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\begin{center}
{\Large{\textbf{THE PROBABLE ERROR OF A MEAN}}}

\bigskip

{\large{\textsc{By STUDENT}}}
\end{center}

\begin{center}
\section*{Introduction}
\end{center}

{\textsc Any} experiment may he regarded as forming an
individual of a population'' of experiments which might he performed
under the same conditions.  A series of experiments is a sample drawn
from this population.

Now any series of experiments is only of value in so far as it enables
us to form a judgment as to the statistical constants of the population
to which the experiments belong. In a greater number of cases
the question finally turns on the value of a mean, either directly, or
as the mean difference between the two quantities.

If the number of experiments be very large, we may have precise
information as to the value of the mean, but if our sample be small, we
have two sources of uncertainty: (1) owing to the error of random
sampling'' the mean of our series of experiments deviates more or less
widely from the mean of the population, and (2) the sample is not
sufficiently large to determine what is the law of distribution of
individuals. It is usual, however, to assume a normal distribution,
because, in a very large number of cases, this gives an approximation
so close that a small sample will give no real information as to the
manner in which the population deviates from normality: since some law
of distribution must he assumed it is better to work with a curve whose
area and ordinates are tabled, and whose properties are well known.
This assumption is accordingly made in the present paper, so that its
conclusions are not strictly applicable to populations known not to be
normally distributed; yet it appears probable that the deviation from
normality must be very extreme to load to serious error. We are
concerned here solely with the first of these two sources of
uncertainty.

The usual method of determining the probability that the mean of the
population lies within a given distance of the mean of the sample is to
assume a normal distribution about the mean of the sample with a
standard deviation equal to $s/\sqrt{n}$, where $s$ is the standard
deviation of the sample, and to use the tables of the probability
integral.

But, as we decrease the number of experiments, the value of the
standard deviation found from the sample of experiments becomes
itself subject to an increasing error, until judgments reached in this
way may become altogether misleading.

In routine work there are two ways of dealing with this difficulty:
(1) an experiment may he repeated many times, until such a long series
is obtained that the standard deviation is determined once and for all
with sufficient accuracy.  This value can then he used for subsequent
shorter series of similar experiments. (2) Where experiments are done
in duplicate in the natural course of the work, the mean square of the
difference between corresponding pairs is equal to the standard
deviation of the population multiplied by $\sqrt{2}$. We call thus combine
together several series of experiments for the purpose of determining
the standard deviation.  Owing however to secular change, the value
obtained is nearly always too low, successive experiments being
positively correlated.

There are other experiments, however, which cannot easily be repeated
very often; in such cases it is sometimes necessary to judge of
the certainty of the results from a very small sample, which
itself affords the only indication of the variability. Some chemical,
many biological, and most agricultural and large-scale experiments
belong to this class, which has hitherto been almost outside the
range of statistical inquiry.

Again, although it is well known that the method of using the normal
curve is only trustworthy when the sample is large'', no one has
yet told us very clearly where the limit between large'' and small''
samples is to be drawn.

The aim of the present paper is to determine the point at which we
may use the tables of the probability integral in judging of the
significance of the mean of a series of experiments, and to furnish
alternative tables for use when the number of experiments is
too few.

The paper is divided into the following nine sections:

\medskip

\noindent
I. The equation is determined of the curve which represents the
frequency distribution of standard deviations of samples drawn from a
normal population.

\medskip

\noindent
II. There is shown to be no kind of correlation between the mean and
the standard deviation of such a sample.

\medskip

\noindent
III. The equation is determined of the curve representing the
frequency distribution of a quantity $z$, which is obtained by dividing
the distance between the mean of a sample and the mean of the population
by the standard deviation of the sample.

\medskip

\noindent
IV. The curve found in I is discussed.

\medskip

\noindent
V.  The curve found in III is discussed.

\medskip

\noindent
VI. The two curves are compared with some actual distributions.

\medskip

\noindent
VII. Tables of the curves found in III are given for samples of different
size.

\medskip

\noindent
VIII and IX. The tables are explained and some instances are given of their
use.

\medskip

\noindent
X.   Conclusions.

\bigskip

%\begin{center}
\section*{Section 1}
%\end{center}

Samples of $n$ individuals are drawn out of a population distributed
normally, to find an equation which shall represent the frequency of
the standard deviations of these samples.

If $s$ be the standard deviation found from a sample $x_1x_2\dots x_n$ (all
these being measured from the mean of the population), then
$s^2=\frac{S(x_1^2)}{n}-\left(\frac{S(x_1)}{n}\right)^2= \frac{S(x_1^2)}{n}-\frac{S(x_1^2)}{n^2}-\frac{2S(x_1x_2)}{n^2}.$

Summing for all samples and dividing by the number of samples we get
the moan value of $s^2$, which we will write $\bar s^2$:
$\bar s^2=\frac{n\mu_2}{n}-\frac{n\mu_2}{n^2}=\frac{\mu_2(n-1)}{n},$
where $\mu_2$ is the second moment coefficient in the original normal
distribution of $x$: since $x_1$, $x_2$, etc.\ are not correlated and the
distribution is normal, products involving odd powers of $x_1$ vanish
on summing, so that $\frac{2S(x_1x_2)}{n^2}$ is equal to 0.

If ${M'}_R$ represent the $R$th moment coefficient of the distribution of $s^2$
about the end of the range where $s^2=0$,
$M_1'=\mu_2\frac{(n-1)}{n}.$

Again
\begin{align*}
s^4 &= \left\{\frac{S(x_1^2)}{n}-\left(\frac{S(x_1)}{n}\right)\right\}^2\\
&=\left(\frac{S(x_1^2)}{n}\right)^2
-\frac{2S(x_1^2)}{n}\left(\frac{S(x_1)}{n}\right)^2
+\left(\frac{S(x_1)}{n}\right)^4 \\
&=\frac{S(x_1^4)}{n^2}+\frac{2S(x_1^2x_2^2)}{n^2}
-\frac{2S(X_1^4)}{n^3}-\frac{4S(x_1^2x_2^2)}{n^3}+\frac{S(x_1^4)}{n^4}\\
&+\frac{6S(x_1^2x_2^2)}{n^4}+
\text{other terms involving odd powers of $x_1$, etc.\ which} \\
\end{align*}

Now $S(x_1^4)$ has $n$ terms, but$S(x_1^2x_2^2)$ has $\frac{1}{2}n(n-1)$, hence
summing for all samples and dividing by the number of samples, we get
\begin{align*}
M_2' &=\frac{\mu_4}{n}+\mu_2^2\frac{(n-1)}{n}-\frac{2\mu_4}{n^2}
-2\mu_2^2\frac{(n-1)}{n^2}+\frac{\mu_4}{n^3}+3\mu_2^2\frac{(n-1)}{n^3}\\
&=\mu_4{n^3}\{n^2-2n+1\}+\frac{\mu_2^2}{n^3}(n-1)\{n^2-2n+3\}.
\end{align*}

Now since the distribution of $x$ is normal, $\mu_4=3\mu_2^2$, hence
$M_2'=\mu_2^2\frac{(n-1)}{n^3}\{3n-3+n^2-2n+3\} =\mu_2^2\frac{(n-1)(n+1)}{n^2}.$
In a similar tedious way I find
$M_3'=\mu_2^3\frac{(n-1)(n+1)(n+3)}{n^3}$
and
$M_4'=\mu_2^4\frac{(n-1)(n+1)(n+3)(n+5)}{n^4}.$

The law of formation of these moment coefficients appears to be a
simple one, but I have not seen my way to a general proof.

If now $M_R$ be the $R$th moment coefficient of $s^2$ about its mean, we have
\begin{align*}
M_2&=\mu_2^2\frac{(n-1)}{n^3}\{(n+1)-(n-1)\}
=2\mu_2^2\frac{(n-1)}{n^2}. \\
M_3 &=\mu_2^3\left\{\frac{(n-1)(n+1)(n+3)}{n^3}
-\frac{3(n-1)}{n}.\frac{(2(n-1)}{n^2}-\frac{(n-1)^3}{n^3}\right\} \\
&=\mu_2^3\frac{(n-1)}{n^3}\{n^2+4n+3-6n+6-n^2+2n-1\}
=8\mu_2^3\frac{(n-1)}{n^3}, \\
M_4 &=\frac{\mu_2^4}{n^4}\left\{(n-1)(n+1)(n+3)(n+5)
-32(n-1)^2-12(n-1)^3-(n-1)^4\right\} \\
&=\frac{\mu_2^4(n-1)}{n^4}
\{n^3+9n^2+23n+15-32n+32 \\
&\phantom{=\mu\_2^4(n-1)\{}\ -12n^2+24n-12-n^3+3n^2-3n+1\} \\
&=\frac{12\mu_2^4(n-1)(n+3)}{n^4}.
\end{align*}

Hence
\begin{gather*}
\beta_1=\frac{M_3^2}{M_2^3}=\frac{8}{n-1},
\therefore 2\beta_2-3\beta_1-6=\frac{1}{n-1}\{6(n+3)-24-6(n-1)\}=0.
\end{gather*}

Consequently a curve of Prof.\ Pearson's Type III may he expected to fit
the distribution of $s^2$.

