Download 3 randomized complete block design (rcbd) and more Exams Design in PDF only on Docsity!
3 RANDOMIZED COMPLETE BLOCK DESIGN (RCBD)
- The experimenter is concerned with studying the effects of a single factor on a response of interest.
However, variability from another factor that is not of interest is expected.
- The goal is to control the effects of a variable not of interest by bringing experimental units that are
similar into a group called a “block”. The treatments are then randomly applied to the experimental
units within each block. The experimental units are assumed to be homogeneous within each block.
- By using blocks to control a source of variability, the mean square error (MSE) will be reduced. A
smaller MSE makes it easier to detect significant results for the factor of interest.
- Assume there are a treatments and b blocks. If we have one observation per treatment within each
block, and if treatments are randomized to the experimental units within each block, then we have a
randomized complete block design (RCBD). Because randomization only occurs within blocks,
this is an example of restricted randomization.
3.1 RCBD Notation
- Assume μ is the baseline mean, τi is the i
th treatment effect, βj is the j
th block effect, and
ij is the random error of the observation. The statistical model for a RCBD is
yij = μ + τi + βj + ij and ij ∼ IIDN (0, σ
2 ). (6)
- μ, τi (i = 1, 2 ,... , a), and βj (j = 1, 2 ,... , b) are not uniquely estimable. Constraints must be
imposed. To be able to calculate estimates μ̂ , ̂τi, and β̂j , we need to impose two constraints.
- Initially, we will assume the textbook constraints:
∑^ a
i=
τi = 0 and
∑^ b
j=
βj = 0.
- These are not the default SAS constraints (τa = 0, βb = 0) or R constraints (τ 1 = 0, β 1 = 0).
- Applying these constraints, will yield least-squares estimates
μ̂ = ̂τi = and β̂j =
where ¯yi· is the mean for treatment i, and ¯y·j is the mean for block j.
- Substitution of the estimates into the model yields:
yij = μ̂ + ̂τi + β̂j + eij
= y¯·· + (¯yi· − ¯y··) + (¯y·j − ¯y··) + eij
where eij = ̂ij is the residual of an observation yij from a RCBD. The value of eij is
eij = yij − (¯yi· − y¯··) − (¯y·j − y¯··) − y¯·· =
- The total sum of squares (SStotal) for the RCBD is partitioned into 3 components:
∑^ a
i=
∑^ b
j=
(yij − y¯··)
2
∑^ a
i=
∑^ b
j=
(¯yi· − y¯··)
2
∑^ b
j=
∑^ a
i=
(¯y·j − y¯··)
2
∑^ a
i=
∑^ b
j=
(yij − y¯i· − ¯y·j + ¯y··)
2
= b
∑^ a
i=
(¯yi· − y¯··)
2
∑^ b
j=
(¯y·b − ¯y··)
2
∑^ a
i=
∑^ b
j=
(yij − y¯i· − y¯·j + ¯y··)
2
= b
∑^ a
i=
∑^ b
j=
∑^ a
i=
∑^ b
j=
OR SST otal = SST rt + SSBlock + SSE
- Alternate formulas to calculate SST otal, SST rt and SSBlock.
SST otal =
∑^ a
i=
∑^ b
j=
y
2 ij −^
y
2 ··
ab
SST rt =
∑^ a
i=
y
2 i·
b
y
2 ··
ab
SSBlock =
∑^ b
j=
y
2 ·j
a
y
2 ··
ab
SSE = SST otal − SST rt − SSBlock where
y
2 ··
ab
is the correction factor.
3.2 Cotton Fiber Breaking Strength Experiment
An agricultural experiment considered the effects of K 2 O (potash) on the breaking strength of cotton
fibers. Five K 2 O levels were used (36, 54, 72, 108, 144 lbs/acre). A sample of cotton was taken from each
plot, and a strength measurement was taken. The experiment was arranged in 3 blocks of 5 plots each.
K 2 O lbs/acre (treatment)
Block 36 54 72 108 144 Totals
1 7.62 8.14 7.76 7.17 7.46 y· 1 =38.
2 8.00 8.15 7.73 7.57 7.68 y· 2 =39.
