4.1. Introduction
Randomized Complete Block Design (RCBD) is generally used to improve the ability to detect true differences between the treatments we are trying by eliminating the influence of other known variations (groups) of experimental errors. If the idea is applied to eliminate two sources of diversity by grouping in two directions, then the design is called the Latin Square Design (LSD). Thus, LSD is an experimental design with two grouping directions , namely row and column. The number of treatments is equal to the number of replications so that each row and column will contain all treatments. In this design, randomization is limited by grouping it into rows and columns, so that each row and column will only get one treatment.
Introduction
RCBD is generally used to improve the ability to detect actual differences between the treatments we are trying to eliminate the effect of other variance that we know (groups) of trial errors. If the idea is applied to eliminate two sources of variance by grouping in two directions, then the design is called LATIN. Thus, the latin square design is an experimental design with two directions of grouping, namely rows and columns. The number of treatments is equal to the number of repeats so that each row and column will contain all the treatments. In this design, randomization is limited by grouping them into rows as well as columns, so that each row and column will get only one treatment.
In biological experiments, observations are often made repeatedly on the same experimental unit. In such cases, it is possible that some treatments will generate different effects during the experiment, consequently, it may be that it will affect the responses observed at different periods. So, the response may be a function of the treatment at a certain period. One way to eliminate such experiment errors is to include the time/period of observation into the treatment. Thus, here there are two directions of grouping (double blocking), firstly based on the group on the basic experiment, and secondly the group from the time of observation, so that the design becomes LATIN.
LATIN where the treatment in numerical or alphabetical order is placed in the first row and column is called the Basic Design (standard). The following figure is the basic design for 2x2, 3x3, and 4x4 sizes. Rapidly increasing the size (treatment) will increase the number of possible basic designs. The number of basic designs that may be formed is (# of standard squares) (K!) (K - 1)! where k is the number of treatments. For example, if k =4, then the possible basic design that can be formed is (4). (4!). (3!) = 576 possibilities.
|
|
|
|
|
|
|
| 4x4 |
|
|
|
|
|
|
|
| A | B | C | D |
| A | B | C | D |
2x2 |
|
| B | A | D | C |
| B | C | D | A | |
A | B |
|
| C | D | B | A |
| C | D | A | B |
B | A |
|
| D | C | A | B |
| D | A | B | C |
|
|
|
|
|
|
|
|
|
|
|
|
|
| 3x3 |
|
| A | B | C | D |
| A | B | C | D |
A | B | C |
| B | D | A | C |
| B | A | D | C |
B | C | A |
| C | A | D | B |
| C | D | A | B |
C | A | B |
| D | C | B | A |
| D | C | B | A |
Figure 6. LATIN basic design for 2x2, 3x3, and 4x4 sizes.
Advantages of latin square design:
- Reducing error variance using two clustering
- Treatment effects can be performed for small-scale experiments
- Relatively easy analysis
- Rows or columns can also be used to increase coverage in conclusions
Disadvantages of latin square design:
- The number of rows, columns and treatments should be the same, so that the more treatment, the more units of experiments required are also more.
- As the number of groups increases in size, the error of experiments per unit of experiment also tends to increase.
- The assumption of the model is very binding, that is, that there is no interaction between any two or all of the criteria, that is, rows, columns and treatments.
- The randomization required is a bit more complicated than the randomization of previous designs.
- Its smaller degree of error-free compared to other similarly sized designs will decrease the level of accuracy, especially if the number of treatments is small.
- Requires basic knowledge/understanding in compiling effective experimental units.
- If there is missing data, even if there are not too many of them, then the results of the analysis are doubtful because the treatment becomes unbalanced.
Randomization of the Latin Square Design
Each treatment appears once in each row and once in each column. First, select the basic design that corresponds to its size, then perform randomization in the direction of the rows, and then randomization in the direction of the columns.
For example, there are 4 treatments A, B, C, D.
