completely randomized block design - wordpress.com · 2012. 10. 24. · • a randomized complete...
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COMPLETELY RANDOMIZED BLOCK DESIGN
• A randomized complete block design is an experimental design for comparing ttreatments in b blocks.
• The blocks consist of t homogeneous experimental units.
• Treatments are randomly assigned to experimental units within a block, with each treatment appearing exactly once in every block.
DEFINITION
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Motivation: to deal with non-homogenous groups (experimental material, area or time) subdivide experimental units into homogenous groupsBLOCKS : controlling the variability of experimental material, area or time that are not homogenous.
Design:number of experimental units in a block = number (or multiple) of studied treatmentsTreatments assigned at random to experimental units within block
Objective: by blocking isolate, remove from the error term variation attributable to blocks
Examples of blocks: breed of animal, age, laboratory, day, etc.
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ADVANTAGES OF THE RANDOMIZED COMPLETE BLOCK DESIGN
1. The design is useful for comparing t treatment means in the presence of a single extraneous source of variability.
2. This design to be more accurate than CRD because the elimination of SS block from SS error usually results in a decrease in the MS error.
3. The statistical analysis is simple.
4. The design is easy to construct.
5. It can be used to accommodate any number of treatments in any number of blocks.
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DISADVANTAGES OF THE RANDOMIZED COMPLETE BLOCK DESIGN
1. Because the experimental units within a block must be homogeneous, the design is best suited for a relatively small number of treatments.
2. This design controls for only one extraneous source of variability (due to blocks). Additional extraneous sources of variability tend to increase the error term, making it more difficult to detect treatment differences.
3. The effect of each treatment on the response must be approximately the same from block to block.
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RANDOMISASI RCBD
Blok I
Blok II
Blok III
Blok IV
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RANDOMIZED COMPLETE BLOCK DESIGN, LAY OUT
Location
1 2 3 4
P2 P1P3P4
P2P4P1P3
P1P3P4P2
P1P2P4P3
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BAK UKURAN 1 X 1 M2
20 ikan
30 ikan30 ikan40 ikan
40 ikan
20 ikan40 ikan30 ikan
30 ikan
40 ikan20 ikan
20 ikan
Blok IIBlok III
Blok I
Blok IV
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PEMBUATAN ROTI DENGAN 4 MACAM ADONAN A, B, C DAN D
Oven lamaModel A
Oven baruModel B
Oven baruModel A
Oven lamaModel B
Awas panasnya tidak sama!
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Blok I
Blok 2
Blok 3
POPULATION PARAMETERS 2T BY 2B –NO INTERACTION
B =1 B=2 A = 1
11µ 12µ A = 2
21µ 22µ
ijjiji eX += µ
22211211 µµµµ −=−Subject to:
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FIGURE : TREATMENT MEANS IN A RANDOMIZED BLOCK DESIGN
100
90
80
70
60
50
40
1 2 3 Treatment
ijµ
34
1
234
1
2
34
1
2
Plot of Treatment Mean by Treatment
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THE HYPOTHESES FOR TESTING TREATMENT MEAN DIFFERENCES
1 2: ...: At least one differs from the rest
o t
a i
HH
µ µ µµ
= = =
The null hypothesis : is no difference among treatment means
versus
The research hypothesis: treatment means differ.
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BlocksTreatment 1 2 3 … r Total Mean1 x11 x12 x13 … x1r x1. x1.2 x21 x22 x23 … x2r x2. x2.3 x31 x32 x33 … x3r x3. x3.. . . . . . . .. . . . . . . .. . . . . . . .t xt1 xt2 xt3 … xtr xt. xt.
Total x.1 x.2 x.3 x.r x..
Mean x.1 x.2 x.3 x.r x..
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µ CONSTANTβj BLOCK EFFECT τi TREATMENT EFFECTeij ERROR TERM: sources of variation other than treatment and
block
Constant
Treatment effect
Error term
Block effect
ijjiji eX +++= βτµ
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Two-way ANOVA: observation analyzed under two criteria, the block and the treatment group to which it belongs
Assumptions:1. Xij is a random independent sample from one of txr populations2. Each of txr populations ~ N(µij,σ2) => eij are independent, N(0, σ2)3. Block and treatment effects are additive, i.e. there is no interaction
between treatments and block
Block-treatment effect = block effect + treatment effect
Its violation => misleading results. Concern when: largest mean>50% smallest
When assumptions hold-> τj and βi are fixed constants -> fixed-effects
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Hypotheses: Ho: τj=0, j=1,2,…,k vs. Ha: not all τj=0
No block effect because: (1) primary interest is on treatment effect and (2) blocks are obtained non-randomly
SST=SSBl+SSTr+SSE
∑∑=
−=t
i
r
jji xxSStotal
1
2..)( ∑=
−=t
iiitreat xxrSS
1
2..)(.
