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MA181 INTRODUCTION TO STATISTICAL MODELLING
HYPOTHESIS TESTING
\end{center}
Suppose $X$ is a random variable that follows a distribution
dependent on a parameter $\theta$ and it is desired to test a
hypothesis about $\theta$ on the basis of a random sample of
observations $\mathbf{X}=(X_1,X_2,\ldots,X_n)$ of $X$.
\begin{description}
\item[Definitions]
The hypothesis under test is called the\textit{null hypothesis}
and is denoted by $H_0$. If the test leads us to reject $H_0$,
then we accept an \textit{alternative hypothesis} $H_1$.
Consequently, $H_0$ and $H_1$ must, between them, cover all the
possible values of the parameter.
Usually, $H_0$ is of the form $H_0:\theta=\theta_0$ while $H_1$
takes one of the three forms $H_1:\theta>\theta_0,\
H_1:\theta<\theta_0$ and $H_1:\theta\neq\theta_0$.
The problem of hypothesis testing reduces to that of dividing the
sample space into two regions, that for which $H_0$ is rejected,
called the \textit{critical region}(C) (or \textit{rejection
region}), and that for which $H_0$ is accepted (or more properly
not rejected), called the \textit{acceptance region}.
When an experiment is carried out, there are two possible states
of nature, $H_0$ true and $H_0$ false, and two possible decisions
which might result, reject $H_0$ and accept $H_0$. The effects of
this set up can be described by the following table.
\begin{center}
\begin{tabular}{c|c|c|}
&Accept $H_0$&Reject $H_0$\\ \hline $H_0$ true&$\sqrt{}$&Type I
error\\ \hline $H_0$ false&Type II error&$\sqrt{}$\\ \hline
\end{tabular}
\end{center}
Let
$$\alpha=P(\textrm{Type I error})=P(\textrm{Reject $H_0|H_0$
true})=P(\mathbf{X}\in C|H_0).$$
Then $\alpha$ is called the \textit{size} of the test or the
\textit{significance level}.
Let
$$\beta=P(\textrm{Type II error})=P(\textrm{Accept $H_0|H_1$
true}).$$
Then
$$1-\beta=P(\textrm{Reject }H_0|H_1)=P(\mathbf{X}\in C|H_1)$$
is called the \textit{power} of the test, or the \textit{power
function} when regarded as a function of the parameter.
A good test would have small values of both $\alpha$ and $\beta$,
i.e. a small size and a large power. It is usually impossible to
achieve both of these simultaneously, so the standard procedure is
to fix $\alpha$ at some acceptable level and then find the test
that maximises $1-\beta$. The commonly used value of $\alpha$ is
0.05 but a value of 0.01 can be used for a more stringent test and
a value of 0.001 for a very stringent test.
\item[Normal distribution]
\textit{One-tailed test}
Suppose $X\sim N(\mu,\sigma^2)$ where, for the moment, $\sigma$ is
assumed known and we wish to test the null hypothesis
$H_0:\mu=\mu_0$ against the alternative $H_1:\mu>\mu_0$. A random
sample $X_1,X_2,\ldots,X_n$ is taken and yields observed values
$x_1,x_2,\ldots,x_n$. It can be shown that the most powerful test
of $H_0$ against $H_1$ leads us to reject $H_0$ is the sample mean
$\overline{x}$ is large or, equivalently, if the critical region
is defined by
$$z=\frac{\overline{x}-\mu_0}{\frac{\sigma}{\sqrt{n}}}>c,$$
where $z$ is the standardised form of $\overline{x}$ under $H_0$.
The value of $c$ is found by considering the fact that, under
$H_0$,
$$Z=\frac{\overline{X}-\mu_0}{\frac{\sigma}{\sqrt{n}}}\sim
N(0,1).$$
Hence, if a test of size (significance level) $\alpha$ is desired,
$c$ is given by
$$\alpha=P(Z>c|\mu=\mu_0)=1-\Phi(c),$$
where $\Phi(.)$ is the cumulative distribution of the standard
normal distribution. The commonly used values of $\alpha$ and $c$
are given in the following table to sufficient accuracy for
practical purposes:
\begin{center}
\begin{tabular}{|c|c|}
\hline $\alpha$&$c$\\ \hline 0.05&1.645\\ 0.01&2.326\\
0.001&3.090\\ \hline
\end{tabular}
\end{center}
We tend to state that, if $z>1.645$, the test is significant to
5\% and there is \lq\lq some'' evidence to reject $H_0$. If
$z>2.326$, there is \lq\lq strong'' evidence to reject and, if
$z>3.090$, there is \lq\lq very strong'' evidence to reject.
