chapter5a. crd 15march2012 final

7/28/2019 Chapter5a. CRD 15March2012 Final

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By

H.M. Edi Armanto and Swaditya Rizki Faculty of Agrotechnology and Food Science, UMT - Malaysia



1) Experimental Classification2) Assumptions3) CRD (2 Slides)

4) Field Layout (4 Slides)5) The Linear Model6) Hypothesis7) One-Way-ANOVA (3 Slides)8) Post Hoc Test9) Example 1: Diet Medicines10) Example 2: Pesticide Formulas



Single Factor Experiment

(One way ANOVA):

1) Completely Randomized

Design (CRD)

2) Randomized Complete

Block Design (RCBD)

3) Latin Square Design (LS)

Two or More Factor Experiment

(Two ways ANOVA):

1) Factorial Experiment in CRD

2) Factorial Experiment in RCBD3) Factorial Experiment in LS

4) Split Plot Design (SPD)

5) Strip Block Design (SBD)



1. Randomness & Independence of Errors

• Independent random samples are drawn for each

condition2. Normality

• Populations (for each condition) are Normally Distributed

3. Homogeneity of Variance• Populations (for each condition) have equal variances



1) Experimental Units (object) are assigned randomly totreatments (subjects are assumed homogeneous)

2) Experimental units receive the same treatment

3) One factor of treatment (2 or more treatment levels )4) Analyzed by One-Way ANOVA

5) Appropriate for homogenous experimental unit, i.e.laboratory and not suitable in the Agricultural fields

6) Environmental effects are relatively easy to control

7) It is assumed that there is no interaction

CRD (1)



Treatment Replications1 2 3 4

Experimental units are randomly assigned

to treatments



The Linier Model

i = 1,2,…, t j = 1,2,…, r

X ij = the observation in i th

treatment and the j th

replication

= overall mean

t i

= the effect of the i th

treatment ij = random error

Xij =μ + ti +εij



H0: 1 = 2 = 3 = ... = t

• All Population Means are Equal

• No Treatment Effect

H1: Not All i Are Equal• At Least 1 Pop. Mean is Different

• Treatment Effect

NOT 1 2 ... t



H0: 1 = 2 = 3 = ... = t

• All Population Means are

Equal

• No Treatment Effect

H1 : Not All i Are Equal

• At Least 1 Pop. Mean is

Different

• Treatment Effect

NOT 1 2 ... t

X

f(X)

1 = 2 = 3

X

f(X)

1 = 2 3



1. Compares 2 Types of Variation to Test

Equality of Means

2. Comparison Basis Is Ratio of Variances3. If Treatment Variation Is Significantly Greater

Than Random Variation, then Means Are Not

Equal4. Variation Measures Are Obtained by

‘Partitioning’ Total Variation



Sum of Squares Within Sum of Squares Error

(SSE)

Within Groups Variation

Sum of Squares Among Sum of Squares Between

Sum of SquaresTreatment (SST)

Among Groups Variation

Variation due to

treatment

Variation due to

random sampling

Total variation



The One-Way ANOVA procedure produces

a one-way analysis of variance for a

quantitative dependent variable by a single

factor (independent) variable. Analysis of

variance is used to test the hypothesis that

several means are equal. This technique is

an extension of the two-sample t test.



In addition to determining that differences exist

among the means, you may want to know which

means differ. There are two types of tests for

comparing means: a priori contrasts and post-hoc tests. Contrasts are tests set up before

running the experiment, and pos t hoc tests are

run after the experiment has been conducted.

You can also test for trends across categories.



Once you have determined that differences exist

among the means, post hoc range tests and

pairwise multiple comparisons can determine

which means differ. Range tests identifyhomogeneous subsets of means that are not

different from each other. Pairwise multiple

comparisons test the difference between each

pair of means and yield a matrix where asterisksindicate significantly different group means at an

alpha level of 0.05.



Bonferroni. Uses t tests to perform pairwise

comparisons between group means, but controls

overall error rate by setting the error rate for

each test to the experiment wise error ratedivided by the total number of tests. Hence, the

observed significance level is adjusted for the

fact that multiple comparisons are being made.



Tukey. Uses the Studentized range statistic to make all of

the pairwise comparisons between groups. Sets the

experimentwise error rate at the error rate for the collection

for all pairwise comparisons. Duncan. Makes pairwise comparisons using a stepwise

order of comparisons identical to the order used by the

Student-Newman-Keuls test, but sets a protection level for

the error rate for the collection of tests, rather than an error rate for individual tests. Uses the Studentized range statistic.

