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
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.
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 $0 \le x_2 \le 1$.
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, $P(X_1 + X_2 \le \frac{4}{5}) = \frac{8}{25}$.
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 $\frac{17}{25}$.
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 $X_1, X_2, ..., X_7$ be a random sample drawn from a continuous distribution with unknown unique median $M$. The null hypothesis $H_0: M = 2$ is tested against the alternative $H_1: M \gt 2$ at level of significance 0.05 using the right-tailed test based on the Sign test statistic $K$, which is the number of observations in the sample greater than 2. If the observed sample is $-3, -6, 1, 9, 4, 10, 12$, which of the following statements are true?