\documentclass[a4paper,12pt]{article}

\begin{document}

\parindent=0pt

\begin{center}

MA181 INTRODUCTION TO STATISTICAL MODELLING

CONFIDENCE INTERVALS

\end{center}
Suppose the random variable $X$ follows a normal distribution with
mean $\mu$ and variance $\sigma^2$, and it is desired to estimate
$\mu$ from a random sample of observations. The best point
estimator to use is, in many respects, the sample mean
$\overline{X}$. However, a value $\overline{x}$ given by this
estimator carries with it no measure of its precision; a value of
13.6 from a sample of 10 values looks just like a value of 13.6
from a sample of 100, yet the latter is more precise, or reliable,
than the former. One way to remedy this is to construct an
interval around $\overline{x}$, the length of which  reflects its
reliability or, in other words, our confidence about its
containing $\mu$.

The starting point for the construction of this interval is the
distribution of $\overline{X}$, which is $N(\mu,\sigma^2/n)$,
where $n$ is the sample size. On standardising this result, we
obtain

\begin{equation}
Z=\frac{\overline{X}-\mu}{\frac{\sigma}{\sqrt{n}}}\sim N(0,1).
\end{equation}

\begin{description}

\item[Variance known]
If the variance $\sigma^2$ is known, the distribution of $Z$ given
at (1), i.e. $N(0,1)$, can be used to construct the required
interval. For a given probability $\alpha$, let $c$ be the point
such that $\frac{\alpha}{2}=P(Z<-c)=P(Z>c)$. Then, from (1), we
have

$$1-\alpha=P\left(-c\leq\frac{\overline{X}-\mu}{\frac{\sigma}{\sqrt{n}}}\leq
c\right)=P(\overline{X}-c\sigma/\sqrt{n}\leq\mu\leq\overline{X}+c\sigma/\sqrt{n}).$$

This is not a probability statement about the population mean
$\mu$, which remains constant throughout, but about the interval
$[\overline{X}-c\sigma/\sqrt{n},\overline{X}+c\sigma/\sqrt{n}]$,
which in repeated sampling contains $\mu$ with probability
$1-\alpha$. For a single sample, with observed mean
$\overline{x}$, the interval
$[\overline{x}-c\sigma/\sqrt{n},\overline{x}+c\sigma/\sqrt{n}]$ is
called a $100(1-\alpha)\%$ \textit{confidence interval} for $\mu$.
Our level of confidence springs from the fact that this interval
is either one  of the $100(1-\alpha)\%$ of intervals that would
contain $\mu$ if we continued drawing samples of size $n$, or one
of the $100\alpha\%$ of intervals that would fail to contain it.
The end-points of the interval, $\overline{x}-c\sigma/\sqrt{n}$
and $\overline{x}+c\sigma/\sqrt{n}$, are known as the
\textit{confidence limits} to $\mu$, and the fraction
$100(1-\alpha)\%$ as the \textit{confidence coefficient}.

As is common with confidence interval construction, a value of
$\alpha=0.05$ is chosen for general use, leading to a 95\%
interval, while a value of $\alpha=0.01$, which leads to a longer
99\% interval, is chosen for increased confidence that the
interval contains $\mu$. Occasionally $\alpha=0.1$ is used.

Note that a $100(1-\alpha)\%$ confidence interval for $\mu$
contains those values that would not be rejected by a two-sided
test with significance level $\alpha$.

\item[Examples]

\begin{description}

\item[(i)]
The height of an adult male is normally distributed with standard
deviation 2.56 inches, and a random sample of 120 men yields a
mean height of 68.21 inches. For a 95\% confidence interval for
$\mu$, the mean height of men in the population from which the
sample is drawn, we need to set $c=1.9600$. Then the confidence
limits are found to be

$$\overline{x}\mp
c\sigma/\sqrt{n}=68.21\mp1.9600(2.56)/\sqrt{120}=68.21\mp0/46.$$

Hence the 95\% confidence interval for $\mu$ is
$\left[67.75,68,67\right].$

For a confidence coefficient of 99\%, $c=2.5758$, which leads to
the interval $\left[67.61,68.81\right]$.

Not surprisingly, the desire to be more confident about the
interval's containing $\mu$ results in an increase in its length.