3 7.93 7.87 7.74 7.80 7.21 y· 3 =38.
y 1 · y 2 · y 3 · y 4 · y 5 ·
Totals 23.55 24.16 23.23 22.54 22.35 y··=115.
Treatment Means y¯ 1 · = 7. 850 ¯y 2 · = 8. 053 y¯ 3 · = 7. 743 y¯ 4 · = 7. 513 y¯ 5 · = 7. 450
Block Means y¯· 1 = 7. 630 ¯y· 2 = 7. 826 y¯· 3 = 7. 710
Grand Mean y¯ = 7. 723
Uncorrected Sum of Squares =
∑a
i=
∑b
j=1 y
2 ij =
Correction factor = y
2 ··/ab^ = 115.^83
2 /15 =
a ∑
i=
y
2 i·
b
2
2
2
2
2
∑^ b
j=
y
2 ·j
a
2
2
2
SST otal = 895. 6183 − 894 .4393 =
SST rt = 895. 1717 − 894 .4393 =
SSBlock = 894. 5364 − 894 .4393 =
SSE = 1. 1790 − 0. 7324 − 0 .0971 =
Analysis of Variance (ANOVA) Table
Source of Sum of Mean F
Variation Squares d.f. Square Ratio p-value
K 2 O lbs/acre .18311.
Blocks .04856 —–
Error .043685 ——
Total 14 —— ——
Dependent Variable: strength
Fit Diagnostics for strength
Adj R-Square 0.
R-Square 0.
MSE 0.
Error DF 8
Parameters 7
Observations 15
Proportion Less
0.0 0.4 0.
Residual
0.0 0.4 0.
Fit–Mean
-0.
-0.
-0.48 -0.24 0 0.24 0.
Residual
0
10
20
30
Percent
0 5 10 15
Observation
Cook's D
7.2 7.4 7.6 7.8 8.0 8.
Predicted Value
strength
-2 -1 0 1 2
Quantile
-0.
Residual
0.5 0.6 0.7 0.8 0.
Leverage
0
1
2
RStudent
7.4 7.6 7.8 8.0 8.
Predicted Value
0
1
2
RStudent
7.4 7.6 7.8 8.0 8.
Predicted Value
-0.
-0.
Residual
ANOVA RESULTS FOR STRENGTH BY TREATMENT
The GLM Procedure
36 54 72 108 144
k2O
strength
block^123
Interaction Plot for strength
ANOVA RESULTS FOR STRENGTH BY TREATMENT
The GLM Procedure
ANOVA RESULTS FOR STRENGTH BY TREATMENT
The GLM Procedure
7.2^15
strength
1 2 3
block
Distribution of strength
strength
Level of block N Mean Std Dev
1 5 7.63000000 0.
2 5 7.82600000 0.
3 5 7.71000000 0.
ANOVA RESULTS FOR STRENGTH BY TREATMENT
The GLM Procedure
Dependent Variable: strength
ANOVA RESULTS FOR STRENGTH BY TREATMENT
The GLM Procedure
Dependent Variable: strength
Parameter Estimate
Standard Error t Value Pr > |t|
K2O=36 0.12800000 0.10793208 1.19 0.
K2O=54 0.33133333 0.10793208 3.07 0.
K2O=72 0.02133333 0.10793208 0.20 0.
K2O=108 -0.20866667 0.10793208 -1.93 0.
K2O=144 -0.27200000 0.10793208 -2.52 0.
ANOVA RESULTS FOR STRENGTH BY TREATMENT
The GLM Procedure
Tukey's Studentized Range (HSD) Test for strength
ANOVA RESULTS FOR STRENGTH BY TREATMENT
The GLM Procedure
Tukey's Studentized Range (HSD) Test for strength
Note: This test controls the Type I experimentwise error rate, but it generally has a higher Type II error rate than REGWQ.
Alpha 0.
Error Degrees of Freedom 8
Error Mean Square 0.
Critical Value of Studentized Range 4.
Minimum Significant Difference 0.
Means with the same letter are not significantly different.
Tukey Grouping Mean N k2O
A 8.0533 3 54
A
B A 7.8500 3 36
B A
B A 7.7433 3 72
B A
B A 7.5133 3 108
B
B 7.4500 3 144
Critical Value of Studentized Range 4.
