Sidebar Menu

Applied Example of a Split Plot Design Experiment:

This article is a continuation of the Split Plot Design

Suppose there is a study that wants to examine the effect of combination of NPK fertilization and rice genotype on rice yield (kg/plot). The combined effect of NPK fertilization (A) consisted of 6 levels placed as the main plot (main plot) and the rice genotype (B) consisted of 2 levels placed as subplots (subplot). The main plots were arranged using the basic RAK design with 3 replications. The experimental data and the steps for calculating the analysis of variance followed by Post Hoc test: Fisher's LSD can be studied in the document below. 

You can learn the tutorial analyse data using SmartstatXL and SPSS Software  on the following link:

Applied Example

Experiment: Effect of the combination of NPK fertilization and rice genotype on rice yield (kg/plot).  The effect of the NPK fertilization combination (A) consists of 6 levels placed as the main plot and the rice genotype (B) consists of 2 levels placed as sub plot (subplot).  The main plots are arranged using the basic design of the RCBD.  The experiment was repeated 3 times.  The data from the experiment are given in the following table.

Table 4.    Effect of the combination of NPK fertilization and rice genotype on rice yield (kg/plot)

Fertilizer (A)

Genotype (B)

Block (K)

1

2

3

4

Control

IR-64

20.7

32.1

29.5

37.7

 

S-969

27.7

33.0

26.3

37.7

PK

IR-64

30.0

30.7

25.5

36.9

 

S-969

36.6

33.8

27.0

39.0

N

IR-64

39.9

41.5

46.4

44.5

 

S-969

37.4

41.2

45.4

44.6

NP

IR-64

40.8

43.5

43.3

43.4

 

S-969

42.2

46.0

45.9

46.2

NK

IR-64

42.4

45.6

44.8

47.0

 

S-969

39.8

39.5

40.9

44.0

NPK

IR-64

48.6

49.8

42.6

46.6

 

S-969

42.9

45.9

43.9

45.6

Calculation by Hand:

Step 1: Calculate the Correction Factor

 $$ CF=\frac{Y...^2}{abr}=\frac{(1906.3)^2}{6\times2\times4}=75707.9102$$

Step 2: Calculate the Sum of The Total Squares

 $$\begin{matrix}SSTOT=\sum_{i,j,k}{Y_{ijk}}^2-CF\\=(20.7)^2+(32.1)^2+...+(45.6)^2-75707.9102\\=2273.93979\\\end{matrix}$$

Create a Fertilizer Table x Blocks:

Fertilizer (A)

Block (K)

Total Fertilizer

1

2

3

4

(Σai)

Control

48.4

65.1

55.8

75.4

244.7

PK

66.6

64.5

52.5

75.9

259.5

N

77.3

82.7

91.8

89.1

340.9

NP

83.0

89.5

89.2

89.6

351.3

NK

82.2

85.1

85.7

91.0

344.0

NPK

91.5

95.7

86.5

92.2

365.9

Total Block (Σrk)

449.0

482.6

461.5

513.2

1906.3

Step 3: Calculate the Sum of Squares of Blocks

 $$\begin{matrix}SSR=\frac{\sum_{k}{(r_k)^2}}{ab}-CF\\=\frac{(449)^2+(482.6)^2+(461.5)^2+(513.2)^2}{6\times2}-75707.9102\\=197.110625\\\end{matrix}$$

Step 4: Calculate the Sum of Squares of Factor A

 $$\begin{matrix}SS(A)=\frac{\sum_{i}{(a_i)^2}}{rb}-CF\\=\frac{(244.7)^2+(259.5)^2+...+(365.9)^2}{4\times2}-75707.9102\\=1674.79604\\\end{matrix}$$

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

 $$\begin{matrix}SS(Ea)=\frac{\sum_{i,k}{(a_ir_k)^2}}{b}-CF-SSR-SS(A)\\=\frac{(48.4)^2+(65.1)^2+...+(86.5)^2+(92.2)^2}{2}-75707.9102-197.110625-1674.79604\\=267.728125\\\end{matrix}$$

Create A Table For Total Treatment:

Fertilizer (A)

Genotype (B)

Total A

IR-64

S-969

(Σai)

Control

120.0

124.7

244.7

PK

123.1

136.4

259.5

N

172.3

168.6

340.9

NP

171.0

180.3

351.3

NK

179.8

164.2

344.0

NPK

187.6

178.3

365.9

Total B (Σbj)

953.8

952.5

1906.3

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

 $$\begin{matrix}SS(B)=\frac{\sum_{j}{(b_j)^2}}{ra}-CF\\=\frac{(953.8)^2+(952.5)^2}{4\times6}-75707.9102\\=0.03520833\\\end{matrix}$$

Step 7: Calculate the Sum of Squares of AB Interactions

 $$\begin{matrix}SS(AB)=\frac{\sum_{i,j}{(a_ib_j)^2}}{r}-CF-SS(A)-SS(B)\\=\frac{(120.0)^2+(124.7)^2+...+(187.6)^2+(178.3)^2}{4}-75707.9102-1674.79604-0.03520833\\=78.5910417\\\end{matrix}$$

