nama kursus : rekabentuk dan analisis eksperimen...

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NAMA KURSUS : REKABENTUK DAN ANALISIS EKSPERIMEN (Experimental Design and Analysis) KOD KURSUS : PRT 3202 KREDIT : 3 (2+1) JUMLAH JAM PEMBELAJARAN PELAJAR : 120 jam per semester PRASYARAT : Tiada HASIL PEMBELAJARAN : Pelajar dapat:

1. merumuskan kesimpulan daripada eksperimen (C5, LL) 2. mengorganisasikan eksperimen berasaskan objektif dan

keadaan persekitaran (P4, CTPS) 3. memilih kaedah sesuai untuk analisis data (A3)

SINOPSIS : Kursus ini meliputi kaedah melaksanakan eksperimen yang merangkumi rekabentuk eksperimen, persampelan, analisis dan intepretasi data, dan membuat rumusan. (This course covers the methods of conducting experiments including experimental design, sampling, data analysis and interpretation, and formulating conclusions.)

KANDUNGAN

Jam Pembelajaran

Bersemuka

KULIAH : 1. 2.

Prinsip asas - jenis-jenis eksperimen - pemilihan tapak eksperimen - keseragaman kawasan - langkah-langkah dalam melaksana

eksperimen - jenis data untuk dikumpulkan - ujian hipotesis Petak ekperimen lapangan - kawalan ralat eksperimen - saiz petak - keseragaman petak - kesan sempadan petak

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3

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3. Reka bentuk eksperimen asas 4 - kepentingan merawak

- reka bentuk rawak lengkap (CRD) - konsep replikasi dan blok - reka bentuk rawak blok lengkap

(RCBD)

4.

Analisis varians (ANOVA) - andaian untuk memenuhi syarat

ANOVA - penjelmaan data - ANOVA untuk CRD dan RCBD

4

5. Perbandingan min rawatan - perbandingan min antara rawatan - regresi dalam ANOVA

3

6. Analisis untuk variabel tak normal - analisis untuk bilangan - kaedah Non-Parametrik

3

7. Eksperimen faktoran - kesan utama dan interaksi - kontras berortogon

4

8. Eksperimen dengan saiz petak berbeza - reka bentuk petak belahan (Split

Plot) - reka bentuk petak jaluran (Strip Pot) - pengukuran berulang

4

Jumlah 28

Jam Pembelajaran

Bersemuka

AMALI : 1. Menggunakan perisian lembaran untuk menangani data

6

2. Menggunakan perisian statistik untuk

meganalisis data 6

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3. Melaksanakan ANOVA sehala (CRD) 3 4. Melaksanakan ANOVA dua hala

(RCBD) 3

5. Melaksanakan perbandingan antara

rawatan 3

6. Melaksanakan penjelmaan data 3 7. Melaksanakan kaedah non-parametrik 6 8. Menganalisis eksperimen faktoran 6 9. Melaksanakan kontras berortogon 3 10. Melaksanakan ANOVA untuk reka

bentuk petak belahan 3

Jumlah 42

PENILAIAN : Kerja Kursus 60 % Peperiksaan Akhir 40 % RUJUKAN : 1.

2.

Casella, G. (2008). Statistical Design. New York: Springer Clewer, A.G. & Scarisbrick, D. H.. (2001). Practical Statistics and Experimental Design for Plant and Crop Science. New York: John Wiley & Sons.

3. Gomez, K.A. & Gomez, A.G. (2005). Statistical Procedures for Agricultural Research (4th Edition). New York: John Wiley & Sons.

4. Hinkelmann, K. & Kempthorne, O. (2007). Design and Analysis of Experiments, Introduction to Experimental Design (2nd Edition). New York: Wiley-Interscience.

5. Petersen, R. G. (1994). Agricultural Field Experiments: Design and Analysis. New York: Marcel Dekker.

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3. Reka bentuk eksperimen asas 4 - kepentingan merawak

- reka bentuk rawak lengkap (CRD) - konsep replikasi dan blok - reka bentuk rawak blok lengkap

(RCBD)

4.

Analisis varians (ANOVA) - andaian untuk memenuhi syarat

ANOVA - penjelmaan data - ANOVA untuk CRD dan RCBD

4

5. Perbandingan min rawatan - perbandingan min antara rawatan - regresi dalam ANOVA

3

6. Analisis untuk variabel tak normal - analisis untuk bilangan - kaedah Non-Parametrik

3

7. Eksperimen faktoran - kesan utama dan interaksi - kontras berortogon

4

8. Eksperimen dengan saiz petak berbeza - reka bentuk petak belahan (Split

Plot) - reka bentuk petak jaluran (Strip Pot) - pengukuran berulang

4

Jumlah 28

Jam Pembelajaran

Bersemuka

AMALI : 1. Menggunakan perisian lembaran untuk menangani data

6

2. Menggunakan perisian statistik untuk

meganalisis data 6

iii

3. Melaksanakan ANOVA sehala (CRD) 3 4. Melaksanakan ANOVA dua hala

(RCBD) 3

5. Melaksanakan perbandingan antara

rawatan 3

6. Melaksanakan penjelmaan data 3 7. Melaksanakan kaedah non-parametrik 6 8. Menganalisis eksperimen faktoran 6 9. Melaksanakan kontras berortogon 3 10. Melaksanakan ANOVA untuk reka

bentuk petak belahan 3

Jumlah 42

PENILAIAN : Kerja Kursus 60 % Peperiksaan Akhir 40 % RUJUKAN : 1.

2.

Casella, G. (2008). Statistical Design. New York: Springer Clewer, A.G. & Scarisbrick, D. H.. (2001). Practical Statistics and Experimental Design for Plant and Crop Science. New York: John Wiley & Sons.

3. Gomez, K.A. & Gomez, A.G. (2005). Statistical Procedures for Agricultural Research (4th Edition). New York: John Wiley & Sons.

4. Hinkelmann, K. & Kempthorne, O. (2007). Design and Analysis of Experiments, Introduction to Experimental Design (2nd Edition). New York: Wiley-Interscience.

5. Petersen, R. G. (1994). Agricultural Field Experiments: Design and Analysis. New York: Marcel Dekker.

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EEXXPPEERRIIMMEENNTTAALL DDEESSIIGGNN AANNDD AANNAALLYYSSIISS

PPRRTT 33220022

DR. ANUAR ABD RAHIM Department of Land Management

Faculty of Agriculture Universiti Putra Malaysia

43400 UPM Serdang Selangor Darul Ehsan

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COURSE INTRODUCTION

a. Course Information

Department : Land Management

Course Name : EExxppeerriimmeennttaall DDeessiiggnn aanndd AAnnaallyyssiiss

CCoouurrssee CCooddee : PRT 3202

Credit Hours : 3 (2+1)

Course Description and Summary

The course consists of 2 hours lecture and 3 hours laboratory per week. To fulfill the

laboratory requirement, student need to submit individual laboratory assignment

each week that will be given through the lecturer’s Putra Learning Management

System (LMS) or email consisting of about 42 hours per semester. Four laboratory

meetings will be conducted per semester. The laboratory assignment is self-learning

due to the distance learning that permit limited meetings.

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EEXXPPEERRIIMMEENNTTAALL DDEESSIIGGNN AANNDD AANNAALLYYSSIISS

PPRRTT 33220022

DR. ANUAR ABD RAHIM Department of Land Management

Faculty of Agriculture Universiti Putra Malaysia

43400 UPM Serdang Selangor Darul Ehsan

v

COURSE INTRODUCTION

a. Course Information

Department : Land Management

Course Name : EExxppeerriimmeennttaall DDeessiiggnn aanndd AAnnaallyyssiiss

CCoouurrssee CCooddee : PRT 3202

Credit Hours : 3 (2+1)

Course Description and Summary

The course consists of 2 hours lecture and 3 hours laboratory per week. To fulfill the

laboratory requirement, student need to submit individual laboratory assignment

each week that will be given through the lecturer’s Putra Learning Management

System (LMS) or email consisting of about 42 hours per semester. Four laboratory

meetings will be conducted per semester. The laboratory assignment is self-learning

due to the distance learning that permit limited meetings.

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vi

b. Writer Information

Name : Anuar Abd Rahim, PhD

Address : Department of Land Management,

Faculty of Agriculture, Universiti Putra Malaysia

43400 UPM Serdang Selangor

Telephone No. : 03-89474857

Fax No. : 03-89408316

e-mail : [email protected]

c. Course Objectives/ Learning Outcomes Student will be able to:

Formulate summary and conclusion from experiments (C5, LL)

1. Organize experiments based on the objectives and real situation

of the environment (P4, CTPS)

2. Select suitable methodology in analyzing data (A3)

d. Course Synopsis This course covers the methods of conducting experiments including

experimental design, data sampling and analysis, and formulating conclusions

from the analysis.

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e. Course Content - Principles of Experimental Design

- Experimental Designs

- Analysis of Variances

- Comparison of Treatment Means

- Data Transformation

- Non-parametric Tests

- Factorial Experiments

- Experiments with Different Size of Experimental Plots

f. Instructions for Laboratory Assignments

Laboratory assignments will be displayed each week through the lecturer’s

Putra Learning Management System (LMS) or email. These assignments have

to be completed manually and using statistical software. The reports should

be sent within one week after being introduced via email or any convenience

way to the assigned lecturer.

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b. Writer Information

Name : Anuar Abd Rahim, PhD

Address : Department of Land Management,

Faculty of Agriculture, Universiti Putra Malaysia

43400 UPM Serdang Selangor

Telephone No. : 03-89474857

Fax No. : 03-89408316

e-mail : [email protected]

c. Course Objectives/ Learning Outcomes Student will be able to:

Formulate summary and conclusion from experiments (C5, LL)

1. Organize experiments based on the objectives and real situation

of the environment (P4, CTPS)

2. Select suitable methodology in analyzing data (A3)

d. Course Synopsis This course covers the methods of conducting experiments including

experimental design, data sampling and analysis, and formulating conclusions

from the analysis.

vii

e. Course Content - Principles of Experimental Design

- Experimental Designs

- Analysis of Variances

- Comparison of Treatment Means

- Data Transformation

- Non-parametric Tests

- Factorial Experiments

- Experiments with Different Size of Experimental Plots

f. Instructions for Laboratory Assignments

Laboratory assignments will be displayed each week through the lecturer’s

Putra Learning Management System (LMS) or email. These assignments have

to be completed manually and using statistical software. The reports should

be sent within one week after being introduced via email or any convenience

way to the assigned lecturer.

8

viii

Table 1: Title of Unit and Suggested Lecture’s Hours

Unit Title Lecture’s Hours

1. Principles of Experimental Design

4

2. Experimental Designs

3

3. Analysis of Variances

3

4. Comparison of Treatment Means

4

5. Data Transformation

3

6. Non-parametric Tests

3

7. Factorial Experiments

4

8. Experiments with Different Size of Experimental Plots

4

(** Pembelajaran maya : 1 jam kuliah bersamaan dengan 3 jam pembelajaran

Kendiri)

ix

A field experiment was conducted in Ladang Puchong, UPM to evaluate

the effect of nitrogen fertilizer rate (0, 50, 100, 150, and 200 kg N ha-1)

with 4 replications on the yield of maize cobs using CRD. The data are as

the following:

N rate 0 50 100 150 200 (kg ha-1) Replication 1 4.2 5.2 6.9 8.2 7.3 2 4.6 5.0 6.6 9.4 7.6 3 5.6 6.7 11.8 13.3 11.6 4 5.8 7.1 11.5 13.4 12.0

1. Complete the ANOVA table for the above experiment.

2. Is there any block effect in this experiment?

3. Use contrast to compare the effect of fertilizers with the control.

4. Which rate do you recommend from this experiment? Give reason(s)

5. Sketch by using graph to show the result of this experiment.

Examples of Laboratory Assignment

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viii

Table 1: Title of Unit and Suggested Lecture’s Hours

Unit Title Lecture’s Hours

1. Principles of Experimental Design

4

2. Experimental Designs

3

3. Analysis of Variances

3

4. Comparison of Treatment Means

4

5. Data Transformation

3

6. Non-parametric Tests

3

7. Factorial Experiments

4

8. Experiments with Different Size of Experimental Plots

4

(** Pembelajaran maya : 1 jam kuliah bersamaan dengan 3 jam pembelajaran

Kendiri)

ix

A field experiment was conducted in Ladang Puchong, UPM to evaluate

the effect of nitrogen fertilizer rate (0, 50, 100, 150, and 200 kg N ha-1)

with 4 replications on the yield of maize cobs using CRD. The data are as

the following:

N rate 0 50 100 150 200 (kg ha-1) Replication 1 4.2 5.2 6.9 8.2 7.3 2 4.6 5.0 6.6 9.4 7.6 3 5.6 6.7 11.8 13.3 11.6 4 5.8 7.1 11.5 13.4 12.0

1. Complete the ANOVA table for the above experiment.

2. Is there any block effect in this experiment?

3. Use contrast to compare the effect of fertilizers with the control.

4. Which rate do you recommend from this experiment? Give reason(s)

5. Sketch by using graph to show the result of this experiment.

Examples of Laboratory Assignment

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g. Course Evaluation

Evaluation for this course is divided into:

(i) Coursework

Laboratory assignments (individual) 40 %

(ii) Mid-Term Examination 20%

(i) + (ii) 60% (ii) Final Examination 40%

Total 100% ** Course evaluation can be changed from time to time depending to the lecturer.

Suggested Learning Activities

1. Lecture 8 hours

2. Self Learning 45 hours per week

3. Tutorials (4 sessions) 12 hours

4. Online/Email/Telephone/LMS/Maya Class with Lecturer 10 hours

5. Laboratory Assignments 42 hours Total Hours 117 hours

xi

h. Mid-term Examination

Mid-term examination has to be taken by the student. Questions are based on

the modules that presented and in the form of subjective. This examination will

consists of units 1 to 4, and it can be adjusted in terms of form and number

of questions, topics and marks, and will be discussed during direct

meeting between the students and the lecturer concern

i. Final Examination

Questions will consist of all the units and topics that are presented in the module.

Students can consult the lecturer when concern for update details of

the examination. Questions will be asked in the form of subjective and

applications

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g. Course Evaluation

Evaluation for this course is divided into:

(i) Coursework

Laboratory assignments (individual) 40 %

(ii) Mid-Term Examination 20%

(i) + (ii) 60% (ii) Final Examination 40%

Total 100% ** Course evaluation can be changed from time to time depending to the lecturer.

Suggested Learning Activities

1. Lecture 8 hours

2. Self Learning 45 hours per week

3. Tutorials (4 sessions) 12 hours

4. Online/Email/Telephone/LMS/Maya Class with Lecturer 10 hours

5. Laboratory Assignments 42 hours Total Hours 117 hours

xi

h. Mid-term Examination

Mid-term examination has to be taken by the student. Questions are based on

the modules that presented and in the form of subjective. This examination will

consists of units 1 to 4, and it can be adjusted in terms of form and number

of questions, topics and marks, and will be discussed during direct

meeting between the students and the lecturer concern

i. Final Examination

Questions will consist of all the units and topics that are presented in the module.

Students can consult the lecturer when concern for update details of

the examination. Questions will be asked in the form of subjective and

applications

12

xii

.

SECTION A Answer all questions in this section

1. Draw the arrangement of all the experimental units for the following:

(a) Source of Variation df

Block 4

Treatments 3 Error 12

Total 19 (5 marks)

(b)

Source of Variation df

Column 4

Row 4

Treatments 4

Error 12

Total 24 (5 marks)

Examples of mid-term and final examinations

:

xiii

j. Main References

1. Casella, G. (2008). Statistical Design. New York: Springer 2 Clewer, A.G. and Scarisbrick, D. H. (2001). Practical Statistics

and Experimental Design for Plant and Crop Science. New York: John Wiley and Sons.

2. A study was conducted on the effect of 4 types of media with 4 replications on the

leaf width of crop Y. Sum of squares [SS] for Between media is 10.1596 and

Mean Square for Within is 0.809. The means of the leaf width for media A, B, C

and D are 2.39 cm, 3.25 cm, 1.93 cm and 3.69 cm respectively.

i. Is there a significant effect of media in this experiment? (5 marks)

ii. Recommend a suitable media for crop Y. (5 marks)

13

xii

.

SECTION A Answer all questions in this section

1. Draw the arrangement of all the experimental units for the following:

(a) Source of Variation df

Block 4

Treatments 3 Error 12

Total 19 (5 marks)

(b)


Column 4

Row 4

Treatments 4

Error 12

Total 24 (5 marks)

Examples of mid-term and final examinations

:

xiii

j. Main References

1. Casella, G. (2008). Statistical Design. New York: Springer 2 Clewer, A.G. and Scarisbrick, D. H. (2001). Practical Statistics

and Experimental Design for Plant and Crop Science. New York: John Wiley and Sons.

2. A study was conducted on the effect of 4 types of media with 4 replications on the

leaf width of crop Y. Sum of squares [SS] for Between media is 10.1596 and

Mean Square for Within is 0.809. The means of the leaf width for media A, B, C

and D are 2.39 cm, 3.25 cm, 1.93 cm and 3.69 cm respectively.

i. Is there a significant effect of media in this experiment? (5 marks)

ii. Recommend a suitable media for crop Y. (5 marks)

14

xiv

k. Extra References

1. Gomez, K. A. and Gomez, A.G. (2005). Statistical Procedures for Agricultural Research (4th Edition). New York: John Wiley and Sons. .

2. Hinkelmann, K. and Kempthorne, O. (2007). Design and Analysis

of Experiments. Introduction to Experimental Design (2nd). New York: Wiley-Interscience.

