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\begin{document}
\section{Random Samples from Normal Distributions}

Statistical theory for random samples drawn from normal distributions is very
important, partly because a great deal is known about its various associated
distributions and partly because the central limit theorem suggests that for
large samples a normal approximation may be appropriate.

\subsection{\smallskip Useful theoretical results}

\textbf{Theorem 3.1}\quad If $X_{1},\,X_{2\,},\ldots,X_{n}$ are independent
random variables with $X_{j\,}\sim N(\mu_{j},\sigma_{j}^{2})$, then
$Y=\sum_{j=1}^{n}\alpha_{j}X_{j}$ is also normally distributed with mean
$\sum_{j=1}^{n}\alpha_{j}\mu_{j}$ and variance $\sum_{j=1}^{n}\alpha_{j}%
^{2}\sigma_{j}^{2}$.\newline$\square\medskip$\newline\textbf{Proof}\quad Using
moment generating functions - which you have just heard about:
\[%
\begin{array}
[c]{ll}%
M_{X_{j}}\left(  t\right)   & =\exp\left(  \mu_{j}t+\frac{1}{2}\sigma_{j}%
^{2}t^{2}\right)  \\
M_{Y}\left(  t\right)   & =E\left(  e^{t\sum_{j}\alpha_{j}X_{j}}\right)  \\
& =\prod\limits_{j=1}^{n}M_{X_{j}}(\alpha_{j}t)\,,\text{ by independence,}\\
& =\exp\left(  t%
%TCIMACRO{\tsum \limits_{j=1}^{n}}%
%BeginExpansion
{\textstyle\sum\limits_{j=1}^{n}}
%EndExpansion
\alpha_{j}\mu_{j}+\frac{1}{2}t^{2}%
%TCIMACRO{\tsum \limits_{j=1}^{n}}%
%BeginExpansion
{\textstyle\sum\limits_{j=1}^{n}}
%EndExpansion
\alpha_{j}^{2}\sigma_{j}^{2}\right)  .
\end{array}
\]
By uniqueness of the moment generating function the result follows.\smallskip
\newline$\blacksquare\medskip$\newline\textbf{Definition 3.1\quad}A chi-square
distribution with $r$ degrees of freedom is a gamma distribution: $\chi
^{2}(r)$ is the same as $\Gamma\left(  \dfrac{r}{2},\dfrac{1}{2}\right)  .$
\smallskip\ \newline Note that the moment generating function is
\[
M(t)=\left(  \frac{1}{1-2t}\right)  ^{\frac{r}{2}}.
\]
$\square\medskip$\newline The density function $\chi^{2}(1)$ is given by
\[
f(x)=\frac{1}{\sqrt{2\pi}}\,x^{-\frac{1}{2}}\,e^{-\frac{1}{2}x},\quad x>0.
\]
Using the above moment generating function, we see that $E\left(  X\right)
=M_{X}^{\prime}(0)=1,$ $V\left(  X\right)  =2.$ You can see from the graphs
below that $\chi^{2}$-distributions have a distinctively skewed shape.
\newline%
%TCIMACRO{\FRAME{dhFU}{3.7291in}{2.4241in}{0pt}{\Qcb{\QTR{bf}{Figure 3.1}\quad
