\magnification=\magstep1
\centerline{MONT 105N -- Analyzing Environmental Data}
\centerline{Solutions for Midterm Exam Practice Problems}
\centerline{March 26, 2013}

\bigskip
\noindent
A.   A.   Waterbuck are large mammals in the deer family native to 
eastern South Africa.  In 1976, biologists launched a three-year
study of waterbuck ecology and determined the annual survival and
fertility rates for females given in the following table:
\bigskip
$$\matrix{& {\rm A} = {\rm calves} & {\rm B} = {\rm yearlings} & {\rm C} = 
{\rm adults}\cr
{\rm fertility} & 0 & 0.048 & 0.081\cr
{\rm survival} & 0.94 & 1.00 & 0.75\cr}$$ 
\item{1.}  Express the information given above as a life-cycle graph.
\medskip
\item{} {\it Solution:}  The life cycle graph should show the three groups $A,B,C$, with 
arrows 
\itemitem{$\bullet$}  labeled $0.94$ from $A$ to $B$, 
\itemitem{$\bullet$}  labeled $1.00$ from $B$ to $C$,
\itemitem{$\bullet$}  labeled $.75$ from $C$ to itself,
\itemitem{$\bullet$}  labeled $.048$ from $B$ back to $A$ (new female births from yearling mothers)
\itemitem{$\bullet$}  labeled $.081$ from $C$ back to $A$ (new female births from adult mothers)
\medskip
\item{2.}  Express the information given above as a system of difference equations (time $n$ in years).
\medskip
\item{}  {\it Solution:}  The corresponding difference equations would be
$$\eqalign{A(n) &= .048 B(n-1) + .081 C(n-1)\cr
           B(n) &= .94 A(n-1)\cr
           C(n) &=  B(n-1) + .75 C(n-1)\cr}$$
\item{3.}  If the initial conditions are $A(0) = 100$, $B(0) = 2000$, $C(0) = 3000$, find the 
population of each group in years $n = 1, 2, 3, 4$.  
\medskip
\item{} {\it Solution:}  From the initial values, we compute
$$\eqalign{
A(1) &= (.048)(2000) + (.081)(3000) = 339\cr
B(1) &= (.94)(100) = 94\cr
C(1) &= 2000 + (.75)(3000) = 4250\cr}$$
Then the successive years are computed similarly yielding the following table of 
results (all values rounded to the nearest whole number):
$$\matrix{{\rm year} & A & B & C\cr
             0 & 100 & 2000 & 3000 \cr
             1 & 339 & 94 & 4250\cr
             2 & 349 & 319 & 3282\cr
             3 & 281 & 328 & 2780\cr
             4 & 241 & 264 & 2413\cr}$$
(If you continue for several more years it becomes clear that the overall population is
in a steep decline.)
\bigskip
\noindent
B.  The following data set has $n = 9$:
$23,28,40,44,47,50,51,54,55$
\medskip
\item{1.}  Find the ``5-number'' summary for this
data set.
\medskip
\item{} {\it Solution:}  Min = $23$, $Q1 = 40$ (median of ``lower half'' -- including
the 47 in the middle), Median = $47$, $Q3 = 51$ (median of ``upper half''), Max = $55$
\medskip
\item{2.}  Draw the corresponding box plot.
\medskip
\item{} {\it Solution:}  Show a box from $Q1$ to $Q3$, with the position of the
Median marked by a vertical line, plus ``whiskers'' out to the Min and Max on
either side.  
\medskip
\item{3.}  Compute the (Bowley) measure of skewness.  Does this seem
reasonable from the box plot?
\medskip
\item{} {\it Solution:}  
$${\rm skewness} = {Q3 - 2\times {\rm Median} + Q1\over Q3 - Q1} = 
{51 - 2\cdot 47 + 40\over 51 - 40} \doteq -0.27 $$
The boxplot shows the negative skew since the ``whisker'' on the left
and the distance from $Q1$ to the Median are larger than the 
distance from the median to $Q3$ and the length of the right ``whisker.''
