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% Title, subtitle, and other information
\title{Stat472/572 Sampling: Theory and Practice}
\author[Yan Lu]{Instructor: Yan Lu}
\institute[UNM]{University of New Mexico}
\date{}

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    \centering
    \textbf{\Large Chapter 5: Cluster Sampling with Equal Probabilities} % Emphasize chapter title

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\section{Chapter 5}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
    \frametitle{Motivation: Sampling Students in High School}

    \textbf{Motivation:}
    \begin{itemize}
        \item \textbf{Simple Random Sampling (SRS):}  
        \begin{itemize}
            \item Randomly sample individual students directly from the entire school or an area.   
            \item This requires a complete list of all students, which can be difficult and labor-intensive to obtain.  
            \item Additionally, it involves locating and measuring students from various classes or schools, increasing logistical complexity and cost.
        \end{itemize}
        \item \textbf{Cluster Sampling as an Alternative:}  
        \begin{itemize}
            \item Take a random sample of \(n\) classes (the classes are called the primary sampling units (psus) or clusters).  
            \item Measure all students in the selected classes (the students within the classes are called the secondary sampling units (ssus)).  
            \item Often, the ssus are the elements of the population.  
            \item In the design of experiments, this approach is called a nested design.  
        \end{itemize}
    \end{itemize}
\end{frame}










%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
{\bf Cluster Sampling:}
Definition:
\bi
\item Primary Sampling Units (psu) or Cluster: a grouping of
the members of the population, usually naturally occurring units\\
Example: classes, blocks, nest of bees
\item Secondary Sampling
Units (ssu): units in the psu. Often the ssu's are the elements in
the population
\item One stage cluster sampling:
\bi
\item Stage 1: Randomly select $n$ clusters
\item Stage 2: Survey all units in the selected clusters
\ei
\item
Two stage cluster sampling:
\bi
\item Stage 1: Randomly select $n$ clusters
\item Stage 2: Survey partial of the units in the selected clusters
\ei
\ei
\end{frame}

%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\begin{figure}[!htp]
       \centering
       \includegraphics[width=0.90\textwidth,
       height=1.2\textheight]{./plots/clustersampling.pdf}
\caption{page 183}
       \label{archers}
     \end{figure}

\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Comments:}
\bi
\item Students in the selected classes are not as likely
to mirror the diversity of the high school as well as students
chosen at random\\
----But it is much cheaper and easier to interview all students in the same
class than students selected at random from the high school\\
---- Cluster sampling may result in more information per dollar spent
\item Cluster sampling complicates design and analysis and it
usually decreases precision
\ei
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Why use cluster sampling?}
\bi
\item Constructing a sampling frame list of observation units may be
difficult, expensive, or impossible\\
-----can't list all honeybees in a region or customers in a store\\
-----possible to list all individuals
in a city, but it is time-consuming and expensive, since in a general case, we only
have a list of housing units or a list of the phone numbers
\ei
\end{frame}

\begin{frame}
    

    \textbf{Reasons for Using Cluster Sampling:}  
    \begin{itemize}
        \item Constructing a sampling frame of individual observation units can be:  
        \begin{itemize}
            \item Difficult, expensive, or even impossible.  
            \item For example:
            \begin{itemize}
                \item It is impractical to list all honeybees in a region.  
                \item Listing all customers in a store is often unfeasible.  
            \end{itemize}
        
        \item While it may be technically possible to list all individuals in a city, doing so is often:  
        \begin{itemize}
            \item Time-consuming and expensive.  
            \item Typically, only lists of housing units or phone numbers are available.  
        \end{itemize}
    \end{itemize}
    \end{itemize}
    \end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\bi
\item The population may be widely distributed geographically or may
occur in natural clusters such as households or schools\\
{\bf Example 1:} Want to interview residents of nursing homes in the
United States
\item[] It is much cheaper to sample nursing homes and
interview every resident in the selected homes than to interview an
SRS of nursing home residents
\item[] With an SRS of residents, you
might have to travel to a nursing home just to interview one
resident
\item[] {\bf Example 2:} In an archaeological survey, you would examine
all artifacts found in a region instead of choosing points at random
and examine only artifacts found at those isolated points
\ei
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
{\bf Comparing Cluster Sampling with Stratification}

{\footnotesize
\begin{tabular}{|c|c|}
\hline
{\bf Stratified Sampling} & {\bf Cluster Sampling}\\
\hline
Each element of the population& Each element of the population\\
 is in exactly one stratum &  is in exactly one cluster\\
 \hline
 Population of $H$ strata & Population of $N$ clusters\\
 \hline
 Take an SRS from each stratum & Take an SRS of clusters\\
 \hline
 Variance of the estimator of $\bar y_U$  & Variance of the estimator of $\bar y_U$ \\
depends on  the variability of  & depends primarily
 on the variability\\
values within
 strata &  between cluster means\\
 \hline
 For great precision,want & For great precision, want\\
 similar values within each stratum &different
 values within each cluster\\
 stratum means differ from each other & cluster means are similar to
 one another\\
 \hline
\end{tabular}
}
\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\begin{figure}[!htp]
       \centering
       \includegraphics[width=0.90\textwidth,
       height=1.5\textheight]{./plots/comparison.pdf}
\caption{Example 3.10}
       \label{archers}
     \end{figure}

\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Notation}

{\normalsize 
\bi
\item[] $y_{ij}$: measurement for $j$th element in the $i$th psu\\
-----psu level
\item $N$: number of psus in the population
\item $M_i$: number of ssus in the psu $i$
\item $M_0=\sum\limits_{i=1}^N M_i$: total number of ssus in the population
\item $t_i = \sum\limits_{j=1}^{M_i} y_{ij}$: total in psu $i$.
\item $t=\sum\limits_{i=1}^N t_i = \sum\limits_{i=1}^N \sum\limits_{j=1}^{M_i} y_{ij}$: population total.
\item $S_t^2 = \ds\frac 1 {N-1} \sum\limits_{i=1}^N {\left(t_i -\ds\frac t N\right)^2}$ : population variance of the psu totals (between
cluster variation).
\ei
}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
-----ssu level
\bi
\item $\bar y_U = \sum\limits_{i=1}^N \sum\limits_{j=1}^{M_i} \ds\frac {y_{ij}}{M_0}$: population mean
\item $\bar y_{iU}=\sum\limits_{j=1}^{M_i} \ds\frac{y_{ij}}{M_i} = \ds\frac{t_i}{M_i}$: population mean in psu $i$
\item $S^2=\sum\limits_{i=1}^N \sum\limits_{j=1}^{M_i} \ds\frac{(y_{ij} - \bar y_U)^2}{M_0-1}$: population variance (per ssu)
\item $S_i^2=\sum\limits_{j=1}^{M_i} \ds\frac{(y_{ij} - \bar y_{iU})^2}{M_i -1}$:  population variance within the psu $i$.
\ei
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\small
-----Sample values
\bi
\item $n$ : number of psus in the sample
\item $m_i$ : number of elements in the sample for the $i$th psu
\item $\bar y_i = \sum\limits_{j\in S_i} \ds\frac{y_{ij}}{m_i}$ : sample mean (per ssu) for psu $i$
\item $\hat t_i = M_i \bar y_i = M_i \cdot \ds\frac{\sum\limits_{j\in S_i} y_{ij}}{m_i}$ : estimated total for psu $i$
\item $\hat t_{\mbox{unb}} = N \bar t = N \cdot \ds\frac {\sum\limits_{i\in S}\hat t_i} n $ : unbiased estimator of $t$ (population total)
\item $s_t^2 = \ds\frac 1{n-1} \sum_{i\in S} (\hat t_i - \bar t)^2 =\ds\frac 1{n-1} \sum_{i\in S} \left(\hat t_i - \ds\frac {\hat t_{\mbox{unb}}}{N}\right)^2$: estimated variance of psu totals
\item $s_i^2 = \sum\limits_{j\in S_i} \ds\frac{(y_{ij} - \bar y_i)^2}{m_i -1}$ :sample variance within psu
$i$
\ei
}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
{\bf One-stage cluster sampling with equal sizes: $M_i = m_i =
M$}