%P.d.f.s of $\chi^{2}\left(  1\right)  $ and $\chi^{2}\left(  3\right)  $
%distributions}}{}{a5norm_f1.eps}{\special{ language "Scientific Word";
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%filename 'a5norm_f1.eps';file-properties "XNPEU";}}}%
%BeginExpansion
\begin{center}
\includegraphics[
height=2.4241in,
width=3.7291in
]%
{a5norm_f1.eps}%
\\
\textbf{Figure 3.1}\quad P.d.f.s of $\chi^{2}\left(  1\right)  $ and $\chi
^{2}\left(  3\right)  $ distributions
\end{center}
%EndExpansion
\newline The solid line is $\chi^{2}(1)$ and the dotted line is $\chi^{2}%
(3)$.\bigskip\newline\textbf{Lemma 3.1\quad}If $X\sim N(0,1),\,$then
$Y=X^{2}\sim\chi^{2}(1)$.\newline$\square\medskip$\newline\textbf{Proof}\quad
Let $Y=X^{2}.$ Then%
\[
F_{Y}\left(  y\right)  =P\left(  Y\leq y\right)  =P\left(  X^{2}\leq y\right)
=P\left(  -\sqrt{y}\leq X\leq\sqrt{y}\right)  =F_{X}\left(  \sqrt{y}\right)
-F_{X}\left(  -\sqrt{y}\right)
\]
so that%
\[
f_{Y}\left(  y\right)  =F_{Y}^{\prime}\left(  y\right)  =\frac{1}{2\sqrt{y}%
}\left(  f_{X}\left(  \sqrt{y}\right)  +f_{X}\left(  -\sqrt{y}\right)
\right)  =\frac{1}{\sqrt{2\pi y}}\,e^{-\frac{1}{2}y},\quad y>0.
\]
$\blacksquare\bigskip$\newline\textbf{Theorem 3.2} \quad If $Y_{1}%
,\,Y_{2\,},\ldots,Y_{r}$ are an independent random sample, each with a
$\chi^{2}(1)$ distribution then
\[%
%TCIMACRO{\tsum \limits_{j=1}^{r}}%
%BeginExpansion
{\textstyle\sum\limits_{j=1}^{r}}
%EndExpansion
Y_{j}\sim\chi^{2}(r).
\]
$\square\medskip$\newline\textbf{Proof} \quad Just consider the moment
generating functions. The m.g.f of a $\chi^{2}(r)$ random variable is $\left(
1-2t\right)  ^{-r/2}$, so the m.g.f. of a $\chi^{2}(1)$ random variable is
$\left(  1-2t\right)  ^{-1/2}$. But the m.g.f. of a sum of $r$ independent
identically distributed random variables, each with m.g.f. $M\left(  t\right)
$ is $\left[  M\left(  t\right)  \right]  ^{r}$, so the m.g.f. of $\sum
_{j=1}^{r}Y_{j}$ is $\left(  1-2t\right)  ^{-r/2}$, which is the m.g.f. of a
$\chi^{2}(r)$ random variable. Therefore, by the uniqueness theorem for
m.g.f.s,
\[%
%TCIMACRO{\tsum \limits_{j=1}^{r}}%
%BeginExpansion
{\textstyle\sum\limits_{j=1}^{r}}
%EndExpansion
Y_{j}\sim\chi^{2}(r).
\]
\newline$\blacksquare\bigskip$\newline\textbf{Corollary\quad}If $Y_{1}\sim
\chi^{2}(r)$ and $Y_{2}\sim\chi^{2}(s)$ , and are independent, then
\[
Y_{1}+Y_{2}\sim\chi^{2}(r+s).
\]
$\blacksquare$