Minimum Significant Difference 0.
**Comparisons significant at the 0.05 level are indicated by *.
k2O Comparison
Difference Between Means
Simultaneous 95% Confidence Limits
54 - 36 0.2033 -0.3862 0.
54 - 72 0.3100 -0.2796 0.
54 - 108 0.5400 -0.0496 1.
54 - 144 0.6033 0.0138 1.1929 ***
36 - 54 -0.2033 -0.7929 0.
36 - 72 0.1067 -0.4829 0.
36 - 108 0.3367 -0.2529 0.
36 - 144 0.4000 -0.1896 0.
72 - 54 -0.3100 -0.8996 0.
72 - 36 -0.1067 -0.6962 0.
72 - 108 0.2300 -0.3596 0.
72 - 144 0.2933 -0.2962 0.
108 - 54 -0.5400 -1.1296 0.
108 - 36 -0.3367 -0.9262 0.
108 - 72 -0.2300 -0.8196 0.
108 - 144 0.0633 -0.5262 0.
144 - 54 -0.6033 -1.1929 -0.0138 ***
144 - 36 -0.4000 -0.9896 0.
144 - 72 -0.2933 -0.8829 0.
144 - 108 -0.0633 -0.6529 0.
3.3 SAS Code for Cotton Fiber Breaking Strength RCBD
DM ’LOG; CLEAR; OUT; CLEAR;’;
OPTIONS NODATE NONUMBER LS=76;
ODS GRAPHICS ON;
ODS PRINTER PDF file=’C:\COURSES\ST541\RCBD.PDF’;
*** A RANDOMIZED COMPLETE BLOCK DESIGN ***;
DATA in; INPUT k2O block strength @@; CARDS;
36 1 7.62 36 2 8.00 36 3 7.
54 1 8.14 54 2 8.15 54 3 7. 72 1 7.76 72 2 7.73 72 3 7.
108 1 7.17 108 2 7.57 108 3 7.
144 1 7.46 144 2 7.68 144 3 7.
PROC GLM DATA=in PLOTS = (ALL);
CLASS k2O block; MODEL strength = k2O block / SS3 SOLUTION;
MEANS block;
MEANS k2O / TUKEY CLDIFF LINES;
ESTIMATE ’K2O=36’ K2O 4 -1 -1 -1 -1 / DIVISOR=5;
ESTIMATE ’K2O=54’ K2O -1 4 -1 -1 -1 / DIVISOR=5;
ESTIMATE ’K2O=72’ K2O -1 -1 4 -1 -1 / DIVISOR=5;
ESTIMATE ’K2O=108’ K2O -1 -1 -1 4 -1 / DIVISOR=5; ESTIMATE ’K2O=144’ K2O -1 -1 -1 -1 4 / DIVISOR=5;
TITLE ’ANOVA RESULTS FOR STRENGTH BY TREATMENT’;
RUN;
- In both the fixed and random block cases, the ANOVA F -tests associated with treatment effects are
identical. You use F 0 = M Strt/M SE to test
H 0 : τ 1 = · · · = τa = 0 against H 1 : not all of the τis are equal (9)
- The EMS for the RCBD indicates that the correct denominator EMS for testing for a significant
block effect (either fixed or random) is the EMS for the restriction error. The problem is that this is
not estimable from the data.
- Under these circumstances, the test of the hypothesis involving the combination of the block effects
and the restriction error in (10) would be appropriate to test for a ‘general’ blocking effect.
- The statistic F = M Sblocks/M SE is actually a test of
H 0 : σ
2 δ +^ φ(β) = 0^ against^ H^1 :^ σ
2 δ +^ φ(β)^6 = 0^ (10)
Note that even if β 1 = β 2 = · · · = βb = 0 (fixed) or σ
2 β = 0 (random)^ is true, we still have the
restriction error in the EMS which prevents it from matching the error EMS = σ
2 .
- Because of the restriction on randomization, A&M claim that there is no F test for blocks. That
is, there is no test for H 0 : σ
2 β
= 0 if blocks are random and no test for H 0 : β 1 = β 2 = · · · = βb if
blocks are fixed.
