mean.
D. There is no change in the old mean.

43
4. The median score for a group of 19 students was 58. A 20th student had a score
of 45. What is the new median score?
A. 10.5
B. 45.0
C. 58.0
D. It cannot be determined

5. In an end-of-semester examination, a mean of 60.0 was obtained by a statistics

class. The median score was 54.0. The performance of the class can be said to be
A. Normally distributed.
B. Skewed to the left.
C. Skewed to the right.
D. Unimodal.

6. A group of 20 students earned a class mean of 30 on a quiz. A second group of

30 students had a mean score of 45 on the same test. What is the mean score of
the 50 students?
A. 30.0
B. 37.5
C. 41.0
D. 45.0

7. In a class quiz, a mean of 62 was obtained with a median of 48. How would the
performance of the class be described?
A. Above average
B. Average
C. High
D. Low

8. Scores on a statistics quiz are as follows:

32 15 20 54 45 37 22
10 25 30 19 42 28 18

The median for this distribution is

A. 7.5
B. 25.0
C. 26.5
D. 28.0
44
 UNIT 4

MEASURES OF VARIABILITY
These are also called measures of variation, dispersion or scatter. The main
measures that are used mainly in education are:

1. The range
2. The Variance
3. The Standard Deviation
4. The Quartile Deviation (Semi-interquartile range)
They are used as single scores to describe individual differences in terms of
achievement.
For example: 48, 51, 47, 50 Total = 196 Mean = 49 …..(i)
30, 72, 90, 4 Total = 196 Mean = 49 …..(ii)
However, a closer look at the two sets of data shows that the distribution within each
set is not the same. Where the scores cluster around the mean, performance is said to
be homogeneous as in (i). Where the scores move away from the mean,
performance is said to be heterogeneous as in (ii).

THE RANGE
It is the difference between the highest and the lowest values in a set of data.
e.g.: 48, 51, 47, 50 Total = 196 Mean = 49 …..(i) Range: 51 – 47 = 4
30, 72, 90, 4 Total = 196 Mean = 49 …..(ii) Range: 90 – 4 = 86

Features
1. It is easy to compute.
2. It is easy to interpret.
3. It is a crude measure of dispersion and does not take into account all the
data/scores.
4. It ignores the spread of all the scores.
5. It uses only two values and does not consider how the other scores relate to each
other.
6. The range does not consider the typical observations in the distribution but
concentrates only on the extreme values.
7. It can give a distorted picture of the variation within a set of data.
8. Different distributions can have the same range which would give misleading
conclusions.

45
 Uses
1. When data is too scanty or too scattered to justify the computation of a more
precise measure.
2. When knowledge of extreme scores or total spread is all that is needed.

VARIANCE & STANDARD DEVIATION

The variance is always considered together with the standard deviation. It is the
square of the standard deviation. Both variance and standard deviation are computed
for both ungrouped and grouped data. Microsoft Excel is also useful in obtaining the
variance and standard deviations.

Ungrouped data
This is based on raw data. It is computed by using the following formulae.
Variance (S2,  )
2

 X  X  X
2 2

1. Var ( S
2
) 2. Var ( S 2
)  X 2 3.
n n
X X
2

2

Var ( S 2
)   

n  n 

Standard Deviation (S, )

 X  X  X
2 2

Std .Dev ( S )  or Std .Dev ( S )   X 2 or

n n

X X
2
2

Std .Dev ( S )    

n  N 

Given a set of data as 48 51 50 47 and the mean of the distribution as 49, the
variance and the standard deviation could be computed as follows:

46
 196
X   49
4
X XX X  X  2
X2
48 -1 1 2304
51 2 4 2601
47 -2 4 2209
50 1 1 2500
Total 10 9614

 X  X 
2
10
SD    2.5  1.58 Variance = 1.582 = 2.5
n 4

OR

SD 
 X2  2

9614
 49 2  2403 .5  2401 .0  2.5  1.58 Var. =
n X 4
1.582 = 2.5

OR

2
SD 
 X 2    X  
9614  196 

2
  2403 .5  2401 .0  2.5  1.58
n  n  4  4 


Grouped data:
This is based on a frequency distribution of the scores.

 f X  X   f X  X 
2 2
Long method: SD  Var 
n n

   fX
2

 fX 2   fX fX 2
2
 
Short method: SD    
 Var    

n  n  n  n 

47
 2
 fd 2

  fd 
Coding Method SD  i  This is useful with equal class
n
 n

intervals.

Using the short method

Scores Midpoint Freq

X2 fX fX2
X f
46 – 50 48 4 2304 192 9216
41 – 45 43 6 1849 258 11094
36 – 40 38 10 1444 380 14440
31 – 35 33 12 1089 396 13068
26 – 30 28 8 784 224 6272
21 – 25 23 7 529 161 3703
16 – 20 18 3 324 54 972
Total 50 1665 58765

Short method:

 fX 2   fX
2
 58765  1665 
2

SD        1175 .3  1108 .89  8.15

n  n  50  50 
 

Using the coding method

Scores Midpoint Freq code

d2 fd fd2
X f d
46 – 50 48 4 3 9 12 36
41 – 45 43 6 2 4 12 24
36 – 40 38 10 1 1 10 10
31 – 35 33 12 0 0 0 0
26 – 30 28 8 -1 1 -8 8
21 – 25 23 7 -2 4 -14 28
16 – 20 18 3 -3 9 -9 27
Total 50 3 133

48
Coding Method

2
 fd 2   fd

 133  3 
2

SD  i  5     5 2.66  0.0036  8.15

50  50 
n
 n

OPTIONAL
Using Microsoft Excel

Standard Deviation
1. Open Excel
2. Type in data to be used in one column, if data is not yet entered.
3. Click an empty cell where you want the result to be and type in Std. Dev.
4. Click the empty cell directly below where you typed Std. Dev.
5. Click white space to the right of the fx symbol.
6. Type in =STDEVPA(cell number where data begins from:cell number
where data ends at). E.g. =STDEVPA(B2:B32). This means that data begins
at cell B2 and ends at cell B32.
7. Press Enter. (The standard deviation is given in the empty cell clicked.

