method and procedure of the study -...

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CHAPTER- III

METHOD AND PROCEDURE OF THE STUDYThe methods and procedure of conducting a research study are, by and large,

determined by the design of the study and realization of its objectives, stipulated

purposes and testing the variables involved. This would imply sub-heads like:

3.1 Method of research

3.2 Design of the study

3.3 Population and sample

3.4 Procedure followed

3.5 Statistical analysis

3.6 Precautions observed

3.7 Constraints and difficulties

3.1 METHOD OF RESEARCH

According to Hillway (1964, p.138) ----- ‘to describe in detail the specific

method being used, incidentally, constitutes a very good way of determining whether

the method chosen has been worked out properly and is likely to prove effective. If

the scholar cannot describe his method, the chances are that it is too vague and

general to yield him satisfactory results’.

Broudy (1963) stated that “Method refers to the formal structure of the

sequence of acts commonly denoted by instruction. The term method covers both

strategy and tactics of teaching and involves the choice of what is to be taught and the

order in which it is to be taught.

George J. Mouly has classified research methods into three basic types:

survey, historical and experimental methods.

Methods to conduct a research study differ in their nature and intent. Choice of

the methods of research is determined by the nature of the problem.

The present study is an attempt to study the effect of ICT on the students’

academic achievements. It is obvious that the effect of ICT cannot be studied through

survey or historical method. It needs an experimental setting. Keeping this in mind,

the investigator used pre-test, post-test experimental method to conduct this study .

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3.2 DESIGN OF THE STUDY

A design is used to structure the research, to show how all the major parts of

the research projects the sample or groups, measures, treatments or programmes and

method of assignments work together to try to address the central research question.

Winer (1971) compared the design of an experiment to an architect’s plan for

the structure of a building. The designer of experiments performs a role similar to that

of the architect. The prospective owner of a building gives his basic requirements to

the architect, who then exercising his ingenuity prepares a plan or a blueprint

outlining the final shape of the structure. Similarly, the designer of the experiment has

to do the planning of the experiment so that the experiment on completion fulfils the

objectives of research.

In the present study, students in the ICT group were taught using a power

point programme saved to CD-ROM. The power point presentation included animated

pictures, video clips. Students in the traditional group were taught using a chalkboard,

textbooks, models and charts. Experimental classes housed a ceiling-mounted LCD

projector that was connected to a computer and classroom projector projected onto a

interactive whiteboard. The power point presentation was presented on a ceiling-

mounted LCD projector. The presentation expanded each lesson by providing extra

examples and examples from the homework. Students were able to solve example

problems and then instantly see the answer on a large screen in the classroom. This

presented them immediate feedback.

In the present study, pre-test post-test control group quasi experimental,

design was employed with a purposive sample in the form of intact sections of class

IX of the same school.

The study included a control group (34 students) and an experimental group

(34 students). The experimental group was taught through ICT used teaching and the

control group through traditional method.

The intact sections were equated on intelligence and socio-economic status. A

figurative representation of the design is given in Table 3.1.

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Table 3.1 Design of the Study

Groups Pre-Test Independent variable Post-test

Experimental Y1 ICT-used teaching Y2

control Y1 Traditional teaching Y2

The study involved three operational stages as identification stage, treatment

stage and post-testing stage. The first stage involved pre-testing of all the students of

both groups on intelligence, socio-economic status, and achievement and confidence

level in mathematics. The second stage involved the experimental treatment, which

consisted of ten subunits of IX grade mathematics taught through ICT used teaching

and through traditional teaching to control group. The third stage dealt with post

testing of the control and experimental group using the achievement test in

mathematics. A schematic view of the phases of experiment is presented in Table3.2.

Table 3.2 Phases of the study

Stage Control Group Experimental Group

I Pre-testing 1.Measurement of intelligence of

pupils

2.Measurement of socio-economic

status of pupils

3. Measurement of achievement in

mathematics

4. Measurement of confidence

level of pupils

1.Measurement of intelligence of

pupils

2.Measurement of socio-

economic status of pupils

3. Measurement of achievement

in mathematics


level of pupils

II. Treatment Teaching mathematics through

conventional method

Teaching mathematics through

ICT used method

III.Post-testing 1. Measurement of achievement in

mathematics.


level in answering test questions of

pupils

1.Measurement of achievement

in mathematics


level in answering test questions

of pupils

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Variables under Study

In an experimental research, the relationship between two types of variables,

namely independent and dependent variables, is studied. Independent variables are the

causes, while dependent ones are effects. Another category of variables, which is

equally important, is of the intervening variables. The three kinds of variables,

identified for the study are:

- Independent Variables

These variables are manipulated in order to see their effect on the learning

outcome of students. In this study ‘Treatments’ acted as an independent variable. The

treatments involved the two approaches of teaching viz., ICT-used teaching and

traditional teaching. The experimental group was taught through ICT- used teaching

and the control group was taught through the traditional teaching. Thus, ICT-used

teaching and traditional teachings were the two independent variables for the study.

- Dependent Variables

Achievement in mathematics and confidence level were taken as dependent

variables, which were measured twice during the course of the study. First, before

beginning the experimental treatment, i.e., at the pre-test stage and then, after

completing the experimental treatment, i.e., at the post-test stage.

- Intervening Variables

There are certain variables known as intervening variables which have their

effect on the learning outcomes, and influence both independent and dependent

variables. Intervening variables such as nature of school, grade level, subject to be

taught, intelligence of pupils, socio-economic status of pupils, previous knowledge of

pupils etc. were successfully controlled experimentally.

Control Employed

It is necessary to control all those variables that may significantly affect the

dependent variables. Hence, such intervening variables were controlled by employing

suitable controls.

1. Nature of school

The sample was selected from a single English medium public school (M.M.

Public School, Assandh) situated in Karnal district.

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2. Grade Level

IXclass students were selected for the study and grade level was thus kept

constant during the study.

3. Teacher behaviour

The investigator herself taught the content to both experimental and control

group i.e., inter-teacher variation was eliminated. She herself outlined the entry level

behaviour, prepared achievement test, lesson plans etc. Hence there was equal

familiarity with all the treatments.

4. Subject

The two groups were taught same units of mathematics.

5. Socio-Economic Status

The experimental group and the control group were given S.E.S. Test. t-test

was applied to find out the difference between S.E.S. test scores of the two groups.

