Question

You are given a margin of error as 3%age points in a confidence level 99%, if the sample percentage from a recent post 35% find the minimum sample size to estimate a population proportion given n is 2500 and p is 0.86 find the margin of error e that corresponds to a 99% confidence level

Answer 1

Use the standard formulas for a CI for a proportion.

1) Minimum sample size with e = 3% (0.03), confidence 99%, observed p̂ = 0.35:
- z for 99% ≈ 2.5758
- n = (z^2 p(1−p)) / e^2
= (2.5758^2 · 0.35·0.65) / 0.03^2
≈ (6.6349 · 0.2275) / 0.0009
≈ 1.50945 / 0.0009 ≈ 1677.17
Round up: minimum n = 1,678.

(If p were unknown and you use the conservative p = 0.5, n ≈ 1,844.)

2) Margin of error with n = 2,500 and p = 0.86 at 99%:
- e = z · sqrt[ p(1−p) / n ]
= 2.5758 · sqrt(0.86·0.14 / 2500)
= 2.5758 · sqrt(0.1204 / 2500)
= 2.5758 · sqrt(0.00004816)
≈ 2.5758 · 0.006942 ≈ 0.0179

So e ≈ 0.0179 = 1.79 percentage points.