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\begin{center}
\large\bf
36-303: Sampling, Surveys and Society\\
Exam 1\\
Thu Feb 18, 2010
\end{center}

\vspace*{0.5in}

\begin{itemize}\parsep=0pt\parskip=0pt\itemsep=0pt
\item You have 80 minutes for this exam.
\item The exam is closed-book, closed notes.
\item A calculator is allowed.
\item {\bf A formula sheet is provided on the next page for your convenience.}
\item Please write all your answers on the exam itself; your work must
  be your own.
\end{itemize}

\vspace*{0.1in}


\begin{center}
\begin{tabular}{|c|c|c|}
\hline
\bf Question & \bf Points Possible & \bf Points Earned \\ \hline
1 & 20 & \\ 
  &    & \\ \hline
2 & 18 & \\ 
  &    & \\ \hline
3 & 18 & \\ 
  &    & \\ \hline
4 & 20 & \\ 
  &    & \\ \hline
5 & 24 & \\ 
  &    & \\ \hline
Total & 100 & \\ 
  & & \\ \hline
\end{tabular}
\end{center}

\vspace*{\fill}

\noindent
Name: ~~~~~\rule{4.25in}{0.1pt}

\vspace*{\fill}

\noindent
Signature: \rule{4.25in}{0.1pt}

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\begin{center}
\large\bf Some Useful Formulas From the Statistics of Survey Sampling
% \\
% \normalsize Do not bring this to the exam.  A copy will be provided for you.
\end{center}

\noindent
{\bf Equally-Likely Outcomes \& Counting}

\begin{itemize}

\item If $K$ outcomes $O_1, \ldots, O_K$ are equally likely, then the
  probability of any one of them is $1/K$. 

\item Consider taking a sample of $n$ objects from a population of $N$
  objects. 
\begin{itemize}
\item Sampling with replacement, there are $N^n$ possible samples of
  size $n$; the probability of any one of them is $1/N^n$.
\item Sampling without replacement, there are 
$
	{N \choose n} = \frac{N!}{n!(N-n)!}
$ 
possible samples of size $n$ [where
$N!=N\cdot(N-1)\cdot(N-2)\cdots3\cdot2\cdot1$], so the probability of
any one of them is $1\left/{N\choose n}\right.$.
\end{itemize}

\end{itemize}

\noindent
{\bf Discrete Random Variables}

Let $X$ and $Y$ be random variables with sample spaces
$\{x_1,\ldots,x_K\}$ and $\{y_1,\ldots,y_K\}$ and distributions
\begin{eqnarray*}
  P[X=x_i,Y=y_j] ~ = ~ p_{ij} ~~,~~~
  P[X=x_i]      ~  = ~ p_{i\cdot} ~ = ~ \sum_{j=1}^K p_{ij} ~~,~~~
  P[Y=y_j]       ~ = ~ p_{\cdot j} ~ = ~ \sum_{i=1}^K p_{ij}
\end{eqnarray*}
Then, for example
\begin{eqnarray*}
  E[X] ~ = ~ \sum_{i=1}^K x_i p_{i\cdot} ~~,~~~
  Var(X) ~ = ~ \sum_{i=1}^K (x_i  - E[X])^2p_{i\cdot} &,~&
  Cov(X,Y) ~ = ~ \sum_{i=1}^K (x_i  - E[X])(y_i-E[Y])p_{ij}\\
  P[X=x_i|Y=y_j] ~ = ~ p_{ij}/p_{\cdot j}~~,~~~
  E[X|Y=y_j] ~ = ~ \sum_{i=1}^K x_i P[X=x_i|Y=y_j] &,~&
  E[aX + bY + c] ~ = ~ aE[X] + bE[Y] + c
\end{eqnarray*}

\noindent
{\bf Random Sampling From a Finite Population}

Consider a population of size $N$ and a sample of size $n$.  Let $y_i$
be the (fixed) values of some variable of interest in the population
(such as a person's age, or whether they would vote for Obama).  Let
\[
	Z_i = \left\{\begin{array}{l}
        1, \mbox{~if $i$ is in the sample}\\
        0, \mbox{~else}
        \end{array}\right.
\]
be the random sample inclusion indicators, and let $Y_i$ be the random
observations in the sample.  Then the sample average can be written
% \def\cond{\parbox{2.9em}{\scriptsize\em $i$ is in the sample}}
\[
    \bar Y ~ = ~ \frac1n\sum_{i=1}^n Y_i 
    ~ = ~ \frac1n\sum_{i=1}^N Z_i y_i
\]
The $Z_i$'s are Bernoulli random variables with
\begin{eqnarray*}
  E[Z_i] ~ = ~ \frac{n}{N} ~~,~~~ Var(Z_i) =
  \frac{n}{N}\left(1-\frac{n}{N}\right) 
  ~~,~~~ Cov(Z_i,Z_j) = - \frac{1}{N-1} \frac{n}{N}\left(1-\frac{n}{N}\right) 
\end{eqnarray*}