\subsection{Independence of $\overline{X\text{ }}$\ and $S^{2}$\ for normal
samples.\smallskip}

One of the key results for normal random samples is the independence of
$\overline{X\text{ }}=\dfrac{1}{n}\sum_{i=1}^{n}X_{i}$ and $S^{2}=\dfrac
{1}{n-1}\sum_{i=1}^{n}\left(  X_{i}-\overline{X\text{ }}\right)  ^{2}$, and
their relationship to the mean and variance parameters of a normal
distribution.\bigskip\ \newline\textbf{Theorem 3.3}\quad If $X_{1}%
,\,X_{2\,},\;.\;.\;,\,X_{n}$ are independent, identically distributed random
variables with normal distribution $N\left(  \mu,\sigma^{2}\right)  ,$ then
$\overline{X\text{ }}$ and $S^{2}$ are independent with distributions

\begin{enumerate}
\item[(i)] $\overline{X\text{ }}\sim N\left(  \mu,\dfrac{\sigma^{2}}%
{n}\right)  ;$

\item[(ii)] $\dfrac{\left(  n-1\right)  S^{2}}{\sigma^{2}}\sim\chi^{2}(n-1). $
\end{enumerate}

\noindent$\square\medskip$\newline\textbf{Proof}\quad There are various
methods of proof. We will use one which delivers both independence and
distribution within the same argument. \newline%
\[
X_{i}\sim N\left(  \mu,\sigma^{2}\right)  \Longrightarrow Z_{i}=\frac
{X_{i}-\mu}{\sigma}\sim N\left(  0,1\right)
\]
\newline Now, from Theorem 3.1, we know that, if $\mathbf{Z}$ is a vector of
normal random variables and $\mathbf{L}$ is a linear transformation, then
$\mathbf{Y}=\mathbf{LZ}$ is also a vector of normal random variables. Suppose
that $\mathbf{L}$ is orthogonal so that $\mathbf{L}^{T}\mathbf{L}=\mathbf{I}$.
Then
\[
\mathbf{Y}^{T}\mathbf{Y}=\mathbf{Z}^{T}\mathbf{L}^{T}\mathbf{LZ}%
=\mathbf{Z}^{T}\mathbf{Z}\quad\text{or}\quad\sum_{i=1}^{n}Y_{i}^{2}=\sum
_{i=1}^{n}Z_{i}^{2}.
\]
Thus, if the joint p.d.f. of independent $N\left(  0,1\right)  $ variables
$Z_{i}$, $i=1,\ldots,n$ is%
\[
f_{Z}\left(  \mathbf{z}\right)  =\frac{1}{\sqrt{2\pi}}\exp\left(  -\frac{1}%
{2}\sum_{i=1}^{n}z_{i}^{2}\right)  ,\quad\mathbf{Z}\in\mathbb{R}^{n},
\]
then joint p.d.f. of the $Y_{i}$, $i=1,\ldots,n$ is%
\[
f_{Y}\left(  \mathbf{y}\right)  =\frac{1}{\sqrt{2\pi}}\exp\left(  -\frac{1}%
{2}\sum_{i=1}^{n}y_{i}^{2}\right)  ,\quad\mathbf{Y}\in\mathbb{R}^{n},
\]
and the $Y_{i}$ variables are also independent and distributed as $N\left(
0,1\right)  $.\medskip\newline Now suppose we choose $\mathbf{L}$ such that
its first row is
\[
\left(  \frac{1}{\sqrt{n}}\,,\,\frac{1}{\sqrt{n}}\,,\ldots,\frac{1}{\sqrt{n}%
}\right)  .
\]
Then $Y_{1}=\dfrac{1}{\sqrt{n}}\sum_{i=1}^{n}Z_{i}=\sqrt{n}\overline{Z\text{
}}$, and%
\[
\sum_{i=1}^{n}\left(  Z_{i}-\overline{Z\text{ }}\right)  ^{2}=\sum_{i=1}%
^{n}Z_{i}^{2}-n\overline{Z\text{ }}^{2}=\sum_{i=1}^{n}Y_{i}^{2}-Y_{1}^{2}%
=\sum_{i=2}^{n}Y_{i}^{2},
\]
which is independent of $Y_{1}$. Thus%
\[
\sum_{i=1}^{n}\left(  Z_{i}-\overline{Z\text{ }}\right)  ^{2}\text{ is
independent of }\overline{Z\text{ }}\quad\Rightarrow\quad\sum_{i=1}^{n}\left(
X_{i}-\overline{X\text{ }}\right)  ^{2}\text{ is independent of }%
\overline{X\text{ }}%
\]
since $Z_{i}=\dfrac{X_{i}-\mu}{\sigma}$. The independence of $\overline
{X\text{ }}$ and $S^{2}$ is therefore proved.

\begin{enumerate}
\item[(i)] $Y_{1}\sim N\left(  0,1\right)  \quad\Rightarrow\quad\sqrt
{n}\overline{Z}\sim N\left(  0,1\right)  \quad\Rightarrow\quad\sqrt{n}%
\dfrac{\left(  \overline{X\text{ }}-\mu\right)  }{\sigma}\sim N\left(
0,1\right)  \quad\Rightarrow\quad\overline{X\text{ }}\sim N\left(  \mu
,\sigma^{2}/n\right)  ;$

\item[(ii)] From Theorem 3.2,%
\begin{align*}
\sum_{i=2}^{n}Y_{i}^{2}  & \sim\chi^{2}\left(  n-1\right)  \quad
\Rightarrow\quad\frac{\sum_{i=1}^{n}\left(  X_{i}-\overline{X\text{ }}\right)
^{2}}{\sigma^{2}}\sim\chi^{2}\left(  n-1\right) \\
& \Rightarrow\quad\dfrac{\left(  n-1\right)  S^{2}}{\sigma^{2}}\sim\chi
^{2}(n-1).
\end{align*}
$\blacksquare$
\end{enumerate}