3. Peterson, R. G. (1994). Agricultural Field Experiments: Design and Analysis. New York: Marcel Dekker.

4. Adele. M. Holman (1989). Family Assessment: Tools for Understanding and Invention. New York: Sage Publication.

l. Information of icon in the module

a)

Objective Objective of module, unit or topic

b)

Introduction Introduction to unit, topic or sub-topic

c)

Important Content

Important content in unit or topic

d)

Attention

This symbol is used for information to be given attention by the students

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Content

Unit Title

Page

1

Principles of Experimental Design

Topic 1: Experiment

Topic 2: Treatment

Topic 3: Experimental Unit

Topic 4: Sample

Topic 5: Replication

Topic 6: Randomization

Topic 7. Variables

Topic 8: Control

Topic 9: Responses

Topic 10: Experimental Error

Topic 11: Types of Experiment

Topic 12: Selection of Test Site

Topic 13: Uniformity of Experiment Site

Topic 14: Procedure in Planning an Experiment

Topic 15: Types of Measurement /Data

Topic 16: Hypothesis Testing

Topic 17: Methods of Error Control in Experiment

Topic 18: Plot Size and Shape

Topic 19: Uniformity of Experimental Plot

1

2

3

4

5

6

7

8

8

9

10

11

12

13

13

14

15

16

16

2 Experimental Designs

Topic 1: Complete Randomized Design

Topic 2: Randomized Complete Block Design

Topic 3: Latin Square Design

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20

22

15

xiv

k. Extra References

1. Gomez, K. A. and Gomez, A.G. (2005). Statistical Procedures for Agricultural Research (4th Edition). New York: John Wiley and Sons. .

2. Hinkelmann, K. and Kempthorne, O. (2007). Design and Analysis

of Experiments. Introduction to Experimental Design (2nd). New York: Wiley-Interscience.

3. Peterson, R. G. (1994). Agricultural Field Experiments: Design and Analysis. New York: Marcel Dekker.

4. Adele. M. Holman (1989). Family Assessment: Tools for Understanding and Invention. New York: Sage Publication.

l. Information of icon in the module

a)

Objective Objective of module, unit or topic

b)

Introduction Introduction to unit, topic or sub-topic

c)

Important Content

Important content in unit or topic

d)

Attention

This symbol is used for information to be given attention by the students

xv

Content

Unit Title

Page

1

Principles of Experimental Design

Topic 1: Experiment

Topic 2: Treatment

Topic 3: Experimental Unit

Topic 4: Sample

Topic 5: Replication

Topic 6: Randomization

Topic 7. Variables

Topic 8: Control

Topic 9: Responses

Topic 10: Experimental Error

Topic 11: Types of Experiment

Topic 12: Selection of Test Site

Topic 13: Uniformity of Experiment Site

Topic 14: Procedure in Planning an Experiment

Topic 15: Types of Measurement /Data

Topic 16: Hypothesis Testing

Topic 17: Methods of Error Control in Experiment

Topic 18: Plot Size and Shape

Topic 19: Uniformity of Experimental Plot

1

2

3

4

5

6

7

8

8

9

10

11

12

13

13

14

15

16

16

2 Experimental Designs

Topic 1: Complete Randomized Design

Topic 2: Randomized Complete Block Design

Topic 3: Latin Square Design

18

20

22

16

xvi

Unit Title

Page

3 Analysis of Variances (ANOVA) Topic 1: F Distribution

Topic 2: ANOVA for One Factor Experiment

Topic 3: ANOVA for Various Designs

24

25

26

4 Comparison of Treatment Means Topic 1: Least Significant Difference (LSD)

Topic 2: Duncan’s Multiple Range Test

Topic 3: Contrast

34

37

39

5 Data Transformation

Topic 1: Log Transformation

Topic 2: Square-Root Transformation

Topic 3: Arc-sine Transformation

43

44

45

6 Non-parametric Tests

Topic 1: One Sample Sign Test

Topic 2: Paired Data Sign Test

Topic 3: Wilcoxon-Mann-Whitney

Topic 4: Chi-square Test

46

48

49

51

7 Factorial Experiment

Topic 1: Effect of Main Factor

Topic 2: Interaction Effect

57

59

8 Experiment with Different Sizes of Experimental Units Topic 1: Split Plot Design

Topic 2: Experiment with Repeated Data

66

74

1

UNIT 1

PRINCIPLES OF EXPERIMENTAL DESIGN

Introduction

In designing an experiment, it is essential to state the objectives of the experiment as

to answer the questions, stated hypothesis to be tested and the effect to be

estimated. Experimental design is how treatments under investigation are arranged

such that their effect are revealed and are accurately measured. All designs are

characterized by experimental units classified by treatments, but in some cases they

are further classified into blocks, rows, columns main plots and so on. An

experimental design can be complex or simple.

Objective To evaluate the information in the principles of the experimental design. TOPIC 1: EXPERIMENT

Important Content Experiment is an investigation to obtain

a) new information

b) proving the result of an earlier experiment

c) it is conducted to answer question(s)

17

xvi

Unit Title

Page

3 Analysis of Variances (ANOVA) Topic 1: F Distribution

Topic 2: ANOVA for One Factor Experiment

Topic 3: ANOVA for Various Designs

24

25

26

4 Comparison of Treatment Means Topic 1: Least Significant Difference (LSD)

Topic 2: Duncan’s Multiple Range Test

Topic 3: Contrast

34

37

39

5 Data Transformation

Topic 1: Log Transformation

Topic 2: Square-Root Transformation

Topic 3: Arc-sine Transformation

43

44

45

6 Non-parametric Tests

Topic 1: One Sample Sign Test

Topic 2: Paired Data Sign Test

Topic 3: Wilcoxon-Mann-Whitney

Topic 4: Chi-square Test

46

48

49

51

7 Factorial Experiment

Topic 1: Effect of Main Factor

Topic 2: Interaction Effect

57

59

8 Experiment with Different Sizes of Experimental Units Topic 1: Split Plot Design

Topic 2: Experiment with Repeated Data

66

74

1

UNIT 1

PRINCIPLES OF EXPERIMENTAL DESIGN

Introduction

In designing an experiment, it is essential to state the objectives of the experiment as

to answer the questions, stated hypothesis to be tested and the effect to be

estimated. Experimental design is how treatments under investigation are arranged

such that their effect are revealed and are accurately measured. All designs are

characterized by experimental units classified by treatments, but in some cases they

are further classified into blocks, rows, columns main plots and so on. An

experimental design can be complex or simple.

Objective To evaluate the information in the principles of the experimental design. TOPIC 1: EXPERIMENT

Important Content Experiment is an investigation to obtain

a) new information

b) proving the result of an earlier experiment

c) it is conducted to answer question(s)

18

PRT 3202 Experimental Design and Analysis PJJ UPM / UPMET

2

Laboratory Exercise

Laboratory exercise for topic 1 unit 1 will be delivered during week 1 through

Putra Learning Management System (LMS) or email.

TOPIC 2: TREATMENT

Important Content Procedure whose effect of a material to be tested and compared with other

treatments

Example 1: types of fertilizer: NPK Blue and NPK yellow

Example 2: fertilizer rates: 10, 20 and 30 kg N ha-1

Laboratory Exercise


LMS or email.


3

TOPIC 3: EXPERIMENTAL UNIT

Important Content This is the unit of material that receives a treatment or where treatment is given. The

unit may be a single plant, a single animal, a leaf, ten plants in a square meter plot or

so on. For a field experiments, a decision on the size and shape of the experimental

unit has to be made. In situations where non-uniformity in experimental plot is

anticipated, the plots should be reasonably long and narrow. Effort should be made

to control the influence of each adjacent unit on the other. This can be achieved by

randomizing treatments and also by making use of guard rows.

Example: a plant

an animal

area of land

a square meter plot

Example 1: Effect of different type of fertilizers on sunflower

Fertilizer NPK Blue Fertilizer NPK Yellow

Experimental unit

(a plant)

Example 2: Effect of fertilizer rates on a plant

10 kg N ha-1 20 kg N ha-1

Experimental unit

(an area)

19


2

Laboratory Exercise


Putra Learning Management System (LMS) or email.

TOPIC 2: TREATMENT

Important Content Procedure whose effect of a material to be tested and compared with other

treatments

Example 1: types of fertilizer: NPK Blue and NPK yellow

Example 2: fertilizer rates: 10, 20 and 30 kg N ha-1

Laboratory Exercise


LMS or email.


3

TOPIC 3: EXPERIMENTAL UNIT

Important Content This is the unit of material that receives a treatment or where treatment is given. The

unit may be a single plant, a single animal, a leaf, ten plants in a square meter plot or

so on. For a field experiments, a decision on the size and shape of the experimental

unit has to be made. In situations where non-uniformity in experimental plot is

anticipated, the plots should be reasonably long and narrow. Effort should be made

to control the influence of each adjacent unit on the other. This can be achieved by

randomizing treatments and also by making use of guard rows.

Example: a plant

an animal

area of land

a square meter plot

Example 1: Effect of different type of fertilizers on sunflower

Fertilizer NPK Blue Fertilizer NPK Yellow

Experimental unit

(a plant)

Example 2: Effect of fertilizer rates on a plant


Experimental unit

(an area)

20


4

Laboratory Exercise


LMS or email.

TOPIC 4: SAMPLE

Important Content Part of experimental unit where the effect of treatment is measured.


Sample (only these part is measured)

Laboratory Exercise


LMS or email.


5

TOPIC 5: REPLICATION

Important Content Repetition or appearance of a treatment more than once in an experiment is referred

to as replication. Replication is the sole means of measuring the validity of a

conclusion drawn from the experiment, the number of replications should be chosen

such that the required precision of the treatment estimate is produced. Several

factors affect the number of replications for an experiment; perhaps the most

important of all is the degree of precision required. When a treatment effect is small

and requires high precision to be detected or measured, the greater the number of

replicates the better.


Replication 1


Replication 2

Purpose of replication:

• to calculate the mean of the treatment

• to improve the accuracy of the experiment

• to measure the experimental error

21


4

Laboratory Exercise


LMS or email.

TOPIC 4: SAMPLE

Important Content Part of experimental unit where the effect of treatment is measured.


Sample (only these part is measured)

Laboratory Exercise


LMS or email.


5

TOPIC 5: REPLICATION

Important Content Repetition or appearance of a treatment more than once in an experiment is referred

to as replication. Replication is the sole means of measuring the validity of a

conclusion drawn from the experiment, the number of replications should be chosen

such that the required precision of the treatment estimate is produced. Several

factors affect the number of replications for an experiment; perhaps the most

important of all is the degree of precision required. When a treatment effect is small

and requires high precision to be detected or measured, the greater the number of

replicates the better.


Replication 1


Replication 2

Purpose of replication:

• to calculate the mean of the treatment

• to improve the accuracy of the experiment

• to measure the experimental error

22


6

Laboratory Exercise


LMS or email.

TOPIC 6: RANDOMIZATION

Important Content

Arrangement of treatments of experimental unit so as that each experimental unit

has the same chance to be selected to receive a treatment

Example : Effect of 4 types of fertilizer with 2 replications.

REPLICATION 1

REPLICATION 2

Laboratory Exercise


LMS or email.

A B

C D

D C

A B


7

TOPIC 7: VARIABLES

Important Content Characteristics of the experimental unit that can be measured:

• Yield

• Height of a plant

• Soil pH

• Number of insects

2 types of variables

• Quantitative

• Qualitative

Laboratory Exercise


LMS or email.

TOPIC 8: CONTROL

Important Content A standard treatment that is used as a baseline or basis of comparison for the other

treatments. The control treatment does not receive the treatment or the experimental

manipulation that the experimental treatments receive.

23


6

Laboratory Exercise


LMS or email.

TOPIC 6: RANDOMIZATION

Important Content

Arrangement of treatments of experimental unit so as that each experimental unit

has the same chance to be selected to receive a treatment

Example : Effect of 4 types of fertilizer with 2 replications.

REPLICATION 1

REPLICATION 2

Laboratory Exercise


LMS or email.

A B

C D

D C

A B


7

TOPIC 7: VARIABLES

Important Content Characteristics of the experimental unit that can be measured:

• Yield

• Height of a plant

• Soil pH

• Number of insects

2 types of variables

• Quantitative

• Qualitative

Laboratory Exercise


LMS or email.

TOPIC 8: CONTROL

Important Content A standard treatment that is used as a baseline or basis of comparison for the other

treatments. The control treatment does not receive the treatment or the experimental

manipulation that the experimental treatments receive.

24


8

Laboratory Exercise


LMS or email.

TOPIC 9: RESPONSES

Important Content Outcomes that are being observed after applying a treatment to an experimental unit

Example: Treatment :- application of 3 types of nitrogen fertilizer

Response :- nitrogen content or biomass of corn plants

Laboratory Exercise


LMS or email.


9

TOPIC 10: EXPERIMENTAL ERROR

Important Content The random variation present in all experimental results. Errors can be minimized by

having a large sample size as well as by replications and blocking.

Laboratory Exercise

Laboratory exercise for topic 10 unit 1 will be delivered during week 2

through LMS or email.

TOPIC 11: TYPES OF EXPERIMENT

Important Content 1. Manipulative experiment – one or more conditions are varied while all other

conditions are maintained, perform under controlled conditions such as in a laboratory.

2. Field experiment – similar to manipulative experiment, but it is carried out in an open area where environmental factors and extraneous variables are present.

Laboratory Exercise



25


8

Laboratory Exercise


LMS or email.

TOPIC 9: RESPONSES

Important Content Outcomes that are being observed after applying a treatment to an experimental unit

Example: Treatment :- application of 3 types of nitrogen fertilizer

Response :- nitrogen content or biomass of corn plants

Laboratory Exercise


LMS or email.


9

TOPIC 10: EXPERIMENTAL ERROR

Important Content The random variation present in all experimental results. Errors can be minimized by

having a large sample size as well as by replications and blocking.

Laboratory Exercise



TOPIC 11: TYPES OF EXPERIMENT

Important Content 1. Manipulative experiment – one or more conditions are varied while all other

conditions are maintained, perform under controlled conditions such as in a laboratory.

2. Field experiment – similar to manipulative experiment, but it is carried out in an open area where environmental factors and extraneous variables are present.

Laboratory Exercise



26


10

TOPIC 12: SELECTION OF TEST SITE

Important Content Selection of sites where the trial is to be conducted. Selection procedures:

1. Clearly specify the desired test environment and identify the sources of variability e.g. soil, climate, topography, water regime.

2. Select a large area that is homogenous and satisfies those selected features mentioned above.

3. Choose an area/field that is large enough to accommodate the

experiment.

Laboratory Exercise


through LMS email.


11

TOPIC 13: UNIFORMITY OF EXPERIMENTAL SITE

Important Content 1. Slope - Fertility gradients are more pronounced in sloping areas. Ideally,

experiments should be conducted in areas with no slopes (level). If this not avoidable, proper blocking is needed.

2. Areas used for experiments in previous cropping - When the area to be used for a future experiment has been used in a previous experiment, study the nature of the previous study to determine if it will have any direct or serious effect on the outcome of the new experiment.

3. Presence of large trees, poles, and structures - Areas with surrounding

permanent structures should be avoided, not only because of the direct effect of shading but also the nature of the soil near the structure.

Laboratory Exercise



27


10

TOPIC 12: SELECTION OF TEST SITE

Important Content Selection of sites where the trial is to be conducted. Selection procedures:

1. Clearly specify the desired test environment and identify the sources of variability e.g. soil, climate, topography, water regime.

2. Select a large area that is homogenous and satisfies those selected features mentioned above.

3. Choose an area/field that is large enough to accommodate the

experiment.

Laboratory Exercise


through LMS email.


11

TOPIC 13: UNIFORMITY OF EXPERIMENTAL SITE

Important Content 1. Slope - Fertility gradients are more pronounced in sloping areas. Ideally,

experiments should be conducted in areas with no slopes (level). If this not avoidable, proper blocking is needed.

2. Areas used for experiments in previous cropping - When the area to be used for a future experiment has been used in a previous experiment, study the nature of the previous study to determine if it will have any direct or serious effect on the outcome of the new experiment.

3. Presence of large trees, poles, and structures - Areas with surrounding

permanent structures should be avoided, not only because of the direct effect of shading but also the nature of the soil near the structure.

Laboratory Exercise



28


12

TOPIC 14: PROCEDURES IN PLANNING AN EXPERIMENT

Important Content 1. Statement of the objectives of the experiment. 2. Identification of the resources available for the experiment. 3. Assessment of the location and the conditions under which the experiment to

be conducted. 4. Identification of the population of subjects that are to be tested. 5. Consideration of the amount of variability that is likely to arise within samples. 6. Identification of the type of observation/measurements that are to be made. 7. Identification of the most appropriate technique for analyzing data. 8. Identification of treatment groups and assignment of treatments.

Laboratory Exercise



TOPIC 15: TYPES OF MEASUREMENT/DATA

Important Content A response or dependent variable that really provides information about the problem under study Primary observations = grain yield Explanatory observations = number of tillers, panicle number, spikelet number Covariate observations = percent infestation (if the plants were infested by disease)


13

Laboratory Exercise



TOPIC 16: HYPOTHESIS TESTING

Important Content It is a statistical test used to the objective Two types of hypotheses: Null hypothesis (H0) Alternative hypothesis (HA) Significance testing is achieved based on the critical level of probability which is commonly set to 5% or α=0.05, written as (P≤ 0.05).

H0 = there is no difference between the sample means µ1= µ2 = µn HA = there is a difference between the sample means µ1 ≠ µ2 ≠ µn

If the value of P is smaller than (or equal to) the critical value (α=0.05), H0 is rejected while HA is accepted. If the value of P is larger than the critical value, H0 is accepted while HA is rejected.

Laboratory Exercise



29


12

TOPIC 14: PROCEDURES IN PLANNING AN EXPERIMENT

Important Content 1. Statement of the objectives of the experiment. 2. Identification of the resources available for the experiment. 3. Assessment of the location and the conditions under which the experiment to

be conducted. 4. Identification of the population of subjects that are to be tested. 5. Consideration of the amount of variability that is likely to arise within samples. 6. Identification of the type of observation/measurements that are to be made. 7. Identification of the most appropriate technique for analyzing data. 8. Identification of treatment groups and assignment of treatments.