Variance
1. Open Excel
2. Type in data to be used in one column, if data is not yet entered.
3. Click an empty cell where you want the result to be and type in Variance.
4. Click the empty cell directly below where you typed Variance.
5. Click white space to the right of the fx symbol.
6. Type in =VARPA(cell number where data begins from:cell number where
data ends at). E.g. =VARPA(B2:B32). This means that data begins at cell
B2 and ends at cell B32.
7. Press Enter. The variance is given in the empty cell clicked.

49
An example is below:

Features of the variance/standard deviation

1. The standard deviation/variance of a constant is zero.

Eg. 8 8 8 8 8 8 8 8 8 8 has a standard deviation/variance of zero
2. It is not resistant. It is affected by extreme scores or outliers. For example, given
the scores, 5, 8, 10, 6, 7, SD = 1.9 and Var = 3.7. However, if the scores become,
5, 8, 10, 6, 24, SD = 7.7 and Var = 59.8
3. The standard deviation/variance is independent of change of origin. If each score
in a set of data is reduced or increased by the same amount, the standard
deviation/variance of the new set of data does not change. For example, given
the data 48 51 47 50 with a standard deviation of 1.58. If 10 points is added to
each score to obtain 58 61 57 60, the standard deviation/variance remains
unchanged.
4. The standard deviation/variance is not independent of change of scale. If each
score in a set of data is multiplied or divided by the same amount, say a constant
k, the resulting standard deviation equals k multiplied by the old standard
deviation but the variance equals k2 multiplied by the old variance. For example,
given the data 48 51 47 50 with a standard deviation of 1.58. If each score is
50
 multiplied by 10 points to obtain 480 510 470 500, the standard deviation
becomes 1.58 x 10 = 15.8 and the variance becomes 102 x 2.5=250.
5. It uses every value in the distribution.
6. It is difficult to calculate for open-ended distributions.
7. It is affected by extreme values. It gives more weight to extreme values.

USES
1. It is used as the most appropriate measure of variation/dispersion when there is
reason to believe that the distribution is normal.
2. It helps to find out the variation in achievement among a group of students. i.e. it
determines if a group is homogeneous or heterogeneous.
Where the standard deviation is relatively small, the group is believed to be
homogeneous i.e. performing at the about the same level. On the other hand,
where the standard deviation is relatively large, the group is believed to be
heterogeneous, i.e. performing at different levels.
To be more precise, the coefficient of variation (CV) is computed.


CV = x 100 If the value of CV is greater than 33, the group is
x
heterogeneous, otherwise it is homogeneous.
With this information, the teacher has to adopt a teaching method to suit each
group.
3. It is helpful in computing other statistics e.g. standard scores, correlation
coefficients.
4. It is useful in determining the reliability of test scores. The split-half correlation
method or internal consistency methods use the standard deviation of the scores.

In most score interpretations in education and for descriptive statistics, the

standard deviation is preferred to variance because
1. the standard deviation (S), is the natural measure of spread or variation for
normal distributions
2. the variance (S2) involves squaring the deviations and does not have the same
unit of measurement as the original observations.

QUARTILE DEVIATION

It is also called the semi-inter quartile range and it depends on quartiles.

Quartiles divide distributions into 4 equal parts. Practically there are 3 quartiles.
The QD is half the distance between the first quartile (Q1) and the third quartile
(Q3).
51
 Q3  Q1
Method: QD =
2

Computing Quartiles from Ungrouped data

There are two methods – the median method and the formula method.

The median method

1. First arrange the scores in a sequential order.

 n  1
2. Find the median (i.e. score at the   position) for the data set. The
 2 
median divides the distribution in two equal parts.
3. Find the median for the first half/part. This median becomes Q1, the first
quartile.
4. Find the median for the second half/part. This median becomes the Q3, third
quartile.

Example.
Given the following scores, 8, 10, 12, 7, 6, 13, 18, 25, 4, 22, 9.

Arrange in ascending order as, 4, 6, 7, 8, 9, 10, 12, 13, 18, 22, 25

Q1 Median Q3

The Formula method

1
Q1 is the score at the (n+1)th position.
4
3
Q3 is the score at the (n+1)th position.
4

Given the following scores, 8, 10, 12, 7, 6, 13, 18, 25, 4, 22, 9, after arranging them
in ascending order as,4, 6, 7, 8, 9, 10, 12, 13, 18, 22, 25
1 1
Q1 = (n+1)th position → (12) = 3rd position
4 4
3 3
Q3 = (n+1)th position → (12) = 9th position
4 4
52
 Q3  Q1 18  7 11
QD     5.5
2 2 2

Computing Quartiles from grouped data.

N 
 4  cf 
Q1 = LQ1   i where
 fQ1 
 
LQ1 is the lower class boundary of the lower quartile class
N is the total frequency
cf is the cumulative frequency of the class just below the lower quartile class
i is the class size/width
fQ1 is the frequency of the lower quartile class

 3N 
 4  cf 
Q3 = LQ3   i where
 fQ3 
 

LQ3 is the lower class boundary of the upper quartile class

N is the total frequency
cf is the cumulative frequency of the class just below the upper quartile class
i is the class size/width
fQ3 is the frequency of the upper quartile class

Given the distribution below:

Example
Classes Midpoint Freq Cum Freq
X f cf
46 – 50 48 4 50
41 – 45 43 6 46
36 – 40 38 10 40
31 – 35 33 12 30
26 – 30 28 8 18
21 – 25 23 7 10
16 – 20 18 3 3 __
Total 50 __
53
Step 1. Identify the quartile class. It is the class that will contain the quartile of
N 3N
interest. Find the value of , for the lower quartile and for the upper quartile
4 4
(where N is the total score) as positions. Checking from the cumulative frequency
column, find the number equal to the position or the smallest number that is greater
N 50
than the position. From the table above,   12.5 , therefore the number is 18
4 4
3N 150
and   37.5 therefore the number is 40. The classes that these numbers
4 4
belong to are the quartile classes. From the table above, the lower quartile class is 26
– 30 and the upper quartile class is 36 – 40
.
Step 2. Use the formulae below to obtain the lower and upper quartiles.

