Interval estimation - Estimation - Statistics Previous Year Questions

Let \( X_1, ..., X_n \) be a random sample from normal distribution with mean \( \mu \) and variance 1. Let \( \Phi \) be the cumulative distribution function of the standard normal distribution. Given \( \Phi(1.96) = 0.975 \), the minimum sample size required such that the length of the 95% confidence interval for \( \mu \) does NOT exceed 2 is ...

Let \(X_1, \ldots, X_{10}\) be a random sample from a Weibull distribution with the probability density function \(f(x; \theta) = \begin{cases} 3\theta x^2 e^{-\theta x^3}, & x > 0 \\ 0, & \text{otherwise} \end{cases}\) where \(\theta \in (0, \infty)\). For any positive integer \(v\) and any \(\alpha \in (0, 1)\), let \(\chi^2_{v, \alpha}\) denote the \((1 - \alpha)\)-th quantile of the central chi-square distribution with \(v\) degrees of freedom. Then, a 90% confidence interval for \(\theta\) is

Let \(\{x_1, x_2, \ldots, x_n\}\) be a realization of a random sample of size \(n\) (≥ 2) from a \(N(\mu, \sigma^2)\) distribution, where \(-\infty < \mu < \infty\) and \(\sigma > 0\). Which of the following statements is/are true? P : 95% confidence interval of \(\mu\) based on \(\{x_1, x_2, \ldots, x_n\}\) is unique when \(\sigma\) is known. Q : 95% confidence interval of \(\mu\) based on \(\{x_1, x_2, \ldots, x_n\}\) is NOT unique when \(\sigma\) is unknown.

Let \(X_1, X_2, \ldots, X_{18}\) be a random sample from the distribution \[f(x; \theta) = \[\begin{cases}\] \[\frac{2x}{\theta} e^{-x^2/\theta}, & x > 0, \\\] 0, & x \leq 0. \end{cases}\] Let \(\chi_{\alpha, n}^2\) denote the value of a Chi-square random variable \(Y\) with \(n\) degrees of freedom such that \(P(Y > \chi_{\alpha, n}^2) = \alpha\). If \(x_1, x_2, \ldots, x_{18}\) is a realization of this random sample, then, based on the sufficient statistic \(\sum_{i=1}^{18} X_i^2\), which one of the following is a 98% confidence interval for \(\theta\)?

A random sample \( X \) of size one is taken from a distribution with the probability density function \[ f(x; \theta) = \begin{cases} \frac{2x}{\theta^2}, & 0 < x < \theta, \\ 0, & \text{elsewhere}. \end{cases} \] If \( \frac{X}{\theta} \) is used as a pivot for obtaining the confidence interval for \( \theta \), then which one of the following is an 80% confidence interval (confidence limits rounded off to three decimal places) for \( \theta \) based on the observed sample value \( x = 10 \)?

Let X₁, X₂, ..., X₁₀ be a random sample from the following probability density function
f(x) = { 2(x - μ) e^{-(x-μ)²} if x > μ,
0 otherwise,
where μ ∈ (-∞, ∞) is an unknown parameter. It is given that the observed value of min{X₁, X₂, ..., X₁₀} is 1. Using the pivot min{X₁, X₂, ..., X₁₀} - μ, suppose a 95% confidence interval of μ is of the form (c, 1), then c equals __________ (rounded off to two decimal places).