Step 8: Calculate the Sum of Squares of Sub plot Errors (Eb)

 $$\begin{matrix}SS(Eb)=SSTOT\ -\ SS(Other)\ \\=SSTOT\ -\ SSB\ -\ SS(A)\ -\ SSEa-SS(B)\ -SS(AB)\\=2273.93979-197.110625-1674.79604-267.728125-0.03520833-78.5910417\\=55.67875\\\end{matrix}$$

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

Table 5      Anova Table of Rice Yield

Variety Sources

DF

SS

MS

F-stat

F .05

Main Plot

         

Block (K)

3

197.110625

65.7035417

3.68 *

3.287

Fertilizer (A)

5

1674.79604

334.959208

18.77 **

2.901

Error(a)

15

267.728125

17.8485417

-

 

Sub Plot

         

Genotype (B)

1

0.03520833

0.03520833

0.01 ns

4.414

AxB

5

78.5910417

15.7182083

5.08 **

2.773

Error(b)

18

55.67875

3.09326389

-

 

Total

47

2273.93979

 

 

 

CV (a) = 10.64 %;  CV (b) = 4.43 %; 

     

 $$\begin{matrix}CV(a)=\frac{\sqrt{MS(Ea)}}{\bar{Y}...}=\frac{\sqrt{17.8485}}{39.715}\\=10.64\%\\\end{matrix}$$

 $$\begin{matrix}CV(b)=\frac{\sqrt{MS(Eb)}}{\bar{Y}...}=\frac{\sqrt{3.09326}}{39.715}\\=4.43\%\\\end{matrix}$$

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

Since Fstat (5.08) > 2,773 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 analysis of variance, 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:

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

Comparison of The Average of Sub Plot (between two rice genotypes on a specific fertilization combination):

Calculate the appropriate Comparative Value (LSD)

  1. To compare two sub plot's averages (between rice genotypes) on the same main plot treatment (a combination of specific fertilizations), it is necessary to first determine the standard error (sy) of the SPD using the formula: 

 $$s_{\bar{y}}=\sqrt{\frac{2MS(Eb)}{r}}$$

  • Determine the t-student value:


 $$\begin{matrix}LSD=t_{0.05/2;18}\cdot s_{\bar{Y}}\\=t_{0.05/2;18}\cdot\sqrt{\frac{2MS(Eb)}{r}}\\=2.101\times\sqrt{\frac{2(3.0933)}{4}}\\=2.6129\ \ kg\\\end{matrix}$$

Compare the difference in the average treatment with the LSD value = 2.6129. State it differently if the average difference is greater than the LSD value.  Since the level of rice genotypes is only two, a simple examination of the effect on the ratio of two averages of sub plot can be simplified.  Give the same letter (a) to the two genotypes when the average difference is ≤ LSD and a different letter when the average difference is > LSD.  The result is as follows:

 

Fertilizer

 

Control

PK

N

NP

NK

NPK

IR-64

30.00 a

30.78 a

43.08 a

42.75 a

44.95 b

46.90 a

S-969

31.18 a

34.10 b

42.15 a

45.08 a

41.05 a

44.58 a.

Difference

1.18

3.33 *

0.93

2.33

3.90 *

2.33

Comparison of Main Plot Mean (between two fertilization combinations on the same or different genotypes):

Calculate the appropriate Comparative Value (LSD)

To compare two plot averages (average pairs of fertilization combinations) on the same or different main plot treatments, the standard error is calculated using the formula:

 $$s_{\bar{Y}}=2[(b-1)MS(Eb)+MS(Ea)]rb$$

From the formula, it can be seen that to compare the two average values of the main plots on the same or different sub plot treatment, two types of MS (Error) are used, namely MS (Ea) and MS (Eb).  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 ta and tb are successively the values of t obtained from the table with a certain significant degree at error-degree of freedom and error-degree of freedom b, then the weighted value of t is:

 $$ t\prime=\frac{(b-1)(MS\ \ Eb)(t_b)+(MS\ \ Ea)(t_a)}{(b-1)(MS\ \ Eb)+(MS\ \ Ea)}$$

ta = t(0.05/2.15) = 2.131

tb = t(0.05/2.18) = 2.101

b = 2 (subplot/genotype level)

MS(Ea) = 17.8485

MS(Eb) = 3.0933

so that:

 $$\begin{matrix}t'=\frac{(b-1)(MS\ \ Eb)(t_b)+(MS\ \ Ea)(t_a)}{(b-1)(MS\ \ Eb)+(MS\ \ Ea)}\\=\frac{(2-1)(3.0933)(2.101)+(17.8485)(2.131)}{(2-1)(3.0933)+(17.8485)}\\=2.1266\\\end{matrix}$$

and

 $$sY=2[(b-1)MS(Eb)+MS(Ea)]rb=2[(18-1)(3.0933)+17.8485]4×2=2.288111$$

So:

 $$\begin{matrix}LSD=t' \times s_Y\\=2.1266\times2.2881\\=4.8667\ \ kg\\\end{matrix}$$