\noindent
{\bf Confidence Intervals and Sample Size}

\begin{itemize}

\item[(a)] A CLT-based $100(1-\alpha)\%$ confidence interval for the
  population mean is $(\bar Y - z_{\alpha/2}SE~,~~\bar Y +
  z_{\alpha/2}SE)$.
\item[(b)] For sampling with replacement from an infinite population, $SE =
  SD/\sqrt{n}$.
\item[(c)] For sampling without replacement from a finite population, the
  SE has to be multiplied by the finite population correction (FPC).
\item[(d)] For a given margin of error (ME, half the width of the CI) and
  confidence level $1-\alpha$, we can find the sample size by solving
\[
      z_{\alpha/2}SE < ME
\]
for $n$.  The same approach works for both SRS with replacement (using
the SE in (b)) and SRS without replacement (using the SE in (c)).

\end{itemize}
% {\em You should memorize the formula for the FPC, and the two formulas
%   for sample size calculations (one for sampling with replacement, one
%   for sampling without replacement).}

}

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\begin{enumerate}

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\clearpage

\item\pts{20} {\em Multiple Choice (4 parts).  For each part, circle the
  roman numeral of the one best answer.}

\begin{enumerate}

\item\pts5 Which of the following statements is \ul{\bf most correct}:
\begin{enumerate}

\item If you do not have a random sample, it cannot be a
  representative sample.

\item The only valid random sampling for surveys is sampling without
  replacement.

\item It is possible to construct a representative sample without
  random sampling, but it is more difficult to argue that it is really
  representative.

\item If the sample is random, it is representative, regardless of the
  response rate.

\end{enumerate}

\item\pts5 Requiring researchers to obtain {\em Informed Consent} from
  participants in surveys and other human subjects research is
  considered an ethical obligation (and sometimes a legal one).  Which
  one of the following is the \ul{\bf best} justification for informed
  consent for survey participants, according to our textbook:
\begin{enumerate}

\item To protect respondents from harm.

\item To give respondents meaningful control over information about
  themselves.

\item To give respondents a chance to understand the research they
  will be participating in.

\item To make sure that the respondents benefit from the survey by
  getting a copy of the final report.

\end{enumerate}

\item\pts5 Two important fractions in sample surveys are the {\em sampling
  fraction} $n/N$ and the {\em response rate} $r/n$ (where $N$ is the
  population size, $n$ is the intended sample size, and $r$ is the
  number in the sample that actually responded).  Which of the
  following is \ul{\bf not} true, for a simple random sample with
  replacement from the target population?

\begin{enumerate}

\item You can get a more representative sample by increasing $n$,
  regardless of the response rate. 

\item To decrease variability in sample estimates, increase the
  sampling fraction.

\item To decrease possible bias in sample estimates, increase the
  response rate.

\item You can force the standard error of sample estimates to be zero
 by making the sampling fraction large enough.

\end{enumerate}

\item\pts5 When making a public report on a survey, which of the following
  is \ul{\bf not} required?
\begin{enumerate}

\item Who sponsored it, who carried it out.

\item Sample size and precision (SE) of estimates.

\item The name of the statistical package used to do the analyses.

\item Target population, sampling frame, sampling method, response
  rates.

\end{enumerate}

\end{enumerate}

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\clearpage

\item\pts{18} According to the blog {\tt http://nullspace2.blogspot.com}
  (about economic issues in the greater Pittsburgh area), there are 
  2,505 full-time police officers employed by all Allegeny local
  governments (the county government, the city of Pittsburgh, and
  surrounding municipalities such as Monroeville, Mount Lebanon,
  Wilkinsburg, etc.).  Your survey team is going to survey police
  attitudes toward their work.  One of your survey questions is
\begin{quote}
\em Should persons taken into custody always be told their Miranda
rights (`You have the right to remain silent; anything you say can and
will be used against you in a court of law [etc.]') ~~~ Yes~/~No''.
\end{quote}
Answer the following questions (3 parts).
\begin{enumerate}
\item\pts6 Let $p$ be the proportion of police officers that 
  would say yes.  Assuming you are doing SRS without replacement, how
  large must your sample size be, to estimate $p$ within a margin of
  error of $\pm0.07$, with a 95\% confidence interval?

\vspace*{\fill}

\clearpage

\item\pts6 It turns out you only have enough resources (time and money
  to collect data and process the responses, mainly) to do a phone
  survey of 50 police officers, chosen at random from a complete list
  of all 2,505 officers.  Amazingly, all 50 officers reply, and of
  those, 20 respond ``yes'' to the question above.  Compute a 95\%
  confidence interval for $p$.

