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\textbf{College of the Holy Cross, Spring Semester, 2018} \\
\textbf{MONT 101N -- Analyzing Environmental Data, Final Exam Solutions} \\
\textbf{Tuesday, May 15}
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\vskip 10pt
\noindent
I. {\it Briefly} identify \emph{any 4} of the following.
If you submit answers for more than 4, the best 4 will be counted.  

\begin{enumerate}
\item[(A)] (5) The Central Limit Theorem
\item[]  \emph{Solution:}   The Central Limit Theorem says that if independent and identically 
distributed values $Y_i$, $i = 1, \ldots, N$ are sampled from any distribution with finite mean
$\mu$ and SD $\sigma$, then the distribution of the \emph{sample means} 
$$\overline{Y} = \frac{1}{N}(Y_1 + \cdots + Y_N)$$
becomes more and more normal as $N \to \infty$.  Precisely, the distribution of 
$$\frac{\sqrt{N}(\overline{Y} - \mu)}{\sigma}$$
approaches a standard normal distribution as $N\to \infty$.
\item[(B)] (5) A $95\%$ confidence interval for a population mean
\item[] \emph{Solution:}  The $95\%$ confidence interval (for a sample size $N > 30$)
is the interval 
$\overline{Y} \pm 1.96 \frac{S}{\sqrt{N}}$.  The idea is that if we repeat the sampling process many
times, we would expect about 95\% of the intervals constructed this way to 
contain the corresponding population mean.  If $N < 30$, then the factor $1.96$ would be replaced
by the $.025$ percentage point for the $t$-distribution with $N - 1$ degrees of freedom.       
\item[(C)] (5) A $t$-distribution
\item[] \emph{Solution:} When sampling from a normal population with mean $\mu$, 
the $t$ distribution describes the ``standardized sample means'' where the population $\sigma$
is estimated by the sample $S$:
$\displaystyle\frac{\overline{Y} - \mu}{S/\sqrt{N}}$.  This gives the $t$-distribution
with $N - 1$ degrees of freedom.  
\item[(D)] (5) Standard normal distribution
\item[] \emph{Solution:}  The normal distribution with $\mu = 0$ and $\sigma = 1$.  
\item[(E)] (5) The box-and-whisker plot of a data set.  
\item[] \emph{Solution:}  A plot showing the minimum, first quartile, median, third quartile
and maximum of a data set.  The ``box'' is drawn between the first quartile and the third quartile;
the whiskers are thinner lines extending to the left and right (or down and up) to the minimum
and maximum, respectively.   
\item[(F)] (5) The $p$-value of a test of hypotheses
\item[] \emph{Solution:}
The $p$-value is the \emph{smallest value of the Type I error 
probability $\alpha$ for which the null hypothesis would be rejected with the observed
data.}  In more intuitive terms, it is the 
chance that the observed test statistic, or a ``more extreme''
value, would be observed from a sample \emph{if the null hypothesis is true}.  The
smaller the $p$-value is, the stronger the evidence is for rejecting the null hypothesis.  
\end{enumerate}
\vskip 10pt
II.  Short answer
\begin{enumerate}
\item[(A)] (5)  If you want to get polling results with comparable accuracy, do you need a bigger
sample size if the polling is done in Texas than you do if the polling is done
in Delaware?  Explain by using formula for the standard error for a percentage.
\item \emph{Solution:} The answer is NO.  The formula standard error (SE) for the
percent shows why.  The SE for the percent  is always bounded above by 
$\sqrt{\frac{(.5)(.5)}{N}}$ -- the SE for the 
percent depends only on the sample size, not on the size of the whole population.
(This \emph{is assuming} that the size of the population is
enough larger than the size of the sample that the difference between sampling with 
replacement and sampling without replacement can be ignored, but that would be 
true for any population the size of a U.S. state.)
\item[(B)] (5)  True/False and explain:  If the $p$-value of a hypothesis test works
out to $p = .034$ then we can say that the null hypothesis was \emph{definitely false}.  
\item[] \emph{Solution:}  This is FALSE.  The $p$-value gives the probability of 
getting the observed test statistic, or something even more extreme, under the 
assumption the null hypothesis is true.  We cannot say the null hypothesis is