- We choose the basic design size 4x4.
Rows\Columns | 1 | 2 | 3 | 4 |
1 | A | B | C | D |
2 | B | A | D | C |
3 | C | D | B | A |
4 | D | C | A | B |
- Randomization at the position of the row. For example, randomization using the Rand() function in MS Excel (how to randomize in detail with the help of MS Excel can be seen in CRD randomization). From the randomization process we get a new sequence of 4, 3, 1, 2. That is, the row whose origin is in the 1st position (original order) changes to position 4 (Randomization order), the 2nd row becomes the 3rd row, etc.
From the results of randomization at the position of the row we get the following results:
Rows\Columns | 1 | 2 | 3 | 4 |
4 | D | C | A | B |
3 | C | D | B | A |
1 | A | B | C | D |
2 | B | A | D | C |
- In the same way, we do randomization for the position of the column. Suppose we get the randomization order: 2, 1, 4, 3. That is, the 1st column becomes the 2nd column, the 2nd column becomes the 1st column, etc. From the results of randomization at the position of the column we get the following results:
Rows\Columns | 2 | 1 | 4 | 3 |
4 | C | D | B | A |
3 | D | C | A | B |
1 | B | A | D | C |
2 | A | B | C | D |
When you're done randomizing the row and column directions, then create the layout of the design as shown below.
C | D | B | A |
D | C | A | B |
B | A | D | C |
A | B | C | D |
Figure 7. Design Layout for LATIN experiments
Linear Model of Latin Square Design
The linear model of the latin square design is:
$$Y_{ijk}=\mu+\beta_i+\kappa_j+\tau_k+\varepsilon_{ijk}$$
μ = general average
βi = effect of the i-th row
κj = effect of the j-th column
τk = kth treatment effect
εijk = random effect of i-th row, k-th column and k-th treatment
i = 1, 2, ... ,r ; j = 1, 2, ... ,r ; k = 1, 2, ... ,r
Assumption:
The effect of fixed treatment | Effect of random treatment |
$$\sum{\beta_i\ \ =\ \sum{\kappa_j=}\sum\tau_k}\ =\ 0\ ;\ \ \ \ \ \varepsilon_{ijk}\buildrel~\over~bsiN(0,\sigma^2)$$ | $$\begin{matrix}\beta_i\buildrel~\over~bsiN(0,{\sigma_\beta}^2);\ \ \ \ \ \ \kappa_j\buildrel~\over~bsiN(0,{\sigma_j}^2);\ \ \ \ \ \\\tau_k\buildrel~\over~bsiN(0,{\sigma_\tau}^2);\ \ \ \ \ \ \ \ \ \varepsilon_{ijk}\buildrel~\over~bsiN(0,\sigma^2)\\\end{matrix}$$ |
Hypothesis:
Hypotheses to Be Tested: | The effect of fixed model | Effect of random model |
H0 | All τk = 0 (k = 1, 2, ..., r) | στ2 = 0 (no variance in the treatment population) |
H1 | Not all τk = 0 (k = 1, 2, ..., r) | στ2 > 0 (There is variance in the treatment population) |
Analysis of variance:
Parameters | Estimators |
$$ \mu\$$ | $$\hat{\mu}=\ \bar{Y}..$$ |
$$\beta_i$$ | $${\hat{\beta}}_{i\ }\ =\ Y_{i.}-\bar{Y}..$$ |