$$\hat t = N \bar t = \ds\frac N n \sum_{i\in S} t_i$$
$$V(\hat t) = N^2 \left(1-\ds\frac n N\right) \ds\frac {S_t^2}{n}, \quad \hat V(\hat t) = N^2 \left(1-\ds\frac n N\right) \ds\frac {s_t^2}{n}$$
$$\hat {\bar y} = \ds\frac {\hat t}{NM}$$
$$V(\hat {\bar y}) = \left(1-\ds\frac n N\right) \ds\frac
{S_t^2}{nM^2}, \quad \hat V(\hat {\bar y}) = \left(1-\ds\frac n N\right) \ds\frac
{s_t^2}{nM^2}$$
$S_t^2 = \ds\frac 1{N-1} \sum\limits_{i=1}^N\left(t_i -\ds\frac t N
\right)^2$\\
$S_t^2$ is estimated by $s_t^2$ with\\
 $s_t^2 = \ds\frac 1 {n-1}
\sum_{i\in S} (t_i - \bar t)^2=\ds\frac 1 {n-1}
\sum_{i\in S} \left(t_i - \ds\frac {\hat t} N\right)^2$\\
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\begin{center}
\begin{table}
\caption{ANOVA Table}
\begin{tabular}{c|c|c|c}
\hline Source& df& Sum of Squares & Mean Squares\\
\hline Between& $N-1$ & SSB=  & MSB\\
psu's & & $\sum\limits_{i=1}^N
\sum\limits_{j=1}^M(\bar y_{iU} - \bar y_U)^2$& \\
\hline Within& $N(M-1)$ & SSW= & MSW\\
psu's & & $\sum\limits_{i=1}^N \sum\limits_{j=1}^M( y_{ij} - \bar
y_{iU})^2$& \\
\hline
Total & NM-1 & SSTO=& $S^2$\\
& &$\sum\limits_{i=1}^{N} \sum\limits_{j=1}^M (y_{ij} - \bar
y_U)^2$&\\
 \hline

\end{tabular}
\end{table}
\end{center}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Example 5.2:} A student wants to estimate the average grade point average
(GPA) in his dormitory. Instead of obtaining a listing of all
students in the dorm and conducting a simple random sample, he
notices
\bi
\item the dorm consist of 100 suites, each with 4
students;
\item  he chooses 5 of those suites at random, and asks
every person in the
5 suites what her or his GPA is. \\
The results are as follows:
\ei
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[fragile]
\begin{verbatim}
> print.data.frame(gpadata)
   suite  gpa wt
1      1 3.08 20
2      1 2.60 20
3      1 3.44 20
4      1 3.04 20
5      2 2.36 20
6      2 3.04 20
7      2 3.28 20
8      2 2.68 20
9      3 2.00 20
10     3 2.56 20
11     3 2.52 20
12     3 1.88 20
13     4 3.00 20
14     4 2.88 20
15     4 3.44 20
.......
20     5 3.20 20
\end{verbatim}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\begin{tabular}{c|ccccc}
\hline
Person& suite1&suite2& suite3 & suite4& suite5\\
\hline 1& 3.08 & 2.36& 2.00& 3.00& 2.68\\
2 & 2.60& 3.04& 2.56& 2.88& 1.92\\
3& 3.44&3.28& 2.52& 3.44& 3.28\\
4& 3.04& 2.68& 1.88& 3.64& 3.20\\
\hline
 Total & 12.16& 11.36& 8.96& 12.96& 11.08\\
\hline
\end{tabular}

The psu's are the suites, $N=100$, $n=5$, and $M=4$.
$$\bar t = (12.16+11.36+8.96+12.96+11.08)/5 = 11.304$$
$$\hat t = 100 \bar t =1130.4$$
and
\begin{eqnarray*}
s_t^2 &= &\ds\frac 1{5-1} \left[(12.16 -11.304)^2 + \cdots +
(11.08-11.304)^2\right]\\
& =& 2.256
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

\begin{eqnarray*}
\hat V(\hat t) &=& N^2\left(1-\ds\frac n N\right)\ds\frac {s_t^2} n\\
&=& 65.4706
\end{eqnarray*}
$$\hat {\bar y} = 1130.4/400 = 2.826$$
$$SE(\hat {\bar y}) = \sqrt {\left(1-\ds\frac 5 {100}\right) \ds\frac
{2.256}{(5)(4)^2}} = .164$$
A 95\% CI for the mean is given by
$$2.826 \pm 2.776*(0.164) = [2.37, 3.28]$$
where $2.776$ is the percentile from a $t$ distribution with $(n-1) = 4$ df.
 Note: Only the ``total" column of the
data table is used, the individual GPAs are only used for their
contribution to the suite total.
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
ANOVA Table\\
\begin{tabular}{c|ccc}
Source& df& SS&MS\\
\hline Between Suites & 4& 2.2557 & .56392\\
Within suites& 15& 2.7756& .18504\\
 \hline
 Total & 19 & 5.0313 & .2648\\
\end{tabular}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[fragile]
\begin{verbatim}
> dgpa<-svydesign(id=~suite,weights=~wt,fpc=~rep(100,20),
data=gpadata) 
> dgpa
1 - level Cluster Sampling design
With (5) clusters.
svydesign(id = ~suite, weights = ~wt, fpc = ~rep(100, 20), 
data = gpadata)
> gpamean<-svymean(~gpa,dgpa)
> gpamean
     mean     SE
gpa 2.826 0.1637
> degf(dgpa)
[1] 4
> confint(gpamean,level=.95,df=4)
       2.5 %   97.5 %
gpa 2.371593 3.280407
\end{verbatim}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[fragile]
\begin{verbatim}
> gpatotal<-svytotal(~gpa,dgpa)
> gpatotal
     total     SE
gpa 1130.4 65.466
> confint(gpatotal,level=.95,df=4)
       2.5 %   97.5 %
gpa 948.6374 1312.163

\end{verbatim}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[fragile]
\begin{verbatim}
> suitesum<-tapply(gpadata$gpa,gpadata$suite,sum) 
 #sum gpa for each suite
> suitesum
    1     2     3     4     5 
12.16 11.36  8.96 12.96 11.08 
> # variability comes from among the suites
> st2<-var(suitesum) 
> st2
[1] 2.25568
> # SE of t-hat, formula (5.3) of SDA
> vthat <-100^2*(1-5/100)*st2/5
> sqrt(vthat) 
[1] 65.46596
> # SE of ybar, formula (5.6) of SDA
> sqrt(vthat)/(4*100)
[1] 0.1636649
\end{verbatim}
\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Weight:} One-stage cluster sampling with an SRS of psu's produces a
self-weighting sample. The weight for each observation unit is
$$w_{ij} = \ds\frac 1 {P\{\mbox{ssu } j \mbox{ from psu }i\mbox{ is in sample}\}} = \ds\frac
N n$$

\begin{eqnarray*}
\hat t &=& \sum_{i \in S} \sum_{j \in S_i} w_{ij} y_{ij}\\
&=& \ds\frac N n (3.08+ 2.60+\cdots + 3.28+3.20)\\
&=& \ds\frac {100}{5} (56.52)\\
&=& 1130.4
\end{eqnarray*}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{frame}
%Example: Take an SRS of classes and sampled all students for
%reading score within the selected class, we would likely to find
%that average reading scores varied from class to class
%\bi
%\item An excellent reading teacher might raise the reading scores for the
%entire class
%\item A class of students from an area with much
%poverty might tend to score poorly in reading
%\item Unmeasured
%factors, such as teaching skill or poverty, can affect the overall
%mean for a cluster and thus cause MSB to be large
%\item Within a
%class, students' reading scores vary. If the clusters are relatively
%homogeneous---for example, students in the same class have similar
%scores-the MSW will be small.
%\ei
%\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
{\bf Comparing One-stage Cluster Sampling with SRS with $nM$ elements}