Laboratory Exercise



TOPIC 15: TYPES OF MEASUREMENT/DATA

Important Content A response or dependent variable that really provides information about the problem under study Primary observations = grain yield Explanatory observations = number of tillers, panicle number, spikelet number Covariate observations = percent infestation (if the plants were infested by disease)


13

Laboratory Exercise



TOPIC 16: HYPOTHESIS TESTING

Important Content It is a statistical test used to the objective Two types of hypotheses: Null hypothesis (H0) Alternative hypothesis (HA) Significance testing is achieved based on the critical level of probability which is commonly set to 5% or α=0.05, written as (P≤ 0.05).

H0 = there is no difference between the sample means µ1= µ2 = µn HA = there is a difference between the sample means µ1 ≠ µ2 ≠ µn

If the value of P is smaller than (or equal to) the critical value (α=0.05), H0 is rejected while HA is accepted. If the value of P is larger than the critical value, H0 is accepted while HA is rejected.

Laboratory Exercise



30


14

TOPIC 17: METHODS OF ERROR CONTROL IN EXPERIMENT

Important Content

1. Blocking 2. Proper plot technique 3. Covariance analysis

Laboratory Exercise



TOPIC 18: PLOT SIZE AND SHAPE

Important Content An experiment conducted on soils of high variability require a small plot size with increased number of replications will minimize or reduce experimental error because the distance between any farthest points in each block will be shorter than when using large plots. As a result, the variability within each block is minimized. If the plot size cannot be reduced and it is suspected that the soil is highly variable with unknown direction, the use of square-shaped blocks is recommended. The distance between any two farthest points in a square block is shorter that those in a long and narrow block.

Laboratory Exercise Laboratory exercise for topic 18 unit 1 will be delivered during week 3 through LMS or email.


15

TOPIC 19: UNIFORMITY OF EXPERIMENT PLOT

Important Content In a plot, each block must be of the same size and shape with equal numbers of experimental units arranged randomly according to the specified design. Except for split plot design, the size of the main plot is bigger than the sub-plot size.

Laboratory Exercise



31


14

TOPIC 17: METHODS OF ERROR CONTROL IN EXPERIMENT

Important Content

1. Blocking 2. Proper plot technique 3. Covariance analysis

Laboratory Exercise



TOPIC 18: PLOT SIZE AND SHAPE

Important Content An experiment conducted on soils of high variability require a small plot size with increased number of replications will minimize or reduce experimental error because the distance between any farthest points in each block will be shorter than when using large plots. As a result, the variability within each block is minimized. If the plot size cannot be reduced and it is suspected that the soil is highly variable with unknown direction, the use of square-shaped blocks is recommended. The distance between any two farthest points in a square block is shorter that those in a long and narrow block.

Laboratory Exercise Laboratory exercise for topic 18 unit 1 will be delivered during week 3 through LMS or email.


15

TOPIC 19: UNIFORMITY OF EXPERIMENT PLOT

Important Content In a plot, each block must be of the same size and shape with equal numbers of experimental units arranged randomly according to the specified design. Except for split plot design, the size of the main plot is bigger than the sub-plot size.

Laboratory Exercise



32


16

UNIT 2

EXPERIMENTAL DESIGNS

Introduction Arrangement of experimental unit that contains treatments and replications into

various designs to estimate and control experimental error so as to interpret results

accurately. The major difference among experimental designs is the way in which

experimental units are classified or grouped. An experimental design can be simple

or complex. It is, however, advisable to choose a less complicated design that best

provides the desired precision.

Objective To estimate and control experimental error for accurate interpretation,

TOPIC 1: COMPLETE RANDOMIZED DESIGN

Important Content

It is used when an area or location or experimental materials are

homogeneous. For completely randomized design (CRD), each experimental

unit has the same chance of receiving a treatment in completely randomized

manner.


17

Example: Testing 4 varieties (V1, V2, V3 and V4) in a homogeneous field.

The soil is homogeneous so the varieties can be

located at any of the compartment without any

effect of the soil.

All the 4 compartments have the same soil fertility.

..... effect of block is neglected or is not considered

.... easy placement as the treatment can be placed in any of the compartments

.... easy to arrange experimental unit due to lack of block effects

Disadvantage: difficult to obtain homogeneity in the field.

Example: Testing of yield of 4 crop varieties with 4 replications.

Varieties: V1, V2, V3, V4 (control)

Replications R1, R2, R3, R4

V1 R3 V2 R2 V1 R4 V4 R1

V2 R4 V1 R2 V3 R1 V4 R4

V2 R1 V2 R3 V4 R2 V3 R4

V3 R2 V1 R1 V4 R3 V3 R3

All the varieties with 4 replications can be placed at any of the compartments

Each compartment the soil fertility is the same.

Laboratory Exercise


LMS or email.

V1 V2

V3 V4

33


16

UNIT 2

EXPERIMENTAL DESIGNS

Introduction Arrangement of experimental unit that contains treatments and replications into

various designs to estimate and control experimental error so as to interpret results

accurately. The major difference among experimental designs is the way in which

experimental units are classified or grouped. An experimental design can be simple

or complex. It is, however, advisable to choose a less complicated design that best

provides the desired precision.

Objective To estimate and control experimental error for accurate interpretation,

TOPIC 1: COMPLETE RANDOMIZED DESIGN

Important Content

It is used when an area or location or experimental materials are

homogeneous. For completely randomized design (CRD), each experimental

unit has the same chance of receiving a treatment in completely randomized

manner.


17

Example: Testing 4 varieties (V1, V2, V3 and V4) in a homogeneous field.

The soil is homogeneous so the varieties can be

located at any of the compartment without any

effect of the soil.

All the 4 compartments have the same soil fertility.

..... effect of block is neglected or is not considered

.... easy placement as the treatment can be placed in any of the compartments

.... easy to arrange experimental unit due to lack of block effects

Disadvantage: difficult to obtain homogeneity in the field.

Example: Testing of yield of 4 crop varieties with 4 replications.


Replications R1, R2, R3, R4

V1 R3 V2 R2 V1 R4 V4 R1

V2 R4 V1 R2 V3 R1 V4 R4

V2 R1 V2 R3 V4 R2 V3 R4

V3 R2 V1 R1 V4 R3 V3 R3

All the varieties with 4 replications can be placed at any of the compartments

Each compartment the soil fertility is the same.

Laboratory Exercise


LMS or email.

V1 V2

V3 V4

34


18

TOPIC 2: RANDOMIZED COMPLETE BLOCK DESIGN

Important Content In this design treatments are assigned at random to a group of experimental units

called the block. A block consists of uniform experimental units. The main aim of this

design is to keep the variability among experimental units within a block as small as

possible and to maximize differences among the blocks.

.... it is used for an area or location or materials that are heterogeneous

.... group of treatments is placed randomly in a block or replication

.... block or replication is created to reduce the heterogeneity of the experimental

unit

.... each block containing homogenous experimental unit

.... treatments are arranged in each block or replication

.... effect of block is considered in the calculation of ANOVA

Method of blocking in a field

a) One directional gradient

Arrange the block at right angles with the gradient

High Fertility Low Fertility

b) Two ways gradients: 1 strong, 1 less in strength

Moderate

Arrange block perpendicular to the gradient



19

c) Two ways gradients same strength

Square blocking as much as possible

Example : Testing 4 crop varieties with 4 replications.


Replication: R1, R2, R3, R4

R1 R2 R3 R4

V1 V2 V3 V4

V4 V1 V1 V2

V3 V4 V4 V3

V2 V3 V2 V1

Fertility Gradient

Laboratory Exercise


LMS or email.

35


18

TOPIC 2: RANDOMIZED COMPLETE BLOCK DESIGN

Important Content In this design treatments are assigned at random to a group of experimental units

called the block. A block consists of uniform experimental units. The main aim of this

design is to keep the variability among experimental units within a block as small as

possible and to maximize differences among the blocks.

.... it is used for an area or location or materials that are heterogeneous

.... group of treatments is placed randomly in a block or replication

.... block or replication is created to reduce the heterogeneity of the experimental

unit

.... each block containing homogenous experimental unit

.... treatments are arranged in each block or replication

.... effect of block is considered in the calculation of ANOVA

Method of blocking in a field

a) One directional gradient

Arrange the block at right angles with the gradient


b) Two ways gradients: 1 strong, 1 less in strength

Moderate

Arrange block perpendicular to the gradient



19

c) Two ways gradients same strength

Square blocking as much as possible

Example : Testing 4 crop varieties with 4 replications.


Replication: R1, R2, R3, R4

R1 R2 R3 R4

V1 V2 V3 V4

V4 V1 V1 V2

V3 V4 V4 V3

V2 V3 V2 V1

Fertility Gradient

Laboratory Exercise


LMS or email.

36


20

TOPIC 3: LATIN SQUARE DESIGN

Important Content

Latin square design handles two known sources of variation among experimental

units simultaneously. It treats the sources as two independent blocking criteria: row-

blocking and column-blocking. This is achieved by making sure that every treatment

occurs only once in each row-block and once in each column-block. This helps to

remove variability from the experimental error associated with both these effects.

• Treatments are arranged in row and column

• Error is being reduced due to two ways heterogeneity (row and column)

• More efficient than RCBD when there is two ways heterogeneity

• Number of replication should be equal to number of treatment

• Usually such arrangement is suitable for 4 to 8 treatments

STEPS IN ARRANGING TREATMENTS WITH RANDOMIZATION IN A LATIN SQURE DESIGN

Example: Effect of 6 different fertilizer N treatments (A, B, C, D, E, and F) on the

yield of corn.

1. Arrange each treatment so that it occurs once in a row and once in a column

only.

Row

Column

B D E F A C

C E A D F B

A F C B E D

D A F C B E

F B D E C A

E C B A D F


21

1. Use random number table, assign numbering for row and column randomly.

Row Column

4 2 5 1 3 6

1 B D E F A C

3 C E A D F B

5 A F C B E D

4 D A F C B E

2 F B D E C A

6 E C B A D F

2. Arrange the treatments in the field based on the arrangement in the above

table 2.

Row Column

1 2 3 4 5 6

1 F D A B E C

3 E B C F D A

5 D E F C A B

4 C A B D F E

2 B F E A C D

6 A C D E B F

Laboratory Exercise


LMS or email

37


20

TOPIC 3: LATIN SQUARE DESIGN

Important Content

Latin square design handles two known sources of variation among experimental

units simultaneously. It treats the sources as two independent blocking criteria: row-

blocking and column-blocking. This is achieved by making sure that every treatment

occurs only once in each row-block and once in each column-block. This helps to

remove variability from the experimental error associated with both these effects.

• Treatments are arranged in row and column

• Error is being reduced due to two ways heterogeneity (row and column)

• More efficient than RCBD when there is two ways heterogeneity

• Number of replication should be equal to number of treatment

• Usually such arrangement is suitable for 4 to 8 treatments

STEPS IN ARRANGING TREATMENTS WITH RANDOMIZATION IN A LATIN SQURE DESIGN

Example: Effect of 6 different fertilizer N treatments (A, B, C, D, E, and F) on the

yield of corn.

1. Arrange each treatment so that it occurs once in a row and once in a column

only.

Row

Column

B D E F A C

C E A D F B

A F C B E D

D A F C B E

F B D E C A

E C B A D F


21

1. Use random number table, assign numbering for row and column randomly.

Row Column

4 2 5 1 3 6

1 B D E F A C

3 C E A D F B

5 A F C B E D

4 D A F C B E

2 F B D E C A

6 E C B A D F

2. Arrange the treatments in the field based on the arrangement in the above

table 2.

Row Column

1 2 3 4 5 6

1 F D A B E C

3 E B C F D A

5 D E F C A B

4 C A B D F E

2 B F E A C D

6 A C D E B F

Laboratory Exercise


LMS or email

38


22

UNIT 3

ANALYSIS OF VARIANCE

Introduction Analysis of variance (ANOVA) is to determine the ratio of between samples to the

variance of within samples that is the F distribution. The value of F is used to reject

or accept the null hypothesis. It is used to analyze the variances of treatments or

events for significant differences between treatment variances, particularly in

situations where more than two treatments are involved. ANOVA can only be used to

ascertain if the treatment differences are significant or not.

Objective To accept or reject the null hypothesis where more than two treatments are involved.

TOPIC 1: F DISTRIBUTION

Important Content F value is used to test the significant difference between more than two treatment

means

F = s2, calculated from sample mean

s2, calculate from variance between individual sample

= sa2 (variance between samples)

sd2 (variance within samples)

df (numerator) = n -1, where n = number of samples

df (denominator) = n(r – 1), where r = size of samples


23

Laboratory Exercise


LMS or email.

TOPIC 2: ANOVA FOR ONE FACTOR EXPERIMENT

Important Content

- Using the same data, F can be calculated using Table of ANOVA:

1. Table of ANOVA

Source df Sum of Squares Mean Square F of Variation (SS) (MS)

Treatment n-1 SS treatment SStreatment /n-1 SStreatment/MSerror

Error n(r-1) SSerror SSerror /n(r-1)

Total rn-1 SSTotal

Laboratory Exercise



39


22

UNIT 3

ANALYSIS OF VARIANCE

Introduction Analysis of variance (ANOVA) is to determine the ratio of between samples to the

variance of within samples that is the F distribution. The value of F is used to reject

or accept the null hypothesis. It is used to analyze the variances of treatments or

events for significant differences between treatment variances, particularly in

situations where more than two treatments are involved. ANOVA can only be used to

ascertain if the treatment differences are significant or not.

Objective To accept or reject the null hypothesis where more than two treatments are involved.

TOPIC 1: F DISTRIBUTION

Important Content F value is used to test the significant difference between more than two treatment

means

F = s2, calculated from sample mean

s2, calculate from variance between individual sample

= sa2 (variance between samples)

sd2 (variance within samples)

df (numerator) = n -1, where n = number of samples

df (denominator) = n(r – 1), where r = size of samples


23

Laboratory Exercise


LMS or email.

TOPIC 2: ANOVA FOR ONE FACTOR EXPERIMENT

Important Content

- Using the same data, F can be calculated using Table of ANOVA:

1. Table of ANOVA

Source df Sum of Squares Mean Square F of Variation (SS) (MS)

Treatment n-1 SS treatment SStreatment /n-1 SStreatment/MSerror

Error n(r-1) SSerror SSerror /n(r-1)

Total rn-1 SSTotal

Laboratory Exercise



40


24

TOPIC 3: ANOVA FOR VARIOUS DESIGNS

Important Content

a. Complete Randomized Design Example: Testing yield of 4 varieties with 4 replications.

Varieties: V1, V2, V3, V4 (Control)

Replications: R1, R2, R3, R4

V1R3 (50) V2R2 (69) V1R4 (54) V4R1(51)

V2 R4 (57) V1R2 (67) V3R1 (65) V4R4 (62)

V2R1 (57) V2R3 (53) V4R2 (52) V3R4 (74)

V3R2 (54) V1R1 (57) V4R3 (47) V3R3 (59)

Calculation:

Arrange the data according to treatment and replication

Variety Replication Total Mean

1 2 3 4

V1 57 67 50 54 228 57

V2 57 69 53 57 236 59

V3 65 54 59 74 252 63

V4 51 52 47 62 212 53

Total

928


25

ANOVA

HO: no significant difference in yield between the varieties.

Table of ANOVA

Source of df SS MS F FTable

Variation (p=0.05)

Variety 3 208 69.3 1.29 3.49

Error 12 646 53.8

Total 15 854

Calculation

1. Degree of Freedom, df

df (total) = vr – 1 = 4(4) – 1 = 15

df (variety) = v – 1 = 4 – 1 = 3

df (error) = df(total) – df(variety) = 15 – 3 = 12

or

df (error) = v(r – 1) = 4(4 – 1) = 12

2. Correction factor, CF

CF = Y..2 / rv = (928)2/(4×4) = 53824

3. Sum Square Total (SST)

SST = ƩYij2 – CF

= (572 + 672 + ... + 472 + 622) – 53824

= 54678 – 53824 = 854

4. Sum Square Variety (SSV)

SSV = (ƩY.j2/ r) – CF

= (2282 + 2362 + 2522 + 2122)/ 4 – 53824

= 54032 – 53824 = 208

5. Sum Square Error (SSE)

SSE = SST – SSV

= 854 – 208 = 646

41


24

TOPIC 3: ANOVA FOR VARIOUS DESIGNS

Important Content

a. Complete Randomized Design Example: Testing yield of 4 varieties with 4 replications.

Varieties: V1, V2, V3, V4 (Control)

Replications: R1, R2, R3, R4

V1R3 (50) V2R2 (69) V1R4 (54) V4R1(51)

V2 R4 (57) V1R2 (67) V3R1 (65) V4R4 (62)

V2R1 (57) V2R3 (53) V4R2 (52) V3R4 (74)

V3R2 (54) V1R1 (57) V4R3 (47) V3R3 (59)

Calculation:

Arrange the data according to treatment and replication

Variety Replication Total Mean

1 2 3 4

V1 57 67 50 54 228 57

V2 57 69 53 57 236 59

V3 65 54 59 74 252 63

V4 51 52 47 62 212 53

Total

928


25

ANOVA

HO: no significant difference in yield between the varieties.