N 
 4  cf 
Q1 = LQ1   i =
 fQ1 
 
 50 
 4  10  12.5  10   2 .5 
25.5   5  25.5  5  25.5   5  25.5  1.5625  27.06
 8   8   8 
 

 3N 
 4  cf 
Q3 = LQ3   i =
 fQ3 
 
 150 
 4  30   37.5  30   7 .5 
35.5   5  35.5  5  35.5   5  35.5  3.75  39.25
 10   10   10 
 

Q3  Q1 39.25  27.06 12.19

QD     6.095
2 2 2

54
OPTIONAL
Using Microsoft Excel

First/Lower Quartile

1. Open Excel
2. Type in data to be used in one column, if data is not yet entered.
3. Click an empty cell where you want the result to be and type in Q1(Lower
Quartile).
4. Click the empty cell directly below where you typed Q1.
5. Click white space to the right of the fx symbol.
6. Type in =QUARTILE(cell number where data begins from:cell number
where data
ends at,1). E.g. = QUARTILE(B2:B32,1). This means that data begins at
cell B2 and
ends at cell B32 and 1 means first or lower quartile) .
7. Press Enter. The Q1, first/lower quartile is given in the empty cell clicked.

Third/Upper Quartile

1. Open Excel
2. Type in data to be used in one column, if data is not yet entered.
3. Click an empty cell where you want the result to be and type in Q3 (Upper
Quartile).
4. Click the empty cell directly below where you typed Q3.
5. Click white space to the right of the fx symbol.
6. Type in =QUARTILE(cell number where data begins from:cell number
where data
ends at,3). E.g. = QUARTILE(B2:B32,3). This means that data begins at
cell B2 and
ends at cell B32 and 3 means third or upper quartile) .
7. Press Enter. The Q3, third/upper quartile is given in the empty cell clicked.

55
An example is shown below.

Features of the Quartile Deviation

1. For skewed distributions, where the median is used as a measure of location
the quartile deviation is a better measure of variability.
2. The quartile deviation is a measure of individual differences. It helps to find
out the variation in achievement among a group of students. . i.e. it
determines if a group is homogeneous or heterogeneous.
Where the quartile deviation is relatively small, the group is believed to be
homogeneous i.e. performing at the about the same level. On the other hand,
where the quartile deviation is relatively large, the group is believed to be
heterogeneous, i.e. performing at the different levels.
To be more precise, the coefficient of variation (CV) is computed.

QD
CV = x 100 If the value of CV is greater than 33, the group is
Mdn
heterogeneous, otherwise it is homogeneous.
With this information, the teacher has to adopt a teaching method to suit
each group.

56
 3. It does not make use of all the information provided by the scores.

Class Exercise
Distribution of examination scores for Statistics students

Classes Frequency
46 - 50 12
41 - 45 14
36 - 40 12
31 – 35 6
26 – 30 5
21 – 25 1
Total 50

1. Calculate the standard deviation of the scores.

2. Calculate the quartile deviation of the scores.

3. The following scores were obtained by students in a Statistics Class:

44 78 65 67 89 45 57 80 50 54
60 85 82 78 89 75 85 88 70 58
63 48 50 56 84 82 75 70 73 63
52 58 72 80 45 48 70 75 68 55
48 52 80 75 86 45 46 64 52 40
Using Microsoft Excel, obtain the variance, lower quartile and upper quartile.
(Ans: Var. = 208.25; Q1 = 52: Q3 = 78)

57
 Practice Exercises
1. The standard deviation is a measure of the
A. Center of a distribution.
B. Center of the mean deviation.
C. Validity of a measurement.
D. Variability of a distribution.

2. Scores on a Statistics examination have a mean of 64 and standard deviation of

9. The variance of these scores is
A. 3.0
B. 8.0
C. 64.0
D. 81.0

3. The range of the following set of scores is

25 42 20 58 32 35 44 28 15 10
A. 10
B. 43
C. 48
D. 58

4. The first quartile in the following distribution is

8 10 1 3 12 7 4 6
A. 3.0
B. 3.5
C. 6.5
D. 9.0

5. When a distribution is highly skewed to the right, the most appropriate measure
of variability is the
A. first quartile.
B. mean deviation.
C. quartile deviation.
D. standard deviation.

58
6. The first quartile in a distribution of scores is 10.0. The third quartile in the same
distribution is 25.0. What is the value of the semi-interquartile range?
A. 7.5
B. 10.0
C. 15.0
D. It cannot be determined.

7. In a Psychology course quiz, the mean was 45 with a standard deviation of 15.
The instructor later added 10 points to every student‟s score. What is the new
standard deviation?
A. 10
B. 15
C. 25
D. It cannot be determined

8. What is the upper quartile in the following distribution?

81 89 76 78 68 52 60 90
A. 87
B. 81
C. 73
D. 64

59
 UNIT 5

MEASURES OF RELATIVE POSITION

There are two main measures. These are Percentiles and Percentile Ranks, Z scores
and T scores. Z scores and T scores are often referred to as standard scores.
The main purpose of these measures is to describe an individual‟s position in relation
to a known group or the norm group.

PERCENTILES
Definition: They are points in a distribution below which a given percent, P, of the
cases lie.
 There are 99 percentiles that divide a distribution into 100 equal parts.
 Percentiles are individual scores.
Notation: P40 = 60. Sixty is the score below which 40% of the scores lie in a
specific group after the scores have been arranged sequentially. This
means that a student who obtains a score of 60 has done better than
40% of the members in the specific group.
P75 = 50. Fifty is the score below which 75% of the scores lie in a
specific group after the scores have been arranged sequentially. This
means that a student who obtains a score of 50 has done better than
75% of the members in the specific group.
 A score in one group may be a different percentile in another group.
For example, in Statistics Quiz 1, a student with a score of 15 may be at P90 in the
Social Science group but the same score may put the student at P85 in the Home
Economics group.
 P50 is the same as the median. P25 is the first quartile and P75 is the third
quartile.

