Nine of ten correct, the exact tail

Question

A trader makes 10 independent directional calls and gets 9 correct. Under the null of guessing, each call is right with probability $0.5$ .

Compute the exact one-sided p-value (probability of 9 or more correct by guessing) and interpret it.

Accepted Answer

Set up. Let X \sim 	ext{Binomial}(10, 0.5). The one-sided p-value is p = \mathbb{P}(X \geq 9) = \frac{\binom{10}{9} + \binom{10}{10}}{2^{10}}. Count and compute. Here \binom{10}{9} = 10 and \binom{10}{10} = 1, summing to 11, over 2^{10} = 1024: p = \frac{11}{1024} \approx 0.0107 \approx 0.011. Interpretation. A pure guesser scores 9 or more out of 10 only about 1.1% of the time, so this record is significant at the 5% level and even at 2%. Notice the sharp jump from the previous case: going from 8 correct (p \approx 0.055) to 9 correct (p \approx 0.011) drops the p-value fivefold, because th…

Nine of ten correct, the exact tail

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