The equation referred to an origin at the zero end of the curve will be
$y=Cx^pe^{-\gamma x},$
where
$\gamma=2\frac{M_2}{M_3}=\frac{4\mu_2^2(n-1)n^3}{8n^2\mu_2^2(n-1)} =\frac{n}{2\mu_2}$
and
$p=\frac{4}{\beta_1}-1=\frac{n-1}{2}-1=\frac{n-3}{2}.$

Consequently the equation becomes
$y=Cx^{\frac{n-3}{2}}e^{-\frac{nx}{2\mu_2}},$
which will give the distribution of $s^2$.

The area of this curve is
$C\int_0^{\infty} x^{\frac{n-3}{2}}e^{-\frac{nx}{2\mu_2}}dx=I$ (say).
The first moment coefficient about the end of the range will therefore be
$\frac{C\int_0^{\infty} x^{\frac{n-1}{2}}e^{-\frac{nx}{2\mu_2}}dx}{I} =\frac{\left[ C\frac{-2\mu_2}{n}x^{\frac{n-1}{2}}e^{-\frac{nx}{2\mu_2}} \right]_{x=0}^{x=\infty}}{I}+ \frac{C\int_0^{\infty} \frac{n-1}{n}\mu_2x^{\frac{n-3}{2}}e^{-\frac{nx}{2\mu_2}}dx}{I}.$

The first part vanishes at each limit and the second is equal to
$\frac{\frac{n-1}{n}\mu_2I}{I}=\frac{n-1}{n}\mu_2.$
and we see that the higher moment coefficients will he formed by
multiplying successively by $\frac{n+1}{n}\mu_2$, $\frac{n+3}{n}\mu_2$
etc., just as appeared to he the law of formation of $M_2'$, $M_3'$,
$M_4'$, etc.

Hence it is probable that the curve found represents the theoretical
distribution of $s^2$; so that although we have no actual proof we
shall assume it to do so in what follows.

The distribution of $s$ may he found from this, since the frequency of $s$
is equal to that of $s^2$ and all that we must do is to compress the base
line suitably.

Now if\qquad\,$y_1=\phi(s^2)$ be the frequency curve of $s^2$\newline
and\qquad\qquad\quad$y_2=\psi(s)$ be the frequency curve of $s$,\newline
then
\begin{align*}
y_1d(s^2)     &=y_2ds, \\
y_2ds         &=2y_1sds, \\
\therefore y_2&=2sy_1.
\end{align*}

Hence
$y_2=2Cs(s^2)^{\frac{n-3}{2}}e^{-\frac{ns^2}{2\mu_2}}.$
is the distribution of $s$.

This reduces to
$y_2=2Cs^{n-2}e^{-\frac{ns^2}{2\sigma^2}}.$

Hence $y = Ax^{n-2}e^{-\frac{s^2}{2\mu_2}}$ will give the frequency
distribution of standard deviations of samples of $n$, taken out of
a population distributed normally with standard deviation $\sigma^2$.  The
constant $A$ may he found by equating the area of the curve as
follows:
$\text{Area}=A\int_0^{\infty}x^{n-2}e^{-\frac{nx^2}{2\sigma^2}}dx.\quad \left(\text{Let I_p represent\ } \int_0^{\infty}x^pe^{-\frac{-nx^2}{2\sigma^2}}dx.\right)$
Then
\begin{align*}
I_p&=\frac{\sigma^2}{n}\int_0^{\infty}x^{p-1}\frac{d}{dx}
\left(-e^{-\frac{nx^2}{2\sigma^2}}\right)dx \\
&=\frac{\sigma^2}{n}\left[-x^{p-1}e^{-\frac{nx^2}{2\sigma^2}}
\right]_{x=0}^{x=\infty}
+\frac{\sigma_2}{n}(p-1)\int_0^{\infty}x^{p-2}e^{-\frac{x^2}{2\sigma^2}}
dx \\
&=\frac{\sigma^2}{n}(p-1)I_{p-2},
\end{align*}
since the first part vanishes at both limits.

By continuing this process we find
$I_{n-2}=\left(\frac{\sigma^2}{n}\right)^{\frac{n-2}{2}}(n-3)(n-5)\dots 3.1 I_0$
or
$I_{n-2}=\left(\frac{\sigma^2}{n}\right)^{\frac{n-2}{2}}(n-3)(n-5)\dots 4.2 I_1$
according $n$ is even or odd.

But $I_0$ is
$\int_0^{\infty}e^{-\frac{nx^2}{2\sigma^2}}dx =\sqrt{\left(\frac{\pi}{2n}\right)}\sigma,$
and $I_1$ is
$\int_0^{\infty}xe^{-\frac{nx^2}{2sigma^2}}dx =\left[-\frac{\sigma^2}{n} e^{-\frac{nx^2}{2\sigma^2}}\right]_{x=0}^{x=\infty} =\frac{\sigma^2}{n}.$

Hence if $n$ be even,
$A=\frac{\text{Area}} {(n-3)(n-5)\dots3.1\sqrt{\left(\frac{\pi}{2}\right)} \left(\frac{\sigma^2}{n}\right)^{\frac{n-1}{2}}},$
while is $n$ be odd
$A=\frac{\text{Area}} {(n-3)(n-5)\dots4.2 \left(\frac{\sigma^2}{n}\right)^{\frac{n-1}{2}}}.$

Hence the equation may be written
$y=\frac{N}{(n-3)(n-5)\dots3.1} \sqrt{\left(\frac{2}{\pi}\right)} \left(\frac{n}{\sigma^2}\right)^{\frac{n-1}{2}} x^{n-2}e^{-\frac{nx^2}{2\sigma^2}}\text{\ (n even)}$
or
$y=\frac{N}{(n-3)(n-5)\dots4.2} \left(\frac{n}{\sigma^2}\right)^{\frac{n-1}{2}} x^{n-2}e^{-\frac{nx^2}{2\sigma^2}}\text{\ (n odd)}$
where $N$ as usual represents the total frequency.

\begin{center}
\section*{Section II}
\end{center}

To show that there is no correlation between ($a$) the distance of the
mean of a sample from the mean of the population and ($b$) the
standard deviation of a sample with normal distribution.

(1) Clearly positive and negative positions of the mean of the
sample are equally likely, and hence there cannot be correlation
between the absolute value of the distance of the mean from the mean
of the population and the standard deviation, but (2) there might be
correlation between the square of the distance and the square of the
standard deviation.
Let
$u^2=\left(\frac{S(x_1)}{n}\right)^2\quad\text{and}\quad s^2=\frac{S(x_1^2)}{n}-\left(\frac{S(x_1)}{n}\right)^2.$
Then if $m_1'$, $M_1'$ be the mean values of $u^2$ and $s^z$, we have by the
preceding part
$M_1'=\mu_2\frac{(n-1)}{n}\quad\text{and}\quad m_1'=\frac{\mu_2}{n}.$

Now
\begin{align*}
u^2s^2 &= \frac{S(x_1^2)}{n}\left(\frac{S(x_1)}{n}\right)^2
-\left(\frac{S(x_1)}{n}\right)^4\\
&=\left(\frac{S(x_1^2)}{n}\right)^2+
2\frac{S(x_1x_2).S(x_1^2)}{n^3}
-\frac{S(x_1^4)}{n^4}-\frac{6S(x_1^2x_2^2)}{n_4} \\
&\quad-\text{other terms of odd order which will vanish on summation.}
\end{align*}

Summing for all values and dividing by the number of cases we get
$R_{u^2s_2}\sigma_{u^2}\sigma_{s^2}+m_1M_1= \frac{\mu_4}{n_2}+\mu_2^2\frac{(n-1)}{n^2} -\frac{\mu_4}{n_3}-3\mu_2^2\frac{(n-1)}{n^3},$
where $R_{u^2s^2}$ is the correlation between $u^2$ and $s^2$.
$R_{u^2s_2}\sigma_{u^2}\sigma_{s^2}+\mu_2^2\frac{(n-1)}{n^2} =\mu_2^2\frac{(n-1)}{n^3}\{3+n-3\}=\mu_2^2\frac{(n-1)}{n^2}.$

Hence $R_{u^2s_2}\sigma_{u^2}\sigma_{s^2}=0$, or there is no correlation
between $u^2$ and $s^2$.

\bigskip

%\begin{center}
\section*{Section III}
%\end{center}

To find the equation representing the frequency distribution of
the means of samples of $n$ drawn from a normal population, the mean
being expressed in terms of the standard deviation of the
sample.

We have $y=\frac{C}{\sigma^{n-1}}s^{n-2}e^{-\frac{nx^2}{2\sigma^2}}$ as
the equation representing the distribution of $s$, the standard
deviation  of a sample of $n$, when the samples are drawn from a normal
population with standard deviation $s$.

Now the means of these samples of $n$ are distributed according to the
equation\footnote{Airy, \textit{Theory of Errors of Observations},
Part II, \S6.}
$y=\frac{\sqrt{(n)}N}{\sqrt{(2\pi)}\sigma} e^{-\frac{nx^2}{2\sigma^2}},$
and we have shown that there is no correlation between $x$, the
distance of the mean of the sample, and $s$, the standard deviation
of the sample.

