Question

He's a sample n equals 500 p equals 0.90 and a 90 percent confidence level to construct a confidence interval estimate of the population proportion p

Answer 1

Use the normal approximation. For a 90% CI, z* = 1.645.

Sample proportion p̂ = 0.90, n = 500.
Standard error = sqrt[p̂(1 − p̂)/n] = sqrt(0.90·0.10/500) = sqrt(0.00018) ≈ 0.01342.
Margin of error = z* · SE = 1.645 × 0.01342 ≈ 0.02207.

Confidence interval: 0.90 ± 0.02207 → (0.8779, 0.9221).

Rounded: approximately (0.878, 0.922).

Interpretation: We are 90% confident the true population proportion p is between about 0.878 and 0.922.