The results are given in the Table 3.3

Table 3.3‘t’-Value of S.E.S. test scores of Experimental groupand Control Group

Group N Mean SD t-Value

Control Group

Experimental

Group

34

34

19.265

19.853

4.166

3.823

.607(NS)

(NS) Not significant at 0.01 level

Table3.3. shows that the t-value between the groups is (0.607) which is not

significant at 0.01 levels. It means that no significant differences existed between

the S.E.S. of the two groups, indicating that they belonged almost to the same kind of

a socio -economic milieu.

6. Intelligence of Pupils:

To eliminate the initial variability of the pupils statistically in the two groups,

they were measured on general mental ability, through Cattell,S Culture Fair

Intelligence Test.General mental ability is an index of intelligence which might have

effected the independent variables.

t-value was computed to analyse the difference between intelligence test scores of the

two groups. The results are given in Table 3.4.

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Table 3.4‘t’-VALUE OF INTELLIGENCE TEST SCORES OF

EXPERIMENTAL GROUP AND CONTROL GROUP

Group N Mean SD t-Value

Control Group

Experimental

Group

34

34

101.294

101.764

11.68

7.61

0..21(NS)

(NS) Not significant at 0.01 level

Table3.4. shows that the t-value between the two groups is (0.21) which is

not significant at 0.01 level. It means that no significant difference existed between

the intelligence of the two groups.

Initially, general mental ability was thought to be controlled statistically

through covariance but since the two groups selected did not differ on general mental

ability at the pre-test stage, there was no need to control covariate.

The independent variables, dependent variables, control variables and the kind of

control employed in the study are summarized in Table 3.5.

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Table3.5 Control Employed to Variables

Independent

Variables

Dependent

Variables

Control

Variables

Control

Employed

Method

of

Teaching

1. Achievement

In Mathematics

2.Confidence level of

pupils in answering

test questions

1. Nature

of school

2. Grade Level

3. Teacher

4. Subject to be

taught

5. Duration of

treatment

6. Pupils’ socio-

economic status

7. Pupils’

intelligence

1. Administrative

(Single School)

2. Administrative

(Only IX class chosen

as sample and taught)

3. Both the groups

were taught by

the same teacher

(investigator hereby)

4. Administrative

(Same units of

Mathematics taught in

both groups)

5. The two groups

taught for 33 days, 40

minutes each period

daily.

6. Belonged to the

same milieu.

7. No need.

Specific events and factors like anxiety, home environment, adjustment, and

social maturity could have only a marginal effect upon the experiment, so these

factors were not taken into account.

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3.3 POPULATION AND SAMPLE

The term ‘Population’ is used in research to describe any group of individuals,

events or observations in which the researcher is interested. In the present study, the

term population refers to class IX students studying in English Medium Public School

of Karnal district.

A sample is a finite part of a statistical population whose properties are studied

to gain information about the whole (Webster, 1985). When dealing with people, it

can be defined as the set of respondents (people) selected from a larger population for

the purpose of a survey. In majority of the studies, it is just not feasible to collect data

from each and every subject. In addition, to work on a sample saves time, labour and

money.

Sampling makes it possible to draw valid generalisation by studying a

relatively small proportion of the population selected for observation and analysis. In

the present investigation, Karnal district of Haryana was the field of study. The

sample of the study comprised 34 pupils each studying in two sections of the IX class

of M.M. public School, Assandh situated in Karnal district. One section formed the

control group and the other section formed the experimental group, as shown in

Table3.6.

TABLE 3.6 SAMLE OF STUDY

S. NO. Groups Total No. of Students

1 Experimental group 34

2 Control Group 34

Total 68

No doubt, the sample is small for the result of the study to be generalized, an

experimental study is normally more suitable on a small sample, as is evident from

earlier investigations conducted through experimental design, which used small

samples only.Krulger (1999) and Angrist and Lavy (2004) provide evidence in favour

of positive and significant effect of small classes in experimental studies.

Arias and Walker (2004) conducted an experiment to test the relationship

between class size and student performance. They controlled variation in instruction,

lecture material, and topic coverage by using the same instructors. Their results were

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statistically significant showing that small class size had a positive impact on student

performance.

3.4 PROCEDURE FOLLOWED

Procedure of the experiment comprised of two main stages, that is, selection

of the sample and conducting the experiment.

Stage1: Selection of the sample

The sample of the study comprised of 68 students of class IX (34 as control

group and 34 as experimental group) studying in M.M. Public School, Assandh distt.

Karnal.

Selection of Experimental Group:

For the experimental group, a total of 34 learners studying in IX standard,

section A was chosen from M.M. Public School Assandh distt. Karnal.

Selection of Control Group:

The control group consisted of 34 learners studying in IX standard, section B

of the same school. The group was exposed to traditional method of instruction. No

novel treatment was given to the control group of students.

Stage2: Conducting the experiment

The experiment was conducted in three phases:

Phase I: Administration of the Pre-test

Before the start of the experiment, the sample subjects were contacted and

rapport was established with them. They were oriented about the tests to be used.

Three pre-tests i.e., S.E.S., Intelligence, Achievement Test were administered to the

students of two groups by the researcher herself. Cooperation of the class teacher was

sought for administering the tests properly. The instructions pertaining to the tests

were explained verbally in clear terms to the students before administering the test.

The administration of the tests was carried out as per norms and instructions

contained in respective test manuals.

After this, the students of both the groups were provided orientation and

instructions about the treatment to be allotted to them to get over the anxiety and

curiosity of the students. The students of the experimental group were given a trial of

their respective materials, which helped them in getting over the curiosity and anxiety

around via the electronic system being applied in the classroom setting. The students

of the control group were also made familiar about the objectives, etc, of the tests to

elicit their cooperation in the conduct of the study.

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Phase II: Conducting the Instructional programme

The second phase of the experiment was addressed to the real execution of the

experiment. In this phase, the experimental group students were taught by ICT-used

teaching and the control group students were taught by traditional method of teaching.

The instructional treatment was given about 40 days to the experimental

group, where as the control group was taught by the traditional method. Same content

was taught to both the groups.

Phase III: Administration of Post-test

Immediately after the instructional treatment was over, the researcher tested

the subjects of experimental group and control group on the dependent variables

(Academic Achievement and Confidence level ).

