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The type of cluster design we have already discussed is called a one-stage cluster sampling design, which can be described as having the following steps.
A two-stage cluster sampling design can be described as follows.
Additional notation is necessary for two-stage cluster sampling designs. Let \(M_i\) denote the number of elements in the \(i\)-th cluster, and let \(m_i\) denote the number of sampled elements in the \(i\)-th cluster. So \(m_i \le M_i\) for all clusters, and \(m_i < M_i\) for at least some of the clusters for a two-stage cluster sampling design.
Note: Do not confuse “two-stage” sampling with “two-phase” sampling — the latter is sometimes used describe double sampling designs. While there are some similarities between two-stage and two-phase sampling these are largely superficial.
Example: Consider the following sampling design. The population, clusters, and their respective sizes are as follows. \[ \mathcal{P} = \{ \underbrace{\mathcal{E}_1, \mathcal{E}_2, \mathcal{E}_3}_{\mathcal{U}_1}, \underbrace{\mathcal{E}_4, \mathcal{E}_5, \mathcal{E}_6, \mathcal{E}_7}_{\mathcal{U}_2}, \underbrace{\mathcal{E}_8,\mathcal{E}_9,\mathcal{E}_{10},\mathcal{E}_{11},\mathcal{E}_{12}}_{\mathcal{U}_3}\}, \ N=3, \ M=12 \] \[\begin{align*} \mathcal{U}_1 & = \{\mathcal{E}_1,\mathcal{E}_2,\mathcal{E}_3\}, \ M_1 = 3 \\ \mathcal{U}_2 & = \{\mathcal{E}_4,\mathcal{E}_5,\mathcal{E}_6,\mathcal{E}_7\}, \ M_2 = 4 \\ \mathcal{U}_3 & = \{\mathcal{E}_8,\mathcal{E}_9,\mathcal{E}_{10},\mathcal{E}_{11},\mathcal{E}_{12}\}, \ M_3 = 5 \end{align*}\] Note that there are actually two kinds of sampling units here, the sampling units for the first stage (i.e., clusters), and the sampling units for the second stage (i.e., the elements within the clusters). These are called the primary and secondary sampling units, respectively. These are denoted above as \(\mathcal{U}\) and \(\mathcal{E}\), respectively.
Example: What would be some possible samples for a two-stage cluster sampling design with \(n\) = 2 and \(m_1\) = 2, \(m_2\) = 2, \(m_3\) = 3 (assuming SRS for both stages) using the population shown above?
| Clusters | Elements |
|---|---|
| \(\mathcal{U}_{1},\mathcal{U}_{3}\) | \(\mathcal{E}_{2},\mathcal{E}_{3};\mathcal{E}_{9},\mathcal{E}_{10},\mathcal{E}_{11}\) |
| \(\mathcal{U}_{1},\mathcal{U}_{2}\) | \(\mathcal{E}_{1},\mathcal{E}_{2};\mathcal{E}_{5},\mathcal{E}_{7}\) |
| \(\mathcal{U}_{1},\mathcal{U}_{2}\) | \(\mathcal{E}_{2},\mathcal{E}_{3};\mathcal{E}_{4},\mathcal{E}_{6}\) |
| \(\mathcal{U}_{1},\mathcal{U}_{3}\) | \(\mathcal{E}_{1},\mathcal{E}_{3};\mathcal{E}_{9},\mathcal{E}_{10},\mathcal{E}_{12}\) |
| \(\mathcal{U}_{2},\mathcal{U}_{3}\) | \(\mathcal{E}_{4},\mathcal{E}_{5};\mathcal{E}_{9},\mathcal{E}_{10},\mathcal{E}_{11}\) |
| \(\mathcal{U}_{1},\mathcal{U}_{3}\) | \(\mathcal{E}_{1},\mathcal{E}_{3};\mathcal{E}_{9},\mathcal{E}_{10},\mathcal{E}_{11}\) |
| \(\mathcal{U}_{1},\mathcal{U}_{3}\) | \(\mathcal{E}_{1},\mathcal{E}_{3};\mathcal{E}_{8},\mathcal{E}_{9},\mathcal{E}_{12}\) |
| \(\mathcal{U}_{1},\mathcal{U}_{2}\) | \(\mathcal{E}_{1},\mathcal{E}_{3};\mathcal{E}_{4},\mathcal{E}_{5}\) |
| \(\mathcal{U}_{1},\mathcal{U}_{3}\) | \(\mathcal{E}_{1},\mathcal{E}_{2};\mathcal{E}_{8},\mathcal{E}_{10},\mathcal{E}_{12}\) |
Note that the table above is not all possible samples (there are 108 possible samples for this design).