Table of ANOVA


Variation (p=0.05)

Variety 3 208 69.3 1.29 3.49

Error 12 646 53.8

Total 15 854

Calculation

1. Degree of Freedom, df

df (total) = vr – 1 = 4(4) – 1 = 15

df (variety) = v – 1 = 4 – 1 = 3

df (error) = df(total) – df(variety) = 15 – 3 = 12

or

df (error) = v(r – 1) = 4(4 – 1) = 12

2. Correction factor, CF

CF = Y..2 / rv = (928)2/(4×4) = 53824


SST = ƩYij2 – CF

= (572 + 672 + ... + 472 + 622) – 53824

= 54678 – 53824 = 854


SSV = (ƩY.j2/ r) – CF

= (2282 + 2362 + 2522 + 2122)/ 4 – 53824

= 54032 – 53824 = 208


SSE = SST – SSV

= 854 – 208 = 646

42


26

6. Mean Square Variety (MSV)

MSV = SSV/dfv = 208/3 = 69.3

7. Mean Square Error (MSE)

MSE = SSE/dfe = 646/ 12 = 53.8

8. F value Variety

F = MSV/MSE = 69.3/53.8 = 1.29

9. F Table

dfv = 3, dfe = 12

At P = 0.05, F = 3.49

10. Conclusion

1.29 < 3.49 → accept HO, no significant difference of yield between the

varieties.

b. Randomized Complete Block Design

Arrange data according to treatments and replications.

Variety R1 R2 R3 R4 Total Mean

V1 57 67 50 54 228 57

V2 57 69 53 57 236 59

V3 65 54 59 74 252 63

V4 51 52 47 62 212 53

Total 230 242 209 247 928

Mean 57.50 60.50 52.25 61.75


27

Calculation:

HO: no significant difference of yield between varieties.

Table of ANOVA


Variation

(p=0.05)

Block (Rep) 3 214.5 71.5 1.49 3.86

Variety 3 208 69.3 1.45 3.86

Error 9 431.5 47.94

Total 15 854

Calculation

1. Degree of Freedom (df)

df (total) = vr – 1 = 4(4) – 1 = 15

df (block) = r – 1 = 4 – 1 = 3

df (variety) = v – 1 = 4 – 1 = 3

df (error) = df (total) – df(block) – df(variety)

= 15 – 3 – 3 = 9 Or

df (error) = (v-1)(r-1) = (4-1)(4-1) = 9

2. Correction Factor, CF

CF = Y..2/rv = (928)2/ (4×4) = 53824


SST = ƩYij2 – CF

= (572 + 672 + ... + 472 + 622) – 53824

= 54678 – 53824 = 854

4. Sum Square Block (SSB)

SSB = (Ʃy.j2/v) – CF

= (2302 + 2422 + 2092 + 2472)/ 4 – 53824

= 54038.5 – 53824 = 214.5

43


26


MSV = SSV/dfv = 208/3 = 69.3


MSE = SSE/dfe = 646/ 12 = 53.8

8. F value Variety

F = MSV/MSE = 69.3/53.8 = 1.29

9. F Table

dfv = 3, dfe = 12

At P = 0.05, F = 3.49

10. Conclusion

1.29 < 3.49 → accept HO, no significant difference of yield between the

varieties.

b. Randomized Complete Block Design

Arrange data according to treatments and replications.

Variety R1 R2 R3 R4 Total Mean

V1 57 67 50 54 228 57

V2 57 69 53 57 236 59

V3 65 54 59 74 252 63

V4 51 52 47 62 212 53

Total 230 242 209 247 928

Mean 57.50 60.50 52.25 61.75


27

Calculation:

HO: no significant difference of yield between varieties.

Table of ANOVA


Variation

(p=0.05)

Block (Rep) 3 214.5 71.5 1.49 3.86

Variety 3 208 69.3 1.45 3.86

Error 9 431.5 47.94

Total 15 854

Calculation

1. Degree of Freedom (df)

df (total) = vr – 1 = 4(4) – 1 = 15

df (block) = r – 1 = 4 – 1 = 3

df (variety) = v – 1 = 4 – 1 = 3

df (error) = df (total) – df(block) – df(variety)

= 15 – 3 – 3 = 9 Or

df (error) = (v-1)(r-1) = (4-1)(4-1) = 9


CF = Y..2/rv = (928)2/ (4×4) = 53824


SST = ƩYij2 – CF

= (572 + 672 + ... + 472 + 622) – 53824

= 54678 – 53824 = 854

4. Sum Square Block (SSB)

SSB = (Ʃy.j2/v) – CF

= (2302 + 2422 + 2092 + 2472)/ 4 – 53824

= 54038.5 – 53824 = 214.5

44


28


SSV = (ƩYi.2/r) – CF

= (2282 + 2362 + 2522 + 2122)/4 – 53824

= 54032 – 53824 = 208


SSE = SST – SSB – SSV

= 854 – 214.5 – 208 = 431.5

7. Mean Square Blok (MSB)

MSB = SSB/dfb = 214.5/3 = 71.5


MSV = SSV/dfv = 208/3 = 69.3


MSE = SSE/dfe = 431.5/9 = 47.94

10. F value

F value (block) = MSB/MSE = 71.5/47.94 = 1.49

F (variety) = MSV/MSE = 69.3/47.94 = 1.45

11. F Table

Block: dfb = 3, dfe = 9, at p = 0.05, F = 3.86

Variety: dfv = 3, dfe = 9, at p = 0.05, F = 8.91

12. Conclusion

Variety: 1.45 < 3.86 → accept HO, there is no significant different between varieties

on yield.

Block: 1.49 < 3.86 → accept HO, there is no significant effect of block on the yield.


29

c. Latin Square Design 1. Arrange the data according to treatment as the arrangement in the

experiment whereby the treatment occur once in the column and once in the

row.

ROW COLUMN

1 2 3 4 5 6 Total

1 A(32.1) B(33.1) C(32.4) D(29.1) E(31.1) F(28.2) 186.0

2 F(24.8) A(30.6) B(29.5) C(29.4) D(33.0) E(31.0) 178.3

3 E(28.8) F(21.7) A(31.9) B(30.1) C(30.8) D(30.6) 173.9

4 D(31.4) E(31.9) F(26.7) A(30.4) B(28.8) C(33.1) 182.3

5 C(33.5) D(32.3) E(30.3) F(25.8) A(30.3) B(30.7) 182.9

6 B(30.7) C(29.7) D(27.4) E(29.1) F(21.4) A(30.8) 169.10

Total 181.3 179.3 178.2 173.9 175.4 184.40

2. Calculate ANOVA.

Source of

Variation df SS MS F F(0.05)

Row 5 33.20 6.640 2.63 2.71

Column 5 12.29 2.458 0.98 2.71

Treatment 5 186.78 37.356 14.81 2.71

Error 20 50.43 2.521

Total 35 282.70

3. Degree Freedom (df)

dfT (total) = rc -1 = 6(6)-1 = 35

dfR (row) = r – 1 = 6-1 = 5

dfC (column) = c -1 = 6-1 = 5

dfV (treatment) = b -1 = 6-1 = 5

dfE (error) = DfT – DfR – DfC – DfB = 20

or = (r-1)(c-1) – (v-1) = (5)(5) – 5 = 20

45


28


SSV = (ƩYi.2/r) – CF

= (2282 + 2362 + 2522 + 2122)/4 – 53824

= 54032 – 53824 = 208


SSE = SST – SSB – SSV

= 854 – 214.5 – 208 = 431.5

7. Mean Square Blok (MSB)

MSB = SSB/dfb = 214.5/3 = 71.5


MSV = SSV/dfv = 208/3 = 69.3


MSE = SSE/dfe = 431.5/9 = 47.94

10. F value

F value (block) = MSB/MSE = 71.5/47.94 = 1.49

F (variety) = MSV/MSE = 69.3/47.94 = 1.45

11. F Table

Block: dfb = 3, dfe = 9, at p = 0.05, F = 3.86

Variety: dfv = 3, dfe = 9, at p = 0.05, F = 8.91

12. Conclusion

Variety: 1.45 < 3.86 → accept HO, there is no significant different between varieties

on yield.

Block: 1.49 < 3.86 → accept HO, there is no significant effect of block on the yield.


29

c. Latin Square Design 1. Arrange the data according to treatment as the arrangement in the

experiment whereby the treatment occur once in the column and once in the

row.

ROW COLUMN

1 2 3 4 5 6 Total

1 A(32.1) B(33.1) C(32.4) D(29.1) E(31.1) F(28.2) 186.0

2 F(24.8) A(30.6) B(29.5) C(29.4) D(33.0) E(31.0) 178.3

3 E(28.8) F(21.7) A(31.9) B(30.1) C(30.8) D(30.6) 173.9

4 D(31.4) E(31.9) F(26.7) A(30.4) B(28.8) C(33.1) 182.3

5 C(33.5) D(32.3) E(30.3) F(25.8) A(30.3) B(30.7) 182.9

6 B(30.7) C(29.7) D(27.4) E(29.1) F(21.4) A(30.8) 169.10

Total 181.3 179.3 178.2 173.9 175.4 184.40

2. Calculate ANOVA.

Source of

Variation df SS MS F F(0.05)

Row 5 33.20 6.640 2.63 2.71

Column 5 12.29 2.458 0.98 2.71

Treatment 5 186.78 37.356 14.81 2.71

Error 20 50.43 2.521

Total 35 282.70

3. Degree Freedom (df)

dfT (total) = rc -1 = 6(6)-1 = 35

dfR (row) = r – 1 = 6-1 = 5

dfC (column) = c -1 = 6-1 = 5

dfV (treatment) = b -1 = 6-1 = 5

dfE (error) = DfT – DfR – DfC – DfB = 20

or = (r-1)(c-1) – (v-1) = (5)(5) – 5 = 20

46


30


CF = Y...2 / rc

= (1072.5)2 / 6(6)

= 31951.56

5. Sum of Square (SS) and Mean Square (MS)

Total SST = ƩY2… – CF

= 32.12 + 33.12 + .... + 21.42 + 30.82 – 31951.56

= 282.70

Row SSR = Ʃyi..2/c – CF

= (186.02 +.... + 169.12)/6 – 31903.91

= 33..20

MSR = SSR/DfR = 33.20/5

= 6.64

Column SSC = Ʃy.j.2/r – CF

= (181.32 + ... + 184.42)/6 – 31951.56

= 12.29

MSC = SSC/DfC = 12.29/5

= 2.458

Treatment (V)

SSV = Ʃy..K2/rep – CF

= (186.12 + ... + 148.62)/6 – 31951.56

= 186.78

MSB = SSV/DfV = 186.78/5

= 37.356

Error SSE = SST – SSR – SSC – SSV

= 282.70 – 33.20 – 12.29 – 186.78

= 50.43

MSE = SSE/DFE = 50.43/20

= 2.521


31

6. F value

F (row) = MSR/MSE = 6.640/2.521 = 2.63

F (Column) = MSC/MSE = 2.458/2.521 = 0.98

F (Treatment) = MSV/MSE = 37.356/2.521 =14.81

7. Table F value

Rows: dfR = 5, dfE = 20, p = 0.05 F = 2.71

Column:dfC = 5, dfE = 20, p = 0.05 F = 2.71

Treatment:dfV = 5, dfE = 20, p = 0.05 F = 2.71

Conclusion Treatment: F (14.81) > F table (2.71) Reject HO, there is at least one

significant difference between the treatments.

Row: F value (2.63) < F table (2.71) accept HO, there is no significant

difference between the rows.

Column: F value (0.98) < F table (2.71) accept HO, there is no significant

different between columns.

Laboratory Exercise


LMS or email.

47


30


CF = Y...2 / rc

= (1072.5)2 / 6(6)

= 31951.56

5. Sum of Square (SS) and Mean Square (MS)

Total SST = ƩY2… – CF

= 32.12 + 33.12 + .... + 21.42 + 30.82 – 31951.56

= 282.70

Row SSR = Ʃyi..2/c – CF

= (186.02 +.... + 169.12)/6 – 31903.91

= 33..20

MSR = SSR/DfR = 33.20/5

= 6.64

Column SSC = Ʃy.j.2/r – CF

= (181.32 + ... + 184.42)/6 – 31951.56

= 12.29

MSC = SSC/DfC = 12.29/5

= 2.458

Treatment (V)

SSV = Ʃy..K2/rep – CF

= (186.12 + ... + 148.62)/6 – 31951.56

= 186.78

MSB = SSV/DfV = 186.78/5

= 37.356

Error SSE = SST – SSR – SSC – SSV

= 282.70 – 33.20 – 12.29 – 186.78

= 50.43

MSE = SSE/DFE = 50.43/20

= 2.521


31

6. F value

F (row) = MSR/MSE = 6.640/2.521 = 2.63

F (Column) = MSC/MSE = 2.458/2.521 = 0.98

F (Treatment) = MSV/MSE = 37.356/2.521 =14.81

7. Table F value

Rows: dfR = 5, dfE = 20, p = 0.05 F = 2.71

Column:dfC = 5, dfE = 20, p = 0.05 F = 2.71

Treatment:dfV = 5, dfE = 20, p = 0.05 F = 2.71

Conclusion Treatment: F (14.81) > F table (2.71) Reject HO, there is at least one

significant difference between the treatments.

Row: F value (2.63) < F table (2.71) accept HO, there is no significant

difference between the rows.

Column: F value (0.98) < F table (2.71) accept HO, there is no significant

different between columns.

Laboratory Exercise


LMS or email.

48


32

UNIT 4

COMPARISON OF TREATMENT MEANS

Introduction

Comparison of means is conducted when the null hypothesis (HO ) is being rejected

during the process of ANOVA. When HO is rejected, there is at least one significant

difference between the treatment means. There are various methods to compare for

significant difference between the treatments means. The means of more than two

means are often compared for significant difference using Least Significant

Difference (LSD) test, Duncan’s New Multiple Range (DMRT) test, Tukey’s test,

Scheffe’s test, Student –Newman-Keul’s test (SNK), Dunnett’s test and Contrast.

However, more often than not, such tests are misused. One of the main reasons for

this is the lack of clear understanding of what pair and group comparisons as well as

what the structure of treatments under investigation are. There are two types of pair

comparison namely planned and unplanned pair.

Objective To compare between the treatment means after rejecting the HO from ANOVA.


33

TOPIC 1: Least Significant Difference (LSD)

Important Content It is a t-test and usually suitable to compare between two means.

Example:

Source of df SS MS F Ftable

Variation

(p=0.05)

Block (Rep) 3 576 192 24.7 3.86

Variety 3 208 69.3 8.9 3.86

Error 9 70 7.78

Total 15 854

Arrange the means from low to high or from high to low

Variety V4 V1 V2 V3

Mean 53 57 59 63

Calculation

T = (d - µd) / sd

D = Y1 – Y2

Assume the mean are from the same population, so µd = 0

t = d/sd

t = LSD/ sd

LSD = t. sd

sd = √2MSE/r

sd = √2(7.78) /4 = 1.972

obtain t value from table df = dfe = 9, p = 0.05

t = 2.262

LSD = 2.262 × 1.972 = 4.46 t ha-1

49


32

UNIT 4

COMPARISON OF TREATMENT MEANS

Introduction

Comparison of means is conducted when the null hypothesis (HO ) is being rejected

during the process of ANOVA. When HO is rejected, there is at least one significant

difference between the treatment means. There are various methods to compare for

significant difference between the treatments means. The means of more than two

means are often compared for significant difference using Least Significant

Difference (LSD) test, Duncan’s New Multiple Range (DMRT) test, Tukey’s test,

Scheffe’s test, Student –Newman-Keul’s test (SNK), Dunnett’s test and Contrast.

However, more often than not, such tests are misused. One of the main reasons for

this is the lack of clear understanding of what pair and group comparisons as well as

what the structure of treatments under investigation are. There are two types of pair

comparison namely planned and unplanned pair.

Objective To compare between the treatment means after rejecting the HO from ANOVA.


33

TOPIC 1: Least Significant Difference (LSD)

Important Content It is a t-test and usually suitable to compare between two means.

Example:

Source of df SS MS F Ftable

Variation

(p=0.05)

Block (Rep) 3 576 192 24.7 3.86

Variety 3 208 69.3 8.9 3.86

Error 9 70 7.78

Total 15 854

Arrange the means from low to high or from high to low

Variety V4 V1 V2 V3

Mean 53 57 59 63

Calculation

T = (d - µd) / sd

D = Y1 – Y2

Assume the mean are from the same population, so µd = 0

t = d/sd

t = LSD/ sd

LSD = t. sd

sd = √2MSE/r

sd = √2(7.78) /4 = 1.972

obtain t value from table df = dfe = 9, p = 0.05

t = 2.262

LSD = 2.262 × 1.972 = 4.46 t ha-1

50


34

Compare the difference of two means and compare with the LSD value,

Higher than LSD value → significant different

Lower than LSD value →no significant difference

V3 – V2 = 63 - 59 = 4, < 4.46 → no significant difference

V3 – V1 = 63 – 57 = 6, > 4.46 → significant different

V3 – V4 = 63 – 53 =10, >4.46 → significant different

V2 – V1 = 59 – 57 = 2, <4.46 → no significant difference

V2 – V4 = 59 – 53 = 6, >4.46 → significant different

V1 – V4 = 57 – 53 = 4, < 4.46 → no significant difference

or can be present as the following:

Variety Yield (t ha-1)

V3 63 a

V2 59 ab

V1 57 bc

V4 53 c

LSD0.05 4.46

Laboratory Exercise


LMS or email.


35

TOPIC 2: DUNCAN’S MULTIPLE RANGE TEST

Important Content To compare between treatments means for multiple comparison.

Calculation

1. Calculate LSD value

LSD0.05 = t √2MSE/r = 2.262√2(7.78)/4 = 4.46

2. Calculate D value

D = R(LSD)

R from table, that is up to 4 levels of comparison

Level of Comparison 2 3 4

R0.05 value 1 1.04 1.07

D = R(LSD) 4.46 4.64 4.77

2. Arrange the means from small to high

V4 V1 V2 V3

53 57 59 63

a

b

c

Compare the means of the highest and the lowest first, the difference is compared

with D. The value of D will depend on the level of the means

Example:

V3 – V4, compare with D at level 4 -1 + 1 =4, D = 4.77


V3 vs V4 = 63-53 = 10 > 4.77 → significant different


V3 vs V2 = 63-59 = 4 < 4.46 → no significant different

51


34

Compare the difference of two means and compare with the LSD value,

Higher than LSD value → significant different

Lower than LSD value →no significant difference

V3 – V2 = 63 - 59 = 4, < 4.46 → no significant difference

V3 – V1 = 63 – 57 = 6, > 4.46 → significant different

V3 – V4 = 63 – 53 =10, >4.46 → significant different

V2 – V1 = 59 – 57 = 2, <4.46 → no significant difference

V2 – V4 = 59 – 53 = 6, >4.46 → significant different

V1 – V4 = 57 – 53 = 4, < 4.46 → no significant difference

or can be present as the following:


V3 63 a

V2 59 ab

V1 57 bc

V4 53 c

LSD0.05 4.46

Laboratory Exercise


LMS or email.


