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\begin{center}
\bf
36-303: Sampling, Surveys and Society\\
Exam 2 --- SOLUTIONS\\
% Tue Apr 17, 2012
\end{center}

\begin{enumerate}

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

\begin{enumerate}\def\tick{\hspace*{-2em}\fbox{\phantom{X~}}~}

\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 $N = N_R + N_M$.\\[-0.75em]

Which statement below is false (or, circle iv. if all are OK)?
\begin{enumerate}\small
\item\tick 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 meetinags
  attended by nonresponders, vs the mean number attended by
  responders, the bigger the bias due to missing responses.
\item\tick The larger your SRS, the better you can estimate $\bar Y_{POP}$.
\item All of the above statements are true.
\end{enumerate}
{\em NOTE: (iii) was the intended answer but due to poor editing on my
  part, either (i) or (iii) are acceptable answers. --BJ}

\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\tick 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}


\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\tick 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\tick All of the above {\em are} usually part of post-survey processing!
\end{enumerate}



\end{enumerate}


% \vspace*{\fill}

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% \clearpage

\item\pts{24} {\em Stratified sampling\ldots}\\ % based on #19, pp108-9, Lohr 2nd ed
%
{
\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}  $\bar y_{srs}$ and
  $\sqrt{\Var(\bar y_{srs})}$, using $\hat p(1-\hat p)$\ldots

\begin{itemize}
 
\item $\bar y_{srs} = 759/899 = \fbox{0.84}$

\item $\Var(\bar y_{srs}) = (1-f) s^2_y/n = (1 - n/N)s^2_y/n =
  (1-899/3681)(0.84)(1-0.84)/899 = 0.0001129876$, so the SE is
  $\sqrt{0.0001129876} = \fbox{0.0106}$

\end{itemize}



\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.]}

\begin{itemize}

\item Stratum weights and sampling fractions are 
\begin{center}
\begin{tabular}{c|cccc}
$h$          & 1 & 2 & 3 & 4  \\
\hline
$N_h$        & 1374 & 1960 & 252 & 95 \\
$n_h$        & 232 & 514 & 67 & 86 \\
$W_h =N_h/N$ & 0.37 & 0.53 & 0.07 & 0.03 \\
$f_h= n_h/N_h$& 0.17&  0.26 & 0.27 & 0.91 
\end{tabular} 
\end{center}
where $N=N_1+N_2+N_3+N_4 = 3681$.

\item $\bar y_{st} = \sum_{h=1}^H W_h \bar y_h$ $=$ (0.37)(0.72) $+$
  (0.53)(0.89) $+$ (0.07)(0.87) $+$ (0.03)(0.87) $=$ \fbox{0.8251}

\item $\Var(\bar y_{st}) = \sum_{h=1}^H W_h^2(1-f_h)s^2_h/n_h$ $=$
$(0.37)^2(1-0.17)(0.2016)/232$ $+$ $(0.53)^2(1-0.26)(0.0979)/514$ $+$
  $(0.07)^2(1-0.27)(0.1131)/67$ $+$ $(0.03)^2(1-0.91)(0.1131)/86$
  $=$ $0.0001444743$, so the SE is $\sqrt{0.0001444743} = \fbox{0.0120}$.

\end{itemize}


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

\begin{itemize}

\item DEFF = $\Var(\bar y_{st})/\Var(\bar y_{srs}) =
  0.0001444743/0.0001129876 = \fbox{1.28}$

\item In this case it was \fbox{\em not worth it} to stratify since
  the DEFF is bigger than 1 (and when stratification is working it is
  more efficient than SRS, i.e. DEFF$<$1!).

\end{itemize}

\item\pts{6} Clearly not everyone who received a survey responded.
\begin{itemize}
\item The response rates in each stratum are as follows:
\begin{center}
\begin{tabular}{c|cccc}
$h$ & 1 & 2 & 3 & 4 \\
\hline
$n_{resp}$    &  232 & 514 &  67 & 86 \\
$n_{sampled}$ & 500 & 653 &  74 &  95 \\
resp rate = $n_{resp}/n_{sampled}$ &  \fbox{0.46} & \fbox{0.79} &
\fbox{0.91} & \fbox{0.91}  
\end{tabular}
\end{center}

\item I would expect \fbox{more nonresponse bias in the SRS
  estimates}.
%
(A reason does not have to be given but my reasoning is this:  The
  response rates differ dramatically across strata and so do the
  proportions of employees who would want the school closed over
  winter break in the future.  When the response rate is related to
  the response, we expect to find nonresponse bias.)

