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 $\theta$?
This question requires us to find a 95% confidence interval for the parameter \(\theta\) of a given probability density function (PDF). The PDF provided is:
\(f(x|\theta) = \begin{cases} \frac{2}{\theta^2}(\theta - x), & \text{if } 0 < x < \theta \\ 0, & \text{otherwise.} \end{cases}\)
Where \(x\) is a sample point and \(\theta > 0\) is unknown.
To find the confidence interval for \(\theta\), we follow these steps:
- Given the range \(0 < x < \theta\), the maximum likelihood estimator (MLE) for \(\theta\) is based on the largest possible value that \(x\) can take, which is the observation itself \(X\).
- The distribution is a form of the uniform distribution over \((0, \theta)\), where the maximum value \(\theta\) provides a natural estimate \(X\).
- For confidence interval calculation, the pivotal quantity method can be applied. The pivotal quantity is \(T = \frac{X}{\theta}\), and for this distribution, it follows a known distribution whose behavior can be utilized for interval estimation.
- To construct the interval, note that \(T\) covers interval \((0, 1)\). For a confidence level of 95%, if the lower bound stands from sample proportion, the higher bound of \(\theta\) corresponds to \(\frac{X}{1-\sqrt{0.05}}\).
- Concretely, find the point \(\alpha\) such that \(P(T > \alpha) = 0.05\). This involves solving for \(\alpha\) in terms of probability cutoff.
Following this logic, the correct 95% confidence interval is given by:
The interval \((X, \frac{X}{1-\sqrt{0.05}})\) ensures that the coverage probability is approximately 95%.
Therefore, among the provided options, the correct interval for \(\theta\) is:
Correct Answer: \((X, \frac{X}{1-\sqrt{0.05}})\)
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 $1$?
- Consider a finite population of size $N = 100$. Let $T_1$ be the sample mean of a study variable based on a sample of size $n$ ($1 < n < N$) under simple random sampling with replacement scheme. Let $T_2$ be the sample mean of the same study variable based on a sample of size $n$ under simple random sampling without replacement scheme. If $Var(T_1) = 9Var(T_2)$, then the sample size $n$ equals
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 $X_1, X_2,..., X_{15}$ 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?