$$\kappa_j$$ | $${\hat{\kappa}}_j\ =\ {\bar{Y}}_{.j}-\bar{Y}..$$ |
$$\tau_k$$ | $${\hat{\tau}}_k\ =\ {\bar{Y}}_k-\bar{Y}..$$ |
$$\varepsilon_{ij(k)}$$ | $${\hat{\varepsilon}}_{ij}=Y_{ij}-{\overline{Y}}_{i.}-{\overline{Y}}_{.j}-{\overline{Y}}_k+2{\overline{Y}}_{..}$$ |
The Sum of squares equation for the linear model Yij(k) = μ + βi + κj + τ(k) + εij is as follows:
$$\sum_{i,j}^{r}{({\overline{Y}}_{ij}-{\overline{Y}}_{..})^2=r\sum_{i=1}^{r}{({\overline{Y}}_{i.}-{\overline{Y}}_{..})^2+r\sum_{j=1}^{r}{({\overline{Y}}_{.j}-{\overline{Y}}_{..})^2}+r\sum_{k=1}^{r}{({\overline{Y}}_k-{\overline{Y}}_{..})^2}+\sum_{i,j}^{k}{(Y_{ij}-{\overline{Y}}_{i.}-{\overline{Y}}_{.j}+-{\overline{Y}}_k+2{\overline{Y}}_{..})^2}}}$$
or, SSTOT = SSRow + SS Column + SST + SSE
The formula of the sum of squares in the latin square design is given below :
| Definition | Calculation by Hand |
CF | $$\frac{Y..^2}{r^2}$$ | $$\frac{Y..^2}{r^2}$$ |
SSTOT | $$\sum_{i,j}{\left(Y_{ij}-\bar{Y}..\right)^2=\sum_{i,j}{{Y_{ij}}^2-\frac{Y..^2}{r^2}}}$$ | $$\sum_{i,j}{{Y_{ij}}^2-CF}$$ |
SSRow | $$ r\sum_{i}\left({\bar{Y}}_{i.}-\bar{Y}..\right)^2=\sum_{i}\frac{{Y_{i.}}^2}{r}-\frac{Y..^2}{r^2}$$ | $$\sum_{i}\frac{{Y_{i.}}^2}{r}-CF$$ |
SSColumns | $$ r\sum_{j}\left({\bar{Y}}_{.j}-\bar{Y}..\right)^2=\sum_{j}\frac{{Y_{.j}}^2}{r}-\frac{Y..^2}{r^2}$$ | $$\sum_{j}\frac{{Y_{.j}}^2}{r}-CF$$ |
SST | $$ r\sum_{k}\left({\bar{Y}}_k-\bar{Y}..\right)^2=\sum_{k}\frac{{Y_k}^2}{r}-\frac{Y..^2}{r^2}$$ | $$\sum_{k}\frac{{Y_k}^2}{r}-CF$$ |
SSE | $$\sum_{i,j}{(Y_{ij}-{\bar{Y}}_{i..}-{\bar{Y}}_{.j.}-{\bar{Y}}_k+2{\bar{Y}}_{..})^2}$$ | SSTOT – SSRow – SSColumn – SST |
Table 18. Anova Table of Latin Square Designs
Sources of Variance | Degree of freedom | Sum of Squares | Mean Square | Fstat | E(MS) | |
| (DF) | (SS) | (MS) |
| All fixed factors | All random factors |
Row | r-1 | SSROW | SSROW/(r-1) | $$\frac{MSROW}{MSE}$$ | $$\sigma^2+r\frac{\sum{\beta_i}^2}{r-1}$$ | $$\sigma^2+r{\sigma^2}_\alpha$$ |
Column | r-1 | SSCOL | MSCOL/(t-1) | $$\frac{MSCOL}{MSE}$$ | $$\sigma^2+r\frac{\sum{\kappa_j}^2}{r-1}$$ | $$\sigma^2+r{\sigma^2}_\beta$$ |
Treatment | r-1 | SST | SST/(r-1) | $$\frac{MST}{MSE}$$ | $$\sigma^2+r\frac{\sum{\tau_k}^2}{r-1}$$ | $$\sigma^2+{\sigma^2}_\tau$$ |
Error | (r-1) (r-2) | SSE | SSE/(r-1)(r-2) |
| $$\sigma^2$$ | $$\sigma^2$$ |
Total | r2 –1 |
|
|
|
|
|
Test statistics suitable for testing the above hypothesis : $ F=\frac{MST}{MSE}$
The rule of decision if the error of type I is α, if F≤ Ftable [α; r-1, (r-1)(r-2)] then the decision is to accept H0 and vice versa. Ftable [α; r-1, (r-1)(r-2)] is the value of the broad table F to its right by α with the numerator-degree of freedom r-1 and the denominator-degree of freedom (r-1)(r-2).