\end{itemize}

\end{enumerate}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\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 
\[
   \hat p = 112/134 = 0.83582
\]
with 
\[
   \Var(\hat p) = 0.83582(1-0.83582)/134 = 0.00102
\]
%
{\setlength{\unitlength}{1in}
\begin{picture}(0,0)
\put(4.5,0.10){\fbox{\begin{tabular}{|p{1.5in}|}
\hline
\em The use of 112 instead of 113 in these formulas was just a typo and
not relevant to the statistical issues that this problem was inteded
to address.\\
\hline
\end{tabular}
}}
\end{picture}
}%
\begin{enumerate}

\item\pts{6} \\
\begin{itemize}
\item The {\bf target population} is \fbox{registered voters in the suburb}.
\item The {\bf sampling frame} is \fbox{households listed in the phone
  book}. {\em Or you could say:
\begin{itemize}
\item The phone book, or
\item Voters in households listed in the phone book.
\end{itemize}}
\end{itemize}



\item\pts{6}\\
\begin{itemize}
\item The psu's (primary sampling units) are the \hbox to
  0pt{~households}\rule{2in}{0.01in}~.

\item The ssu's (secondary sampling units) are the \hbox to
  0pt{~registered voters
  in each household}\rule{2in}{0.01in}~.
\end{itemize}

\item\pts{6} \\
\begin{itemize}
\item The the estimates $\hat p$ and $\Var(\hat p)$ given in
  the problem statement are \fbox{not valid}. 
\item REASON(S): {\em (only one is needed)}
\begin{itemize}
\item \fbox{A cluster sample is being treated as an SRS.} The estimate of
  $\hat p$ is approximately correct, but the variance will likely be
  too small (since, likely, DEFF$>$1). {\em (this is the better answer)}
\item \fbox{There may be nonresponse bias} since we do not know why 23
  of the voters refused to answer. 
\end{itemize}
\end{itemize}

\end{enumerate}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


% Groves 2nd ed p 367 #1, sort of
\item\pts{18}
Give at least one strength and one weakness of each of the following
strategies for identifying and correcting such coding errors.

\begin{itemize}

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

\begin{itemize}
\item {\bf Strength:} Fast, doesn't take much effort, will identify
  observations which seem too  high or too low relative to the rest of
  the data (like 1750 in the example).


\item {\bf Weakness:} Will probably not identify observations that are
  miscoded and in the middle of the distribution (like 15,799 or
  17,588 in the example).


\end{itemize}

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

\begin{itemize}
\item {\bf Strength:} Fast, doesn't take much effort, will find any
  kind of miscoding in the sample (whether it is an outlier like 1750
  or a non-outlier case like 15,799 or 17,588).


\item {\bf Weakness:} 90\% of the data will not be checked for errors!


\end{itemize}

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

\begin{itemize}
\item {\bf Strength:} Every coding error will be found and corrected.


\item {\bf Weakness:} Takes a great deal of time and effort!


\end{itemize}

\end{itemize}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\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.

\begin{tabular}{|p{5in}|}
\hline
{\em If the response to question \#17 is missing, fill in the mean of
  all the responses to \#17 that you have from other survey
  participants who did answer that question.}\\
\hline
\end{tabular}

\item\pts{3} This approach only works under \fbox{MCAR}.

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

{\em 
Only one answer is required:
\begin{itemize}
\item \fbox{MCAR might not be true}
\item \fbox{Whatever the mean is, it probably isn't what this person would
  have responded.}
\end{itemize}
}

\end{enumerate}

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

\begin{tabular}{|p{5in}|}
\hline
{\em If the response to question \#17 is missing, build a regression
  equation for answers to \#17 using demographic information,
  answers to other survey questions, etc., using those survey
  participants that did answer \#17.  Then use the regression equation
  with this person's demographic data \& answers to other survey
  questions, to predict what this person would have answered to \#17.}\\
\hline
\end{tabular}

\item\pts{3} This works under \fbox{MAR}.

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

\begin{tabular}{|p{5in}|}
\hline
{\em
MAR means, in this case, that all people with the same demographic
informaion and the same answers to other survey questions would be
nonresponders by chance only, and not because of what they would have
said.  If we don't get the right variables into the regression
equation, then we can't argue that MAR holds.
}\\
\hline
\end{tabular}

Other answers may be acceptable here, but the above is the best
answer.

~\\

{\em Note: A typo on the test mixed had ``hot-deck imputation''
  instead of ``regression imputation'' for one part of this question.
  But the correction should have been announced during the exam.}


\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}

