Question

Solve by using R program or any other statistical program(python,R)

2. Three types of ferilizer are to be tested to see which one yields more corn crop. Forty similar plots of land were available for testing purposes. The 40 plots are divided at random into four groups, 10 plots in Group 1. Similarly, Fertilizers 2 and 3 were applied to the plots in Group 2 and 3, respectively. The corn plants in Group 4 were not given any fertilizer; it will serve as the control group. Table 2 gives the corn yield yii for each of the 40 plots. Please answer the following questions. You may use R or other program language if needed. (a) Create three indicator variables F1,F2,F3, one for each of the three fertilizer groups. (b) Fit the model yij = Ho + HiFi + H2F12 +13 Fi3 + tij. (c) Test the hypothesis that, on the average, none of the three types of fertilizer has an effect on corn crops. Specify the hypothesis to be tested, the test used, and your conclusions at the 5% significance level (d) Test the hypothesis that, on the average, none of the three types of fertilizer have equal effects on corn crop but different from that of the control group. Specify the hypothesis to be tested, the test used, and your conclusions at the 5% significance level. (e) Which of the three fertilizers has the greatest effects on corn yield?

Answer 1

• The complete R snippet is as follows

  # read the data into R dataframe
  data.df<- read.csv("fertilizer.csv",header=TRUE)
  str(data.df)

  data.df$Fertilizer <- as.factor(data.df$Fertilizer)

  # perform anova analysis
  a<- aov(lm(Yield~ Fertilizer,data=data.df))

  #summarise the results
  summary(a)

  plot(a$residuals,pch=16,col="violetred2")

  
  colr<-c(
  "yellow3","palegreen1" ,"orangered" ,"magenta4" )
  # plots

  boxplot(Yield~ Fertilizer, data=data.df,ylab="Values",
  main="Boxplots of the Data",col=colr,horizontal=TRUE)

  # Plot Means with Error Bars
  library(gplots)
  attach(data.df)
  plotmeans(Yield~ Fertilizer,xlab="Types",
  ylab="Value", main="Mean Plot\nwith 95% CI")

• The results are

  > summary(a)
  Df Sum Sq Mean Sq F value Pr(>F)   
  Fertilizer 3 362.6 120.87 5.144 0.0046 ** ## as the p value is less than 0.05 , hence the results are statistically significant
  Residuals 36 845.8 23.49
  ---
  Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

  

• as the result is signficant , we can conduct Tukey HSD to perform the test in R

  ## Tukey

  t<- TukeyHSD(a)
  t
  t$Fertilizer[,4]

  plot(t,col = ifelse(t$Fertilizer[,4] < 0.05,"darkgreen","orange"))

  

• Tukey multiple comparisons of means
  95% family-wise confidence level

  Fit: aov(formula = lm(Yield ~ Fertilizer, data = data.df))

  $Fertilizer
  diff lwr upr p adj
  2-1 -6.7 -12.5380872 -0.8619128 0.0191978
  3-1 -1.7 -7.5380872 4.1380872 0.8610988
  4-1 -6.8 -12.6380872 -0.9619128 0.0170864
  3-2 5.0 -0.8380872 10.8380872 0.1154045
  4-2 -0.1 -5.9380872 5.7380872 0.9999640
  4-3 -5.1 -10.9380872 0.7380872 0.1050237

  all the pairwise comparisons whose p value is less than 0.05 are significant