PERCENTILE RANKS
Definition: The percentage of cases falling below a given point on the measurement
scale. It is the position on a scale of 100 to which an individual score
lies.
Notation: PR of 60 = 75. Seventy-five is the position for a score of 60 when the
distribution is divided into 100 parts. This means that a student who
obtains a score of 60 has 75% of the scores falling below him/her in
the group.

60
The easiest way to obtain percentiles and percentile ranks is to use the ogive
(cumulative percentage graph).

100
90
80
70
60
50
40
30
20
10
0
0 5 10 15 20 25 30 35 40 45

Scores
From the ogive, P60 = 34. PR of a score of 26 is 40.

STANDARD SCORES (Z, T)

It indicates the number of standard deviation units an individual score is above or
below the mean of each group. It represents an individual score that has been
transformed into a common standard using the mean and the standard deviation.

XX
Formula: Z  , T = 50 + 10Z, where mean is 50 and standard deviation
s
is 10.

Example. Given that a student obtained 15 in a quiz with a mean of 12 and a

standard deviation of 2. The Z and T scores become
15  12
Z  1.5 T = 50 + 10(1.5) = 65
2
 For z scores, 0 is the mean score. Positive scores are scores above the mean
(average) and negative scores are scores below the mean (average).
 An individual‟s performance can be described as far above average, above
average, just above average, just below average and far below average.
 In case of T scores, 50 is the mean score. Scores greater than 50 are above
average and scores less than 50 are below average.
 Z scores range between ─ 4 and + 4 while T scores are between 10 and 90.
61
Computational examples

1. A student had a Z score of 2.5. The mean for the class was 60 with a standard
deviation of 4.0. What was the student‟s observed score?

XX X  60
Z → 2.5  → 10  X  60 →X = 10 + 60 = 70
s 4

2. A student obtained a raw score of 70 in an examination. If the raw score gives

her a Z-score of 3.5, what would be the class mean if it is known that the standard
deviation is 5.0?

XX 70  X
Z → 3 .5  → 17.5  70  X → X = 70 − 17.5 = 52.5
s 5

USES

1. It helps the teacher to know an individual‟s position in relation to the rest of

the class. A student with a Z score of 3.2 is performing far above average.
2. It enables the teacher to compare student‟s performances in different subjects
to know individual strengths and weaknesses.

Eg. Mathematics Social Studies

Mean 50 60
Standard deviation 2.5 4.0
Observed Score 55 55
Salome‟s Z score 2.0 -1.25

Salome has done better in Mathematics than Social Studies, considering the class
performance.
3. It helps the teacher to guide and counsel the student to choose the correct
course for a future career and vocation

English Maths Pre-Tech

Mean 80 70 75
Standard Deviation 6.0 2.0 4.0
Observed Score 85 76 80
George‟s Z score 0.42 3.0 1.25

George is more likely to succeed in Maths-related course.

62
 Practice Exercises
1. Scores on an SSSCE Social Studies paper had a mean of 46. Joana obtained a
score of 70, giving her a standard score of 3.0. What was the standard deviation
of the scores?
A. 3
B. 8
C. 24
D. 72
Use the ogive below to answer questions 2 & 3.

100
%age
Cum.
Freq. 80

60

40

20

0 10 20 30 40 50 60 70 80 90
Scores

2. What is the percentile rank of a score of 60?

A. 29.
B. 60.
C. 85.
D. 90.

3. What score is at P40 ?

A. 80.
B. 78.
C. 29.
D. 19.
63
4. Scores on the Teacher Training College entrance exam had a mean of 50 and
standard deviation of 5. Martin‟s standard score on the examination was Z = 2.0.
What was Martin‟s actual examination score?
A. 40
B. 50
C. 52
D. 60

5. Joyce‟s percentile rank in an end of year examination was 80. Her actual
examination score was 95. This information means that she performed better
than _____ of the students in the class.
A. 5%
B. 20%
C. 80%
D. 95%

6. Paul‟s score in a final examination is at the 80th percentile of the scores in the
class. Paul‟s score lies
A. above the third quartile.
B. at the median.
C. below the first quartile.
D. between the median and the third quartile.

7. Which of the following is NOT a measure of the variability of scores in an

examination?
A. Quartile deviation
B. Standard deviation
C. Upper quartile
D. Variance

8. Vida‟s percentile rank in an end-of-year examination was 75. Her actual

examination score was 80. This information means that she performed worse
than ______ of the students in the class.
A. 20%
B. 25%
C. 75%
D. 80%
64
 UNIT 6

MEASURES OF RELATIONSHIPS
Concept

Natural relationships exist in the world. Parents and children as well as twins have
things in common. Males are normally attracted to females and rain results in good
harvest.

In education, absenteeism tends to go with performance in class tests and

examinations. Studies have also shown that females generally do better than men in
the reading subjects while males generally tend to do better than females in the
science-related subjects like Physics, Chemistry and Mathematics.

The concept of correlation provides information about the extent of the relationship
between two variables. Two variables are correlated if they tend to „go together‟.
For example, if high scores on one variable tend to be associated with high scores on
a second variable, then both variables are correlated.

Correlations aim at identifying relationships between variables and also to be able to

predict performances based on known results.

The statistical summary of the degree and direction of the linear relationship or
association between any two variables is given by the coefficient of correlation.
Correlation coefficients range between -1.0 and +1.0. Correlation coefficients are
normally represented by the symbols, r and ρ (rho).

Scatter plots
A scatter plot or scatter diagram shows the nature of the relationship between any 2
variables. To obtain a scatter plot, marks are made on a graph representing the
intersection of the two variables. Scatter plots could either be linear or curvilinear.

