A stratified random sample with $2$ strata of Statisticians (Stratum-1) and Biologists (Stratum-2), where $12$ Statisticians and $15$ Biologists are drawn from Stratum-1 and Stratum-2, respectively, using SRSWOR scheme.
Denote $\bar{y}_S, \bar{y}_B$, and $\bar{y}$ as the mean of the variable among the Statistician sample, Biologist sample, and the combined sample, respectively. Which of the following is an unbiased estimator of $\bar{Y}$?
Stratified Sample Overview
The question asks us to find an unbiased estimator for the population mean ($\bar{Y}$) of a variable for a finite population. The population consists of $N = 85$ individuals, divided into two distinct groups, or strata: Statisticians ($N_1 = 34$) and Biologists ($N_2 = 51$). We are using a stratified random sampling method. In this method, we draw samples independently from each stratum using Simple Random Sampling Without Replacement (SRSWOR).
Specifically, a sample of $n_1 = 12$ Statisticians is drawn from Stratum-1, and a sample of $n_2 = 15$ Biologists is drawn from Stratum-2. Let $\bar{y}_S$ denote the sample mean for the Statisticians and $\bar{y}_B$ denote the sample mean for the Biologists.
Calculating Population Proportions
In stratified sampling, the overall population mean $\bar{Y}$ is estimated by taking a weighted average of the sample means from each stratum. The weights used in this average are the proportions of the total population that each stratum represents.
First, let's calculate the population proportion for each stratum:
- Stratum-1 (Statisticians): The population size is $N_1 = 34$ and the total population size is $N = 85$. The population proportion (weight) for Stratum-1 is:
$W_1 = \frac{N_1}{N} = \frac{34}{85}$
Simplifying this fraction by dividing the numerator and denominator by their greatest common divisor (17):
$W_1 = \frac{34 \div 17}{85 \div 17} = \frac{2}{5}$
- Stratum-2 (Biologists): The population size is $N_2 = 51$ and the total population size is $N = 85$. The population proportion (weight) for Stratum-2 is:
$W_2 = \frac{N_2}{N} = \frac{51}{85}$
Simplifying this fraction by dividing the numerator and denominator by their greatest common divisor (17):
$W_2 = \frac{51 \div 17}{85 \div 17} = \frac{3}{5}$
Notice that the sum of the weights is $W_1 + W_2 = \frac{2}{5} + \frac{3}{5} = \frac{5}{5} = 1$, as expected.
Unbiased Estimator Formula Application
For a two-strata population, the unbiased estimator for the population mean ($\bar{Y}$) in stratified random sampling is given by the formula:
$\hat{\bar{Y}}_{st} = W_1 \bar{y}_1 + W_2 \bar{y}_2$
Where $\bar{y}_1$ is the sample mean of Stratum-1 and $\bar{y}_2$ is the sample mean of Stratum-2.
In our case, $\bar{y}_1 = \bar{y}_S$ (mean for Statisticians) and $\bar{y}_2 = \bar{y}_B$ (mean for Biologists).
Substituting the calculated population proportions (weights) $W_1 = \frac{2}{5}$ and $W_2 = \frac{3}{5}$ into the formula, we get:
$\hat{\bar{Y}}_{st} = \frac{2}{5} \bar{y}_S + \frac{3}{5} \bar{y}_B$
To combine these terms into a single fraction, we find a common denominator, which is 5:
$\hat{\bar{Y}}_{st} = \frac{2\bar{y}_S}{5} + \frac{3\bar{y}_B}{5} = \frac{2\bar{y}_S + 3\bar{y}_B}{5}$
This formula provides an unbiased estimate of the population mean $\bar{Y}$.
Evaluating the Options
Let's compare our derived estimator with the given options:
- Option 1: $\bar{y}$ is the simple mean of the combined sample, which is generally not unbiased in stratified sampling unless specific conditions like proportional allocation are met, which is not the case here ($n_1/N_1 \neq n_2/N_2$).
- Option 2: $\frac{2\bar{y}_S + 3\bar{y}_B}{5}$. This matches our derived unbiased estimator using the population proportions as weights.
- Option 3: $\frac{4\bar{y}_S + 5\bar{y}_B}{9}$. This uses arbitrary weights (4 and 5) that do not correspond to the population proportions.
- Option 4: $\frac{\frac{1}{12}\bar{y}_S + \frac{1}{15}\bar{y}_B}{\frac{1}{12}+ \frac{1}{15}}$. This uses weights related to the inverse of the sample sizes ($\frac{1}{n_1}, \frac{1}{n_2}$), not the population proportions ($W_h$).
Therefore, the correct unbiased estimator for the population mean $\bar{Y}$ is $\frac{2\bar{y}_S + 3\bar{y}_B}{5}$.
Important Questions from Random Variables
- A mobile manufacturing company uses two brands of batteries for its mobiles. The life (in years) of batteries of Brand I follows an exponential distribution with the probability density function
$ f(x) = \begin{cases} e^{-x}, & \text{if } x>0, \\ 0, & \text{otherwise,} \end{cases} $
and that of Brand II follows a gamma distribution with the probability density function
$ g(x) = \begin{cases} \frac{x}{4} e^{-x/2}, & \text{if } x>0, \\ 0, & \text{otherwise.} \end{cases} $
The company uses the batteries of Brands I and II in proportion of $20\%$ and $80\%$ respectively, in its mobiles. The probability that a randomly selected mobile has the battery life more that $2$ years is - Consider a discrete random variable $X$ with the probability mass function
$ P(X = 0) = \frac{\theta}{3}, \ P(X = 1) = 1 - \frac{\theta}{2}, \ P(X = 2) = \frac{\theta}{6}, $
where $\theta \in (0,1)$ is an unknown parameter. In a random sample of size $90$ from this distribution, the observed counts for $X = 0, 1$ and $2$ are $20, 60$ and $10$, respectively. Then, the maximum likelihood estimate of $\theta$ is - Let $X$ be a random sample of size $1$ from the probability density function
$ f(x|\theta) = \begin{cases} \frac{3}{\theta^3} (\theta - x)^2, & \text{if } 0<x<\theta, \\ 0, & \text{otherwise.} \end{cases} $
If $ \left(\frac{X}{1-\lambda_1}, \frac{X}{1-\lambda_2}\right) $ is a confidence interval for $\theta$ with confidence coefficient $1 - \alpha$, where $\lambda_i \in (0,1), \ i = 1,2, \ \lambda_1<\lambda_2$, and $\alpha \in (0,1)$, then which of the following statements is true? - Let $X_1, X_2, . . ., X_n$ be a random sample from a continuous distribution with the common probability density function
$ f(x|\theta) = \begin{cases} \frac{2\theta^2}{x^{\theta+1}}, & \text{if } x>2, \\ 0, & \text{otherwise,} \end{cases} $
where $\theta (> 0)$ is an unknown parameter. Suppose $P(Y>\chi^2_{m,\beta}) = \beta$, where $Y \sim \chi^2_m$. For testing $H_0: \theta = 1$ against $H_1 : \theta>1$, a uniformly most powerful test of size $\alpha, \ 0<\alpha<1$, will reject $H_0$ if Let $X$ be a random variable with the cumulative distribution function
$F_X(x) = \begin{cases} 0, & \text{if } x<0, \\ \frac{x+2}{5}, & \text{if } 0 \le x<2, \\ 1, & \text{if } x \ge 2. \end{cases}$
Then, which of the following statements are true?