Now let us suppose $x$ measured in terms of $s$, i.e.\ let us find the
distribution of $z=x/s$.

If we have $y_1=\phi(x)$ and $y_2=\psi(z)$ as the equations
representing the frequency of $x$ and of $z$ respectively, then
\begin{gather*}
y_1dx=y_2dz=y_3\frac{dx}{s},\\
\therefore y_2=sy_1.
\end{gather*}
Hence
$y=\frac{N\sqrt{(n)}s}{\sqrt{(2\pi)}\sigma}e^{-\frac{ns^2z^2}{2\sigma^2}}$
is the equation representing the distribution of $z$ for samples of $n$
with standard deviation $s$.

Now the chance that $s$ lies between $s$ and $s + ds$ is
$\frac {\int_s^{s+ds}\frac{C}{\sigma^{n-1}}s^{n-2}e^{-\frac{ns^2}{2\sigma^2}}ds} {\int_0^{\infty}\frac{C}{\sigma^{n-1}}s^{n-2}e^{-\frac{ns^2}{2\sigma^2}}ds}$
which represents the $N$ in the above equation.

Hence the distribution of $z$ due to values of $s$ which lie between $s$ and
$s+ds$ is
$y=\frac {\int_s^{s+ds}\frac{C}{\sigma^n}\sqrt{\left(\frac{n}{2\pi}\right)}s^{n-1} e^{-\frac{ns^2(1+z^2)}{2\sigma^2}}ds} {\int_0^{\infty}\frac{C}{\sigma^{n-1}}s^{n-2}e^{-\frac{ns^2}{2\sigma^2}}ds} =\sqrt{\left(\frac{n}{2\pi}\right)}\frac {\int_s^{s+ds}\frac{C}{\sigma^n}s^{n-1} e^{-\frac{ns^2(1+z^2)}{2\sigma^2}}ds} {\int_0^{\infty}\frac{C}{\sigma^{n-2}}s^{n-2}e^{-\frac{ns^2}{2\sigma^2}}ds}$
and summing for all values of $s$ we have as an equation giving the
distribution of $z$
$y=\frac{\sqrt{\left(\frac{n}{2\pi}\right)}}{\sigma} \frac {\int_s^{s+ds}\frac{C}{\sigma^n}s^{n-1} e^{-\frac{ns^2(1+z^2)}{2\sigma^2}}ds} {\int_0^{\infty}\frac{C}{\sigma^{n-2}}s^{n-2}e^{-\frac{ns^2}{2\sigma^2}}ds}.$
By what we have already proved this reduces to
$y=\frac{1}{2}\frac{n-2}{n-3}.\frac{n-4}{n-5}\dots\frac{5}{4}.\frac{3}{2} (1+z^2)^{-\frac{1}{2}n},\quad \text{if n be odd}$
and to
$y=\frac{1}{2}\frac{n-2}{n-3}.\frac{n-4}{n-5}\dots\frac{4}{3}.\frac{2}{21} (1+z^2)^{-\frac{1}{2}n},\quad \text{if n be even}$

Since this equation is independent of $\sigma$ it will give the
distribution of the distance of the mean of a sample from the mean of
the population expressed in terms of the standard deviation of the
sample for any normal population.

\begin{center}
\section*{Section IV.  Some Properties of the Standard\\
Deviation Frequency Curve}
\end{center}

By a similar method to that adopted for finding the constant we may
find the mean and moments: thus the mean is at $I_{n-1}/I_{n-2}$,\newline
which is equal to
$\frac{n-2}{n-3}.\frac{n-4}{n-5}\dots\frac{2}{1} \sqrt{\left(\frac{2}{\pi}\right)}\frac{\sigma}{\sqrt{n}}, \quad\text{if n be even,}$
or
$\frac{n-2}{n-3}.\frac{n-4}{n-5}\dots\frac{3}{2} \sqrt{\left(\frac{\pi}{2}\right)}\frac{\sigma}{\sqrt{n}}, \quad\text{if n be odd\phantom{e}.}$

The second moment about the end of the range is
$\frac{I_n}{I_{n-2}}=\frac{(n-1)\sigma^2}{n}.$

The third moment about the end of the range is equal to
\begin{align*}
\frac{I_{n+1}}{I_{n-2}}&=\frac{I_{n+1}}{I_{n-1}}.\frac{I{n-1}}{I_{n-2}} \\
&=\sigma^2\times\text{the mean}.
\end{align*}

The fourth moment about the end of the range is equal to
$\frac{I_{n+2}}{I_{n-2}}=\frac{(n-1)(n+1)}{n^2}\sigma^4.$

If we write the distance of the mean from the end of the range
$D\sigma/\sqrt{n}$ and the moments about the end of the range
$\nu_1$, $\nu_2$, etc.,\newline
then
$\nu_1=\frac{D\sigma}{\sqrt{n}},\quad \nu_2=\frac{n-1}{n}\sigma_2,\quad \nu_3=\frac{D\sigma^3}{\sqrt{n}},\quad \nu_4=\frac{N^2-1}{n}\sigma^4.$

From this we get the moments about the mean:
\begin{align*}
\mu_2 &= \frac{\sigma^2}{n}(n-1-D^2), \\
\mu_3 &=\frac{\sigma^3}{n\sqrt{n}}\{nD-3(n-1)D+2D^2\}
=\frac{\sigma^3D}{n\sqrt{n}}\{2D^2-2n+3\}, \\
\mu_4 &= \frac{\sigma^2}{n^2}\{n^2-1-4D^2n+6(n-1)D^2-3D^4\} \\
&=\frac{\sigma^4}{n^2}\{n^2-1-D^2(3D^2-2n+6)\}.
\end{align*}
It is of interest to find out what these become when $n$ is large.

In order to do this we must find out what is the value of $D$.

Now Wallis's expression for $\pi$ derived from the infinite product
value of $\sin x$ is
$\frac{\pi}{2}(2n+1)=\frac{2^2.4^2.6^2\dots(2n)^2} {1^23^25^2\dots(2n-1)^2}.$

If we assume a quantity $\theta\left(=a_0+\frac{a_1}{n}+\text{etc.}\right)$
which we may add to the $2n+1$ in order to make the expression approximate
more rapidly to the truth, it is easy to show that
$\theta=-\frac{1}{2}+\frac{1}{16n}-$etc., and we get\footnote{This
expression will be found to give a much closer approximation to $\pi$
than Wallis's}
$\frac{\pi}{2}\left(2n+\frac{1}{2}+\frac{1}{16n}\right) =\frac{2^2.4^2.6^2\dots(2n)^2} {1^23^25^2\dots(2n-1)^2}.$

From this we find that whether $n$ be even or odd $D^2$ approximates to
$n-\frac{3}{2}+\frac{1}{8n}$ when $n$ is large.

Substituting this value of $D$ we get
$\mu_2=\frac{\sigma^2}{2n}\left(1-\frac{1}{4n}\right),\ \mu_2=\frac{\sigma^3\sqrt{\left(1-\frac{3}{2n}+\frac{1}{16n^2}\right)}}{4n^2},\ \mu_4=\frac{3\sigma_4}{4n^2}\left(1+\frac{1}{2n}-\frac{1}{16n^2}\right).$

Consequently the value of the standard deviation of a standard deviation
which we have found
$\left(\frac{\sigma}{\sqrt{(2n)}\sqrt{\{1-(1/4n)\}}}\right)$
becomes the same as that found for the normal curve by Prof.\ Pearson
$\{\sigma/(2n)\}$ when $n$ is large enough to neglect
the $1/4n$ in comparison with 1.

Neglecting terms of lower order than $1/n$, we find
$\beta_1=\frac{2n-3}{n(4n-3)},\quad \beta)_2=3\left(1-\frac{1}{2n}\right)\left(1+\frac{1}{2n}\right).$

Consequently, as $n$ increases, $\beta_2$ very soon approaches the
value 3 of the normal curve, but $\beta_1$ vanishes more slowly, so that
the curve remains slightly skew.

Diagram I shows the theoretical distribution of the standard
deviations found from samples of 10.

\begin{figure}
\begin{center}
\epsfig{file=student_diag1.eps,width=12cm,height=9cm,angle=0,clip=}
\end{center}
\end{figure}

\bigskip

%\begin{center}
\section*{Section V.  Some Properties of the Curve}
%\end{center}

Writing $z=\tan\theta$ the equation becomes $y=\frac{n-2}{n-3}.\frac{n-4}{n-5} \dots\text{etc.}\times\cos^n\theta$, which affords an easy way of drawing
the curve. Also $dz=d\theta/\cos^2\theta$.

Hence to find the area of the curve between any limits we must find
\begin{align*}
\times\int\cos^{n-2}\theta d\theta \\
&=\frac{n-2}{n-3}.\frac{n-4}{n-5}\dots\text{etc.}
\left\{\frac{n-3}{n-2}\int\cos^{n-4}\theta d\theta+
\left[\frac{cos^{n-3}\theta\sin\theta}{n-2}\right]\right\} \\
&=\frac{n-2}{n-3}.\frac{n-4}{n-5}\dots\text{etc.}
\int\cos^{n-4}\theta d\theta+
\frac{1}{n-3}\frac{n-4}{n-5}\dots\text{etc.}
[\cos^{n-3}\theta\sin\theta],
\end{align*}
and by continuing the process the integral may he evaluated.

