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  {Department of Mathematical Sciences \hspace{\fill}\rightmark \hspace{\fill}\thepage}{}%
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\markright{ 2000 Spring Retreat}

\slideframe{Oval}

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\begin{document}

%slide 1%
\begin{slide}
\heading{Numerical Descriptive Statistics}

Given {\em n} observations $x_{1},x_{2},\ldots,x_{n}$,
$\clubsuit${\bf {\em Arithmetic} Mean \hspace{0.25in}} $\overline{x}$ = 
$\displaystyle \frac{1}{n}\sum_{i=1}^{n}x_{i}$ = 
$\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}$
$\clubsuit${\bf Mode \hspace{0.25in}} A mode, note that there can be 
several, of a set of observations of size {\em n} is the observation 
that appears most often, or with the greatest frequency. Note, that 
it 
is not necessary that a mode exists, nor is it necessary that there is 
only one mode.
$\clubsuit${\bf Median ($M$, $Q_{2}$, $\tilde{x}$) \hspace{0.25in}} Arrange 
all observations in order of 
size, from smallest to largest. 

If the number {\em n} of observations is even, 
the median {\em M} is the average of the two center
observations in the ordered list. The location
of the median is again $(n + 1)/2$ from the 
bottom of the list. 

If the number {\em n} of observations is odd,
the median {\em M} is the center observation in 
the ordered list. 
\end{slide}
\begin{slide}
    $\clubsuit${[\bf Quantiles]} For a data set consisting of {\em n} values 
that 
when ordered are  $x_{(1)}~\leq~x_{(2)}\leq~\ldots~\leq~x_{(n)}$,

$\clubsuit$ for any number {\bf p} of the form $\frac{i-0.5}{n}$, where 
{\em i}
is an integer from 1 to {\em n}, the {\bf p} quantile of the data set 
will be
taken to be $x_{i}$. (The {\em i}th smallest data point will be 
called the $\frac{i-0.5}{n}$ quantile.)
$\clubsuit$ for any number {\bf p} between $\frac{0.5}{n}$ and 
$\frac{n-0.5}{n}$ that is not of the form $\frac{i-0.5}{n}$,
the {\bf p} quantile of the data set will be obtained by linear 
interpolation between the quantiles 
corresponding to the two values of $\frac{i-0.5}{n}$ that bracket 
{\bf p}.

In both cases, the notation Q({\bf p}) will be used to symbolize the {\bf 
p} quantile.

\end{slide}
\begin{slide}
$\clubsuit${[\bf Quartiles $Q_{1},Q_{2},Q_{3}$:]} \\
$Q_{2}$ = Median. 
\begin{eqnarray*}
{\rm If\:\:the\:\:number\:\:} n {\rm \:\:is \:\:even \:\:}&:&
\left \{ \begin{array}{ll} 
Q_{1}& {\rm 
\:\:is\:\:the\:\:median\:\:of\:\:the\:\:smallest\:\:}n/2{\rm 
\:\:observations}\\
Q_{3}& {\rm 
\:\:is\:\:the\:\:median\:\:of\:\:the\:\:largest\:\:}n/2{\rm\:\:observations}\\

\end{array}
\right. \\&~&\\
{\rm If\:\:the\:\:number\:\:} n {\rm\:\:is \:\:odd\:\:}&:&
\left \{ \begin{array}{ll} 
Q_{1}& {\rm \:\:is\:\:the\:\:median\:\:of\:\:the\:\:smallest\:\:}(n - 
1)/2{\rm \:\:observations}\\
Q_{3}& {\rm \:\:is\:\:the\:\:median\:\:of\:\:the\:\:largest\:\:}(n - 
1)/2{\rm \:\:observations}\\
\end{array}
\right.
\end{eqnarray*}
\end{slide}
\begin{slide}