Two-stage cluster sampling designs involve two stages/levels of sampling units: primary sampling units (PSU) and secondary sampling units (SSU). The primary sampling units are the clusters. The secondary sampling units are the elements within the clusters.
| Primary | Secondary | Target Variable |
|---|---|---|
| box | widget | weight |
| block | household | income |
| county | farm | acres of wheat |
| classroom | student | test score |
| hour | minute | number of fish |
| plot | tree | volume |
Consider describing a two-stage cluster sampling design in terms of five quantities:
How is a one-stage cluster sampling design a special case?
How is a stratified sampling design a special case?
What are the advantages and disadvantages of two-stage cluster sampling relative to simple random sampling?
What advantage does two-stage cluster sampling having relative to one-stage cluster sampling?
Assuming we are using simple random sampling at both stages, how might we estimate \(\tau\) and \(\mu\)?
Recall that with a one-stage cluster sampling design, the “unbiased estimator” of \(\tau\) is \[ \hat\tau = \frac{N}{n}\sum_{i \in \mathcal{S}} y_i, \] where \(y_i\) is the total of the target variable for all the elements in the \(i\)-th cluster. In a two-stage cluster sampling design we do not know \(y_i\) if we do not sample all elements in the cluster, but we can estimate it with the “expansion estimator” \(M_i\bar{y}\), where \(\bar{y}_i\) is the mean of the target variable for the sampled elements from the \(i\)-th cluster. So the “unbiased” estimator of \(\tau\) for a two-stage cluster sampling design is \[ \hat\tau = \frac{N}{n}\sum_{i \in \mathcal{S}} M_i\bar{y}_i. \] The “unbiased” estimator of \(\mu\) is then obtained by dividing this estimator by \(M\) to get \[ \hat\mu = \frac{N}{Mn}\sum_{i \in \mathcal{S}} M_i\bar{y}_i. \] Note that the set \(\mathcal{S}\) denotes the indices of the set of sampled clusters.
Example: A two-stage cluster sampling design selects \(n\) = 3 boxes using simple random sampling. These boxes each contain 10, 12, and 5 widgets, but we select 2, 3, and 1 widget(s) from these boxes, respectively, using simple random sampling. The mean weight of the sampled widgets from each box are 2.2, 1.9, and 2.1, respectively. Assume that there are \(N\) = 100 boxes of widgets in the warehouse, containing a total of \(M\) = 425 widgets. What are the estimates of \(\tau\) and \(\mu\) using the unbiased estimators?
The variance of \(\hat\tau\) is \[ V(\hat\tau) = \underbrace{N^2\left(1-\frac{n}{N}\right)\frac{\sigma_c^2}{n}}_{\text{first stage of sampling}} + \underbrace{\frac{N}{n}\sum_{i=1}^N M_i^2\left(1-\frac{m_i}{M_i}\right)\frac{\sigma_i^2}{m_i}}_{\text{second stage of sampling}} \] where \(\sigma_c^2\) and \(\sigma_i^2\) are the variances of the cluster totals and within the \(i\)-th cluster, respectively, defined as \[ \sigma_c^2 = \frac{1}{N-1}\sum_{i=1}^N(y_i - \tau/N)^2 \ \ \ \text{and} \ \ \ \sigma_i^2 = \frac{1}{M_i-1}\sum_{j=1}^{M_i}(y_{ij} - \mu_i)^2, \] where \(\mu_i\) is the mean of all the elements in the \(i\)-th cluster. The variance of \(\hat\mu\) is \(V(\hat\mu) = V(\hat\tau)/M^2\). Modified formulas (not shown) can be used to estimate these variances.
Recall that for a one-stage cluster sampling design, the “ratio estimator” of \(\tau\) is \[ \hat\tau = M\frac{\sum_{i \in \mathcal{S}} y_i}{\sum_{i \in \mathcal{S}} m_i} \] where \(y_i\) is again the total of the target variable of the elements in the \(i\)-th cluster, and \(m_i\) is the number of elements in the \(i\)-th cluster (which we would call \(M_i\) in the context of two-stage cluster sampling). For a two-stage cluster sampling design we estimate \(y_i\) with \(M_i\bar{y}_i\) and replace \(m_i\) with \(M_i\) to get \[ \hat\tau = M\frac{\sum_{i \in \mathcal{S}} M_i\bar{y}_i}{\sum_{i \in \mathcal{S}} M_i}. \] The corresponding estimator of \(\mu\) is then \[ \hat\mu = \frac{\sum_{i \in \mathcal{S}} M_i\bar{y}_i}{\sum_{i \in \mathcal{S}} M_i}. \] Example: A two-stage cluster sampling design selects \(n\) = 3 boxes using simple random sampling. These boxes each contain 10, 12, and 5 widgets, but we select 2, 3, and 1 widget(s) from these boxes, respectively, using simple random sampling. The mean weight of the sampled widgets from each box are 2.2, 1.9, and 2.1, respectively. Assume that there are \(N\) = 100 boxes of widgets in the warehouse, containing a total of \(M\) = 425 widgets. What are the estimates of \(\tau\) and \(\mu\) using the ratio estimators?