65
Examples

100
80
Chemistry

60
Linear relationship
40
20
0
0 20 40 60 80 100
Mathematics

100
80
English

60 Curvilinear relationship
40
20
0
0 20 40 60 80 100
Accounts

Assumptions

1. The variables are random. Neither the values of X nor Y are predetermined.
2. The relationship between the variables is linear.
3. The probability distribution of X‟s, given a fixed Y, is normal, i.e. the sample
is drawn from a joint normal distribution.
4. The standard deviation of X‟s, given each value of Y is assumed to be the
same, just as the standard deviation of Y‟s given each value of X is the same.

66
Assume the following scores in two tests.
Student 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
X 14 16 15 10 9 18 18 14 12 13 15 18 10 12 16 20 15 12 14 10
Y 10 12 15 10 12 15 15 12 14 14 14 10 12 15 10 12 15 15 10 14

Y = 10, X = 14, 10, 18, 16, 14; Distribution of X is normal

Y = 12, X = 16, 9, 14, 10, 20; Distribution of X is normal
Y = 14, X = 12, 13, 15, 10; Distribution of X is normal
Y = 15, X = 15, 18, 18, 12, 15, 12; Distribution of X is normal

Nature of the linear relationship

The relationship is described by direction and degree.

(a) Direction: Positive, (+) High values go with high values and low values go
with low values.
Negative (─) High values go with low values and low values go with
high values.

(b) Degree: High (strong) r > 0.60 r < −0.60

Moderate (mild) 0.40 ≤ r ≤ 0.60 −0.40 ≥ r ≥ −0.60
Low (weak) r < 0.40 r > −0.40
Perfect r = 1.0 r = −1.0
Zero r = 0.0

67
Some examples

Perfect linear positive

correlation

Perfect linear negative

correlation

Zero linear correlation

High linear positive

correlation

68
 Moderate linear positive
correlation

Low linear positive correlation

Commonly Used Types

1. Pearson Product Moment correlation coefficient (r). This is applicable when

both variables are continuous in nature. It uses interval and ratio scale data.
For example the relationship between test scores and age of students.
2. Spearman‟s rank correlation coefficient (ρ). This is suitable for variables that
are both continuous and ranked. It uses ordinal scale data. For example ranks
in terms of school attendance and age
3. Phi coefficient (φ). This is used when both variables are natural dichotomies.
It is also applicable for nominal data. For example the relationship between
gender and political party affiliation.
4. Point biserial correlation coefficient (rpb). This is applicable when one
variable is continuous and the other is a natural dichotomy. It combines
nominal scale data with either interval or ratio scale data. For example the
relationship between gender and test scores
69
Coefficient of Determination (r2)
It is the square of the correlation coefficient. It is the proportion of the variance in Y
accounted for by X. An r of 0.71 gives r2 to be 0.50. This means that 50% of the
variance in Y is associated with variability in X. For example, if the correlation
between class attendance and performance in Statistics is 0.8, then class attendance
explains 64% of the variation in the scores in performance.

Causation and correlation

The presence of a correlation between two variables does not necessarily mean that
there exists a causal relationship between the two variables. A very strong or high
relationship between two variable does not imply that one causes the other. No cause
and effect relationship is determined purely by correlation coefficients.

Computational examples

1. The Pearson product moment r (for interval and ratio scales)

r=
Co var iance ( X , Y )
=
 ( X  X )(Y  Y ) =
S X .S Y n S x.S Y

 ( X  X )(Y  Y ) ……...................….(1)
 ( X  X ) . (Y  Y )
2 2

n  XY   X  Y )
r= .....(2)
[ n  X 2  ( X ) 2 ][ n  Y 2  ( Y ) 2 ]

70
Student Quiz Quiz
XX (X  X )2 Y Y (Y  Y ) 2 ( X  X )( Y  Y )
1 2
No. X Y
1 4 6 -2 4 -1 1 2
2 8 8 2 4 1 1 2
3 10 9 4 16 2 4 8
4 7 7 1 1 0 0 0
5 6 8 0 0 1 1 0
6 3 2 -3 9 -5 25 15
7 8 9 2 4 2 4 4
8 5 10 -1 1 3 9 -3
9 5 6 -1 1 -1 1 1
10 4 5 -2 4 -2 2 4
Total 60 70 44 50 33

Note: X = 6 and Y  7

Using Formula 1:

33 33
r= = = 0.7
( 44)(50) 46.9

2. The Spearman rank correlation coefficient (ρ): For ordinal scale variables

6 d 2
ρ = 1
N N 2  1

71
Given the following scores:

Student Quiz 1 Quiz 2 Quiz 1 Quiz 2 d d2

No. X Y Ranks Ranks Q1-Q2 ranks
1 44 16 8.5 7.5 1.0 1
2 48 18 2.5 4.5 -2.0
4.0
3 50 19 1 2.5 -1.5
2.25
4 47 17 4 6 4.0-2
5 46 18 5 4.5 0.250.5
6 43 12 10 10 0.0 0
7 48 19 2.5 2.5 0.0 0
8 45 20 6.5 1 30.255.5
9 45 16 6.5 7.5 -1.0
1.0
10 44 15 8.5 9 -0.5
0.25
43.00
________________________________________________________

6 d 2 643 258
ρ = 1 1 1  1  0.26  0.74

N N 1
2
 10100  1 990

The Phi Coefficient (φ) For nominal scale variables

2
Φ=
n
This is used when there are only two sub-categories for rows as well as columns i.e.
2x2

The Contingency Coefficient (C) For nominal scale variables

2
C=
n 2
This is used when there is at least more than two sub-categories for either row or
column. i.e. 2x3, 3x3, 2x4, 3x4, etc.

The formula for calculating the χ2 value is as follows.

