# 9.2 - Two Stages with Primary Units Selected by Probability Proportional to Size and Secondary Units Selected with S.R.S.

9.2 - Two Stages with Primary Units Selected by Probability Proportional to Size and Secondary Units Selected with S.R.S.

Multi-stage design with primary units selected with p.p.s. and secondary units selected with simple random sampling.

Using the Hansen-Hurwitz estimator, we get the following:

To estimate the population total:

$$\hat{\tau}_p=\dfrac{M}{n}\sum\limits_{i=1}^n \dfrac{\hat{y}_i}{M_i}=M \dfrac{\sum \bar{y}_i}{n}$$, where $$\bar{y}_i=\dfrac{\hat{y}_i}{M_i}$$

$$\hat{V}ar(\hat{\tau}_p)=\dfrac{M^2}{n(n-1)} \sum (\bar{y}_i-\hat{\mu}_p)^2$$

To estimate the population mean:

$$\hat{\mu}_p=\dfrac{\sum \bar{y}_i}{n}$$

[since $$\hat{\mu}_p=\left(\dfrac{\hat{\tau}_p}{M}\right)$$ and thus it becomes $$\dfrac{\sum \bar{y}_i}{n}$$]

$$\hat{V}ar(\hat{\mu}_p)=\dfrac{1}{n(n-1)}\sum (\bar{y}_i-\hat{\mu}_p)^2$$

## Example

There are 36 departments in a small liberal arts college. One wants to estimate the average amount of money the students spent on textbooks last semester. Since the size of each department varies very much, a two-stage cluster sampling using probability proportional to size for the primary unit is carried out. The results are listed in the table below.

Department $$\mathbf{M_i}$$ $$\mathbf{m_i}$$ Textbook expenses in \$ for last semester
1 10 4 326, 400, 423, 443
2 20 8 278, 312, 450, 350, 227, 438, 512, 403
3 30 12 512, 256, 332, 402, 512, 309, 411, 610, 422, 630, 550, 470
4 15 6 426, 312, 512, 440, 342, 533

### Minitab output:

#### Descriptive Statistics: dept1, dept2, dept3, dept4

Variable Mean SE Mean StDev Variance
dept1 398.0 25.6 51.1 2612.7
dept2 371.3 34.1 96.3 9277.4
dept3 451.3 33.9 117.6 13828.8
dept4 427.5 36.1 88.4 7815.9

## Try it!

Estimate the population mean using probability proportional to the size estimator (Hansen-Hurwitz) and estimate the variance of that estimator.

$$\hat{\mu}_p=\dfrac{\sum \bar{y}_i}{n}=\dfrac{398+371.3+451.3+427.5}{4}=412.025$$

\begin{align}
\hat{V}ar(\hat{\mu}_p) &= \dfrac{1}{n(n-1)}\sum (\bar{y}_i-\hat{\mu}_p)^2\\
&= \dfrac{1}{4\times 3}\left[(398-412.025)^2+(371.3-412.025)^2+(451.3-412.025)^2+(427.5-412.025)^2\right]\\
&= 303.12\\
\end{align}

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