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\begin{center}
\large\bf
36-303: Sampling, Surveys and Society\\
Exam 2\\
Tue Apr 17, 2012
\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 Two formula sheets are provided  for your
  convenience.}
\item Please write all your answers on the exam itself; your work must
  be your own.
\item If you need more room, continue onto the back of the same page
  as the\\ question you are answering {\em (and let us know that is what you
  are doing!)}.
\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 & 24 & \\ 
  &    & \\ \hline
3 & 18 & \\ 
  &    & \\ \hline
4 & 18 & \\ 
  &    & \\ \hline
5 & 20 & \\ 
  &    & \\ \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, I
% \\
% \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{center}
\large\bf Some Useful Formulas From the Statistics of Survey Sampling, II
% \\
% \normalsize Do not bring this to the exam.  A copy will be provided for you.
\end{center}

\noindent
{\bf Stratified Sampling}

\noindent
Consider $H$ strata with population counts $N = \sum_{h=1}^H N_h$ and
sample counts $n=\sum_{h=1}^H n_h$.  Let $f_h = n_h/N_h$; $W_h =
N_h/N$; and $\bar y_h = \frac{1}{n_h}\sum_{i=1}^{n_h} y_{ih}$ in each
stratum, and let $s_h^2=\frac{1}{n_h-1}\sum_i(y_{ih}-\bar y_h)^2$ be the
sample variance in each stratum.  Then
\begin{eqnarray*}
\overline y_{st} ~ = ~ \sum_{h=1}^H W_h \overline y_h ~~,~~~~
\Var(\overline y_{st}) 
	~ \approx ~  \sum_{h=1}^H W_h^2 (1-f_h) \frac{s_h^2}{n_h} ~~,~~~~
DEFF ~ = ~ \frac{\Var(\overline y_{st})}{\Var(\overline y_{srs})}
	=\frac{\sum_{h=1}^H W_h^2 (1-f_h) \frac{s_h^2}{n_h}}
	{(1-f)\frac{s^2}{n}}
\end{eqnarray*}



\noindent
{\bf Cluster Sampling}

\noindent
Consider a population of $N$ clusters.  We take an SRS ${\cal S}$ of
$n$ clusters, and all units within each sampled cluster (one-stage
clustering).  Assume clusters all have same size $M$.  Let $\bar y_i
= \frac{1}{M}\sum_{j=1}^M y_{ij}$ in each cluster.  Then
\begin{eqnarray*}
  \bar y_{cl} ~ = ~ \frac{1}{n} \sum_{i\in {\cal S}} \bar y_i
~~,~~~~
   \Var(\bar y_{cl}) 
  ~ \approx ~  \left(1-\frac{n}{N}\right) \frac1n s^2_{\bar y_i}
  ~ = ~ \left(1-\frac{n}{N}\right) \frac1n  \left[\frac1{n-1} 
     \sum_{i\in{\cal S}} (\bar y_i - \bar y_{cl})^2\right]
\end{eqnarray*}
and 
\[
 DEFF = \frac{\Var(\bar y_{cl})}{\Var(\bar y_{srs})}
    =  \frac{ M s^2_{\bar y_i}}{s^2_{y_{ij}}}
    \approx 1 + (M-1)\rho 
\]
where $s^2_{\bar y_i}$ is the sample varance of the cluster means,
$s^2_{y_{ij}}$ is the sample variance of the individual observations,
and $\rho$ is the intraclass (intracluster) correlation, or ICC.

\noindent 
{\bf Post-Stratification Weights and Means}

\noindent
As part of survey data collection it is a good idea to get general
demographic information (e.g. in our surveys: sex, age, class, major,
hometown, etc.).  After data collection we compare the proportions in
each of these categories in our sample with the same proportions in
the population. If they agree, great.  If not, calculate
\[
    w_i = (N_h/N)/(n_h/n) \mbox{~for each $i$ in post-stratum $h$}~~,~~~
    \mbox{\em and ~~} \overline y_w = \frac{\sum_i w_i y_i}{\sum_i w_i} 
\]