$\clubsuit${[\bf Variability:(sample), unbiased ]} $s^{2}$ = 
$\displaystyle \frac{1}{n-1}\sum_{i=1}^{n}\left (x_{i}^{2} - 
\overline{x} \right 
)^{2}$ = $\displaystyle \frac{\sum_{i=1}^{n}x_{i}^{2} - \frac{\left ( 
\sum_{i=1}^{n}x_{i} \right )^{2}}{n}}{n-1}$
$\clubsuit${[\bf Variability:(sample), biased ]} $\tilde{s}^{2}$= 
$\displaystyle \frac{1}{n}\sum_{i=1}^{n}\left (x_{i}^{2} - 
\overline{x} \right 
)^{2}$= $\displaystyle \frac{\sum_{i=1}^{n}x_{i}^{2} - \frac{\left ( 
\sum_{i=1}^{n}x_{i} \right )^{2}}{n}}{n}$
$\clubsuit${[\bf Standard Deviation:]} \\ $s$= 
$\sqrt{s^{2}}$(unbiased variance) or 
$\tilde{s}$= $\sqrt{\tilde{s}^{2}}$(biased variance)
\end{slide}
\begin{slide}
\end{slide}
\begin{slide}
$\clubsuit${[\bf Standard Error of $\overline{X}$ as an estimate of the 
population mean:]  } \\
s.e $\left (\overline{x} \right )$= 
$s_{\overline{x}}$ = s/$\sqrt{n}$
$\clubsuit${[\bf Range:]} {\em R} = 
$\max \{ x_{1},x_{2},\ldots,x_{n}\}$ - 
$\min\{x_{1},x_{2},\ldots,x_{(n)}\}$ = 
$x_{(n)}$ - $x_{(1)}$
$\clubsuit${[\bf Interquartile Range:]} {\em IQR} = $Q_{3}$ - 
$Q_{1}$
$\clubsuit${[\bf Fences useful for Box Plots in determining outliers:]}
\[
\begin{array}{lll}
\mbox{Inner Fences:}& Q_{1}-1.5IQR,& Q_{3}+1.5IQR\\
\mbox{Outer Fences:}& Q_{1}-3IQR,& Q_{3}+3IQR\\
\end{array}
\]

\end{slide}
\begin{slide}
$\clubsuit${[\bf Linear Transformations:] }
Let $y_{i}$ = a$x_{i}$+ b, then $\overline{y}$ = a$\overline{x}$ + b, 
$s_{y}^{2}$ = a$^{2}s_{x}^{2}$, $s_{y}$ = $\mid$ a $\mid$ $s_{x}$\\ 
Important notes: Linear transformations do not change the shape of 
 the data (distribution).
\end{slide}
\begin{slide}
    \heading{Calculating the standard deviation}

The standard deviation is a measure of the variability of a data set. 
As we will see in our in class discussion, it will give us 
information about how the data is distributed.
Consider the two methods for calculating the variance for the first 
die set:
\begin{table}[h!]
\centering
\begin{tabular}{cccc}
x$_{i}$ & $\left(x_{i} - \overline{x}\right)$ & $\left(x_{i} - 
\overline{x}\right)^{2}$ & x$^{2}_{i}$ \\\hline
18.9 & -1.44 & 2.0736& 357.21\\
19.3 & -1.04 & 1.0816&372.49\\
19.5 & -0.84 & 0.7056&380.25\\
20 & -0.34 & 0.1156&400\\
20.5 & 0.16 & 0.0256&420.25\\
20.6 & 0.26 & 0.0676&424.36\\
20.7 & 0.36 & 0.1296&428.49\\
20.8 & 0.46 & 0.2116&432.64\\
21 & 0.66 & 0.4356&441\\
22.1 & 1.76 & 3.0976&488.41\\ \hline
$\displaystyle \sum_{i=1}^{n}x_{i}$=203.4 & 0 & $\displaystyle \sum_{i=1}^{n}\left (x_{i} - 
\overline{x} \right )^{2}$=7.944 & $\displaystyle \sum_{i=1}^{n}x_{i}^{2}$=4145.1 \\ &&&\\
$\overline{x}$=20.34 & $s^{2}$ = $\displaystyle 
\frac{1}{n-1}\displaystyle \sum_{i=1}^{n}\left (x_{i} - 
\overline{x} \right )^{2}$ = 0.88266667\\
\end{tabular}
\end{table}

Hence $\sqrt{s^{2}}$  =  s = 0.9395.

