\documentclass[12pt]{article}
\usepackage{verbatim, amssymb, amsmath, float, fullpage}
\begin{document}
\title{36-303 Sampling, Surveys \& Society \\ Homework 05 Solutions}
\maketitle
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  PROB1  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{True/False}
(a) True: we aim for homogeneity within strata, so that we don't introduce additional error by taking a sample from within each stratum. \\
(b) False: we aim for heterogeneity within clusters, so that we don't introduce additional error by taking a sample of the clusters.

\section{Stratifying Variables}
Answers may vary; an acceptable set follows. \\
(a) gender, income level, reported race/ethnicity, reported party affiliation \\
(b) class year, major \\
(c) specific time slot, type of program (sitcom, drama, news, etc.) 

\section{Stratified Sampling}

(a) In order to estimate the total number of bushels of clams in the area, we will estimate the average number in a tow's worth of area, $\bar{y}_{st}$ and then multiply afterwards by the number of tows that would be needed to cover the entire area. We calculate some intermediate quantities from the data: $N_{h}$, N, $W_{h}$, and $f_{h}$, as defined on page 2 of the stratification handout. \\

Below is a table with the relevant quantities for the calculation. Note that when I did my calculation, I didn't round until the end, but some of the quantities below might be reported rounded off for space reasons. \\ \\
\centerline{\begin{tabular}{c c c c c c c c}
Stratum              & Area & $N_{h}$ & $W_{h}$ & $n_{h}$ & $\bar{y}_{h}$ & $f_{h}$ & ${s^2_{h}}$\\
\hline
1 & 222.81 & 5704 & 0.428 & 4 & 0.44 & 0.0007 & 0.068 \\
2 & 49.61 & 1270 & 0.095 & 6 & 1.17 & 0.0047 & 0.042  \\
3 & 50.25 & 1286 & 0.097 & 3 & 3.92 &  0.0023 & 2.146 \\
4 & 197.81 & 5064 & 0.380 & 5 & 1.80 &  0.0010 & 0.794 \\
Total & 520.48 & 13324 & 1 & 18 & ? & 0.0014 & ? \\
\end{tabular}} \\ 

To estimate the average number of bushels in a tow's worth of area, using the information in the table, we have
\begin{eqnarray}
\bar{y}_{st} = \sum_{h=1}^{H} W_{h}\bar{y}_{h} = 1.36
\end{eqnarray}
To get the total number of bushels in the area, we multiply this estimate by N $\approx$ 13324, yielding $\approx$ 18153. \\ \\ 
For the standard error of our estimate, we first calculate the variance of the average number of bushels in a tow's worth of area.
\begin{eqnarray}
Var(\bar{y}_{st}) = \sum_{h=1}^{H} W^2_{h}(1-f_h)\frac{s^2_h}{n_h} = 0.033
\end{eqnarray}
Note that for the total number of bushels, 
\begin{eqnarray}
\sqrt[]{Var(N\bar{y}_{st})} = \sqrt[]{N^2Var(\bar{y}_{st})} =  N\sqrt[]{Var(\bar{y}_{st})} \approx 2411
\end{eqnarray}
\\
(b) We are given that $s^2$ = 1.084. 
\begin{eqnarray}
Var(\bar{y}_{SRS}) = (1-f)\frac{s^2}{n} = (1-\frac{18}{13324})\frac{1.084}{18} \approx 0.060
\end{eqnarray}.

Thus the design effect, $\frac{Var(\bar{y}_{st})}{Var(\bar{y}_{SRS})}$, is $\approx$ 0.544, so it was indeed better to do a stratified sample; it decreased our standard error of estimation by about 46\%. 

\section{Clustered Sampling}
(a) This is a clustered sample because an SRS of 3-packs (clusters, psu's) was taken, instead of a sample of cans (units, ssu's). \\
(b) Each cluster's mean is given below. \\

\centerline{
\begin{tabular}{ c c c }
  Cluster & Mean \\
  \hline
  1 & 4.33 \\
  2 & 3.33\\
  3 & 1 \\
  4 & 5 \\
  5 & 7 \\
  6 & 3.33 \\
  7 & 3.67 \\
  8 & 1.67 \\
  9 & 5 \\
  10 & 2.67 \\
  11 & 6.67 \\
  12 & 0 \\
  \hline
  Overall & 3.64 \\
\end{tabular}
}

For the clustered sample, we have 
\begin{eqnarray}
\bar{y}_{cl} = \frac{1}{12}\sum_{i=1}^{12} \bar{y}_{i} = 3.64
\end{eqnarray}
and
\begin{eqnarray}
s^2_{\bar{y}_{i}} = \frac{1}{12-1}\sum_{i=1}^{12}(\bar{y}_{i} - \bar{y}_{cl})^2 \\ 
= \frac{1}{11}[(4.33 - 3.64)^2 + (3.33 - 3.64)^2 + ... + (0 - 3.64)^2] \approx\ 4.53
\end{eqnarray}

Hence, 
\begin{eqnarray}
Var(\bar{y}_{cl}) = (1 - \frac{n}{N})\frac{1}{n}s^2_{\bar{y}_{i}} = (1 - \frac{12}{580})\frac{1}{12}4.53 \approx 0.37
\end{eqnarray}
So the standard error of our estimate, the square root of $Var(\bar{y}_{cl})$, is $\approx$ 0.61.
\\ (c) We have that the design effect

\begin{eqnarray}
\frac{Var(\bar{y}_{st})}{Var(\bar{y}_{SRS})} = \frac{Ms^2_{\bar{y}_{i}}}{s^2_{\bar{y}_{ij}}}
\end{eqnarray}
The simple sample variance of the 36 cans is 
\begin{eqnarray}
s^2_{\bar{y}_{ij}} = \frac{1}{36-1}\sum_{i=1}^{36}(\bar{y}_{i} - \bar{y}_{SRS})^2 \approx 7.38
\end{eqnarray}
Thus, the design effect is 
\begin{eqnarray}
\frac{Ms^2_{\bar{y}_{i}}}{s^2_{\bar{y}_{ij}}} \approx \frac{3 * 4.53}{7.38} \approx 1.84
\end{eqnarray}
(d) We have that the the intracluster correlation $\rho$, approximately satisfies $DEFF \approx 1 + (M-1)\rho$. Thus we can estimate $\rho$ by solving
\begin{eqnarray}
\rho \approx \frac{DEFF - 1}{M-1} = \frac{1.84 - 1}{3 - 1} = 0.42.
\end{eqnarray}
\end{document}