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\begin{center}
\large\bf
36-303: Sampling, Surveys and Society\\
Exam 2\\
Tue Apr 15, 2008
\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.
\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 & 20 & \\ 
  &    & \\ \hline
3 & 18 & \\ 
  &    & \\ \hline
4 & 24 & \\ 
  &    & \\ \hline
5 & 18 & \\ 
  &    & \\ \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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\clearpage

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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} You have conducted a survey in which you have collected
  data on respondents' ages and incomes; all respondents are between
  20 and 60 years old, and earn between \$25,000 and \$250,000.  Some
  cases (respondents) did not report their incomes: older people were
  less likely to report their income, and among people who did report
  incomes, older people tended to earn more than younger people.\\[-0.75em]

  Which imputation method below is {\bf not} recommended (or, circle
  iv. if all are OK)?
\begin{enumerate}\small
% \item Delete cases from the data set for which there is no income
%   data.
\item For each missing income, fill in the mean of all of the incomes
  you do have.
\item Build a regression model predicting income from age, using the
  respondents who have answered both questions, and use the model to
  predict what the missing incomes should be, based on the
  respondents' ages.
\item For each person who did not report income, find all of the other
  people in the sample of the same age who did report an income, and
  fill in the average of those persons' incomes.
\item All of the above methods are OK.
\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 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{5} Which of the following is {\bf not} one of the
  recommended things to work on, to reduce the tendency of survey
  subjects to not respond?
\begin{enumerate}\small
\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{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}


\end{enumerate}

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\clearpage

\item\pts{20} {\em Cluster Sampling (3 parts).}

One of the joys of summer is driving by a roadside stand and buying
bunches of fresh sweetcorn to take home and cook.  Of course, roadside
stands have to satisfy health inspection requirements like all other
food sellers.

One roadside stand sells corn in bunches of 3 ears each; they have 580
such bunches for sale.  A food inspector samples 12 of these bunches
and counts the number of worm fragments he finds on each ear in all 12
bunches, to determine the average number of worm fragments per ear of
corn.  The inspector's data are as follows:

\begin{center}
\begin{tabular}{c|*{12}{c}}
 \multicolumn{1}{c}{}& \multicolumn{12}{c}{Bunch} \\
&  01 &  02 &  03 &  04 &  05 &  06 &  07 &  08 &  09 &  10 &  11 &  12 \\
\hline
\hline
Ear 1 & 1 & 4 & 0 & 3 & 4 & 0 & 5 & 3 & 7 & 3 & 4 & 0 \\
Ear 2 & 5 & 2 & 1 & 6 & 9 & 7 & 5 & 0 & 3 & 1 & 7 & 0 \\
Ear 3 & 7 & 4 & 2 & 6 & 8 & 3 & 1 & 2 & 5 & 4 & 9 & 0 \\
\hline
mean  & 4.33 & 3.33 & 1.00 & 5.00 & 7.00 & 3.33 & 3.67 & 1.67 & 5.00 & 2.67 & 6.67 & 0.00 
\end{tabular}
\end{center}
If we treat this data as arising from a one-stage cluster sampling design,
the sample variance between cluster means is  
\[
   s^2_{\bar y_i} = \frac{1}{12-1}\sum_{i=1}^{12} (\bar y_i - \bar
   y_{cl})^2 = 4.53
\]
and the sample variance of all 36 individual observations is 
\[
  s^2_{y_{ij}} = \frac{1}{36-1} \sum_{i=1}^{12}\sum_{j=1}^3 (y_{ij} -
  \bar y_{srs})^2 = 7.38 
\]


\begin{enumerate}


\item\pts{6}  Compute $\bar y_{cl}$ and $\bar y_{srs}$.  If they
  should be the same, explain why.  If they should be different,
  explain why {\em [use the back of this page to show work, if you
      need to]}.

\vspace*{\fill}

\clearpage


\item\pts{6}  Compute SE$(\bar y_{cl})$ and use this to create an
  approximate 95\% confidence interval for the mean number of worm
  fragments per ear in the entire population of $580\times3=1740$ ears
  of corn.