Date Schedule of the Instructional Phase for both the groups:

Phase 1: Pre-test Stage

9th November 2009-Administration of Achievement Test in Mathematics

Phase2: Instructional Programme

Table3.7 Date Schedule of the Instructional Phase for both the groups

S.NO. Name of Sub-unit Date Schedule of the

Instructional Phase

1 Surface Area of cuboids 10th-13th NOV.2009

2 Surface Area of cube 14th-18thNOV.2009

3 Surface Area of right circular cylinder 19th-23rd NOV.2009

4 Surface Area of right circular cone 24th-27th NOV.2009

5 Surface Area of Sphere 28NOV.-2nd DEC.2009

6. Volume of cuboids 3rd.-5th DEC.2009

7 Volume of cube 7th.-10th DEC.2009

8. Volume of right circular cylinder 11th-16th DEC.2009

9. volume of a right circular cone 17th-19th DEC.2009

10 Volume of a Sphere 21st-24thDEC.2009

25th December2009- Achievement Test in Mathematics

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As many as 354 slides were prepared (Appendix-I) and demonstrated before

the students in the experimental classroom situation to explain and elaborate the

teaching of different topics listed above to enrich the teaching-learning process and to

improve the quality of schooling. Their details are briefly summarized lesson –wise as

follows as part of the delivery of demonstration and teaching topic-wise and day-wise

to facilitate understanding the merit of the use of ICT in the classroom.

Day-wise details of the Instructional Programme

S.NO Name of Sub-unit Date Schedule of the

Instructional Phase

1 Surface Area of cuboids 10th-13th NOV.2009

(1) Topic- ‘Surface Area of cuboids’ (10 th Nov.2009)

Material used:

Power-point CD with LCD projector

Method :

Step 1: Asked some questions from the students to know their previous

knowledge and to motivate them (Slide-1).

After that announced the topic ‘Surface Area of cuboids’ and showed it on

screen (slide2-3).

Step 2: Explained about its Geometrical Representation and also told

them that in cuboids there were 8 vertexes, 12 edges and 6 rectangular faces

(slide4-5).

Concept of a cuboid was made clear by showing the examples related to daily

life, i.e.,picture of a match box and picture of a book (slide6-8)

Step 3: Explained the procedure about deriving the formula for finding the

‘.Surface Area of cuboids’ (slide 10).

\To explain it, followed the following steps:

Explained that pair of opposite rectangular faces is of same area which were of

different colors (slide9-12).

With the help of MS Animation Clips explained that to find the ‘Surface

Area of cuboids’ we have to add the surface areas of all its 6 faces (slide13-

15).

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Step 4: Also explained the steps for solution of a problem based on the

‘Surface Area of cuboids’ (slide16-17).

Step 5: Gave the students one sum based on the above topic to solve in their

note books (slide-18)

and also check their note books.


note books at their (slide-19).

(2) Topic- ‘Lateral Surface Area of cuboids’ (11 th Nov.2009)

Material used: -

Power point CD with LCD projector

Method:


knowledge and motivate them and then announced the topic ‘Lateral

Surface Area of cuboids’ and showed it on screen.(slide-20)


‘Lateral Surface Area of cuboids’. To explain it, the researcher followed the

following steps

Explained that pair of opposite rectangular faces is of same areas which were

of different colors .With the help of MS Animation Clips, explained that to

find the ‘Lateral Surface Area of cuboids’ we have to add the surface areas

of all its 4faces leaving the bottom and top faces. (slide21-23).


‘Lateral Surface Area of cuboids’ (slide24-25).


note books (slide-26)

and also checked their note books.

Step 5: Gave the students two sums based on the above topic to solve in their

note books at their homes (slide-27).

3) Topic- ‘Solutions of the Problem based on ‘Surface Area of cuboids’

(12 Th Nov.2009)

Material used:

Power - point CD with LCD projector

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Method:


knowledge and to motivate them.

Step 2: Explained the steps for solution of a problem based on the ‘Surface

Area of cuboids’ (slide-28).

Step 3: In Question No.1 (slide-30).

Also told the formula to calculate the area of sheet required for making the

box as well as the steps to be followed for its calculation.


- Explained the formula for calculating the area of four walls of the

rooms as well as the steps for its calculation.

- Also told them to add the areas of four walls of the rooms and area of

the ceiling.

- Explained that to calculate total cost of white washing we have to

multiply the Total Area by Cost per meter square.


- Explained the formula for calculating the area of four walls of the

rooms as well as the steps for its calculation.

- Explained the formula for calculating the Perimeter of the hall as well

as the steps for its calculation.

- Also told them the procedure of calculating the height of the hall.


note books (slide-33) and also check their note books.


note books at their homes (slide-34).

1. Surface Area of cube 13th-18th NOV.2009

(1) Topic- ‘Surface Area of cube’ (13 th Nov.2009)

Material used:


Method :

Step1: Asked some questions from the students to know their previous

knowledge and motivate them.

After that announced the topic ‘Surface Area of a cube’ and showed it on

screen. (Slide35-37)

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Step 2: Explained about its Geometrical Representation and also told them

that in cuboids there were 8 vertexes, 12 edges and 6squre faces (slide38-40).

Concept of a cube was made clear by showing the examples related to daily

life i.e. Picture of a Ruby cube and picture of ice cubes (slide45-46)


‘.Surface Area of cuboids’ (slide41- 42).

To explain it , followed the following steps:-

- Explained that pair of opposite squared faces are of same area which

were of same colors (slide44).

- With the help of MS Animation Clips explained that to find the

‘Surface Area of a cube’ we have to add the surface areas of all its 6 faces

(slide45-47).


‘Surface Area of a cube’’ (slide-49).

Step 5: Gave to the students one sum based on the above topic to solve in

their note books (slide-50) and also checked their note books.


their note books at their home (slide-51).

(2) Topic- ‘Lateral Surface Area of a cube’ (14 th Nov.2009)

Material used:


Method:

Step 1: Aked some questions from the students to know their previous

knowledge and to motivate them. After that, announced the topic ‘Lateral

Surface Area of a cube’ and showed it on screen.(slide-52)


‘Lateral Surface Area of a cube’. To explain it, followed the following steps

- Explained that pair of opposite square faces is of same areas which

were of same colors.

- With the help of MS Animations Clips, explained that to find the ‘Lateral

Surface Area ofa cube’ we have to add the surface areas of all its 4faces

leaving the bottom and top faces. (slide53-55).