Instead of taking a cluster sample of $M$ elements in each of $n$
clusters, we had taken an SRS with $nM$ observations, the variance
of the estimated total would have been
\begin{eqnarray*}
V(\hat t_{\mbox{srs}}) &=& (NM)^2 \left(1-\ds\frac{nM}{NM}\right)\ds\frac{S^2}{nM}\\
& =& N^2\left(1-\ds\frac n N\right) \ds\frac{MS^2}{n}
\end{eqnarray*}
$$S_t^2 = \sum\limits_{i=1}^N \ds\frac{(t_i -\bar t_U)^2}{N-1}
= \sum\limits_{i=1}^N \ds\frac{M^2 (\bar y_{iU} -\bar y_U)^2}{N-1}
=M(MSB)$$

$$V(\hat t_{\mbox{cluster}}) = N^2
\left(1-\ds\frac n N\right) \ds\frac{M(\mbox{MSB})}{n}$$ If
$\mbox{MSB}
> S^2$, cluster sampling is less efficient than simple random
sampling
\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Intraclass Correlation Coefficient (ICC)}

For clusters of equal sizes
$$\mbox{ICC} = 1- \ds\frac M {M-1} \cdot \ds\frac {\mbox{SSW}}{\mbox{SSTO}}$$
\bi
\item ICC is defined to be the Pearson correlation coefficient for
the $NM(M-1)$ pairs $(y_{ij},y_{ik})$ for $i$ between $1$ and $N$ and
$j\neq k$

\item $-\ds\frac 1 {M-1} \le \mbox{ICC} \le 1$, since $0\le \mbox{SSW/SSTO} \le 1$
\item If the clusters are perfectly homogeneous $\mbox{SSW}=0$, $\mbox{ICC} = 1$
\item ICC tells us how similar elements in the same cluster are, or provides a measure of homogeneity within the clusters
\ei
\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
Recall: If $\mbox{MSB} > S^2$, cluster sampling is less
efficient than simple random sampling
\bi
\item $\mbox{MSB} = \ds\frac{NM-1}{M(N-1)} S^2 [1+(M-1)\mbox{ICC}]$
\item ICC is positive if elements within a psu tend to be similar.
If the ICC is positive, cluster sampling is less efficient than
simple random sampling\\
----If the clusters occur naturally in the population, ICC is
usually positive\\
----Elements within the same cluster tend to be more similar than
elements selected at random from the population. This occurs
because the elements in a cluster share a similar environment
\ei
\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\bi
\item ICC is negative if elements within a cluster are dispersed more
than a randomly chosen group would be. This force the cluster means
to be very nearly equal\\
-----If $\mbox{ICC}<0$, cluster sampling is more efficient than simple random
sampling of elements\\
----- The $\mbox{ICC}$ is rarely negative in naturally occurring clusters;
negative values can occur in some systematic samples or artificial
clusters
\ei
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}


{\bf Design Effect (deff)}\\
$\mbox{deff(plan, statistic)}$
$$= \ds\frac{V\mbox{(estimator from a sampling
plan)}} {V\mbox{(estimator from an SRS with same number of observation
units)}}$$
\begin{eqnarray*}
 \ds\frac {V(\hat t_{\mbox{cluster}})}{V(\hat t_{\mbox{srs}})} &= &\ds\frac
{MSB}{S^2}\\
 &=& \ds\frac{NM-1}{M(N-1)} [1+(M-1)\mbox{ICC}]\\
&\approx& 1+(M-1)\mbox{ICC}
\end{eqnarray*}
$1+(M-1)\mbox{ICC}$ ssus, taken in a one-stage cluster sample, gives
approximately the same amount of information as one ssu from an
SRS
\end{frame}

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

\begin{frame}

    {\bf{Adjusted $R^2$}}
  
  Adjusted $R^2$ is an alternative measure of homogeneity in general populations:  
        \[
        R_a^2 = 1 - \frac{\text{MSW}}{S^2}
        \]
         \begin{itemize}
         \item It represents the relative amount of variability in the population explained by cluster means, adjusted for the degrees of freedom.  
        \item When clusters are highly homogeneous:  
        \begin{itemize}
            \item Cluster means exhibit high variability relative to within-cluster variation.  
            \item $R_a^2$ will be large, reflecting greater homogeneity.  
        \end{itemize}
        \item Recall:  
        \[
        \text{ICC} = 1 - \frac{M}{M-1} \cdot \frac{\text{SSW}}{\text{SSTO}}
        \]
        \item Adjusted $R^2$ is a numerically adjusted approximation of the ICC, and their values are typically very close. 
    \end{itemize}
\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Example:} Comparison of Two Artificial Populations\\
 
Each population contains three clusters with three elements per cluster.

\begin{tabular}{c|ccc|ccc}
\hline
& \multicolumn{3}{|c|}{Population $A$} & \multicolumn{3}{|c}{Population $B$} \\
\hline
Cluster 1 & 10 & 20 & 30 & 9 & 10 & 11 \\
Cluster 2 & 11 & 20 & 32 & 17 & 20 & 20 \\
Cluster 3 & 9 & 17 & 31 & 31 & 32 & 30 \\
\hline
\end{tabular}

\bi
\item Both populations share the same mean, $\bar y_U = 20$, and variance, $S^2 = 84.5$.  
\item In Population $A$, most of the variability is **within clusters**.  
\item In Population $B$, most of the variability is **between clusters**.  
\ei

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\begin{tabular}{c|cc|cc}
\hline
 & \multicolumn{2}{|c|}{Population
$A$}&\multicolumn{2}{|c}{Population $B$}\\
& $\bar y_{iU}$ & $S_i^2$ & $\bar y_{iU}$ & $S_i^2$\\
\hline
Cluster 1 & 20 & 100 & 10 & 1\\
Cluster 2 & 21 & 111 & 19 & 3\\
Cluster 3 & 19 & 124 & 31 &1\\
\hline
\end{tabular}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
ANOVA Table for population $A$:\\
\begin{tabular}{ccccc}
Source & df & SS & MS & F\\
\hline Between clusters & 2 & 6 &3 & .03\\
Within clusters & 6 & 670& 111.67 & \\
\hline Total & 8 & 676 & 84.5 &
\end{tabular}

\vspace{0.1in}

ANOVA Table for population $B$:\\
\begin{tabular}{ccccc}
Source & df & SS & MS & F\\
\hline Between clusters & 2 & 666& 333 & 199.8 \\
Within clusters & 6 & 10 & 1.67 & \\
\hline Total & 8 & 676 & 84.5
\end{tabular}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
Population $A$:
 $$R_a^2 = -.3215$$
and $$\mbox{ICC} = 1- \ds\frac 3 2\cdot \ds\frac{670}{676} =-.4867$$
\bi
\item Population $A$ has much variation among elements within the
clusters but little variation among the cluster means
\item Elements in the same cluster are actually less similar than randomly selected
elements from the whole population
\item Cluster sampling is more efficient than simple random sampling
\ei
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 Population $B$:
$$R_a^2 = .9803$$
and $$\mbox{ICC} = 1- \ds\frac 3 2\cdot \ds\frac{10}{676} = .9778$$
\bi
\item Most of the variability occurs between clusters, and the clusters
themselves are relatively homogeneous
\item The $\mbox{ICC}$ and $R_a^2$
are very close to $1$, indicating that little new information would
be gained by sampling more than one element in a cluster
\item One-stage cluster sampling is much less efficient than simple random
sampling
\ei
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
{\bf Comments:}
\bi
\item Most real life populations fall somewhere between the
above two extremes
\item The ICC is usually positive but not overly
close to $1$
\item There is a penalty in efficiency for using
cluster sampling, and that decreased efficiency should be offset by
cost savings
\item In general, for a given sample size,
Cluster sampling will produce estimates with the largest variance.
SRS will be intermediate. Stratification will give the smallest
variance.
\ei
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
{\bf Unequal PSU Size}

\begin{tabular}{ccccc}
\hline
psu Totals & $t_1$ & $t_2$ & $\cdots$ & $t_N$ \\
psu Sizes & $M_1$ & $M_2$ & $\cdots$ & $M_N$ \\
\hline
\end{tabular}

\bi
\item Take a Simple Random Sample (SRS) of $n$ psus.  
\item Estimated Population Total:
\[
\hat{t}_{\text{unb}} = N \bar{t} = \ds\frac{N}{n} \sum\limits_{i \in S} t_i
\]
\item Standard Error (SE):
\[
\mbox{SE}(\hat{t}_{\text{unb}}) = N \sqrt{\left(1 - \ds\frac{n}{N}\right) \ds\frac{s_t^2}{n}}
\]
\item Note: $s_t^2$ can be large if $t_i \propto M_i$, i.e., when psu totals are proportional to psu sizes.
\ei

\vspace{0.3cm}
Example:
- The number of physicians in different areas is often proportional to the area's size or population.  \\
- Larger areas or populations tend to have more physicians, resulting in greater variability in $t_i$.

