\inputencoding{latin1}
\HeaderA{crimtab}{Student's 3000 Criminals Data}{crimtab}
\keyword{datasets}{crimtab}
\begin{Description}\relax
Data of 3000 male criminals over 20 years old undergoing their
sentences in the chief prisons of England and Wales.
\end{Description}
\begin{Usage}
\begin{verbatim}data(crimtab)\end{verbatim}
\end{Usage}
\begin{Format}\relax
A \code{\LinkA{table}{table}} object of  \code{\LinkA{integer}{integer}} counts, of dimension
\eqn{42 \times 22}{} with a total count, \code{sum(crimtab)} of
3000.

The 42 \code{\LinkA{rownames}{rownames}} (\code{"9.4"}, \code{"9.5"}, \dots)
correspond to midpoints of intervals of finger lengths
whereas the 22 column names (\code{\LinkA{colnames}{colnames}})
(\code{"142.24"}, \code{"144.78"}, \dots) correspond to (body) heights
of 3000 criminals, see also below.
\end{Format}
\begin{Details}\relax
Student is the pseudonym of William Sealy Gosset.
In his 1908 paper he wrote (on page 13) at the beginning of section VI
entitled \emph{Practical Test of the forgoing Equations}:

\dQuote{Before I had succeeded in solving my problem analytically,
I had endeavoured to do so empirically.  The material used was a
correlation table containing the height and left middle finger
measurements of 3000 criminals, from a paper by W. R. MacDonell
(\emph{Biometrika}, Vol. I., p. 219).  The measurements were written
out on 3000 pieces of cardboard, which were then very thoroughly
shuffled and drawn at random.  As each card was drawn its numbers
were written down in a book, which thus contains the measurements of
3000 criminals in a random order.  Finally, each consecutive set of
4 was taken as a sample---750 in all---and the mean, standard
deviation, and correlation of each sample determined.  The
difference between the mean of each sample and the mean of the
population was then divided by the standard deviation of the sample,
giving us the \emph{z} of Section III.}

The table is in fact page 216 and not page 219 in MacDonell(1902).
In the MacDonell table, the middle finger lengths were given in mm
and the heights in feet/inches intervals, they are both converted into
cm here.  The midpoints of intervals were used, e.g., where MacDonell
has \eqn{4' 7''9/16 -- 8''9/16}{}, we have 142.24 which is 2.54*56 =
2.54*(\eqn{4' 8''}{}).

MacDonell credited the source of data (page 178) as follows:
\emph{The data on which the memoir is based were obtained, through the
kindness of Dr Garson, from the Central Metric Office, New Scotland Yard...}
He pointed out on page 179 that : \emph{The forms were drawn at random
from the mass on the office shelves; we are therefore dealing with a
random sampling.}
\end{Details}
\begin{Source}\relax
\url{http://pbil.univ-lyon1.fr/R/donnees/criminals1902.txt}
thanks to Jean R. Lobry and Anne-Béatrice Dufour.
\end{Source}
\begin{References}\relax
Garson, J.G. (1900)
The metric system of identification of criminals, as used in in Great
Britain and Ireland.
\emph{The Journal of the Anthropological Institute of Great Britain
and Ireland} \bold{30}, 161--198.

MacDonell, W.R. (1902)
On criminal anthropometry and the identification of criminals.
\emph{Biometrika} \bold{1}, 2, 177--227.

Student (1908) The probable error of a mean.
\emph{Biometrika} \bold{6}, 1--25.
\end{References}
\begin{Examples}
\begin{ExampleCode}
require(stats)
dim(crimtab)
utils::str(crimtab)
## for nicer printing:
local({cT <- crimtab
       colnames(cT) <- substring(colnames(cT), 2,3)
       print(cT, zero.print = " ")
})

## Repeat Student's experiment:

# 1) Reconstitute 3000 raw data for heights in inches and rounded to
#    nearest integer as in Student's paper:

(heIn <- round(as.numeric(colnames(crimtab)) / 2.54))
d.hei <- data.frame(height = rep(heIn, colSums(crimtab)))

# 2) shuffle the data:

set.seed(1)
d.hei <- d.hei[sample(1:3000), , drop = FALSE]

# 3) Make 750 samples each of size 4:

d.hei$sample <- as.factor(rep(1:750, each = 4))

# 4) Compute the means and standard deviations (n) for the 750 samples:

h.mean <- with(d.hei, tapply(height, sample, FUN = mean))
h.sd   <- with(d.hei, tapply(height, sample, FUN = sd)) * sqrt(3/4)

# 5) Compute the difference between the mean of each sample and
#    the mean of the population and then divide by the
#    standard deviation of the sample:

zobs <- (h.mean - mean(d.hei[,"height"]))/h.sd

# 6) Replace infinite values by +/- 6 as in Student's paper:

zobs[infZ <- is.infinite(zobs)] # 3 of them
zobs[infZ] <- 6 * sign(zobs[infZ])

# 7) Plot the distribution:

require(grDevices); require(graphics)
hist(x = zobs, probability = TRUE, xlab = "Student's z",
     col = grey(0.8), border = grey(0.5),
     main = "Distribution of Student's z score  for 'crimtab' data")
\end{ExampleCode}
\end{Examples}

