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\centerline{MONT 107N -- Understanding Randomness}
\centerline{Sample Question Solutions for Midterm Exam}
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\noindent
({\it Note:}  The actual exam will be somewhat shorter than this;
there are parts here to illustrate most of the different kinds of 
questions that might appear.)
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\noindent
I. Assume 125 draws with replacement are made from the following box:
$$                                    1\quad    3\quad    5\quad   7$$
\item{A)} What are the EV and SE for the sum of numbers on the drawn tickets?
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\item{} {\it Solution -- } The average of the box is $4$ and the SD of the box is
$\sqrt{5} \doteq 2.24$.  So with $125$ draws, ${\rm EV} = 125 \times 4 = 500$ and
the ${\rm SE} = \sqrt{125} \times \sqrt{5} = 25$.  
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\item{B)} What are the chances that sum of the numbers will be 480 or larger?
\item{} {\it Solution -- }  In standard units, 480 corresponds to 
$$z = {480 - 500\over 25} = -.8$$  
The area under the normal curve with $z \ge -.8$ is the central area,
plus the upper tail area:
$$A(.8) + {1\over 2}(100\% - A(.8)) = 50\% + 28.815\% \doteq 78.815\%.$$  
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\noindent
II. Consider the following box.
$$0\quad    1\quad     2\quad    2\quad   2\quad   3\quad   3\quad    4\quad   4\quad    4$$
\item{A)} If a single draw is made from the box, construct the probability
histogram for the number that is drawn.
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\item{} {\it Solution --} The probability histogram should have 5 boxes: the
first from -.5 to .5 with height $1/10 \times 100\% = 10\%$
representing the chance of drawing a 0, second from .5 to 1.5 with height
$10\%$, third from 1.5 to 2.5 with height $30\%$, fourth from 2.5 to 
3.5 with height $20\%$, last from 4 to 4.5 with height $30\%$ (chance of drawing a 4).  Note
the total area is $100\%$ as always for us.
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\item{B)} If 1000 draws with replacement were made from this box, 
draw an approximate probability histogram for the value of the sum.  
Show carefully the location of the EV and the SE for the sum relative
to your histogram.
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\item{}{\it Solution -- }  The average of this box is ${25\over 10} = 2.5$ and the 
SD is about $1.28$.   
From the Central Limit Theorem (since every integer between
0 and 4000 can be obtained as a sum of tickets from this box), we would expect to 
see something very close to a shifted and scaled {\it normal curve} with peak at the EV for the sum:
$1000 \times 2.5 = 2500$, and inflection points (concavity changes) at
$$2500 \pm {\rm SE\ for\ sum} \doteq 2500 \pm (\sqrt{1000} \times 1.28) \doteq 2500 \pm 40.62.$$
(Note: A completely to-scale drawing should also show the correct heights so that
the area is exactly $100\%$.  We did not really discuss how
those would be obtained.  So, any recognizably normal graph would be 
OK here!)    
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\item{C)} Now, assume 200 draws with replacement are made from this box.
What box model would you use in order to model {\it the number of 4s} that are chosen?
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\item{} {\it Solution -- } We create a 0/1 box by replacing all non-4s with 0s
and all 4's with 1s:
$$0, 0, 0, 0, 0, 0, 0, 1, 1, 1.$$
\item{D)} Exactly 200 draws are made and 4s are drawn a total of 67 times.  What is the difference
between for the number of 4s for these 200 draws and the expected value of 
the number of 4s (that is, the {\it chance error} for this sample)?
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\item{}{\it Solution -- }  The EV for the sum of 200 draws from 
the 0/1 box is $200 \times .3 = 60$.  This is the same as the expected number
of 4s.  The chance error is $67 - 60 = 7$.    
\item{E)}  Would you consider the chance error from part D to be a large chance error, roughly
an average chance error, or a small chance error? Explain your answer. (Hint:
You may want to do some additional calculations to decide on your answer.)
\medskip
\item{} {\it Solution -- }
To decide, we need to know the SE, which is $\sqrt{200} \times \sqrt{(.3)(.7)} = 6.48$
(using the ``short-cut rule'' which applies for 0/1 boxes).  The chance error in part D
is an {\it average chance error}, since we expect most sums will be 
within 1 SE of the EV.  
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\noindent
III. On March 3, 2010, the {\it New York Times} reported results of a survey in 
which researchers contacted 600 people who graduated 
from a variety of high schools across the country in the last 4 to 12 years. The 
study was sponsored by the Bill and Melinda Gates Foundation, which has sought to 
shed light on low completion rates at both the high school and college levels. 
For the purposes of this question, assume the 600 survey respondents were a simple 
random sample of all recent high school graduates.
In this sample, $402$ people indicated that they thought their high school
guidance counselor was ``poor'' or ``fair'' in helping them decide what colleges or
technical schools would be good for them.