72
 Oij  Eij 
r c 2

i 1 j 1
Eij
where Oij is the observed count in each cell and Eij is

the expected count in each cell. It is obtained by the formula:

Eij = [(ith row total) (jth column total)]/Grand total

Example 1: Association between Gender and Passing Driving Test (2x2)

Using the Phi Coefficient

Gender
Male Female
Result Total

Pass 150 100 250

(125) (125)
Fail 50 100 150
(75) (75)
Total 200 200 400
The figures in bold and in bracket are the expected counts in each cell.


r c Oij  Eij  2 150  125 2 100  125 2 50  75 2 100  75 2
  
  = 125 125 75 75
i 1 j 1 Eij

625 625 625 625

=    = 5 + 5 +8.33 + 8.33 = 26.66
125 125 75 75

2 26.66
Φ= = = 0.365
n 200

The result shows that there is a weak positive association between gender and
passing a driving test.

73
Example 2: Association between Halls of Residence and Region of Birth (3x3)

Using the Contingency Coefficient

Hall of Residence
Region of Birth Hall 1 Hall 2 Hall 3 Total
Region 1 40 30 30 100
(30) (30) (40)
Region 2 50 40 60 150
(45) (45) (60)
Region 3 30 50 70 150
(45) (45) (60)
Total 120 120 160 400
The figures in bold and in bracket are the expected counts in each cell.

 
r c 2
Oij  Eij

i 1 j 1
Eij
=

40  30  2
30  30 2
 30  40
2
 
50  45
2
 
40  45
2

    
30 30 40 45 45

60  60  2
30  45 2
 50  45
2
 
70  60
2

  
60 45 45 60
100 0 100 25 25 0 225 25 100
=        
30 30 40 45 45 60 45 45 60

= 3.3+0.0+2.5+0.56+0.56+0.0+5.0+0.56+1.67 = 14.15

2 14.15 14.15
C=    0.0342  0.185
n 2 400  14.15 414 .15

The result shows that there is a very weak positive association between gender and
passing a driving test.

74
 Uses of correlation in education

1. It is useful for selection and placement. For example, if mathematics scores

relate well with scores in chemistry, then mathematics scores can be used for
selection into a chemistry class without conducting a chemistry selection
examination.
2. It is used to determine the reliability of standardized and classroom tests. The
Spearman-Brown split-half method uses correlation coefficients.
3. It aids in the provision of evidences for the validity of assessment instruments.
Construct and criterion-related validity evidences are obtained through the
computation of the correlation between two variables.
4. It puts the teacher in a position to predict the future performance of a student.
An established relationship between two subjects is often used as the basis for
predicting performance, but not with 100% certainty. For example, if those
with aggregate 6, from WASHSCE have been found in the University of Cape
Coast to be obtaining First Class degrees, then it can be predicted that any one
with WASHSCE aggregate 6, would do well in the University.
5. It is useful for research purposes. A study of the relationship between study
habits and the academic performance of students in the University of Cape
Coast would use correlations.

Practice Exercises
1. The Phi Coefficient (Φ) is the most appropriate measure of linear relationship
when two variables are:
A. Both continuous.
B. Both natural dichotomies.
C. Continuous and natural dichotomy.
D. Continuous and artificial dichotomy.

2. A University professor wishes to find the relationship between the age of the
students in her Statistics class and the scores in a quiz. What is the most
appropriate measure of relationship to use?
A. Pearson‟s product moment correlation coefficient
B. Phi coefficient
C. Point-biserial correlation coefficient
D. Spearman‟s rank correlation coefficient

75
3. Which of the following correlation coefficients indicates the strongest
relationship?
A. –0.6
B. 0.07
C. 0.25
D. 0.55
4. A teacher wishes to find the relationship between the spelling ability of the
students in her class and their performance in the end-of-semester examination.
What is the most appropriate measure of relationship to use?
A. Pearson‟s product moment correlation coefficient
B. Phi coefficient
C. Point-biserial correlation coefficient
D. Spearman‟s rank correlation coefficient

5. The correlation between study habits and achievement in Statistics has been
found to be 0.92 in a study. The study implies that a student with a
A. high score in study habits is more likely to score low in Statistics.
B. low score in study habits is more likely to obtain a moderate score in
Statistics.
C. low score in study habits is more likely to score low in Statistics.
D. moderate score in study habits is more likely to score high in Statistics.
6. A teacher wishes to find the relationship between the gender of students in his
class and their performance in the end-of-semester examination. What is the
most appropriate measure of relationship to use?
A. Phi coefficient
B. Spearman‟s rank correlation coefficient.
C. Point-biserial correlation coefficient.
D. Pearson‟s product moment correlation coefficient.

76
The graph below shows the relationship between achievement in Mathematics and
English. The correlation of the relationship is approximately

60

50

40
English

30

20

10

0
0 10 20 30 40 50

Mathematics

A. –0.8
B. –0.2
C. 0.3
D. 0.8

The scatter plot shows that students with low college entrance examination scores are
more likely to have
A. low college GPA.
B. high college GPA.
C. perfect college GPA.
D. zero college GPA.

77
 UNIT 7

SIMPLE REGRESSION ANALYSIS

Purpose
Simple regression is concerned with the prediction of the value of a dependent
random variable e.g. (Y) on the basis of known measurement of an independent
controlled variable (X). For example UCC may predict final degree classification
(1st class, 2nd Upper, 2nd Lower), Y, from SSSCE grades, X. A teacher may predict
the performance of a student in a final examination from performance in a class quiz,
X.
Prediction and correlation are closely related. The degree of correlation between any
two variables determines the usefulness of prediction. A correlation of 0.9 between
any two variables would produce better prediction results than a correlation of 0.6.

Conditions/Assumptions
1. The possible values of the independent variable, X, are fixed in advance.
2. The true relationship between the variables, X and Y, is linear and expressed by
the equation, Y = a + bX +ei known as the regression equation. a and b are
parameters of the population and are estimated while ei is the random error. The
equation is the line of regression of Y on X. a is the Y intercept and b is the
regression coefficient or the slope of the regression line.
3. The probability distribution of Y‟s, given a fixed X, is normal.

Estimating the parameters, a, b

The most common method of obtaining the parameters is the Least Squares Method.
The least squares method is named such because the sum of squares of the vertical
deviations of the points from this line is less than the sum of squares of the vertical
deviations from any other line.