For example, if we wish to find the area between 0 and $\theta$ for $n=8$
we have
\begin{align*}
\text{Area}&=\frac{6}{5}.\frac{4}{3}.\frac{2}{1}.\frac{1}{\pi}
\int_0^{\theta}\cos^6\theta d\theta \\
&=\frac{4}{3}.\frac{2}{\pi}\int_0^{\theta}\cos^4\theta d\theta
+\frac{1}{5}.\frac{4}{3}.\frac{2}{\pi}\cos^5\theta\sin\theta\\
&=\frac{\theta}{\pi}+\frac{1}{\pi}\cos\theta\sin\theta+
\frac{1}{3}.\frac{2}{\pi}\cos^3\theta\sin\theta
+\frac{1}{5}.\frac{4}{3}.\frac{2}{\pi}\cos^5\theta\sin\theta
\end{align*}
and it will be noticed that for $n=10$ we shall merely have to add
to this same expression the term
$\frac{1}{7}.\frac{6}{5}.\frac{4}{3}.\frac{2}{\pi}\cos^7\theta\sin\theta$.

The tables at the end of the paper give the area between $-\infty$ and $z$
$\left(\text{or\ }\theta=-\frac{\pi}{2}\text{\ and\ }\theta=\tan^{-1}z\right).$

This is the same as $0.5 + \text{the}$ area between $\theta=0$, and
$\theta=\tan^{-1}z$, and as the whole area of the curve is equal to 1,
the tables give the probability that the mean of the sample does not
differ by more than $z$ times the standard deviation of the sample from
the mean of the population.

The whole area of the curve is equal to
$\frac{n-2}{n-3}.\frac{n-4}{n-5}\dots\text{etc.} \times\int_{-\frac{1}{2}\pi}^{+\frac{1}{2}\pi}\cos^{n-2}\theta d\theta$
and since all the parts between the limits vanish at both limits this
reduces to 1.

Similarly, the second moment coefficient is equal to
\begin{align*}
&\frac{n-2}{n-3}.\frac{n-4}{n-5}\dots\text{etc.}
\times\int_{-\frac{1}{2}\pi}^{+\frac{1}{2}\pi}
\cos^{n-2}\theta\tan^2\theta d\theta \\
\times\int_{-\frac{1}{2}\pi}^{+\frac{1}{2}\pi}
(\cos^{n-4}\theta-\cos^{n-2}\theta)d\theta \\
\end{align*}

Hence the standard deviation of the curve is $1/\sqrt{(n-3)}$.  The fourth
moment coefficient is equal to
\begin{align*}
&\frac{n-2}{n-3}.\frac{n-4}{n-5}\dots\text{etc.}
\times\int_{-\frac{1}{2}\pi}^{+\frac{1}{2}\pi}
\cos^{n-2}\theta\tan^4\theta d\theta \\
\times\int_{-\frac{1}{2}\pi}^{+\frac{1}{2}\pi}
(\cos^{n-6}\theta-2\cos^{n-4}\theta+\cos^{n-2}\theta)d\theta \\
-\frac{2(n-2)}{n-3}+1=\frac{3}{(n-3)(n-5)}.
\end{align*}

The odd moments are of course zero, an the curve is symmetrical, so
$\beta_1=0,\quad\beta_2=\frac{3(n-3)}{n-5}=3+\frac{6}{n-5}.$

Hence as it increases the curve approaches the normal curve whose
standard deviation is $1/\sqrt{(n-3)}$.

$\beta_2$, however, is always greater than 3, indicating that large
deviations are mere common than in the normal curve.

\begin{figure}
\begin{center}
\epsfig{file=student_diag2.eps,width=12cm,height=9cm,angle=0,clip=}
\end{center}
\end{figure}

I have tabled the area for the normal curve with standard deviation
$1/\sqrt{7}$ so as to compare, with my curve for $n=10$\footnote{See p.\ 29}.
It will be seen that odds laid according  to either table would not
seriously differ till we reach $z = 0.8$, where the odds are about
50 to 1 that the mean is within that limit: beyond that the normal
curve gives a false feeling of security, for example, according to the
normal curve it is 99,986 to 14 (say 7000 to 1) that the mean of the
population lies between $-\infty$ and $+1.3s$, whereas the real odds are
only 99,819 to 181 (about 550 to 1).

Now 50 to 1 corresponds to three times the probable error in the normal
curve and for most purposes it would be considered significant; for this
reason I have only tabled my curves for values of $n$ not greater than 10,
but have given the $n=9$ and $n=10$ tables to one further place
of decimals. They can he used as foundations for finding values
for larger samples.\footnote{E.g.\ if $n=11$, to the corresponding value
for $n=9$, we add
$\frac{7}{8}\times\frac{5}{6}\times\frac{3}{4}\times\frac{1}{2}\times \frac{1}{2}\cos^8\theta\sin\theta$: if $n=13$ we add as well
$\frac{9}{10}\times\frac{7}{8}\times\frac{5}{6}\times\frac{3}{4}\times \frac{1}{2}\times\frac{1}{2}\cos^{10}\theta\sin\theta$, and so on.}

The table for $n=2$ can be readily constructed by looking out
$\theta=\tan^{-1}z$ in Chambers's tables and then $0.5+\theta/\pi$
gives the corresponding value.

Similarly $\frac{1}{2}\sin\theta+0.5$ gives the values when $n=3$.

There are two points of interest in the $n = 2$ curve. Here $s$ is
equal to half the distance between the two observations,
$\tan^{-1}\frac{s}{s}=\frac{\pi}{4}$, so that between $+s$ and
$-z$ lies $2\times\frac{\pi}{4}\times\frac{1}{\pi}$ or half the probability,
i.e.\ if two observations have been made and we have no other
information, it is an even chance that the mean of the (normal)
population will lie between them. On the other hand the second
moment coefficient is
$\frac{1}{\pi}\int_{=-\frac{1}{2}\pi}^{+\frac{1}{2}\pi} \tan^2\theta d\theta= \frac{1}{\pi}\left[\tan\theta -\theta\right]_{=-\frac{1}{2}\pi}^{+\frac{1}{2}\pi=\infty} =\infty,$
or the standard deviation is infinite while the probable error is
finite.

\begin{center}
\section*{Section VI. Practical Test of the foregoing Equations}
\end{center}

Before I bad 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.~Macdonnell (\textit{Biometrika}, \textsc{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\footnote{I hope to publish the
results of the correlation work shortly.} 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 $z$ of Section III.

This provides us with two sets of 750 standard deviations and two sets
of 750 $z$'s on which to test the theoretical results arrived at. The
height and left middle finger correlation table was chosen because the
distribution of both was approximately normal and the correlation was
fairly high. Both frequency curves, however, deviate slightly from
normality, the constants being for height $\beta_1 = 0.0026$,
$\beta_2=3.176$, and for left middle finger lengths $\beta_1 = 0.0030$,
$\beta_2= 3.140$, and in consequence there is a tendency for a certain
number of larger standard deviations to occur than if the distributions
wore normal. This, however, appears to make very little difference to
the distribution of $z$.

Another thing which interferes with the comparison is the
comparatively large groups in which the observations occur. The
heights are arranged in 1 inch groups, the standard deviation
being only 2.54 inches. while, the finger lengths wore originally
grouped in millimetres, but unfortunately I did not at the time see
the importance of having a smaller unit and condensed them into
2 millimetre groups, in terms of which the standard deviation is
2.74.

Several curious results follow from taking samples of 4 from
material disposed in such wide groups. The following points may
be noticed:

(1) The means only occur as multiples of 0.25. (2) The standard
deviations occur as the square roots of the following types
of numbers: $n$, $n + 0.10$, $n + 0.25$, $n + 0.50$, $n + 0.69$, $2n+0.75$.

(3) A standard deviation belonging to one of these groups can only be
associated with a mean of a particular kind; thus a standard  deviation
of $\sqrt{2}$ can only occur if the mean differs by a whole number
from the group we take as origin, while $\sqrt{1.69}$ will only occur when
the mean is at $n\pm0.25$.

(4) All the four individuals of the sample will occasionally come from
the same group, giving a zero value for the standard deviation. Now this
leads to an infinite value of $z$ and is clearly due to too wide a
grouping, for although two men may have the same height when measured by
inches, yet the finer the measurements the more seldom will they he
identical, till finally the chance that four men will have
\textit{exactly} the same height is infinitely small. If we had smaller
grouping the zero values of the standard deviation might be expected to
increase, and a similar consideration will show that the smaller values
of the standard deviation would also be likely to increase, such as
0.436, when 3 fall in one group and 1 in an adjacent group, or 0.50 when
2 fall in two adjacent groups. On the other hand, when the individuals
of the sample lie far apart, the argument of Sheppard's correction will
apply, the real value of the standard deviation being more likely to he
smaller than that found owing to the frequency in any group being
greater on the side nearer the mode.

These two effects of grouping will tend to neutralize the effect on the
mean value of the standard deviation, but both will increase the
variability.

