How to conduct LSD test with interactions in R? - r

Question

I have a field data
sowing_date<- rep(c("Early" ,"Normal"), each=12)
herbicide<- rep (rep(c("No" ,"Yes"), each=6),2)
nitrogen<- rep (rep(c("No" ,"Yes"), each=3),4)
Block<- rep(c("Block 1" ,"Block 2", "Block 3"), times=8)
Yield<- c(30,27,25,40,41,42,37,38,40,48,47,46,25,27,26,
41,41,42,38,39,42,57,59,60)
DataA<- data.frame(sowing_date,herbicide,nitrogen,Block,Yield)
and I conducted 3-way ANOVA
anova3way <- aov (Yield ~ sowing_date + herbicide + nitrogen +
sowing_date:herbicide + sowing_date:nitrogen +
herbicide:nitrogen + sowing_date:herbicide:nitrogen +
factor(Block), data=DataA)
summary(anova3way)
There is a 3-way interaction among 3 factors. So, I'd like to see which combination shows the greatest yield.
I know how to compare mean difference with single factor like below, but in case of interactions, how can I do that?
library(agricolae)
LSD_Test<- LSD.test(anova3way,"sowing_date")
LSD_Test
For example, I'd like to check the mean difference under 3 way interaction, and also interaction between each two factors.
For example, I'd like to get this LSD result in R
Could you tell me how can I do that?
Many thanks,

Answer 1

One way which does take some manual work is to encode the experimental parameters as -1 and 1 in order to properly separate the 2 and 3 parameter interactions.
Once you have values encoded you can pull the residual degree of freedoms and the sum of the error square from the ANOVA model and pass it to the LSD.test function.
See Example below:
sowing_date<- rep(c("Early" ,"Normal"), each=12)
herbicide<- rep (rep(c("No" ,"Yes"), each=6),2)
nitrogen<- rep (rep(c("No" ,"Yes"), each=3),4)
Block<- rep(c("Block 1" ,"Block 2", "Block 3"), times=8)
Yield<- c(30,27,25,40,41,42,37,38,40,48,47,46,25,27,26,
41,41,42,38,39,42,57,59,60)
DataA<- data.frame(sowing_date,herbicide,nitrogen,Block,Yield)
anova3way <- aov (Yield ~ sowing_date * herbicide * nitrogen +
factor(Block), data=DataA)
summary(anova3way)
#Encode the experiment's parameters as -1 and 1
DataA$codeSD <- ifelse(DataA$sowing_date == "Early", -1, 1)
DataA$codeherb <- ifelse(DataA$herbicide == "No", -1, 1)
DataA$codeN2 <- ifelse(DataA$nitrogen == "No", -1, 1)
library(agricolae)
LSD_Test<- LSD.test(anova3way, c("sowing_date"))
LSD_Test
#Manually defining the treatment and specifying the
# degrees of freedom and Sum of the squares (Frin the resduals from the ANOVA)
print(LSD.test(y=DataA$Yield, trt=DataA$sowing_date, DFerror=14, MSerror=34.3))
#Example for a two parameter value
print(LSD.test(y=DataA$Yield, trt=(DataA$codeSD*DataA$codeherb), DFerror=14, MSerror=34.3))
print(LSD.test(y=DataA$Yield, trt=(DataA$codeSD*DataA$codeherb*DataA$codeN2), DFerror=14, MSerror=34.3))
#calaculate the means and std (as a check)
#DataA %>% group_by(sowing_date) %>% summarize(mean=mean(Yield), sd=sd(Yield))
#DataA %>% group_by(codeSD*codeherb*codeN2) %>% summarize(mean=mean(Yield), sd=sd(Yield))
You will need to manually track which run/condition goes with the -1 and 1 in the final report.
Edit:
So my answer above with show the overall effect based on interactions. For example how does the interaction of herbicide and nitrogen effect yield.
Based on your comment where you want to determine which combination provides the greatest yield, you the use the LSD.test() function again but passing a vector of parameter names.
LSD_Test<- LSD.test(anova3way, c("sowing_date", "herbicide", "nitrogen"))
LSD_Test
From the groups part of the out put you can see Normal, Yes and Yes is the optimal yield. The "groups" column will identify the unique clusters of results. For example the last 2 rows provide a similar yield.
...
$groups
Yield groups
Normal:Yes:Yes 58.66667 a
Early:Yes:Yes 47.00000 b
Normal:No:Yes 41.33333 c
Early:No:Yes 41.00000 cd
Normal:Yes:No 39.66667 cd
Early:Yes:No 38.33333 d
Early:No:No 27.33333 e
Normal:No:No 26.00000 e
...