Type II error of a coin-fairness test

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You flip a coin n = 100 times to test H0: p = 0.5 against H1: p 0.5. The decision rule is: reject H0 (declare the coin biased toward heads) if the number of heads is 58 or more. Suppose the coin is actually biased with true p = 0.6. What is the probability of a type II error (failing to reject, so missing the real bias)?

Accepted Answer

A type II error means the coin is truly biased (p = 0.6) but you get 57 heads or fewer, so you fail to reject. Model the count under H1. With p = 0.6 and n = 100, the number of heads has mean np = 60 and standard deviation \sqrt{np(1-p)} = \sqrt{100 	imes 0.6 	imes 0.4} = \sqrt{24} \approx 4.90. Apply the normal approximation with a continuity correction. "Fail to reject" is X \le 57, so use the boundary 57.5: \beta = P(X \le 57.5 \mid p = 0.6) = \Phi\!\left(\frac{57.5 - 60}{4.90}ight) = \Phi(-0.51) \approx 0.305. So the test misses a real bias of this size about 30% of the time; its powe…

Type II error of a coin-fairness test

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