\end{slide}
\begin{slide}
Also
\begin{eqnarray*}
s^{2}
 &=& \displaystyle \frac{\displaystyle \sum_{i=1}^{n}x_{i}^{2} - \frac{\left ( 
\displaystyle \sum_{i=1}^{n}x_{i} \right )^{2}}{n}}{n-1}\\
& = & \displaystyle \frac{4145.1 - \frac{\left ( 
203.4\right )^{2}}{10}}{9}\\
&=&0.8827
\end{eqnarray*}
for die type 1. You should verify the others variances and standard 
deviations, both biased and unbiased.
\end{slide}
\begin{slide}

\heading{Quantiles}
\begin{center}
{\large \bf Quantiles via Linear Interpolation \\ No More Rounding!}
\end{center}
Suppose we have the following ordered stock values and we would like to know
what the {\bf exact} quartiles are:
\begin{table}[h!]
\centering
\begin{tabular}{lcccccc}
Raw Data: & 10.3 & 4.9 & 8.9 & 11.7 & 6.3 & 7.7\\
Ordered: & 4.9 & 6.3 & 7.7 & 8.9 & 10.3 & 11.7\\
Position: & 1 & 2 & 3 & 4 & 5 & 6\\
\end{tabular}
\end{table}
\begin{eqnarray*}
Q_{1}-Position&=& \frac{1\cdot\left(n+1\right)}{4}\\
&=& \frac{1\cdot\left(7\right)}{4}{\rm\:\:}={\rm\:\:}1.75.
\end{eqnarray*}
which we rounded to 2. From now on we will not be rounding.
To find Q$_{1}$ we are going to use a technique called ``linear
interpolation''. This is how it's done.

\end{slide}
\begin{slide}
In order to find the number in the 1.75$^{th}$ position, we have
4.9 in the 1$^{st}$ position and 6.3 in the 2$^{nd}$ position, the
difference in the 1$^{st}$ and 1.75$^{th}$ is  0.75$^{th}$ of a position.
So we would take 4.9, the number in the  1$^{st}$ position and add .75 $^{th}$
of a position. A position for 4.9 and 6.3 has width (6.3-4.9)~=~1.4.
We take this difference in the numbers and multiply by .75. Then add
this value to 4.9. Mathematically, this is,

\begin{eqnarray*}
Q_{1} &=& 4.9 +  0.75 \times \left(6.3 - 4.9 \right )   \\
& = & 4.9 + 0.75 \times 1.4   \\
& = & 5.95.
\end{eqnarray*}
\end{slide}
\begin{slide}

Let's get a complete understanding of this. What's so special about 
the 0.75$^{th}$ position ? Nothing. So we consider the other 
possibility, the difference between the 1.75$^{th}$ position and the
2$^{nd}$ position, which is .25$^{th}$ of a position. Again, we can
do a little math by taking the difference of the two numbers in the 
1$^{st}$ position and the 2$^{nd}$ position, multiply by 0.25 an then 
subtract that number from 6.3. Mathematically, this is,
\end{slide}
\begin{slide}

\begin{eqnarray*}
Q_{1} &=& 6.3 - 0.25 \times \left(6.3 - 4.9 \right )   \\
& = & 6.3 - 0.35\\
& = & 5.95.
\end{eqnarray*}

\begin{eqnarray*}
Q_{2}-Position&=& \frac{2\cdot\left(n+1\right)}{4}\\
&=& \frac{2\cdot\left(7\right)}{4}{\rm\:\:}={\rm\:\:}3.5.
\end{eqnarray*}

The median is trivial. How about Q$_{3}$~?
\begin{eqnarray*}
Q_{3}-Position&=& \frac{3\cdot\left(n+1\right)}{4}\\
&=& \frac{3\cdot\left(7\right)}{4}{\rm\:\:}={\rm\:\:}5.25.
\end{eqnarray*}
which we rounded to 5. Again we will not be rounding.
To find Q$_{3}$ we are going to use the same method as before.
\end{slide}
\begin{slide}