\noindent
{\bf Post-Stratification Variance Calculations}

\noindent
\ul{\em Taylor series}:
\[
\Var_{TS}(\overline y_w) \approx 
	\frac{1}{\left(\sum_i w_i\right)^2}
	\left[
	\Var(\sum_i w_i y_i) - 2\overline y_w \Cov(\sum_i w_i y_i, \sum_i w_i)
	+ (\overline y_w)^2\Var(\sum_i w_i)
	\right]
\]
where $\bar y_w$ is as above, $\bar w=\frac1n\sum_i w_i$,~~
$\bar{wy}=\frac1n\sum_i w_i y_i$,~~\\
\begin{center}
$\Var(\sum_{i=1}^n w_i) 
	 \approx  n  \cdot \frac{1}{n-1}\sum_{i=1}^n (w_i - \bar w)^2$,~~
$\Var(\sum_{i=1}^n  y_i w_i) 
	~ \approx ~ n \cdot \frac{1}{n-1} \sum_{i=1}^n (w_i y_i - \bar{wy})^2$,~~\\[1em]
% \begin{eqnarray*}
%     \Var(\sum_{i=1}^n w_i) 
%	~ \approx ~ n  \cdot \frac{1}{n-1}\sum_{i=1}^n (w_i - \bar w)^2 
%~~,~~~
%	\Var(\sum_{i=1}^n  y_i w_i) 
%	~ \approx ~ n \cdot \frac{1}{n-1} \sum_{i=1}^n (w_i y_i - \bar{wy})^2
%\\
$	\Cov(\sum_{i=1}^n y_i w_i, \sum_{i=1}^n w_i) 
	 \approx 
	n \cdot \frac{1}{n-1} \sum_{i=1}^n (w_i y_i - \bar{wy}) (w_i - \bar w)
$
%\end{eqnarray*}
\end{center}
\noindent
\ul{\em Jackknife}:
\begin{itemize}
\item Replicate $n$ times (by removing one obs. each time and
  recalculating weights):
\[
	\overline y_w^{(r)} = \frac{\sum_{i=1}^n w_i^{(r)}
          y_i^{(r)}}{\sum_{i=1}^n w_i^{(r)}} 
\]
\item Calculate
\begin{eqnarray*}
	\bar y_{JK} ~ = ~ \frac1n\sum_{r=1}^n \bar y_w^{(r)}
~~,~~
	Var_{JK}(\bar y_w) 
	~ \approx ~ \frac{n-1}{n} \sum_{r=1}^n (\bar y_w^{(r)} - \bar y_{jk})^2
\end{eqnarray*}
\end{itemize}

}


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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\pts{5} Let $Y_i$ be the number of monthly neighborhood watch
  meetings attended by the $i^{th}$ resident in a neighborhood, in the
  last year.  You are going to conduct a survey, using a SRS to
  estimate $\bar Y_{pop}$, the population mean number of meetings
  attended by neighborhood residents.  Among the $N_R$ residents who
  {\em would respond} to your survey, the mean number of meetings is
  $\bar Y_R$, and among the $N_M$ number of residents who would {\em
    not respond}, the mean number of meetings attended is $\bar Y_M$.
  In class we showed that the {\em bias} between $\bar Y_R$ and $\bar
  Y_{pop}$, due to missing responses, is
\[
    \bar Y_R - \bar Y_{pop} = \frac{N_M}{N} (\bar Y_R - \bar Y_M)
\]
where $N = N_R + N_M$.  Assume that $\bar Y_R\neq\bar Y_M$.\\[-0.75em]

Which statement below is false (or, circle iv. if all are OK)?
\begin{enumerate}\small
\item The more people who respond to the survey, the smaller the bias
  due to missing responses.
\item The bigger the difference between mean number of meetings
  attended by nonresponders, vs the mean number attended by
  responders, the bigger the bias due to missing responses.
% \item The larger your SRS, the better you can estimate $\bar
% Y_{POP}$.
\item The larger the fraction of nonrespondents, the bigger the bias
  due to missing responses.
\item All of the above statements are true.
\end{enumerate}

\item\pts{5} Suppose we divide a sampling frame into groups, which we
  may treat as either strata for stratified sampling, or clusters for
  cluster sampling.  If we make the groups so that {\em observations
    {\bf within groups} are more {\bf similar} to each other}, and
  {\em observations {\bf between groups} are more {\bf different} from
    each other}, then, all other things being equal, we expect
\begin{enumerate}\small
\item The variance of the stratified sample mean $\bar y_{st}$ will go
  {\bf down} and the variance of the cluster sample mean $\bar y_{cl}$
  will go {\bf up}.
\item The variance of the stratified sample mean $\bar y_{st}$ will go
  {\bf up} and the variance of the cluster sample mean $\bar y_{cl}$
  will go {\bf down}.
\item Both variances will go {\bf down}.
\item Both variances will go {\bf up}.
\end{enumerate}