Sometimes we want to test whether or not there is an effect of grouping changers. If the change in grouping is fixed, then the hypothesis test is as follows:
Test the hypothesis for changing the grouping of rows :
H0 :All βi = 0
H1 :Not all βi = 0
with test statistics $ F=\frac{MSROW}{MSE}$
Hypothesis Test for changing column groupings:
H0 :All κj = 0
H1 :Not all κj = 0
Test statistics : $ F=\frac{MSCOL}{MSE}$
Decision rules for the effect of rows and columns : if F≤ Ftable [α; r-1, (r-1)(r-2)] accept H0 and vice versa.
Standard Error
The standard error for the difference among treatment averages is calculated by the following formula:
$$S_{\bar{Y}}=\sqrt{\frac{2MSE}{t}}$$
Example of application:
To facilitate the understanding of the procedure for calculating anova of LATIN design, the following are presented case examples along with the calculation of fingerprints. The following table is the Layout and data of the results of the 4x4 LATIN experiment for hybrid corn flat yield data (A, B, and D) and testers (C) (Gomez & Gomez, 1995 p. 34).
No Row | Yield (t ha-1) | Sum of Rows | |||
1 | 2 | 3 | 4 | ||
1 | 1,640 (B) | 1,210 (D) | 1,425 (C) | 1,345 (A) | 5.620 |
2 | 1,475 (C) | 1,185 (A) | 1,400 (D) | 1,290 (B) | 5.350 |
3 | 1,670 (A) | 0.710 (C) | 1,665 (B) | 1,180 (D) | 5.225 |
4 | 1,565 (D) | 1,290 (B) | 1,655 (A) | 0.660 (C) | 5.170 |
Sum of Columns | 6.350 | 4.395 | 6.145 | 4.475 |
|
Total |
|
|
|
| 21.365 |
Steps in compiling the Anova Calculation:
- Arrange the data as in the table above (according to the layout of the experiment in the field), also include an explanation of the treatment code.
- Count the number of rows (B) and columns (B) and the number of General (G) as in the example table above.
- Calculate the Amount and Average for each Treatment.
| Sum | Average |
A | 5.855 | 1.464 |
B | 5.885 | 1.471 |
C | 4.270 | 1.068 |
D | 5.355 | 1.339 |
- Calculate Sum Squares for all sources of variance.
- Arrange the Anova Table and F-table Values:
Sources of Variance | DF | SS | MS | Fstat | F0.05 |
Row | 3 | 0.030154 | 0.010051 | 0.465393 | 4.757 |
Column | 3 | 0.827342 | 0.275781 | 12.7691* | 4.757 |
Treatment | 3 | 0.426842 | 0.142281 | 6.58783* | 4.757 |
Error | 6 | 0.129585 | 0.021598 |
|
|
Total | 15 | 1.413923 |
|
|
|
Post-Hoc (Tukey HSD)
- Step 1: Calculate the HSD value:
- Determine the value of the MSE and its degree of freedom obtained from the Variance Analysis Table.
- MSE = 0.021598
- ν = df = 6
- Determine the critical value from the table of the student's significant region.
- There are three parameters needed to determine the value of q, namely the significant level (α), p = the number of treatments to be compared, and the error-degree of freedom (df). In this example, p=4, the value df=6 (see error df in its Variance Analysis table) and α=0.05. Next, specify the value of q0.05(4, 6).
- To find the value of q0.05(6, 24) we can see it in the table Studentized range distribution at a significant level α = 0.05 with p = 4 and degree of freedom (v)= 6. Consider the following image to specify the q-table.
