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\begin{center}
\large\bf
36-303: Sampling, Surveys and Society\\
Exam 2\\
Tue Apr 12, 2011
\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 & 24 & \\ 
  &    & \\ \hline
2 & 26 & \\ 
  &    & \\ \hline
3 & 24 & \\ 
  &    & \\ \hline
4 & 26 & \\ 
  &    & \\ \hline
% 5 & 18 FIX & \\ 
%   &    & \\ \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{24} {\em Multiple Choice (4 parts).  For each part, circle the
  roman numeral of the one best answer.}

\begin{enumerate}

\item\pts{6} Which of the following is {\em not} a usual part of
  post-survey processing?
\begin{enumerate}\small
\item Data entry
\item Sample size 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}


\item\pts{6} 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 up} and the variance of the cluster sample mean $\bar y_{cl}$
  will go {\bf down}.
\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 Both variances will go {\bf up}.
\item Both variances will go {\bf down}.
\end{enumerate}

\item\pts{6} Which of the following is {\em not} one of the recommended
  things to work on, to reduce the tendency of survey subjects to not
  respond?
\begin{enumerate}
\item  Followup.
\item Choice of stratified or cluster sampling.
\item Amount of effort it takes respondents to undeerstand/respond to
  questions.
\item Assurance of confidentiality, especially for sensitive
  questions.
\end{enumerate}

\item\pts{6} Weights can be calculated and applied to individual
  observations for a variety of reasons.  Circle the reason below that
  is {\em not} appropriate.
\begin{enumerate}
\item Weights may be calculated in designing a stratified sample
  designs.
\item Weights may be calculated in designing certain kinds of surveys
  in which not every respondent has an equal chance of being selected.
\item Weights may be calculated after the survey to compensate for
  some kinds of informative (non-ignorable) missingness.
\item Weights may be calculated after the survey to adjsut sample
  proportions in various post-strata to equal the population
  proportions.
\item All of the above {\em are} appropriate reasons to compute weights! 
\end{enumerate}



\end{enumerate}

\comment{

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

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\item\pts{26} {\em Cluster sampling}.
% From Groves, pp 142--143, \#10, but modified since (a) it's a bit
% long, and (b) it's not clear how to deal with the ``assumption of
% simple random sampling of households'' in part (a) of his question.
A survey is conducted to find out the proportion of cell phone users
in a certain city.  From a population of 2500 residential blocks, 10
are sampled at random without replacement, and each person residing on
that block is asked whether they use a cell phone.  We will assume
there are exactly 40 people living in each block\footnote{This is a
  reasonable approximation as long as all the blocks have close to 40
  residents.}.  This yields the following table of data:
\begin{center}
\begin{tabular}{cccc}
\hline
        & Total \# of & \# of Cell & Proportion of Cell\\
Block $h$   & People $M$ & Phone Users $c_h$ & Phone Users $p_h$ \\
\hline
 1 & 40 & 10 & 0.25 \\ 
 2 & 40 &  8 & 0.20 \\
 3 & 40 & 16 & 0.40 \\
 4 & 40 & 15 & 0.38 \\
 5 & 40 & 24 & 0.60 \\
 6 & 40 & 17 & 0.42 \\
 7 & 40 & 12 & 0.30 \\
 8 & 40 & 13 & 0.32 \\
 9 & 40 & 16 & 0.40 \\
10 & 40 & 13 & 0.32 \\
\hline
Total & 400 & 144 & 
\end{tabular}
\end{center}

\begin{enumerate}

\vfill

\item\pts{6} This is an example of one-stage clustered sampling.
  Circle one word in each pair of choices in the following sentences:
\begin{itemize} 
\item The primary sampling
  units (psu's) are the {\bf (blocks, residents)}.
\item The secondary
  sampling units (ssu's) are the {\bf (blocks, residents)}.
\end{itemize}

\vfill

\hfil{\sl[continued on next page]}

\clearpage

\item\pts{4} Ignoring the clustering and treating this as an SRS of 400
  residents, estimate the proportion of cell phone users and its
  standard error.

