Running tally — estimated s and 95% CI across rounds
What you just did has a name
Every round reduced to the same calculation:
Δψ = ln(pt/(1−pt)) − ln(p0/(1−p0)),
and ŝ = Δψ / t. The point estimate only depends on
p0, pt, and t. The
confidence depends on Ne·t·p̄(1−p̄),
because the drift variance of Δψ is 1/(2Ne·p̄(1−p̄))
per generation.
Rounds 1 and 2 had identical endpoints but different t — so the same observed change implied very different s. Round 3 used the same change in a small population: the CI spanned zero, so selection was not distinguishable from drift. Round 4 (LTEE-like) had a huge product of Ne·t and gave a razor-tight CI. Round 5 (FSJ-like) had a small product and gave a CI that spanned zero — the observed Δp is consistent with pure drift.
Estimating s requires all three: Δp, t, and Ne. Give up any one and the answer is either unidentified or has no CI. In your own work, when you want to report "how strong is selection here?", you must report all three alongside.