\magnification=\magstep1
\centerline{MONT 106N -- Identifying Patterns}
\centerline{Final Exam Solutions -- December 17, 2009}\bigskip
\noindent
I.  The City University of New York has about 200,000 students
on 21 campuses, whose ages are distributed as in the following
table:
$$\matrix{{\rm Ages} & \qquad & {\rm Percent}\cr
           & \qquad &\cr
           \le 19 & \qquad & 21.1\cr
           20 {\rm\ to\ } 22 & \qquad & 26.2\cr
           23 {\rm\ to\ } 24 & \qquad & 10.8\cr
           25 {\rm\ to\ } 29 & \qquad & 15.5\cr
           30 {\rm\ to\ } 44 & \qquad & 20.2\cr
           45 {\rm\ to\ } 64 & \qquad & 4.6\cr
            \ge 65 & \qquad & 1.6\cr}$$
\item{A)}  (10)  On the graph paper supplied, construct
a histogram for this data.  To do this in a standard way, think of
each year of age as corresponding to the interval starting at that year
and going all the way to the next whole number of years.  Thus, for instance, the
age range 20 to 22 will correspond to a histogram bar with base
from 20 to 23.  Make the first bar extend from 15 to 20 and the last
one extend from 65 to 75.
\medskip
\item{} {\it Solution:}  Following the directions above, the first
bar in the histogram extends from 15 to 20, so its height is $21.1/5 \doteq 4.22$,
the second has height $26.2/3 \doteq 8.73$, the third has height $10.8/2 = 5.4$,
the fourth has height $15.5/5 = 3.1$, the fifth has height $20.2/15 \doteq 1.35$,
the sixth has height $4.6/20 = .23$, and the last one has height $1.6/10 = .16$.
\medskip
\item{B)}  (10)  Is the {\it average} age or the {\it median} age larger
for the CUNY student population?
\medskip
\item{} {\it Solution:}  Since the distribution of ages has a long right tail,
the average is larger than the median.
\bigskip
\noindent
II.  Suppose you are going to design a study to test the effectiveness of
teaching elementary German by two different methods:  (a) in
a traditional classroom, with spoken lessons, text readings, etc.;
(b) in a media lab with language tapes, music videos,
cultural films, etc.  There are 50 students who will be starting
elementary German at your school next year and one section will
be taught by each method.
\medskip
\item{A)}  (10) {\it From the point of view of the reliability of
the results of the study},
would it be a good idea to have the same professor assigned to teach
both sections?  Explain.
\medskip
\item{} {\it Solution:}  Yes.  This would be one way to 
control for possible differences in the effectiveness of different
teachers.
\medskip
\item{B)}  (10) {\it From the point of view of the reliability of
the results of the study}, is there a problem with letting
the students select which section they register for?  Explain.
\medskip
\item{} {\it Solution:}  Yes.  If students select which section
they register for, then we might expect to see some bias from confounding
factors since the students who learn better in the traditional classroom might
chose that option and vice versa.  Note that this is one place
where educational objectives might clash with the requirements
of good experimental design(!)
\bigskip
\noindent
III.  The scatter plot for the first- and second-year
GPA's of a large number of university students is football-shaped, with
a positive $r$-value.
\medskip
\item{A)} (10)  True or False:  On average, students who had GPAs above
average in their first year also had GPAs above average in their second year.
Explain, {\it briefly}.
\medskip
\item{} {\it Solution:}  True.  Since $r$ is positive, the regression line
slopes up to the right, so an above-average GPA in the first year would
lead to a predicted above-average GPA in the second year.
\medskip
\item{B)} (10)  True or False:  A student who was at the 40th percentile
of the first year GPAs is also likely to be at the 40th percentile
of the second year GPAs.  Explain, {\it briefly}.
\medskip
\item{} {\it Solution:}  False.  From the regression effect, we would
expect the second-year GPA to be somewhat higher than the 40th percentile.
\bigskip
\noindent
IV. (Hypothetical) The grades in a large university statistics class were normally
distributed with an average of 65 (out of 100) and an SD of 10.
\medskip
\item{A)} (10)  Draw a smoothed histogram for the data, showing
the locations of the average and ${\rm ave} \pm SD$ on your plot.
\medskip
\item{} {\it Solution:}  The graph should be a normal-shaped 
curve with maximum at ${\rm ave} = 65$, and inflection points
(changes in concavity) at $65 - 10 = 55$ and $65 + 10 = 75$.
\medskip
\item{B)} (10)  If 250 students took the test, estimate how
many of them scored between 70 and 75.
\medskip
\item{} {\it Solution:}  The interval from $70$ to $75$ is 
$z = 0.5$ to $z = 1.0$ in standard units.  The area under
the normal curve between these $z$-values is 
$${1\over 2}(A(1.0) - A(0.5)) = {1\over 2}(68.26\% - 38.30\%) \doteq 15\%$$
\item{C)} (10)  What exam score was at the 90th percentile
of the distribution?
\medskip
\item{} {\it Solution:}  The $90$th percentile is the 
score $z$ where $A(z) = 80\%$.  This happens when $z \doteq 1.28$ (with
interpolation) or $z \doteq 1.30$ if you just estimate from the 
entries in our table.  The standard unit value $z = 1.28$ corresponds
to the score $65 + 1.3\times 10 \doteq 78$.  
\bigskip
\noindent
V.  All parts of this problem deal with the following data set:
$$\matrix{ x & \qquad & y\cr
             & \qquad & \cr
           1 &\qquad & 1\cr
           8 &\qquad & 4\cr
           10 &\qquad & 6\cr
           10 &\qquad & 12\cr
           14 &\qquad & 12\cr
           17 &\qquad & 7\cr}$$
\item{A)} (10)  On the graph paper provided, draw the scatter plot.