In order to find the number in the 5.25$^{th}$ position, we have
10.3 in the 5$^{th}$ position and 11.7 in the 6$^{th}$ position, the
difference in the 5$^{th}$ and 5.25$^{th}$ is  0.25$^{th}$ of a position.
So we would take 4.9, the number in the  5$^{th}$ position and add .25 $^{th}$
of a position. A position for 10.3 and 11.7 has width ( 11.7-10.3)= 1.4, this is only 
coincidental.
We take this difference in the numbers and multiply by .25. Then add
this value to 10.3. Mathematically, this is,

\begin{eqnarray*}
Q_{1} &=& 10.3 +  0.25 \times \left(11.7 - 10.3 \right )   \\
& = & 10.3 + 0.25 \times 1.4   \\
& = & 10.65.
\end{eqnarray*}
\end{slide}
\begin{slide}

Let's get a complete understanding of this. What's so special about 
the 0.25$^{th}$ position ? Nothing. So we consider the other 
possibility, the difference between the 5.25$^{th}$ position and the
6$^{th}$ position, which is .75$^{th}$ of a position. Again, we can
do a little math by taking the difference of the two numbers in the 
5$^{th}$ position and the 6$^{th}$ position, multiply by 0.75 an then 
subtract that number from 11.7. Mathematically, this is,

\begin{eqnarray*}
Q_{1} &=& 11.7 - 0.25 \times \left(11.7 - 10.3 \right )   \\
& = & 11.7 - 1.05\\
& = & 10.65.
\end{eqnarray*}

\end{slide}
\begin{slide}

\heading{Box-plots}
A box-plot or box-and-whisker plot is a graphical representation of 
data in which a rectangle is used to summarize the data distribution.
The top and the bottom, sometimes the left and right, of the 
rectangle represent the {\bf third} and {\bf first} quartiles, 
respectively. The line inside the rectangle 
represents the median. The lines extending
from the top and bottom of the rectangle 
represent either the actual limits of the
data, or the limits of the bulk of the
data (with unusual observations, sometimes referred to as {\bf outliers} see below,  being 
represented by individual symbols [``flagged"] 
if they are further out, {\bf modified box plot}). The box-plot is
particularly useful for comparing the location and variability of several batches 
of data, as boxes can be plotted side-by-side on one plot.

% \heading{Construction of a Box-plot}
% 
\end{slide}
\begin{slide}
Suppose we have a set of observations, say the tensile strengths given before.
The following plot shows box-plots for the four type to die sets. Notice how
different these are, with die sets 2 and 4 the more {\bf \em similar} and die sets 3 and
4 the most {\bf \em dissimilar}.

\begin{figure}[h!]
\centerline{\psfig{figure=Pearson:Oztex:EPS:tensilebox.eps,height=2.75in,width=2.75in}}
\end{figure}
\end{slide}
\begin{slide}

Instead of considering all of these at once, let's consider only two of these box plots. 
Let's take the observations 
from die types 3 and 4. On the following page is a plot of the
observations for die type 3.
Recall the inner fences given before:

\[
\begin{array}{lll}
\mbox{Inner Fences:}& Q_{1}-1.5IQR,& Q_{3}+1.5IQR\\
% \mbox{Outer Fences:}& Q_{1}-3IQR,& Q_{3}+3IQR\\
\end{array}
\]
\end{slide}
\begin{slide}

For die type 3, we have
\begin{eqnarray*}
Q_{1}-1.5IQR & = & 21.3-1.5\left(21.9-21.3\right) {\rm\:\:\:}={\rm\:\:\:}20.4\\
Q_{3}+1.5IQR & = & 21.9+1.5\left(21.9-21.3\right) {\rm\:\:\:}={\rm\:\:\:}22.8\\
\end{eqnarray*}
So for die type three, the interval produced by the inner fences is 
[20.4,~22.8].
\end{slide}
\begin{slide}