\Step 3: Also explained the steps for solution of a problem based on the

‘Lateral Surface Area of a cube’ (slide56-57).

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Step4: Gave to the students one sum based on the above topic to solve in

their note books (slide-58)

and also checked their note books.

Step 5: Gave to the students two sums based on the above topic to solve in

their note books at their home (slide-59 ).

(3) Topic- ‘Solutions of the Problem based (16 Th Nov.2009)

on ‘Surface Area of cube’

Material used:


Method:

Step 1: Asked some questions to the students to know their previous



‘Surface Area of cube’ (slide-60).

Step 3: Question No.4 (slide-61).

The researcher explained the formula to calculate the area of a brick and also

told them the procedure of finding the nos. of bricks required when the total

surface area as well as surface area of a brick was given.

Step 4: In Question No.5 (slide62-63).

- Explained the formula for calculating the Lateral Surface Area of a

cube and of a cubical box.

- Also told them to subtract the Lateral Surface Areas of a cubical box

from a cube.

- Explained the formula for calculating the Total Surface Area of a cube

and of a cubical box.

- Also told them to subs tract the Total Surface Areas of a cubical box

from a cube.


their note books and also checked their note books (slide64-65).



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(4) Topic- ‘Solutions of the Problem based on‘Surface Area of cuboids’

(17 Th Nov.2009)

Material used:


Method:

Step 1 Asked some questions to the students to knowtheir previous



‘Surface Area of cuboids’ (slide-67).


- Explained the students that the surface area of a glass can be calculated

by using the surface areas of cuboids.

- The researcher also explained the students that length of tape can be

calculated by calculating the length of 12 edges.


their note books and also checked their note books. (slide-69).


their note books at their homes (slide-70).

(5) Topic- ‘Solutions of the Problem based on (18 Th Nov.2009)

‘Surface Area of cuboids’

Material used:


Method:




‘Surface Area of cuboids’ .

Step 3: In Question No.7 (slide72-75).

- Explained to the students that the surface area of a bigger box can be

calculated by using the surface areas of cuboids.

- Explained to the students that the surface area of a 250 bigger boxes

can be calculated by multiplying the surface area of a bigger box by 250.

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- Also explained t0 the students that the surface area of a smaller box

can be calculated by using the surface areas of cuboids.

- Explained to the students that the surface area of a 250 smaller boxes

can be calculated by multiplying the surface area of a smaller box by 250.

- Told the students to add the areas of both type of boxes

- Explained to the students about the area of sheet required for making

250 boxes of including extra required area of 5% for overlaps.

- Explained to the students about total cost of sheet at the rate of Rs.4 for

1000cm2.

Step4: Question No.8 (slide76-77).

- Explained to the students that the area of trapallin required to cover

four sides and the roof can be calculated by the formula i.e. lxb+2(l+b) x h in

which L=4m, b=3m, h=2.5m.


their note books and also checked their note books. (Slide-78).

Step 6 Gave to the students two sums based on the above topic to solve in


3. Surface Area of right circular cylinder 19th-23rd NOV.2009

(1) Topic- ‘Surface Area of right circular cylinder’ (19 th Nov.2009)

Material used:

Power -point CD with LCD projector

Method :


knowledge and to motivate them. After that announced the topic ‘Surface

Area of a right circular cylinder’ and showed it on screen.(slide-80)

Step 2: With the help of MS Animation Clips, explained that in a right

circular cylinder the angle between radius and height is of 900.(slide81-82)

Step 3: Explained about its Geometrical Representation (slide-84) and also

told them that in a right circular cylinder there were 3 surfaces (slide-83) and

also explained them the shapes of these three surfaces (slide-84).

Step 4 Also explained the definition of a right circular cylinder (slide-85).

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Step 5: Explained the procedure about the deriving the formula for finding

the‘.Surface Area of a right circular cylinder (slide 86-94).

To explain it, followed the following steps

- Explained that in a right circular cylinder areas of top and bottom

circles are of same area.

With the help of MS Animation Clips explained that to find the ‘Surface Area

of right circular cylinder’ we have to add the surface areas of all its 3

surfaces.

Step 6: Also explained the steps for finding the area of middle part which is

of curved shape. With the help of MS Animation Clips, explained the

students that area of curved surface is equal to the area of a rectangle of

length 2Π r and breadth h.

Step 7: Also told them to calculate the total surface area of a right circular

cylinder.


‘Surface Area of a right circular cylinder’ (slide95).


their note books (slide-96) and also checked their note books.


note books at their home (slide97-98).

(2) Topic- ‘Solutions of the Problem based on (20 Th Nov.2009)

‘Surface Area of right circular cylinder’

Material used:


Method:

Step 1: Asked some questions from the students to know the previous



‘Surface Area of right circular cylinder.

Step 3: Question No.2 (slide100).

-Explained the students that the surface area of a cylindrical tank can be

calculated by using the formula of surface areas of right circular cylinder.

85

Step 4: Question No.3(slide101-102)

- Explained to the students that the inner curved surface area of a metal

pipe can be calculated by using the formula 2Πrh in which ‘r’ is 2 cm and ‘h’

is 77 cm.

- Also explained the students that the outer curved surface area of a pipe

can be calculated by using the surface area of right circular cylinder in which

‘R’ is 2.2 cm and height ‘77’ cm.

- Also told the students that the total surface area can be calculated by

adding the curved surface area of a hollow pipe and area of inner and outer

curved surface areas.

Step 5: Question No.4 (slide103)

-Explained the students that the area covered by one revolution can be

calculated by using the formula of curved surface area of a right circular

cylinder in which height is 120 cm and radius is 42 cm.

- Told the students to multiply the area covered by one revolution by

500 to find area of the playground.


their note books (slide-104) and also check their note books.


their note books at their home (slide-105-106).

(3) Topic- ‘Solutions of the Problem based on (21st Nov.2009)


Material used: -


Method:

Step 1: Asked some questions from to the students to know the previous



‘Surface Area of right circular cylinder.


- Explained the students that the curved surface area of a cylindrical pillar can

be calculated by using the formula of surface areas of right circular cylinder in

which ‘r’ is 25 cm and ‘h’ is 3.5 cm.

86

- Explained them that total cost of painting the curved surface of the pillar can

be calculated by multiplying its curved surface area by cost of paint per m2.