\medskip
\item{} {\it Solution:}  Plot omitted.
\medskip
\item{B)} (10)  Compute the averages and SD's of the $x$ and $y$ data.
\medskip
\item{} {\it Solution:}  We compute:
$${\rm ave}_x = {1 + 8 + 10 + 10 + 14 + 17\over 6} = {60\over 6} = 10$$
and 
$${\rm ave}_y = {1 + 4 + 6 + 12 + 12 + 7\over 6} = {42\over 6} = 7.$$
Then
$$SD_x = \sqrt{{1\over 6}((-9)^2 + (-2)^2 + 0 + 0 + 4^2 + 7^2)} = 5$$
and
$$SD_y = \sqrt{{1\over 6}((-6)^2 + (-3)^2 + (-1)^2 + 5^2 + 5^2 + 0)} = 4.$$
\item{C)} (10)  Compute the correlation coefficient, $r$.
\medskip 
\item{}  We proceed as usual.  The standard unit equivalents of the 
$xy$ data are
$$\matrix{ z & \qquad & w\cr
             & \qquad & \cr
           -9/5 &\qquad & -6/4\cr
           -2/5 &\qquad & -3/4\cr
           0 &\qquad & -1/4\cr
           0 &\qquad & 5/4\cr
           4/5 &\qquad & 5/4\cr
           7/5 &\qquad & 0\cr}$$
The average of the products $z_i w_i$ is 
$$r = {1\over 6}(54/20 + 6/20 + 0 + 0 + 20/20 + 0) = {80\over 120} = {2\over 3} \doteq .67$$ 
\item{D)} (10)  On your scatter plot, draw in the regression line
for estimating values of $y$, given values of $x$.
\medskip
\item{} {\it Solution:}  The equation of the regression line is 
$y - 7 = {4\over 5}\times {2\over 3} (x - 10)$, or
$$y = {8\over 15}x + {5\over 3}.$$
\bigskip
\noindent
VI. (Hypothetical) At a college, a large group of incoming
students took language placement tests in {\it both} Spanish and French.
The average on the Spanish test was 70 (out of 100) with an SD of 10.
The average on the French test as 75, with an SD of 12.
The correlation coefficient between the Spanish and French test scores
was $r = .6$.  The scatter plot was football-shaped (homoscedastic).
\medskip
\item{A)} (10)  Estimate the average French score for the students who scored
80 on the Spanish exam.
\medskip
\item{} {\it Solution:} Call $x$ the Spanish exam score and $y$ the 
French exam score.   The desired estimate can be found from the equation of
the regression line:  $y - 75 = {12 \times .6\over 10}(x - 70)$.  With
$x = 80$ the estimate is
$$75 + {12 \times .6\over 10}(10) = 82.2$$
The estimated average French score for these students is about $82$.
\medskip  
\item{B)} (10)  For the students who scored 80 on the Spanish test,
estimate the SD of their scores on the French exam.
\medskip
\item{} {\it Solution:}  Since we are given that the scatter diagram
is football-shaped, this SD would be estimated by the rms error
for regression, which is 
$$\sqrt{1 - r^2} \times SD_y = \sqrt{1 - (.6)^2} \times 12 = .8 \times 12 = 9.6 \doteq 10$$
\item{C)} (10)  Using your answers for parts A and B, among
the students who scored an 80 on the Spanish test, estimate
the percentage who scored above 85 on the French test.
\medskip
\item{} {\it Solution:}  Using the estimate of 10 for the rms error of 
regression from part B and the estimate of 82 for the average from 
part A,  we want the percentage whose scores were $.3$ standard units
or more above the average.  The area from $z=-.3$ to $z = .3$ under
the normal curve is $A(.3) \doteq 23.58\%$.  We want the 
area in the upper tail which is $50 - {23.58\over 2} \doteq 38.21$. 
So about $38\%$ scored above $85$ on the French exam.  
\bigskip
\noindent
VII.  Three cards are dealt (without replacement) off
the top of a well-shuffled standard deck (with 52 cards --
ace through king in each of the four suits: clubs, diamonds,
hearts, spades).
\item{A)} (10)  What is the chance that you get only kings?
\medskip
\item{} {\it Solution:}  The probability is
$${4\over 52} \times {3\over 51} \times {2\over 50}.$$
\item{B)} (10)  What is the chance that you get no kings?
\medskip
\item{} {\it Solution:}  There are 48 non-kings to 
start, then 47 non-kings left provided that the first draw
is a non-king, then 46 non-kings left for the last, provided
that the first two are non-kings.  The probability is 
$${48\over 52}\times {47\over 51} \times {46\over 50}.$$
\bigskip
\noindent
VIII.  A box contains 8 red marbles and 3 green marbles.
\bigskip
\item{A)} (10)  If you draw six marbles at random, record
the color, replace the marble each time and mix up the box,
what is the chance of getting 4 red and 2 green marbles?
\medskip 
\item{} {\it Solution:}  This is given by the binomial formula:
$${6\choose 4} \left({8\over 11}\right)^4 \left({3\over 11}\right)^2 
= {6!\over 4! 2!}\left({8\over 11}\right)^4 \left({3\over 11}\right)^2 .$$
(This is a chance of about $31.2\%$.)  
\item{B)} (10) Is the chance the same if you set aside
each marble as it is drawn (without replacing it in the box)?
Explain why or why not.
\medskip
\item{} {\it Solution:}  No, the probability would not be the 
same.  If you do not replace the marbles, then the probabilities
of drawing reds and greens change according to how many of
each color have been drawn.  
\bigskip
\noindent
\centerline{\it Have a peaceful and joyous holiday season!}
\bye