\noindent As we have just seen,$X_{1},\,X_{2\,},\ldots,\,X_{n}$ is a random
sample from $N\left(  \mu,\sigma^{2}\right)  $ then
\[
\sqrt{n}\frac{\left(  \overline{X\text{ }}-\mu\right)  }{\sigma}\sim N\left(
0,1\right)  .
\]
When trying to make inferences from normal data we are often interested in
$\mu,$ the location parameter, but the variance $\sigma^{2}$ is unknown. We
need to find an estimator for the mean $\mu$ which does not contain the
unknown variance parameter. To do this we are going to need to define
something new and do a little more theory.\smallskip\ \bigskip\newline%
\textbf{Definition 3.2} \quad If $U\sim N\left(  0,1\right)  $ and $V\sim
\chi^{2}(r)$ are independent, then
\[
T=\frac{U}{\sqrt{V/r}}\sim t(r)
\]
has a $t$-distribution with $r$ degrees of freedom. This defines the
distribution $t(r)$.\newline$\square$\newline Figure 3.2 shows a graph of
$t\left(  7\right)  $ compared with a standard $N\left(  0,1\right)  $
distribution. You can see that it is very similar, but has fatter tails.%
%TCIMACRO{\FRAME{dhFU}{4.6622in}{3.6002in}{0pt}{\Qcb{\QTR{bf}{Figure 3.2}\quad
%Normal and \QTR{it}{t}(7) distributions}}{}{a5norm_f2.eps}%
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%cropbottom "0";  filename 'a5norm_f2.EPS';file-properties "XNPEU";}}}%
%BeginExpansion
\begin{center}
\includegraphics[
height=3.6002in,
width=4.6622in
]%
{a5norm_f2.eps}%
\\
\textbf{Figure 3.2}\quad Normal and \textit{t}(7) distributions
\end{center}
%EndExpansion
\smallskip\ \newline\textbf{Properties of the t-distribution}

\begin{enumerate}
\item[(i)] The pdf of the $t(r)$ distribution is
\[
f(t)=\frac{\Gamma\left(  \frac{r+1}{2}\right)  }{\sqrt{\left(  \pi r\right)
}\Gamma\left(  \frac{r}{2}\right)  }\;\left(  1+\frac{t^{2}}{r}\right)
^{-\frac{r+1}{2}},\;t\in\mathbb{R}.
\]


\item[(ii)] If $r=1$ then $t(1)$ is a Cauchy distribution without finite mean
or variance.

\item[(iii)] As $r\rightarrow\infty$ then $t(r)\rightarrow\mathrm{N}\left(
0,1\right)  .$
\end{enumerate}

Referring again to Figure 3.2, the distributions are not so very different in
shape, and the higher the number of degrees of freedom the closer the
$t$-distribution approaches to the standard normal distribution.\medskip
\ \newline These results are all leading to the fact that, if we replace
$\sigma$ by $S $, then we know the distribution of the resulting estimator and
the mean, variance and all quantiles are tabulated in either tables or any
statistical package, including R.\bigskip\newline\textbf{Theorem 3.4}\quad If
$X_{1},\,X_{2\,},\ldots,\,X_{n}$ are a normal random sample then
\[
\frac{\sqrt{n}(\overline{X\text{ }}-\mu)}{S}\sim t(n-1).
\]
$\square\medskip$\newline\textbf{Proof}$\quad\overline{X}$ and $S^{2}$ are
independent and
\[
\frac{\sqrt{n}(\overline{X\text{ }}-\mu)}{\sigma}\sim N\left(  0,1\right)
\;,\;\frac{\left(  n-1\right)  S^{2}}{\sigma^{2}}\sim\chi^{2}(n-1).
\]
Using the definition of the t-distribution, the result follows with the
unknown $\sigma$ cancelling.\newline$\blacksquare$