Step 4: Question No.6 (slide-109)

-Explained the students that the inner curved surface area of a well can be

calculated by using the formula 2Πrh in which ‘2r’ is 3.5 m and ‘h’ is10m.


-Also told the students that the total cost of plastering this curved surface area

can be calculated by multiplying its curved surface area by cost of plastering

per m2.


-Explained the students that the total radiating surface area can be calculated

by using the curved surface area of a cylinder in which r=5/2 cm and h= 28 m.

-Gave to the students two sums based on the above topic to solve in their note

books at their home (slide-112).

Step 7 Gave to the students one sum based on the above topic to solve in their

note books at their home (slide-113).

(4) Topic- ‘Solutions of the Problem based on (23rd Nov.2009)


Material used:


Method:

Step 1: Asked some questions to the students to know the previous

knowledge and to motivate them. The researcher also explained the steps for

solution of a problem based on the ‘Lateral Surface Area of right circular

cylinder.


-Explained to the students that the lateral surface area of a closed cylindrical

tank can be calculated by using the formula of surface areas of right circular

cylinder in which ‘2r’ is 4.2 m and ‘h’ is 4.5 m.

-Explained to the students that the total surface area of a closed cylindrical

tank can be calculated by using the formula of surface areas of right circular

cylinder in which ‘2r’ is 4.2 m and ‘h’ is 4.5 m.

87


-Explained to the students that the total height of a frame of a Lampshade is

35cm, i.e., 30+2.5+2.5 and the length of the cloth required for covering the

lampshade can be calculated by using the formula 2Πrh in which ‘r’ is 10cm

and ‘h’ is 35 cm.

Step 4: Question No.11 (slide117-118)

- Explained the students the cardboard required for one penholder can be

calculated by using the formula (r2+2Πrh) in which ‘r’ is 3 cm and ‘h’ 10.5

cm.

-Explained the students the cardboard required for 35 penholders can be

calculated by multiplying cardboard required for one penholder by 35.





4 Surface Area of right circular cone 24th-27th NOV.2009

(1) Topic- ‘Surface Area of right circular cone’ (24 th Nov.2009)

Material used:


Method :



After that Announced the topic ‘Surface Area of a right circular cone’ and

showed it on screen.(slide-121-122)

Step 2: Also explained the definition of a right circular

cone (slide-123).

Step 3: Explained about its Geometrical

Representation (slide124-126) and also told them about ‘vertex’

‘height ‘radius’ and ‘slant height’.

Step 4: With the help of MS Animation Clips, explained that there are two

surfaces in a right circular cylinder .one is flat base circular in shape and

second is a lateral surface curved in shape (slide-127).


‘.Surface Area of a right circular cone (slide 128-140).

88


- Explained that in a right circular cone area of base is Πr2.

- With the help of MS Animation Clips, explained that to find the

‘Surface Area of right circular cone’ we have to add the surface areas of all

its 2 surfaces.

Step 6: Also explained the steps for finding the area of curved surface. With

the help of MS Animation Clips, explained to the students that area of curved

surface is equal to half of the length of entire curved

boundary.

Step 7: Also told them to calculate the total surface area of a right circular

cone.

Step 8: Also explained the steps for solution of a

problem based on the ‘Surface Area of a right circular cone’ (slide-141) .


their note books (slide-142) and also check their note books.


their note books at their home (slide-143-144).

2) Topic- Solutions of the Problem based on (25th Nov.2009)

‘Surface Area of right circular cone’

Material used:


Method:

Step 1: Asked some questions to the students to know

the previous knowledge and to motivate them.

Step 2: Also explained the steps for solution of a problem

based on the ‘Surface Area of right circular cone.

Step 3: Question No. 2 (slide146).

- Explained to the students that the total surface area of a cone can

be calculated by using the formula of surface areas of right circular cylinder in

which ‘r’ is 12 cm and ‘l’ is 21 cm.


- Explained to the students that the radius of a cone can be obtained by

dividing its curved surface area with Π times its slant height.

89

- Also told the students that the total surface area can be calculated by

substituting the value of ‘r’ and ‘h’.


- Explained, the students that the slant height of a right circular cone

can be calculated by using the Pythagoras theorem.

- Also told the students that area of canvas can be calculated by curved

surface area of a right circular cone.

Step 6: Also told the students that cost of canvas can be calculated by

multiplying its curved surface area by Rs.70.


- Explained to the students that the slant height of a right circular cone


- Also told the students that area of canvas can be calculated by curved

surface area of a right circular cone.

Step 8: Also told the students that cost of canvas can be calculated by

multiplying its curved surface area by Rs.70.

Step 9: Explained to the students that the length of canvas

required can be calculated by adding its curved surface area and

stitching margins.

Step10: Gave to the students one sum based on the above topic to solve in


Step 11 Gave to the students two sums based on the above topic to solve in


3) Topic- Solutions of the Problem based on (26th Nov.2009)

‘Surface Area of right circular cone’

Material used:


Method:

Step 1: Asked some questions to the students to know

the previous knowledge and to motivate them.


90

- Explained to the students that the curved surface area of a cone can be

calculated by using its formula in which ‘r’ are 7m and ‘l’ is 25 cm.

- Also explained to the students that the cost of white

washing can be calculated by multiplying its curved surface by Rs.210 per 100

m2.


- Explained to the students that the area of the sheet required to make

one can be calculated by using curved surface area of a right circular cone.

- Also told the students that sheet required for 10 caps can be calculated

multiplying curved surface area of a right circular cone by 10.


- Explained to the students that the slant height of a right circular cone


- Explained to the students that the curved surface area of 50 cones can

be calculated multiplying curved surface area of a right circular cone by 50.

- Also told the students that cost of painting all these cones can be

calculated multiplying curved surface area of a right circular cone by Rs.12

per m2 Step 5: Gave to the students one sum based on the above topic

to solve in their note books at their home (slide-158).

Step 6: Gave the students one sum based on the above

topic to solve in their note books at their home (slide-159).

4 Surface Area of Sphere 28 th NOV. _ 2nd DEC.2009

(1) Topic- ‘Surface Area of Sphere’ (28 th NOV.2009)

Material used:


Method :


knowledge and to motivate After that, announced the topic ‘Surface Area of

a Sphere’ and showed it on screen (slide-160) .

Step 2: Also explained the definition of a right circular cone (slide-162).