Accordingly, we find that the mean value of the standard deviation is
quite close to that calculated, while in each case the variability is
sensibly greater.  The fit of the curve is not good, both for this
reason and because the frequency is not evenly distributed owing to
effects (2) and (3) of grouping. On the other hand, the fit of the
curve giving the frequency of $z$ is very good, and as that is the only
practical point the comparison may he considered satisfactory.

The following are the figures for height:
\begin{center}
\begin{tabular}{lll}
Mean value of standard deviations: &Calculated&$\phantom{-}2.027\pm±0.02$ \\
&Observed  &$\underline{\phantom{-}2.026}$ \\
& Difference =& $-0.001$ \\
Standard deviation of standard deviations:& Calculated
& $\phantom{-}0.8558\pm0.015$ \\
& Observed & $\underline{\phantom{-}0.9066}$\\
& Difference & $+0.0510$
\end{tabular}
\end{center}

{\small
\begin{center}
\textit{Comparison of Fit. Theoretical Equation:
$y=\frac{16\times750}{\sqrt{(2\pi)}\sigma^2}x^2e^{-\frac{2x^2}{\sigma^2}}$}\\
\begin{tabular}{|c@{\ }c@{\ }c@{\ }c@{\ }c@{\ }c@{\ }
c@{\ }c@{\ }c@{\ }c@{\ }c@{\ }c@{\ }
c@{\ }c@{\ }c@{\ }c@{\ }c@{\ }c|}
\hline
\multicolumn{18}{|l|}{Scale in terms of standard deviations of population}\\
\hline
\multicolumn{18}{|l|}{Calculated frequency}\\
\hline
$1\frac{1}{2}$&$10\frac{1}{2}$&27&$45\frac{1}{2}$&$64\frac{1}{2}$&
$78\frac{1}{2}$&
87&88&$81\frac{1}{2}$&71&58&45&33&23&15&$9\frac{1}{2}$&$5\frac{1}{2}$&7\\
\hline
\multicolumn{18}{|l|}{Observed frequency}\\
3&$14\frac{1}{2}$&$24\frac{1}{2}$&$37\frac{1}{2}$&107&67&73&77&
$77\frac{1}{2}$&
64&$\frac{1}{2}$&$49\frac{1}{2}$&35&28&$12\frac{1}{2}$&9&$11\frac{1}{2}$&7\\
\hline
\multicolumn{18}{|l|}{Difference}\\
$+1\frac{1}{2}$&$+4$&$-2\frac{1}{2}$&$-8$&$+42\frac{1}{2}$&
$-11\frac{1}{2}$&$-14$&$-11$&
$-4$&$-7$&$-5\frac{1}{2}$&$+4\frac{1}{2}$&$+2$&$+5$&$-2\frac{1}{2}$&
$-\frac{1}{2}$&$+6$&0 \\
\hline
\end{tabular}\\
Whence $\chi^2=48.06$, $P=0.00006$ (about).
\end{center}
}

In tabling the  observed frequency, values between 0.0125 and 0.0875
were included in one group, while between 0.0875 and 0.012.5 they were
divided over the two groups. As an instance of the irregularity due to
grouping I may mention that there were 31 cases of standard deviations
1.30 (in terms of the grouping) which is 0.5117 in terms of the
standard deviation of the population, and they wore therefore divided
over the groups 0.4 to 0.5 and 0.5 to 0.6. Had they all been
counted in groups 0.5 to 0.6 $\chi^2$ would have fallen to 20.85 and $P$
would have risen to 0.03.  The $\chi^2$ test presupposes random sampling
from a frequency following the given law, but this we have not got
owing to the interference of the grouping.

When, however, we test the $z$'s where the grouping has not had so much
effect, we find a close correspondence between the theory and the actual
result.

There were three cases of infinite values of $z$ which, for the
reasons given above, were given the next largest values which
occurred, namely $+6$ or $-6$. The rest were divided into
groups of 0.1; 0.04, 0.05 and 0.06, being divided between the two
groups on either side.

The calculated value for the standard deviation of the frequency curve
was 1 ($\pm0.0171$), while the observed was 1.030.  The value of the
standard deviation is really infinite, as the fourth moment
coefficient is infinite, but as we have arbitrarily limited the
infinite cases we may take as an approximation $1/\sqrt{1500}$ from
which the value of the probable error given above is obtained.
The fit of the curve is as follows:
{\small
\begin{center}
\textit{Comparison of Fit.  Theoretical Equation:
$y=\frac{2N}{\pi}\cos^4\theta$, $z=\tan\theta$} \\
\begin{tabular}{|c@{\ }c@{\ }c@{\ }c@{\ }c@{\ }
c@{\ }c@{\ }c@{\ }c@{\ }c@{\ }
c@{\ }c@{\ }c@{\ }c@{\ }c|}
\hline
\multicolumn{15}{|l|}{Scale of $z$} \\
\hline
\multicolumn{15}{|l|}{Calculated frequency} \\
\hline
5&$9\frac{1}{2}$&$13\frac{1}{2}$&$34\frac{1}{2}$&$44\frac{1}{2}$&
$78\frac{1}{2}$&119&141&
$78\frac{1}{2}$&$44\frac{1}{2}$&$34\frac{1}{2}$&$13\frac{1}{2}$&
$13\frac{1}{2}$&
$9\frac{1}{2}$&5 \\
\hline
\multicolumn{15}{|l|}{Observed frequency} \\
9&$14\frac{1}{2}$&$11\frac{1}{2}$&33&$43\frac{1}{2}$&$70\frac{1}{2}$&
$119\frac{1}{2}$&
$151\frac{1}{2}$&122&$67\frac{1}{2}$&49&$26\frac{1}{2}$&16&10&6 \\
\hline
\multicolumn{15}{|l|}{Difference} \\
$+4$&$+4$&$-2$&$-2$&$-1\frac{1}{2}$&$-1$&$-8$&$+\frac{1}{2}$&
$+10\frac{1}{2}$&$+3$&$-11$&$+4\frac{1}{2}$&$-8$&$+2\frac{1}{2}$&
$+\frac{1}{2}$ \\
\hline
\end{tabular} \\
Whence $\chi^2=12.44$, $P=0.56$.
\end{center}
}

This is very satisfactory, especially when we consider that as a rule
observations are tested against curves fitted from the mean and one or
more other moments of the observations, so that considerable
correspondence is only to ])c expected; while this curve is
exposed to the full errors of random sampling, its constants
having been calculated quite apart from the observations.

The left middle finger samples show much the same features as
those of the height, but as the grouping is not so large
compared to the variability the curves fit the observations more
closely. Diagrams III\footnote{There are three small mistakes in
plotting the observed values in Diagram III, which make the fit appear
worse than it really is} and IV give the standard deviations of the
$z$'s for the set of samples.  The results are as follows:
\begin{center}
\begin{tabular}{lll}
Mean value of standard deviations: & Calculated  & $\phantom{-}2.186\pm0.023$ \\
& Observed    & $\underline{\phantom{-}2.179}$ \\
& Difference =& $-0.007$ \\
Standard deviation of standard deviations:&Calculated
&$\phantom{-}0.9224\pm0.016$ \\
& Observed    & $\underline{\phantom{-}0.9802}$ \\
& Difference =& $+0.0578$
\end{tabular}
\end{center}

\begin{figure}
\begin{center}
\epsfig{file=student_diag3.eps,width=5cm,height=15cm,clip=}
\ \
\epsfig{file=student_diag4.eps,width=5cm,height=15cm,clip=}
\end{center}
\end{figure}

{\scriptsize
\begin{center}
\textit{Comparison of Fit.  Theoretical Equation:}
$y=\frac{16\times750}{\sqrt{(2\pi)}\sigma^2}x^2e^{-\frac{2x^2}{\sigma^2}}$\\
\begin{tabular}{|c@{\ }c@{\ }c@{\ }c@{\ }c@{\ }c@{\ }
c@{\ }c@{\ }c@{\ }c@{\ }c@{\ }c@{\ }
c@{\ }c@{\ }c@{\ }c@{\ }c@{\ }c|}
\hline
\multicolumn{18}{|l|}{Scale in terms of standard deviations of population}\\
$1\frac{1}{2}$&$10\frac{1}{2}$&27&$45\frac{1}{2}$&$64\frac{1}{2}$&
$78\frac{1}{2}$&87&88&
$81\frac{1}{2}$&71&58&45&33&23&15&$9\frac{1}{2}$&$5\frac{1}{2}$&7\\
\hline
\multicolumn{18}{|l|}{Calculated frequency} \\
2&14&$27\frac{1}{2}$&51&$64\frac{1}{2}$&91&$94\frac{1}{2}$&$68\frac{1}{2}$&
$65\frac{1}{2}$&73&$48\frac{1}{2}$&$40\frac{1}{2}$&$42\frac{1}{2}$&20&
$22\frac{1}{2}$&
12&5&$7\frac{1}{2}$ \\
\hline
\multicolumn{18}{|l|}{Observed frequency} \\
$+\frac{1}{2}$&$+3\frac{1}{2}$&$+\frac{1}{2}$&$+5\frac{1}{2}$&---&
$+12\frac{1}{2}$&$+7\frac{1}{2}$&$-19\frac{1}{2}$&$-16$&$+2$&
$-9\frac{1}{2}$&$-4\frac{1}{2}$&$+9\frac{1}{2}$&$-3$&$+7\frac{1}{2}$&
$+2\frac{1}{2}$&$-\frac{1}{2}$&$+\frac{1}{2}$ \\
\hline
\end{tabular} \\
Whence $\chi^2=21.80$, $P=0.19$.

