Expected Distinct Units Calculation
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.
Formula for Expected Distinct Units
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 $n = 2$.
Given Probabilities
- For units $i = 1, \cdots, 10$ (10 units): $p_i = \frac{1}{40}$.
- For units $i = 11, \cdots, 20$ (10 units): $p_i = \frac{3}{40}$.
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.
Calculation Steps
We need to calculate the term $\sum_{i=1}^{N} (1 - p_i)^n$. Since $n=2$, we calculate $(1 - p_i)^2$.
- Calculate $(1 - p_i)^2$ for the first group of 10 units:
For $p_i = \frac{1}{40}$, $(1 - p_i)^2 = \left(1 - \frac{1}{40}\right)^2 = \left(\frac{39}{40}\right)^2 = \frac{1521}{1600}$. - Calculate $(1 - p_i)^2$ for the second group of 10 units:
For $p_i = \frac{3}{40}$, $(1 - p_i)^2 = \left(1 - \frac{3}{40}\right)^2 = \left(\frac{37}{40}\right)^2 = \frac{1369}{1600}$. - Calculate the sum $\sum_{i=1}^{20} (1 - p_i)^2$:
The sum consists of 10 terms of the first type and 10 terms of the second type.
$ \sum_{i=1}^{20} (1 - p_i)^2 = 10 \times \left(\frac{39}{40}\right)^2 + 10 \times \left(\frac{37}{40}\right)^2 $ $ = 10 \times \frac{1521}{1600} + 10 \times \frac{1369}{1600} $ $ = \frac{15210}{1600} + \frac{13690}{1600} = \frac{28900}{1600} = \frac{289}{16} $ - Calculate the Expected Number of Distinct Units $E(V)$:
$ E(V) = N - \sum_{i=1}^{20} (1 - p_i)^2 $ $ E(V) = 20 - \frac{289}{16} $ $ E(V) = \frac{20 \times 16}{16} - \frac{289}{16} = \frac{320}{16} - \frac{289}{16} = \frac{320 - 289}{16} $ $ E(V) = \frac{31}{16} $
Final Answer
The expected number of distinct units drawn is $\frac{31}{16}$.
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 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?
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?
