# 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:**

##### Decriptive 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!

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

\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}