Probability of Unit Inclusion/Exclusion in Simple Random Sampling
This solution calculates the probability that a specific unit (the 4th) is included in a simple random sample (SRS) of size $n$ drawn without replacement from a population of size $N$, while another specific unit (the 6th) is excluded.
Calculating the Probability
We use the concept of conditional probability for sampling without replacement.
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Define Events:
- Let $E_{4, \text{in}}$ be the event that the 4th population unit is included in the sample.
- Let $E_{6, \text{out}}$ be the event that the 6th population unit is NOT included in the sample.
- We need to find the probability $P(E_{4, \text{in}} \cap E_{6, \text{out}})$.
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Probability of 4th Unit Inclusion:
In an SRS of size $n$ from a population of size $N$, the probability that any specific unit is included is $\frac{n}{N}$.
$ P(E_{4, \text{in}}) = \frac{n}{N} $ -
Conditional Probability of 6th Unit Exclusion:
Given that the 4th unit is already included in the sample, there are $n-1$ remaining spots to be filled from the remaining $N-1$ population units. The 6th unit is one of these $N-1$ units.
The probability that the 6th unit IS included among the remaining $n-1$ selections is $\frac{n-1}{N-1}$.
Therefore, the probability that the 6th unit is NOT included, given the 4th was included, is:
$ P(E_{6, \text{out}} | E_{4, \text{in}}) = 1 - P(\text{6th included} | E_{4, \text{in}}) = 1 - \frac{n-1}{N-1} $ $ P(E_{6, \text{out}} | E_{4, \text{in}}) = \frac{(N-1) - (n-1)}{N-1} = \frac{N-n}{N-1} $ -
Combined Probability:
Using the multiplication rule for conditional probability:
$ P(E_{4, \text{in}} \cap E_{6, \text{out}}) = P(E_{4, \text{in}}) \times P(E_{6, \text{out}} | E_{4, \text{in}}) $ $ P(E_{4, \text{in}} \cap E_{6, \text{out}}) = \frac{n}{N} \times \frac{N-n}{N-1} $ $ P(E_{4, \text{in}} \cap E_{6, \text{out}}) = \frac{n(N-n)}{N(N-1)} $
This result matches option B.
Important Questions from Ratio And Regression
- 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?
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 we draw a random sample of size $n$ from a population of size $N$, where $1 < n < N$, using simple random sampling without replacement scheme. Let $P$ be the population proportion of units possessing a particular attribute and $p$ be the corresponding sample proportion. Which of the following is an unbiased estimator for $P(1 - P)$?
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?