2
1..).( xxtSS
r
jjblocks −= ∑
=
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df= (tr-1)
df= (r-1)
df= (t-1)
df= (t-1)(r-1)
Xij - µ= i + βj + eij, I , βj & eij independenΣ Σ (Xij - µ)2 = Σ Σ ( i + βj + eij)2
Σ Σ (Xij - µ)2 = Σ Σ i2 + Σ Σ βj
2 + Σ Σ eij2
JK total JK perlakuan JK Blok JK sesatan
Σ Σ (Xij - µ)2= Σ Σ Xij2 + trµ2 – 2trµ2
= Σ Σ Xij2 - trµ2
= Σ Σ Xij2 – tr(Σ Σ Xij)2/(tr)2
= Σ Σ Xij2 – (Σ Σ Xij)2/tr
∑∑∑∑
= =
= =
−=t
i
r
j
t
i
r
j
tr
XijXijJKtotal
1 1
2
1 1218
ijjiji eX +++= βτµ
τ τττ
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JK PERLAKUANJK plk =r Σ i
2
∑ −= 2..)( XXr i
22
..
.)(
2
+−= ∑∑∑∑∑
rtX
trrtX
rX jijii
2
1 11
2
.
.
rt
X
r
XJKplk
t
i
r
jij
t
ii
−=∑∑∑= ==
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τ
JK BLOKJK blok =t Σ βj
2
∑ −= 2. ..)( XXt j
22
..
.)(
2
+−= ∑∑∑∑∑
rtX
trrtX
tX jijij
2
1 11
2.
.rt
X
t
XJKblok
t
i
r
jij
r
jj
−=∑∑∑= ==
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21
∑∑∑∑= =
∧
= =
−−−==t
i
r
jjiji
t
i
r
jijSesa xeJK
1 1
2
1 1
2tan )ˆˆ( βτµ
tr
XCF
t
i
r
jij
2
1 1)(∑∑
= ==
JKbloknJKperlakuaJKtotalJKError −−=
2
1 11
2.
1
2.
1 1 .tan
rt
X
t
X
r
XXJKsesa
t
i
r
jij
r
jj
t
iit
i
r
jij
+−−=∑∑∑∑
∑∑ = ===
= =
TOTAL SUM OF SQUARES
2
2
.
..
rt
X
r
XSStreat
t
i
r
j
t
ii ji
−=∑∑∑
∑∑∑∑
= =
= =
−=t
i
r
j
t
i
r
j
tr
XijijXSStotal
1 1
2
1 12
2
2
.
..
rt
X
t
XSSblocks
t
i
r
j
t
ij ji
−=∑∑∑
Partition of TSS:
∑∑=
−=t
i
r
jji xxSStotal
1
2..)(
∑∑==
=−=t
iii
t
iiitreat rxxrSS
1
2
1
2..)(. τ
∑∑==
=−=r
jj
r
jjblocks txxtSS
1
22
1..).( β
∑∑∑∑= =
∧
= =
−−−==t
i
r
jjiji
t
i
r
jijError xeSS
1 1
2
1 1
2 )ˆˆ( βτµ
SSblockSStreatSStotalSSerror −−= .tr
XCF
t
i
r
jij
2
1 1)(∑∑
= ==
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ANALYSIS OF VARIANCE TABLE FOR A RANDOMIZED COMPLETE BLOCK DESIGN
Statistical Decision: when Ho is true,
MSTr/MSE ~ Fα;(t-1);,(t-1)(r-1)
if F stat ≥ F reject
Source df SS MS F
Blocks r-1 SSB MSB=SSB/(r-1) MSB/MSE
Treatment t-1 SST MST=SST/(t-1) MST/MSE
Error (t-1)(r-1) SSE MSE=SSE/(t-1)(r-1)
Total tr-1 SSTot
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EXPECTED MEAN SQUARES
When is true, both MST and MSE are unbiased estimates of , the variance of the experimental error.