- Fortunately this is not a problem because most of the time the experimenter is only interested in
whether or not blocking had been effective in reducing the M SE for improved testing of the effects
of the treatment of interest.
3.5 Example of an Analysis With and Without Blocks
Three different disinfecting solutions are being compared to study their effectiveness in stopping the growth
of bacteria in milk containers. The analysis is done in a laboratory, and only three trials can be run on
any day. Because days could represent a potential source of variability, the experimenter decides to use
a randomized block design with days as blocks. Observations are taken for four days. The inside of the
milk containers are covered with a certain amount of bacteria. The response is the percentage of bacteria
remaining after rinsing the container with a disinfecting solution.
Day
Solution 1 2 3 4
- The data were analyzed assuming two different models. The first model does not include blocks.
The second model includes blocks. The SAS analysis for both models is on the next page. Here are
important results:
Without blocks With blocks
R
M SE
p-value
- Note that we would fail to reject H 0 if blocks were not in the model because there is large variability
across blocks (M Sday = 368.97).
- If the SSday = 1106.92 and dfday = 3 is pooled with the the SSE = 41.83 and dfE = 6 in the model
with days (blocks), then it forms the SSE = 1158.75 and dfE = 9 for the model without days (blocks).
SAS Code for RCBD Analyses With and Without Blocks
DM ’LOG;CLEAR;OUT;CLEAR’;
ODS GRAPHICS ON;
- ODS PRINTER PDF file=’C:\COURSES\ST541\RCBD2.PDF’;
OPTIONS NODATE NONUMBER LS=76 PS=54;
*** RCBD ANALYSES WITH AND WITHOUT BLOCKS ***;
DATA IN;
DO solution = 1 TO 3;
DO day = 1 TO 4;
INPUT growth @@; OUTPUT;
END; END;
LINES;
13 22 18 39 16 24 17 44 5 4 1 22
;
*******************************************************;
*** RUN AN ANOVA WITH SOLUTION ONLY, NO DAY BLOCKS ***;
*******************************************************;
PROC GLM DATA=IN;
CLASS solution;
MODEL growth = solution / ss3;
TITLE ’RCBD WITHOUT DAYS (BLOCKS) IN THE MODEL’;
*** RUN AN ANOVA WITH DAYS AS BLOCKS ***;
PROC GLM DATA=IN;
CLASS day solution;
MODEL growth = solution day / ss3;
TITLE ’RCBD WITH DAYS (BLOCKS) IN THE MODEL’;
RUN;
3.6 Type I vs Type III Analyses
- Without the /ss3 option in the MODEL statement, SAS will contain two ANOVA tables: ANOVA
for Type I sum of squares and ANOVA for Type III sum of squares.
- If there are no missing observations, the Type I and Type III analyses are identical.
- If there are missing observations, the Type I and Type III analyses are different. To see how they
differ we will first look at the Type I analysis.
3.6.1 Type I Analysis
- The Type I analysis is based on sequentially fitting the data to the model one factor at a time. It is
often referred to as the sequential sum of squares method.
- For the RCBD there are two possibilities that I will refer to as
- Version 1 (V1) when fitting treatments before blocks.
- Version 2 (V2) when fitting blocks before treatments.
- Let RSSi be the error sum of squares (SSE ) after fitting the model in the i
th step.
- The steps for determining the ANOVA SS for V1 are:
- Fit yij = μ + ij and obtain RSS 1 = SStotal.
- Fit yij = μ + τi + ij and obtain RSS 2 = SSE for the model with treatments only.
- Fit yij = μ + τi + βj + ij and obtain RSS 3 = SSE for the model with treatments and blocks.
- The steps for determining the ANOVA SS for V2 are:
- Fit yij = μ + ij and obtain RSS 1 = SStotal.
′
. Fit yij = μ + βj + ij and obtain RSS
∗ 2 =^ SSE^ for the model with blocks only.
- Fit yij = μ + τi + βj + ij and obtain RSS 3 = SSE for the model with blocks and treatments..