\vspace*{\fill}

\item\pts6 Now suppose that you call a random sample of 200 officers
  from the complete list of 2,505 officers, but only 50 agree to be
  interviewed over the phone.  Of those 50, 20 respond ``yes''.  You
  calculate the same confidence interval as in part (b).  \\
% 
%   Compute the 95\% confidence interval for $p$.
% 
% \vspace*{\fill}
%
%\item\pts5 

Which of the confidence intervals (this one, or the one in part (b))
is harder to believe as evidence about the whole population of
Allegheny full-time police officers?  Why?

\vspace*{\fill}

\end{enumerate}

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\clearpage

\item\pts{18} {\em Simple Random Sampling (3 parts).}

\begin{enumerate}


\item\pts6 Carefully define {\em Simple Random Sampling (SRS) \ul{\bf with}
  replacement}, and give an example.  Give enough detail that it is
  obvious that this is a good example.

\vspace*{\fill}

\item\pts6 Carefully define {\em Simple Random Sampling (SRS) \ul{\bf without}
  replacement}, and give an example.  Give enough detail that it is
  obvious that this is a good example.

\vspace*{\fill}

\clearpage

\item\pts6 Suppose we take a SRS without replacement of $n$
  individuals from a population of $N$.  Let $y_i$ be the height of
  the $i^{th}$ person.  Let
\[
	\bar Y ~ = ~ \frac1n\sum_{i=1}^n Y_i ~ = ~ \frac1n\sum_{i=1}^N Z_iy_i
\]
be the sample average height, and  let
\[
        \mu ~ = ~ \frac1N\sum_{i=1}^N y_i
\]
be the mean height in the population.  Show that $\hat Y$ is an unbiased
estimator for $\mu$.  In other words, show that
\[
	E[\hat Y] ~ = ~ \mu ~ .
\]

\end{enumerate}

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\clearpage

% Groves, problem 11.3, p 375

\item\pts{20} As part of their course assignment, undergraduate
  students in introductory psychology classes are sent to various
  areas in New York City to ask 1,520 passersby one of a variety of
  simple requests.  They are told to ask for different kinds of help
  (e.g. asking for directions, asking for change for a \$1 bill, etc.)
  and to ask for it in different ways.  The varying responses to their
  requests provide a first approximation to answering questions about
  the prevalence of ``altruistic compliance'' (roughly, willingness to
  help with a stranger's requests) and the factors influencing it.

Answer the following questions (3 parts).
\begin{enumerate}

\item\pts6 What, if any, ethical principles do you think this study
  violates? (circle the roman numeral of the one best answer)

\begin{enumerate}

% \item Avoiding Fabrication, Falsification, or Plagiarism

\item Beneficence

\item Justice

\item Respect for Persons; Confidentiality 

\item Informed Consent

\item None of the above; it's just fine.

\item Not listed above; I think~ \rule{3.8in}{1pt}\\[2em]
\rule{5.4in}{1pt}~.

\end{enumerate}

\item\pts8 Do you think the results of this survey will be
  representative of the degree of altruistic compliance of all New
  Yorkers? 
\begin{itemize}
\item[]\rule{1em}{1pt} {\bf Yes, representative.}  If you select this
  answer, use the space below to explain why it is representative.
\item[]\rule{1em}{1pt} {\bf No, not representative.}  If you select
  this answer, use the space below to suggest one or more
  modifications of the survey design that would make it more
  representative.
\end{itemize}
{\em Be clear and complete, not just a word or two.}  Continue onto
the next page if you need to.

\vspace*{\fill}

\clearpage

\item[] {\em (more room for your answer to (b))}

\vspace*{\fill}

\item\pts6 Sometimes a study that is not perfect can still be useful.
  List one or two possible advantages of doing the study as designed,
  rather than modifying it in any way.  {\em Be clear and complete,
    not just a word or two.}

\vspace*{\fill}



\end{enumerate}


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\clearpage

\item\pts{24} Below are several survey questions.  For each question:
  (i) indicate a potential problem with the question {\em using
  specific ideas from the lecture notes on question writing}; (ii)
  suggest a way to rewrite it (as one or more questions, by providing
  more information, by improving grammar, etc.) that gets at the same
  thing while avoiding the problem you raised.

\def\ans{
\begin{enumerate}
\item\pts3 A Potential Problem:
\vspace{1.25in}
\item\pts3 Suggestion(s) For Rewrite:
\vspace{1.5in}
\end{enumerate}
}

\begin{enumerate}

\item {\em ``Most people think that abortion is immoral.  Do you agree?''}  
\ans

\item {\em ``Do you agree or disagree that NCLB should be abolished
  and replaced with merit pay for teachers?''}  
\ans

\item {\em ``How many times have you taken a PAT bus to go shopping
  since the start of classes at CMU in August? Answer here:
  \rule{3in}{1pt}''} \ans

\item {\em ``What kind of exercise do you do?
\begin{itemize}
\item[]\rule{1em}{1pt} Walking, jogging or running
\item[]\rule{1em}{1pt} Exercise machines or weights''
\end{itemize}
}
\ans

\end{enumerate}

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\end{enumerate}

\end{document}

