9.9%, plus or minus what?

dropped_out on clinical-trial is true in 9.9% of 900 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 1.00 percentage points, so the 95% interval runs 7.9% to 11.8% — a width of 3.9 points.

import math

n, k = 900, 89
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 wide enough that a change of a point or two means nothing, and it will be reported as a change anyway unless someone puts the bounds next to it.

Getting the interval down to ±0.5 points would need about 13,693 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.