\vspace*{\fill}

\item\pts{8} Estimate
\begin{itemize}
\item The DEFF (design effect) for this design; and
\item The ICC (intra-cluster correlation, $\rho$).
\end{itemize}

\vspace*{\fill}

% \item\pts{??} Explain what the ICC is, and why it is nonzero.
% 
% \vspace*{\fill}
%
% Putting this question in the M/C part...

\end{enumerate}

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\clearpage

\item\pts{18} {\em Yale Work \& Life Survey, Part I (3 parts).}

On September 20th, 2005, the New York Times ran a front-page article,
{\em Many Women at Elite Colleges Set Career Path to Motherhood},
alleging that undergraduate women at elite colleges such as Yale plan
to choose motherhood over their careers.  Dr. Victoria Brescoll and
the Yale Women's Center set out to test this claim with a
comprehensive study.  A summary of her survey results are available at
http://www.yale.edu/wc/worklifesurvey.  According to
http://www.yale.edu/oir/facts05.html, there were 2,707 men and 2,609
women enrolled at Yale in 2005--2006, the academic year in which
Brescoll did her study.

In Brescoll's survey, 154 men responded, of which 134 planned to
become parents someday, and 315 women responded, of which 247 planned
to be parents someday\footnote{Other Likert-type questions more
  directly addressed the tradeoff between career and family, but we
  will just work with this yes/no question.}.

\begin{enumerate}

\item\pts{6} Treating the sample of men as an SRS w/o replacement,
  compute $\hat p_{men}$, the proportion of men at Yale who plan on
  becoming parents, and its standard error, ${\rm SE}(\hat p_{men})$.

\vspace*{\fill}

\item\pts{6} Do the same for women: compute $\hat p_{women}$ and
  ${\rm SE}(\hat p_{women})$.

\vspace*{\fill}

\clearpage

\item\pts{6} A two-sample $z$-test for whether there is a significant
  difference between men and women on this question can be based on
  the test statistic
\[
        z = \frac{\hat p_{men} - \hat p_{women}}
        {\sqrt{{\rm SE}(\hat p_{men})^2 + {\rm SE}(\hat p_{women})^2}}~,
\]
which follows a Normal distribution with mean 0 and variance 1 under
the Null Hypothesis that the population difference $p_{men}-p_{women}$
is really zero.\\[-0.75em]

Perform this test, using your answers to parts (a) and (b).\\[-0.75em]

What conclusion do you draw?

\vspace*{\fill}

\end{enumerate}

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\clearpage

\item\pts{24} {\em Yale Work \& Life Survey, Part II (4 parts).}

\begin{enumerate}

\item\pts{6} Combining men and women, and treating the survey as an
  SRS w/o replacement, estimate the overall proportion of students at
  Yale who plan on becoming parents.  (You do not have to compute the
  SE).

\vspace*{\fill}

\item\pts{6} We really do not know if Brescoll's survey was an SRS,
  what the response rate was, etc., but it seems clear there is an
  imbalance among the respondents between men \& women, vs. the Yale
  student population.  Using sex to define two post-strata, compute
  post stratification weights to apply to this survey.

\vspace*{\fill}

\clearpage

\item\pts{6} Using the weights you computed in part (b), compute the
  weighted proportion of students at Yale who plan on becoming parents
  (you do not have to compute the SE).

\vspace*{\fill}

\item\pts{6} Use the values of the weights to explain\footnote{If you
  were unable to complete part (c), use the weights to make a
  prediction about whether your answer to part (c) would be higher or
  lower than your answer to part (a), and explain why.} the difference
  between your answers in parts (a) and (c).

\vspace*{\fill}


\end{enumerate}



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\clearpage

% Examples from lecture 09...

\item\pts{18} A 36-303 group wants to do a survey of student attitudes
  toward Carnegie Mellon sports teams and sporting events.  In their
  survey proposal they specify the {\em target population} to be all
  currently enrolled undergraduates at Cargnegie Mellon, and they
  indicate that to collect data they will advertise for volunteers to
  take the survey on FaceBook.com, with a link to a formal survey
  instrument at QuestionPro.com.
\begin{enumerate}

\item\pts{6} 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{6} Write down \ul{\bf two} very likely sources of {\em coverage
  error} for this survey as currently proposed.

\vspace*{\fill}

\item\pts{6} Make one big suggestion that this group could implement
  to improve their sampling plan.

\vspace*{\fill}

\end{enumerate}


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

\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}