- The ANOVA sum of squares for V1 and V2 are summarized in the following table:
Step V1 Source Fit df Type I SS for V
1 Total μ N − 1 RSS 1
2 Treatment τi a − 1 R(τ |μ) = RSS 1 − RSS 2
3 Blocks βj b − 1 R(β|τ, μ) = RSS 2 − RSS 3
3 Error ij N − a − b + 1 RSS 3
Step V2 Source Fit df Type I SS for V
1 Total μ N − 1 RSS 1
′ Blocks βj b − 1 R(β|μ) = RSS 1 − RSS
∗ 2
3 Treatment τi a − 1 R(τ |β, μ) = RSS
∗ 2 −^ RSS
∗ 3
3 Error ij N − a − b + 1 RSS 3
- In V1, the quantity R(τ |μ) is called the reduction in SS due to τ adjusted for μ and R(β|τ, μ)
is called the reduction in SS for β adjusted for τ and μ.
- In V2, the quantity R(β|μ) is called the reduction in SS due to β adjusted for μ and R(τ |β, μ)
is called the reduction in SS for τ adjusted for β and μ.
3.6.2 Type III Analysis
- The Type III analysis is referred to as the marginal means or the Yates weighted squares of
means analysis.
- For a RCBD, the Type III SStrt and SSblocks are computed using the following procedure:
- Fit the model with treatments only: yij = μ + τi + ij. Then RSS 2 = SSE for this model.
- Fit the model with blocks only: yij = μ + βj + ij. Then RSS
∗ 2 =^ SSE^ for this model.
- Fit the model yij = μ + τi + βj + ij. Then RSS 3 = SSE and RSS 1 = SStotal for the model
with both treatments and blocks.
Step Source Fit df Type III SS
1 Total μ N − 1 RSS 1
2 Treatment τi a − 1 R(τ |β, μ) = RSS
∗ 2 −^ RSS^3
3 Blocks βj b − 1 R(β|τ, μ) = RSS 2 − RSS 3
1 Error ij N − a − b + 1 RSS 3
- If any yij values are missing, then SStrt + SSblocks + SSE 6 = SStotal for a Type III analysis.
3.6.3 RCBD Analysis with a Missing Observation
See the example in Section 3.5 for the description of the experiment. Suppose y 23 was missing from the
RCBD. The RCBD data table is:
Day
Solution 1 2 3 4
- Let us examine the Type I and Type III sums of squares. The next page contains the SAS output.
- The top of the next page contains the Type I (V1) sum of squares and the bottom of the page contains
the Type I (V2) sum of squares. Note the difference in sums of squares, mean squares, F-statistics,
and p-values for the Type I analyses.
- The reason for the difference between the V1 and V2 Type I sum of squares is that a Type I analysis
is sequential so the order in which terms enter the model is important.
- The Type III analysis is the same for both analyses Type III sums of squares are not calculated
sequentially. That is, the order in which terms enter the model is not important.
- The following page contains the two analyses with only one effect in each model. I included these
analyses so you can see how RSS 2 and RSS
∗ 2 are calculated.
RSS 2 is the SSE for the model with only treatments and no blocks.
ANOVA RESULTS FOR THE MODEL WITH SOLUTION (TREATMENTS) ONLY
ANOVA RESULTS (SOLUTION ONLY)
The GLM Procedure
Variable: growth
ANOVA RESULTS (SOLUTION ONLY)
The GLM Procedure
Variable: growth
Source DF
Sum of
Squares Mean Square F Value Pr > F
Model 2 790.909091 395.454545 2.96 0.
Error 8 1068.000000 133.
Corrected Total 10 1858.
R-Square Coeff Var Root MSE growth Mean
0.425469 61.10405 11.55422 18.
Source DF Type I SS Mean Square F Value Pr > F
solution 2 790.9090909 395.4545455 2.96 0.
Source DF Type III SS Mean Square F Value Pr > F
solution 2 790.9090909 395.4545455 2.96 0.
0
10
20
30
40
growth
1 2 3
solution
Prob > F 0.
F 2.
Distribution of growth
0
10
20
30
40
growth
1 2 3
solution
Prob > F 0.
F 2.
Distribution of growth
RSS
∗ 2 is the^ SSE^ for the model with only blocks and no treatments.