\subsection{Estimators and confidence intervals}

\noindent$\mathbf{Z}$\textbf{-statistics }\smallskip\ \newline Given a set of
data which we regard as plausibly normal we might wish to find a point
estimate of the mean $\mu.$ The previous section suggests that $\overline{X}$
is an obvious candidate. We also need to know what is the likely error range.
If we had a different estimate, how close to the estimator $\overline{X}$
should it lie to be regarded as a plausible estimator of $\mu$? For example
consider the radiocarbon dating sample in Data Set 2.4. Without the outlier we
have seen that it is plausible that the data is a normal random sample. With
the outlier, $\overline{x}=2622$, without $\overline{x}=2505.9$. In either
case, how reliable is our estimate, can we trust it? To within what error
bounds? Should we include or exclude the outlier? If we exclude the outlier
then the data is plausibly normal and we have a possible measure of spread in
terms of the sample standard deviation $s$, although the variance $\sigma^{2}$
is unknown and is often referred to as a nuisance parameter. We need some
theory, making use of the previous sections.\bigskip\ \newline%
\textbf{Definition 3.3}\quad An \textit{estimator} of a parameter $\theta$ is
a statistic, say a function $A\left(  X_{1},\,X_{2\,},\ldots,\,X_{n}\right)  $
of the random sample, which does not depend on any unknown parameters in the
model and which we use to give a point estimate of the parameter from the
data.\newline$\square$ \smallskip\ \newline An example of this is the way we
use $\overline{X}$ to estimate the mean of a distribution. \newline If the
estimator is to have any use at all, it should have some nice properties. For
example, we know that $\overline{X}\overset{P}{\rightarrow}\mu$ by the weak
law of large numbers, ensuring that $\overline{X}$ is a sensible estimator for
$\mu$. \smallskip\ \newline A starting point for considering the likely error
using the normal distribution is given by
\[
\frac{\sqrt{n}(\overline{X\text{ }}-\mu)}{\sigma}\sim N\left(  0,1\right)  .
\]
\newpage

\noindent\textbf{Definition 3.4}\quad A \textit{Z-statistic} is a statistic
with a standard normal distribution (as above).\newline$\square\medskip
$\newline The main use of $Z$-statistics stems from the facts that, for a
general distribution, the Central Limit Theorem implies asymptotically that
\[
\frac{\sqrt{n}(\overline{X\text{ }}-\mu)}{\sigma}\sim N\left(  0,1\right)  ,
\]
and that the standard normal distribution involves no unknown parameters: it
can be (and is) tabulated. \medskip\ \newline We can use the $Z$-statistic to
calculate a range of plausible values for $\mu$, under the assumption that
$\sigma^{2}$ is known.\bigskip\newline\textbf{Definition 3.5\quad
}\emph{Confidence interval\medskip\newline} Let $\mathbf{X}$ represent a
vector of random variables with entries $X_{i} $. If $\left(  a(\mathbf{X}%