\medskip
\item{4.}  Compute the SD of the data set.  How many of the 
points lie within two SD's of the mean?  Is Chebyshev's Rule
satisfied here?   (Say what that rule says, and determine whether
or not it is satisfied.)
\medskip
\item{} {\it Solution:}  Mean: $\overline{x} \doteq 43.6$, so we will 
compute SD using the formula
$$SD = \sqrt{{(23 - 43.6)^2 + (28 - 43.6)^2 + \cdots + (55 - 43.6)^2\over 8}}$$
(see the class notes for an explanation why it's {\it not} divided
by $9$ in the square root).  This gives an approximate value
$SD = 11.3$.  The interval $\overline{x} \pm 2 SD$ is $(20.9,66.2)$.
{\it All of the data points lie in this range}.  That is consistent with 
Chebyshev's rule, because that says {\it at least 75\%} of the data points
should lie in this interval.  (Note:  Chebyshev's rule is an honest-to-God
rule -- it is {\it always} valid(!))
\bigskip
\noindent
C.  (Short answer) Suppose that a researcher collects 80 individuals of
the Atlantic surf clam.  These clams
can be found at levels down to about a meter in the sand, 
and larger clams tend to live at deeper levels.  
The researcher finds an average shell width 
$10.2$ cm.  Think of this as a sampling process.
\medskip
\item{1.}  What is the population?  What is the sample?
\medskip
\item{} {\it Solution:}  The population is the collection of widths of
all Atlantic surf clam shells.  The sample is the $n = 80$ width measurements
described in the problem.
\medskip
\item{2.}  Is the $10.2$ a statistic or a parameter of the population?
\medskip
\item{} {\it Solution:} It is a {\it statistic} -- something computed from the measured 
widths in the sample.
The parameter would be the population mean -- the average width of all the shells.
\medskip
\item{3.}  Would the researcher {\it know} the population mean in
this circumstance?
\medskip
\item{} {\it Solution:} No.  (Research like this is usually aimed at 
estimating population parameters like the mean shell width, but those
values are never
known exactly.)
\medskip
\item{4.}  What additional information would the researcher need in order
to find a {\it confidence interval} for the population mean?  Describe
how that would be determined and how that would be interpreted.
\medskip
\item{}  {\it Solution:}  The additional information needed to derive
the confidence interval is just the SD of the measured values.  Since
$n =80 \ge 30$, the confidence interval would be computed using 
$$10.2 \pm 1.96\cdot {SD\over \sqrt{80}}$$
\item{5.}   If you knew that reasearcher was being lazy about digging 
and the clams he collected were all taken from sand levels no deeper than
$10$cm, would that be a {\it simple random sample}?  
\medskip
\item{} {\it Solution:} No.  A simple random sample is produced by a 
process that would find any given collection of 80 shell widths from the population 
with equal probability.  Here, the sampling process will be missing
clams from the lower sand levels.  (The results will be skewed toward
smaller clams.)  
\bigskip
\noindent
D.  Suppose that a large data set of air temperature readings is normally 
distributed with  $\overline{x} = 18.6^\circ$C and $SD = .2^\circ$ C.
\medskip
\item{1.}  What would be the $z$-score of a reading of $17.9^\circ$?
\medskip
\item{} {\it Solution:}  $z = {17.9 - 18.6\over .2} = -3.5$.
\medskip
\item{2.}  What temperature reading would correspond to a $z$-score
of $1.4$?
\medskip
\item{} {\it Solution:}  ${x - 18.6\over .2} = 1.4$ when $x = 18.6 + (1.4)(.2) = 18.88$.
\medskip
\item{3.}  Based on this information, if a temperature reading $T$ 
is selected at random from the data set, 
what is the probability that $18.2^\circ \le T \le 18.9^\circ$?
