Question

You were giving me equals 0.020 and a confidence level 90% and a sample proportion of 85% from a recent survey find the minimum sample size to estimate a population proportion

Find z(under/a/2)a=0.01 (also how do I write the z thing?

Answer 1

Step 1 — identify values
- Desired margin of error: E = 0.020
- Sample proportion: p̂ = 0.85 (so p̂(1−p̂) = 0.85·0.15 = 0.1275)
- Confidence level: 90% ⇒ α = 1 − 0.90 = 0.10 ⇒ α/2 = 0.05

Notation: the critical z-value is written z_{α/2} (or z_{(α/2)}). It is the z with upper-tail area α/2, equivalently the z for cumulative area 1 − α/2. For 90% CI, z_{α/2} = z_{0.05} = z_{0.95} ≈ 1.645.

Step 2 — sample size formula for a proportion
n = (z_{α/2}^2 · p̂(1−p̂)) / E^2

Plug in numbers:
- z_{α/2} ≈ 1.645
- p̂(1−p̂) = 0.1275
- E^2 = 0.02^2 = 0.0004

n = (1.645^2 · 0.1275) / 0.0004 ≈ 0.34502 / 0.0004 ≈ 862.545

Round up to whole people: n = 863.

Note: if you didn’t have a prior p̂ you would use p̂ = 0.5 (gives the largest required n). Also, if you meant α = 0.01 (i.e., 99% confidence) instead, z_{α/2} ≈ 2.576 and the required n would be about 2,115.