\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
Population mean: $$\bar y_U = \ds\frac{\sum_{i=1}^N
t_i}{\sum_{i=1}^N M_i} =\ds\frac{t}{M_0}$$ where $t_i$ and $M_i$ are
typically positively correlated. Thus, we can consider $\bar y_U = B$ as a ratio estimation by using $M_i$ as the auxiliary
variable.
\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\bi
\item {\bf Unbiased estimator} of overall mean $\bar y_U$: 
$$\hat {\bar y}_U = \hat t_{unb}/M_0={\hat
t_{unb}}/{\sum\limits_{i=1}^N M_i},$$
 but $\sum\limits_{i=1}^N M_i$ may not
be available
\item {\bf Ratio estimator:} $$\hat {\bar y}_r = \ds\frac{\hat t_{\mbox{unb}}}{\hat M_0} = \ds\frac{\sum\limits_{i\in S}
t_i}{\sum\limits_{i\in S} M_i},$$
where the denominator $\sum_{i \in S} M_i$ is a random
quantity that depends on which particular psus are included in the sample. 
$$\hat {\bar y}_r = \ds\frac{\hat t_{\mbox{unb}}}{\hat M_0} =\ds\frac{\sum\limits_{i\in S}
 \sum\limits_{j\in S_i} w_{ij} y_{ij}}{\sum\limits_{i\in S}
 \sum\limits_{j \in S_i} w_{ij}}$$
\ei
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\bi
\item
Let $e_i = t_i - M_i \hat {\bar
y}_r = M_i (\bar y_i - \hat {\bar
y}_r)$
\begin{eqnarray*}
\mbox{SE}(\hat {\bar y}_r)& =& \sqrt{\left(1- \ds\frac n N\right)
\ds\frac{s_e^2}{n\bar
M^2}}\\
&=& \sqrt{\left(1- \ds\frac n N\right)\ds\frac 1 {n \bar M^2}
\ds\frac{ \sum\limits_{i\in S} (t_i - \hat {\bar
y}_r M_i)^2}{n-1}}\\
&=& \sqrt{\left(1- \ds\frac n N\right)\ds\frac 1 {n \bar
M^2}\ds\frac {\sum\limits_{i\in S} M_i^2 (\bar y_i -\hat {\bar
y}_r)^2}{n-1}}
\end{eqnarray*}
$\mbox{SE}(\hat {\bar y}_r)$ depends on variability between $\mbox{psu}$ means
\ei
\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\small
\begin{table}\caption{one stage cluster sampling}
\begin{tabular}{cccc}
\hline
& equal size & unequal size & unequal size\\
& &unbiased estimator & ratio estimator\\
\hline
$\hat t$ & $N \sum\limits_{i\in S} t_i/n$ & $N
\sum\limits_{i\in
S} t_i/n$ & $M_0 \hat {\bar y}_{r}$\\
\hline $\hat{\bar{y}}_U$
 & ${\sum\limits_{i \in S}
t_i}/{nM}$&$\left({N  \sum\limits_{i \in S}
t_i}/n\right)/{\sum\limits_{i=1}^N M_i}$ &
${\sum\limits_{i\in S} t_i}/{\sum\limits_{i\in S}M_i}$\\
\hline
$V(\hat {\bar{y}}_{U})$ & $(1- n/ N){S_t^2}/{nM^2}$ &$N^2 (1-
n/N){S_t^2}/{n}M_0^2$ & $(1-n/N)S_e^2 /n \bar M^2$\\
\hline
\end{tabular}
\end{table}
}
Notes: If all $M_i$'s are equal, the unbiased estimator is in fact
the same as the ratio estimator; If the $M_i$'s vary, the unbiased
estimator often performs poorly
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Example 5.6: One-Stage Cluster Sampling}

\bi \item One-stage cluster sampling is commonly used in educational studies because students are naturally grouped into clusters such as classrooms or schools.
\item Consider a population of 187 high school algebra classes ($N = 187$) in a city.
\item An investigator randomly selects a simple random sample (SRS) of 12 classes ($n = 12$) and administers a test on function knowledge to all students in the selected classes.
\item The (hypothetical) data for this study are provided in the file {\tt algebra.dat}, along with the following summary statistics.
\ei

\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[fragile]
\begin{verbatim}
> nrow(algebra)
[1] 299
> head(algebra)
  class    Mi score
1    23    20    57
2    23    20    90
3    23    20    56
4    23    20    57
5    23    20    46
6    23    20    55
\end{verbatim}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\begin{table}\caption{Example 5.6}
\begin{tabular}{ccccc}
\hline  Class number & $M_i$ & $\bar y_i$ & $t_i$ & $M_i^2
(\bar y_i - \hat{\bar y}_r)^2$\\
\hline
23 & 20& 61.5& 1,230& 456.7298 \\
37 &26 &64.2 &1,670 &1,867.7428 \\
\vdots & & & & \\
108 &26 &67.2 &1,746 &14212.7867 \\
\hline
Total &299 & &18,708 &194,827.0387 \\
\hline
\end{tabular}
\end{table}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
$$\hat {\bar y}_r = \ds\frac{\sum_{i\in S} t_i}{\sum_{i \in S} M_i} =\ds\frac{18,708}{299} = 62.57$$
\begin{eqnarray*}
\mbox{SE}(\hat {\bar y}_r)& =& \sqrt{\left(1- \ds\frac n
N\right)\cdot \ds\frac 1 {n \bar M^2}\cdot \ds\frac
{\sum\limits_{i\in S} M_i^2
(\bar y_i -\hat {\bar y}_r)^2}{n-1}}\\
&=& \sqrt{\left(1-\ds\frac{12}{187}\right)\cdot \ds\frac 1{ 12\cdot
24.92^2} \cdot \ds\frac{194,827}{11}} \\
&=& 1.49
\end{eqnarray*}
A 95\% CI is given by
$$62.57 \pm 2.20*(1.49) = [59.29, 65.85],$$
where 2.20 is the percentile from a $t$ distribution with $n-1 = 12-1 = 11$ df.
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[fragile]
\begin{verbatim}
> algebra$sampwt<-rep(187/12,299)
> dalg<-svydesign(id=~class,weights=~sampwt,
fpc=~rep(187,299), data=algebra) 
> dalg
1 - level Cluster Sampling design
With (12) clusters.

> svymean(~score,dalg)
        mean     SE
score 62.569 1.4916
> degf(dalg)
[1] 11
> confint(svymean(~score,dalg),level=.95,df=11)
         2.5 %  97.5 %
score 59.28562 65.8515
\end{verbatim}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
{\bf Two-Stage Cluster Sampling}

If items within a cluster are highly similar, measuring all of them may be unnecessary. Instead, a more efficient approach is to take a simple random sample (SRS) of units within each selected primary sampling unit (psu).

\bi
\item Stage 1: Select an SRS of $n$ psus from the population of $N$ psus.
\item Stage 2: From each selected psu, draw an SRS of $m_i$ units.
\ei

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}  

{\bf Unbiased Estimator}  

To estimate $t_i$, we use:  
$$\hat{t}_i = M_i \bar{y}_i, \quad \text{where } \bar{y}_i = \frac{1}{m_i} \sum_{j \in S_i} y_{ij}.$$  

The estimated total is:  
$$\hat{t}_{\text{unb}} = N \bar{t} = \frac{N}{n} \sum_{i \in S} \hat{t}_i = \frac{N}{n} \sum_{i \in S} M_i \bar{y}_i = \sum_{i \in S} \sum_{j \in S_i} \frac{N}{n} \cdot \frac{M_i}{m_i} y_{ij}.$$  

The probability of selecting the $j$th ssu within the $i$th psu is:  
\begin{eqnarray*}
& & p (\mbox{jth ssu in ith psu is selected})\\
 &=& p(\mbox{ith psu selected}) \times
p(\mbox{jth ssu selected}|\mbox{ith psu
selected})\\
&=& \ds\frac n N \cdot \ds\frac{m_i}{M_i}
\end{eqnarray*}

