Let X be a random sample of size 1 from a population with cumulative distribution function $$ F(x) = \begin{cases} 0 & \text{if } x

Question

Let X be a random sample of size 1 from a population with cumulative distribution function $$ F(x) = \begin{cases} 0 & \text{if } x < 0, \ 1 - (1 - x)^\theta & \text{if } 0 \leq x < 1, \ 1 & \text{if } x \geq 1 \end{cases} $$ where $ \theta > 0 $ is an unknown parameter. To test $ H_0 : \theta = 1 $ against $ H_1 : \theta = 2 $, consider using the critical region $ \{ x \in \mathbb{R} : x < 0.5 \} $. If $ \alpha $ and $ \beta $ denote the level and power of the test, respectively, then $ \alpha + \beta $ (rounded off to two decimal places) equals:

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The critical region is $ \{ x \in \mathbb{R} : x<0.5 \} $, meaning we reject $ H_0 $ if $ x<0.5 $. - The level $ \alpha $ is the probability of rejecting $ H_0 $ when $ H_0 $ is true. This is $ P(X<0.5 | \theta = 1) $. For $ \theta = 1 $, the CDF is: $$ F(x) = 1 - (1 - x)^\theta $$ Substituting $ \theta = 1 $: $$ F(x) = 1 - (1 - x) $$ Thus, $$ P(X<0.5 | \theta = 1) = F(0.5) = 1 - (1 - 0.5) = 0.5 $$ Therefore, $ \alpha = 0.5 $. - The power $ \beta $ is the probability of rejecting $ H_0 $ when $ H_1 $ is true. This is $ P(X<0.5 | \theta = 2) $. For $ \theta = 2 $, the CDF is: $$ F(x) = 1 - (1 - x)^2 $$ Thus, $$ P(X<0.5 | \theta = 2) = F(0.5) = 1 - (1 - 0.5)^2 = 1 - 0.25 = 0.75 $$ Therefore, the power $ \beta = 0.75 $. Thus, $ \alpha + \beta = 0.5 + 0.75 = 1.20 $.

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