Example of Split Block Design (Strip Plot)

Application Example

Suppose, the same data as the example on the split-plot but designed using a split-block design. Combination of NPK Fertilizer (Vertical Factor, A) and Rice Genotype (Horizontal Factor, B).

Table 13. The effect of applying a combination of fertilizers and rice genotypes on rice yields.

Calculation by Hand:

Step 1: Calculate the Correction Factor

$$ CF=\frac{Y...^2}{abr}=\frac{(1906.3)^2}{6\times2\times4}=75707.91$$

Step 2: Calculate the Sum of The Total Squares

$$\begin{matrix}SSTOT=\sum_{i,j,k}{Y_{ijk}}^2-CF\\=(20.7)^2+(32.1)^2+...+(45.6)^2-75707.91\\=2273.94\\\end{matrix}$$

Create Vertical Direction Table (Factor A x Block)

Step 3: Calculate the Sum of Squares of Blocks

$$\begin{matrix}SSR=\frac{\sum_{k}{(r_k)^2}}{ab}-CF\\=\frac{(449)^2+(482.6)^2+(461.5)^2+(513.2)^2}{6\times2}-75707.91\\=197.11\\\end{matrix}$$

Step 4: Calculate the Sum of Squares of Factor A

$$\begin{matrix}SS(A)=\frac{\sum_{i}{(a_i)^2}}{rb}-CF\\=\frac{(244.7)^2+(259.5)^2+...+(365.9)^2}{4\times2}-75707.91\\=1674.80\\\end{matrix}$$

Step 5: Calculate the Sum of Squares of Main Plot Errors (Ea)

$$\begin{matrix}SS(Ea)=\frac{\sum_{i,k}{(a_ir_k)^2}}{b}-CF-SSR-SS(A)\\=\frac{(48.4)^2+(65.1)^2+...+(86.5)^2+(92.2)^2}{2}-75707.91-197.11-1674.80\\=267.73\\\end{matrix}$$

Create a Horizontal Direction Table (Factor B x Block):

Step 6: Calculate the Sum of The Squares of Factor B

$$\begin{matrix}SS(B)=\frac{\sum_{j}{(b_j)^2}}{ra}-CF\\=\frac{(953.8)^2+(952.5)^2}{4\times6}-75707.91\\=0.035\\\end{matrix}$$

$$\begin{matrix}SS(Eb)==\frac{\sum_{j,k}{(b_lr_k)^2}}{a}-CF-SSR-SS(B)\\=\frac{(222.4)^2+(243.2)^2+...+(229.4)^2+(257.1)^2}{6}-75707.91-197.11-0.035\\=3.33\\\end{matrix}$$

Create A Table For Total Treatment:

Step 7: Calculate the Sum of Squares of AB Interactions

$$\begin{matrix}SS(AB)=\frac{\sum_{i,j}{(a_ib_j)^2}}{r}-CF-SS(A)-SS(B)\\=\frac{(120.0)^2+(124.7)^2+...+(187.6)^2+(178.3)^2}{4}-75707.91-1674.80-0.035\\=78.59\\\end{matrix}$$

Step 8: Calculate the Sum of Squares of Error C (SSEc)

$$\begin{matrix}SSEc=SSTOT\ -\ SS(Other)\ \\=SSTOT\ -\ SSB\ -\ SS(A)\ -\ SSEa-SS(B)-SSEb-SS(AB)\\=2273.94-197.114-1674.80-267.73-0.035-3.33-78.59\\=52.35\\\end{matrix}$$

Step 9: Create an Analysis of variance Table with its F-Values

Table 14. Anova Table of Rice Yield

$$\begin{matrix}CV(a)=\frac{\sqrt{MS(Ea)}}{\bar{Y}...}=\frac{\sqrt{17.849}}{39.715}\\=10.64\%\\\end{matrix}$$

$$\begin{matrix}CV(b)=\frac{\sqrt{MS(Eb)}}{\bar{Y}...}=\frac{\sqrt{1.110}}{39.715}\\=2.65\%\\\end{matrix}$$

$$\begin{matrix}CV(c)=\frac{\sqrt{MS(Ec)}}{\bar{Y}...}=\frac{\sqrt{3.490}}{39.715}\\=4.70\%\\\end{matrix}$$

Step 10: Make a Conclusion

First, we check whether the Interaction Effect is significant or not? If it is significant, then examine the simple effect of the interaction, and ignore the main effect, even if the main effect is significant! Why? Take another look at the discussion about the effect of interactions and main effects! Main effect testing (if significant) is only performed when the effect of the interaction is not significant.

Effect of AB Interactions

Because Fstat (4.50) > 2. 901 then we reject H₀: μ₁ = μ₂ = ... at a confidence level of 95% (usually given one asterisk mark (*), which indicates a noticeable difference)

Main Effects

Since the effect of the interaction is significant, then its main effect does not need to be discussed further.

Post Hoc

Based on the anova, the effect of the interaction is significant so that the testing of the main effect of the treatment of the combination of fertilizers and the two rice genotypes does not need to be carried out. The next step is to examine its simple effect because the interaction between the two factors is significant.