All values for die type 3 that falls outside of this interval would be considered
an outlier. Hence, the points 19.9 and 20.2 would be considered outliers as well as
the point 23.0. Notice that the minimum value that is not an outlier is 21.3 and the 
maximum value that is not an outlier is 22.5.

\begin{figure}[h!]
\centerline{\psfig{figure=Pearson:Oztex:EPS:die3.eps,height=2.75in,width=2.75in}}
\end{figure}

For die type 4, we have
\begin{eqnarray*}
Q_{1}-1.5IQR & = & 16.8-1.5\left(17.9-16.8\right) {\rm\:\:\:}={\rm\:\:\:}15.15\\
Q_{3}+1.5IQR & = & 17.9+1.5\left(17.9-16.8\right) {\rm\:\:\:}={\rm\:\:\:}19.55\\
\end{eqnarray*}
\end{slide}
\begin{slide}

Hence, for die type 4, the interval produced by the inner fences is 
[15.15, 19.55]. Since all of the values fall within this interval, none of the
observations would be considered outliers. Notice that the minimum value that 
is not an outlier is 15.9 and the maximum value that is not an outlier is 19.0.
Note that this implies that there are no outliers for this set.

\begin{figure}[h!]
\centerline{\psfig{figure=Pearson:Oztex:EPS:die4.eps,height=2.75in,width=2.75in}}
\end{figure}
\end{slide}
\begin{slide}

\heading {Creating a frequency distribution}
\begin{itemize}
\item$\clubsuit$ Order the observations in ascending order. Recall that 
this is an important first step in describing the data.
\item$\clubsuit$ Choose the number of classes you wish to use. (For
our example I will choose 10 for die set 3. Since we have a small set of observation,
it would generally be best to choose a smaller number, but I choose
10 for simplicity of illustration.)
In general you would choose a number between 5-20.
\item$\clubsuit$ Determine the class widths. To do this divide the 
range of the numbers by the number of classes. ( In our
 case, that's
$\frac{23.0 - 19.9}{10} \approx 0.31$). Hence our class widths should be 0.31 units
wide. Let's choose our ``anchor" point to be the first point in the data set for die
set 3, that number is 19.9. So the first class will range from 19.9 to 20.21, the second from
20.21 to 20.52, etc.
\item$\clubsuit$ Create your classes, these classes must be mutually 
exclusive, which means a data point can not fall in more than
one class, and these classes must be mutually exhaustive, every
point must fall in one of the classes.
\end{itemize}
\end{slide}
\begin{slide}
Hence we can construct the following table of frequencies, relative frequencies, and cumulative frequencies.

\begin{table}[h!]
\centering
\small
\begin{tabular}{ccccc} \\ \hline
 &  &  &   & cumulative\\  
&  &  relative &  cumulative & relative\\  
Class & frequency & frequency &  frequency & frequency\\ \hline
(19.9~-~20.21) & 2 & $\frac{2}{10}$& 2 & $\frac{2}{10}$ \\&&&& \\
(20.21~-~20.52) & 0 & $\frac{0}{10}$& 2 & $\frac{2}{10}$ \\&&&& \\
(20.52~-~20.83) &0  & $\frac{0}{10}$& 2& $\frac{2}{10}$ \\&&&& \\
(20.83~-~21.14) & 0 & $\frac{0}{10}$& 2 & $\frac{2}{10}$\\&&&& \\
(21.14~-21.45) & 1 & $\frac{1}{10}$& 3 & $\frac{3}{10}$\\&&&& \\
(21.45~-21.76) & 2 & $\frac{2}{10}$& 5 & $\frac{5}{10}$ \\&&&& \\
(21.76~-22.07) & 3 & $\frac{3}{10}$& 8 & $\frac{8}{10}$ \\&&&& \\
(22.07~-22.38) & 0 & $\frac{0}{10}$& 8 & $\frac{8}{10}$ \\&&&& \\
(22.38~-22.69) & 1 & $\frac{1}{10}$& 9 & $\frac{9}{10}$ \\&&&& \\
(22.69~-23.00) & 1 & $\frac{1}{10}$& 10 & $\frac{10}{10}$=1 \\ \hline
\end{tabular}
\end{table}