35

TOPIC 2: DUNCAN’S MULTIPLE RANGE TEST

Important Content To compare between treatments means for multiple comparison.

Calculation

1. Calculate LSD value

LSD0.05 = t √2MSE/r = 2.262√2(7.78)/4 = 4.46

2. Calculate D value

D = R(LSD)

R from table, that is up to 4 levels of comparison

Level of Comparison 2 3 4

R0.05 value 1 1.04 1.07

D = R(LSD) 4.46 4.64 4.77

2. Arrange the means from small to high

V4 V1 V2 V3

53 57 59 63

a

b

c

Compare the means of the highest and the lowest first, the difference is compared

with D. The value of D will depend on the level of the means

Example:





V3 vs V2 = 63-59 = 4 < 4.46 → no significant different

52


36

Use the same letter for non-significant difference between the means and then

continues in comparing between the second highest mean with the lowest and so on.

Present the result as the following:


V3 63 a

V2 59 ab

V1 57 bc

V4 53 c

Means with the same letter is not significantly different according to DNMRT at p=

0.05.

Laboratory Exercise


LMS or email.

TOPIC 3: CONTRAST

Important Content

One of the methods of comparison between treatments means and also between

groups of means with df value of 1.


37

Example 4 varieties of sweet potato are tested using a RCBD design with 4

replications.

Variety Type of plant Leaf

V1 Creeper (M) Wide (L)

V2 M Small (K)

V3 Cluster (R) L

V4 R K

Results

Variety Rep 1 Rep 2 Rep 3 Rep 4 Total

V1 47 52 62 51 212

V2 50 64 67 57 228

V3 57 53 69 57 236

V4 54 65 74 59 252

Total

928

Table of ANOVA

Source of

Variation df SS MS F Ftable

Replication 3 576 192 24.7 3.86

Variety 3 208 69.3 8.9 3.86

Error 9 70 7.78

Total 15 854

Comparison

1. Is creeping variety and cluster variety produced the same yield?

2. Are variety with wide and small leaves produced the same yield?

3. Is variety with wide leaf suitable with creeping than variety with small leaf and

cluster?

I

This can be answered with orthogonal contrast, how?

1. Create table for orthogonal coefficient (c)

53


36

Use the same letter for non-significant difference between the means and then

continues in comparing between the second highest mean with the lowest and so on.

Present the result as the following:


V3 63 a

V2 59 ab

V1 57 bc

V4 53 c

Means with the same letter is not significantly different according to DNMRT at p=

0.05.

Laboratory Exercise


LMS or email.

TOPIC 3: CONTRAST

Important Content

One of the methods of comparison between treatments means and also between

groups of means with df value of 1.


37

Example 4 varieties of sweet potato are tested using a RCBD design with 4

replications.

Variety Type of plant Leaf

V1 Creeper (M) Wide (L)

V2 M Small (K)

V3 Cluster (R) L

V4 R K

Results

Variety Rep 1 Rep 2 Rep 3 Rep 4 Total

V1 47 52 62 51 212

V2 50 64 67 57 228

V3 57 53 69 57 236

V4 54 65 74 59 252

Total

928

Table of ANOVA

Source of

Variation df SS MS F Ftable

Replication 3 576 192 24.7 3.86

Variety 3 208 69.3 8.9 3.86

Error 9 70 7.78

Total 15 854

Comparison

1. Is creeping variety and cluster variety produced the same yield?

2. Are variety with wide and small leaves produced the same yield?

3. Is variety with wide leaf suitable with creeping than variety with small leaf and

cluster?

I

This can be answered with orthogonal contrast, how?

1. Create table for orthogonal coefficient (c)

54


38

Comparison V1 V2 V3 V4

ML MK RL RK

Creeping vs cluster +1 +1 -1 -1

Wide vs Small leaf +1 -1 +1 -1

(M vs R) (L vs K) +1 -1 +1 -1

Comparison of ‘orthogonal coefficient’ must fulfill following principles:

a) Total coefficient (c) for each comparison must be zero (0)

Example: M vs R: +1 +1 -1 -1 = 0

b) Result of multiply between the two comparison must be zero = 0

Example: M vs R and L vs K

= [+1(+1)] + [+1(-1)] + [-1(+1)] + [-1(-1)]

= +1-1-1+1 = 0

2. Calculate ANOVA

Source of Variation Df

Replication 3

Variety 3

Creeping vs Cluster 1

Wide vs small 1

(M vs R) (L vs K) 1

Error 9

Total 15

a) Sum of Square (SS) and Mean Square (MS)

Type of plant (SSP): Creeping vs Cluster SSP = (∑ci Yi.)2

r ∑ci2

= [1(212) + 1(228) – 1(236) – 1(252)]2 4[(+1)2 + (+1)2 + (-1)2 + (-1)2]

= (-48)2 = 144, or 4(4)


39

SSP = [(212 + 228)2 + (236 + 252)2 – CF 4(2)

CF = (928)2/4(4) = 53824 = 144

MSP = SSP/dfP = 144/1 = 144

Size of leaf (SSL): Wide vs Small

SSD = (∑ci Yi.)2

r ∑ci2

= [1(212) - 1(228) + 1(236) – 1(252)]2

4[(+1)2 + (-1)2 + (+1)2 + (-1)2]

= 64, Or

= (212 + 236)2 + (228 + 252)2 – CF

4(2)

= 64

MSD = SSD/dfD = 64/1 = 64

Interaction between type of plant and size of leaf (SSI): (M vs R)(L vs K) SSI = (∑ci Yi.)2 r ∑ci2

= [1(212) - 1(228) – 1(236) + 1(252)]2 4[(+1)2 + (-1)2 + (-1)2 + (+1)2] = 0, or = (212 + 252)2 + (228 + 236)2 – CF 4(2) = 0 MSI = SSI/dfI = 0/1 = 0 F value

Type of plant: F = MSP/MSE = 144/7.78 = 18.5

Size of leaf: F = MSD/MSE = 64/7.78 = 8.2

Interaction: F = MSI/ dfI = 0/1 = 0

55


38

Comparison V1 V2 V3 V4

ML MK RL RK

Creeping vs cluster +1 +1 -1 -1

Wide vs Small leaf +1 -1 +1 -1

(M vs R) (L vs K) +1 -1 +1 -1

Comparison of ‘orthogonal coefficient’ must fulfill following principles:

a) Total coefficient (c) for each comparison must be zero (0)

Example: M vs R: +1 +1 -1 -1 = 0

b) Result of multiply between the two comparison must be zero = 0

Example: M vs R and L vs K

= [+1(+1)] + [+1(-1)] + [-1(+1)] + [-1(-1)]

= +1-1-1+1 = 0

2. Calculate ANOVA

Source of Variation Df

Replication 3

Variety 3

Creeping vs Cluster 1

Wide vs small 1

(M vs R) (L vs K) 1

Error 9

Total 15

a) Sum of Square (SS) and Mean Square (MS)

Type of plant (SSP): Creeping vs Cluster SSP = (∑ci Yi.)2

r ∑ci2

= [1(212) + 1(228) – 1(236) – 1(252)]2 4[(+1)2 + (+1)2 + (-1)2 + (-1)2]

= (-48)2 = 144, or 4(4)


39

SSP = [(212 + 228)2 + (236 + 252)2 – CF 4(2)

CF = (928)2/4(4) = 53824 = 144

MSP = SSP/dfP = 144/1 = 144

Size of leaf (SSL): Wide vs Small

SSD = (∑ci Yi.)2

r ∑ci2

= [1(212) - 1(228) + 1(236) – 1(252)]2

4[(+1)2 + (-1)2 + (+1)2 + (-1)2]

= 64, Or

= (212 + 236)2 + (228 + 252)2 – CF

4(2)

= 64

MSD = SSD/dfD = 64/1 = 64

Interaction between type of plant and size of leaf (SSI): (M vs R)(L vs K) SSI = (∑ci Yi.)2 r ∑ci2

= [1(212) - 1(228) – 1(236) + 1(252)]2 4[(+1)2 + (-1)2 + (-1)2 + (+1)2] = 0, or = (212 + 252)2 + (228 + 236)2 – CF 4(2) = 0 MSI = SSI/dfI = 0/1 = 0 F value

Type of plant: F = MSP/MSE = 144/7.78 = 18.5

Size of leaf: F = MSD/MSE = 64/7.78 = 8.2

Interaction: F = MSI/ dfI = 0/1 = 0

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Source of Variation df SS MS F F(0.05)

Replication 3 576 192.00 24.7 3.86

Variety 3 208 69.30 8.9 3.86

M vs R 1 144 144.00 18.5 5.12

L vs K 1 64 64.00 8.2 5.12

(M vs R)(L vs K) 1 0 0.00 0.0 5.12

Error 9 70 7.78

Total 15

Conclusion

1. There is significant effect (18.5 > 5.12) of type of plant on yield. Variety of

cluster (61) is significantly higher compared to the creeping plant (55).

2. There is significant effect (8.2 > 5.12) of leaf size on yield. Variety with small

leaf produce significant higher yield than with wide leaf.

3. There is no interaction between type of plant and size of leaf on yield.


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41

UNIT 5

DATA TRANSFORMATION

Introduction

Data that are not conformed to normal distribution need to be transformed to

normalize the data. Usually discrete data are required to be transformed so that

various statistical analyses can be carried out.

Objective To transform data using various methods to normalize the data.

TOPIC 1: LOG TRANSFORMATION

Important Content Log transformation is conducted when the variance or standard deviation increase

proportional to the mean that is an increasing standard deviation with a

corresponding increasing mean.

Example :

- Number of insects per plot

- Number of eggs of insect per plant

- Number of leaves per plant

If there is zero (0), convert all the data to lo log (x + 1).

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Source of Variation df SS MS F F(0.05)

Replication 3 576 192.00 24.7 3.86

Variety 3 208 69.30 8.9 3.86

M vs R 1 144 144.00 18.5 5.12

L vs K 1 64 64.00 8.2 5.12

(M vs R)(L vs K) 1 0 0.00 0.0 5.12

Error 9 70 7.78

Total 15

Conclusion

1. There is significant effect (18.5 > 5.12) of type of plant on yield. Variety of

cluster (61) is significantly higher compared to the creeping plant (55).

2. There is significant effect (8.2 > 5.12) of leaf size on yield. Variety with small

leaf produce significant higher yield than with wide leaf.

3. There is no interaction between type of plant and size of leaf on yield.


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41

UNIT 5

DATA TRANSFORMATION

Introduction

Data that are not conformed to normal distribution need to be transformed to

normalize the data. Usually discrete data are required to be transformed so that

various statistical analyses can be carried out.

Objective To transform data using various methods to normalize the data.

TOPIC 1: LOG TRANSFORMATION

Important Content Log transformation is conducted when the variance or standard deviation increase

proportional to the mean that is an increasing standard deviation with a

corresponding increasing mean.

Example :

- Number of insects per plot

- Number of eggs of insect per plant

- Number of leaves per plant

If there is zero (0), convert all the data to lo log (x + 1).

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42


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TOPIC 2 : SQUARE ROOT TRANSFORMATION

Important Content Square root transformation is conducted for low value data or occurrence of a unique

/ weird situation. The square-root transformation is also suitable for percentage with

a range of 0 – 30% and 70 – 100%.

Example:

- Number of plants with diseases

- Number of weed per plot

- Number of infested plants in a plot

If there is zero (0), use square root of (x + 0.5)


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TOPIC 3 : ARCSINE TRANSFORMATION

Important Content

Arcsine transformation is conducted for ratio, data obtained from counting and

data expressed as decimal fractions or percentages.

For percentage:

It is necessary to transform to arcsine when the percentage range is more than 40

(that is between the smallest data and the highest data).

Example:

Smallest data = 2% and highest data = 82%

Difference = 82 – 2 = 80%, so arcsine is necessary

or Criteria 1: If percentage data fall between 30 – 70, no transformation required Criteria 2: If percentages fall between 0 – 30 or 70 – 100, use square root transformation. Criteria 3: If it did not qualifies for criteria 1 and 3, use arcsine When there is zero (0), change to (1/4n) and when there is100 change to [100 – (1/4n)] before changing all the data to arcsine.


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TOPIC 2 : SQUARE ROOT TRANSFORMATION

Important Content Square root transformation is conducted for low value data or occurrence of a unique

/ weird situation. The square-root transformation is also suitable for percentage with

a range of 0 – 30% and 70 – 100%.

Example:

- Number of plants with diseases

- Number of weed per plot

- Number of infested plants in a plot

If there is zero (0), use square root of (x + 0.5)


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TOPIC 3 : ARCSINE TRANSFORMATION

Important Content

Arcsine transformation is conducted for ratio, data obtained from counting and

data expressed as decimal fractions or percentages.

For percentage:

It is necessary to transform to arcsine when the percentage range is more than 40

(that is between the smallest data and the highest data).

Example:

Smallest data = 2% and highest data = 82%

Difference = 82 – 2 = 80%, so arcsine is necessary

or Criteria 1: If percentage data fall between 30 – 70, no transformation required Criteria 2: If percentages fall between 0 – 30 or 70 – 100, use square root transformation. Criteria 3: If it did not qualifies for criteria 1 and 3, use arcsine When there is zero (0), change to (1/4n) and when there is100 change to [100 – (1/4n)] before changing all the data to arcsine.


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44

UNIT 6

NON-PARAMETRIC TEST

Introduction

Even after transformation the data are still not normalized, and then non-parametric

is used for data analysis.

Objective To use various methods of non-parametric test for analyzing the data.

TOPIC 1: ONE SAMPLE SIGN TEST

Important Content Example: 15 random samples of the height (cm) of corn collected after 2 months of

planting.

97.5 95.2 97.3 96.0 93.2

96.8 100.3 97.4 95.3 99.1

96.1 97.6 98.2 98.5 94.9

Test Ho : µ = 98.5

Ha : µ < 98.5

at p = 0.05


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Methodology

1. Assign negative sign (-) for value < 98.5 and plus sign (+) for value > 98.5

and no sign if the value is 98.5

97.5 -

95.2 -

97.3 -

96.0 -

93.2 -

96.8 -

100.3 +

97.4 -

95.3 -

99.1 +

96.1 -

97.6 -

98.2 -

98.5

94.9 -

2. Determine X that is sign with plus sign (+) X = 2

3. From table of ‘Binomial Probabilities’, determine the probabilities X ≤ 2

at n = 14 (number of sign) and P = 0.5, the probability (p)

X ≤ 2 is ( 0 + 0.001 + 0.006) = 0.007 Conclusion

p = 0.007 < 0.05, reject Ho, height of corn is less than 98.5 cm.


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44

UNIT 6

NON-PARAMETRIC TEST

Introduction

Even after transformation the data are still not normalized, and then non-parametric

is used for data analysis.

Objective To use various methods of non-parametric test for analyzing the data.

TOPIC 1: ONE SAMPLE SIGN TEST

Important Content Example: 15 random samples of the height (cm) of corn collected after 2 months of

planting.

97.5 95.2 97.3 96.0 93.2

96.8 100.3 97.4 95.3 99.1

96.1 97.6 98.2 98.5 94.9

Test Ho : µ = 98.5

Ha : µ < 98.5

at p = 0.05


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Methodology

1. Assign negative sign (-) for value < 98.5 and plus sign (+) for value > 98.5

and no sign if the value is 98.5

97.5 -

95.2 -

97.3 -

96.0 -

93.2 -

96.8 -

100.3 +

97.4 -

95.3 -

99.1 +

96.1 -

97.6 -

98.2 -

98.5

94.9 -

2. Determine X that is sign with plus sign (+) X = 2

3. From table of ‘Binomial Probabilities’, determine the probabilities X ≤ 2

at n = 14 (number of sign) and P = 0.5, the probability (p)

X ≤ 2 is ( 0 + 0.001 + 0.006) = 0.007 Conclusion

p = 0.007 < 0.05, reject Ho, height of corn is less than 98.5 cm.


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TOPIC 2 : PAIRED DATA SIGN TEST

Important Content Test for paired data.

Example: number of rhizomes per plant of two varieties planted at 10 locations.

Location Variety A Variety B Sign

1 3 1 +

2 2 0 +

3 4 2 +

4 3 2 +

5 2 3 -

6 6 3 +

7 5 2 +

8 1 2 -

9 3 3

10 1 0 +

Test Ho : number of rhizomes variety A = number of rhizomes variety B

Ha : number of rhizomes variety A > number of rhizomes variety B

at p = 0.05

Methodology

1. Assign (+) when variety A > variety B

Assign (-) when variety A < variety B

No sign when variety A=variety B

2. Determine X, that is number of (+), X = 7

3. Determine

The probability (p) for X ≥ 7

at n = 9 (number of sign), and

P = 0.50, p for X ≥ 7 is (0.070 + 0.018 + 0.002) = 0.090

Conclusion p = 0.090 > 0.05, accept Ho

Variety A is the same as variety B.


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TOPIC 3: WILCOXON-MANN-WHITNEY

Important Content For large samples (n = 10 or more)

Z = x – npo____

√ npo (1- po)

where po is 0.5, for normal distribution.

Example : Number of seeds per fruit of Mengkudu for variety Bosa.

Test Ho : µ = 43

Ha : µ ≠ 43

at p = 0.05

Fruit Number of Seeds Sign

1 46 +

2 48 +

3 52 +

4 39 -

5 47 +

6 46 +

7 52 +

8 41 -

9 54 +

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TOPIC 2 : PAIRED DATA SIGN TEST

Important Content Test for paired data.