- From the table we get the value of the value of q0.05(6, 24) = 4.90
- Calculate the HSD value by using the following formula:
- $\begin{matrix}\omega=q_\alpha(p,\nu)\sqrt{\frac{MSE}{t}}\\=4.90\times\sqrt{\frac{0.021598}{4}}\\=0.36\\\end{matrix}$
- Test criteria:
- Compare the absolute value of the difference between the two averages that we will see the difference with the HSD value with the following test criteria:
$$ if\ \ \left|\mu_i-\mu_j\right|\ \ \left\langle\ \ \begin{matrix}>HSD_{0.05}\ \mathbf{reject}\ H_0,\ two\ means\ are\ significantly\ different\\\le HSD_{0.05}\ \mathbf{accept}\ H_0,\ two\ means\ are\ not\ significantly\ different.\\\end{matrix}\right.$$
- Step 2: Sort the average table of treatment from small to large or vice versa. Create a Matrix table of differences between treatment mean and compare them with comparator values (Tukey HSD = 0.36).
(C) | (D) | (A) | (B) | Notation | ||
Treatment | Average | 1.068 | 1.339 | 1.464 | 1.471 |
|
(C) | 1.068 | 0.000 | a | |||
(D) | 1.339 | 0.271 ns | 0.000 | ab | ||
(A) | 1.464 | 0.396 * | 0.125 ns | 0.000 | b | |
(B) | 1.471 | 0.404 * | 0.133 ns | 0.007 ns | 0.000 | b |
Description: ignore the red row, because it is already represented by the second row (b)
Calculation using SmartstatXL Excel Add-In
The result of the calculation using SPSS Software V.16.
No Row | Yield results (t ha-1) | |||
1 | 2 | 3 | 4 | |
1 | 1,640 (B) | 1,210 (D) | 1,425 (C) | 1,345 (A) |
2 | 1,475 (C) | 1,185 (A) | 1,400 (D) | 1,290 (B) |
3 | 1,670 (A) | 0.710 (C) | 1,665 (B) | 1,180 (D) |
4 | 1,565 (D) | 1,290 (B) | 1,655 (A) | 0.660 (C) |
Create a list of data as follows:
Row | Column | Treatment | Result |
1 | 1 | (B) | 1.640 |
2 | 1 | (C) | 1.475 |
3 | 1 | (A) | 1.670 |
4 | 1 | (D) | 1.565 |
1 | 2 | (D) | 1.210 |
2 | 2 | (A) | 1.185 |
3 | 2 | (C) | 0.710 |
4 | 2 | (B) | 1.290 |
1 | 3 | (C) | 1.425 |
2 | 3 | (D) | 1.400 |
3 | 3 | (B) | 1.665 |
4 | 3 | (A) | 1.655 |
1 | 4 | (A) | 1.345 |
2 | 4 | (B) | 1.290 |
3 | 4 | (D) | 1.180 |
4 | 4 | (C) | 0.660 |
Type:
UNIANOVA Results BY Row Column Treatment
/METHOD=SSTYPE(3)
/INTERCEPT=INCLUDE
/POSTHOC=treatment(TUKEY LSD)
/CRITERIA=ALPHA(0.05)
/DESIGN=Row Column Treatment.