\end{slide}
\begin{slide}
Similarly for die set 4,  we would get the range being 19-15.9 = 0.31 (very coincidental).
we can construct the following table of frequencies, relative frequencies, and cumulative frequencies.

\begin{table}[h!]
\centering
\small
\begin{tabular}{ccccc} \\ \hline
 &  &  &   & cumulative\\  
&  &  relative &  cumulative & relative\\  
Class & frequency & frequency &  frequency & frequency\\ \hline
(15.9~-~16.21)  & 2 & $\frac{2}{10}$& 2 & $\frac{2}{10}$ \\&&&& \\
(16.21~-~16.52) & 0 & $\frac{0}{10}$& 2 & $\frac{2}{10}$ \\&&&& \\
(16.52~-~16.83) & 1 & $\frac{1}{10}$& 3 & $\frac{3}{10}$ \\&&&& \\
(16.83~-~17.14) & 0 & $\frac{0}{10}$& 3 & $\frac{3}{10}$\\&&&& \\
(17.14~-~17.45) & 2 & $\frac{2}{10}$& 5 & $\frac{5}{10}$\\&&&& \\
(17.45~-~17.76) & 2 & $\frac{2}{10}$& 7 & $\frac{7}{10}$ \\&&&& \\
(17.76~-~18.07) & 1 & $\frac{1}{10}$& 8 & $\frac{8}{10}$ \\&&&& \\
(18.07~-~18.38) & 1 & $\frac{1}{10}$& 9 & $\frac{9}{10}$ \\&&&& \\
(18.38~-~18.69) & 0 & $\frac{0}{10}$& 9 & $\frac{9}{10}$ \\ &&&& \\
(18.69~-~19.00) & 1 & $\frac{1}{10}$& 10 & $\frac{10}{10}$=1 \\ \hline
\end{tabular}
\end{table}
\end{slide}
\begin{slide}
\heading{Histograms} 

A histogram is another graphical representation of the distribution 
of a batch of data. 
The data values are usually grouped into mutually exclusive and 
exhaustive intervals of equal width, and the number of observations
in each interval is determined and represented by a vertical bar.
In some variations the widths of the intervals are varied, 
resulting in potentially different appearances in the plot. 
\end{slide}
\begin{slide}
Histogram of the tensile strength data for die types 3 and 4 are
given below.

\begin{figure}[h!]
\centerline{\psfig{figure=Pearson:Oztex:EPS:die3hist.eps,height=2.75in,width=2.75in}}
\end{figure}

\begin{figure}[h!]
\centerline{\psfig{figure=Pearson:Oztex:EPS:die4hist.eps,height=2.75in,width=2.75in}}
\end{figure}
\end{slide}
\begin{slide}

\heading{Stem and leaf plot (Stemplot)}

A stem-and-leaf display is another graphical 
representation of the distribution of a batch of data. 
Very similar to a histogram, it is often accompanied by additional 
information about the data, such as cumulative frequencies and the
 position of the median. The plot represents the data values by 
 their numerical values, providing additional information over 
 the histogram, but the grouping intervals are usually chosen 
 based on using round numbers, rather than in an attempt to 
 provide the most effective plot. If were wanted to consider a single set of observations,
we could construct a stem and leaf plot. If we wanted to compare two sets of
observations, we could construct a back-to-back stem and leaf plot to
compare the distributions (location and spread) of the observations.
Consider the observations from die set 1. If we let the stem be the tens unit and
the leaf be the ones unit, stem and leaf plot for die type one would be:
\end{slide}
\begin{slide}

\[
\begin{tabular}{c|l}
1&8 9 9\\
2&0 0 0 0 0 1 2\\
\end{tabular}
\]
\end{slide}
\begin{slide}