1 2: ...o tH α α α= = =2εσ
( ) ( )2 2MST MSEE Eε εσ σ= =
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BERBAGAI NILAI DUGA DENGAN MODEL
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∑∑∑∑= =
∧
= =
−−−=t
i
r
jjiji
t
i
r
jij Xe
1 1
2
1 1
2 )ˆˆ( βτµ
..ˆ XtrXij
=ΣΣ=µ
.... XX
rXi
ii −=−Σ=∧∧
µτ
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....
1XX
tjX
j
r
jj −=−=
∧
=
∧
∑ µβ
.... XXXXe
Xe
jiijij
jiijij
+−−=
−−−=
∧
∧∧∧∧
βτµ
EXAMPLE: PROTEIN CONTENT (%) OF THE SOYBEAN TREATED BY FOLIAGE FERTILIZER A, B, C AND D IN THE FIELD WITH DIFFERENT FERTILITY
A=18 C=10 B=8 D=12
B=8 D=8 A=13 C=7
C=8 A=11 B=5 D=12
D=8 B=7 C=11 A=14
What is the experimental design ?
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SS => Linear model
i j Xij µ ζi βJ eij1 1 18 10 4 2 21 2 13 10 4 -1 01 3 11 10 4 -1 -21 4 14 10 4 0 02 1 8 10 -3 2 -12 2 8 10 -3 -1 22 3 5 10 -3 -1 -12 4 7 10 -3 0 03 1 10 10 -1 2 -13 2 7 10 -1 -1 -13 3 8 10 -1 -1 03 4 11 10 -1 0 24 1 12 10 0 2 04 2 8 10 0 -1 -14 3 12 10 0 -1 34 4 8 10 0 0 -2Σ ( ….) 160 160 0 0 0Σ( …..)2 1762 1600 104 24 34
TSS CF SST SSB SSE10/22/2012 28
ALTERNATE METHOD
160044
1602
==x
CF
1044
40362856 2222
=−+++
= CFSST
244
40363648 2222
=−+++
= CFSSB
16216008....1318 222 =−+++=TSS
3424104162 =−−=SSE
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Analysis of Variance ProcedureClass Level Information
Class Levels Values
TREAT 4 1 2 3 4
BLOCK 4 1 2 3 4
Number of observations in data set = 16
Analysis of Variance Procedure
Dependent Variable: PROTEIN
Source DF Sum of Squares Mean Square F Value Pr > FModel 6 128.00000000 21.33333333 5.65 0.0109Error 9 34.00000000 3.77777778Corrected Total 15 162.00000000
R-Square C.V. Root MSE PROTEIN Mean0.790123 19.43651 1.94365063 10.00000000
Source DF Anova SS Mean Square F Value Pr > F
BLOCK 3 24.00000000 8.00000000 2.12 0.1681TREAT 3 104.00000000 34.66666667 9.18 0.0042
SAS Program for RCBD
data agus; Iinput treat block protein; cards;1 1 181 2 13……4 3 124 4 8;proc print uniform;proc anova;class treat block;model protein=block treat;run; quit;
Tell me the conclusion!30
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Model linear:
Xijk= Data perlakuan ke-i, blokn ke-j, sampel ke-kµ= rerata umumµ = pengaruh perlakuan ke-iβj=pengaruh blok ke-jεij = pengaruh blok ke-j pada perlakuan ke-iδijk = pengaruh sesatan ke-k pada perlakuan ke-i dan
blok ke-ji = perlakuan ke-ij = blok ke-jk = sampel ke-ki = 1, 2, 3, ……., tj = 1, 2, 3, ……, rk = 1, 2, 3, ……. , s
ijkijjijkiX δεβτµ ++++=
iτ
RCBD DG BEBERAPA PENGAMATAN TIAPEXPERIMENTAL UNIT
Perlakuan 1 Perlakuan 2
Blok 1 Blok 2 Blok 3 Blok 1 Blok 2 Blok 3
Sample 1 X111 X121 X131 X211 X221 X231Sample 2 X112 X122 X132 X212 X222 X232Sample 3 X113 X123 X133 X213 X223 X233Sample 4 X114 X124 X134 X214 X224 X234
X11. X12. X13. X21. X22. X23.
