Calculating Sample Size for Sampling Schemes
This solution explains how to find the sample size ($n$) using the relationship between the variances of sample means obtained from Simple Random Sampling With Replacement (SRSWR) and Simple Random Sampling Without Replacement (SRSWOR).
Understanding the Problem
- Population size $N = 100$.
- $T_1$: Sample mean using SRSWR.
- $T_2$: Sample mean using SRSWOR.
- Sample size $n$, where $1 < n < 100$.
- Given condition: $Var(T_1) = 9 \times Var(T_2)$.
- Goal: Find the value of $n$.
Formulas for Variance
- The variance of the sample mean ($T_1$) under SRSWR is:
$ Var(T_1) = \frac{\sigma^2}{n} $
where $\sigma^2$ is the population variance. - The variance of the sample mean ($T_2$) under SRSWOR is:
$ Var(T_2) = \frac{\sigma^2}{n} \left( \frac{N-n}{N-1} \right) $
where $\left( \frac{N-n}{N-1} \right)$ is the Finite Population Correction (FPC) factor.
Deriving the Sample Size
We use the given relationship $Var(T_1) = 9Var(T_2)$ and substitute the variance formulas:
$ \frac{\sigma^2}{n} = 9 \times \left( \frac{\sigma^2}{n} \left( \frac{N-n}{N-1} \right) \right) $
Assuming $\sigma^2 \neq 0$ and $n > 1$, we can cancel $\frac{\sigma^2}{n}$ from both sides:
$ 1 = 9 \left( \frac{N-n}{N-1} \right) $
Now, solve for $n$:
- Rearrange the equation:
$ \frac{1}{9} = \frac{N-n}{N-1} $
- Cross-multiply:
$ N-1 = 9(N-n) $
- Expand:
$ N-1 = 9N - 9n $
- Isolate the term with $n$:
$ 9n = 9N - N + 1 $
$ 9n = 8N + 1 $
- Substitute the population size $N=100$:
$ 9n = 8(100) + 1 $
$ 9n = 800 + 1 $
$ 9n = 801 $
- Calculate $n$:
$ n = \frac{801}{9} $
$ n = 89 $
The calculated sample size $n = 89$ fits the condition $1 < n < N$. Therefore, the sample size is 89.
Important Questions from Random Variables
Let $X$, $Y$, and $Z$ be independent Normal random variables with means $-1$, $0$, and $1$, respectively, and variances $1$, $1$, and $3$, respectively. Which of the following random variables has a Cauchy distribution with location parameter $0$ and scale parameter $1$?
Let $X_1$ and $X_2$ be a random sample from Uniform$[0, \theta]$ distribution, where $\theta > 0$. For testing the hypothesis
$H_0: \theta = 1$ against $H_1: \theta = 2$,
consider a test which rejects $H_0$ if $X_1 + X_2 > \frac{4}{5}$. Then, the probability of type-I error is- Let $\{Y_n: n \ge 1\}$ be a sequence of independent and identically distributed random variables, where $Y_1 \sim \text{Bernoulli}(\frac{1}{2})$. Define $Z = \sum_{n=1}^\infty \frac{4Y_n}{5^n}$. Then, which of the following statements is true?
Let $X$ be a single sample from an absolutely continuous distribution with probability density function
$f(x|\theta) = \begin{cases} \frac{2}{\theta^2}(\theta - x), & \text{if } 0 < x < \theta \\ 0, & \text{otherwise,} \end{cases}$
where $\theta > 0$ is unknown. Which of the following intervals is a $95\%$ confidence interval for $\theta$?Let $X_1, X_2,..., X_{15}$ be a random sample from an Exponential distribution with the probability density function
$f(x) = \begin{cases} \frac{1}{\sigma} \exp \left(-\frac{x}{\sigma}\right) & \text{if } x > 0, \\ 0, & \text{elsewhere,} \end{cases}$
where the unknown parameter $\sigma$ is positive. Let $\bar{X} = \frac{1}{15}\sum_{i=1}^{15} X_i$. Suppose that $\phi$ denotes the likelihood ratio test for testing $H_0: \sigma \le 1$ against $H_1 : \sigma > 1$ at level $\alpha = 0.1$. It is given that $\chi^2_{15,0.1} = 22.307$, $\chi^2_{15,0.9} = 8.547$, $\chi^2_{30,0.1} = 40.256$, $\chi^2_{30,0.9} = 20.599$, where $P(W > \chi^2_{n,\alpha}) = \alpha$ and $W \sim \chi^2_n$. Then which of the following statements are true?