Example: number of rhizomes per plant of two varieties planted at 10 locations.

Location Variety A Variety B Sign

1 3 1 +

2 2 0 +

3 4 2 +

4 3 2 +

5 2 3 -

6 6 3 +

7 5 2 +

8 1 2 -

9 3 3

10 1 0 +

Test Ho : number of rhizomes variety A = number of rhizomes variety B

Ha : number of rhizomes variety A > number of rhizomes variety B

at p = 0.05

Methodology

1. Assign (+) when variety A > variety B

Assign (-) when variety A < variety B

No sign when variety A=variety B

2. Determine X, that is number of (+), X = 7

3. Determine

The probability (p) for X ≥ 7

at n = 9 (number of sign), and

P = 0.50, p for X ≥ 7 is (0.070 + 0.018 + 0.002) = 0.090

Conclusion p = 0.090 > 0.05, accept Ho

Variety A is the same as variety B.


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TOPIC 3: WILCOXON-MANN-WHITNEY

Important Content For large samples (n = 10 or more)

Z = x – npo____

√ npo (1- po)

where po is 0.5, for normal distribution.

Example : Number of seeds per fruit of Mengkudu for variety Bosa.

Test Ho : µ = 43

Ha : µ ≠ 43

at p = 0.05

Fruit Number of Seeds Sign

1 46 +

2 48 +

3 52 +

4 39 -

5 47 +

6 46 +

7 52 +

8 41 -

9 54 +

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10 45 +

11 52 +

12 44 +

13 49 +

14 42 -

15 45 +

Methodology At p=0.05, Z = ± 1.96

1. Assign (+) for > 43 and (-) for < 43

2. Determine X = number of (+)

X = 12

3. Calculate Z

Z = x – npo_______

√ npo (1- po)

Z = 12 – 15(0.5)

√ 15 (0.5) (0.5)

= 2.32

Conclusion

2.32 > 1.96. Reject Ho, number seeds is not equal to 43.


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TOPIC 4: CHI SQUARE

Important Content Analysis for data involving number or frequency Example : number of students in a class Male 19 Female 12 Is number of male and female in the ratio of 1:!? Formula X2 =∑ (O – E)2/E where O = observed value; E= expected value Uses: 1. Test of Goodness-of-fit

Hypothesis testing that is not determined by the data but from the ratio 9:3:3:1

Example: Segregation of corn seeds at F2 follows the ratio of 9:3:3:1. Characteristics Expected Observe Expected Ratio Value Value yellow + flint 9 496 450 yellow + sweet 3 158 150 white + flint 3 112 150 white + sweet 1 34 50 Total 16 800 800 Calculation : Expected for yellow + flint = (9/16) x 800 = 450 Expected for yellow + sweet = (3/16) x 800= 150 Expected for white + flint = (3/16) x 800 = 150 Expected for white + sweet = (1/16) x 800 = 50 X2 =∑(O–E)2/E=(496–450)2/450+(158 -150)2/150 +(112 – 150)2/150+(34 – 50)2/50 = 19.88 df = n – 1 = 4 – 1 = 3 X2 table df =3, p = 0.05 is 7.815 19,88 >7.815; reject Ho, segregation corn seed did not follow the ratio of 9:3:3:1.

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10 45 +

11 52 +

12 44 +

13 49 +

14 42 -

15 45 +

Methodology At p=0.05, Z = ± 1.96

1. Assign (+) for > 43 and (-) for < 43

2. Determine X = number of (+)

X = 12

3. Calculate Z

Z = x – npo_______

√ npo (1- po)

Z = 12 – 15(0.5)

√ 15 (0.5) (0.5)

= 2.32

Conclusion

2.32 > 1.96. Reject Ho, number seeds is not equal to 43.


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TOPIC 4: CHI SQUARE

Important Content Analysis for data involving number or frequency Example : number of students in a class Male 19 Female 12 Is number of male and female in the ratio of 1:!? Formula X2 =∑ (O – E)2/E where O = observed value; E= expected value Uses: 1. Test of Goodness-of-fit

Hypothesis testing that is not determined by the data but from the ratio 9:3:3:1

Example: Segregation of corn seeds at F2 follows the ratio of 9:3:3:1. Characteristics Expected Observe Expected Ratio Value Value yellow + flint 9 496 450 yellow + sweet 3 158 150 white + flint 3 112 150 white + sweet 1 34 50 Total 16 800 800 Calculation : Expected for yellow + flint = (9/16) x 800 = 450 Expected for yellow + sweet = (3/16) x 800= 150 Expected for white + flint = (3/16) x 800 = 150 Expected for white + sweet = (1/16) x 800 = 50 X2 =∑(O–E)2/E=(496–450)2/450+(158 -150)2/150 +(112 – 150)2/150+(34 – 50)2/50 = 19.88 df = n – 1 = 4 – 1 = 3 X2 table df =3, p = 0.05 is 7.815 19,88 >7.815; reject Ho, segregation corn seed did not follow the ratio of 9:3:3:1.

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2. Test of Independence To determine the relationship between two or more characteristics. X2 =∑(O–E)2/E if df > 1 X2 =∑(|O–E|) – 0.5)2/E if df = 1. Example: Segregation of maturity and deficiency of chlorophyll content.

Ho : no relationship between the two characteristics. Characteristics Chlorophyll deficiency Normal Viersen Total Late O 3470 910 4380 E 3457.9 922.1 Early O 1030 290 1320 E 1041.1 277.9 Total 4500 1200 5700 Example: Calculation of expected value (E) Late Normal = (4380/5700) x 4500 = 3457.9 Late Viersen = (4380/5700) x 1200 = 922.1 Early Normal = (1320/5700) x 4500 = 1042.1 Early Viersen = (1320/5700) x 1200 = 277.9 df = (r -1)(c-1) = 1 X2 =∑(|O–E|) – 0.5)2/E

= (|3470 – 3457.9| – 0.5)2/3457.9 + (|1030 – 1042.1| - 0.5)2/1042.1 + (|910 – 922.1| - 0.5)2/922.1 + (|290 – 277.9| - 0.5)2/277.9 = 0.798 Determine X2 table X2 table , df = 1, p= 0.05 is 3.84


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0.798 < 3.84, accept Ho : Maturity and chlorophyll deficiency did not influence each other. Segregation occurs independently 3. Test of heterogeneity

- To determine whether the samples are homogeneous and coming from the

same population. Example: Segregation of colour characteristics of a legume seed for 8 progeny. Progeny df Yellow Green X2 (3:1) Compare X2 table 1 1 315 85 3.00 (follow 3:1) 2 1 602 170 3.65 (follow 3:1) 3 1 868 252 3.73 (follow 3:1) 4 1 174 42 3.56 (follow 3:1) 5 1 192 48 3.20 (follow 3:1) 6 1 165 39 3.76 (follow 3:1) 7 1 161 43 1.67 (follow 3:1) 8 1 629 175 4.48 (did not follow 3:1) Total 8 27.05 (did not follow 3:1) Pooled 1 3106 854 24.91 (did not follow 3:1) Heterogeneity 7 2.14 homogenous Ho : all progeny are homogenous. X2 table (df = 7, p = 0.05) is 14.07 2.14 > 14.07, accept Ho : samples are homogenous and coming from the same

population with segregation ratio of 3:1.

So, the best method to determine whether the progeny following the ratio is to use

the pooled data (total plants for all progenies), that is:

24.91> 3.84 so segregation of progeny does not follow the ratio 3:1.

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2. Test of Independence To determine the relationship between two or more characteristics. X2 =∑(O–E)2/E if df > 1 X2 =∑(|O–E|) – 0.5)2/E if df = 1. Example: Segregation of maturity and deficiency of chlorophyll content.

Ho : no relationship between the two characteristics. Characteristics Chlorophyll deficiency Normal Viersen Total Late O 3470 910 4380 E 3457.9 922.1 Early O 1030 290 1320 E 1041.1 277.9 Total 4500 1200 5700 Example: Calculation of expected value (E) Late Normal = (4380/5700) x 4500 = 3457.9 Late Viersen = (4380/5700) x 1200 = 922.1 Early Normal = (1320/5700) x 4500 = 1042.1 Early Viersen = (1320/5700) x 1200 = 277.9 df = (r -1)(c-1) = 1 X2 =∑(|O–E|) – 0.5)2/E

= (|3470 – 3457.9| – 0.5)2/3457.9 + (|1030 – 1042.1| - 0.5)2/1042.1 + (|910 – 922.1| - 0.5)2/922.1 + (|290 – 277.9| - 0.5)2/277.9 = 0.798 Determine X2 table X2 table , df = 1, p= 0.05 is 3.84


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0.798 < 3.84, accept Ho : Maturity and chlorophyll deficiency did not influence each other. Segregation occurs independently 3. Test of heterogeneity

- To determine whether the samples are homogeneous and coming from the

same population. Example: Segregation of colour characteristics of a legume seed for 8 progeny. Progeny df Yellow Green X2 (3:1) Compare X2 table 1 1 315 85 3.00 (follow 3:1) 2 1 602 170 3.65 (follow 3:1) 3 1 868 252 3.73 (follow 3:1) 4 1 174 42 3.56 (follow 3:1) 5 1 192 48 3.20 (follow 3:1) 6 1 165 39 3.76 (follow 3:1) 7 1 161 43 1.67 (follow 3:1) 8 1 629 175 4.48 (did not follow 3:1) Total 8 27.05 (did not follow 3:1) Pooled 1 3106 854 24.91 (did not follow 3:1) Heterogeneity 7 2.14 homogenous Ho : all progeny are homogenous. X2 table (df = 7, p = 0.05) is 14.07 2.14 > 14.07, accept Ho : samples are homogenous and coming from the same

population with segregation ratio of 3:1.

So, the best method to determine whether the progeny following the ratio is to use

the pooled data (total plants for all progenies), that is:

24.91> 3.84 so segregation of progeny does not follow the ratio 3:1.

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Example 2. Segregation of color characteristics of corn seeds for 8 progeny. Progeny df Yellow Green X2 (3:1) Compare X2 table 1 1 285 115 3.00 (follow 3:1) 2 1 556 216 3.65 (follow 3:1) 3 1 812 308 3.73 (follow 3:1) 4 1 150 66 3.56 (follow 3:1) 5 1 192 48 3.20 (follow 3:1) 6 1 165 39 3.76 (follow 3:1) 7 1 161 43 1.67 (follow 3:1) 8 1 629 175 4.48 (did not follow 3:1) Total 8 27.05 (did not follow 3:1) Pooled 1 2950 1010 0.54 (follow 3:1) Heterogeneity 7 26.51 heterogeneous Ho : all progeny are homogenous X2 table (df = 7, p = 0.05) is 14.07 26.51 > 14.07, reject Ho, sample are heterogeneous So to determine whether there is segregation of progeny following the ratio 3:1, pooled data (total), cannot be used. 24.91> 3.84 so segregation of progeny does not follow the ratio 3:1. using pooled data (total of three for all progeny) is the best way to evaluate whether progeny is according the ratio or not. Separate analysis is needed for each progeny. All progenies follow the ratio 3:1 accept progeny 8.


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53

UNIT 7

FACTORIAL EXPERIMENT

Introduction Factorial experiment is conducted for more than one factor with the intention to

check not only the effect of each factor but whether there is interaction or not among

the factors. It is one in which the treatments consists of all possible combinations of

the selected levels of two or more factors.

Objective

To observe for the main and interaction effects between the factors

TOPIC 1: EFFECT OF MAIN FACTOR

Important Content

An experiment which consists of more than one factor where each factor

contain various levels

Example : Factorial experiment 2 x 2

- factor 1 = variety ==> two levels: variety A1, variety A2

- factor 2 = fertilizer ==> two levels: 60 kg ha-1 (B1), 100kg ha-1 (B2)

- effect of various factor can be evaluated simultaneously

- interaction between factors can be determined

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Example 2. Segregation of color characteristics of corn seeds for 8 progeny. Progeny df Yellow Green X2 (3:1) Compare X2 table 1 1 285 115 3.00 (follow 3:1) 2 1 556 216 3.65 (follow 3:1) 3 1 812 308 3.73 (follow 3:1) 4 1 150 66 3.56 (follow 3:1) 5 1 192 48 3.20 (follow 3:1) 6 1 165 39 3.76 (follow 3:1) 7 1 161 43 1.67 (follow 3:1) 8 1 629 175 4.48 (did not follow 3:1) Total 8 27.05 (did not follow 3:1) Pooled 1 2950 1010 0.54 (follow 3:1) Heterogeneity 7 26.51 heterogeneous Ho : all progeny are homogenous X2 table (df = 7, p = 0.05) is 14.07 26.51 > 14.07, reject Ho, sample are heterogeneous So to determine whether there is segregation of progeny following the ratio 3:1, pooled data (total), cannot be used. 24.91> 3.84 so segregation of progeny does not follow the ratio 3:1. using pooled data (total of three for all progeny) is the best way to evaluate whether progeny is according the ratio or not. Separate analysis is needed for each progeny. All progenies follow the ratio 3:1 accept progeny 8.


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53

UNIT 7

FACTORIAL EXPERIMENT

Introduction Factorial experiment is conducted for more than one factor with the intention to

check not only the effect of each factor but whether there is interaction or not among

the factors. It is one in which the treatments consists of all possible combinations of

the selected levels of two or more factors.

Objective

To observe for the main and interaction effects between the factors

TOPIC 1: EFFECT OF MAIN FACTOR

Important Content

An experiment which consists of more than one factor where each factor

contain various levels

Example : Factorial experiment 2 x 2

- factor 1 = variety ==> two levels: variety A1, variety A2

- factor 2 = fertilizer ==> two levels: 60 kg ha-1 (B1), 100kg ha-1 (B2)

- effect of various factor can be evaluated simultaneously

- interaction between factors can be determined

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Treatment combination: between levels of factors

Example: 2 x 2 = 4 combination that is A1B1, A1B2, A2B1, A2B2

Treatment effects

.. Factor A

B Level a1 a2

b1 a1b1 a2b1

b2 a1b2 a2b2

Simple effect

.. Effect of A (a2 – a1) at each level of B (b1 or b2) and

.. Effect of B (b2 - b1) at each level of A (a1 or a2)

Main effect

.. A = ½ [(a2b2 – a1b2) + (a2b1 – a1b1)]

= ½ [(a2b2 + a2b1) – (a1b2 + a1b1)]

B = ½ [(a2b2 – a2b1) + (a1b2 – a1b1)]

= ½ [(a2b2 + a1b2) – (a2b1 + a1b1)]


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TOPIC 2: INTERACTION EFFECT

Important Content Interaction effect

.. differences of a factor at different levels of different factor

AB = ½ [(a2b2 – a1b2) – (a2b1 – a1b1)]

= ½ [(a2b2 + a1b1) – (a1b2 + a2b1)]


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Example: Factorial Experiment 2 (a1,a2) X 2 (b1,b2)

Factor A

B Level a1 a2 Mean a2 – a1

b1 30 32 31 2

b2 36 44 40 8

Mean 33 38 35.5 5

b2-b1 6 12 9

Simple effect: A at level b1 = 32 – 30 = 2

B at level a2 = 44 – 32 = 12

Main effect: A = ½ [(44-36) + (32 – 30)] = 5, or

= ½ [(44 + 32) – (36 + 30)] = 5

Interaction effect AB = ½ [(44 – 36) – (32 – 30)] = 3, or

= ½ [(44 + 30) – (36 + 32)] = 3

Interaction

b1

a1 a2

b2

b2

b1

a2

a1

No interaction

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Treatment combination: between levels of factors

Example: 2 x 2 = 4 combination that is A1B1, A1B2, A2B1, A2B2

Treatment effects

.. Factor A

B Level a1 a2

b1 a1b1 a2b1

b2 a1b2 a2b2

Simple effect

.. Effect of A (a2 – a1) at each level of B (b1 or b2) and

.. Effect of B (b2 - b1) at each level of A (a1 or a2)

Main effect

.. A = ½ [(a2b2 – a1b2) + (a2b1 – a1b1)]

= ½ [(a2b2 + a2b1) – (a1b2 + a1b1)]

B = ½ [(a2b2 – a2b1) + (a1b2 – a1b1)]

= ½ [(a2b2 + a1b2) – (a2b1 + a1b1)]


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TOPIC 2: INTERACTION EFFECT

Important Content Interaction effect

.. differences of a factor at different levels of different factor

AB = ½ [(a2b2 – a1b2) – (a2b1 – a1b1)]

= ½ [(a2b2 + a1b1) – (a1b2 + a2b1)]


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Example: Factorial Experiment 2 (a1,a2) X 2 (b1,b2)

Factor A

B Level a1 a2 Mean a2 – a1

b1 30 32 31 2

b2 36 44 40 8

Mean 33 38 35.5 5

b2-b1 6 12 9

Simple effect: A at level b1 = 32 – 30 = 2

B at level a2 = 44 – 32 = 12

Main effect: A = ½ [(44-36) + (32 – 30)] = 5, or

= ½ [(44 + 32) – (36 + 30)] = 5

Interaction effect AB = ½ [(44 – 36) – (32 – 30)] = 3, or

= ½ [(44 + 30) – (36 + 32)] = 3

Interaction

b1

a1 a2

b2

b2

b1

a2

a1

No interaction

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Calculation of ANOVA

Example: Factorial experiment 2 x 2, with 5 replications.

a) CRD

Source of Variation

Df SS MS F

Treatment ab-1 = 2(2) – 1 = 3

A a – 1 = 2 – 1 = 1

B b – 1 = 2 – 1 = 1

AB (a - 1)(b - 1) = (2 – 1)(2 – 1) = 1

a1

a2

Interaction

b2

b1


57

Error ab(r – 1) = 2(2)(5 – 1) = 16

Total rab -1 = 2(2)(5) – 1 = 19

b) RCBD

Source of Variation

Df SS MS F

Replication R – 1 = 5 – 1 = 4

Treatment ab -1 = 2(2) – 1 = 3

A a – 1 = 2 – 1 = 1

B b – 1 = 2 – 1 = 1

AB (a - 1)(b - 1) = (2 – 1)(2 – 1) = 1

Error (ab-1)(r – 1) = (4 – 1)(5 – 1) = 12

Total rab -1 = 2(2)(5) – 1 = 19

Calculation

2 x 2, 5 replications, CRD: Effect of factor A and B on yield of corn.

