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\hfill October 3, 2006
\vskip 12 pt
\centerline{\bf Exam I --- Math 102 / Core 143 CX}
\vskip 12 pt
\noindent
Points are in parentheses.  Show your work to receive partial credit;
an answer like $[1-(6/7)^3]/4$ is worth more than $0.09257$, because
it displays your reasoning more clearly.

\begin{itemize}
\item[1.] Identify the histogram below that best matches the data
below.  If none of the histograms match, sketch one that gives a
better match than any given and explain.\begin{itemize}
\item[(a)] Years of education from a large group of Americans
\item[(b)] Numbers on the caller's markers from an old bingo game,
missing several markers
\item[(c)] Heights of a large group of nine-year-old boys
\item[(d)] Numbers of ways to get a given number on a roll of a
pair of dice
\end{itemize}
\centerline{\includegraphics{e1f06n1.png}}

\item[2,] Find, for the list of 10 numbers
$$
-4, -1, -1, -1, 0, 0, 0, 2, 2, 3:
$$
\begin{center}
(a) the average \qquad\qquad (b) the standard deviation \qquad\qquad
(c) the median
\end{center}
\begin{center}
(d) the 35th percentile \qquad\qquad (e) the IQR
\end{center}

\item[3.] Over a two-year period, world oil production (in Mbbl/day)
averaged 5, with a standard deviation of 2; and gasoline prices (in
\$/gal) averaged 1.5 with a standard deviation of 0.8.  The
correlation was $-0.5$.\begin{itemize}
\item[(a)] Is the sign of the correlation reasonable?  Explain.
\item[(b)] Estimate the gas price when the oil production is 6~Mbbl/day.
\item[(c)] About how far should you expect your estimate in (b) to be 
off?
\item[(d)] If you were to estimate the oil production when the 
gas price is 2.3, would your estimate be closest to 1, 4, 6 or 9?
\end{itemize}

\item[4.] \begin{itemize}
\item[(a)] For a long list of numbers that has an average of 140 and
a standard deviation of 20, use normal approximation to estimate
the 80-th percentile. 
\item[(b)] For the same data as in (a) estimate the percentile rank 
of the datum 130.
\item[(c)] The list of 20 numbers
$$
-14,-13,-12,-11,-10,-9,-8,-7,-6,-5,5,6,7,8,9,10,11,12,13,14
$$
has an average of 0 (that should be clear from the symmetry) and a
standard deviation of not quite 10; so only 50\% of the data is within
one standard deviation of the average.  How does this relate to the
``Rule of Thumb'' in the text?
\item[(d)] For the data in (c), without doing arithmetic, would you
guess that the standard deviations of the shorter lists
$$
-14,-13,-12,-11,-10,-9,-8,-7,-6,-5\qquad{\rm and}\qquad 
5,6,7,8,9,10,11,12,13,14
$$
are equal to, less than, or greater than the standard deviation 
in (c)?
\end{itemize}

\item[5.] You want to know whether vitamin~C alleviates the symptoms
of a cold.  Give a quick comment on each of the following methods of
gathering data to answer the question.\begin{itemize}
\item[(a)] Put a questionnaire on the web asking for responses from
those who have taken vitamin~C for cold symptoms.
\item[(b)] Get a list of people who have purchased vitamin~C, and
interview as many as possible, including in your study the ones who
say they took it for cold symptoms.
\item[(c)] Get a list of cold sufferers, give them vitamin~C and
interview them for results.
\item[(d)] Get a list of cold sufferers, give half of them (chosen at
random) vitamin~C, and interview all of them for results.
\end{itemize}

\item[6.] Related to the article, ``The median isn't the message",
by Stephen Jay Gould: Explain why Gould feels that his knowledge of
statistics played a major role in saving his life.

\end{itemize}

\vskip 12 pt
\hrule
\vskip 12 pt
\noindent
Some possibly useful formulas:
$$
s = \sqrt{\frac{\sum(x-\overline{x})^2}n}  \qquad\qquad 
z_x = \frac{x-\overline{x}}s \qquad\qquad 
r = \frac{\sum z_xz_y}n  
$$
$$
{\rm RMS\ error} = \sigma_y\sqrt{1-r^2} \qquad\qquad
y - \overline{y} = r\frac{s_y}{s_x}(x-\overline{x}) \qquad\qquad
y - \overline{y} = ({\rm sign\ of\ }r)\frac{s_y}{s_x}(x-\overline{x}) 
$$