Step 3: Explained about its Geometrical

Representation and also told them about its’ radius’ and ‘centre’

(slide163).

91

Step 4: Concept of a Sphere was made clear by showin examples related to

daily life, i.e. , Marble Ball,Ball,footbal globe.(slide164-168)


the ‘.Surface Area of a Sphere. (slide 169-17

To explained it, followed the following steps

- With the help of MS Animation Clips, explained that the ‘Surface

Area of Sphere is equal to the four times area of a circle having radius ‘r’.


‘Surface Area of a Sphere’.


Explained the students that the surface area of a Sphere cone can be calculated

by substituting the value ‘r’ in the formula (4Π r2).


Explained to the students that the surface area of aSphere cone can be

calculated by substituting the value ‘2r’ in the formula a (4Π r2).

Step 8: In Question No.4 (slide180)

Explained to the students that the ratio of two surface areas of spherical

balloon can be calculated by dividing their surface area.





(2) Topic- ‘Curved Surface Area of a Hemisphere’ (30th NOV.2009)

- With the help of MS Animation Clips, explained that the ‘ Total

Surface Area of a Hemisphere is sum of the areas of both surfaces (area of

Base and area of curved surface).


‘Surface Area of a Hemisphere’.


Explained to the students that the surface area of a Hemisphere cone can be

calculated by substituting the value ‘r’ in the formula (3Π r2).

Step 6: Question No.5(slide191)

- Explained to the students that the inner surface area of a Bowl can be

calculated by substituting the value ‘2r’ in the formula (2Π r2).

92

- Also explained to the students that the cost of tin painting can be

calculated multiplying its inner surface area by Rs.16 per 100cm2.



Step 8: Gave to the students two sums based on the above topic to solve

in their note books at their home (slide-193).

(2) Topic- Solutions of the Problem based on 1st DEC.2009

‘Surface Area of Sphere’

Material used:


Method :




Surface Area of a Sphere’.


Explained to the students that the radius ’r’ of a Sphere can be calculated by

substituting the values in the formula (4Π r2).


Explained to the students that the ratio of the surface area of moon and earth

can be calculated by dividing their surface areas.

Step 5 : Question No.8 (slide197)

- Explained to the students that the radius of outer radius can be calculated

by adding radius of inner radius and thickness of steel sheet.

- Also explained the students that the outer curved surface area of a

spherical bowl can be calculated by using formula 2ΠR2.


-Explained to the students that the radius of a right circular cylinder is equal to

that of a Sphere (r).and height of a cylinder is twice that the height of a

Sphere.

-Explained to the students that the ratio of the surface area of a Sphere to that

of curved surface area can be calculated by dividing them.



93



6 . Volume of cube 3 rd DEC. - 5th DEC.2009

(1) Topic- ‘Volume of cube’ (3 rd DEC.2009)

Material used:


Method :


knowledge and to motivate them. After that , announced the topic

‘Volume of cube’ and showed it on screen. (slide-203)


the ‘Volume of cube’ (slide 204-205).

Step 3: Asked objective type questions related to the

topic to know whether they understood or not (slide 206).



(2) Topic-Solution of problem based (4 th DEC.2009)

on ‘Volume of cube’

Material used:


Method :

Step 1: Asked some questions from the students to knowtheir previous


Step 2: Also explained the steps for solution of problems based on the

‘Volume of cube’.

Step 3: Question No. 8 (slide-209)

Explained to the students that the length of the side of the

new cube can be calculated by equating the volume of eight cubes

with the volume of a cube whose edge is ‘a’.





94

(3) Topic-Solution of problem based (5 th DEC.2009)

On ‘Volume of cube’

Material used:


Method :



Also explained the steps for solution of a problem based on the ‘Volume of

cube’.

Step 2: Question (slide-213)

Explained to the students that the edge cube can be calculated by equating the

volume of cube with the (edge)3.





7. Volume of cuboids 7 th DEC. - 10th DEC.2009

1) Topic- ‘Volume of a cuboids’ (7 th DEC.2009)

Material used:


Method :



After that , announced the topic ‘Volume of a cuboids’ and showed it on

screen (slide-218) .

Step 2: Explained the procedure about the deriving the formula for

finding the ‘Volume of a cuboids’ (slide 219-224).


-With the help of MS Animation Clips, explained that to find the ‘Volume of

a cuboids’ fill the bottom of the cuboids with a rectangle until the cuboids is

completely filled.

95

- With the help of MS Animation Clips, Also explained that the volume of a

cuboids can be calculated multiplying the area of the plane region occupied by

each rectangle by height.

- Also told them to calculate the volume of a cuboids.

- Also explained the steps for solution of a problem based on the

‘Volume of a cuboids’.


- Explained to the students that the volume of a packet containing

one matchbox be calculated by using the volume of a cube.

- Explained to the students that the volume of 12 packets

can be calculated by multiplying the volume of one matchbox by 12.


Explained to the students that the capacity of water tank

calculated by using the volume of a cube.

Step 5: Gave to the students one sum based on the above




2) Topic-Solution of problems based on (8 th DEC.2009)

Volume of a cuboids’

Material used:


Method :



Step 2: Also explained the steps for solution of a problem

based on the ‘Volume of a cuboids’.


Explained to the students that the high of a cubical vessel can

be calculated by using the volume of cuboids having its length 10 m and

wide 8 m.

Step 4: Question No. 4 (slide-231)

Explained to the students that the internal space of the pit can

96

be calculated by using the volume of a cuboids in which ‘l’ is 8m,’b’

is 6m and ‘h’ is3m.

Step 5: Explained to the students that the cost of digging a

cubical pit can be calculated its volume by Rs.30.


Explained to the students that the breadth can be calculated

dividing its volume by the product of length and depth.


topic to solve in their note books. (slide-233).




‘Volume of a cuboids’

Material used:

Power- point CD with LCD projector

Method :

Step 1: Asked some questions from the students to know

their previous knowledge and to motivate them.


‘Volume of a cuboids’.


Explained to the students that the number water of days can be calculated by

dividing the capacity of tank by quantity of water per day.


Explained to the students that the maximum number Of wooden crates can be

calculated by dividing the capacity of godown by volume of one wood crate.


Explained to the students that the side of a new cube can be calculated by

equating the volumes of eight cubes with the volume of a

cube having edge ‘12’.


topic to solve in their note books. (Slide-239).