ANOVA RESULTS FOR THE MODEL WITH DAYS (BLOCKS) ONLY
ANOVA RESULTS (DAY ONLY)
The GLM Procedure
Variable: growth
ANOVA RESULTS (DAY ONLY)
The GLM Procedure
Variable: growth
Source DF
Sum of
Squares Mean Square F Value Pr > F
Model 3 1141.075758 380.358586 3.71 0.
Error 7 717.833333 102.
Corrected Total 10 1858.
R-Square Coeff Var Root MSE growth Mean
0.613842 53.55403 10.12658 18.
Source DF Type I SS Mean Square F Value Pr > F
day 3 1141.075758 380.358586 3.71 0.
Source DF Type III SS Mean Square F Value Pr > F
day 3 1141.075758 380.358586 3.71 0.
40
Prob > F 0.
F 3.
Distribution of growth
40
Prob > F 0.
F 3.
All of these calculations are done automatically in the RCBD analyses for the two models Distribution of growth
on the previous page.
Type I SS (V1) Summary
RSS 1 = 1858. 91 R(μ) = RSS 1 = 1858.
RSS 2 = 1068. 00 R(τ |μ) = RSS 1 − RSS 2 = 790.
RSS 3 = 47. 33 R(β|τ, μ) = RSS 2 − RSS 3 = 1020.
Type I SS (V2) Summary
RSS 1 = 1858. 91 R(μ) = RSS 1 = 1858.
RSS
∗
2 = 717.^83 R(β|μ) =^ RSS^1 −^ RSS
∗
RSS 3 = 47. 33 R(τ |β, μ) = RSS
∗
2 −^ RSS^3 = 670.
Type III SS Summary
RSS 1 = 1858. 91 R(μ) = RSS 1 = 1858.
RSS 3 = 47. 33
RSS
∗
2 = 717.^83 R(β|τ, μ) =^ RSS
∗
2 −^ RSS^1 = 1020.
RSS 2 = 1068. 00 R(τ |β, μ) = RSS 2 − RSS 1 = 670.
DM ’LOG; CLEAR; OUT; CLEAR;’;
ODS GRAPHICS ON;
ODS PRINTER PDF file=’C:\COURSES\ST541\RCBDMISS.PDF’;
OPTIONS NODATE NONUMBER;
***************************************;
*** RCBD WITH A MISSING OBSERVATION ***;
***************************************;
DATA IN;
DO solution = 1 TO 3; DO day = 1 TO 4; INPUT growth @@; OUTPUT; END; END;
CARDS;
13 22 18 39 16 24. 44 5 4 1 22
;
***************************************************;
*** RUN AN ANOVA WITH SOLUTION APPEARING FIRST ***;
***************************************************;
PROC GLM DATA=IN;
CLASS solution day; MODEL growth = solution day;
TITLE ’ANOVA RESULTS (SOLUTION THEN DAY)’;
**********************************************;
*** RUN AN ANOVA WITH DAY APPEARING FIRST ***;
**********************************************;
PROC GLM DATA=IN;
CLASS day solution; MODEL growth = day solution;
TITLE ’ANOVA RESULTS (DAY THEN SOLUTION)’;
****************************************;
*** RUN AN ANOVA WITH SOLUTION ONLY ***;
****************************************;
PROC GLM DATA=IN;
CLASS solution; MODEL growth = solution;
TITLE ’ANOVA RESULTS (SOLUTION ONLY)’;
***********************************;
*** RUN AN ANOVA WITH DAY ONLY ***;
***********************************;
PROC GLM DATA=IN;
CLASS day; MODEL growth = day;
TITLE ’ANOVA RESULTS (DAY ONLY)’;
RUN;
3.6.4 Type I vs Type III Hypotheses
- Because of differences between Type I and Type III SS, there will be differences in the hypotheses
associated with the F -tests (assuming the restriction on randomization is ignored).
- Let μij = μ + τi + βj be the i
th treatment, j
th block mean.
Hypotheses for Type III and Type I (V2) Sum of Squares
H 0 : μ 1 · = μ 2 · = · · · = μa·
H 1 : μi· 6 = μi∗· for some i 6 = i
∗ and μi· =
∑^ b
j=
μij
(^) /b.