X1.. X2.. X…
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BERBAGAI NILAI DUGA DENGAN MODEL
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∑∑∑∑∑∑= =
∧∧
== = =
−−−−=t
i
r
jijjijki
s
k
t
i
r
jijk
s
kX
1 1
2
11 1
2
1)ˆˆ( εβτµδ
...ˆ XtrsXijk
=ΣΣΣ=µ
....... XX
rsXi
ii −=−Σ=∧∧
µτ
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....... XX
tsjX
jj −=−Σ=∧∧
µβ
.ijijkijk
ijjiijkijk
XX
X
−=
−−−−=
∧
∧∧∧∧∧
δ
εβτµδ
........
.
XXXX
X
jiijij
jiijij
+−−=
−−−=
∧
∧∧∧∧
ε
βτµε
TOTAL SUM OF SQUARES
2
2
..
...
srt
X
rs
XSStreat
t
i
r
j
s
k
t
ii ji
−=∑∑∑∑
∑∑∑∑∑
∑= =
= = =
=
−=t
i
r
j
t
i
r
j
s
kijk
s
k trs
XijkXSStotal
1 1
2
1 1 12
1
2
2.
...
..
srt
X
st
XSSblocks
t
i
r
j
s
k
t
ij jki
−=∑∑∑∑
∑∑∑= = =
−=t
i
r
jjki
s
kXXSStotal
1 1
2
1...)(
∑∑==
=−=t
iii
t
iijitreat sjrXXsrSS
1
2
1
2...)..(. τ
∑∑==
=−=r
jj
r
jjblocks tsXXtsSS
1
22
1. ...).( β
∑∑∑∑∑∑= =
∧∧
== = =
−−−−==t
i
r
jijjiji
s
k
t
i
r
jijk
s
kError kXSS
1 1
2
11 1
2
1)ˆˆ( εβτµδ
.. atSSblokxtreSSblockSStreatSStotalSSerror −−−=trs
XCF
t
i
r
jijk
s
k
2
1 1 1)(∑∑∑
= = ==
35
∑∑∑∑====
=−=r
jij
t
i
r
jji
t
ikstreatxbloc sXXsSS
1
2
1
2
1.
1...)( ε
2
2..
22
..
..
srt
X
ts
X
rs
X
s
XokSStreatxbl
t
i
r
j
s
k
r
jj
t
ii
r
jij
t
iji
+−−=∑∑∑∑∑∑∑
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ANALYSIS OF VARIANCE TABLE FOR A RCBD SUBSAMPLE
Statistical Decision: when Ho is true,
MSTr/MSE ~ Fα;(t-1);,(s-1)tr
if VR≥F reject
Source df SS MS F
Blocks r-1 SSB MSB=SSB/(r-1) MSB/MSE
Treatment t-1 SST MST=SST/(t-1) MST/MSE
BlocksXTreat (r-1)(t-1) SSTXB MSBxT=SSBxT/(r-1)(t-1) MSBxT/MSE
Error (s-1)tr SSE MSE=SSE/(s-1)tr
Total trs-1 SSTot
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SOAL LATIHAN
BLOK SAMPLETreatment
1 2 3 4 5
1 5 6 9 10 12Blok I 2 4 7 9 8 10
3 2 6 8 10 111 2 4 9 9 9
Blok II 2 3 6 8 8 113 2 6 10 10 111 3 5 9 10 9
Blok III 2 3 6 8 9 103 4 6 9 9 11
Buatlah analisis varian dengan α = 5%Apabila terdapat perbedaan pengaruh yang nyata, lakukan uji lanjut dg DMRT α = 5%
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BLOK SAMPLETREATMENT
1 2 3 4 5 Total
1 5 6 9 10 12 42
Blok I 2 4 7 9 8 10 38
3 2 6 8 10 11 37
1 2 4 9 9 9 33
Blok II 2 3 6 8 8 11 36
3 2 6 10 10 11 39
1 3 5 9 10 9 36
Blok III 2 3 6 8 9 10 36
3 4 6 9 9 11 39
Total 28 52 79 83 94
Buatlah analisis varian dengan α = 5%Apabila terdapat perbedaan pengaruh yang nyata, lakukan uji lanjut dg DMRT α = 5%
SOAL LATIHAN
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