Type II error from a fixed decision threshold

Question

A detector reads a value X. When there is no signal (H0), X \sim 	ext{Normal}(	ext{mean } 0,\ 	ext{sd } 1). When a signal is present (H1), X \sim 	ext{Normal}(	ext{mean } 3,\ 	ext{sd } 1). The decision rule is: declare "signal" whenever X 2. What is the probability this rule makes a type II error (fails to declare a signal when one is truly present)?

Accepted Answer

A type II error means a real signal is present but X lands at or below the cutoff of 2, so the detector stays quiet. Compute \beta. Under H1, X \sim 	ext{Normal}(3, 1). Standardize the cutoff: \beta = P(X \le 2 \mid H1) = \Phi\!\left(\frac{2 - 3}{1}ight) = \Phi(-1) \approx 0.159. So the rule misses a real signal about 16% of the time; its power is 1 - \beta \approx 0.84. For context, the type I error. Under H0, X \sim 	ext{Normal}(0,1), and a false alarm happens when X 2: \alpha = P(X 2 \mid H0) = 1 - \Phi(2) \approx 0.023. Notice the trade-off baked into the single cutoff of 2: it sits t…

Type II error from a fixed decision threshold

Your answer

Discussion

Learn the concepts

Related questions