Output
Tests of Between-Subjects Effects | |||||
Dependent Variable: result |
|
|
|
| |
Source | Type III Sum of Squares | Df | Mean Square | F | Sig. |
Corrected Model | 1,284a | 9 | .143 | 6.607 | .016 |
Intercept | 28.529 | 1 | 28.529 | 1320.944 | .000 |
row | .030 | 3 | .010 | .465 | .717 |
column | .827 | 3 | .276 | 12.769 | .005 |
Treatment | .427 | 3 | .142 | 6.588 | .025 |
Error | .130 | 6 | .022 |
|
|
Total | 29.943 | 16 |
|
|
|
Corrected Total | 1.414 | 15 |
|
|
|
a. R Squared = .908 (Adjusted R Squared = .771) |
|
| |||
Post Hoc
result | ||||
| Treatment | N | Subset | |
| 1 | 2 | ||
Tukey HSDa | (C) | 4 | 1.0675000000E0 |
|
(D) | 4 | 1.3387500000E0 | 1.3387500000E0 | |
(A) | 4 |
| 1.4637500000E0 | |
(B) | 4 |
| 1.4712500000E0 | |
Sig. |
| .138 | .608 | |
Means for groups in homogeneous subsets are displayed. Based on observed means. The error term is Mean Square(Error) = .022. | ||||
a. Uses Harmonic Mean Sample Size = 4,000. | ||||
Exercise
Petersons et. al (1951) conducted an experiment on the Turnip Green plant using the Latin Square Design. Weighing water content loss (I, II, III, IV, V) is carried out at a certain time / period. The data produced in the following table is water content-80 (percent) (ST, p. 225).
If we look at it, in the experiment there were two groups, first by plant, and secondly by time of measurement of water content.
Plants (Row) | Leaf size (A=smallest, B=largest) (Column) | ||||
A | B | C | D | E | |
1 | 6.67(V) | 7.15(IV) | 8.29(I) | 8.95(III) | 9.62(II) |
2 | 5.40(II) | 4.77(V) | 5.40(IV) | 7.54(I) | 6.93(III) |
3 | 7.32(III) | 8.53(II) | 8.50(V) | 9.99(IV) | 9.68(I) |
4 | 4.92(I) | 5.00(III) | 7.29(II) | 7.85(V) | 7.08(IV) |
5 | 4.88(IV) | 6.16(I) | 7.83(III) | 5.83(II) | 8.51(V) |
Data processing using Statistics software V. 7.0:
Variance Analysis Table
Degree of Freedom (df) | Water Content SS | Water Content MS | Water Content F | Water Content p | |
Intercept | 1 | 1297.296 | 1297.296 | 1924.799 | 1.28E-14 |
Plant | 4 | 28.8853 | 7.221324 | 10.71428 | 0.000623 |
Leaf Size | 4 | 23.70814 | 5.927034 | 8.793941 | 0.001483 |
Time | 4 | 0.627256 | 0.156814 | 0.232665 | 0.914655 |
Error | 12 | 8.087888 | 0.673991 | ||
Total | 24 | 61.30858 |
Post-Hoc (Tukey HSD)
Tukey HSD test; variable Water Content (LATIN_Torrie in ExampleData.stw) Homogenous Groups, alpha= .05000 Error: Between MS = .67399, df = 12.000 | |||
Treatment | Water Content | 1 | |
5 | IV | 6.900000 | **** |
3 | III | 7.206000 | **** |
1 | V | 7.260000 | **** |
4 | I | 7.318000 | **** |
2 | II | 7.334000 | **** |
Tukey HSD test; variable Water Content (LATIN_Torrie in ExampleData.stw) Homogenous Groups, alpha= .05000 Error: Between MS = .67399, df = 12.000 | |||||
Plant | Water Content | 1 | 2 | 3 | |
2 | 2 | 6.008000 | **** | ||
4 | 4 | 6.428000 | **** | ||
5 | 5 | 6.642000 | **** | **** | |
1 | 1 | 8.136000 | **** | **** | |
3 | 3 | 8.804000 | **** | ||
Tukey HSD test; variable Water Content (LATIN_Torrie in ExampleData.stw) Homogenous Groups, alpha= .05000 Error: Between MS = .67399, df = 12.000 | ||||
Leaf Size | Water Content | 1 | 2 | |
1 | A | 5.838000 | **** | |
2 | B | 6.322000 | **** | |
3 | C | 7.462000 | **** | **** |
4 | D | 8.032000 | **** | |
5 | E | 8.364000 | **** | |
