Analysis of Probability Proportional to Size (PPS) Sampling Statements
Evaluating Statement A: Unbiased Estimator
The question concerns Probability Proportional to Size (PPS) sampling with replacement for a sample size $n=2$. A standard estimator for the population total $Y = \sum_{i=1}^{N} y_i$ in this design is:
$ \hat{Y} = \frac{1}{n} \sum_{i \in s} \frac{y_i}{p_i} $
Where $s$ is the sample, $y_i$ is the value of the study variable for unit $i$, and $p_i$ is the probability of selecting unit $i$ in a single draw. The estimator $T$ proposed is:
$ T = \left(\frac{1}{2}\right) \sum_{i \in s} \frac{y_i}{p_i} $
With $n=2$, this matches the form $\hat{Y}$. This estimator is known to be unbiased for the population total $Y$ under PPS sampling with replacement.
Conclusion: Statement A is correct.
Evaluating Statement B: Inclusion Probabilities
The inclusion probability $\pi_i$ for unit $i$ in a sample of size $n=2$ drawn with replacement is the probability that unit $i$ is selected in at least one of the two draws. It is calculated as:
$ \pi_i = 1 - P(\text{unit } i \text{ not selected in draw 1}) \times P(\text{unit } i \text{ not selected in draw 2}) $
$ \pi_i = 1 - (1 - p_i) \times (1 - p_i) = 1 - (1 - p_i)^2 $
Using the given probabilities $p_1 = 0.2$ and $p_2 = 0.3$:
- For unit 1: $\pi_1 = 1 - (1 - 0.2)^2 = 1 - (0.8)^2 = 1 - 0.64 = 0.36$.
- For unit 2: $\pi_2 = 1 - (1 - 0.3)^2 = 1 - (0.7)^2 = 1 - 0.49 = 0.51$.
These calculated values match the statement.
Conclusion: Statement B is correct.
Evaluating Statement C: Joint Inclusion Probability
The joint inclusion probability $\pi_{ij}$ is the probability that both unit $i$ and unit $j$ ($i \ne j$) are included in the sample of size $n=2$ drawn with replacement. This occurs if unit $i$ is drawn first and unit $j$ second, OR if unit $j$ is drawn first and unit $i$ second.
$ \pi_{ij} = P(\text{draw 1}=i, \text{draw 2}=j) + P(\text{draw 1}=j, \text{draw 2}=i) $
Due to independence in sampling with replacement:
$ \pi_{ij} = p_i p_j + p_j p_i = 2 p_i p_j $
Using $p_1 = 0.2$ and $p_2 = 0.3$:
$ \pi_{12} = 2 \times 0.2 \times 0.3 = 2 \times 0.06 = 0.12 $
This calculated value matches the statement.
Conclusion: Statement C is correct.
Evaluating Statement D: Sum of Inclusion Probabilities
We need to calculate the inclusion probabilities for units 3 and 4 using the formula $\pi_i = 1 - (1 - p_i)^2$. Given $p_3 = 0.1$ and $p_4 = 0.4$:
- $\pi_3 = 1 - (1 - 0.1)^2 = 1 - (0.9)^2 = 1 - 0.81 = 0.19$.
- $\pi_4 = 1 - (1 - 0.4)^2 = 1 - (0.6)^2 = 1 - 0.36 = 0.64$.
The sum of all inclusion probabilities is:
$ \sum_{i=1}^4 \pi_i = \pi_1 + \pi_2 + \pi_3 + \pi_4 = 0.36 + 0.51 + 0.19 + 0.64 = 1.70 $
The statement claims the sum is $2$. Since $1.70 \ne 2$, the statement is false.
Conclusion: Statement D is incorrect.
Overall Conclusion
Statements A, B, and C were found to be correct based on the principles of PPS sampling with replacement.
Important Questions from Ratio And Regression
- A simple random sample (without replacement) of size $n$ is drawn from a finite population of size $N (\ge 7)$. What is the probability that the $4^{\text{th}}$ population unit is included in the sample but the $6^{\text{th}}$ population unit is not included in the sample?
- For a data set $(x_1, y_1), (x_2, y_2), \dots, (x_n, y_n)$ the following two models were fitted using least square method.
Model 1: $y_i = \beta_0 + \beta_1 x_i \quad i = 1, 2, \dots n$
Model 2: $y_i = \beta_0^* + \beta_1^* x_i + \beta_2^* x_i^2 \quad i = 1, 2, \dots n$
Let $\hat{\beta}_0, \hat{\beta}_1$ be least square estimates of $\beta_0, \beta_1$ from model 1 and $\hat{\beta}_0^*, \hat{\beta}_1^*, \hat{\beta}_2^*$ be the least square estimates from model 2.
Let $A = \sum_1^n \left(y_i - (\hat{\beta}_0 + \hat{\beta}_1 x_i)\right)^2$,
$B = \sum_1^n \left(y_i - (\hat{\beta}_0^* + \hat{\beta}_1^* x_i + \hat{\beta}_2^* x_i^2)\right)^2$
Then - Suppose $\bar{Y}$ is the sample mean of the study variables corresponding to a sample of size n using simple random sampling with replacement scheme and $\bar{Y}_{st}$ is the sample mean of the study variables corresponding to a sample of size n using stratified random sampling with replacement scheme under proportional allocation. Which of the following is/are sufficient condition/conditions for $Var(\bar{Y}) = Var(\bar{Y}_{st})$?
- Consider the problem of drawing a sample of size 2 from a finite population of size 20. The sampling is done with replacement using probability proportional to size sampling scheme. The normed size measures $p_1, \cdots, p_{20}$ are given by $p_i = \frac{1}{40}$, $i = 1, \cdots, 10, \; p_i = \frac{3}{40}$, $i = 11, \cdots, 20$. The expected number of distinct units drawn is
- Consider a finite population of size $N$. Let $T_1$ be the sample mean based on a sample of size $n$ under simple random sampling with replacement (SRSWR) scheme. Let $T_2$ be the sample mean based on a stratified random sample of size $n$ where the samples are drawn from each of 4 strata using SRSWR scheme under proportional allocation. Then which of the following are sufficient conditions for $\text{Var}(T_1) = \text{Var}(T_2)$ to hold?