\medskip

\begin{tabular}{llr}
Value of standard deviation: & Calculated   & 1($\pm0.017$) \\
& Observed     & \underline{0.982} \\
& Difference = & $-0.018$
\end{tabular} \\

\medskip

\textit{Comparison of Fit.  Theoretical Equation:
$y=\frac{2N}{\pi}\cos^4\theta$, $z=\tan\theta$} \\
\begin{tabular}{c@{\ }c@{\ }c@{\ }c@{\ }c@{\ }c@{\ }c@{\ }
c@{\ }c@{\ }c@{\ }c@{\ }c@{\ }c@{\ }c@{\ }c}
\multicolumn{15}{l}{Scale of $z$} \\
\multicolumn{15}{l}{Calculated frequency} \\
5&$9\frac{1}{2}$&$13\frac{1}{2}$&$34\frac{1}{2}$&$44\frac{1}{2}$&
$78\frac{1}{2}$&119&
141&119&$78\frac{1}{2}$&$44\frac{1}{2}$&$34\frac{1}{2}$&$13\frac{1}{2}$&
$9\frac{1}{2}$&5 \\
\multicolumn{15}{l}{Observed frequency} \\
4&$15\frac{1}{2}$&18&$33\frac{1}{2}$&44&75&122&138&$120\frac{1}{2}$&71&
$46\frac{1}{2}$&36&11&9&6 \\
\multicolumn{15}{l}{Difference} \\
$-1$&$+6$&$+4\frac{1}{2}$&$-1$&$-\frac{1}{2}$&$-3\frac{1}{2}$&$+3$&
$-3$&$+1\frac{1}{2}$&$-7\frac{1}{2}$&$+2$&$+1\frac{1}{2}$&
$-2\frac{1}{2}$&$-\frac{1}{2}$&$+1$
\end{tabular} \\
Whence $\chi^2=7.39$, $P=0.92$.
\end{center}
}

A very close fit.

We see then that if the distribution is approximately normal our
theory gives us a satisfactory measure of the certainty to be derived
from a small sample in both the cases we have tested; but we have
an indication that a fine grouping is of advantage.  If the distribution
is not normal, the mean and the standard deviation of a sample will be
positively correlated, so although both will have greater variability,
yet they will tend to counteract one another, a mean deriving largely
from the general mean tending to be divided by a larger standard
deviation.  Consequently, I believe that the table
given in Section VII below may be used in estimating the degree of
certainty arrived at by the mean of a few experiments, in the case
of most laboratory or biological work where the distributions
are as a rule of a cocked hat'' type and so sufficiently nearly
normal

\newpage

\begin{center}
\section*{Section VII.  Tables of \\ \ \\
$\frac{n-2}{n-3}.\frac{n-4}{n-5}\dots\left( \begin{array}{l} \frac{3}{2}.\frac{1}{2}\,n\text{\ odd} \\ \frac{2}{1}.\frac{1}{\pi}\,n\text{\ even} \end{array}\right) \int_{-\frac{1}{2}\pi}^{\tan^{-1}z}\cos^{n-2}\theta d\theta$
\\ \ \\
for values of $n$ from 4 to 10 inclusive}
\end{center}

\begin{center}
\textit{Together with $\frac{\sqrt{7}}{\sqrt{(2\pi)}} \int_{-\infty}^x e^{-\frac{7x^2}{2}}dx$ for comparison when $n=10$} \\
\begin{tabular}{ccccccccc}
\end{tabular}
\end{center}

{\footnotesize
\begin{center}
\begin{tabular}{|c@{\ }c@{\ }c@{\ }c@{\ }c@{\ }
c@{\ }c@{\ }c@{\ }c|}
\hline
$z\left(=\frac{x}{s}\right)$&
$n=4$&$n=5$&$n=6$&$n=7$&$n=8$&$n=9$&$n=10$&For comparison \\
&      &        &        &        &        &         &         &
$\left(\frac{\sqrt{7}}{\sqrt{(2\pi)}}\int_{-\infty}^{x} e^{-\frac{7x^2}{2}}dx\right)$ \\
\hline
0.1 &
0.5633 & 0.5745 & 0.5841 & 0.5928 & 0.6006 & 0.60787 & 0.61462 & 0.60411\\
0.2 &
0.6241 & 0.6458 & 0.6634 & 0.6798 & 0.6936 & 0.70705 & 0.71846 & 0.70159\\
0.3 &
0.6804 & 0.7096 & 0.7340 & 0.7549 & 0.7733 & 0.78961 & 0.80423 & 0.78641\\
0.4 &
0.7309 & 0.7657 & 0.7939 & 0.8175 & 0.8376 & 0.85465 & 0.86970 & 0.85520\\
0.5 &
0.7749 & 0.8131 & 0.8428 & 0.8667 & 0.8863 & 0.90251 & 0.91609 & 0.90691\\
0.6 &
0.8125 & 0.8518 & 0.8813 & 0.9040 & 0.9218 & 0.93600 & 0.94732 & 0.94375\\
0.7 &
0.8440 & 0.8830 & 0.9109 & 0.9314 & 0.9468 & 0.95851 & 0.96747 & 0.96799\\
0.8 &
0.8701 & 0.9076 & 0.9332 & 0.9512 & 0.9640 & 0.97328 & 0.98007 & 0.98253\\
0.9 &
0.8915 & 0.9269 & 0.9498 & 0.9652 & 0.9756 & 0.98279 & 0.98780 & 0.99137\\
1.0 &
0.9092 & 0.9419 & 0.9622 & 0.9751 & 0.9834 & 0.98890 & 0.99252 & 0.99820\\
\hline
1.1 &
0.9236 & 0.9537 & 0.9714 & 0.9821 & 0.9887 & 0.99280 & 0.99539 & 0.99926\\
1.2 &
0.9354 & 0.9628 & 0.9782 & 0.9870 & 0.9922 & 0.99528 & 0.99713 & 0.99971\\
1.3 &
0.9451 & 0.9700 & 0.9832 & 0.9905 & 0.9946 & 0.99688 & 0.99819 & 0.99986\\
1.4 &
0.9451 & 0.9756 & 0.9870 & 0.9930 & 0.9962 & 0.99791 & 0.99885 & 0.99989\\
1.5 &
0.9598 & 0.9800 & 0.9899 & 0.9948 & 0.9973 & 0.99859 & 0.99926 & 0.99999\\
1.6 &
0.9653 & 0.9836 & 0.9920 & 0.9961 & 0.9981 & 0.99903 & 0.99951 & \\
1.7 &
0.9699 & 0.9864 & 0.9937 & 0.9970 & 0.9986 & 0.99933 & 0.99968 & \\
1.8 &
0.9737 & 0.9886 & 0.9950 & 0.9977 & 0.9990 & 0.99953 & 0.99978 & \\
1.9 &
0.9970 & 0.9904 & 0.9959 & 0.9983 & 0.9992 & 0.99967 & 0.99985 & \\
2.0 &
0.9797 & 0.9919 & 0.9967 & 0.9986 & 0.9994 & 0.99976 & 0.99990 & \\
\hline
2.1 &
0.9821 & 0.9931 & 0.9973 & 0.9989 & 0.9996 & 0.99983 & 0.99993 & \\
2.2 &
0.9841 & 0.9941 & 0.9978 & 0.9992 & 0.9997 & 0.99987 & 0.99995 & \\
2.3 &
0.9858 & 0.9950 & 0.9982 & 0.9993 & 0.9998 & 0.99991 & 0.99996 & \\
2.4 &
0.9873 & 0.9957 & 0.9985 & 0.9995 & 0.9998 & 0.99993 & 0.99997 & \\
2.5 &
0.9886 & 0.9963 & 0.9987 & 0.9996 & 0.9998 & 0.99995 & 0.99998 & \\
2.6 &
0.9898 & 0.9967 & 0.9989 & 0.9996 & 0.9999 & 0.99996 & 0.99999 & \\
2.7 &
0.9908 & 0.9972 & 0.9989 & 0.9997 & 0.9999 & 0.99997 & 0.99999 & \\
2.8 &
0.9916 & 0.9975 & 0.9989 & 0.9998 & 0.9999 & 0.99998 & 0.99999 & \\
2.9 &
0.9924 & 0.9978 & 0.9989 & 0.9998 & 0.9999 & 0.99998 & 0.99999 & \\
3.0 &
0.9931 & 0.9981 & 0.9989 & 0.9998 &   ---  & 0.99999 &   ---   & \\
\hline
\end{tabular}
\end{center}
}

\begin{center}
\section*{Explanation of Tables}
\end{center}

The tables give the probability that the value of the mean, measured
from the mean of the population, in terms of the standard deviation of
the sample, will lie between $-\infty$ and $z$. Thus, to take the
table for samples of 6, the probability of the mean of the
population lying between $-\infty$ and once the standard deviation
of the sample is 0.9622, the odds are about 24 to 1 that the mean of
the population lies between these limits.