\centerline{\begin{tabular}{|cc|cc|cc|cc|cc|}
\hline\multicolumn{10}{|c|}{Normal table (Area between $-z$ and $z$)}\\\hline
$z$ & Area(\%) & $z$ & Area(\%) & $z$ & Area(\%) & $z$ & Area(\%) & $z$ & Area(\%)\\
\hline
0.0 & 0.0 & 0.9 & 63.19 & 1.8 & 92.81 & 2.7 & 99.31 & 3.6 & 99.968 \\
0.05 & 3.99 & 0.95 & 65.79 & 1.85 & 93.57 & 2.75 & 99.4 & 3.65 & 99.974 \\
0.1 & 7.97 & 1 & 68.27 & 1.9 & 94.26 & 2.8 & 99.49 & 3.7 & 99.978 \\
0.15 & 11.92 & 1.05 & 70.63 & 1.95 & 94.88 & 2.85 & 99.56 & 3.75 & 99.982 \\
0.2 & 15.85 & 1.1 & 72.87 & 2 & 95.45 & 2.9 & 99.63 & 3.8 & 99.986 \\
0.25 & 19.74 & 1.15 & 74.99 & 2.05 & 95.96 & 2.95 & 99.68 & 3.85 & 99.988 \\
0.3 & 23.58 & 1.2 & 76.99 & 2.1 & 96.43 & 3 & 99.73 & 3.9 & 99.99 \\
0.35 & 27.37 & 1.25 & 78.87 & 2.15 & 96.84 & 3.05 & 99.771 & 3.95 & 99.992 \\
0.4 & 31.08 & 1.3 & 80.64 & 2.2 & 97.22 & 3.1 & 99.806 & 4 & 99.9937 \\
0.45 & 34.73 & 1.35 & 82.3 & 2.25 & 97.56 & 3.15 & 99.837 & 4.05 & 99.9949 \\
0.5 & 38.29 & 1.4 & 83.85 & 2.3 & 97.86 & 3.2 & 99.863 & 4.1 & 99.9959 \\
0.55 & 41.77 & 1.45 & 85.29 & 2.35 & 98.12 & 3.25 & 99.885 & 4.15 & 99.9967 \\
0.6 & 45.15 & 1.5 & 86.64 & 2.4 & 98.36 & 3.3 & 99.903 & 4.2 & 99.9973 \\
0.65 & 48.43 & 1.55 & 87.89 & 2.45 & 98.57 & 3.35 & 99.919 & 4.25 & 99.9979 \\
0.7 & 51.61 & 1.6 & 89.04 & 2.5 & 98.76 & 3.4 & 99.933 & 4.3 & 99.9983 \\
0.75 & 54.67 & 1.65 & 90.11 & 2.55 & 98.92 & 3.45 & 99.944 & 4.35 & 99.9986 \\
0.8 & 57.63 & 1.7 & 91.09 & 2.6 & 99.07 & 3.5 & 99.953 & 4.4 & 99.9989 \\
0.85 & 60.47 & 1.75 & 91.99 & 2.65 & 99.2 & 3.55 & 99.961 & 4.45 & 99.9991 \\
\hline
\end{tabular}}


\newpage

\noindent
{\bf Solutions to Exam I}

\begin{itemize}

\item[1.] (a)~V --- the humps come from the many people who finish
high school or college.  (b)~I --- each number from 1 to 75 appears
at most once, but some are missing.  (c)~III.  (d)~IV.

\item[2.]\begin{itemize}
\item[(a)] $[-4+3(-1)+3(0)+2(2)+3]/10 = 0$
\item[(b)] $\sqrt{[(-4-0)^2+3(-1-0)^2+3(0-0)^2+2(2-0)^2+(3-0)^2]/10} 
\approx 1.9$
\item[(c)] 0, between the fifth and sixth numbers in the list
\item[(d)] $-1$, between the third and fourth numbers in the list
\item[(e)] The third quartile may be 0 or 2 or anything between, but the
first quartile is surely
$-1$, so their difference, the $IQR$, could be anything from 1 to 3.
\end{itemize}

\item[3.]\begin{itemize}
\item[(a)] Yes: As oil production decreases, gas prices probably go up.
\item[(b)] 6 Mbbl/day is $(6-5)/2 = 0.5$ standard deviations above average,
so gas prices are projected to be $(-0.5)(0.5) = -0.25$ standard deviations
above average, or $1.5 + (-0.25)(0.8) = 1.3$~\$/gal.
\item[(c)] We should expect it to be off by a ``root mean square error for
regression'', $.8\sqrt{1-(.5)^2} \approx .69$
\item[(d)] A gas price of 2.3 is one SD above the average, so we should
expect oil production to be $.5$ SD's below average, or $5-(.5)(2) = 4$ 
(in other words, below average, but not \underbar{as much} below average
as the gas price was above its average, both measured in standard units). 
\end{itemize}

\item[4.]\begin{itemize}
\item[(a)] The 80-th percentile has, between it and its corresponding 
value on the opposite side of the average, 60\% of the data.  From a 
normal table, that value, in standard units, is about .84, so in the original
values it is about $140+.84(20) = 156.8$. 
\item[(b)] 130 is half an SD below average, and from the normal table, 
the area below $-.5$ is $(100-38)/2 = 31$ percent.  So 130 is about the 
thirty-first percentile.
\item[(c)] The data aren't normally distributed and so need not (and clearly
do not) follow the Rule of Thumb.
\item[(d)] It should be smaller.  The pieces each have the same SD, but their
averages are different, so the joined group would have, in general, higher 
deviations from the new average of 0 than from their old averages (of $-9.5$
and 9.5 respectively). 
\end{itemize}

\item[5.] \begin{itemize}
\item[(a)] Volunteer samples are useless.
\item[(b)] It's an observational study, and your subjects are probably
biased in favor of vitamin~C, but it's a start toward more reliable data.
\item[(c)] Better still, but a control group would tell you whether
symptoms improved because of the vitamin~C or just because they were
recovering from the cold.
\item[(d)] Good!  A controlled experiment with random assignment to
groups (hopefully conducted double-blind, but we have no info about
that) can establish that vitamin~C actually causes an improvement
in cold symptoms. But no scientific result is ever final.
\end{itemize}

\item[6.] Gould feels that his knowledge of statistics helped him
to realize that, though the prognosis for most victims of his type of
cancer is not good, he was among the people (younger than average, otherwise
healthy and with access to good care) who had a better chance for long
survival (in the right ``tail'' of survivors).  The positive
outlook this gave him helped to survive longer and finally to beat that
cancer (though he died of another kind of cancer not long ago).

\end{itemize}

\end{document}

\centerline{\includegraphics{.png}}