definitely false even with a $p$-value as small as $p = .034$.  
\end{enumerate}
\noindent
III.  Let $X$ represent a normal random variable with $\mu = 4.3$ and $\sigma = 1.4$.  
\begin{enumerate}
\item[(A)] (10) Find $P(3.2 < X < 5.0)$
\item[] \emph{Solution:}  This probability is the same as 
$$P\left(\frac{3.2 - 4.3}{1.4} < Z < \frac{5.0 - 4.3}{1.4}\right) = P(-0.79 < Z < 0.5) = .2852 + .1915 = 
.4767$$
\item[(B)] (10) Suppose that values of $X$ are sampled $n = 10$ times independently.
Give the \emph{formula for computing} the probability that exactly $5$ out of the $10$ values satisfy 
$3.2 < X < 5.0$.  (The number the correct formula gives is roughly $.2434$ 
if you want to check your answer by computing the value.)
\item[] \emph{Solution:}  This is a binomial experiment.  The probability that
exactly $5$ out of the $10$ values are in the range is
$$\binom{10}{5} (.4767)^5 (1-.4767)^5 =  252 (.4767)^5 (.5233)^5 \doteq .2434 $$
\end{enumerate}
\vskip 10pt
\noindent
IV.  A Gallup poll released on April 3, 2017 surveyed a random sample of 706 adults nationwide.
It reported ``that $59\%$ of those sampled say the environment should be prioritized over energy 
production.''  
\begin{enumerate}
\item[(A)] (20) Find the 95\% confidence interval for the actual percentage of Americans
who feel this way.  (Use the conservative method for estimating the standard error.)
\item[] \emph{Solution:}  The confidence interval is 
$$59\% \pm 2 (\frac{\frac{1}{2}}{\sqrt{706}} \times 100\%) = 59\% \pm 3.8\%$$
This extends from $55.2\%$ to $62.8\%$
\item[(B)] (5) Based on your answer to part (A), is $55\%$ a ``believable'' value for the
percentage of all Americans who feel that way?
\item[] \emph{Solution:}  \emph{Probably not}.  The lower endpoint of the confidence 
interval is $55.2 > 55$, so $55$ lies outside the most probable range of values
for the population percentage.    
\end{enumerate} 
\vskip 10pt
\noindent
V. (25) A study was conducted by the Florida Game and Fish Commission to assess the amounts
of chemical residues found in the brain tissue of brown pelicans.  In a test for the 
pesticide DDT, random samples of $N_1 = 10$ juveniles (young birds foraging independently)
and $N_2 = 13$ nestlings (birds still being fed by their parents) produced the following
data (DDT measurements in parts per million).  Test the alternative hypothesis that mean amounts
of DDT found in juveniles and nestlings differ versus the null hypothesis that they do not
differ.  Use $\alpha = 0.05$.  
\[
\begin{array}{cc}\hline
\text{Juveniles} &  \text{Nestlings}\\ \hline
       N_1 = 10  &  N_2 = 13\\
       \overline{Y}_1 = .041 & \overline{Y}_2 = .026\\
             S_1 = .017 & S_2 = .006\\
\end{array}
\]
(Notes: This data has important implications regarding the accumulation
of chemical residues over time. You may assume that the population SD's are equal to the sample
SD's in deciding how to set up the test.)
\vskip 10pt
\noindent
\emph{Solution:}  Since $N_1 = 10$ and $N_2 = 13$ are both less than 30, we want to use
the ``small sample'' test for equality of population means from the fifth line in our
summary table about hypothesis tests.  We have
\begin{itemize}
\item $H_0:$  $\mu_1 = \mu_2$
\item $H_a:$  $\mu_1 \ne \mu_2$ (a two-tail test, indicated by the way the question is phrased:
`` Test the alternative hypothesis that mean amounts
of DDT found in juveniles and nestlings differ versus the null hypothesis that they do not
differ''
\end{itemize}
The test statistic will have a $t(9)$ distribution (the number of degrees of freedom
is $9 = $ the smaller of $10 - 1$ and $13 - 1$), so 
the rejection region for the two-tail test will be $t < -2.26126$ and $t > 2.26126$ (from the 
$.025$ column since the $\alpha = .05$ is split even over the upper and lower tails).  

We compute 
$$t = \frac{.041 - .026}{\sqrt{\frac{(.017)^2}{10} + \frac{(.006)^2}{13}}} \doteq 2.665$$
This is in the rejection region, so we reject the null hypothesis that there is no
difference in the DDT levels in the brains of the juvenile birds and the nestlings.  

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