Hypotheses for Type I (V1) Sum of Squares
H 0 :
n 1 ·
∑^ b
j=
n 1 j μ 1 j =
n 2 ·
∑^ b
j=
n 2 j μ 2 j = · · · =
na·
∑^ b
j=
naj μaj
H 1 :
ni·
∑^ b
j=
nij μij 6 =
ni∗·
∑^ b
j=
ni∗j μi∗j for some i 6 = i∗.
where ni· = the number of nonmissing yij values for the i
th treatment, and nij = 1 if yij is not missing
and nij = 0 if yij is missing.
- The Type III hypotheses are comparing the treamtment means average across the blocks (and are
the ones I want to test.) Therefore I recommend using the p-values from a Type III analysis.
- If there are no missing yij values, the Type I and Type III hypotheses are the same.
3.7 RCBD Normal Equations
- For model yij = μ + τi + βj + ij , the error is ij = yij − μ − τi − βj
- Substituting in estimates produces the residual ̂ij = eij = yij − μ̂ − τ̂i − β̂j.
- Goal: Find μ̂ , ̂τi, and β̂j that minimize L:
L =
a ∑
i=
b ∑
j=
2 ij =
a ∑
i=
b ∑
j=
(yij − μ̂ − ̂τi − β̂j )
2
- Solution: Solve the normal equations
∂L
∂ μ̂
∑^ a
i=
∑^ b
j=
(yij − μ̂ − ̂τi − β̂j ) = 0
∂L
∂ ̂τi
∑^ b
j=
(yij − ̂μ − τ̂i − β̂j ) = 0 for i = 1, 2 ,... , a
∂L
∂ β̂j
∑^ a
i=
(yij − ̂μ − τ̂i − β̂j ) = 0 for j = 1, 2 ,... , b
- After distributing the sum and then simplifying, we get:
(i) y·· = ab μ̂ + b
a ∑
i=
̂ τi + a
b ∑
j=
β̂ j
(ii) yi· = bμ̂ + b ̂τi +
b ∑
j=
β̂ j for^ i^ = 1,^2 ,... , a
(iii) y·j = a μ̂ +
a ∑
i=
̂ τi + aβ̂j for j = 1, 2 ,... , b
- For (i), (ii), and (iii), there is a total of 1 + a + b equations. If you sum the a equations in (ii), you
get (i). If you sum the b equations in (iii), you also get (i). Thus, the rank is a + b − 1 which implies
that μ and each τi and βj are not uniquely estimable. To get estimates of μ and each τi and βj , we
must impose 2 constraints. We will use
∑^ a
i=
τi = 0 and
∑^ b
j=
βj = 0.
- Substitution of these constraints into (i), (ii), and (iii) yields
(1) abμ̂ = y·· (2) bμ̂ + b̂τi = yi· (3) a μ̂ + aβ̂j = y·j
μ̂ =
y··
ab
- Substitution of ̂μ = y·· in (2) yields:
by·· + b̂τi = yi· −→ y·· + ̂τi = yi· −→ ̂τi =
- Substitution of ̂μ = y·· in (3) yields:
ay·· + a β̂j = y·j −→ y·· + β̂j = y·j −→ β̂j =
3.8 Matrix Forms for the RCBD
Example: The goal is to determine whether or not four different tips produce different readings on a
hardness testing machine. The machine operates by pressing the tip into a metal test coupon, and from
the depth of the resulting depression, the hardness of the coupon can be determined. The experimenter
decides to obtain four observations for each tip. Four randomly selected coupons (blocks) were used and
each tip (treatment) was tested on each coupon. The data represent deviations from a desired depth in
0.1 mm units:
Type of Tip
Type of Coupon 1 2 3 4
Alternate Approach: Keeping a + b + 1 Columns
μ τ 1 τ 2 τ 3 τ 4 β 1 β 2 β 3 β 4
X =
y =
X
′ X =
(X
′ X)
− 1
X
′ y =
(X
′ X)
− 1 X
′ y =
μ
̂ τ 1
̂ τ 2
̂ τ 3
̂ τ 4
β̂ 1
β̂ 2
β̂ 3
β̂ 4
