10.3%, plus or minus what?

converted on ab-test-checkout is true in 10.3% of 6,000 rows. That is a point estimate, and on its own it is half a sentence.

The standard error on a proportion is sqrt(p(1-p)/n). Here that is 0.39 percentage points, so the 95% interval runs 9.5% to 11.1% — a width of 1.5 points.

import math

n, k = 6000, 617
p = k / n
se = math.sqrt(p * (1 - p) / n)
print(f"{p:.3%}  [{p - 1.96*se:.3%}, {p + 1.96*se:.3%}]")

That interval is tight, which is what 6,000 rows buys you. It is worth knowing why it is tight, so you recognise the cases where it is not.

Getting the interval down to ±0.5 points would need about 14,177 rows. Precision costs sample size quadratically — halving the width costs four times the data — which is the single most useful fact for anyone about to promise a more precise answer next week.

Normal approximation, and it starts lying at small counts or proportions near the boundaries; the Wilson interval behaves there. The pattern has both.