The variance of \(\hat\tau\) is
\[
V(\hat\tau) \approx
\underbrace{N^2\left(1-\frac{n}{N}\right)\frac{\sigma_r^2}{n}}_{\text{first
stage of sampling}} +
\underbrace{\frac{N}{n}\sum_{i=1}^N
M_i^2\left(1-\frac{m_i}{M_i}\right)\frac{\sigma_i^2}{m_i}}_{\text{second
stage of sampling}},
\] where \[
\sigma_r^2 = \frac{1}{N-1}\sum_{i=1}^N(y_i - M_i\mu)^2.
\]
The variance of \(\hat\mu\) is \(V(\hat\mu) =V(\hat\tau)/M^2\). Modified
formulas (not shown) can be used to estimate these
variances.
We have two estimators of \(\tau\), the “unbiased” and “ratio” estimators which are \[ \hat\tau_u = \frac{N}{n}\sum_{i \in \mathcal{S}} M_i\bar{y}_i \ \ \ \text{and} \ \ \ \hat\tau_r = M\frac{\sum_{i \in \mathcal{S}} M_i\bar{y}_i}{\sum_{i \in \mathcal{S}} M_i}, \] respectively. Why might we choose one over the other?
We have two estimators of \(\mu\), the “unbiased” and “ratio” estimators which are \[ \hat\mu_u = \frac{N}{Mn}\sum_{i \in \mathcal{S}} M_i\bar{y}_i \ \ \ \text{and} \ \ \ \hat\mu_r = \frac{\sum_{i \in \mathcal{S}} M_i\bar{y}_i}{\sum_{i \in \mathcal{S}} M_i}, \] respectively. How do these compare?
If the clusters are sampled with replacement with selection probabilities proportional to cluster size (i.e., PPS sampling with \(\delta_i = M_i/M\)), then we can use a variant of the Hansen-Hurwitz estimator. Recall that for one-stage cluster sampling this estimator can be written as \[ \hat\tau = \frac{M}{n}\sum_{i \in \mathcal{S}}\frac{y_i}{m_i}, \] where \(y_i\) is the total of the target variable for the elements in the \(i\)-th cluster. To derive the estimator for a two-stage cluster sampling design, replace \(y_i\) with the estimator \(M_i\bar{y}_i\) and \(m_i\) with \(M_i\) to get \[ \hat\tau = \frac{M}{n}\sum_{i \in \mathcal{S}}\bar{y}_i, \] where \(\bar{y}_i\) is the mean of the target variable for the sampled elements from the \(i\)-th cluster. The estimator of \(\mu\) is then \[ \hat\mu = \frac{1}{n}\sum_{i \in \mathcal{S}} \bar{y}_i. \] Example: A two-stage cluster sampling design selects \(n\) = 3 boxes using simple random sampling. These boxes each contain 10, 12, and 5 widgets, but we select 2, 3, and 1 widget(s) from these boxes, respectively, using PPS sampling. The mean weight of the sampled widgets from each box are 2.2, 1.9, and 2.1, respectively. Assume that there are \(N\) = 100 boxes of widgets in the warehouse, containing a total of \(M\) = 425 widgets. What are the estimates of \(\tau\) and \(\mu\) using Hansen-Hurwitz estimator?
The variance of \(\hat\tau\) can be written as \[ V(\hat\tau) = \underbrace{\frac{M}{n}\sum_{i \in \mathcal{S}}M_i(\mu_i - \mu)^2}_{\text{first stage of sampling}} + \underbrace{\frac{M}{n}\sum_{i \in \mathcal{S}}M_i\left(1 - \frac{m_i}{M_i}\right)\frac{\sigma_i^2}{m_i}}_{\text{second stage of sampling}}, \] where \(\sigma_i^2\) is as defined earlier, and \(\mu_i\) is the mean of the target variable for all elements in the \(i\)-th cluster. The variance of \(\hat\mu\) is then \(V(\hat\mu) = V(\hat\tau)/M^2\). Interestingly the estimated variance of \(\hat\tau\) is very simple. It can be written as \[ \hat{V}(\hat\tau) = \frac{M^2}{n(n-1)}\sum_{i \in \mathcal{S}}(\bar{y}_i - \hat\mu)^2, \] where \(\hat\mu\) is the Hansen-Hurwitz estimator of \(\mu\). The variance of \(\hat\mu\) is then \[ \hat{V}(\hat\mu) = \frac{1}{n(n-1)}\sum_{i \in \mathcal{S}}(\bar{y}_i - \hat\mu)^2. \]
