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\bf 36-303: Sampling, Surveys \& Society\\
HW05: Hand in on paper Thu Apr 07, 2011
\end{center}

\noindent\ul{\em Reminders:}
\begin{itemize}
\item Please see Lecture 18, Week 09, for an outline of events for the
  rest of the semester.
\item Reading:
\begin{itemize}
\item Hill Handout on Oral Presenations
\item Bem Handout on Writing Papers
\end{itemize}
Examples of past 303 papers, are on the course website, under the link
{\tt Handouts on Writing \& Speaking}.

\item Exercises \#\ref{taylor} and \#\ref{jackknife} depend on
  material I will present in lecture Mar 31 and Apr 05.

\end{itemize}

\noindent\ul{\em Exercises to Turn In (there are 3 exercises):}
\begin{enumerate}

% \item {\bf \#1}.


\item {\bf Cluster Sampling (Lohr, 1999, p. 170).}  An accounting firm
  is interested in estimating the error rate in a compliance audit it
  is conducting.  The population consists of 828 claims and the firm
  audits an SRS of 85 of those claims.  In each of the 85 sampled
  claims, 215 fields are checked for errors.  We can think of the
  claims as psu's (clusters) and the fields as ssu's (units), in a
  single-stage cluster sample.

Several clusters had the same number of errors, so the results of this
single-stage cluster sample can be summarized as follows:
\begin{center}
\def\z{\phantom{0}}
\begin{tabular}{ccc}
Number of claims & Number of errors & $\bar y_i$ \\
(clusters)       & out of 215 fields & $=$ (\#errors)/(\#fields) \\
\hline
\hline
\z1  & 4  & 0.01860\\
\z1  & 3  & 0.01395\\ 
\z4  & 2  & 0.00930\\ 
22 & 1  & 0.00465\\
57 & 0  & 0.00000\\
\hline
\end{tabular}
\end{center}
(data courtesy of Fritz Scheuren).  
\begin{enumerate}
\item Use the ideas about cluster sampling discussed in class and in
  the handout on cluster sampling, to estimate the {\em proportion} of
  all 828$\times$215=178,020 fields with errors in them.  Give a
  standard error for your estimate.

\item Use your answer in part (a) to estimate, and give a standard
  error for, the {\em total} number of errors in all 178,020 fields.

\item Suppose that, instead of taking a cluster sample, the firm takes
  an SRS of $(85)\cdot(215)=18,275$ fields from the $178,020$ fields
  in the population.  If the estimated error rate is the same as in
  part (a), what would the estimated variance be?  How does this
  compare to the estimated variance in part (a)?

\end{enumerate}



\item {\bf Taylor Series Variance Estimates.}\label{taylor} At a
  certain university there are 1998 undergraduates.  An SRS without
  replacement, of size $n=20$ is taken, to find out what is the
  average number of hours per week the students study.  Some
  demographic data was collected also.  Here is the data from 20
  respondents in the survey:

\begin{center}
\begin{tabular}{@{}c@{\hspace*{4em}}c@{}}
\begin{tabular}{ccc}
Sex & College & Hrs/Wk \\
\hline
\hline
M & Eng & 28 \\
M & Eng & 29 \\
M & Eng & 23 \\
M & Eng & 35 \\
M & Eng & 29 \\
M & Eng & 30 \\
M & Eng & 34 \\
M & Eng & 31 \\
F & Eng & 30 \\
F & Eng & 31 \\
\hline
\end{tabular} 
&
\begin{tabular}{ccc}
Sex & College & Hrs/Wk \\
\hline
\hline
F & Eng & 36 \\
F & Eng & 33 \\
M & Lib & 27 \\
M & Lib & 28 \\
F & Lib & 29 \\
F & Lib & 30 \\
F & Lib & 28 \\
F & Lib & 28 \\
F & Lib & 32 \\
F & Lib & 30 \\
\hline
\end{tabular}
\end{tabular}
\end{center}
\begin{enumerate}
\item Estimate the mean number of hours/wk students study at this
  university, and give the standard error.

\item Administrative records show that the total undergrad enrollment
  at the university is
\begin{center}
\begin{tabular}{c|cc}
Sex & Eng & Lib \\
\hline
M & 617 & 380 \\
F & 450 & 551 \\
\end{tabular}
\end{center}
Make four post-strata by combining sex and college of the students,
compute post-stratum weights to make the sample more representative of
the population, and compute the weighted average number of hours/wk
students study.

\item Use the Taylor Series method discussed in class to compute a
  standard error for your weighted estimate in part (b).  Compare to
  the standard error you computed in part (a).

\end{enumerate}


\item {\bf Jackknife Variance Estimates.}\label{jackknife} Use the
  jackknife method discussed in class to compute a standard error for
  the weighted estimate in problem \ref{taylor}(b).  Compare to the standard
  errors you calculated in problems \ref{taylor}(a) and \ref{taylor}(c).

\item[]
\ul{\em Notes on \#\ref{taylor} and \#\ref{jackknife}} {\em: The
  sample size $n=20$ is small enough that you can do all the
  computations ``by hand'' using Minitab, R, Excel, etc.  If you know
  R and you want to compare your answers with the results you get from
  the R functions I introduced in class, please feel free to do so.}

\comment{

% A quesion not asked (but would be good)  [might need a little
% attention to notation and wording...

\item {\em Jackknifing a simple sample.}  Suppose that we have only
  one post-stratum, with sample size $n$, and suppose the sampled
  observations $y_1, \ldots, y_n$ are an SRS {\bf with} replacement,
  so we can ignore fpc's.
\begin{enumerate}
\item Show that for each leave-one-out sample of size $n-1$, the $n-1$
  post-stratification weights $w_i^{(r)}$ are all equal---and so they
  might as well be equal to 1 in the weighted average:
\[
    \bar y^{(r)}_w = \frac{\sum_{i\neq r} w_i^{(r)} y_i}{\sum_{i\neq r} w_i^{(r)}}
\]
\item Show that in this one-stratum case, the jackknife estimate of the
  population mean equals the SRS estimate:
\[
    \bar y_{JK} = \frac1n\sum_{r=1}^n \bar y^{(r)}_w = \frac1n
    \sum_{i=1}^n y_i = \bar y_{srs}
\]
\item Show that in this one-stratum case, the jackknife estimate of
  the variance equals the SRS estimate:
\[
    s^2_{JK} = \frac{n-1}n \sum_{r=1}^n (\bar y^{(r)}_w - \bar
    y_{jk})^2 = \frac1{n-1} \sum_{i=1}^n (y_i - \bar y_{srs})^2 = s^2_{srs}
\]

\end{enumerate}

}

\end{enumerate}

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