Thus, the unbiased estimator can also be written as:  
$$\hat{t}_{\text{unb}} = \sum_{i \in S} \sum_{j \in S_i} w_{ij} y_{ij}, \quad \text{where } w_{ij} = \frac{N M_i}{n m_i}.$$  

\end{frame}  

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}  

{\bf Variance for Two-Stage Cluster Sampling}  

The variance of $\hat{t}_{\text{unb}}$ consists of two components:  
\bi
\item The variance from one-stage cluster sampling ($S_t^2$).  
\item An additional term to account for the extra variance ($S_i^2$) due to estimating the $\hat{t}_i$'s rather than measuring them directly.  
\ei  

The variance is expressed as:  
$$  
V(\hat{t}_{\text{unb}}) = N^2\left(1-\frac{n}{N}\right)\frac{S_t^2}{n} +  
\frac{N}{n} \sum_{i \in S} \left(1-\frac{m_i}{M_i}\right)M_i^2  
\frac{S_i^2}{m_i}.  
$$  

\end{frame}  

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

\begin{frame}  

{\bf Estimated Variance for Two-Stage Cluster Sampling}  

\textbf{Between-Cluster Variance:}  
\bi  
\item Viewing the $\hat{t}_i$ as an SRS:  
$$  
s_t^2 = \frac{1}{n-1} \sum_{i \in S} \left(\hat{t}_i - \frac{\hat{t}_{\text{unb}}}{N} \right)^2  
$$  
\ei  

\textbf{Within-Cluster Variance:}  
\bi  
\item Viewing the $y_{ij}$ as an SRS.  
\item For cluster $i$,  
$$  
s_i^2 = \frac{1}{m_i-1} \sum_{j \in S_i}(y_{ij} - \bar{y}_i)^2  
$$  

The estimated variance of $\hat{t}_{\text{unb}}$ is given by:  
$$  
\hat{V}(\hat{t}_{\text{unb}}) = N^2\left(1-\frac{n}{N}\right)\frac{s_t^2}{n} +  
\frac{N}{n} \sum_{i \in S} \left(1-\frac{m_i}{M_i}\right)M_i^2 \frac{s_i^2}{m_i}  
$$  
\ei  

\end{frame}  

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Summary of unbiased estimators for Two-Stage Cluster Sampling:}
\bi
\item $\hat t_{\mbox{unb}} = N \bar t = \ds\frac N n \sum_{i \in S} \hat t_i = \ds\frac N n \sum_{i \in S} M_i \bar y_i =\sum_{i\in S} \sum_{j\in S_i} \ds\frac N n \ds\frac{M_i}{m_i} y_{ij}$
\item $\hat V(\hat t_{\mbox{unb}}) = N^2\left(1-\ds\frac n N\right)\ds\frac{s_t^2}{n} +
\ds\frac N n \sum_{i \in S} \left(1-\ds\frac{m_i}{M_i}\right)M_i^2
\ds\frac {s_i^2}{m_i}$
\item $\hat{\bar y}_{\mbox{unb}} = \ds\frac{\hat t_{\mbox{unb}}}{M_0}$
\item $\mbox{SE}(\hat{\bar y}_{\mbox{unb}}) = \ds\frac{\mbox{SE}(\hat t_{\mbox{unb}})}{M_0}$
\ei
\end{frame}

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

\begin{frame}  

{\bf Recall:}  
\begin{align*}  
s_t^2 &= \frac{1}{n-1} \sum_{i \in S} \left(\hat{t}_i - \frac{\hat{t}_{\text{unb}}}{N} \right)^2 \\  
\hat{V}(\hat{t}_{\text{unb}}) &= N^2\left(1-\frac{n}{N}\right)\frac{s_t^2}{n}  
+ \frac{N}{n} \sum_{i \in S} \left(1-\frac{m_i}{M_i}\right)M_i^2 \frac{s_i^2}{m_i}  
\end{align*}  

{\bf Comments:}  
\begin{itemize}  
\item The term $s_t^2$ can be very large, as it is influenced both by variations in the unit sizes ($M_i$) and by variations in the cluster means ($\bar{y}_i$).  
\item When cluster sizes vary significantly, this component becomes dominant, even if the cluster means remain relatively stable.  
\end{itemize}  

\end{frame}  

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

\begin{frame}
{\bf Ratio Estimation}

Let:  
\begin{itemize}
    \item $y$: Cluster totals $t_i$  
    \item $x$: Cluster sizes $M_i$  
\end{itemize}

The ratio estimator of the mean:  
\[
\hat{\bar{y}}_r = \frac{\sum\limits_{i \in S} \hat{t}_i}{\sum\limits_{i \in S} M_i} = \frac{\sum\limits_{i \in S} M_i \bar{y}_i}{\sum\limits_{i \in S} M_i}
\]

Recall: $w_{ij} = \frac{N M_i}{n m_i}$,  

\[
\hat{\bar{y}}_r = \frac{\hat{t}_{\text{unb}}}{\hat{M}_0} = \frac{\sum\limits_{i \in S} \sum\limits_{j \in S_i} w_{ij} y_{ij}}{\sum\limits_{i \in S} \sum\limits_{j \in S_i} w_{ij}}
\]
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
The variance of the ratio estimator:  
\[
\hat{V}(\hat{\bar{y}}_r) = \frac{1}{\bar{M}^2} \left[\left(1-\frac{n}{N}\right) \frac{s_r^2}{n} + \frac{1}{nN} \sum\limits_{i \in S} M_i^2 \left(1-\frac{m_i}{M_i}\right) \frac{s_i^2}{m_i}\right]
\]

where:  
$s_r^2 = \ds\frac{\sum\limits_{i\in S}(M_i \bar y_i - M_i \hat {\bar
y}_r)^2}{n-1}$
 and $\bar M = \hat M_0/N$ is the average size of
the psus in the sample.

{\bf Note:} For $\hat{V}(\hat{t}_{\text{unb}})$ and $\hat{V}(\hat{\bar{y}}_r)$, the second term is typically negligible compared to the first term.  
Most survey software packages calculate the variance using only the first term.  

\end{frame}

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

\begin{frame}

{\bf Example 5.9: The Case of the Six-Legged Puppy}

\begin{itemize}
    \item Goal: Estimate the average number of legs on the healthy puppies in the sampled city's puppy homes.
    \item The sampled city has two puppy homes:
    \begin{itemize}
        \item Puppy Palace (PP) with 30 puppies.
        \item Dog's Life (DL) with 10 puppies.
    \end{itemize}
    \item Sampling procedure:
    \begin{itemize}
        \item Select one puppy home with probability $1/2$.
        \item After a home is selected, randomly select 2 puppies from that home.
    \end{itemize}
    \item Estimation methods:
    \begin{itemize}
        \item Use $\hat{\bar{y}}_{\text{unb}}$ to estimate the average number of legs per puppy.
        \item Use ratio estimation to estimate the average number of legs per puppy.
    \end{itemize}
\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\begin{itemize}
    \item Population: $N = 2$ puppy homes.
    \item Sampling procedure:
    \begin{itemize}
        \item Select $n = 1$ home with probability $1/2$.
        \item Randomly select 2 puppies from the selected home.
    \end{itemize}
\end{itemize}

\textbf{Case 1: Puppy Palace (PP) is selected}
\begin{itemize}
    \item Each of the two sampled puppies has 4 legs.\\
          Estimated total number of legs in PP: 
          $$\hat{t}_{PP} = 30 \times 4 = 120$$
    \item Unbiased estimate of the total number of puppy legs across both homes:
          $$\hat{t}_{\text{unb}} = 2 \times \hat{t}_{PP} = 240$$
    \item Mean number of legs per puppy:
          $$\hat{\bar{y}}_{\text{unb}} = \frac{\hat{t}_{\text{unb}}}{40} = \frac{240}{40} = 6$$
\end{itemize}