The probability is therefore 0.0378 that it is greater than once the
standard deviation and 0.07511 that it lies outside $\pm1.0$ times the
standard deviation.

\bigskip

%\begin{center}
\section*{Illustration of Method}
%\end{center}

\textit{Illustration I.} As an instance of the kind of use which may be
made of the tables, I take the following figures from a table by
A.~R.~Cushny and A.~R.~Peebles in the
\textit{Journal of Physiology} for 1904, showing the different effects
of the optical isomers of hyoscyamine hydrobromide in producing sleep.
The average number of hours' sleep gained by the use of the drug is
tabulated below.

The conclusion arrived at was that in the usual doses 2 was, but 1 was
not, of value as a soporific.

{\small
\begin{center}
\textit{Additional hours' sleep gained by the use of
hyoscyamine hydrobromide} \\
\begin{tabular}{clclclc}
Patient && 1 (Dextro-) && 2 (Laevo-) && Difference ($2-1$) \\
\phantom{0}1 && $+0.7$ && $+1.9$ && $+1.2$ \\
\phantom{0}2 && $-1.6$ && $+0.8$ && $+2.4$ \\
\phantom{0}3 && $-0.2$ && $+1.1$ && $+1.3$ \\
\phantom{0}4 && $-1.2$ && $+0.1$ && $+1.3$ \\
\phantom{0}5 && $-0.1$ && $-0.1$ && $0$ \\
\phantom{0}6 && $+3.4$ && $+4.4$ && $+1.0$ \\
\phantom{0}7 && $+3.7$ && $+5.5$ && $+1.8$ \\
\phantom{0}8 && $+0.8$ && $+1.6$ && $+0.8$ \\
\phantom{0}9 && $0$ && $+4.6$ && $+4.6$ \\
10 && $+2.0$ && $+3.4$ && $+1.4$ \\
&Mean          & $+0.75$ & Mean        & $+2.33$ & Mean        & $+1.58$ \\
&\textsc{s.d.} & $1.70$ &\textsc{s.d.}& $1.90$ &\textsc{s.d.}& $1.17$
\end{tabular}
\end{center}
}

First let us see what is the probability that 1 will on the average
give increase of sleep; i.e.\ what is the chance that the mean
of the population of which these experiments are a sample is
positive.\ $+0.75/1.70 = 0.44$, and looking out $z = 0.44$ in the table for
ten experiments we find by interpolating between 0.8697 and 0.9161
that 0.44 corresponds to 0.8873, or the odds are 0.887 to 0.113
that the mean is positive.

That is about 8 to 1, and would correspond to the normal curve to about
1.8 times the probable error.  It is then very likely that 1 gives an
increase of sleep, but would occasion no surprise if the results were
reversed by further experiments.

If now we consider the chance that 2 is actually a soporific we have the
mean inclrease of sleep $=2.33/1.90$ or 1.23 times the \textsc{s.d.}
From the table the probability corresponding to this is 0.9974, i.e.\
the odds are nearly 400 to 1 that such is the case.  This corresponds to
about 4.15 times the probable error in the normal curve.  But I take it
that the real point of the authors was that 2 is better than 1.  This we
must t4est by making a new series, subtracting 1 from 2.  The mean values
of this series is $+1.38$, while the \textsc{s.d.}\ is 1.17, the mean
value being $+1.35$ times the \textsc{s.d.}  From the table, the
probability is 0.9985, or the odds are about 666 to one that 2 is the
better soporific.  The low value of the \textsc{s.d.}\ is probably due
to the different drugs reacting similarly on the same patient, so that
there is correlation between the results.

Of course odds of this kind make it almost certain that 2 is the better
soporific, and in practical life such a high probability is in most
matters considered as a certainty.

\textit{Illustration II.}  Cases where the tables will be useful are not
uncommon in agricultural work, and they would be more numerous if the
advantages of being able to apply statistical reasoning were borne in
mind when planning the experiments.  I take the following instances from
the accounts of the Woburn farming experiments published yearly by Dr
Voelcker in the \textit{Journal of the Agricultural Soceity}.

A short series of pot culture experiments were conducted in order to
determine the casues which lead to the production of
Hard (glutinous) wheat or Soft (starchy) wheat.  In three successive
years a bulk of seed corn of one variety was picked
over by hand and two samples were selected, one consisting
of hard'' grains avid the other of soft''. Some of each of them
were planted in both heavy and light soil and the resulting crops wore
weighed and examined for hard and soft corn.

The conclusion drawn was that the effect of selecting the seed was
negligible compared with the influence of the soil.

This conclusion was thoroughly justified, theheavy soul
producing in each case nearly 100\% of hard corn, but still the
effect of selecting the seed could just be traced in each year.

But a curious point, to which Dr Voelcker draws attention in the second
year's report, is that the soft seeds produced the higher yield of both
corn and straw. In view of the well-known fact that the
\textit{varieties} which have a high yield tend to produce soft corn, it
is interesting to see how much evidence the experiments afford as to the
correlation between softness and fertility in the same \textit{variety}.

Further, Mr Hooker\footnote{\textit{Journal of the Royal Statistical
Society}, 1897} has shown that the yield of wheat in one year is largely
determined by the weather during the preceding year.  Dr Voelcker's
results may afford a clue as to the way in which the seed id affected,
and would almost justify the selection of particillar soils
for growing wheat.\footnote{And perhaps a few experiments to see whether
there is a correlation between yield and mellowness'' in barley.}

Th figures are as follows, the yields being expressed in
grammes per pot:

{\scriptsize
\begin{center}
\begin{tabular}{|l|@{\ }c@{\ }c@{\ }c@{\ }c@{\ }
c@{\ }c@{\ }c@{\ }c@{\ }c|}
\hline
Year &
\multicolumn{2}{c|}{1899} &
\multicolumn{2}{|c|}{1900} &
\multicolumn{2}{|c|}{1901}
& & Standard & \\
\hline
Soil &
Light & Heavy & Light & Heavy & Light & Heavy & Average & deviation & $z$ \\
\hline
Yield of corn from soft seed &
7.55 & 8.89 & 14.81 & 13.55 & 7.49 & 15.39 & 11.328 & & \\
Yield of corn from hard seed &
7.27 & 8.32 & 13.81 & 13.36 & 7.97 & 13.13 & 10.643 & & \\
\hline
Difference &
$+0.58$&$+0.57$&$+1.00$&$+0.19$&$-0.49$&$+2.26$&$+0.685$&0.778&0.88 \\
Yield of straw from soft seed &
12.81 & 12.87 & 22.22 & 20.21 & 13.97 & 22.57 & 17.442 & & \\
Yield of straw from hard seed &
10.71 & 12.48 & 21.64 & 20.26 & 11.71 & 18.96 & 15.927 & & \\
\hline
Difference &
$+2.10$&$+0.39$&$+0.78$&$-0.05$&$+2.66$&$+3.61$&$+1.515$&1.261&1.20 \\
\hline
\end{tabular}
\end{center}
}

If we wish to laid the odds that the soft seed will give a better yield
of corn on the average, we divide, the average difference by the
standard deviation, giving us
$z=0.88.$
Looking this up in the table for $n = 6$ we find $p=0.9465$ or the odds
are 0.9465 to 0.0535 about 18 to 1.

Similarly for straw $z = 1.20$, $p = 0.9782$, and the odds are about 45 to 1.

In order to see whether such odds are sufficient for a practical man
to draw a definite conclusion, I take another act of experiments in which
Dr Voelcker compares the effects of different artificial manures
used with potatoes on a large scale.

The figures represent the difference between the crops grown with
the rise of sulphate of potash and kailit respectively in both 1904
and 1905:

\begin{center}
\mbox{
$\left. \begin{array}{c@{\ }c@{\ }c@{\ }c@{\ }c@{\ }c@{\ } c@{\ }c@{\ }c@{\ }c@{\ }c@{\ }c} & & \text{cwt.} & \text{qr.} & \text{lb.} & & & \text{ton} & \text{cwt.} & \text{qr.} & \text{lb.} \\ 1904 & + & 10 & 3 & 20 & : & + & 1 & 10 & 1 & 26 \\ 1905 & + & \phantom{0}6 & 0 & \phantom{0}3 & : & + & & 13 & 2 & \phantom{0}8 \end{array} \right\}$
(two experiments in each year)
}
\end{center}

The average gain by the use of sulphate of potash was 15.25 cwt. and the
\textsc{s.d.}\ 9 cwt., whence, if we want the odds that the conclusion
given below is right, $z = 1.7$, corresponding, when $n = 4$,to $p=0.9698$ or odds of 32 to 1; this is midway between the odds in the former
example.  Dr Voelcker says: It may now fairly be concluded that for
the potato crop on light land 1 cwt.\ per acre of sulphate of potash
is a better dressing than kailit.''

Am an example of how the table should be used with caution, I take the
following pot culture experiments to test whether it made any difference
whether large or small seeds were sown.