Y=a+bX a, is the value of Y when X is

zero, (intercept)
b is the regression slope which is
the change in Y by a unit change
in X.

b slope

a
intercept
78
1. The first step is to present the variables on a scatter diagram to be sure that the
relationship between the variables is linear.

2. Normal equations are solved to obtain the equations for the parameter estimates in
raw score form.
Y = a + bX
∑Y = na + b∑X
∑XY = a∑X + b∑X2

Slope (regression coefficient) b =

 XY - nXY OR b = r
Sy
 X  nX
2 2
Sx
The regression coefficient shows the increase in Y, the dependent variable as X the
independent variable increases by 1 unit.

Intercept, a =
 Y - b X OR a = Y  bX
n
The intercept is the point on the Y axis where X, the independent variable has a value
of 0.

Example
The following scores were obtained in Quiz 1 and Final Examination.
Quiz 1 Final
Exam
X Y XY X2
18 75 1350 324
12 55 660 144
10 45 450 100
20 85 1700 400
15 65 975 225
15 65 975 225
14 60 840 196
10 60 600 100
12 50 600 144
11 50 550 121
18 70 1260 324
16 75 1200 256
9 45 405 81
13 60 780 169
17 70 1190 289

79
∑X = 210 ∑Y = 930 ∑XY = 13535 ∑X2 = 3098

r = 0.93 Sy = 11.77 Sx = 3.36 Y  62 X  14

b=
 XY - nXY =
13535 - 15 14 62  13535 - 13025 510
= = =3.23
 X  nX
2 2
3098  15 14 
2
3098  2940 158
OR
 11.77 
b = 0.93   =3.22
 3.36 

a=
 Y - b X =
930  3.23210  251.7
= =16.8 OR a = Y  bX =62-3.23(14) =
n 15 15
16.8

Estimated equation: Yˆ  16.8  3.22 X

Final Exam score = 16.38 + 3.22 (Quiz 1 Score)

Use in prediction
After obtaining the estimates of a and b, the least squares regression line can be
drawn using two values for X (including X = 0 to obtain the intercept).
Corresponding Y values are obtained for the X values and these values are used to
draw the estimated regression line. Values can then be read from the regression line
to obtain the predicted values.

Method 1
Given Yˆ  16.8  3.22 X ,
Select two values say 0 and 10 for x and compute the corresponding Y values.

For example:
X = 0, Y = 16.8 +3.22(0) =16.8
X = 10, Y = 16.8+3.22(10)=49
Plot the values (0, 16.8) and (10, 49) on the graph using a graph sheet and draw a

80
straight line. Then estimate any value of Y given an X value on the regression line.

Yˆ  16.8  3.22 X =

49

16.8
0 10

Method 2
The estimated regression equation can be used by substituting the given X values to
obtain the predicted values for Y.

Given Yˆ  16.8  3.22 X .

i. What would be the exam score for a student who obtains 12.5 in Quiz 1?
ˆ
Y  16.8  3.22(12.5)  57

ii. A student obtained 72 in her exam. However, she did not take part in
Quiz 1. What would be an estimate of her Quiz 1 score?

72  16.8  3.22 X
72-16.8 = 3.22X
3.22X = 72 ─ 16.8
72  16.8
X= =17
3.22

81
 Practice Exercises
The following scores were obtained by 20 students in an intelligent test which was
used to predict final examination scores in Educational Psychology.

Intelligent test (X) Final Score (Y)

100 70
120 80
90 56
115 74
70 50
100 65
120 80
90 70
90 70
85 60
100 70
105 70
120 80
100 65
95 70
90 65
85 60
100 60
105 70
80 55

1. Estimate the regression equation for the regression of Y on X, given that:

Sx. =14.2 and Sy = 8.4 and r = 0.88.

2. A student obtained 98 in the intelligent test. What is his final examination score
in Educational Psychology?

3. A student obtained a final examination score of 75. What was her score in the
intelligence test?

82
 Use the following information to answer questions 4 and 5.
A study gathers data on the outside temperature during February and March, in
degrees Celsius, and the amount of electricity an office complex consumes, in
kilowatts. Call the temperature, x and electricity consumed, y. The office complex
uses an air-conditioner, so x helps explain y. The least-squares regression line for
predicting y from x is
y = -1344 + 19x

4. It can be seen from the equation of the line that as the temperature goes up,
electricity used, y, goes
A. Down because –1344 is less than 19.
B. Down, because the slope 19 is positive.
C. Up, because the intercept 1344 is negative.
D. Up, because the slope 19 is positive.

5. When the temperature goes up 1 degree, the electricity usage predicted by the
regression line goes
A. Down 1 kilowatt.
B. Down 19 kilowatts.
C. Up 1 kilowatt.
D. Up 19 kilowatts.

6. The following estimate of a least squares regression equation was obtained for
headteachers‟ supervision and achievement in a private school.
Y = –5.0 + 1.5X, where Y is the achievement score and X is the supervision
rating.
Mr. Danso obtained a supervision rating of 40. What is the estimate of his score
in achievement?
A. 6.5
B. 55.0
C. 65.0
D. 65.5

83
 UNIT 8

THE NORMAL DISTRIBUTION

Nature It is regarded as the foundation of all statistical distributions. It can be
referred to as the „mother‟ of all distributions. It is often regarded as the
most
important of all the statistical distributions.

The horizontal axis is measured in terms of standard deviation units. The values
decrease to the left and increase to the right from the centre.

Suppose the standard deviation is 4 with a mean of 21. The distribution takes the
form below.

-3  -2  -1  μ 1 2 3
9 13 17 21 25 29 33

84
Symbol

A variable which is distributed normally has the symbol, X ~ N( μ, σ2 ) where

μ, is the mean and σ 2 , the variance. This is read as „the variable, X, is distributed
as norma1 with
a given mean and a given variance.

