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\centerline{MONT 106N -- Identifying Patterns}
\centerline{Solutions for Midterm Exam}
\centerline{October 19, 2009}
\vskip 10pt
\noindent
I. The following quotation is taken from an article that appeared in the 
{\it New York Times} on December 3, 1991.
\begin{enumerate}
\item[] ``... new research also demonstrates that spanking, which remains
parents' main disciplinary weapon, is not only useless in the long run but
also potentially harmful to the child's emerging personality.  In experimental
studies with toddlers placed in a room with their mothers and breakable objects
they were told not to touch, the children who were frequently spanked were
less obedient than the children who were disciplined in nonphysical ways.''
\item[A)] (5)  True or False:  The author is claiming that the studies show that 
children who are spanked frequently tend to become more disobedient 
than children who are disciplined in nonphysical ways.    
\item[] {\it Solution:}  This is True -- note the first sentence says
that the research to which the author is referring ``demonstrates that
spanking [is] ... potentially harmful to the child's emerging personality''
because it causes them to be less obedient.   
\item[B)] (10)  What questions should one ask about the way these 
studies were performed before deciding whether they provide good 
evidence for a relation between spanking and obedience?  Explain.
\item[] {\it Solution:}   This kind of study was almost inevitably 
observational in nature.  It would be very hard, if not impossible,
to design a randomized double-blind trial on the effects of spanking.
So while a randomized double-blind trial would give the strongest 
evidence, it is not really practical in this situation.  Hence
the main issue would be to determine how well the designers of
the study tried to {\it control for} underlying confounders like 
conditions that might predispose children to be less obedient in 
response to instructions like ``don't touch'' \emph{and} that might lead to 
spanking if the parents were inclined to spank.  Some possible 
conditions are hyperactivity, attention 
deficit, and so on.  
\vskip 10pt
The scoring of this problem was done as follows:
\begin{itemize}
\item full 10 points only for a complete discussion of the 
possibility of doing randomized double-blind studies, and for
raising issues of controlling for at least one specific and 
reasonably probable confounder,
\item 7 points for raising questions about randomization or double-blinding and
possible confounders, without a specific confounder 
(this is a good answer but not the best one!)
\item 3 points for extraneous points not addressing the question of the 
experimental design.
\end{itemize}
\end{enumerate}

\begin{figure}
\begin{center}
\includegraphics[height=1.0truein,width=4.0truein]{IIAAll.ps}
\caption{Histogram for II A -- Countywide population}
\end{center}
\end{figure}
\begin{figure}
\begin{center}
\includegraphics[height=1.0truein,width=4.0truein]{IIAJur.ps}
\caption{Histogram for II A -- Grand Jurors}
\end{center}
\end{figure}
\vskip 10pt 
\noindent
II. A study of grand juries in Alameda County, California, compared
the demographic characteristics of grand jurors in one year with the 
general population.  Here are the results for age.  Only persons 21 
years and older are included.
$$\begin{matrix} & {\rm Age} & {\rm Countywide\ Percentages} & {\rm Number\ of\ Grand\ Jurors} & \\
           & 20 < {\rm age} \le 40 &  42 & 5 \\
           & 40 < {\rm age} \le 50 &  23 & 9 \\
           & 50 < {\rm age} \le 60 &  16 & 19\\
           & 60 < {\rm age} \le 80 &  19 & 33\\
           & {\rm Totals}  & 100 \% & 66 {\rm\ jurors}
\end{matrix}
$$
\begin{enumerate}
\item[A)] (15) Construct two age histograms, one for the countywide population
         and one for the actual grand jurors. (Use the graph paper provided.)
\item[] {\it Solution:}  Note that the first column is given in units
of percents, while the second is given in raw numbers of jurors.  To construct
the second histogram, we should start by computing what percentage of the 
total number of jurors falls into each age class interval:
$$\begin{matrix} & {\rm Age} & {\rm Percent\ of\ Grand\ Jurors} & \\
           & 20 < {\rm age} \le 40 & 5/66 \doteq 8\% \\
           & 40 < {\rm age} \le 50 & 9/66 \doteq 14\% \\
           & 50 < {\rm age} \le 60 & 19/66 \doteq 29\%\\
           & 60 < {\rm age} \le 80 & 33/66 = 50\%
\end{matrix}$$
Then the histograms are shown in Figure 1 and 2.  The units on the horizontal
axis are years and the units on the vertical axis are $\%$ per year.  For example,
the histogram for the actual grand jurors has a box of height $2.5$ on the 
interval $60$ to $80$ since $50/20 = 2.5$.  