1) Arrange data following treatment combination.

a1b1 a1b2 a2b1 a2b2 Total

8.53 17.53 39.14 32.00

20.53 21.07 26.20 23.80

12.53 20.80 31.33 28.87

14.00 17.33 45.80 25.06

10.80 20.07 40.20 29.33

Total 66.39 96.80 182.67 139.06 484.92

Mean 13.28 19.36 36.53 27.81 24.25

2) Arrange data following treatments.

Factor A Total Mean

a1 a2

73


56


Example: Factorial experiment 2 x 2, with 5 replications.

a) CRD

Source of Variation

Df SS MS F

Treatment ab-1 = 2(2) – 1 = 3

A a – 1 = 2 – 1 = 1

B b – 1 = 2 – 1 = 1

AB (a - 1)(b - 1) = (2 – 1)(2 – 1) = 1

a1

a2

Interaction

b2

b1


57

Error ab(r – 1) = 2(2)(5 – 1) = 16

Total rab -1 = 2(2)(5) – 1 = 19

b) RCBD

Source of Variation

Df SS MS F

Replication R – 1 = 5 – 1 = 4

Treatment ab -1 = 2(2) – 1 = 3

A a – 1 = 2 – 1 = 1

B b – 1 = 2 – 1 = 1

AB (a - 1)(b - 1) = (2 – 1)(2 – 1) = 1

Error (ab-1)(r – 1) = (4 – 1)(5 – 1) = 12

Total rab -1 = 2(2)(5) – 1 = 19

Calculation

2 x 2, 5 replications, CRD: Effect of factor A and B on yield of corn.

1) Arrange data following treatment combination.

a1b1 a1b2 a2b1 a2b2 Total

8.53 17.53 39.14 32.00

20.53 21.07 26.20 23.80

12.53 20.80 31.33 28.87

14.00 17.33 45.80 25.06

10.80 20.07 40.20 29.33

Total 66.39 96.80 182.67 139.06 484.92

Mean 13.28 19.36 36.53 27.81 24.25

2) Arrange data following treatments.

Factor A Total Mean

a1 a2

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58

B b1 66.39 182.67 249.06 24.9

b2 96.80 139.06 235.86 23.6

Total 163.19 321.73 484.92

Mean

16.3 32.2

3) Calculate SS and MS

CF = (484.92)2 / 2(2)(5) = 11,757.37

SS(Total) = 8.532 + 20.532 + … + 25.062 + 29.332 – CF = 1,919.33

SS(Treatment) = 66.392 + 96.802 + 182.672 + 139.062 - CF = 1,539.41

5

SS(A) = 163.192 + 321.732 – CF

rb

= 163.192 + 321.732 – 11,757.37

5(2)

= 1,256.75

MS(A) = SSA / dfA

= 1,256.75 / 1

= 1,256.75

SS(B) = 249.062 + 235.862 – CF

ra

= 249.062 + 235.862 – 11,757.37

5(2)

MS(B) = SSB / dfB

= 8.71 / 1 = 8.71

SS (AB) = SSTreatment – SSA – SSB

= 1539.41 – 1256.75 – 8.71

= 273.95

or = (66.39+139.06)2 + (96.80+182.67)2 – CF


59

5(2)

= 12,031.32 – 11,757.37

= 273.95

MS(AB) = SS(AB) / df(AB)

= 273.95 / 1 = 273.95

SS(Error) = SS(total) – SS(treatment)

= 379.92

MS(Error) = SS(error) / df(Error)

= 379.92 / 16 = 23.75

F value, A = MS(A) / MS(Error)

= 1256.75 / 23.75

B = MS(B) / MS(Error)

= 8.71 / 23.75

= 0.4

AB = MS(AB) / MS(Error)

= 273.95 / 23.75

= 11.5

F table: df(n)= 1, df(d)= df(Error)= 16,

p= 0.05 F= 4.49; p= 0.01 F= 4.49

Source of Variation

df SS MS F F table

p= 0.05 p= 0.01

Treatment 3 1539.41

A 1 1256.75 1256.75 52.9** 4.49 8.53

B 1 8.71 8.71 0.4 4.49 8.53

AB 1 273.95 273.95 11.5** 4.49 8.53

Error 16 379.92 23.75

Total 19 1919.33

Conclusion

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58

B b1 66.39 182.67 249.06 24.9

b2 96.80 139.06 235.86 23.6

Total 163.19 321.73 484.92

Mean

16.3 32.2

3) Calculate SS and MS

CF = (484.92)2 / 2(2)(5) = 11,757.37

SS(Total) = 8.532 + 20.532 + … + 25.062 + 29.332 – CF = 1,919.33

SS(Treatment) = 66.392 + 96.802 + 182.672 + 139.062 - CF = 1,539.41

5

SS(A) = 163.192 + 321.732 – CF

rb

= 163.192 + 321.732 – 11,757.37

5(2)

= 1,256.75

MS(A) = SSA / dfA

= 1,256.75 / 1

= 1,256.75

SS(B) = 249.062 + 235.862 – CF

ra

= 249.062 + 235.862 – 11,757.37

5(2)

MS(B) = SSB / dfB

= 8.71 / 1 = 8.71

SS (AB) = SSTreatment – SSA – SSB

= 1539.41 – 1256.75 – 8.71

= 273.95

or = (66.39+139.06)2 + (96.80+182.67)2 – CF


59

5(2)

= 12,031.32 – 11,757.37

= 273.95

MS(AB) = SS(AB) / df(AB)

= 273.95 / 1 = 273.95

SS(Error) = SS(total) – SS(treatment)

= 379.92

MS(Error) = SS(error) / df(Error)

= 379.92 / 16 = 23.75

F value, A = MS(A) / MS(Error)

= 1256.75 / 23.75

B = MS(B) / MS(Error)

= 8.71 / 23.75

= 0.4

AB = MS(AB) / MS(Error)

= 273.95 / 23.75

= 11.5

F table: df(n)= 1, df(d)= df(Error)= 16,

p= 0.05 F= 4.49; p= 0.01 F= 4.49

Source of Variation

df SS MS F F table

p= 0.05 p= 0.01

Treatment 3 1539.41

A 1 1256.75 1256.75 52.9** 4.49 8.53

B 1 8.71 8.71 0.4 4.49 8.53

AB 1 273.95 273.95 11.5** 4.49 8.53

Error 16 379.92 23.75

Total 19 1919.33

Conclusion

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60

1. There significant effect of factor A but not factor B.

2. There is significant interaction between factor A and factor B.

3. Effect of factor A depends on the combination with factor B.

a1 a2 mean

b1 13.28 36.53 24.9

b2 19.36 27.81 23.6

mean 16.3 32.2

Mean Comparison

1. Between treatment within factor A

LSD = t√ [2(MSerror) / rb]

where t is t table for df(Error) at p=0.05

LSD = 2.12 x √ [2(23.7)/(5)(2)] = 4.62

Factor A Mean

a1 16.3

a2 32.2

LSD(0.05) 4.62

2. Mean between treatment combinations

3. LSD = t√ [2(MSerror) / r]

where ‘t’ is t table df(Error) at p=0.05

LSD = 2.12 x √ [2(23.75) / 5]

= 6.53

Treatment Combination Mean


61

A1b1 13.28

A1b2 19.36

A2b1 36.53

A2b2 27.81

LSD(0.05) 6.53


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UNIT 8

EXPERIMENT WITH DIFFERENT SIZES OF EXPERIMENTAL UNITS

Introduction

In a certain situation there are experimental designs to overcome the environmental

error.

Objective

77


60

1. There significant effect of factor A but not factor B.

2. There is significant interaction between factor A and factor B.

3. Effect of factor A depends on the combination with factor B.

a1 a2 mean

b1 13.28 36.53 24.9

b2 19.36 27.81 23.6

mean 16.3 32.2

Mean Comparison

1. Between treatment within factor A

LSD = t√ [2(MSerror) / rb]

where t is t table for df(Error) at p=0.05

LSD = 2.12 x √ [2(23.7)/(5)(2)] = 4.62

Factor A Mean

a1 16.3

a2 32.2

LSD(0.05) 4.62

2. Mean between treatment combinations

3. LSD = t√ [2(MSerror) / r]

where ‘t’ is t table df(Error) at p=0.05

LSD = 2.12 x √ [2(23.75) / 5]

= 6.53

Treatment Combination Mean


61

A1b1 13.28

A1b2 19.36

A2b1 36.53

A2b2 27.81

LSD(0.05) 6.53


LMS or email.

UNIT 8

EXPERIMENT WITH DIFFERENT SIZES OF EXPERIMENTAL UNITS

Introduction

In a certain situation there are experimental designs to overcome the environmental

error.

Objective

78


62

To use different experimental designs with different sizes to reduce environmental

error

Important Content TOPIC 1: SPLIT PLOT DESIGN

For factorial experiment with two factors where the experimental materials do not

allow for the treatment combinations to be arranged in the usual manner.

.. Contains main plot and sub-plot. Sub-plot is arranged within the main plot.

.. First factor is arranged in the main plot and the second factor is arranged in

the sub- plot. Treatments in the main plot and sub-plot are arranged

randomly. Precision of main plot < sub-plot and Error term is separated for

main plot and sub-plot.


63

N0

Rep 3

Main Plot

N120

N0

Rep 2

N120

N0 Rep 1

N120

Example: Factorial experiment of 2 x 4, with 3 replications.

2 rates of N: 0 kg ha-1 (N0) and 120 kg ha-1 (N120)

4 organic fertilizer: BV, V, F, B

F 20.5

B 25.4

BV 27.6

V 28.4

B 15.2

V 22.3

BV 19.6

F 13.2

BV 18.3

B 15.0

F 13.5

V 22.7

B 24.2

V 24.8

BV 26.7

F 18.0

V 21.0

F 13.8

BV 18.9

B 15.5

B 22.2

F 19.3

V 25.3

BV 25.9

Sub Plot

79


62

To use different experimental designs with different sizes to reduce environmental

error

Important Content TOPIC 1: SPLIT PLOT DESIGN

For factorial experiment with two factors where the experimental materials do not

allow for the treatment combinations to be arranged in the usual manner.

.. Contains main plot and sub-plot. Sub-plot is arranged within the main plot.

.. First factor is arranged in the main plot and the second factor is arranged in

the sub- plot. Treatments in the main plot and sub-plot are arranged

randomly. Precision of main plot < sub-plot and Error term is separated for

main plot and sub-plot.


63

N0

Rep 3

Main Plot

N120

N0

Rep 2

N120

N0 Rep 1

N120

Example: Factorial experiment of 2 x 4, with 3 replications.

2 rates of N: 0 kg ha-1 (N0) and 120 kg ha-1 (N120)

4 organic fertilizer: BV, V, F, B

F 20.5

B 25.4

BV 27.6

V 28.4

B 15.2

V 22.3

BV 19.6

F 13.2

BV 18.3

B 15.0

F 13.5

V 22.7

B 24.2

V 24.8

BV 26.7

F 18.0

V 21.0

F 13.8

BV 18.9

B 15.5

B 22.2

F 19.3

V 25.3

BV 25.9

Sub Plot

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64


1. Arrange data according to treatments, main plot and blocks (replications)

Treatment Replication Total Mean

Fertilizer N

Organic

Fertilizer

1 2 3

F 13.8 13.5 13.2 40.5 13.5

No B 15.5 15.0 15.2 45.7 15.2

V 21.0 22.7 22.3 66.0 22.0

BV 18.9 18.3 19.6 56.8 18.9

Total Main Plot (Y1j.)

69.2 69.5 70.3 209.0

=Y1

17.4

N120 F 19.3 18.0 20.5 57.8 19.3

B 22.2 24.2 25.4 71.8 23.9

V 25.3 24.8 28.4 78.5 26.2

BV 25.9 26.7 27.6 80.2 26.7

Total sub-plot

(Y2j.)

92.7 93.7 101.9 288.3 =

Y2.

24.0

Total Replication (Y.j.)

161.9 163.2 172.2 497.3 =

Y...

20.7


65

2. Arrange data following treatments and sub-plot.

Organic Fertilizer

F B V BV

Total (Y..k) 98.3 117.5 144.5 137.0

Mean (Y..k) 16.3 19.6 24.1 22.8

3. Arrange source of variation and degree of freedom (df)


Replication r - 1 = 2

Nitrogen (N) n – 1 = 1

Error for main plot (Error a) (r-1)(n-1) = 2

Organic fertilizer (G) g – 1 = 3

N x G (n - 1)(g – 1) = 3

Error for sub-plot (Error b) (r – 1)(g – 1) + (r – 1)(n – 1)(g – 1)=12

Total ngr-1=23

4. Correction factor (CF)

CF = [Y..]2 / ngr = (497.3)2 / 2(4)(3) = 10,304.47

5. SS and MS

Total SS = Ɛ Yijk2 – CF

= (13.82 + 13.52 +... + 26.72 +27.62) – 10,304.47

= 516.12

81


64


1. Arrange data according to treatments, main plot and blocks (replications)

Treatment Replication Total Mean

Fertilizer N

Organic

Fertilizer

1 2 3

F 13.8 13.5 13.2 40.5 13.5

No B 15.5 15.0 15.2 45.7 15.2

V 21.0 22.7 22.3 66.0 22.0

BV 18.9 18.3 19.6 56.8 18.9

Total Main Plot (Y1j.)

69.2 69.5 70.3 209.0

=Y1

17.4

N120 F 19.3 18.0 20.5 57.8 19.3

B 22.2 24.2 25.4 71.8 23.9

V 25.3 24.8 28.4 78.5 26.2

BV 25.9 26.7 27.6 80.2 26.7

Total sub-plot

(Y2j.)

92.7 93.7 101.9 288.3 =

Y2.

24.0

Total Replication (Y.j.)

161.9 163.2 172.2 497.3 =

Y...

20.7


65

2. Arrange data following treatments and sub-plot.

Organic Fertilizer

F B V BV

Total (Y..k) 98.3 117.5 144.5 137.0

Mean (Y..k) 16.3 19.6 24.1 22.8

3. Arrange source of variation and degree of freedom (df)


Replication r - 1 = 2

Nitrogen (N) n – 1 = 1

Error for main plot (Error a) (r-1)(n-1) = 2

Organic fertilizer (G) g – 1 = 3

N x G (n - 1)(g – 1) = 3

Error for sub-plot (Error b) (r – 1)(g – 1) + (r – 1)(n – 1)(g – 1)=12

Total ngr-1=23

4. Correction factor (CF)

CF = [Y..]2 / ngr = (497.3)2 / 2(4)(3) = 10,304.47

5. SS and MS

Total SS = Ɛ Yijk2 – CF

= (13.82 + 13.52 +... + 26.72 +27.62) – 10,304.47

= 516.12

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66

Main plot SS (mp) = Ɛ Yij.2 – CF

g

= (69.22 + ... +101.92) - 10,304.47 = 274.92

4

Replication SS (r) = Ɛ Y.j.2 – CF

ng

= (161.92 + 163.22 +172.22) - 10,304.47 = 7.87

2(4)

MS(r) = SS(r) / DF(r)

= 7.87 / 2 = 3.935

Nitrogen

SS (n) = Ɛ Y.j.2 – CF

rg

= (209.02 + 288.32) - 10,304.47 = 262.02

3(4)

MS(n) = SS(n) / DF(n)

= 262.22 / 1 = 262.02

Main-plot (Error a)

SS(Error a) = SS(mp) – SS(r) – SS(n)

= 274.92 – 7.87 – 262.02 = 5.03

MS(Error a) = SS(Error a) / DF (Error a)

= 5.03 / 2 = 2.515

Organic fertilizer

SS(g) = Ɛ Y..k2 – CF

rn

= (98.32 + … + 137.02 ) -10,304.47 = 215.26

3(2)


67

MS(g) = SS(g) / DF(g)

= 215.26 / 3 = 71.753

Nitrogen x Organic fertilizer

SS(nxg) = Ɛ Yi..k2 – CF – SS(n) – SS(g)

r

= (40.52 + … + 80.22) - 10,304.47 – 262.02 – 215.26 = 18.70

3

MS(nxg) = SS(nxg) / DF(nxg)

= 18.70 / 3 = 6.233

Error sub-plot (Error b)

SS(Error b) = SS(t) – SS(mp) – SS(g) – SS(nxg)

= 516.12 – 274.92 – 215.26 -18.70 = 7.24

MS(Error b) = SS(Error b) / DF(Error b)

= 7.24 / 12 = 0.603

6. F value

Nitrogen

Effect of nitrogen is tested using Error of main plot (Error a)

F(n) = MS(n) / MS(Error a)

= 262.02 / 2.515

= 104.18

F(table): DF(n)= 1, DF(Error a)= 2, p= 0.05 F= 18.51

Organic fertilizer and interaction between nitrogen x organic fertilizer,

use error of sub-plot (Error b),

F(g) = MS(g) / MS(Error b)

= 71.753 / 0.603 = 118.99

83


66

Main plot SS (mp) = Ɛ Yij.2 – CF

g

= (69.22 + ... +101.92) - 10,304.47 = 274.92

4

Replication SS (r) = Ɛ Y.j.2 – CF

ng

= (161.92 + 163.22 +172.22) - 10,304.47 = 7.87

2(4)

MS(r) = SS(r) / DF(r)

= 7.87 / 2 = 3.935

Nitrogen

SS (n) = Ɛ Y.j.2 – CF

rg

= (209.02 + 288.32) - 10,304.47 = 262.02

3(4)

MS(n) = SS(n) / DF(n)

= 262.22 / 1 = 262.02

Main-plot (Error a)

SS(Error a) = SS(mp) – SS(r) – SS(n)

= 274.92 – 7.87 – 262.02 = 5.03

MS(Error a) = SS(Error a) / DF (Error a)

= 5.03 / 2 = 2.515

Organic fertilizer

SS(g) = Ɛ Y..k2 – CF

rn

= (98.32 + … + 137.02 ) -10,304.47 = 215.26

3(2)


67

MS(g) = SS(g) / DF(g)

= 215.26 / 3 = 71.753

Nitrogen x Organic fertilizer

SS(nxg) = Ɛ Yi..k2 – CF – SS(n) – SS(g)

r

= (40.52 + … + 80.22) - 10,304.47 – 262.02 – 215.26 = 18.70

3

MS(nxg) = SS(nxg) / DF(nxg)

= 18.70 / 3 = 6.233

Error sub-plot (Error b)

SS(Error b) = SS(t) – SS(mp) – SS(g) – SS(nxg)

= 516.12 – 274.92 – 215.26 -18.70 = 7.24

MS(Error b) = SS(Error b) / DF(Error b)

= 7.24 / 12 = 0.603

6. F value

Nitrogen

Effect of nitrogen is tested using Error of main plot (Error a)

F(n) = MS(n) / MS(Error a)

= 262.02 / 2.515

= 104.18

F(table): DF(n)= 1, DF(Error a)= 2, p= 0.05 F= 18.51

Organic fertilizer and interaction between nitrogen x organic fertilizer,

use error of sub-plot (Error b),

F(g) = MS(g) / MS(Error b)

= 71.753 / 0.603 = 118.99

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68

F(table): DF(g)= 3, DF(Error b)= 12, p= 0.05 F3.49

F(nxg) = MS(nxg) / MS(Error b)

= 6.233 / 0.603 = 10.34

F(table): DF(nxg)= 3, DF(Error b)= 12, p= 0.05 F3.49

ANOVA

Source of Variation

df SS MS F F(0.05)

Replication 2 7.87 3.935

Nitrogen(N) 1 262.02 262.02 104.18 18.51

Error a 2 5.03 2.515

Organic fertilizer (G)

3 215.26 71.753 118.99 3.49

N x G 3 18.70 6.233

Error b 12 7.24 0.603

Total 23 516.12

Mean Comparison

Between means of treatment of the main plot or nitrogen.

