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
Probability of Type-I Error Calculation
The question asks for the probability of a Type-I error when testing hypotheses about the parameter $\theta$ of a Uniform$[0, \theta]$ distribution. A Type-I error occurs when the null hypothesis ($H_0$) is rejected, but it is actually true.
Hypothesis Setup
- Null Hypothesis ($H_0$): $\theta = 1$
- Alternative Hypothesis ($H_1$): $\theta = 2$
- Rejection Region: The test rejects $H_0$ if the sum of the random sample $X_1 + X_2$ is greater than $\frac{4}{5}$.
- Type-I Error Probability ($\alpha$): This is the probability of rejecting $H_0$ given that $H_0$ is true. Mathematically, $\alpha = P(\text{Reject } H_0 | H_0 \text{ is true})$.
Distribution Under $H_0$
When the null hypothesis $H_0: \theta = 1$ is true, the random variables $X_1$ and $X_2$ are independent and identically distributed (i.i.d.) from the Uniform distribution on the interval $[0, 1]$. The probability density function (PDF) for each $X_i$ is $f(x) = 1$ for $0 \le x \le 1$ and $0$ otherwise. The joint PDF for $X_1$ and $X_2$ is $f(x_1, x_2) = f(x_1)f(x_2) = 1 \times 1 = 1$ over the unit square defined by $0 \le x_1 \le 1$ and $0 \le x_2 \le 1$.
Calculating the Error Probability
The probability of a Type-I error is calculated using the rejection region and the distribution under $H_0$: $ \alpha = P(X_1 + X_2 > \frac{4}{5} | \theta = 1) $ This probability corresponds to the area within the unit square ($0 \le x_1 \le 1$, $0 \le x_2 \le 1$) where the condition $x_1 + x_2 > \frac{4}{5}$ holds. It's often easier to calculate the probability of the complementary event, $P(X_1 + X_2 \le \frac{4}{5})$.
The region defined by $x_1 + x_2 \le \frac{4}{5}$ within the unit square forms a right-angled triangle with vertices at $(0, 0)$, $(\frac{4}{5}, 0)$, and $(0, \frac{4}{5})$.
The area of this triangle is given by: $ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{4}{5} \times \frac{4}{5} = \frac{1}{2} \times \frac{16}{25} = \frac{8}{25} $ Thus, $P(X_1 + X_2 \le \frac{4}{5}) = \frac{8}{25}$.
Final Probability Calculation
The probability of the Type-I error is the complement of this area within the unit square:
$ \alpha = P(X_1 + X_2 > \frac{4}{5}) = 1 - P(X_1 + X_2 \le \frac{4}{5}) $ $ \alpha = 1 - \frac{8}{25} = \frac{25}{25} - \frac{8}{25} = \frac{17}{25} $Therefore, the probability of Type-I error is $\frac{17}{25}$.
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 $\{Y_n: n \ge 1\}$ 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?