97


8 . Volume of right circular cylinder 11 th DEC. - 16th DEC.2009

1) Topic- ‘Volume of right circular cylinder (11 th DEC.2009)

Material used:


Method :



After that, announced the topic ‘Volume of right circular cylinder ’ and

showed it on screen (slide-244) .

Step 2: Explained the procedure about deriving the formula for finding the.

‘Volume of right circular cylinder’ (slide 245-253).



‘Volume of right circular cylinder’ fill the bottom of the cylinder with a

circle of radius ‘r’until the cylinder is completely filled.

- With the help of MS Animation Clips, also explained that the volume

of a cylinder can be calculated multiplying the Base Area by height.

- Also told them to calculate the volume of right circular cylinder.


‘Volume of right circular cylinder’.


- Explained to the students that the value of radius ‘r’ of a cylindrical

vessel can be calculated dividing the diameter by 2.

vessel can be calculated by substituting the values 0f ‘r’ and ‘h’ in volume

formula.


topic to solve in their note books. (Slide-255).




‘Volume of a cylinder’

Material used:


98

Method :



Also explained the steps for solution of a problem based on the ‘Volume of a

cylinder’.


- Explained to the students that the volume of wood can be calculated by

using the formula Π (R2-r2)x h.

- Also explained to the students that the mass of the pipe can be obtained

multiplying the volume of wood by 0.6.


- Explained to the students that the volume of a tin with a rectangular

base can be calculated by using the volume formula of cuboid.

- Also explained to the students that the volume of a plastic cylinder

with circular base can be calculated by using the volume formula of cylinder.

- Also told the students to compare their volumes to find out the

container having the greater capacity and how much.


- Explained to the students that the radius of a cylinder can be calculated

by substituting the values of ‘volume’ and height ‘h’ in the formula and

volume can be calculated by using the values of ‘r’ and height’s’.







Material used:


Method :




cylinder’.

99


- Explained to the students that the inner curved surface area of a

cylindrical vessel can be obtained dividing the total cost by cost of painting

per m2.

- Also told that the radius ‘r’ of the cylindrical vessel can be calculated

by substituting the values of ‘h’ in the curved surface area.

- Also explained that the capacity of the vessel can be obtained by

substituting the value of ‘r’ and ‘h’.


- Explained that the radius ‘r’ of the cylindrical vessel can be calculated

by substituting the values of ‘h’ in its volume formula.

- Gave to the students one sum based on the above topic to solve in their


- Gave to the students one sum based on the above topic to solve in




Material used:


Method :


knowledge and to motivate them. Also explained the steps for solution of a

problem based on the ‘Volume of a cylinder’.


-Explained to the students that the volume of Graphite can be obtained by using

the volume formula of cylinder.

-Also explained the method for the calculation of the volume of wood.


Explained that the capacity of one bowl can be obtained by using the volume

formula of a cylinder.



100

Step 5: Gave to the students one sum based on the above topic to solve in their


8 . Volume of right circular cylinder 17 th DEC. - 19th DEC.2009

1) Topic- ‘Volume of right circular cone (17 th DEC.2009)

Material used:


Method :



After that, announced the topic ‘Volume of right circular cone’ can be

determined from the volume formula for a cylinder and showed it on screen.

(slide274)


the. ‘Volume of right circular cylinder’ (slide275-281).



‘Volume of right circular cone’ take a cylinder and a cone having the equal

heights(h) and radii(r) and fill the cone up to the brim with sand once.

- With the help of MS Animation Clips, also explained them to empty

the sand into the cylinder. It fills up only a part of the cylinder and when the

cone is filled up for the third time, and emptied into the cylinder, it can be seen

that the cylinder is also full to the brim.

- With the help of MS Animation Clips, also explained that the volume

of a cone.

- Also told them to calculate the volume of right circular cylinder.


‘Volume of right circular cylinder’.


Explained to the students that the volume of the right circular cone can be

calculated by substituting the value of radius ‘r’ and its height.


Explained to the students that the capacity of a conical vessel can be

calculated by substituting the values 0f ‘r’ and ‘h’ in volume formula.

Step 6: Gave to the students one sum based on the

101

above topic to solve in their note books at their home (slide-285).

Step 7: Gave to the students one sum based on the above topic to solve in their



‘Volume of a cone’

Material used:


Method :




cone’.


Explained to the students that the radius of the base of a cone can be

calculated by using the formula of a cone having ‘h’ as its height.


Explained the relationship of radius and diameter and also explained them

that the radius of the base of a cone can be calculated by using the formula of

a cone having ‘h’ as its height.


Explained to the students that the capacity of a conical vessel can be

calculated by substituting the values of ‘r’ and ‘h’ in volume formula.






‘Volume of a cone’

Material used:


Method :

Step 1: Asked some questions from the students to know their previou

knowledge and to motivate them. Also explained the steps for solution o

a problem based on the ‘Volume of a cone’.

102


- Explained to the students that the height of a cone can be calculated by

using the formula of a cone having ‘r’as radius.

- Explained to the students that the curved surface area can be calculated by

substituting the value of ‘r’ and ‘h’ in the formula.


- Explained to the students that revolution of a right triangle ABC about

the side 12 cm, will take the shape of a cone and volume of this solid cone can

be calculated by substituting the value of ‘r’ and ‘h’ in the volume formula.


- Explained to the students that revolution of a right triangle ABC about

the side 5 cm it will take the shape of a cone and volume of this solid cone can

be calculated by substituting the value of ‘r’ and ‘h’ in the volume formula.

- Also told them to find the ratio of these volumes.


- Explained to the students that volume of a cone can

be calculated by substituting the value of’r’ and ‘h’ in the formula.

- Explained to the students that the area of canvas can be calculated by

using the curved surface formula of a cone. volume





10) Volume of a Sphere 21st DEC. - 24th DEC.2009

1) Topic- ‘Volume of a Sphere (21st th DEC.2009)

Material used:


Method :


knowledge and to motivate them. After that, announced the topic ‘Volume

of a Sphere’ that can be determined from the volume formula for a cylinder

and showed it on screen.(slide303)


the. ‘Volume of a Sphere’ (slide 304-309).

103


- With the help of MS Animation Clips, explained that to find

the‘Volume of a Sphere ’ take a cylinder and a sphere having the equal and

radii(r) and height of a cylinder is equal to the Diameter of a sphere and fill the

sphere up to the brim with sand once.

