hernandez.nitrogen: Multi-environment trial of corn with nitrogen fertilizer at 5...

In agridat: Agricultural Datasets

Multi-environment trial of corn with nitrogen fertilizer at 5 sites.

Description

Corn response to nitrogen fertilizer at 5 sites.

Format

A data frame with 136 observations on the following 5 variables.

site

site factor, 5 levels

loc

location name

rep

rep, 4 levels

nitro

nitrogen, kg/ha

yield

yield, Mg/ha

Details

Experiment was conducted in 2006 at 5 sites in Minnesota.

Source

Hernandez, J.A. and Mulla, D.J. 2008. Estimating uncertainty of economically optimum fertilizer rates, Agronomy Journal, 100, 1221-1229. https://doi.org/10.2134/agronj2007.0273

Electronic data kindly supplied by Jose Hernandez.

Examples

## Not run: 
  
  library(agridat)
  data(hernandez.nitrogen)
  dat <- hernandez.nitrogen
  cprice <- 118.1 # $118.1/Mg or $3/bu
  nprice <- 0.6615 # $0.66/kg N or $0.30/lb N
  
  # Hernandez optimized yield with a constraint on the ratio of the prices.
  # Simpler to just calculate the income and optimize that.
  dat <- transform(dat, inc = yield * cprice - nitro * nprice)
  libs(lattice)
  xyplot(inc ~ nitro|site, dat, groups=rep, auto.key=list(columns=4),
         xlab="nitrogen", ylab="income", main="hernandez.nitrogen")
  
  # Site 5 only
  dat1 <- subset(dat, site=='S5')
  
  # When we optimize on income, a simple quadratic model works just fine,
  # and matches the results of the nls model below.
  # Note, 'poly(nitro)' gives weird coefs
  lm1 <- lm(inc ~ 1 + nitro + I(nitro^2), data=dat1) 
  c1 <- coef(lm1)
  -c1[2] / (2*c1[3])
  ##    nitro
  ## 191.7198    # Optimum nitrogen is 192 for site 5
  
  
  # Use the delta method to get a conf int
  libs("car")
  del1 <- deltaMethod(lm1, "-b1/(2*b2)", parameterNames= paste("b", 0:2, sep=""))
  # Simple Wald-type conf int for optimum
  del1$Est +  c(-1,1) * del1$SE * qt(1-.1/2, nrow(dat1)-length(coef(lm1)))
  ## 118.9329 264.5067
  
  
  # Nonlinear regression
  # Reparameterize b0 + b1*x + b2*x^2 using th2 = -b1/2b2 so that th2 is optimum
  nls1 <- nls(inc ~ th11- (2*th2*th12)*nitro + th12*nitro^2,
              data = dat1, start = list(th11 = 5, th2 = 150, th12 =-0.1),)
  summary(nls1)
  # Wald conf int
  wald <- function(object, alpha=0.1){
    nobs <- length(resid(object))
    npar <- length(coef(object))
    est <- coef(object)
    stderr <- summary(object)$parameters[,2]
    tval <- qt(1-alpha/2, nobs-npar)
    ci <- cbind(est - tval * stderr, est + tval * stderr)
    colnames(ci) <- paste(round(100*c(alpha/2, 1-alpha/2), 1), "pct", sep= "")
    return(ci)
  }
  round(wald(nls1),2)
  ##          5
  ## th11 936.44 1081.93
  ## th2  118.93  264.51   # th2 is the optimum
  ## th12  -0.03   -0.01
  
  
  # Likelihood conf int
  libs(MASS)
  round(confint(nls1, "th2", level = 0.9),2)
  ##       5
  ## 147.96 401.65
  
  
  # Bootstrap conf int
  libs(boot)
  dat1$fit <- fitted(nls1)
  bootfun <- function(rs, i) { # bootstrap the residuals
    dat1$y <- dat1$fit + rs[i]
    coef(nls(y ~ th11- (2*th2*th12)*nitro + th12*nitro^2, dat1,
             start = coef(nls1) ))
  }
  res1 <- scale(resid(nls1), scale = FALSE) # remove the mean.  Why? It is close to 0.
  set.seed(5) # Sometime the bootstrap fails, but this seed works
  boot1 <- boot(res1, bootfun, R = 500)
  boot.ci(boot1, index = 2, type = c("perc"), conf = 0.9)
  ## Level     Percentile
  ## 90


## End(Not run)

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