\item[B)] (5)  By comparing your histograms, what appears to be true about the age 
distributions of the two groups?
\item{} {\it Solution:} It seems that the grand jurors were older on 
average than the general population.
\end{enumerate}
\vskip 10pt
\noindent
III. The grades in a large university calculus class were normally distributed
with an average of 68 (out of a possible 100 points) and an SD of 10.
\begin{enumerate}
\item[A)] (15) If 400 students took the test, estimate how many scored between 58 and 83 
inclusive.  Show all work and draw normal curves as needed.
\item[] {\it Solution:}  The interval from $58$ to $83$ is $ave - 1 SD$ to $ave + 1.5 SD$,
so it corresponds to the standard unit interval $z = -1$ to $z = 1.5$.  The area of this
region can be computed from the standard normal table:
$$\frac{1}{2}(Area(1) + Area(1.5)) = \frac{1}{2}(68.27 + 86.64) \doteq 74.46\% $$
So about $.7446 \times 400 \doteq 310$ of the students scored in this range.  
\item[B)] (15) The professor decided to set the cutoff numerical grade for an A letter
grade at the 90th percentile of the grade distribution.  
What grade did a student need in order to earn an A in calculus?
\item[] {\it Solution:}  For the $90$th percentile, we want the $z$ so that the
the area between $-z$ and $z$ is $80\%$ (so $10\%$ in both upper and lower tails).  
From the normal table this happens between $z = 1.25$ and $z = 1.3$.  Using
$z = 1.3$ to estimate, we get $ave + 1.3 SD = 68 + 1.3 (10) = 81$.  
\end{enumerate}
\vskip 10pt
\noindent
IV.  The US Department of Justice made a study of civil cases during the year
1992.  Juries awarded money damages to the plaintiffs in 5949 of the cases.
The median damage award was \$50,000 and the average damage award was 
\$450,000.  
\begin{enumerate}
\item[A)]  (10) Sketch a possible (smoothed) histogram of the damage awards.
Does your histogram have a long left tail, a long right tail, or is it symmetric?
\item[] {\it Solution:}  Because the average is so much larger than the 
median, the histogram should show a long right tail (and no area for awards less
than zero).  That is, there had to be a few very large awards to 
make the average come out so large, while the median was relatively small.
Any such histogram is OK.
\item[B)]  (5) Which is larger, the difference between the award at the 
10th percentile and the median award, or the difference between the 
median award and the award at the 90th percentile?  Or are those differences about the same?
Explain briefly.     
\item[]  For any histogram with a long right tail as in part A, the 
difference between the awards at the 90th percentile and the median
would tend to be larger than the difference between the awards at the
10th percentile and the median.  The distribution of awards is skewed to the 
larger values.  
\end{enumerate}
\vskip 10pt
\noindent
V.
\begin{enumerate}
\item[A)] (10)  Find the equation of the line passing through the 
points $P = (1,-2)$ and $Q = (4,3)$ in the plane.
\item[] {\it Solution:}  The slope is $\displaystyle{\frac{3 - (-2)}{4 - 1} = \frac{5}{3}}$.
So the equation of the line is $y + 2 = \frac{5}{3}(x - 1)$ or $y = \frac{5}{3}x - \frac{11}{3}$.
\item[B)] (10) Find the equation of the line parallel to the 
line in part A and passing through the point $R = (1,2)$.
\item[] {\it Solution:}  Parallel lines have the same slope, so the line is
$y - 2 = \frac{5}{3}(x - 1)$, or $y = \frac{5}{3} x + \frac{1}{3}$.
\end{enumerate}

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