This gives us very little information about the actual distribution of the data.
How about this one:

\[
\begin{tabular}{c|l}
18&\\
18&9\\
19&3\\
19&5\\
20&0\\
20&5 6 7 8\\
21&0\\
21&\\
22&1\\
22&\\
\end{tabular}
\]

\end{slide}
\begin{slide}

Which is similar to that produced from STATISTIX:

\end{slide}
\begin{slide}
\begin{verbatim}
STEM AND LEAF PLOT OF DIE1  

   LEAF DIGIT UNIT = 0.1                MINIMUM  18.900                        
   18  9  REPRESENTS 18.9               MEDIAN   20.550                        
                                        MAXIMUM  22.100
          STEM  LEAVES
        1   18  9                                                              
        2   19  3                                                              
        3   19  5                                                              
        4   20  0                                                              
       (4)  20  5678                                                           
        2   21  0                                                              
        1   21                                                                 
        1   22  1                                                              

10 CASES INCLUDED   0 MISSING CASES   
\end{verbatim}
\end{slide}
\begin{slide}
Notice the (4) in the table, this denotes the cell in which the median of the observation
fall. From above, we know that the median is 20.55 or the arithmetic mean of the middle two observations,
20.5 and 20.6.
Here we have used the entire representation of the data. The stem is the whole
number and the leaf the fraction or decimal. Each stem and leaf plot should
have a legend defining its content. In the case of our last stem and leaf plot we would
give the legend as: {\em stem}$\times$ {\em 1 + leaf}.

\end{slide}
\begin{slide}
Often it is the case that we would want to use the stem and leaf plot to compare distributions
as discussed above. Let's construct a back-to-back plot for the first two die
types.

\[
\begin{tabular}{r|c|l}
$D_{1}$&&$D_{2}$\\ \hline
&16&\\
9&16&\\
&17&\\
8 5&17&\\
3 0 &18&\\
8 6&18&9\\
2&19&3\\
9&19&5\\
&20&0\\
&20&5 6 7 8\\
&21&0\\
&21&\\
&22&1\\
&22&\\
\end{tabular}
\]

\end{slide}
\begin{slide}

\heading{A second example:}

The accompanying date are the times to failure (in
trillions of cycles)
of high-speed turbine engine bearings made out of two different compounds.
\[
\begin{array}{lccccccccccc}
\mbox{Compound 1} & 3.03  & 5.53 & 5.60 & 9.30 & 9.92 & 12.51 &
12.95 & 15.21 & 16.04 & 16.84 \\
&&&&&&&&&&&\\
\mbox{Compound 2} & 3.19 & 4.26 & 4.4 & 4.53 & 4.67 & 4.69 &
5.78 & 6.79 & 9.3 & 12.75 \\
&&&&&&&&&&&\\
p & 0.05 & 0.15 & 0.25 & 0.35 & 0.45 & 0.55 &
0.65 & 0.75 & 0.85 & 0.95\\ \hline
\end{array}
\]
\end{slide}
\begin{slide}

Make back-to-back stem-and-leaf plots for comparing the life length
properties of bearings made from Compounds 1 and 2. Compound 1 is on the left and compound 2 on 
the right.
\[
\begin{tabular}{r|c|l}
$C_{1}$&&$C_{2}$\\ \hline
0&3&2\\
 &4&35577\\
65&5&8\\
 &6&8\\
 &7& \\
 &8& \\
93&9&4\\
 &10& \\
 &11& \\
95&12&7\\
 &13& \\
 &14& \\
2&15& \\
80&16& \\
\end{tabular}
\]
\begin{flushleft}
Hundredths place dropped.\\
Stem = Ones Unit\\
Leaf = tenths (rounded) Unit\\
\end{flushleft}
\end{slide}
\begin{slide}
\end{slide}
\begin{slide}
\heading{Scatter plot}

