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?
Understanding the Exponential Distribution Test
The question asks to evaluate statements regarding the rejection of the null hypothesis $H_0: \sigma \le 1$ against the alternative $H_1 : \sigma > 1$ for an Exponential distribution. The test is conducted at a significance level $\alpha = 0.1$ using a random sample of size $n=15$.
Deriving the Test Statistic and Critical Region
For an Exponential distribution with PDF $f(x;\sigma) = \frac{1}{\sigma} \exp \left(-\frac{x}{\sigma}\right)$, the statistic $2n\bar{X}/\sigma$ follows a chi-squared distribution with $2n$ degrees of freedom. Thus, $2n\bar{X}/\sigma \sim \chi^2_{2n}$.
With $n=15$, the statistic becomes $2(15)\bar{X}/\sigma = 30\bar{X}/\sigma$, which follows a $\chi^2_{30}$ distribution.
The likelihood ratio test for $H_0: \sigma \le 1$ vs $H_1: \sigma > 1$ rejects $H_0$ if the test statistic is sufficiently large. The critical value is determined based on the distribution of the statistic under the boundary case of $H_0$, which is $\sigma=1$. The test rejects $H_0$ if:
$P\left(\frac{30\bar{X}}{1} > c\right) = \alpha$
We use the provided chi-squared value for $2n=30$ degrees of freedom and $\alpha=0.1$:
$c = \chi^2_{30, 0.1} = 40.256$
Therefore, the test rejects $H_0$ if $30\bar{X} > 40.256$. Simplifying this inequality gives the critical value for $\bar{X}$:
Decision Rule: Reject $H_0$ if $\bar{X} > \frac{40.256}{30} \approx 1.34187$. Otherwise, do not reject $H_0$.
Evaluating Each Statement
We now apply the decision rule to each observed value of $\bar{X}$.
Statement A: Observed $\bar{X} = 0.6$
Since $0.6 \le 1.34187$, the test does not reject $H_0$. Statement A is true.
Statement B: Observed $\bar{X} = 1.6$
Since $1.6 > 1.34187$, the test rejects $H_0$. Statement B is true.
Statement C: Observed $\bar{X} = 1.3$
Since $1.3 \le 1.34187$, the test does not reject $H_0$. The statement claims rejection, which contradicts the rule. Statement C is false.
Statement D: Observed $\bar{X} = 1.2$
Since $1.2 \le 1.34187$, the test does not reject $H_0$. Statement D is true.
Conclusion
The true statements are A, B, and D.
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 $X$ 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 $\theta$?