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Case 2: Dog's Life (DL) is selected}

\begin{itemize}
    \item Each of the two sampled puppies has 4 legs.\\
          Estimated total number of legs in DL: 
          $$\hat{t}_{DL} = 10 \times 4 = 40$$
    \item Unbiased estimate of the total number of puppy legs across both homes:
          $$\hat{t}_{\text{unb}} = 2 \times \hat{t}_{DL} = 80$$
    \item Mean number of legs per puppy:
          $$\hat{\bar{y}}_{\text{unb}} = \frac{\hat{t}_{\text{unb}}}{40} = \frac{80}{40} = 2$$
\end{itemize}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
Comments:
\bi
\item The estimator is mathematically unbiased: $(6 + 2)/2 = 4$, ensuring that the average over all possible samples yields the correct value
\item $\hat {\bar y}_{\mbox{unb}}$ is unbiased, but exhibits significant variability due to the considerable variation in $M_i$ values (e.g., 30 vs. 10)
\begin{eqnarray*}
V(\hat t_{\mbox{unb}}) &=& \left(1-\ds\frac 1 2\right) 2^2 S_t^2 + 2
\sum\limits_{i=1}^2\left(1-\ds\frac{m_i}{M_i}\right)M_i^2
\ds\frac{S_i^2}{m_i}\\
& =& 6400
\end{eqnarray*}

\item when $M_i$'s unequal, the unbiased estimators
$\hat {\bar y}_{\mbox{unb}}$ are often inefficient
\ei
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Ratio estimators:}
\begin{itemize}
    \item Suppose we select Puppy Palace (PP):
    $$\hat{\bar{y}}_r = \frac{\sum_{i \in S} \hat{t}_i}{\sum_{i \in S} M_i} = \frac{30 \times 4}{30} = 4$$
    
    \item Suppose we select Dog's Life (DL):
    $$\hat{\bar{y}}_r = \frac{10 \times 4}{10} = 4$$
    
    \item The ratio estimates are the same for the two possible samples, implying that 
    $$V(\hat{\bar{y}}_r) = 0.$$
\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Example 5.8}: 
\bi
\item The data coots.dat come from Arnold's (1991) work
on egg size and volume of American coot eggs in Minnedosa, Manitoba
\item In this data set, we look at volumes of a subsample of eggs in
clutches (nests of eggs) with at least two eggs available for
measurement
\item Randomly select 2 eggs in each clutch and measure their volume
\item Want to estimate the mean egg volume
\ei
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[fragile]
\begin{verbatim}
 nrow(coots) #368
[1] 368
> head(coots)
  clutch csize length breadth volume   tmt
1      1    13   44.3    31.1  3.80      1
2      1    13   45.9    32.7  3.93      1
3      2    13   49.2    34.4  4.22      1
4      2    13   48.7    32.7  4.17      1
5      3     6   51.0    34.2  0.932     0
6      3     6   49.4    34.4  0.901     0
> coots$ssu<-rep(1:2,184) 
> coots$relwt<-coots$csize/2
> dcoots<-svydesign(id=~clutch+ssu,weights=~relwt,
data=coots)
> dcoots
2 - level Cluster Sampling design (with replacement)
With (184, 368) clusters.
svydesign(id = ~clutch + ssu, weights = ~relwt, 
data = coots)
\end{verbatim}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[fragile]
\begin{verbatim}
svymean(~volume,dcoots)  #ratio estimator
         mean    SE
volume 2.4908 0.061
> confint(svymean(~volume,dcoots),level=.95,df=183)
          2.5 %   97.5 %
volume 2.370423 2.611134
> dcoots2<-svydesign(id=~clutch,weights=~relwt,data=coots)
> dcoots2
1 - level Cluster Sampling design (with replacement)
With (184) clusters.
> svymean(~volume,dcoots2)  
         mean    SE
volume 2.4908 0.061
\end{verbatim}
SE is the with replacement variance for the first stage sampling. 
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
Suppose there are $N$ clutches, weight for egg $j$ in clutch $i$ is
$$w_{ij}=\ds\frac N n \ds\frac{M_i}{m_i} = \ds\frac N{184} \ds\frac{M_i} 2$$
$$\hat{\bar y}_r = \ds\frac{\sum_{i \in S} \sum_{j\in S_i} w_{ij} y_{ij}} {\sum_{i \in S} \sum_{j\in S_i} w_{ij}} = \ds\frac{4375.947}{1757} = 2.49$$
$$s_r^2 = \ds\frac 1{n-1} \sum_{i\in S} M_i^2 (\bar y_i - \hat{\bar y}_r)^2 = \ds\frac{11438.99}{183} = 62.51$$
$\bar M = \sum_{i\in S} M_i /n = 1758/184 = 9.554$
$$\hat V(\hat{\bar y}_r) = \ds\frac 1{9.554^2} \left[\left(1- \ds\frac{184}{N}\right) \ds\frac{62.51}{184} + \ds\frac 1 N \ds\frac{46.31}{184}\right]$$

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
Now $N$, the total number of clutches in the population, is unknown but presumed to be
large (and known to be larger than 184). 
\bi 
\item take the psu-level fpc to be 1
\item  second term in the estimated variance is small relative to the first
term, and goes to 0 when $N \rightarrow \infty$
\ei

$$\hat V_{WR} (\hat{\bar y}_r) = \ds\frac 1{9.554^2} \ds\frac{62.51}{184}=0.0037,$$
which is the first-stage with replacement variance. 
$$\widehat{ SE}_{WR} (\hat{\bar y}_r) = \sqrt{0.0037} = 0.06082763$$
This is the SE reported from R without the fpc statement. 

\end{frame}

\begin{frame}
Now suppose $N = 10000$, 
\begin{eqnarray*}
\hat V(\hat{\bar y}_r) & =& \ds\frac 1{9.554^2} \left[\left(1- \ds\frac{184}{N}\right) \ds\frac{62.51}{184} + \ds\frac 1 N \ds\frac{46.31}{184}\right]\\
& =& 0.003653388 + 2.757316e-07\\
&=& 0.003653663
\end{eqnarray*}
$$\widehat{ SE}_{WOR} (\hat{\bar y}_r) = \sqrt{0.003653663} = 0.06044554$$
Slightly different from:
$$\widehat{ SE}_{WR} (\hat{\bar y}_r) = \sqrt{0.0037} = 0.06082763$$
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Note on Example 5.8:}
\begin{itemize}
    \item We could not use the unbiased estimator for the mean or population totals because the values of $N$ (the total number of clusters) and $M_0 = \sum_{i=1}^N M_i$ (the total size of all clusters) are unknown.
    \item However, since the $M_i$'s did not vary widely, the unbiased estimator would likely have had a similar coefficient of variation to the ratio estimator.
    \item If all the $M_i$'s are equal, the unbiased estimator coincides with the ratio estimator.
    \item If the $M_i$'s vary significantly, the unbiased estimator often performs poorly.
\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
In {\it svydesign},
\bi
\item the two stages of the cluster sampling are given as id=~clutch+ssu . 
\item when the with-replacement variance is calculated,
however, as is done here, you need only specify the psus
\item the point and variance estimates
are the same whether you specify just the psus or you specify all stages of sampling. 
\item the
weight argument must be included for this design because the weights are unequal.
\item confidence interval uses a $t$ critical value with 183 df (number of psus minus
1). 
\item svydesign does not contain the fpc argument. This is because the total
number of clutches in the population, $N$, is unknown.\\
----- as a result, the svymean does not use
an fpc when calculating estimates.
\item  In general, we recommend omitting the fpc argument
for multi-stage cluster sampling even when $N$ is known. 
\ei
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\bi
\item 
without fpc, it produces a variance estimate whose expectation
is slightly larger than the true variance, but if n/N is small, the difference is negligible.
\item 
if forced to choose between a standard error that is slightly too large and one that is too
small, we usually prefer the former because a too-small standard error leads to claiming
that estimates are more precise than they really are.
\item  ssus in the same psu are usually more homogeneous than randomly selected ssus
from the population. Thus, the essential feature for calculating standard errors is to capture
that homogeneity by including the id=~psuid argument in svydesign. The issue of ``to fpc
or not to fpc"  is minor compared with the effect of clustering.
\ei
\end{frame}

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

\begin{frame}

{\bf Two-Stage Cluster Sampling: Calculate Variance with and without Replacement}\\
{\bf Example 5.7.}

The file {\it schools} contains data from a two-stage sample of students. 
\begin{itemize}
    \item In the first stage of sampling, an SRS of $n = 10$ schools is selected from a population of $N = 75$ schools.
    \item In the second stage, an SRS of $m_i = 20$ students is selected from each sampled school, and assessments for reading and math are administered.
    \item These data are fictional, but the summary statistics are consistent with those typically seen in educational studies.
\end{itemize}