\vfill

\hfil{\sl[Continued on next page\ldots]}

\clearpage

\item\pts{5} In one-stage clustered sampling, the ICC $\rho$ measures
\begin{enumerate}\small
\item The correlation between observations in different clusters.
\item The correlation between the cluster means of different clusters.
\item The correlation between observations in the same cluster.
\item The correlation between the cluster mean and the individual
  observations in the cluster.
\end{enumerate}

\item\pts{5} Which of the following is {\em not} a usual part of
  post-survey processing?
\begin{enumerate}\small
% \item Data entry
\item Coding short-answer or text data
%\item Sample size calculation
\item Variance calculation
\item Imputation
\item Checking post-strata and building weights if needed
\item All of the above {\em are} usually part of post-survey processing!
\end{enumerate}



\end{enumerate}

\vspace*{\fill}

\begin{center}
\bf[This space intentionally left blank]
\end{center}

\vspace*{\fill}

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\clearpage

\item\pts{24} % based on #19, pp108-9, Lohr 2nd ed
%
Many universities in the United States stay open during winter break;
closing during break would save some salary and maintenance costs, but
would be inconvenient for faculty and staff who want to continue
working while students are gone.  In January 1995, the Office of
University Evaluation at Arizon State University surveys faculty and
staff members to find out their reaction the closure of the university
during Winter Break 1994.  Four strata were identified, and surveys
were distributed to an SRS (without replacement) in each stratum.  The
table below gives the results for one of the survey questions, ``{\em
  Would you want to have the university closed during winter break in
  future years?}''.

{
\def\z{\phantom{0}}
\hspace*{-4em}
\begin{tabular}{clccccc}
Stratum & Employee & Population & No. of Surveys  & Number of &
Number of & \\
Number $(h)$ & Type     & Size       & Distributed & Responses
& Yes's & $\bar y_h$\\
\hline
\hline
1     &  Faculty     & 1374       & 500  & 232 & 167 & 0.72\\
2     &  Classified Staff & 1960  & 653  & 514 & 459 & 0.89 \\
3     &  Administrative Staff & \z252 & \z74 & \z67 & \z58 & 0.87 \\
4     & Academic Professional  & \z\z95  & \z95 & \z86 & \z75 & 0.87 \\
\hline
      &                        & 3681 & 1322 & 899 & 759
\end{tabular}
}
\begin{enumerate}

\item\pts{5}\label{srs} Ignoring the strata and treating this as an
  SRS without replacement of size 899, calculate $\bar y_{srs}$ and
  $\sqrt{\Var(\bar y_{srs})}$, the SRS estimate and SE of the
  proportion of faculty and staff that responded ``Yes''.  {\em HINT:
    Use the formula $\hat p(1-\hat p)$ where $\hat p = \bar y_{srs}$,
    in your variance calculation.}
 
\vfill

\hfil{\sl[Continued on next page\ldots]}

\clearpage

\item\pts{9}\label{strat} Now treat this as a pre-stratified sample,
  and calculate $\bar y_{st}$ and $\sqrt{\Var(\bar y_{st})}$, the
  stratified estimate and SE of the proportion of faculty and staff
  that responded ``Yes''.  {\em [HINT: Use the number of responses as
      the sample size within each stratum; and use the same idea as in
      part (\ref{srs}) for the needed variance calculations.]}

\clearpage

\item\pts{4} Calcuate the DEFF for the stratified design and comment
  on whether it was worthwhile to stratify.

\vfill

\item\pts{6} Clearly not everyone who received a survey responded.
\begin{itemize}
\item  Calculate the response rates in each stratum. 
\item Would you expect the bias due to nonresponse to be greater or
  less for the SRS estimates in part (\ref{srs}), vs the stratified estimates
  in part (\ref{strat})?
\end{itemize}

\vfill

\end{enumerate}


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\clearpage

\item\pts{18} % Lohr 2nd ed #1 pp 207-209 
\def\metro{Chicago } % was Phoenix...
% 
The city council of a suburb of \metro wants to know the proportion
of eligible voters that oppose having a \metro garbage incinerator
opened in that suburb.  They randomly select 100 residential phone
numbers from the suburb's telephone book (which contains 3,000
residential numbers in all).  Each selected residence is then called
and asked for (a) the total number of eligible voters in that
household and (b) the number of voters in the household opposed to the
incinerator.  A total of 157 voters were surveyed; of these, 23
refused to answer the question.  Of the remaining 134 voters, 113
opposed to the incinerator, so the council estimates the proportion
opposed as 
\[
   \hat p = 113/134 = 0.8432836
\]
with 
\[
   \Var(\hat p) = 0.8432836(1-0.8432836)/134 = 0.0009862416
\]

\begin{enumerate}

\item\pts{6} What is the {\bf target population}?  What is the {\bf
  sampling frame}?  {\em You do not have to use all the space provided.}
% Will there be much {\em coverage error} with this
%   sampling frame?  {\em Why or why not?}

\vfill

\hfil{\sl[Continued on next page\ldots]}

\clearpage

\item\pts{6} A statistics intern working for the city council argues
  that this is a cluster sample.  Interpreting this as a cluster
  sample, 

~\\
\begin{itemize}
\item The psu's (primary sampling units) are the \rule{2in}{0.01in}~.