Comparison between main plots in sub plot of Genotype IR-64 plots

No Sort

Fertilizer

 

Control

PK

NP

N

NK

NPK

 

   

Average

30.00

30.78

42.75

43.08

44.95

46.90

 

1

Control

30.00

0.00

         

a

2

PK

30.78

0.77

0.00

       

a

4

NP

42.75

12.75

11.98

0.00

     

b

3

N

43.08

13.08

12.30

0.33

0.00

   

b

5

NK

44.95

14.95

14.18

2.20

1.88

0.00

 

b

6

NPK

46.90

16.90

16.13

4.15

3.83

1.95

0.00

b

Comparison between main plots in subplot of Genotype S-969 plots

No Sort

Fertilizer

 

Control

PK

NK

N

NPK

NP

 

     

31.18

34.10

41.05

42.15

44.58

45.08

 

1

Control

31.18

0.00

         

a

2

PK

34.10

2.93

0.00

       

a

5

NK

41.05

9.88

6.95

0.00

     

b

3

N

42.15

10.98

8.05

1.10

0.00

   

b

6

NPK

44.58

13.40

10.48

3.53

2.43

0.00

 

b

4

NP

45.08

13.90

10.98

4.03

2.93

0.50

0.00

b

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.

Fertilizer (P)

Genotype (G)

1

2

Control

30:00 a
(a)

31.18 a
(a)

PK

30.78 a
(a)

34.10 a
(b)

N

43.08 b
(a)

42.15 b
(a)

NK

42.75 b
(a)

45.08 b
(a)

NP

44.95 b
(b)

41.05 b
(a)

NPK

46.90 b
(a)

44.58 b
(a)

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

Interaction (Combined Factor)

Comparison of treatment combinations (???)

If the interaction is significant, a simple effect should be examined. However, if you want to compare the combinations (AxB), it is necessary to first determine the appropriate standard error (sy). In the standard error table for RPT, there is no suitable standard error formula. In the variance table, the error for AxB Interaction is error b, so the standard error that might be used is the standard error of the subplots, with the formula:

 $$\begin{matrix}LSD=t_{0.05/2;18}\cdot s_{\bar{Y}}\\=t_{0.05/2;18}\cdot\sqrt{\frac{2MS(Eb)}{r}}\\=2.101\times\sqrt{\frac{2(3.0933)}{4}}\\=2.6129\ \ kg\\\end{matrix}$$
From the previous calculation, using the standard error, the LSD value = 2.6129 is obtained. 

       

Control

PK

Control

PK

NK

N

NP

N

NPK

NK

NP

NPK

 

       

IR-64

IR-64

S-969

S-969

S-969

S-969

IR-64

IR-64

S-969

IR-64

S-969

IR-64

 

       

30.00

30.78

31.18

34.10

41.05

42.15

42.75

43.08

44.58

44.95

45.08

46.90

 

1

Control

IR-64

30.00

0.00

                     

a

2

PK

IR-64

30.78

0.77

0.00

                   

a

7

Control

S-969

31.18

1.18

0.40

0.00

                 

a

8

PK

S-969

34.10

4.10

3.33

2.93

0.00

               

b

11

NK

S-969

41.05

11.05

10.28

9.88

6.95

0.00

             

c

9

N

S-969

42.15

12.15

11.38

10.98

8.05

1.10

0.00

           

cd

4

NP

IR-64

42.75

12.75

11.98

11.58

8.65

1.70

0.60

0.00

         

cde

3

N

IR-64

43.08

13.08

12.30

11.90

8.98

2.03

0.93

0.33

0.00

       

cde

12

NPK

S-969

44.58

14.58

13.80

13.40

10.48

3.53

2.43

1.83

1.50

0.00

     

def

5

NK

IR-64

44.95

14.95

14.18

13.78

10.85

3.90

2.80

2.20

1.88

0.38

0.00

   

ef

10

NP

S-969

45.08

15.08

14.30

13.90

10.98

4.03

2.93

2.33

2.00

0.50

0.13

0.00

 

ef

6

NPK

IR-64

46.90

16.90

16.13

15.73

12.80

5.85

4.75

4.15

3.83

2.33

1.95

1.83

0.00

f

Calculation by SmartstatXL Excel Add-In

Graphical user interface, application Description automatically generated

Anova

Post Hoc

Main Effect:

Chart Description automatically generated

 

Graphical user interface, text, application Description automatically generated

 

Interaction (Simple Effect)

Chart Description automatically generated with medium confidence

 

Interaction (Combined Factor)

Chart Description automatically generated

 

Anova Assumption:

Table Description automatically generated

 

Chart, scatter chart Description automatically generated