RCBD analysis in R along with LSD and DNMRT test

The blog explains step by step RCBD analysis along with LSD/DNMRT test using R and interpretation of analysis (Reading time 12 min.)
Problem statement
An experiment was conducted in RCBD to study the comparative performance of fodder sorghum lines under rain fed conditions. Data is furnished below. Are all lines same? If not carry out LSD test and Duncan test to compare the lines.

Arrange the data in Excel as shown below

Replication Variety Yield
1 African tall 22.9
1 Co-11 29.5
1 FS-1 28.8
1 K-7 47
1 Co-24 28.9
2 African tall 25.9
2 Co-11 30.4
2 FS-1 24.4
2 K-7 40.9
2 Co-24 20.4
3 African tall 39.1
3 Co-11 35.3
3 FS-1 32.1
3 K-7 42.8
3 Co-24 21.1
4 African tall 33.9
4 Co-11 29.6
4 FS-1 28.6
4 K-7 32.1
4 Co-24 31.8

R Script

#Ho:African tall=Co-11=FS-1=K-7=Co-24, Ha: Atleast one variety is different
#Fitting of linear model
model <-lm(RBD$Yield~ RBD$Replication+RBD$Variety)
#Obtain ANOVA
anova <-anova(model)
anova

#Below codes are used to obtain plots of fitted vs Residuals and Normal QQ plots
par(mfrow=c(1,2))
plot(model, which=1)
plot(model, which=2)

#Load the agricolae package
library("agricolae")

#Duncan test
DNMRT <-duncan.test(RBD$Yield,RBD$Variety,12,29.259)
DNMRT

#LSD test
LSD <-LSD.test(RBD$Yield,RBD$Variety,12,29.259)
LSD

#Save the file in txt
sink("RBD.txt")
print(anova)
print("DNMRT Result")
print(DNMRT$statistics)
print(DNMRT$groups)
print("LSD Result")
print(LSD$statistics)
print(LSD$groups)
sink()

Output from R

Analysis of Variance Table

Response: RBD$Yield
                               Df      Sum Sq      Mean Sq       F value        Pr(>F) 
RBD$Replication   3         80.80          26.934          0.9205        0.46033 
RBD$Variety          4         520.53       130.133         4.4476        0.01958 *
Residuals               12        351.11        29.259                 


[1] "DNMRT Result"
  MSerror Df   Mean       CV
   29.259 12 31.275 17.29547
                    RBD$Yield groups
K-7                  40.700        a
Co-11              31.200        b
African tall      30.450        b
FS-1                28.475         b
Co-24              25.550         b

[1] "LSD Result"
  MSerror Df   Mean       CV  t.value      LSD
   29.259 12 31.275 17.29547 2.178813 8.333639
                RBD$Yield groups
K-7                40.700      a
Co-11             31.200      b
African tall     30.450      b
FS-1               28.475      b
Co-24             25.550      b

Interpretation of the result

From ANOVA: The replication was non-significant so the mean yield for all the four replications is same. The treatment was significant i.e. yield of at least on variety is different from the rest. As treatment is significant we should switch to multiple mean comparison test like LSD or DNMRT test.

From LSD test: The variety K-7 gives highest yield with is significantly different from the rest of the varieties. The performance of variety Co-11 was statistically at  par with African tall, FS-1 and Co-24.

From Dunccan/DNMRT test: Same as the LSD test as both shows same set of sequence of letters.

From plots: The assumptions of ANOVA are not violated.

Love seeing videos instead of reading! Tune to the video

Comments

  1. dear can you explain how Tukey HSD and post hoc calculated .please

    ReplyDelete
    Replies
    1. In R there is a package named agricolae. It has a function named HSD.test(). To know more about it do as follows:
      1) Install agricolae package
      2) Load the package
      3) Run this ?HSD.test
      For above example you can use it as
      HSD.test(RBD$Yield,RBD$Variety,12,29.259)
      Hope you find it useful
      Regards
      RAAJ

      Delete
  2. Many thanks for all great assistance especially to the less privileged Agricultural Scientist in developing countries. Much appreciation from Uganda.
    U.K from Makerere University

    ReplyDelete

Post a Comment

Popular posts from this blog

Completely Randomized Design Analysis in R along with LSD