~\\
\item The ssu's (secondary sampling units) are the \rule{2in}{0.01in}~.
\end{itemize}

\item\pts{6} Are the estimates $\hat p$ and $\Var(\hat p)$ given above
  valid?  Why or why not?  {\em (you do not have to use all the space
    provided)}


\end{enumerate}



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\clearpage

% Groves 2nd ed p 367 #1, sort of
\item\pts{18}
You are examining the data entered on an item measuring the number of
miles driven per year in a travel survey about automobile trips.  You
encounter three cases from the total sample of 20,000:
\begin{itemize}
\item One case reported 17,500 miles driven, but 1750 was entered
\item One case reported 17,599 miles driven, but 17,588 was entered
\item One case reported 17,599 miles driven, but 15,799 was entered
\end{itemize}
In a separate analysis, it was found that the mean number of miles
driven should be around 15,000.

Give at least one strength and one weakness of each of the following
strategies for identifying and correcting such coding errors.

\begin{enumerate}

\item\pts{6} Examining the distribution of the data as it was
  entered, and identifying and eliminating any outliers.

\begin{itemize}
\item {\bf Strength:}

\vfill

\item {\bf Weakness:}

\vfill

\end{itemize}

\hfil{\sl[Continued on next page\ldots]}

\clearpage

\item\pts{6} Drawing a 10\% sample of the 20,000 responses, and
  identifying and correcting any errors found.

\begin{itemize}
\item {\bf Strength:}

\vfill

\item {\bf Weakness:}

\vfill

\end{itemize}

\item\pts{6} Examining all 20,000 responses and correcting any errors
  found.

\begin{itemize}
\item {\bf Strength:}

\vfill

\item {\bf Weakness:}

\vfill

\end{itemize}

\end{enumerate}



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\clearpage

\item\pts{20} {\em Imputation methods.}

\begin{enumerate}

% groves, p 368 \#4
\item One method of imputation for missing responses to
  individual survey items is {\em mean imputation}.
\begin{enumerate}
\item\pts{4} Explain briefly how mean imputation works.

\vfill

\item\pts{3} Under what assumption (MCAR, MAR, MNAR) is mean imputation
  OK?  (Choose one, no explanation needed.)\\

\item\pts{3} Identify a possible problem with mean imputation.

\vfill

\end{enumerate}

\hfil{[continued on next page]}

\clearpage

\item Another method of  imputation for missing responses is {\em
  regression imputation}. 
\begin{enumerate}
\item\pts{4} Explain briefly how regression imputation works.

\vfill

\item\pts{3} Under what assumption (MCAR, MAR, MNAR) is regression imputation
  OK?  (Choose one, no explanation needed.)\\

\item\pts{3} Identify a possible problem with regression imputation.

\vfill

\end{enumerate}

\end{enumerate}


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\comment{

\clearpage

% Examples from lecture 09...

\item\pts{??} Another 36-303 group wants to do a survey of PAT bus
  riders' attitudes toward a proposed schedule of bus fare increases.
  In their survey proposal they specify the {\em target population} to
  be all adult residents of Pittsburgh's East End (Oakland, Shadyside,
  Squirrell Hill \& Point Breeze).  They plan to collect data by
  approaching people waiting at six specific bus stops near where
  these students live, between 8am and 10am for one whole work-week
  (Mon-Fri), for face-to-face interviews.
\begin{enumerate}

\item\pts{??} Did this group specify a {\em sampling frame} for their
  survey?  If so, say what it is.  If not, specify a sampling frame
  that the group could work with.

\vspace*{\fill}

\item\pts{??} Write down two very likely sources of {\em coverage
  error} for this survey as currently proposed.

\vspace*{\fill}

\item\pts{??} Make one big suggestion that this group could implement
  to improve their sampling plan.

\vspace*{\fill}

\end{enumerate}

}

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\end{enumerate}

\end{document}