A scatter-plot a is a graphical method that can be used to study
the joint variation of two variables graphically. Each
observation is represented by a point (x, y) on the plot, indexed
by the values on the axes. Each axis is used for a different
variable. Besides showing how (and whether) two variables are
related to each other, scatter plots also can indicate the
existence of distinct subgroups in the data. Scatter-plots can only be 
used if there are data pairs ($x_{i}, y_{i}$). We will talk in detail 
about scatterplots when we get to chapter 2.
\end{slide}
\begin{slide}
% \heading{Density Curves}
% 
% Density curves are functional and or graphical representations of 
% data that are usually continuous in nature.
\end{slide}
\begin{slide}
\heading{Quantile-plots}
Let's construct a quantile plot for die types 2 and 3. First, reconsider the 
table of values.
\begin{table}[h!]
\centering
\caption{\Large Tensile Strengths from 4 Precision Die Sets}
\begin{tabular}{cccccc} \hline
Observation&p=$\frac{i-0.5}{n}$ & Die 1 & Die 2 & Die 3 & Die 4\\ \hline
1 & 0.05&18.9 &   16.9 &   19.9 &   15.9\\
2 & 0.15&19.3 &   17.5 &   20.2 &   16.0\\
3& 0.25& 19.5 &   17.8 &   21.3 &   16.8\\
4& 0.35&20.0 &   18.0 &   21.5 &   17.2\\
5& 0.45& 20.5 &   18.3 &   21.7 &   17.4\\
6& 0.55& 20.6 &   18.4 &   21.8 &   17.5\\ 
7 & 0.65&20.7 &   18.6 &   21.9 &   17.7\\ 
8& 0.75& 20.8 &   18.8 &   21.9 &   17.9\\ 
9& 0.85& 21.0 &   19.2 &   22.5 &   18.1\\ 
10 & 0.95&22.1 &   19.9 &   23.0 &   19.0\\ \hline
\end{tabular}
\end{table}
A quantile plot is simply a scatterplot of the observation versus it's
quantile. So the quantile plot for the third die type is given by:
\end{slide}
\begin{slide}

\begin{figure}[h!]
\centerline{\psfig{figure=Pearson:Oztex:EPS:qqdie3.eps,height=2.75in,width=2.75in}}
\end{figure}
\end{slide}
\begin{slide}
\heading{Quantile-Quantile Plots}
Quantile-Quantile plots are useful for comparing distributions. They 
are generally used to determine if data in a sample follow a 
particular distribution. In statistics in order to make inferences, 
it is often assumed that data follow the normal distribution. In which 
case the quantiles of the sample are compared to the quantiles of the 
normal distribution. These are generally referred to as normal 
probability plots.
\end{slide}
\begin{slide}

Let's construct a Q-Q plot for die types 2 and 3. First, reconsider the 
table of values.
\begin{table}[h!]
\centering
\caption{Tensile Strengths from 4 Die Sets}
\begin{tabular}{cccccc} \hline
Observation&p=$\frac{i-0.5}{n}$ & Die 1 & Die 2 & Die 3 & Die 4\\ \hline
1 & 0.05&18.9 &   16.9 &   19.9 &   15.9\\
2 & 0.15&19.3 &   17.5 &   20.2 &   16.0\\
3& 0.25& 19.5 &   17.8 &   21.3 &   16.8\\
4& 0.35&20.0 &   18.0 &   21.5 &   17.2\\
5& 0.45& 20.5 &   18.3 &   21.7 &   17.4\\
6& 0.55& 20.6 &   18.4 &   21.8 &   17.5\\ 
7 & 0.65&20.7 &   18.6 &   21.9 &   17.7\\ 
8& 0.75& 20.8 &   18.8 &   21.9 &   17.9\\ 
9& 0.85& 21.0 &   19.2 &   22.5 &   18.1\\ 
10 & 0.95&22.1 &   19.9 &   23.0 &   19.0\\ \hline
\end{tabular}
\end{table}
\end{slide}
\begin{slide}
\begin{figure}[h!]
\centerline{\psfig{figure=Pearson:Oztex:EPS:qq23plot.eps,height=2.75in,width=2.75in}}
\end{figure}
\end{slide}
\begin{slide}

The fact that a significant amount of points do not fall on the line superimposed signifies
that the two sets of observation are different in terms of there distribution.

\end{document}