- With the help of MS Animation Clips, also explained to them to empty

the sand into the cylinder. It fills up only a part of the cylinder.

- With the help of MS Animation Clips, also explained that the volume of

a sphere is equal to 2/3 volume of a cylinder.

- Also told them to calculate the volume of a sphere.


‘Volume of a sphere’.


Explained to the students that the volume of a sphere can be calculated by

substituting the value of radius ‘r’ in its volume formula.


Explained to the students that the amount of water displaced can be calculated

by substituting the values of ‘r’ and ‘h’ in volume formula of a sphere.


Explained to the students that the value of the radius ‘r’of a sphere can be

calculated by using the surface area.

Step 6: Also explained to the students that the volume of a sphere can

be calculated by substituting the value of radius ‘r’ in its volume formula.





2) Topic-Solution of problems based on (22 nd DEC.2009)

‘Volume of a sphere’

Material used:


104

Method :


knowledge and to motivate them. Also explained the steps for solution of a

problem based on the ‘Volume of a Sphere’.


- Explained to the students that the volume of a metallic ball can be calculated

by using the volume formula of a sphere having ‘r’ as radius.

- Explained to the students that the mass of the ball can be calculated by

multiplying volume of a sphere by density of the metal.


- Explained to the students that the shape of both moon and earth are of

sphere and their can be calculated by using the volume formula of a sphere.

- Also told the students to compare the volumes of both moon and earth.


Explained to the students the capacity of the bowl can be

calculated by using the volume formula of a hemisphere, i.e.,(1/2 volume of a

sphere).


- Explained to the students that Radius of outer hemispherical bowl can

be calculated by adding the value of the inner radius and 1/100 m.

- Also explained to the students that volume of outer hemispherical

bowl can be calculated by using its Volume formula.





3) Topic-Solution of problems based on (23 rd DEC.2009)

‘Volume of a sphere’

Material used:


Method :



105


sphere’.


- Explained to the students that radius of a sphere can be calculated by

using its surface area formula.

- Also explained to the students that volume of sphere can be calculated

by using its Volume formula.


- Explained to the students that the inner surface area of hemispherical

dome can be calculated dividing the total cost of white washed by per square

meter.

- Also told the students to calculate the inner radius of hemispherical

dome using surface area of hemisphere.

- Explained to the students that the volume of inside of the dome can be

calculated by using Volume formula of a sphere.


- Explained to the students that the radius of new sphere can be

calculated by equating the volume of new sphere with the Volume of twenty

seven iron sphere.

- Explained to the students to calculate the surface area of both spheres

surface.

- Explained to the students to compare their surface areas.


- Explained to the students that the radius (r’) of new sphere can be

calculated by equating the volume of new sphere with the Volume of twenty

seven iron spheres having radius ’r’.

- Also explained to the students to find surface areas (s) and(s’).

- Explained to the students to find out the ratio of s and s’.


- Explained to the students that capacity of the capsule can be calculated

by using the formula’4/3Π r3, where ‘r’ is 3.5/2 m.

- Gave the students one sum based on the above topic to solve in their


106

- Gave the students one sum based on the above topic to solve in their


3.5 STATISTICAL ANALYSIS

To achieve objectives of the study, the data collected was statistically analysed

using the following techniques:

1. Descriptive statistics such as mean and S.D worked out on the score of

achievement and confidence level in Mathematics.

2. ‘t’ value was computed in order to adjudge pupil’s intelligence and socio-

economic status.

3. ‘t’ test was employed for testing the significance of difference between the

means of pupils’ achievement in mathematics on pre - test, post -test and gain

scores.

4. t’ test was employed for testing the significance of difference between the

means of pupils’ confidence level in answering the test questions in

mathematics on pre - test, post - test and gain scores. The value of‘ ‘t’ was

computed with the help of the following formula:

M1 - M2

t = √σ1 + σ2

NI N2

WhereM1=Mean of first group

M2 = Mean of second group

σ1 = Variance of first group

σ2 = Variance of first group

NI = Number of cases in first group

N2 = Number of cases in second group

Mean scores in respect of achievement in mathematics and confidence level in

answering the test questions were pictorially presented in the form of histograms.

Histograms were drawn in respect of pre-test, post-test and gain scores of

experimental group and control group.

5. Pearson Product moment correlation coefficient was computed to adjudge the

relationship between students’ achievement and their confidence level in answering

the test items on both groups of students

107

3.6 PRECAUTIONS OBSERVED

Following precautions were observed during the course of

experiment (Pre-test –treatment-Post-test) for ensuring effectiveness and high

precision in experimental condition which may have contributed to the results.

No undue stress or control of any kind was imposed on the subjects at any

time during the study and the experiment was conducted in a relaxed natural

sitting.

Both the experimental and control groups were taught by the investigator

herself to avoid any variation.

The effectiveness of the experimental treatment was ensured by establishing

rapport with students and teachers, maintaining natural setting, harmonious

atmosphere, providing sufficient time for various activities in the

experimentation and the like.

It was ensured that the topics on contents of treatment had not been previously

taught to the students in both the experimental and the control groups.

Care was taken to keep importance of content matter during the course of

treatment and it was not underplayed while fitting into the instructional

treatment.

Teaching periods of 40 minutes duration was utilized fully for treatment and

time was not wasted during experimentation.

3.7 CONSTRAINTS AND DIFFICULTIES FACED DURING

EXPERIMENT

It may not be out of place to mention some of the difficulties faced or constraints

of the experiment that needed to taken note of. These, as sorted out by the

researcher were :

Power failure

Infrastructural lapses.

Time-table related difficulties.

Efforts were needed to convince the teachers and the principal about the

experiment and to seek their co-operation in the conduct of the experiment

with in the frame work of the school schedule. The researcher contacted the

school authorities and convinced them about the programme and its usefulness

to ensure that the treatment be fully provided to every student and that the

108

sample groups regularly attended the school during that period. The

experiment was accordingly adjusted as per the time table in vogue for

pursuing a regular course of studies, Making some minor changes in the

regular time-table in consultation with the time-table in charge. The students’

motivation and the helpful attitude of the school authorities encouraged the

researcher to carry out the experiment with full enthusiasm and very smoothly

to study the effectiveness of ICT on students’ achievement in Mathematics.

method and procedure of the study -...

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