\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\small
\begin{frame}[fragile]
\begin{verbatim}
> data(schools)
> print.data.frame(head(schools))
  schoolid gender math reading mathlevel readlevel  Mi finalwt
1        9      F   42      42         2         2 163  61.125
2        9      F   29      30         1         1 163  61.125
3        9      M   31      25         1         1 163  61.125
4        9      F   22      33         1         2 163  61.125
5        9      M   35      36         1         2 163  61.125
6        9      F   30      17         1         1 163  61.125
> unique(schools$schoolid)
 [1]  9 17 18 22 35 43 46 55 62 75
\end{verbatim}
\end{frame}
}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\begin{figure}[!htp]
       \centering
       \includegraphics[width=0.9\textwidth,
       height=0.5\textheight]{./plots/page188.png}
\caption{Example 5.7, direct calculation for mean math scores}
       \label{archers}
     \end{figure}
 \begin{center}
        \textit{\scriptsize Source: Table 5.7 of Sampling: Design and Analysis, 3rd edition, by Sharon L. Lohr}
    \end{center}
\end{frame}

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

\begin{frame}
We use the ratio estimator to estimate the mean math score. From equation (5.30),
$$\hat {\bar y}_r = \frac{\sum\limits_{i\in S} \hat t_i}{\sum\limits_{i\in S} M_i} = \frac{76282.15}{2303} = 33.12.$$
Alternatively, we can use equation (5.31) to calculate $\hat {\bar y}_r$ using the sampling weights (as is done by most software packages). The weight for student $j$ in school $i$ is
$$w_{ij} = \frac{N}{n} \cdot \frac{M_i}{m_i} = \frac{75}{10} \cdot \frac{M_i}{20}.$$
The estimated mean math score is
$$\hat{\bar y}_r = \frac{\sum_{i \in S} \sum_{j \in S_i} w_{ij} y_{ij}} {\sum_{i \in S} \sum_{j \in S_i} w_{ij}} = \frac{572116.1}{17272.5} = 33.12.$$
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf With-replacement variance} 

The weights do not allow us to calculate the standard error directly. We need the clustering information for that. From Table 5.7, 
$$s_r^2 = \frac{1}{n-1} \sum_{i \in S} M_i^2 (\bar y_i - \hat{\bar y}_r)^2 = \frac{14784382}{9} = 1642709.$$
Also, the average size of the sampled clusters is
$$\bar M = \frac{\sum_{i \in S} M_i}{n} = \frac{2303}{10} = 230.3.$$
The with-replacement variance is given by:
$$\hat V(\hat{\bar y}_r) = \frac{s_r^2}{n \bar{M}^2} = \frac{1642709}{10 \times 230.3^2} = 3.097.$$
Finally, the standard error is:
$$SE_{WR}(\hat {\bar y}_r) = \sqrt{3.097} = 1.76.$$
This result matches the R output provided below.

\end{frame}

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

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[fragile]
\begin{verbatim}
# calculate with-replacement variance; no fpc argument
> # include psu variable in id; include weights
> dschools<-svydesign(id=~schoolid,weights=~finalwt,
data=schools)
> dschools
1 - level Cluster Sampling design (with replacement)
With (10) clusters.
svydesign(id = ~schoolid, weights = ~finalwt, 
data = schools)
\end{verbatim}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[fragile]
\begin{verbatim}
mathmean<-svymean(~math,dschools)
> mathmean
       mean     SE
math 33.123 1.7599
> degf(dschools) 
[1] 9
> # use t distribution for confidence intervals because
 there are only 10 psus
> confint(mathmean,df=degf(dschools))
        2.5 %  97.5 %
math 29.14179 37.1041

\end{verbatim}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Without-replacement variance} 

Recall that 
$$\hat V(\hat {\bar y}_r) = \ds\frac 1 {\bar M^2} \left[\left(1-\ds\frac n N\right)
\ds\frac{s_r^2} n + \ds\frac 1 {nN} \sum\limits_{i\in S} M_i^2
\left(1-\ds\frac{m_i}{M_i}\right)\ds\frac{s_i^2}{m_i}\right]$$ where
$s_r^2 = \ds\frac{\sum\limits_{i\in S}(M_i \bar y_i - M_i \hat {\bar
y}_r)^2}{n-1}$
 and $\bar M$ is the average psu size

\begin{eqnarray*}
\hat V(\hat {\bar y}_r) &=& \ds\frac 1 {(230.3)^2} \left[ \left(1-\ds\frac {10}{75}\right) \ds\frac{1642709}{10} +
\ds\frac{1}{75} \ds\frac{2902087}{10}\right]\\
&=& 2.684 + 0.073\\
&=& 2.757
\end{eqnarray*}
$$SE(\hat {\bar y}_r) = \sqrt{2.757} = 1.66$$
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[fragile]
Without-replacement variance
\begin{verbatim}
# create a variable giving each student an id number
> schools$studentid<-1:(nrow(schools))
> # calculate without-replacement variance 
> # specify both stages of the sample in the id argument
> # give both sets of population sizes in the fpc argument
> # do not include the weight argument
> dschoolwor<-svydesign(id=~schoolid+studentid,
fpc=~rep(75,nrow(schools))+Mi, data=schools)
> dschoolwor
2 - level Cluster Sampling design with (10, 200) clusters.
svydesign(id = ~schoolid + studentid, 
fpc = ~rep(75, nrow(schools)) + Mi, data = schools)
\end{verbatim}
\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[fragile]
\begin{verbatim}
mathmeanwor<-svymean(~math,dschoolwor)
> mathmeanwor
       mean     SE
math 33.123 1.6605
> confint(mathmeanwor,df=degf(dschoolwor))
        2.5 %   97.5 %
math 29.36667 36.87923
\end{verbatim}
\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Comparison of with and without replacement variance}

\begin{tabular}{ccccc}
\hline
Variable & DF & Mean & SE & 95\% CI \\
\hline
WR & 9 & 33.123 & 1.76 & [29.14, 37.10] \\
WOR & 9 & 33.123 & 1.66 & [29.37, 36.88] \\
\hline
\end{tabular}

\begin{itemize}
    \item The second term, 0.073, in the without-replacement variance is smaller than the first term, 2.684.
    \item The standard error with replacement, 1.76, is close to the standard error without replacement, 1.66.
    \item The confidence interval from the without-replacement method is narrower than the confidence interval from the with-replacement method.
    \item In general, we recommend omitting the finite population correction (fpc) argument for multi-stage cluster sampling.
\end{itemize}

\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
{\bf Design issues:}
\bi
\item what precision?
\item what are the size of psus?
\item how many ssus per psu?
\item how many psus?
\ei
Goal of designing a survey:
\bi
\item to get the most information
possible for the least cost and inconvenience.
\ei
\end{frame}

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

\begin{frame}

{\bf psu size}
\begin{itemize}
    \item The psu size is often a natural unit. \\
    Example: clutches, farms, classes, schools.
    \item In some surveys, the investigator may have a wide choice for psu size. \\
    Example: Estimating the sex and age ratios of mule deer in a region of Colorado.
    \item[] psu: Designed areas
    \item[] ssu: Might be individual deer or groups of deer in those areas
    \item[] Size of psus might be 1 $\mbox{km}^2$, 2 $\mbox{km}^2$, or 100 $\mbox{km}^2$
    \item[] Usually, the larger the psu size, the more variability you expect to see within a psu. Hence, you expect $R_a^2$ and \text{ICC} to be smaller with a large psu than with a small psu. However, if the psu size is too large, you may lose the cost savings of cluster sampling.
\end{itemize}

\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Comments:}
\bi
\item Bellhouse, D. R. (1984). A review of optimal designs
in survey
sampling. {\it The Canadian Journal of Statistics}, 12, 53-65.\\
------reviews optimal designs for sampling\\
------provides useful guidance for designing a survey
\item There are many ways to ``try out" different psu sizes before taking
a survey\\
------ use different combinations of
$R_a^2$ and M, and compare the costs. \\
-------pilot study, perform an experiment and collect data on
relative costs and variances with different psu sizes.
\ei
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Designing a two-stage cluster survey}

