One-sided p-value from a z of 2.33

Question

A momentum signal is tested with the pre-registered alternative that it beats the benchmark (effect $> 0$ ). The test returns $z = 2.33$ .

Compute the one-sided p-value for $z = 2.33$ , compare it to the two-sided p-value, and note what the pre-registration buys you.

Accepted Answer

The one-sided computation. For an upper-tail alternative, take the area beyond z in the predicted tail only: p{	ext{one}} = 1 - \Phi(2.33) = 1 - 0.9901 = 0.0099 \approx 0.01. Compared to two-sided. Doubling for both tails gives p{	ext{two}} = 2\,[1 - \Phi(2.33)] = 0.0198 \approx 0.02. Both are significant at 5% here, but the one-sided p is exactly half the two-sided p, as always. What pre-registration buys. Because the direction was fixed in advance, the one-sided test is legitimate and more powerful in the predicted direction: it detects real upper-tail effects more easily. The cost is tha…

One-sided p-value from a z of 2.33

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