Yes, as the ingredients do not repeat for any day or batch.
\(x_{ijk} = \mu + \tau_i + \beta_j + \alpha_k + \epsilon_{ijk}\)
library(GAD)
Time<-c(8,7,1,7,3,11,2,7,3,8,4,9,10,1,5,6,8,6,6,10,4,2,3,8,8)
Batch<-rep(c("B1","B2","B3","B4","B5"),each=5)
Day<-rep(c("D1","D2","D3","D4","D5"),5)
Ingredient<-c("A","B","D","C","E","C","E","A","D","B","B","A","C","E","D","D","C","E","B","A","E","D","B","A","C")
Batch<-as.fixed(Batch)
Day<-as.fixed(Day)
Ingredient<-as.fixed(Ingredient)
dat<-data.frame(Time,Batch,Day,Ingredient)
aov.model<-aov(Time~Batch+Day+Ingredient,data=dat)
summary(aov.model)## Df Sum Sq Mean Sq F value Pr(>F)
## Batch 4 15.44 3.86 1.235 0.347618
## Day 4 12.24 3.06 0.979 0.455014
## Ingredient 4 141.44 35.36 11.309 0.000488 ***
## Residuals 12 37.52 3.13
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Only the ingredient has a p-value (.0005) less than the alpha = .05, which makes it the factor that has enough significance to reject the null hypothesis that there is no difference between reaction times across the factors.