\vfill

\hfil{\sl[continued on next page]}

\clearpage

\item\pts{6} Now re-estimate the proportion of cell phone users and
  its standard error, using appropriate cluster sampling methods.
  {\em Hint: to reduce calculation, use the fact that in the table
    above, $s^2_{p_h} = 0.1214333$.}

\vfill


\hfil{\sl[continued on next page]}

\clearpage

\item\pts{6} Calculate the design effect DEFF for this design.

\vfill

\item\pts{4} Calculate $\rho$, the correlation between responses from
  residents on the same block.

\vfill

% \item\pts{} Another way to interpret the design effect is to calculate
%   the {\em equivalent sample size}, that is, the sample size for an
%   SRS w/o replacement that would give the same variance of $\bar y$
%   as the clustered sample.  Use the DEFF to estimate the {\em
%     equivalent sample size} for this clustered sample.

\end{enumerate}

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\clearpage

% groves, p. 213, \#4, with some elaboration.
\item\pts{24} {\em Response rates and missing data.}  You are
  completing a telephone survey of an SRS of 1000 members of a much
  larger professional organization, regarding their level of
  involvement in support of the organization.  Currently the response
  rate is 80\%, with 52.5\% of those responding saying they attend
  every monthly meeting of the local chapter of the organization, and
  47.5\% saying they do not.

\begin{enumerate}

\item\pts{8} Does it seem likely that the 200 (20\% of 1000) who did
  not respond to your survey are missing completely at random (MCAR,
  ignorable missingness) or missing not at random (MNAR, non-ignorable
  missingness)?  Choose MCAR or MNAR and {\em briefly} explain your
  reasoning.

\vfill

\vfill

\vfill

\vfill

\item\pts{8} Which of the following is more likely to be correct
  (circle \ul{one}):
\begin{itemize}
\item An unbiased estimate of the population proportion that attends
  every meeting is 52.5\%
\item An unbiased estimate of the population proportion that attends
  every meeting would probably be less than 52.5\%.
\item An unbiased estimate of the population proportion that attends
  every meeting woudl probably be more than 52.5\%
\end{itemize}

\vfill

\hfil {\sl [continued on next page]}

\clearpage

\item\pts{8} To get the sample size nearer to the target of $n=1000$,
  you could either 
\begin{itemize}
\item Ask the organization to pay for you to call a new SRS of size
  250, hoping that $(0.80)(250)=200$ people choose to respond {\em
    (cost: \$200 because it just involves routine single calls to each
    new phone number)}; or
\item Ask the organization to pay for you to followup with the 200 in
  your original sample who didn't respond yet, to try to get their
  responses {\em (cost: \$800 because it involves repeated call-backs
    until each of the 200 non-respondents either responds or
    refuses)}.
\end{itemize}
Which do you choose, and why {\em (briefly)}?

\vfill

\hrule
\hfil{\bf [The space below this line intentionally left blank]}

\vfill

\end{enumerate}



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\clearpage

% Examples from pp 167-168 of blue Groves book...

\item\pts{26} {\em Imputation methods.}

\begin{enumerate}

% gates, p 368 \#4
\item One method of imputation for missing responses to
  individual survey items is {\em mean imputation}.
\begin{enumerate}
\item\pts{5} Explain briefly how mean imputation works.

\vfill

\item\pts{4} Under what assumption (MCAR, MAR, MNAR) is mean imputation
  OK?  (Choose one, no explanation needed.)\\

\item\pts{4} 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
  hot-deck imputation}. 
\begin{enumerate}
\item\pts{5} Explain briefly how hot-deck imputation works.

\vfill

\item\pts{4} Under what assumption (MCAR, MAR, MNAR) is hot-deck imputation
  OK?  (Choose one, no explanation needed.)\\

\item\pts{4} Identify a possible problem with hot-deck 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}