Features
1. It is a bell-shaped curve.
2. It is unimodal.
3. It is symmetrical.
4. It is asymptotic.
5. The total area under the curve is 1.0.
6. The mean, mode and median are all equal.
7. When the values of a normal distribution have been converted to standard z-
scores, a standard normal curve is obtained. The standard normal curve has a
mean of 0 and a standard deviation/variance of 1.

Symbol: X ~ N (0, 1) read as „The variable, X, is distributed as norma1 with

mean of 0 and variance of 1.

-3 -2 -1 0 1 2 3
Mean = mode = median = 0
The mean of 0 also means that the Z value is 0.

85
8. Areas under the normal curve. Note that these areas are obtained from the table
on normal distributions. Refer to Appendix to follow the areas.

One tail Two tail

1. μ+1σ = 0.3413 (34.13%) μ  1 = 0.6826 (68.26%)

2. μ+2σ = 0.4772 (47.72%) μ  2 = 0.9544 (95.44%)

3. μ+3σ = 0.4987 (49.87%) μ  3 = 0.9974 (99.74%)

4. μ+1.645σ = 0.4500 (45.00%) μ  1.645 = 0.90 (90%) Also 1.65

5. μ+1.96σ = 0.4750 (47.50%) μ  1.96 = 0.95 (95%)

6. μ+2.575σ = 0.4950 (49.50%) μ  2.575 = 0.99 (99%) Also 2.58

Basic applications
Finding Probabilities
1. The distribution for a Statistics examination is normal with a mean of 60 and
variance of 64 (i.e. X ~ N(60, 64). A student is selected at random from the
class. What is the probability that the student selected obtains a score above
68? Above 76? Below 52?
68  60
P (X>68) = P( Z  )
8
8
= P( Z  )
8
= P ( Z 1)

0 1
=0.5000-0.3413
= 0.1587

86
 2. The distribution for a Statistics examination is normal with a mean of 60 and
variance of 64 (i.e. X ~ N(60,64). A student is selected at random from the
class. What is the probability that the student selected obtains a score
between 52 and 76? Between 68 and 76?

52  60 76  60
P (52<X<76) = P( Z  )
8 8
8 16
= P( Z  )
8 8
= P ( 1 Z  2)

─1 0 2
=0.3413 + 0.4773
= 0.8186

Finding performance levels

1. Given that a distribution of scores is normal, with mean 16 and standard

deviation of 2. About what percent of students obtained scores less than 12?
More than 14?

12  16
P (X<12) = P( Z  )
2
4
= P( Z  )
2
= P(Z<−2)

─2 0
=0.5000 − 0.4772
= 0.0228 (About 2%. Actual 2.28%)
87
2. Given that a distribution is normal, with a mean of 50 and a standard deviation of
10. From a class of 2000 students, approximately how many students obtained
scores above 70? Between 40 and 60?

70  50
P (X>70) = P( Z  )
10
20
= P( Z  )
10
= P ( Z  2)
=0.5000-0.4772
= 0.0228
Number of students: 0.0228 x 2000 = 45.6 ≈ 46

3. In a promotion examination, a pass mark was fixed at 40. Given that the
distribution is normal, with a mean of 50 and a standard deviation of 5.1,
approximately how many students failed from a class of 400?

40  50
P (X<40) = P( Z  )
5.1
 10
= P( Z  )
5.1
= P ( Z 1.96)

─1.96 0
=0.5000-0.475
= 0.025
Number of students: 0.025 x 400 = 10

88
 Practice Exercises

1. The distribution of an-end-of term examination scores is normal with mean 55

and standard deviation 10. In a class of 200 students, approximately how
many of them obtained scores greater than 75?
A. 5
B. 10
C. 190
D. 195

2. The distribution of heights of adult men is approximately normal with mean 66

inches and standard deviation 2 inches. About what percent of men are shorter
than 62 inches?
A. 2.5%
B. 5%
C. 95%
D. 99%

3. The distribution of a Mature Students‟ Examination is normal, with mean 45

and standard deviation of 15. The pass mark was 60. In a group of 400
candidates, about how many passed the examination?
A. 64.
B. 128.
C. 272.
D. 336.

4. The mean of a standard normal distribution is

A. 0.
B. 1
C. 2
D. 4

89
5. Which of the following statements is true about the standard normal curve?
A. It is bi-modal.
B. Mean is 1.0.
C. Mean is less than median.
D. Variance is 1.0.

6. The distribution of scores in a final psychology examination was normal with

mean 55 and standard deviation of 10. What percentage of students obtained
scores greater than the mean?
A. 10%.
B. 50%.
C. 68.26%.
D. 95.45%.

7. The distribution of a Mature Students‟ Examination is normal, with

mean 45 and standard deviation of 15. The pass mark was 60. In a
group of 400 candidates, about how many passed the examination?
A. 64
B. 128
C. 272
D. 336

90
 References

Bloomers, P. J. & Forsyth, R. A. (1977). Elementary statistical methods in

psychology and education. Lanham, MD.: University Press of America, Inc.
Ferguson, G. (1981). Statistical analysis in psychology and education. Singapore:
McGraw Hill
Co.
Gordor, B. K. & Howard, N. K. (2000). Elements of statistical analysis. Accra, GH.:
City Printers.
Glass, G. V. & Hopkins, K. D. (1984). Statistical methods in education and
psychology. Englewood Cliffs, NJ: Prentice-Hall, Inc.
Hays, W. (1988). Statistics. New York, NY: Holt, Rinehart and Winston.
Minium, E. W., King, B. M. & Bear, G. (1993). Statistical reasoning in psychology
and
education. New York, NY: John Wiley & Sons, Inc.
Nsowah-Nuamah, N. N. N. (2005). Basic Statistics. Accra, Gh.: Acadec Press.
Pagano, R. R. (1994). Understanding statistics in the behavioural science (4th ed.).
New York, NY: West Publishing Co.
Tamakloe, E. K., Atta, E. T., & Amedahe, F. K. (1996). Principles & methods of
teaching. Accra: Black Mask Ltd.

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