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CI and sample size: overcharged households

Short answer

i) p̂=0.66, SE=0.0212, z₀.₀₀₅=2.576, margin=0.0546, 99% CI (0.605, 0.715). ii) For E=0.025, n = z²p̂q̂/E² = 2.576²×0.66×0.34/0.025² ≈ 2383 households (or 2655 using conservative p=0.5).

In a postal survey of 500 households, 330 said that they thought they were being overcharged for the public services within their area. i) Calculate an approximate 99% confidence interval for the population proportion, p, of households who thought they were being overcharged for public services within their area. ii) Estimate the size of sample required to estimate the value of p to be within 99% confidence limits of ±0.025.

i) 99% confidence interval for the population proportion

n = 500, p̂ = 330/500 = 0.66, q̂ = 0.34
SE = √(p̂q̂/n) = √(0.66×0.34/500) = 0.0212
z (99%) = 2.576
margin = 2.576 × 0.0212 = 0.0546
limits = 0.66 ± 0.0546 = (0.6054, 0.7146)

The 99% interval is (0.605, 0.715).

ii) Sample size for a margin of error E = ±0.025

From E = z·√(pq/n): n = z²·p·q/E²

n = (2.576)² × 0.66 × 0.34 / (0.025)²
  = 6.6358 × 0.2244 / 0.000625
  = 2382.5  → 2383

Answer: i) (0.605, 0.715); ii) n ≈ 2383 households (using p̂ = 0.66 from the survey). If no prior estimate were available, the conservative value p = 0.5 would give n = 6.6358 × 0.25 / 0.000625 = 2654.3, so 2655.

✓ Answer checked by Claude Sonnet 5.5

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