LSD = t√ [2(MSError a) / rg]

where t is t table for df(Error a) at p=0.05

LSD = 4.303 x √ [2(2.515) / 3(4)] = 2.8 t / ha


69

N (kg ha-1) Yield (t ha-1)

0 17.4

120 24.0

LSD 0.05 2.8

Between treatments of the sub-plot or organic fertilizer.

LSD = t√ [2(MSError b) / rn]

where t is t table for df(Error b) at p=0.05

LSD = 2.179 x √ [2(0.603) / 3(2)]

= 1.0 t ha-1

Organic fertilizer Yield (t ha-1)

F 16.4

B 19.6

BV 22.8

V 24.1

LSD0.05 1.0

Conclusion 1. There is significant effect of nitrogen on the yield at p = 0.05

Applying N at the rate of 120 kg ha-1 obtained higher yield than at 0.

2. There is significant effect of organic fertilizer on the yield. Organic fertilizer V the highest yield followed by BV, B and F.


LMS or email.

85


68

F(table): DF(g)= 3, DF(Error b)= 12, p= 0.05 F3.49

F(nxg) = MS(nxg) / MS(Error b)

= 6.233 / 0.603 = 10.34

F(table): DF(nxg)= 3, DF(Error b)= 12, p= 0.05 F3.49

ANOVA

Source of Variation

df SS MS F F(0.05)

Replication 2 7.87 3.935

Nitrogen(N) 1 262.02 262.02 104.18 18.51

Error a 2 5.03 2.515

Organic fertilizer (G)

3 215.26 71.753 118.99 3.49

N x G 3 18.70 6.233

Error b 12 7.24 0.603

Total 23 516.12

Mean Comparison

Between means of treatment of the main plot or nitrogen.

LSD = t√ [2(MSError a) / rg]

where t is t table for df(Error a) at p=0.05

LSD = 4.303 x √ [2(2.515) / 3(4)] = 2.8 t / ha


69

N (kg ha-1) Yield (t ha-1)

0 17.4

120 24.0

LSD 0.05 2.8

Between treatments of the sub-plot or organic fertilizer.

LSD = t√ [2(MSError b) / rn]

where t is t table for df(Error b) at p=0.05

LSD = 2.179 x √ [2(0.603) / 3(2)]

= 1.0 t ha-1

Organic fertilizer Yield (t ha-1)

F 16.4

B 19.6

BV 22.8

V 24.1

LSD0.05 1.0

Conclusion 1. There is significant effect of nitrogen on the yield at p = 0.05

Applying N at the rate of 120 kg ha-1 obtained higher yield than at 0.

2. There is significant effect of organic fertilizer on the yield. Organic fertilizer V the highest yield followed by BV, B and F.


LMS or email.

86


70

TOPIC 2: EXPERIMENT WITH REPEATED DATA

Important Content For perennial crops rubber and oil palm data can be repeated from the same

experimental unit in different years or seasons.

Example: Testing clones using RCBD, data can be taken in several years. The

principle of split plot is used to analyze repeated data assigning the treatments

(clone) as the main plot and the years as the sub-plot.

Example: Yield of alfalfa was tested using RCBD with 5 replications and the data

were collected in the first and second year from the same experiment.

REP 1 REP 2 REP 3 REP 4 REP 5

V2 V3 V2 V1 V4

V4 V2 V1 V3 V1

V1 V1 V3 V4 V2

V3 V4 V4 V2 V4

Results for two years Variety Year REP 1 REP 2 REP 3 REP 4 REP 5 Total

(VT) Mean

1 1 9.02 6.98 10.62 9.52 10.96 47.10 9.42 2 1 8.84 9.83 10.06 9.30 9.50 47.53 9.51 3 1 9.74 9.28 11.74 9.71 10.60 51.07 10.21 4 1 11.40 9.93 10.83 9.65 10.13 51.94 10.39 Total (T1) 39.00 36.02 43.25 38.18 41.19 197.64

(JT1) 9.88

1 2 11.88 11.33 11.81 12.22 10.65 57.89 11.58 2 2 12.15 10.98 12.20 11.30 12.54 59.17 11.83 3 2 12.92 11.95 12.05 11.88 13.19 61.99 12.40 4 2 11.74 11.62 11.54 12.00 11.74 58.64 11.73 Total (T2) 48.69 45.88 47.60 47.40 48.12 237.69

(JT2) 11.88


71

Table of Variety x Replication

Variety REP 1 REP 2 REP 3 REP 4 REP 5 Total (V)

Mean

1 20.90 18.31 22.43 21.74 21.64 105.02 10.50 2 20.99 20.79 22.26 20.60 22.04 106.68 10.67 3 22.66 21.23 23.79 21.59 23.79 113.06 11.31 4 23.14 21.55 22.37 21.65 21.87 110.58 11.06 Total (R) 87.69 81.88 90.85 85.58 89.34 435.34 (JK) Hypothesis: No significant difference between the means of variety.

No significant difference between means of years.

No significant interaction between variety and year.

ANOVA of yield for two years

Source of Variation

df SS MS F F table

5% 1% Replication, R

4 6.13 1.53 4.78 3.26 5.41

Variety, V 3 4.01 1.34 4.18 3.49 5.95 Error (a), VR 12 3.89 0.32 Year , Y 1 39.89 39.89 55.40 4.49 8.53 V x Y 3 1.51 0.50 0.69 3.24 5.29 Error (b), RY+R(VY)

16 11.54 0.72

Total 39 66.95

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TOPIC 2: EXPERIMENT WITH REPEATED DATA

Important Content For perennial crops rubber and oil palm data can be repeated from the same

experimental unit in different years or seasons.

Example: Testing clones using RCBD, data can be taken in several years. The

principle of split plot is used to analyze repeated data assigning the treatments

(clone) as the main plot and the years as the sub-plot.

Example: Yield of alfalfa was tested using RCBD with 5 replications and the data

were collected in the first and second year from the same experiment.

REP 1 REP 2 REP 3 REP 4 REP 5

V2 V3 V2 V1 V4

V4 V2 V1 V3 V1

V1 V1 V3 V4 V2

V3 V4 V4 V2 V4

Results for two years Variety Year REP 1 REP 2 REP 3 REP 4 REP 5 Total

(VT) Mean

1 1 9.02 6.98 10.62 9.52 10.96 47.10 9.42 2 1 8.84 9.83 10.06 9.30 9.50 47.53 9.51 3 1 9.74 9.28 11.74 9.71 10.60 51.07 10.21 4 1 11.40 9.93 10.83 9.65 10.13 51.94 10.39 Total (T1) 39.00 36.02 43.25 38.18 41.19 197.64

(JT1) 9.88

1 2 11.88 11.33 11.81 12.22 10.65 57.89 11.58 2 2 12.15 10.98 12.20 11.30 12.54 59.17 11.83 3 2 12.92 11.95 12.05 11.88 13.19 61.99 12.40 4 2 11.74 11.62 11.54 12.00 11.74 58.64 11.73 Total (T2) 48.69 45.88 47.60 47.40 48.12 237.69

(JT2) 11.88


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Table of Variety x Replication

Variety REP 1 REP 2 REP 3 REP 4 REP 5 Total (V)

Mean

1 20.90 18.31 22.43 21.74 21.64 105.02 10.50 2 20.99 20.79 22.26 20.60 22.04 106.68 10.67 3 22.66 21.23 23.79 21.59 23.79 113.06 11.31 4 23.14 21.55 22.37 21.65 21.87 110.58 11.06 Total (R) 87.69 81.88 90.85 85.58 89.34 435.34 (JK) Hypothesis: No significant difference between the means of variety.

No significant difference between means of years.

No significant interaction between variety and year.

ANOVA of yield for two years

Source of Variation

df SS MS F F table

5% 1% Replication, R

4 6.13 1.53 4.78 3.26 5.41

Variety, V 3 4.01 1.34 4.18 3.49 5.95 Error (a), VR 12 3.89 0.32 Year , Y 1 39.89 39.89 55.40 4.49 8.53 V x Y 3 1.51 0.50 0.69 3.24 5.29 Error (b), RY+R(VY)

16 11.54 0.72

Total 39 66.95

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Calculation 1. df

Total = RVY-1 = (5)(4)(2) – 1 = 39

Replication = R-1 = 5-1 = 4

Variety = V-1 = 4-1 = 3

Error (a) = (R-1)(V-1) = (5-1)(4-1) = 12

Year = Y-1 = 2-1 = 1

Y x V = (Y-1) (V-1) = (2-1)(4-1) = 3

Error (b) = (R-1) (Y-1) + (R-1)(V-1)(Y-1)

= (5-1) (2-1) + (5-1)(4-1)(2-1) = 16

2. CF = (JK)2/ rat = (435.31)2 / (5)(4)(2) = 4738.02 3. SS Total = SSJ = Ʃ(Y)2 – CF

= (9.022+...+11.742) – 4738.02 = 66.95

4. SS Replication = SSR = [Ʃ(R) 2 / vy] – CF = [(87.692+ ... + 89.312) / 4(2)] – 4738.02 = 6.13

5. SSVariety = SSV = [Ʃ(V) 2 / ry] – CF = [(104.992+ ... + 110.582) / 5(2)] – 4738.02 = 4.01

6. SSError (a) = SSE (a) = [Ʃ(VR) 2 / y] – CF – SSR – SSV = [20.902+ ... + 21.872) / 2] – 4738.02 – 6.13 – 4.01 = 3.87

7. SSYear = SST = [Ʃ(Y) 2 / rv] – CF

= [(197.642+ 237.692) / 5(4)] – 4738.02 = 39.89

8. SSVariety x Year = SSVY = [Ʃ(VY) 2 / r] – CF – SSV – SSY = [(47.102+ ... +58.642) / 5] – 4738.02 – 4.01 – 39.89 = 1.51


73

9. SSError (b) = SSE (b) = SSJ – SSR – SSV – SSE(a) – SSY – SSVY

= 66.95 – 6.13 – 4.01 – 3.87 – 39.89 – 1.51 = 11. 54

10. Mean Square MSR = SSR/ DFR = 6.13 / 4 = 1.53 MSV = SSV/ DFV = 4.01 / 3 = 1.34 MST = SST/ DFT = 39.89 / 1 = 39.89 MSVT = SSVT/ DFVT = 1.51 / 3 = 0.50

11. F value Replication F = MSR / MSE (a) = 1.53 / 0.32 = 4.78 Variety F = MSV / MSE (a)

= 1.34 / 0.32 = 4.18

Year F = MSY / MSE (b)

= 39.89 / 0.72 = 55.40 Variety x Year F = MSVY/ MSE (b)

= 0.50 / 0.72 = 0.69

89


72

Calculation 1. df

Total = RVY-1 = (5)(4)(2) – 1 = 39

Replication = R-1 = 5-1 = 4

Variety = V-1 = 4-1 = 3

Error (a) = (R-1)(V-1) = (5-1)(4-1) = 12

Year = Y-1 = 2-1 = 1

Y x V = (Y-1) (V-1) = (2-1)(4-1) = 3

Error (b) = (R-1) (Y-1) + (R-1)(V-1)(Y-1)

= (5-1) (2-1) + (5-1)(4-1)(2-1) = 16

2. CF = (JK)2/ rat = (435.31)2 / (5)(4)(2) = 4738.02 3. SS Total = SSJ = Ʃ(Y)2 – CF

= (9.022+...+11.742) – 4738.02 = 66.95

4. SS Replication = SSR = [Ʃ(R) 2 / vy] – CF = [(87.692+ ... + 89.312) / 4(2)] – 4738.02 = 6.13

5. SSVariety = SSV = [Ʃ(V) 2 / ry] – CF = [(104.992+ ... + 110.582) / 5(2)] – 4738.02 = 4.01

6. SSError (a) = SSE (a) = [Ʃ(VR) 2 / y] – CF – SSR – SSV = [20.902+ ... + 21.872) / 2] – 4738.02 – 6.13 – 4.01 = 3.87

7. SSYear = SST = [Ʃ(Y) 2 / rv] – CF

= [(197.642+ 237.692) / 5(4)] – 4738.02 = 39.89

8. SSVariety x Year = SSVY = [Ʃ(VY) 2 / r] – CF – SSV – SSY = [(47.102+ ... +58.642) / 5] – 4738.02 – 4.01 – 39.89 = 1.51


73

9. SSError (b) = SSE (b) = SSJ – SSR – SSV – SSE(a) – SSY – SSVY

= 66.95 – 6.13 – 4.01 – 3.87 – 39.89 – 1.51 = 11. 54

10. Mean Square MSR = SSR/ DFR = 6.13 / 4 = 1.53 MSV = SSV/ DFV = 4.01 / 3 = 1.34 MST = SST/ DFT = 39.89 / 1 = 39.89 MSVT = SSVT/ DFVT = 1.51 / 3 = 0.50

11. F value Replication F = MSR / MSE (a) = 1.53 / 0.32 = 4.78 Variety F = MSV / MSE (a)

= 1.34 / 0.32 = 4.18

Year F = MSY / MSE (b)

= 39.89 / 0.72 = 55.40 Variety x Year F = MSVY/ MSE (b)

= 0.50 / 0.72 = 0.69

90


74

Table of ANOVA

Source of Variation

df SS MS F F Table

5% 1% Replication,R 4 6.13 1.53 4.78 3.26 5.41 Variety, V 3 4.01 1.34 4.18 3.49 5.95 Error (a), VR 12 3.89 0.32 Year, Y 1 39.89 39.89 55.40 4.49 8.53 V x T 3 1.51 0.50 0.69 3.24 5.29 Error (b), RY+R(VY)

16 11.54 0.72

Total 39 66.95

Conclusion

1. There is a significant difference for means of years and varieties.

2. There is no interaction between year and variety.

Mean Comparison (LSD) Comparison between means of variety

LSD = t [√ (2MSE (a) / ry)]

Where t is value of t table at p = 0.05, dfe (a) = 12 t = 2.179

LSD = 2.179 x √ [2(0.32) / 5(2)] = 0.55


1 10.50

2 10.67

3 11.31

4 11.06

LSD0.05 0.55


75

Comparison between means of years

LSD = t [√ (2MSE (b) / rv)]

LSD = 2.120 x √ [2(0.72) / 5(4)] = 0.57


1 9.88

2 11.88

LSD0.05 0.57


LMS or email.

91


74

Table of ANOVA

Source of Variation

df SS MS F F Table

5% 1% Replication,R 4 6.13 1.53 4.78 3.26 5.41 Variety, V 3 4.01 1.34 4.18 3.49 5.95 Error (a), VR 12 3.89 0.32 Year, Y 1 39.89 39.89 55.40 4.49 8.53 V x T 3 1.51 0.50 0.69 3.24 5.29 Error (b), RY+R(VY)

16 11.54 0.72

Total 39 66.95

Conclusion

1. There is a significant difference for means of years and varieties.

2. There is no interaction between year and variety.

Mean Comparison (LSD) Comparison between means of variety

LSD = t [√ (2MSE (a) / ry)]

Where t is value of t table at p = 0.05, dfe (a) = 12 t = 2.179

LSD = 2.179 x √ [2(0.32) / 5(2)] = 0.55


1 10.50

2 10.67

3 11.31

4 11.06

LSD0.05 0.55


75

Comparison between means of years

LSD = t [√ (2MSE (b) / rv)]

LSD = 2.120 x √ [2(0.72) / 5(4)] = 0.57


1 9.88

2 11.88

LSD0.05 0.57


LMS or email.

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