The problem asks for the expected number of distinct units drawn when sampling 2 units from a population of 20 with replacement, using probability proportional to size (PPS) sampling. The probabilities $p_i$ are given.
For sampling with replacement, the expected number of distinct units ($E(V)$) in a sample of size $n$ from a population of size $N$ is given by:
$E(V) = N - \sum_{i=1}^{N} (1 - p_i)^n$
Here, $N = 20$ and the sample size $n = 2$.
Let's verify the sum of probabilities: $10 \times \frac{1}{40} + 10 \times \frac{3}{40} = \frac{10}{40} + \frac{30}{40} = \frac{40}{40} = 1$. The probabilities are correctly normed.
We need to calculate the term $\sum_{i=1}^{N} (1 - p_i)^n$. Since $n=2$, we calculate $(1 - p_i)^2$.
The expected number of distinct units drawn is $\frac{31}{16}$.
Suppose there are $k$ strata of $N = kM$ units each with size $M$. Draw a sample of size $n_i$ with replacement from the $i^{\text{th}}$ stratum and denote by $\bar{y}_i$ the sample mean of the study variable selected in the $i^{\text{th}}$ stratum, $i = 1, 2, \dots, k$. Define
$$ \bar{y}_s = \frac{1}{k}\sum_{i=1}^k \bar{y}_i \text{ and } \bar{y}_w = \frac{\sum_{i=1}^k n_i \bar{y}_i}{n} $$
Which of the following is necessarily true?
Suppose there are $k$ groups each consisting of $N$ boys. We want to estimate the mean age $\mu$ of these $kN$ boys. Fix $1 < n < N$ and consider the following two sampling schemes.
I. Draw a simple random sample without replacement of size $kn$ out of all $kN$ boys.
II. From each of the $k$ groups draw a simple random sample with replacement of size $n$.
Let $\bar{Y}$ and $\bar{Y}_G$ be the respective sample mean ages for the two schemes. Which of the following are true?