agridat (version 1.23)

kempton.barley.uniformity: Uniformity trial of barley

Description

Uniformity trial of barley at Cambridge, England, 1978.

Format

A data frame with 196 observations on the following 3 variables.

row

row

col

column

yield

grain yield, kg

Details

A uniformity trial of spring barley planted in 1978. Conducted by the Plant Breeding Institute in Cambridge, England.

Each plot is 5 feet wide, 14 feet long.

Field width: 7 plots * 14 feet = 98 feet

Field length: 28 plots * 5 feet = 140 feet

References

McCullagh, P. and Clifford, D., (2006). Evidence for conformal invariance of crop yields, Proceedings of the Royal Society A: Mathematical, Physical and Engineering Science. 462, 2119--2143. https://doi.org/10.1098/rspa.2006.1667

Examples

if (FALSE) {

  library(agridat)
  data(kempton.barley.uniformity)
  dat <- kempton.barley.uniformity

  libs(desplot)
  desplot(dat, yield~col*row,
          aspect=140/98, tick=TRUE, # true aspect
          main="kempton.barley.uniformity")
  
  
  # Kempton estimated auto-regression coefficients b1=0.10, b2=0.91
  
  dat <- transform(dat, xf = factor(col), yf=factor(row))

  # ----------

  if(require("asreml", quietly=TRUE)){
    libs(asreml,lucid)
  
    dat <- transform(dat, xf = factor(col), yf=factor(row))
    m1 <- asreml(yield ~ 1, data=dat, resid = ~ar1(xf):ar1(yf))
  
    # lucid::vc(m1)
    ##       effect component std.error z.ratio bound 
    ##      xf:yf!R    0.1044   0.02197     4.7     P   0
    ## xf:yf!xf!cor    0.2458   0.07484     3.3     U   0
    ## xf:yf!yf!cor    0.8186   0.03821    21       U   0
  
    # asreml estimates auto-regression correlations of 0.25, 0.82
    # Kempton estimated auto-regression coefficients b1=0.10, b2=0.91
  }
  
  # ----------

  if(0){
    # Kempton defines 4 blocks, randomly assigns variety codes 1-49 in each block, fits
    # RCB model, computes mean squares for variety and residual.  Repeat 40 times.
    # Kempton's estimate: variety = 1032, residual = 1013
    # Our estimate: variety = 825, residual = 1080
    fitfun <- function(dat){
      dat <- transform(dat, block=factor(ceiling(row/7)),
                       gen=factor(c(sample(1:49),sample(1:49),sample(1:49),sample(1:49))))
      m2 <- lm(yield*100 ~ block + gen, dat)
      anova(m2)[2:3,'Mean Sq']
    }
    set.seed(251)
    out <- replicate(50, fitfun(dat))
    rowMeans(out) # 826 1079
  }


}