Here are the steps to test Advanced Test by using LSD:

Test criteria:

Compare the absolute value of the difference between the two averages that we will see the difference with the LSD value with the following test criteria:

$$ if\ \ \left|\mu_i-\mu_j\right|\ \ \left\langle\ \ \begin{matrix}>LSD_{0.05}\ \mathbf{reject}\ H_0,\ two\ means\ are\ significantly\ different\\\le LSD_{0.05}\ \mathbf{accept}\ H_0,\ two\ means\ are\ not\ significantly\ different.\\\end{matrix}\right.$$

Comparison of Vertical Factor Mean, A (between two fertilization combinations on the same genotype):

Calculate the appropriate Comparative Value (LSD)

To compare two averages of Factor A (vertical) (the average pair of fertilization combinations) on the same Factor B (horizontal) treatment, the standard error was calculated using the formula:

$$s_{\bar{Y}}=\sqrt{\frac{2\left[(a-1)MS(Ec)+MS(Eb)\right]}{ra}}$$

From the formula, it can be seen that to compare the two average values of Vertical Factor (A) on the same Horizontal Factor (B) treatment, two types of MS (Error) are used, namely MS (Ea) and MS (Ec). The implication is that the ratio of the difference in treatment to standard error does not follow the distribution of t-students, so it needs to be calculated t combined/weighted. If t_a and t_c are successively the values of t obtained from the table with a certain significant degree at error-degree of freedom a and error-degree of freedom c, then the weighted value of t is:

$$ t'=\frac{(b-1)(MS\ \ Ec)(t_c)+(MS\ \ Ea)(t_a)}{(b-1)(MS\ \ Ec)+(MS\ \ Ea)}$$

ta = t_(0.05/2.15) = 2.131

tc = t_(0.05/2.15) = 2.131

b = 2 (Leveling Factor Level, B)

MS(Ea) = 17.8485

MS(Ec) = 3.48999

so that:

$$\begin{matrix}t'=\frac{(b-1)(MS\ \ Ec)(t_c)+(MS\ \ Ea)(t_a)}{(b-1)(MS\ \ Ec)+(MS\ \ Ea)}\\=\frac{(2-1)(3.48999)(2.131+(17.8485)(2.131)}{(2-1)(3.48999)+(17.8485)}\\=2.131\\\end{matrix}$$

and

$$s_{\bar{Y}}=\sqrt{\frac{2\left[(b-1)MS(Ec)+MS(Ea)\right]}{rb}} \\ s_{\bar{Y}}=\sqrt{\frac{2\left[(2-1)3.48999+17.8485\right]}{4\times2}} \\ =2.30968$$

So:

$$\begin{matrix}LSD=t'\times s_Y\\=2.131\times2.3097\\=4.9219\ \ kg\\\end{matrix}$$

Compare the difference in the average treatment with the LSD value = 4.922. State it differently if the average difference is greater than the LSD value.

Comparison between the average fertilization combinations (Factor A) at the level of Genotype IR-64

Comparison between the average fertilization combinations (Factor A) at the level of Genotype S-969

Comparison of Horizontal factors , B (between two rice genotypes on a specific fertilization combination):

Calculate the appropriate Comparative Value (LSD)

To compare the two averages of Factor B (the average pair of rice genotypes) on the same Factor A treatment, the standard error was calculated using the formula:

$$s_{\bar{Y}}=\sqrt{\frac{2\left[(a-1)MS(Ec)+MS(Eb)\right]}{ra}}$$

From the formula, it can be seen that to compare two average values of horizontal factor (B) on the same vertical factor (A) treatment, two types of MS (Error) are used, namely MS (Eb) and MS (Ec ). The implication is that the ratio of the difference in treatment to standard error does not follow the distribution of t-students so it needs to be calculated t combined/weighted. If t_b and t_c are successively the values of t obtained from the student table with a certain significant degree at error-degree of freedom a and error-degree of freedom c, then the weighted value of t is:

$$ t'=\frac{(a-1)(MS\ \ Ec)(t_c)+(MS\ \ Eb)(t_b)}{(a-1)(MS\ \ Ec)+(MS\ \ Eb)}$$

tb = t_(0.05/2.3) = 2.131 (Actually it is no longer feasible, because the error-degree of freedom is less than 6, which is 3)

tc = t_(0.05/2.15) = 2.131

a = 6 (Vertical Factor level, A)

MS(Eb) = 1.10965

MS(Ec) = 3.48999

so that:

$$t'=\frac{(a-1)(MSEc)(t_c)+(MSEa)(t_a)}{(a-1)(MSEc)+(MSEa)}\\=\frac{(6-1)(3.48999)(2.131)+(1.10965)(3.182)}{(6-1)(3.48999)+(1.10965)}\\=2.19384$$

and

$$s_{\bar{Y}}=\sqrt{\frac{2\left[(a-1)MS(Ec)+MS(Eb)\right]}{ra}} \\
s_{\bar{Y}}=\sqrt{\frac{2\left[(6-1)3.48999+1.10965\right]}{4\times6}} \\
=1.24364$$

So:

$$\begin{matrix}LSD=t'\times s_{\bar{Y}}\\=2.19384\times1.24364\\=2.72834\ \ kg\\\end{matrix}$$ 

Compare the difference in the average treatment with the LSD value = 2.728. State it differently if the average difference is greater than the LSD value. The result is as follows:

From the results of further tests of the simple effects above, the results can be summarized in the form of a Table of Fertilizer Interactions x Genotypes as below.

Information:

Letters in brackets are read in a horizontal direction, comparing between 2 G on the same P

Lowercase letters without brackets are read vertically, comparing between 2 P's on the same G

Calculation by SmartstatXL Excel-Add-In

Anova:

Post Hoc:

Interaction (Simple Effect):

Anova Assumption: