agridat (version 1.16)

student.barley: Multi-environment trial of barley

Description

Yield for two varieties of barley grown at 51 locations in the years 1901 to 1906.

Arguments

Format

A data frame with 102 observations on the following 7 variables.

year

year, 1901-1906

farmer

farmer name

place

place (nearest town)

district

district, geographical area

gen

genotype, Archer and Goldthorpe

yield

yield, 'stones' per acre (1 stone = 14 pounds)

income

income per acre in shillings, based on yield and quality

Details

Experiments were conducted for six years by the Department of Agriculture in Ireland. A total of seven varieties were tested, but only Archer and Goldthorpe were tested in all six years (others were dropped after being found inferior, or were added later). Plots were two acres in size. The value of the grain depended on the yield and quality. Quality varied much from farm to farm, but not so much within the same farm.

The phrase "analysis of variance" first appears in the abstract (only) of a 1918 paper by Fisher. The 1923 paper by Student contained the first analysis of variance table (but not for this data).

One stone is 14 pounds. To convert lb/ac to tonnes/ha, multiply by 0.00112085116

Note: The analysis of Student cannot be reproduced exactly. For example, Student states that the maximum income of Goldthorpe is 230 shillings. A quick glance at Table I of Student shows that the maximum income for Goldthorpe is 220 shillings (11 pounds, 0 shillings) in 1901 at Thurles. Also, the results of Kempton could not be reproduced exactly, perhaps due to rounding or the conversion factor that was used.

References

R A Kempton and P N Fox, 1997. Statistical Methods for Plant Variety Evaluation.

Examples

Run this code
# NOT RUN {
data(student.barley)
dat <- student.barley

require(lattice)
bwplot(yield ~ gen|district, dat, main="student.barley - yield")

dat$year <- factor(dat$year)
dat$income <- NULL

# convert to tons/ha
dat <- transform(dat, yield=yield*14 * 0.00112085116)

# Define 'loc' the way that Kempton does
dat$loc <- rep("",nrow(dat))
dat[is.element(dat$farmer, c("Allardyce","Roche","Quinn")),"loc"] <- "1"
dat[is.element(dat$farmer, c("Luttrell","Dooley")), "loc"] <- "2"
dat[is.element(dat$year, c("1904","1905","1906")) & dat$farmer=="Kearney","loc"] <- "2"
dat[dat$farmer=="Mulhall","loc"] <- "3"

dat <- transform(dat, loc=factor(paste(place,loc,sep="")))

if(require(reshape2)){
datm <- melt(dat, measure.var='yield')

# Kempton Table 9.5
round(acast(datm, loc+gen~year),2)

# Kempton Table 9.6
d2 <- dcast(datm, year+loc~gen)
mean(d2$Archer)
mean(d2$Goldthorpe)
mean(d2$Archer-d2$Goldthorpe)
sqrt(var(d2$Archer-d2$Goldthorpe)/51)
cor(d2$Archer,d2$Goldthorpe)
}


# }
# NOT RUN {
  # Kempton Table 9.6b
  require(lme4)
  m2 <- lmer(yield~1 + (1|loc) + (1|year) +
               (1|loc:year) + (1|gen:loc) + (1|gen:year), data=dat,
             control=lmerControl(check.nobs.vs.rankZ="ignore"))
# }
# NOT RUN {
# }

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