Suppose, however, that instead of the above, we wished to test
$H_0:\mu=\mu_0$ against $H_1:\mu<\mu_0$. Then the best critical
region is defined by $z<-c$, where $c$ is given by
$$\alpha=P(Z<-c|\mu=\mu_0)=\Phi(-c).$$
In view of the symmetry of the normal distribution, the values of
$c$ usually required are those given in the above table.
\item[Example]
I.Scribe Inc. manufacture ball point pens and have done so for
many years. The amount of ink in a pen is sufficient to draw a
line the length of which is normally distributed with standard
deviation 0.1 km. The firm tries to keep the ink-filling machine
running so that the mean length of line is 4km, but there is a
suspicion in the mind of the operator that the mean has fallen
somewhat. Consequently, a random sample of eight pens are placed
in a birometer and the lengths of line measured (in km) with
following results:
$$3.81,\ 3.92,\ 3.94,\ 3.93,\ 3.72,\ 4.05,\ 3.88,\ 3.83.$$
It is desired that the test be carried out with $\alpha=0.01$.
We are here testing $H_0:\mu=4$ against $H_1:\mu<4$ so that the
critical region of a 1\% test is $z<-2.326$. Since
$\overline{x}=31.08/8=3.885$, the observed value of $z$ is
$$z=\frac{3.885-4}{0.1/\sqrt{8}}=-3.253.$$
The test is significant at 1\% so that there is strong evidence to
reject $H_0$. (In fact the test would also be significant at
0.1\%.)
We can calculate the power of the test for any alternative value
of $\mu$. If, for instance, $\mu=3.9$, then
\begin{eqnarray*}
1-\beta&=&P\left(\left.\frac{\overline{X}-4}{0.1/\sqrt{8}}<-2.326\right|\mu=3.9\right)\\
&=&P\left(\left.\frac{\overline{X}-3.9}{0.1/\sqrt{8}}<-2.326+\frac{0.1}{0.1/\sqrt{8}}\right|\mu=3.9\right)\\
&=&P(Z<0.502)=0.692
\end{eqnarray*}
\textit{Two-tailed test}
Assume again that $X\sim N(\mu\sigma^2)$ with $\sigma$ assumed
known but that we now wish to test the null hypothesis
$H_0:\mu=\mu_0$ against the alternative $H_1:\mu\neq\mu_0$. The
natural critical region to use is that consisting of both large
and small values of $\overline{x}$ or, equivalently, of
$z=\sqrt{n}(\overline{x}-\mu_0)/\sigma$. In view of the symmetry
of the distribution of $Z$, the critical region takes the form
$|z|>c$ where, for a test of size $\alpha,\ c$ is defined by
$$\alpha=P(|Z|>c|\mu=\mu_0)=P(Z>c|\mu=\mu_0)+P(Z<-c|\mu=\mu_0)=2[1-\Phi(c)]$$
or $1-\Phi(c)=\frac{\alpha}{2}$.
Such a test is called a \textit{two-tailed test}, as opposed to
the one-tailed tests described above, since the critical region is
to be found in the two tails of the distribution of $Z$. The
commonly used values of $\alpha$ and $c$ are given in the
following table:
\begin{center}
\begin{tabular}{|c|c|}
\hline $\alpha$&$c$\\ \hline 0.05&1.960\\ 0.01&2.576\\
0.001&3.291\\ \hline
\end{tabular}
\end{center}
\item[Example]
Milk bottles are vacuum-formed from molten gobs of glass that are
weighed as they fall into moulds. The weight is known to be
normally distributed with standard deviation 2.5 g and it is
important that the bottles mean weight is maintained st 2252; too
low a mean results in too many fragile bottles, too high a mean in
an excess consumption of glass as well as too many bottles having
a low internal volume.
The hypotheses under test here are $H_0:\mu=255$ against
$H_1:\mu\neq255$.
