RPP data: use p-implied SEs, not 1/sqrt(N-3), for Fisher-z effects
In the Reproducibility Project: Psychology data (CenterForOpenScience/rpp, RPPdataConverted.tsv), read the TSV as latin-1; the CSV has a ragged row. With r and N, the naive Fisher-z SE 1/sqrt(N-3) makes only 76 of the 90 originals with reported p < .05 significant. Small-df t/F tests and within-subject designs (e.g. F(2,92) with N=48, F(18,660) with N=12) break it. Better: se_O = atanh(r_O)/Phi^-1(1-p_O/2), using T.pval.USE.O. Then se_R = se_O*sqrt((N_O-3)/(N_R-3)). That reproduces the replications' own p-values (median z ratio 0.99, significance agreement 98%, 33 vs 33). One original reports p = 0 (N = 230,047), so fall back to the naive SE there.
Every number reproduces from RPPdataConverted.tsv (latin-1): 97 complete rows, 90 originals with p < .05, 76 naive-significant, one p = 0 original, median z ratio 0.993, agreement 97.8%, 33 vs 33. T.N.* and N.* columns give the same counts.
The replication check does not distinguish the two methods, though. On the original side, se_O = atanh(r_O)/z(p_O) reproduces p_O by construction, so the replication side is the only independent test. There the naive 1/sqrt(N_R-3) does as well. Median ratio of implied to reported z_R: 0.999 naive vs 0.993 p-implied. Significance agreement: 96.7% vs 97.8%. Within 10% of the reported z_R: 68/89 naive vs 65/89 p-implied. The q10-q90 range of the ratio is 0.83-1.09 naive vs 0.92-1.21 p-implied. So the p-implied SE is the right choice if you want the originals' reported tests honoured. But 'reproduces the replications' p-values' is not evidence that it beats 1/sqrt(N-3) for the replications.