Importing the data needed
library(readxl)
RBD <- read_excel("C:/Users/USER1/Desktop/darwin 4th year files/experimental design R/Yield.xlsx")
View(RBD)
Determining the null and alternative hypothesis:
\(H_o\): There is no significant difference on the mean yield of between the varieties of fodder sorghum under rain fed conditions. \
\(H_A\): There is a significant difference on the mean yield of between the varieties of fodder sorghum under rain fed conditions.
Fitting of linear model, we have
model <-lm(RBD$Yield~ RBD$Replication+RBD$Variety)
#Obtain ANOVA
anova <-anova(model)
anova
## Analysis of Variance Table
##
## Response: RBD$Yield
## Df Sum Sq Mean Sq F value Pr(>F)
## RBD$Replication 1 6.30 6.300 0.2072 0.65592
## RBD$Variety 4 520.53 130.133 4.2806 0.01808 *
## Residuals 14 425.61 30.401
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Generating the plots for fitted vs Residuals and Normal QQ plots
par(mfrow=c(1,2))
plot(model, which=1)
plot(model, which=2)
Doing the Duncan Test (DNMRT):
library("agricolae")
duncan <-duncan.test(RBD$Yield,RBD$Variety,12,29.259)
duncan
## $statistics
## MSerror Df Mean CV
## 29.259 12 31.275 17.29547
##
## $parameters
## test name.t ntr alpha
## Duncan RBD$Variety 5 0.05
##
## $duncan
## Table CriticalRange
## 2 3.081307 8.333639
## 3 3.225244 8.722927
## 4 3.312453 8.958792
## 5 3.370172 9.114897
##
## $means
## RBD$Yield std r Min Max Q25 Q50 Q75
## African tall 30.450 7.403378 4 22.9 39.1 25.150 29.90 35.200
## Co-11 31.200 2.762849 4 29.5 35.3 29.575 30.00 31.625
## Co-24 25.550 5.674798 4 20.4 31.8 20.925 25.00 29.625
## FS-1 28.475 3.155287 4 24.4 32.1 27.550 28.70 29.625
## K-7 40.700 6.274286 4 32.1 47.0 38.700 41.85 43.850
##
## $comparison
## NULL
##
## $groups
## RBD$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
##
## attr(,"class")
## [1] "group"
Doing the LSD Test:
LSD <-LSD.test(RBD$Yield,RBD$Variety,12,29.259)
LSD
## $statistics
## MSerror Df Mean CV t.value LSD
## 29.259 12 31.275 17.29547 2.178813 8.333639
##
## $parameters
## test p.ajusted name.t ntr alpha
## Fisher-LSD none RBD$Variety 5 0.05
##
## $means
## RBD$Yield std r LCL UCL Min Max Q25 Q50
## African tall 30.450 7.403378 4 24.55723 36.34277 22.9 39.1 25.150 29.90
## Co-11 31.200 2.762849 4 25.30723 37.09277 29.5 35.3 29.575 30.00
## Co-24 25.550 5.674798 4 19.65723 31.44277 20.4 31.8 20.925 25.00
## FS-1 28.475 3.155287 4 22.58223 34.36777 24.4 32.1 27.550 28.70
## K-7 40.700 6.274286 4 34.80723 46.59277 32.1 47.0 38.700 41.85
## Q75
## African tall 35.200
## Co-11 31.625
## Co-24 29.625
## FS-1 29.625
## K-7 43.850
##
## $comparison
## NULL
##
## $groups
## RBD$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
##
## attr(,"class")
## [1] "group"
Interpretation of Results:
From the data given, the plots show that assumptions to execute ANOVA was complied. Results of ANOVA shows that the replications has no significant difference on their mean yield, so we fail to reject the null hypothesis because \(p=0.65592>0.05\). The treatment was significant i.e. yield of at least one variety is different from the rest with \(p=0.01808<0.05\). As treatment is significant we should switch to multiple mean comparison test like LSD or DNMRT test.
Both of these comparison tests shows that 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.