And many more…



Three diet medicines (one is commonly used)were tested with 6 replications on the rats. Wewanted to find out which one can be delivered tothe market (the best one according to statistical

analysis). The measured variable is weight lossafter 30-days treatments.

Replications Rat + Treatment

Medicine 1

Rat + Treatment

Medicine 2

Rat + no Medicine

(Used a Control)

1 24 10 7

2 26 12 5

3 11 5 13

4 14 9 10

5 21 17 21

6 17 22 23



Treatments of Diet Medicine (M)Rat+Med 1 Rat+Med 2 Rat+ no Med

Experimental units will be randomly assigned to

treatments by using lottery or other methods

Replications

(6 times)



Note: M: Treatment (Diet Medicine, M1, M2, M3)

R : Rat (Replication, R1, R2, R3, R4, R5,

R6)

Replication

(6 times)

Layout of Experimental Unit for fieldobservations

M1R1

M2R6

M3R2

M2R1

M1R3

M3R4

M2R2

M3R5

M3R3

M2R4

M1R3

M13R6

M1R5

M3R6

M3R1

M1R2

M2R3

M2R5



ReplicationMedicine



1. Determine the data above is normally

distributed and homogeneous.

2. Analyze using one-way ANOVA.

3. Make hypothesis to get a decision

We consider that the data is normally

distributed and homogeneous.



Homogeneous because

0.772 > 050.



050. Significance correction

If the significance obtained > , so the

variance of each sample is homogeneous.

If the significance obtained < , so the

variance of each sample is not homogeneous.



Note: Between Group = Treatments

Within Groups = Error



H0: μ1 = μ2 = μ3

H1: μi not all equal. (i=1,2,3)

Significance correction 050.

If the significance obtained > 0.05 , H0 is received

If the significance obtained < 0.05, H0 is rejected

Conclusion : because of the sig is 0.210, thatmeanssig > 0.05. So all of the medicines have the sameeffect. There is no difference among the medicines.



From the Post Hoc table is obtained that

all of the sig (table) have values more than

0.05 (significance correction). that means,

all of the medicines don’t have differentmean significantly.

Mean difference of medicine 1 2, 13

, 21, 23, 31, 32. All of the meandifference is not significantly.



We can analyze

using Benferroni,

sidak, scheffle,

duncan test, etc likethe Post Hoc box

below, with the

same analysis fromTukey Test.



Three same formulas of pesticides (made in UKM, UPM, UMT). TheUKM pesticide was already standardized (as control). We compare

effectivity of pesticide to kill insects in the 6 experimental stations with

the same doses. The measured variable was amount of killed insects in

the fields. The experimental results are presented in the Table.

Replications UKM (control) UPM UMT

A 50 120 140

B 30 70 125

C 12 70 80D 30 65 90

E 12 90 70

F 30 70 80

G 20 70 80



PesticidesUKM UPM UMT

Experimental units will be randomly assigned to

treatments by using lottery or other methods

Replications(7 times)

A

B

C

D

E

F

G



PesticidesUKM UPM UMT

Replications(7 times)

Note: P: Pesticide (P1: UKM, P2: UPM: P3: UMT)A, B, C, D, E, F, G : Replications

P1A

P2F

P3B

P2A

P1C

P3D

P3G

P2B

P3E

P3C

P2D

P1D

P1F

P1E

P3F

P3A

P1B

P2C

P3E

P2G P1G



Value University



For Post Hoc and Options button is the same like example 1



Note: Between Group = Treatments

Within Groups = Error

Homogeneous because

0.178 > 050.



050. Significance correction

If the significance obtained > , so the

variance of each sample is homogeneous.

If the significance obtained < , so the

variance of each sample is not homogeneous.



H0: μ1 = μ2 = μ3 H1: μi not all equal. (i=1,2,3)

Significance correction 050.

If the significance obtained > 0.05 , H0 is received

If the significance obtained < 0.05, H0 is rejected

Conclusion : because of the sig is 0.000, that

meanssig < 0.05. So H0 is rejected, in other word, H1 isreceived,at least, there is pesticide which is different with other.



1) From the Post Hoc table obtained, for sig value is0.000 (sig < 0.05), that means: two pesticides havethe different impacts. UKM-UPM and UKM-UMT aresignificantly different, that pesticides (made in UPM and

UMT) are better than UKM pesticide. Their causes arethat both pesticides have better active components inthe pesticides.

2) There are sig = 0.348 (sig >0.05). That means: the

two pesticides have the same results, i.e. between UPM – UMT or UMT-UPM have the same capability. Bothpesticides have the same ability to kill insects.



Many thanks for your attention See you in other occasions

chapter5a. crd 15march2012 final

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