Page:Biometrika - Volume 6, Issue 1.djvu/8

Hence

$y={\frac {N{\sqrt {n}}s}{{\sqrt {2\pi }}\sigma }}e^{-{\frac {ns^{2}z^{2}}{2\sigma ^{2}}}}$

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

$y={\frac {\int _{s}^{s+ds}{\frac {C}{\sigma ^{n}}}{\sqrt {\frac {n}{2\pi }}}s^{n-1}e^{-{\frac {ns^{2}(1+z^{2})}{2\sigma ^{2}}}}ds}{\int _{0}^{\infty }{\frac {C}{\sigma ^{n-1}}}s^{n-2}e^{-{\frac {ns^{2}}{2\sigma ^{2}}}}ds}}={\frac {{\sqrt {\frac {n}{2\pi }}}\int _{s}^{s+ds}s^{n-1}e^{-{\frac {ns^{2}(1+z^{2})}{2\sigma ^{2}}}}ds}{\sigma \int _{0}^{\infty }s^{n-2}e^{-{\frac {ns^{2}}{2\sigma ^{2}}}}ds}}$,

and summing for all values of $s$ we have as an equation giving the distribution of $z$

$y={\frac {\sqrt {\frac {n}{2\pi }}}{\sigma }}{\frac {\int _{0}^{\infty }s^{n-1}e^{-{\frac {ns^{2}(1+z^{2})}{2\sigma ^{2}}}}ds}{\int _{0}^{\infty }s^{n-2}e^{-{\frac {ns^{2}}{2\sigma ^{2}}}}ds}}$.

By what we have already proved this reduces to

$y={\frac {1}{2}}{\frac {n-2}{n-3}}\cdot {\frac {n-4}{n-5}}...{\frac {5}{4}}\cdot {\frac {3}{2}}(1+z^{2})^{-{\frac {n}{2}}}$ if $n$ be odd,

and to

$y={\frac {1}{\pi }}{\frac {n-2}{n-3}}\cdot {\frac {n-4}{n-5}}...{\frac {4}{3}}\cdot {\frac {2}{1}}(1+z^{2})^{-{\frac {n}{2}}}$ 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.

Section IV.

Some Properties of the Standard Deviation Frequency Curve.

By a similar method to that adopted for finding the constant we may find the mean and moments: thus the mean is at ${\frac {I_{n-1}}{I_{n-2}}}$, which is equal to

${\frac {(n-2)}{(n-3)}}{\frac {(n-4)}{(n-5)}}...{\frac {2}{1}}{\sqrt {\frac {2}{\pi }}}{\frac {\sigma }{\sqrt {n}}}$ (if $n$ be even),

or

${\frac {(n-2)}{(n-3)}}{\frac {(n-4)}{(n-5)}}...{\frac {3}{2}}{\sqrt {\frac {\pi }{2}}}{\frac {\sigma }{\sqrt {n}}}$ (if $n$ be odd).