\item[(ii)]
The length of a confidence interval is equal to
$2c\sigma/\sqrt{n}$, which does not depend on $\overline{x}$.
Consequently, it is possible to determine the sample size required
to obtain an interval with a maximum specified length. To take an
example, if $\sigma=3$ and we desire a 90\% confidence interval
for $\mu$ with length at most 1.3, then $c=1.6449$ and we need to
satisfy the inequality

$$2c\sigma/\sqrt{n}=2(1.6449)(3)/\sqrt{n}\leq1.3,$$

which leads to $n\geq57.6$. Since $n$ must in practice be an
integer, we need to take a sample of size at least 58.

\end{description}

\item[Variance unknown]
More often than not, the variance $\sigma^2$ of $X$ is not known.
This in no way nullifies the truth of the statement made in (1),
but the random variable $Z$ can no longer be used to construct an
interval estimate of $\mu$ since it depends on $\sigma$. The
obvious solution to this problem is to replace $\sigma$ by an
estimate based on the sample. The estimate usually adopted is the
sample standard deviation defined by

$$s=\left[\sum(x_i-\overline{x})^2/(n-1)\right]^{\frac{1}{2}}.$$

Replacing $\sigma$ by $s$ in (1) leads to the new statistic

\begin{equation}
T=\frac{\overline{X}-\mu}{s/\sqrt{n}}\sim t_{n-1}.
\end{equation}

In repeated sampling, $T$ no longer follows a normal distribution,
in view of its dependence not only on $\overline{X}$ but also on
$s$, which itself varies in value from one sample to another. The
distribution of $T$ was derived in 1908 by W.S.Gosset writing
under the pseudonym Student. Consequently $T$ is said to follow
the Student $t$ distribution (or $t$ distribution for short).
There is, however, not a single $t$ distribution but a family of
them, indexed by the number of degrees of freedom, which is the
number of linearly independent components of $s$. The symbol
$t_{(\nu)}$ is used to denote the $t$ distribution with $\nu$
degrees of freedom, and the distribution of $T$ required in (2) is
the $t$ distribution with $(n-1)$ degrees of freedom, where $n$ in
the sample size.

The probability density function of $t_{(\nu)}$ is similar in
shape to that of the standard normal distribution in that it is
symmetric about zero and bell-shaped. It is, however, somewhat
\lq\lq fatter'' in the tails, and, since inferences about $\mu$
depend on values from the tails of a distribution, it is important
that the correct distribution is used, at least for small sample
sizes. In fact, as $nu\to\infty$, the distribution of $t_{(\nu)}$
converges to that of $N(0,1)$.

A $100(1-\alpha)\%$ confidence interval for $\mu$ can be derived
in a similar manner to before by finding, this time from the $t$
distribution with $(n-1)$ degrees of freedom, the point $c$ such
that $\frac{\alpha}{2}=P(T<-c)=P(T>c)$. Then, from (2), we have

$$1-\alpha=P\left(-c\leq\frac{\overline{X}-\mu}{\frac{s}{\sqrt{n}}}\leq
c\right)=P(\overline{X}-cs/\sqrt{n}\leq\mu\leq\overline{X}+cs/\sqrt{n}).$$

For an observed sample, the interval
$[\overline{x}-c\sigma/\sqrt{n}\leq\mu\leq\overline{x}+c\sigma/\sqrt{n}]$
is a $100(1-\alpha)\%$ confidence interval for $\mu$. Note that
its length, $2cs/\sqrt{n}$, is not a constant as before, but
varies from sample to sample in view of its dependence on $s$.

\item[Example]
In 1928, the LNER ran the locomotive Lemberg with an experimental
boiler pressure of 220lb in five trial runs and measured the coal
consumption in lb per draw-bar horse-power hour. The results were
as follows:

$$3.27,\ 3.17,\ 3.24,\ 2.92,\ 2.99.$$

Denoting the $i$th observation by $x_i$, we have,

$$n=5,\sum x_i=15.59\textrm{ and }\sum x_i^2=48.7059,$$

so that $\overline{x}=15.59/5=3.118$ and
$s=(48.7059-15-59^2/5)/4=0.02407$. For a 95\% confidence interval
for $\mu$, the mean coal consumption of the locomotive, we find,
from the table of the $t_{(4)}$ distribution, that $c=2.776$.
Hence the confidence limits are

$$\overline{x}\mp
cs/\sqrt{n}=3.118\mp2.776\sqrt{(0.0240/5)}=3.118\mp0/193$$

and the 95\% confidence interval for $\mu$ is
$\left[2.925,3.311\right]$.

\end{description}

\end{document}