\textit{Illustration III.} In 1899 and in 1903 head corn'' and tail
corn'' were taken from the same bulks of barley and sown in
pots.  The yields in grammes were as follows:
\begin{center}
\begin{tabular}{lcc}
& 1899 & 1903 \\
Large seed \dots & 13.9 &  7.3 \\
Small seed \dots & \underline{14.4} & \underline{\phantom{0}1.4} \\
&$+0.5$&$+1.4$
\end{tabular}
\end{center}

The average gain is thus 0.95 and the \textsc{s.d.}\ 0.45, giving $z = 2.1$.
Now the table for $n = 2$ is not given, but if we look up the angle
whose tangent is 2.1 in Chambers's tables,
$p=\frac{\tan^{-1}2.1}{180^{\circ}}+0.5=\frac{64^{\circ}39'}{180^{\circ}} = 0.859,$
so that the odds are about 6 to 1 that small corn gives
a better yield than large.  These odds\footnote{[Through a numerical
slip, now corrected, Student had given the odds as 33 to 1 and it is to
this figure that the remarks in this paragraph relate.} are those which
would  be laid, and laid rigidly, by a man whose only knowledge of
the matter was contained in the two experiments. Anyone conversant
with pot culture would however know that the difference
between the two results would generally be greater and would
correspondingly moderate the certainty of his conclusion.  In point of
fact a large-scale experiment confirmed this result, the small corn
yielding shout 15\% more than the large.

I will conclude with an example which comes beyond the range of
the tables, there being eleven experiments.

To test whether it is of advantage to kiln-dry barley seed before
sowing, seven varieties of barley wore sown (both  kiln-dried and not
kiln-dried in 1899 and four in 1900; the results  are given in the
table.

{\tiny
\begin{center}
\begin{tabular}{|c@{\ }c@{\ }c@{\ }c@{\ }c@{\ }c@{\ }
c@{\ }c@{\ }c@{\ }c@{\ }c@{\ }c@{\ }c|}
\hline
& \multicolumn{3}{c} \mbox{Lb.\ head corn per acre}
& \multicolumn{3}{c} \mbox{Price of head corn in}
& \multicolumn{3}{c} \mbox{Cwt.\ straw per acre}
& \multicolumn{3}{c|}{\mbox{Value of crop per acre}} \\
& \multicolumn{3}{l} \mbox{shillings per quarter}
& & & & \multicolumn{3}{l} \mbox{in shillings} & & & \\
\hline
& N.K.D. & N.D. & Diff.
& N.K.D. & N.D. & Diff.
& N.K.D. & N.D. & Diff.
& N.K.D. & N.D. & Diff. \\
& 1903 & 2009 & $+106$
& $26\frac{1}{2}$ & $\frac{1}{2}$ & 0
& $19\frac{1}{2}$ & 25 & $+5\frac{1}{2}$
& $140\frac{1}{2}$ & 152 & $+11\frac{1}{2}$ \\
& 1935 & 1915 & $-\phantom{0}20$
& 28 & $26\frac{1}{2}$ & $-1\frac{1}{2}$
& $22\frac{1}{2}$ & 24 & $+1\frac{1}{2}$
& $152\frac{1}{2}$ & 145 & $-7\frac{1}{2}$ \\
& 1910 & 2011 & $+101$
& $29\frac{1}{2}$ & $28\frac{1}{2}$ & $-1$
& 23 & 24 & $+1$
& $158\frac{1}{2}$ & 161 & $+2\frac{1}{2}$ \\
1899 & 2496 & 2463 & $-\phantom{0}33$
& 30 & 29 & $-1$
& 23 & 28 & $+5$
& $204\frac{1}{2}$ & $199\frac{1}{2}$ & $-5$ \\
& 2108 & 2180 & $+\phantom{0}72$
& $27\frac{1}{2}$ & 27 & $-\frac{1}{2}$
& $22\frac{1}{2}$ & $22\frac{1}{2}$ & 0
& 162 & 142 & $+2$ \\
& 1961 & 1925 & $-36$
& 26 & 26 & 0
& $19\frac{1}{2}$ & $419\frac{1}{2}$ & $-\frac{1}{2}$
& 142 & $139\frac{1}{2}$ & $-2\frac{1}{2}$ \\
& 2060 & 2122 & $+\phantom{0}62$
& 29 & 26 & $-3$
& $24\frac{1}{2}$ & $22\frac{1}{2}$ & $-2\frac{1}{2}$
& 168 & 155 & $-13$ \\
& 1444 & 1482 & $+\phantom{0}38$
& $29\frac{1}{2}$ & $28\frac{1}{2}$ & $-1$
& $15\frac{1}{2}$ & 16 & $+\frac{1}{2}$
& 118 & $117\frac{1}{2}$ & $-\frac{1}{2}$ \\
1900 & 1612 & 1542 & $-\phantom{0}70$
& $28\frac{1}{2}$ & $28$ & $-\frac{1}{2}$
& 18 & $17\frac{1}{2}$ & $-\frac{1}{2}$
& $128\frac{1}{2}$ & 121 & $-7\frac{1}{2}$ \\
& 1316 & 1443 & $+127$
& 30 & 29 & $-1$
& $14\frac{1}{2}$ & $15\frac{1}{2}$ & $+1\frac{1}{2}$
& $109\frac{1}{2}$ & $116\frac{1}{2}$ & $+7$ \\
& 1511 & 1535 & $+\phantom{0}24$
& $28\frac{1}{2}$ & 28 & $-\frac{1}{2}$
& 17 & $17\frac{1}{2}$ & $+\frac{1}{2}$
& 120 & $120\frac{1}{2}$ & $+\frac{1}{2}$ \\
\hline
Average & 1841.5 & 1875.2 & $+33.7$
& 28.45 & 27.55 & $-0.91$
& 19.95 & 21.05 & $+1.10$
& 145.82 & 144.68 & $+1.14$ \\
\hline
Standard & \dots & \dots & 63.1 & \dots & \dots & 0.79
& \dots & \dots & 2.25 & \dots & \dots & 6.67 \\
deviation &      &       &      &       &       &
&       &       &      &       &       & \\
Standard &       &       &      &       &       &
&       &       &      &       &       & \\
deviation & \dots & \dots & 63.1 & \dots & \dots & 0.79
& \dots & \dots & 2.25 & \dots & \dots & 6.67 \\
$\div\sqrt{8}$ & &       &      &       &       &
&       &       &      &       &       & \\
\hline
\end{tabular}
\end{center}
}

It will he noticed that the kiln-dried seed gave on an average the
larger yield. of corn and straw, but that the quality was almost always
inferior. At first sight this might be supposed to be due to superior
germinating power in the kiln-dried seed, but my farming friends tell me
that the effect of this would be that the kiln-dried seed would produce
the better quality barley. Dr Voelcker draws the  conclusion: In such
seasons as 1899 and 1900 there is no particular advantage in kiln-drying
before mowing.''  Our examination completely justifies this and adds
and the quality of the resulting barley is inferior though the yield
may be greater.''

In this case I propose to use the approximation given by the normal
curve with standard deviation $s/\sqrt{n-3}$ and therefore use
Sheppard's tables, looking up the difference divided by $S/\sqrt{8}$.
The probability in the case of yield of corn per acre is given by
looking up $33.7/22.3 = 1.51$ in Sheppard's tables. This
gives $p = 0.934$, or the odds are about 14 to 1 that kiln-dried corn
gives the higher yield.

\setcounter{footnote}{1}

Similarly $0.91/0.28 = 3.25$, corresponding to $p = 0.9994$,\footnote{As
pointed out in Section V, the normal curve gives too large a value for
$p$ when the probability is large.  I find the true value in this case
to be $p=0.9976$.  It matters little, however, to a conclusion of this
kind whether the odds in its favour are 1660 to 1 or merely 416 to 1.}
so that the odds are very great that kiln-dried seed gives barley
of a worse quality than seed which has not been kiln-dried.

Similarly, it is about 11 to 1 that kiln-dried seed gives more
straw and about 2 to 1 that the total value of the crop is less
with kiln-dried seed.

\begin{center}
\section*{Section X. Conclusions}
\end{center}

\hspace{\parindent}\hspace{-\parindent}
1. A curve has been found representing the frequency distribution of
standard deviations of samples drawn from a normal population.

2. A curve has been found representing the frequency distribution of
the means of the such samples, when these values are measured
from the mean of the population in terms of the standard deviation of
the sample.

3. It has been shown that the curve represents the facts fairly
well even when the distribution of the population is not strictly
normal.

4. Tables are given by which it can be judged whether a series of
experiments, however short, have given a result which conforms to any
required standard of accuracy or whether it is necessary to continue the
investigation.

Finally I should like to express my thanks to Prof.\ Karl Pearson,
without whose constant advice and criticism this paper could not have
been written.

\bigskip

\noindent
[\textit{Biometrika}, \textbf{6} (1908), pp.\ 1--25, reprinted on pp.\
11--34 in \textit{Student's'' Collected Papers}, Edited by E.~S.~
Pearson and John Wishart with a Foreword by Launce McMullen, Cambridge
University Press for the Biometrika Trustees, 1942.]

\end{document}

% `