Discussion
Given the frequency function
\(f(x, \theta) = \frac{1}{\theta};\ 0 \le x \le \theta\)
\(= 0;\) otherwise
and that testing the null hypothesis H0 : θ = 1 against its alternative H1 : θ = 2 by means of a single observed value of x. What would be the size of type one error for (0.5 ≤ x) ?
Answer: B
The size of the type I error is 0.50.
The size is the probability of rejecting \(H_0\) when it is true. The critical region is \(x \ge 0.5\), and under \(H_0\) (\(\theta = 1\)) x is uniform on 0 to 1 with density 1:
\[ \begin{aligned} \alpha &= P(x \ge 0.5 \mid \theta = 1) \\ &= \int_{0.5}^{1} 1\,dx = 0.5 \end{aligned} \]
0.25 would be the probability under \(\theta = 2\) of \(x\) lying in 0 to 0.5, not the type I error. 0.75 is the power of the test, \(P(x \ge 0.5 \mid \theta = 2) = 1.5/2\). A size of 1 would mean always rejecting.
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