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\item{A)} Based on this sample, construct a $95\%$ confidence interval
for the the percentage of all recent high school graduates who think
their guidance counselor was
``poor'' or ``fair'' in helping them decide what colleges or
technical schools would be good for them.
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\item{}{\it Solution -- } The observed percent is ${402/600} \times 100\% = 67\%$.
We use the ``bootstrap'' to estimate the SE for the percent:
$${\rm SE} \doteq \sqrt{(.67)\times (.33)\over 600} \times 100\% \doteq 1.9\%$$
So the approximate $95\%$ confidence interval for the population percentage
would be  $67\% \pm 2 \times 1.9\%$, or $62.2\%$ to $70.8\%$.  
\medskip
\item{B)}  The {\it Times} reported that the survey had a sampling
error of $\pm 5\%$.  Does that match your confidence interval?
If not, can you suggest where the $5\%$ figure comes from?
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\item{} {\it Solution -- } Our estimate would say $\pm 3.8\%$.  But of course
that used the bootstrap method, which depended on the observed percent from 
the sample.  The ``conservative'' method we discussed in class would estimate
the SE by $\sqrt{(.5)\times (.5)\over 600} \times 100\% \doteq 2.0\%$, 
so $\pm 2 {\rm SE}$ would be $\pm 4\%$, which is still less than the reported
$\pm 5\%$ error bound.  It might also be the case that the 
researchers were not using a simple random sample, and hence that
they needed to use an even more conservative method to estimate the standard
error for the percent.   
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\noindent
IV.  A sociologist takes a simple random sample of size 300 from 
the 27000 students at a large state university.  In
the sample, $222/300 \times 100\% = 74\%$ are undergraduates.  The
SE for the percentage is calculated as $2.5\%$ and the sociologist
writes down the $95\%$ confidence interval for the percentage
as $74\% \pm 5\%$.  For each part, say whether the statement is true 
or false, and explain:
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\item{A)}  From the information in this sample it is believable
that the actual percentage of undergraduate students is $78\%$.  
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\item{}{\it Solution -- } {\it True} -- any value in the confidence
interval should be considered as a believable value for the population
percentage.
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\item{B)}  The range from $69\%$ to $79\%$ is a $95\%$ confidence
interval for the percentage of undergraduates in the sample.  
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\item{}{\it Solution -- } {\it False} -- the percentage of 
undergradates {\it in the sample} is known ($74\%$).  There
is no chance or uncertainty about that.  This would be
true if ``sample'' was replaced by ``population.'' (Be sure you read
slowly and carefully!) 
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\item{C)}  The $95\%$ figure comes from an area under the
normal curve.  
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\item{} {\it Solution -- } {\it True -- } the bounds
for the confidence interval come from the area between
$-2$ and $+2$ standard units under the normal curve.
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\item{D)}  The probability that the true population percentage
of undergraduates is in the interval $69\%$ to $79\%$ is 
$95\%$.  
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\item{} {\it Solution -- } {\it False -- } This is the 
incorrect interpretation of the confidence interval we
saw in class and in the second lab.  The chance is in the 
sampling process, not in the location of the true
population percentage.  
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\item{E)}  If a second sample of size 300 was taken in the same
semester, and a $95\%$ confidence interval was computed from that 
sample, then the result could be $67\%$ to $79\%$.
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\item{} {\it Solution -- } {\it False -- } (This is a slightly 
tricky question!)  
If this was the confidence interval then
the observed percentage would have to be the midpoint, or 
$73\%$.  That is certainly possible.  But note then that the 
``bootstrap'' formula for the estimated SE would give 
$\sqrt{.73\times .27/300} \times 100\% \doteq 2.6\%$.
The estimated $95\%$ confidence interval would be narrower:
$73\% \pm 5.2\%$.
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\item{F)}  If samples of size 1200 were taken in the 
same semester, then on average, we would expect the 
SE to decrease and the interval to be narrower.  
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\item{} {\it Solution -- }  {\it True -- } On average and 
other things being equal (i.e. assuming the observed proportions $p$
were the same), the larger number of samples will yield
a narrower confidence interval:
$$\sqrt{(p)(1-p)\over 1200} < \sqrt{(p)(1-p)\over 300}.$$
\item{G)}  The process of drawing the sample is like drawing 300
from a box with $27000$ tickets and recording the sum, 
where the ticket for each
first-year student has the value $1$, the ticket for 
each second-year student has the value $2$, the ticket
for each third-year student has the value $3$, the 
ticket for each fourth-year student has the value $4$, and
the ticket for each graduate student has the value $5$.  
\medskip
\item{} {\it Solution -- } {\it False -- } This should be 
a 0/1 box, since we are just counting/classifying.  

\bye