As the manufacturer does not want to stop production and check his
equipment unless there is strong evidence that the mean has
changed, he permits a probability of only 0.01 of making a Type I
error.
The value of $c$ is given by $1-\Phi(c)=\frac{\alpha}{2}=0.005$ so
that $c=2.576$ and the critical region is given by $|z|>2/576$.
A random sample of eight bottles has a mean weight of 253.25g. So
$$z=\frac{\overline{x}-\mu_0}{\frac{\sigma}{\sqrt{n}}}=
\frac{253.25-255}{2.5/\sqrt{8}}=-1.980$$
and $|z|$ does not fall in the critical region. There is therefore
no evidence to reject $H_0$.
(Note that the test would just be significant with $\alpha=0.05$.)
Suppose the population mean weight of the bottles changes to 275g.
Then the power of the test is
\begin{eqnarray*}
1-\beta&=&P\left(\left.\frac{\overline{X}-255}{2.5/\sqrt{8}}>2.576
\right|\mu=257\right)+
P\left(\left.\frac{\overline{X}-255}{2.5/\sqrt{8}}<-2.576\right|
\mu=257\right)\\
&=&P\left(Z>2.576-\frac{2}{2.5/\sqrt{8}}\right)+P\left(Z<-2.576-
\frac{2}{2.5/\sqrt{8}}\right)\\ &=&1-\Phi(0.313)+\Phi(-4.839)\\
&=&0.377+0=0.377
\end{eqnarray*}
\textit{Variance unknown}
In practice, cases where the variance of $X$ is known are
uncommon. Usually $\sigma^2$ is not known. As a result, although
$$Z=\frac{\overline{X}-\mu_0}{\frac{\sigma}{\sqrt{n}}}\sim
N(0,1)$$
remains true, it cannot be used to test hypotheses about $\mu$ as
it depends on $\sigma$. The natural way around this problem is to
estimate $\sigma^2$ by
$$S^2=\frac{\sum(X_i-\overline{X})^2}{n-1}.$$
There are several reasons for setting the divisor equal to $n-1$
(rather than, say, $n$), one of which is to ensure that
$E(S^2)=\sigma^2$.
the use of $S$ leads to a modified statistic defined by
$$T=\frac{\overline{X}-\mu_0}{\frac{S}{\sqrt{n}}}.$$
Under $H_0, T\sim t_{n-1}$, i.e. $T$ follows the (Student) $t$
distribution with $n-1$ degrees of freedom. One-tailed and
two-tailed tests can be carried out as before except that the
critical region is given by $t>c,t<-c$ or $|t|>c$ as appropriate,
where the critical point $c$ must be read from the table of the
$t$ distribution rather than the standardised normal.
\item[Example]
A particular task in a manufacturing industry has long been
scheduled to take 15 minutes on average. The management wants to
introduce a new way of doing the job that should prove more
economical in the tools required and cleaner for the workers
involved. They want, however, to keep the mean time unchanged so
that the task still fits perfectly into the production line
schedule. Consequently, they measure the times for 12 completions
of the task under the new conditions and obtain the following
results, in minutes.
$$13.6,\ 12.3,\ 16.3,\ 15.1,\ 13.8,\ 15.2,\ 14.5,\ 14.0,\ 13.3,\
15.2,\ 16.1,\ 14.1.$$
We will assume that the time to complete the task is normally
distributed and we want to test the null hypothesis $H_0:\mu=15$
against the alternative $H_1:\mu\neq15$. Since $n=12$, the
critical region of a test size $\alpha=0.05$ is given by
$$|t|=\frac{|\overline{x}-15|}{s/\sqrt{12}}>2/201.$$
Now $\sum x_i=173.5$ and $\sum x_i^2=2523.63$.
Hence$\overline{x}=\frac{173.5}{12}=14.458\dot{3}$ and
$s^2=\frac{(2523.63-173.5^2/12)}{11}=\frac{15.1092}{11}=1.3736$.
The observed value of $|T|$ is therefore
$$|t|=\frac{|13.458\dot{3}-15|}{\sqrt{1.3736/12}}=1.601.$$
This does not fall in the critical region so there is no evidence
to reject $H_0$.
\end{description}
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