\item{} {\it Solution:}  $x = 18.2 \leftrightarrow z = -2$ and 
$x = 18.9 \leftrightarrow z = 1.5$.  The probability we want is
the area under the standard normal curve from $z = -2$ to $z = 1.5$.
This can be found from the table and the symmetry of the normal
curve:  ${\rm Area}(2.00) + {\rm Area}(1.50) = .4772 + .4332 = .9104$
(about a $91\%$ chance).  
\medskip
\item{4.}  Based on this information, if a temperature reading $T$ 
is selected at random from the data set, what is the probability that $T > 19.0^\circ$?
\medskip
\item{} {\it Solution:} Again we use the table, but since 
$x = 19.0 \leftrightarrow z = 2$, we want the area to {\it the right of $z = 2$}, 
which is $.5 - {\rm Area(2.00)} = .5 - .4772 = .0228$.  
\bigskip
\noindent
E.  Physicians measured the blood lead levels in $373$ bridge
workers employed by painting contractors in eight states.  The
lead levels had $\overline{x} = 27.2$ micrograms per liter of blood,
with an SD of 16.1 micrograms per liter.
\item{1.}  Determine a $95\%$ confidence interval for the 
average lead level in bridge workers.
\medskip
\item{} {\it Solution:}  Since $n = 373$ we use the value 1.96 in the
margin of error formula and we don't need to consult the $t$-table.  
The confidence interval is 
$$\overline{x} \pm 1.96\cdot {SD\over \sqrt{n}} = 27.2 \pm 1.96\cdot {16.1\over \sqrt{373}}
= 27.2 \pm 1.6.$$
In other words, we are ``96\% confident'' that the mean lead level
is between $27.2 - 1.6 = 25.6$ and $27.2 + 1.6 = 28.8$ micrograms per liter.
\medskip 
\item{2.}  A health objective of a federal regulatory agency was the elimination
of blood lead levels of $28$ micrograms per liter or higher for these
workers.  From the evidence given by your confidence interval, does it
seem that that objective was being met?  Explain, by 
describing the way we interpret the meaning of a confidence interval of
this sort.  
\medskip
\item{} {\it Solution:}  The interval we computed in part 1 contains
both numbers less than the target level of $28$ micrograms per liter and numbers greater
than or equal to that.  All of them should be taken as ``believable''
values for the population average lead level, based on the information
in the sample.  So we cannot say that the objective is being met.
(We cannot say for sure that it is {\it not} being met either.)
\bigskip
\noindent
F.  A study shows that a $95\%$ confidence interval for the average
amount $X$ of hazardous waste generated by a single hospital is
$210 \le X \le 260$ (in units of kg/day).  This interval was computed
using the formulas we have discussed. 
\medskip
\item{1.}  What was the sample mean $\overline{x}$ used
to generate this confidence interval?  What was the margin of error?
\medskip
\item{} {\it Solution:}  Because of the way the formulas work, the 
sample mean is always the {\it midpoint} of the interval -- here
$(210 + 260)/2 = 235$ kg/day.  The margin of error is then the distance
from there to either endpoint:  $260 - 235 = 25$ (and $235 - 210 = 25$), also 
in kg/day.
\medskip
\item{2.}  If the sample size was $n = 100$, what was the SD of the
waste amounts in the sample?  
\medskip
\item{} {\it Solution:}  The SD can be found since we would compute the 
margin of error by the usual formula with $n \ge 30$:  $1.96\cdot {SD \over \sqrt{n}}$, So
$$25 = 1.96 \cdot {SD \over \sqrt{100}} \Rightarrow SD = {250\over 1.96} \doteq 125.6$$
\item{3.}  If the sample size was $n = 16$, what was the SD of the
waste amounts in the sample? 
\medskip
\item{} {\it Solution:}  This is similar to part 2, except that now, since
$n < 30$, we need to use the entry from the $t$-table for $n = 16$:
$$25 = 2.132 \cdot {SD\over \sqrt{16}} \Rightarrow SD = {100\over 2.132} \doteq 46.9.$$
\bye