),b(\mathbf{X})\right)  $ is a random interval such that%
\[
P\left(  a(\mathbf{X})<\mu<b(\mathbf{X})\right)  =1-\alpha,
\]
then a realisation of that interval, $\left(  a(\mathbf{x}),b(\mathbf{x}%
)\right)  $ is said to be a $100\left(  1-\alpha\right)  \%$ confidence
interval for $\mu$.\newline$\square\medskip$\newline It is not easy to get to
grips with what is meant by a confidence interval. Clearly one cannot say that
the parameter $\mu$ has probability $\left(  1-\alpha\right)  $ of lying
within the calculated interval $\left(  a(\mathbf{x}),b(\mathbf{x})\right)  $
because the ends of the interval are fixed numbers, as is $\mu,$ and without
random variables being present, probability statements cannot be made: either
$\mu$ lies between the two numbers or it doesn't, and we have no way of
knowing which. The only viable interpretation is to say that we have used a
procedure which, if repeated over and over again, would give an interval
containing the parameter $100\left(  1-\alpha\right)  \%$ of the time: the
rest of the time we will be unlucky.\textbf{\ }\smallskip\ \newline Central
$100(1-\alpha)\%$ confidence intervals using $Z$-statistics are found as
follows. Remembering that $Z\sim N(0,1),$ choose $z_{\alpha/2}$ such that
\[%
\begin{array}
[c]{rl}%
P\left(  Z\leq z_{\alpha/2}\right)  & =1-\dfrac{\alpha}{2}\\
\Longrightarrow P\left(  -z_{\alpha/2}\leq Z\leq z_{\alpha/2}\right)  &
=1-\alpha.
\end{array}
\]
If \ $Z=\dfrac{\sqrt{n}\left(  \overline{X\text{ }}-\mu\right)  }{\sigma}\;$as
above, then
\[%
\begin{array}
[c]{rl}%
P\left(  -z_{\alpha/2}\leq\dfrac{\sqrt{n}\left(  \overline{X\text{ }}%
-\mu\right)  }{\sigma}\leq z_{\alpha/2}\right)  & =1-\alpha\\
\Longrightarrow\quad P\left(  \overline{X\text{ }}-\dfrac{\sigma}{\sqrt{n}%
}\,z_{\alpha/2}\leq\mu\leq\overline{X\text{ }}+\dfrac{\sigma}{\sqrt{n}%
}\,z_{\alpha/2}\right)  & =1-\alpha.
\end{array}
\]
Hence the appropriate random interval is
\[
\left(  \overline{X\text{ }}-\frac{\sigma}{\sqrt{n}}\,z_{\alpha/2}%
\;,\;\overline{X\text{ }}+\frac{\sigma}{\sqrt{n}}\,z_{\alpha/2}\right)
\]
and the $100\left(  1-\alpha\right)  \%$ confidence interval is%
\[
\left(  \overline{x}-\frac{\sigma}{\sqrt{n}}\,z_{\alpha/2}\;,\;\overline
{x}+\frac{\sigma}{\sqrt{n}}\,z_{\alpha/2}\right)  .
\]
The most common value of $\alpha$ in use is 0.05, in which case $\,\,z_{\alpha
/2}\,=z_{0.025}=1.960.$%
%TCIMACRO{\FRAME{dhFU}{4.088in}{2.994in}{0pt}{\Qcb{\QTR{bf}{Figure
%3.3}\quad95\% interval for $N(0,1)$}}{}{a5norm_f3.eps}%
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%BeginExpansion
\begin{center}
\includegraphics[
height=2.994in,
width=4.088in
]%
{a5norm_f3.eps}%
\\
\textbf{Figure 3.3}\quad95\% interval for $N(0,1)$%
\end{center}
%EndExpansion
\smallskip\ \newline\textbf{Example 3.1}\quad\emph{Radioactive-carbon
dating\medskip\newline} In order to estimate the age of the site, we need to
take the following steps.

