Factorial Randomized Block Design along with LSD test in R
The post explains how to get FRBD ANOVA, Interpretation of ANOVA, R-square, Normality assumption testing, Least significant difference (LSD) test using doebioresearch package in RStudio (Reading time 12 min.)
R-script
#This line will load doebioresearch package
library(doebioresearch)
output
print(output)
sink()
Output and interpretation
Analysis of Variance Table
Response: dependent.var
Df Sum Sq Mean Sq F value Pr(>F)
replicationvector 2 61.06 30.53 1.4112 0.276494
fact.A 1 320.47 320.47 14.8123 0.001772 **
fact.B 1 56.12 56.12 2.5939 0.129584
fact.C 1 154.53 154.53 7.1426 0.018208 *
fact.A:fact.B 1 272.70 272.70 12.6043 0.003199 **
fact.A:fact.C 1 0.22 0.22 0.0102 0.921034
fact.B:fact.C 1 3.60 3.60 0.1666 0.689355
fact.A:fact.B:fact.C 1 3.01 3.01 0.1391 0.714722
Residuals 14 302.90 21.64
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
The ANOVA shows Fact.A (Paddy variety), Fact.C (Irrigation schedule) and Fact.A*Fact.B (Paddy variety*Plant protection schedule) were significant. We will ignore the the LSD results of rest of the combination of the factors and interaction.
[1] "R Square 0.742"
0.742 i.e. 74.20% of variation present in the dependent variable is explained by the model.
[1] "SEm of A: 1.343 , SEd of A: 1.899 , SEm of B: 1.343 , SEd of B 1.899 , SEm of C: 1.343 , SEd of C: 1.899 , SEm of AB: 1.899 , SEd of AB: 2.685 , SEm of AC: 1.899 , SEd of AC: 2.685 , SEm of BC: 1.899 , SEd of BC: 2.685 , SEm of ABC: 2.685 , SEd of ABC: 3.798"
Shapiro-Wilk normality test
data: model$residuals
W = 0.98223, p-value = 0.9332
[1] "Normality assumption is not violated"
[1] "The means of one or more levels of factor A are not same, so go for multiple comparison test"
MSerror Df Mean CV t.value LSD
21.63548 14 119.6458 3.887636 2.144787 4.072787
dependent.var groups
v0 123.3000 a
v1 115.9917 b
Paddy variety vo (ADT-31) gives highest yield which is significantly different from v1 (Vaghai)
[1] "The means of one or more levels of factor C are not same, so go for multiple comparison test"
MSerror Df Mean CV t.value LSD
21.63548 14 119.6458 3.887636 2.144787 4.072787
dependent.var groups
w1 122.1833 a
w0 117.1083 b
The irrigation schedule new practice gives highest yield which is significantly different from the local practice
[1] "The means of levels of interaction between A and B factors are not same, so go for multiple comparison test"
MSerror Df Mean CV t.value LSD
21.63548 14 119.6458 3.887636 2.144787 5.759791
dependent.var groups
v0:p0 128.2000 a
v0:p1 118.4000 b
v1:p1 117.8333 b
v1:p0 114.1500 b
The combination v0p0 (ADT-31*old practice) has highest yield which is significantly different from the rest of three combinations.
Thanks for your good initiation. Please make videos/description of two factor CRD pooled analysis. I am waiting for your positive response.
ReplyDeleteNamste Anuj!
DeletePlease share the no of treatments and replication of CRD design. If possible I will share excel tool for that.
This is the simplest and easiest way to do RBD.
ReplyDeleteCan you please make a video on splitplot and CRD using "doebioresearch"
Split plot analysis using doebioresearch package video
Deletehttps://www.youtube.com/watch?v=j5FRTZDWDlA&t=32s
For CRD you can refer example given in package
Regards
RAAJ
Mr. Raj, you have prepared one of the best packages for the agricultural students. Thank you. Think of including some graphs also.
ReplyDeleteThanks Sirji for the appreciation and suggetions
Delete