{\bf Minimize the variance for a fixed cost}
$$V(\hat t_{\mbox{unb}}) = N^2\left(1-\ds\frac n N\right) \ds\frac{S_t^2}{n} +
\ds\frac N n \sum\limits_{i=1}^N \left(1-
\ds\frac{m_i}{M_i}\right)M_i^2 \ds\frac{S_i^2}{m_i}$$ If $M_i =M,
m_i = m$ for all psus

\begin{equation}\label{eq:variance}
V(\hat{\bar y}_{\mbox{unb}}) = \left(1-\ds\frac n N\right)
\ds\frac{\mbox{MSB}}{nM} +\left(1-\ds \frac m M\right)
\ds\frac{\mbox{MSW}}{nm}
\end{equation}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Choosing Subsampling Sizes}

 $R_a^2 = 1- \ds\frac {\mbox{MSW}}{S^2}$
\bi
\item a measure of homogeneity in general population
\item $\mbox{MSW} =0 \rightarrow R_a^2=1$, all elements within a cluster have the value
of the cluster mean, need only subsample one element
\item for other values of $R_a^2$, optimal allocation depends on the
relative cost of sampling psus and ssus
\ei
\end{frame}

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

\begin{frame}

{\bf Minimum Cost}
\begin{itemize}
    \item One approach to determining sample sizes is to consider costs.
    \item Let $c_1$ be the cost of measuring each psu, and $c_2$ be the cost of measuring each ssu.
    $$\text{Total cost} = C = c_1 n + c_2 nm$$
    \item Minimize equation (\ref{eq:variance}) to get:
    $$n_{\text{opt}} = \frac{C}{c_1 + c_2 m_{\text{opt}}}$$
    $$m_{\text{opt}} = \sqrt{\frac{c_1 M(N-1)(1 - R_a^2)}{c_2 (NM - 1) R_a^2}}$$
\end{itemize}
\end{frame}


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


\begin{frame}

{\bf Example 5.10:}

Recall Example 5.2, where a student wants to estimate the average grade point average (GPA) in his dormitory. He adopted a one-stage cluster sampling plan.
\begin{itemize}
    \item Chooses 5 suites ($n = 5$) from the 100 suites ($N = 100$) at random.
    \item Asks every person (4 students per suite, $M_i = m_i = 4$) in the 5 suites what their GPA is.
\end{itemize}

{\bf Question:} Would subsampling have been more efficient for this case than the one-stage cluster sample used in Example 5.2?

\begin{itemize}
    \item Set total cost $C = 300$.
    \item Set different sets of $(c_1, c_2)$ by $(40, 5)$, $(10, 20)$, and $(20, 10)$.
    \item Consider subsample sizes $m = (1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0)$.
    \begin{itemize}
        \item Calculate the corresponding number of psus by $n_{\text{opt}} = \frac{C}{c_1 + c_2 m}$.
        \item Calculate $\hat V(\hat {\bar y}_{\text{unb}})$.
    \end{itemize}
    \item Plot $\hat V(\hat {\bar y}_{\text{unb}})$ vs $m$.
\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\begin{figure}[!htp]
       \centering
       \includegraphics[width=0.90\textwidth,
       height=1.5\textheight]{./plots/page207.pdf}
\caption{Example 3.10}
       \label{archers}
     \end{figure}


\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Unequal psu Size (Unequal $M_i$s)}

\begin{itemize}
    \item Substitute $\bar M$ for $M$ in the above work, and decide the average subsample size $\bar m$.
    \begin{itemize}
        \item Either take $\bar m$ observations in every cluster.
        \item Or allocate observations so that $\frac{m_i}{M_i} = \text{constant}$.
    \end{itemize}
    \item As long as the $M_i$s do not vary too much, this should produce a reasonable design.
    \item If the $M_i$s are widely variable, and the $t_i$s are correlated with the $M_i$s, a cluster sample with equal probabilities is not necessarily very efficient; an alternative design should be considered.
\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Choosing number of psus}

Assume: clusters are of equal size
\begin{eqnarray*}
V(\hat{\bar y}_{\mbox{unb}}) &=& \left(1-\ds\frac n N\right)
\ds\frac{\mbox{MSB}}{nM}
+\left(1-\ds\frac m M\right) \ds\frac {\mbox{MSW}}{nm}\\
&\le& \ds\frac 1 n \left[\ds\frac{\mbox{MSB}}{M} +\left(1- \ds\frac
m M\right)
\ds\frac{\mbox{MSW}}{m}\right]\\
&=& \ds\frac v n
\end{eqnarray*}
An approximate $100(1-\alpha)$\% CI is $\hat{\bar y}_{\mbox{unb}}
\pm
z_{\alpha/2} \sqrt{\ds\frac 1 n v}$\\
If desired precision is $e$, then $e= z_{\alpha/2} \sqrt{\ds\frac 1
n v}$\\
$n = z_{\alpha/2}^2 v/e^2$, $v$ could be from a prior survey in
literature
\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
{\bf Systematic sampling:}
\bi
\item A special case of cluster
sampling
\item Have a list of $m$ units, take every $k$th one
randomly\\
Example: Want to take a systematic sample of size 3 from a
population that has 12 elements: $1,2,3,4,5,6,7,8,9,10,11,12$
\item Choose a number randomly between 1 and 4
\item Draw that element and every fourth element thereafter
\item The population contains $N=4$ psus\\
$S_1 = \{1,5,9\}, \quad S_2 =\{2,6,10\}$\\
 $S_3 = \{3,7,11\}, \quad S_4 = \{4,8,12\}$
\item Take an SRS of one psu
\ei
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
Consider a population with $NM$ elements.
\begin{itemize}
    \item There are $N$ possible choices for the systematic sample, each of size $M$.
    \item We observe the mean of the one cluster that comprises our systematic sample:
    $$\bar y_i = \bar y_{iU} = \hat{\bar y}_{\mbox{sys}}.$$
\end{itemize}

\newpage

\textbf{Properties of $\hat{\bar y}_{\mbox{sys}}$}
\begin{itemize}
    \item $E[\hat{\bar y}_{\mbox{sys}}] = \bar y_U$.
    \item For a simple systematic sample, select $n = 1$ of the $N$ clusters.
\end{itemize}
\end{frame}

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

\begin{frame}
\begin{eqnarray*}
    V(\hat{\bar y}_{\mbox{sys}}) &=& \left(1 - \frac{1}{N}\right) \frac{S_t^2}{M^2} \\
    &=& \left(1 - \frac{1}{N}\right) \frac{\mbox{MSB}}{M} \\
    &\approx& \frac{S^2}{M} \left[1 + (M - 1) \cdot \text{ICC}\right]
\end{eqnarray*}
$$\mbox{ICC}= 1-\ds\frac {M}{M-1} \ds\frac{\mbox{SSW}}{\mbox{SSTO}}$$
\bi
 \item ICC is a measure of homogeneity within clusters
 \item $\mbox{ICC}>0$ or large, there is little
variation within the systematic samples relative to that in the
population, then the elements in the sample all give similar
information, systematic sampling would be
 expected to have higher variance than an SRS
 \item $\mbox{ICC}<0$, if elements within the systematic sample (psu) are more diverse than SRS would be, systematic
 sampling would be more efficient than an SRS
\ei
\end{frame}


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

\begin{frame}
\textbf{Notes:}
\begin{itemize}
    \item Since $n = 1$ in systematic sampling, we cannot obtain an unbiased estimate of $V(\hat{\bar y})$.
    \item If the sampling frame is in random order, systematic sampling is a good choice.
    \item A potential danger of systematic sampling occurs when the sampling frame follows a regular pattern, such as Male, Female, Male, Female, Male, Female, etc.
    \item Systematic sampling is often used when a researcher wants a representative sample of the population but does not have the resources to construct a complete sampling frame in advance.
\end{itemize}
\end{frame}

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

\begin{frame}
\frametitle{Disclaimer}

Slides are intended for a course based on the book: \textit{Sampling: Design and Analysis, Third Edition} by Sharon L. Lohr

All examples, datasets, and pages directly scanned and included in this chapter are from the textbook.

References to papers or books cited in these slides can be found in the Bibliography section of the textbook.
\end{frame}

\end{document}
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