\begin{enumerate}
\item[(i)] Check that the data are plausibly normal. We did this in Example
2.8 using a normal probability plot. We decided that we should leave out one
point because it was a clear outlier.

\item[(ii)] Estimate the mean of the distribution by the sample mean and write
$\widehat{\mu}=\overline{x}=2505.86.$

\item[(iii)] Use a $Z$-statistic to find a 95\% confidence interval which
gives a range of plausible values for the mean age. This is%
\[
\left(  \overline{x}-\frac{1.96\sigma}{\sqrt{n}}\,\;,\;\overline{x}%
+\frac{1.96\sigma}{\sqrt{n}}\right)  ,
\]
and, putting in $n=7$ and $\overline{x}=2505.86$, we find a central 95\%
confidence interval%
\[
\left(  2505.86-\frac{1.96\sigma}{\sqrt{7}}\,\;,\;2505.86+\frac{1.96\sigma
}{\sqrt{n}}\right)  ,
\]

\end{enumerate}

\noindent Unfortunately we are no better off. We cannot obtain the confidence
interval because we do not know $\sigma$, so what should we do? We would like
to replace $\sigma$ by $s,$ the sample standard deviation, but can we?
$\left[  \text{Recall that }S^{2}=\dfrac{1}{n-1}\sum\left(  x_{i}-\overline
{x}\right)  ^{2}.\right]  $\newline$\blacksquare\bigskip$\newline We know from
Theorem 3.4 that, if $X_{1},\,X_{2\,},\ldots,\,X_{n}$ is a random sample from
a normal distribution $N(\mu,\sigma^{2})$, then
\[
T=\frac{\sqrt{n}\left(  \overline{X\text{ }}-\mu\right)  }{S}\sim t\left(
n-1\right)
\]
We can now look for a confidence interval by replacing the $Z$-statistic with
the $t$-statistic. \smallskip Writing\ $t_{\alpha/2}\left(  n-1\right)  $ for
the $1-\frac{\alpha}{2}$ quantile from the distribution $t(n-1),$%
\[
P\left(  -t_{\alpha/2}\left(  n-1\right)  <\frac{\sqrt{n}\left(
\overline{X\text{ }}-\mu\right)  }{S}<t_{\alpha/2}\left(  n-1\right)  \right)
=1-\alpha.
\]
Re-arranging gives the random interval
\[
\left(  \overline{X\text{ }}-\frac{t_{\alpha/2}}{\sqrt{n}}\,S,\;\overline
{X\text{ }}+\frac{t_{\alpha/2}}{\sqrt{n}}\,S\right)  ,
\]
and the $100\left(  1-\alpha\right)  \%$ confidence interval is the
realisation of this interval.\bigskip\newline\textbf{Example 3.2}%
\quad\emph{Radioactive-carbon dating\medskip\newline} For the carbon-dating
example, $n=7$ and $t_{0.025}(6)=2.447$, from a $t$-distribution with 6
degrees of freedom, $s=56.44$. Plugging these values into the formula results
in a 95\% confidence interval of $(2453.5,\,2558.3)$, thereby giving a range
of plausible values for $\mu$.\newline$\blacksquare$\ 

\subsection{Application of Central Limit Theorem}

The Central Limit Theorem states that, for any random sample $X_{1}%
,\,X_{2\,},\ldots,\,X_{n}$ such that the sample size $n$ is sufficiently
large, we have
\[
\frac{\sqrt{n}\left(  \overline{X\text{ }}-\mu\right)  }{\sigma}\overset
{.}{\sim}N\left(  0,1\right)  .
\]
Notation: $\overset{.}{\sim}$ means `approximately distributed as'. Provided
we are dealing with moderate to large sample sizes we can therefore use the
approximate normality to find confidence intervals, using approximate
Z-statistics.\bigskip\newline\textbf{Example 3.3}\quad\emph{Binomial
Proportion}\medskip\newline In an opinion poll prior to a Staffordshire South
East by-election, of 688 constituents chosen at random 368 said they would
vote Labour (53.5\%). The newspapers are perfectly happy to use these data to
estimate $p,$ the probability that a constituent selected at random would vote
Labour, but they rarely, if ever, give any idea of the quality of the
estimate. Let us see how to obtain a 95\% confidence interval for $p$%
.\medskip\newline First identify the random sample. Constituents questioned
are labelled 1,\ldots, 688. Let
\[
X_{i}=\left\{
\begin{array}
[c]{l}%
1,\quad\text{if }i^{\text{th}}\text{ constituent says "I will vote Labour",}\\
0,\quad\text{otherwise.}%
\end{array}
\right.
\]
Then $X_{i}$ has a Bernoulli distribution $B\left(  1,p\right)  $, the sample
size $n$ is 688, and $E\left(  X_{i}\right)  =p$, $V\left(  X_{i}\right)
=p(1-p) $. We know that $p$ can be estimated by the sample mean $\overline
{x}=\dfrac{368}{688}=0.535$. We can also apply the Central Limit Theorem to
find an approximate confidence interval using the asymptotic normality with
$\mu=p,\;\sigma^{2}=p(1-p).$ Thus%
\[
\frac{\sqrt{n}\left(  \overline{X}-p\right)  }{\sqrt{p\left(  1-p\right)  }%
}\overset{.}{\sim}N\left(  0,1\right)  .
\]
The 95\% random interval is of the form
\[
\left(  \overline{X\text{ }}-\frac{\sigma}{\sqrt{n}}\,\,z_{0.025}%
,\;\overline{X\text{ }}+\frac{\sigma}{\sqrt{n}}\,z_{0.025}\right)
\]
but unfortunately $\sigma$ is a function of $p$. We could solve a quadratic
inequality for $p$, but, since $n=688$ is large, we will replace $\sigma$ by
its estimator $\sqrt{\overline{x}(1-\overline{x})}$. This gives (0.498, 0.572)
as a 95\% confidence interval for $p$, with point estimate 0.535.
\smallskip\ \newline If we required a 99\% confidence interval we would use
$z_{0.005}=2.576$ to replace 1.960, and get a wider interval (0.486, 0.584)
about which we are slightly more confident.\newline$\blacksquare$


\end{document}