---
title: "CI and sample size: overcharged households"
subject: "Probability and Statistics"
chapter: 7
key: "semester_5/probability_and_statistics:postal-survey-overcharged-ci-sample-size"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/postal-survey-overcharged-ci-sample-size
site: BCT Engineering
---

# CI and sample size: overcharged households

Probability and Statistics · Chapter 7: Inference concerning proportion

- Asked in: 2072 Chaitra, 5 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/postal-survey-overcharged-ci-sample-size
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-7

## Question

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.

## 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).

## Answer

#### i) 99% confidence interval for the population proportion

```text
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

```text
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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which z value is used for a 99% confidence interval?
   A. 1.645
   B. 1.960
   C. 2.326
   D. 2.576
   Answer: D. A 99% two-sided interval uses z = 2.576. (Key point: Critical z for 99%)

2. With p-hat = 0.66, n = 500 and SE = 0.0212, what is the 99% interval?
   A. (0.618, 0.702)
   B. (0.605, 0.715)
   C. (0.639, 0.681)
   D. (0.570, 0.750)
   Answer: B. Margin = 2.576 x 0.0212 = 0.0546, so 0.66 +/- 0.0546 = (0.605, 0.715). (Key point: 99% confidence interval)

3. Which formula gives the sample size for margin of error E?
   A. n = z x p x q / E
   B. n = z^2 p q / E^2
   C. n = E^2 / (z^2 p q)
   D. n = z p q / E^2
   Answer: B. Solving E = z sqrt(pq/n) for n gives n = z^2 pq / E^2. (Key point: Sample size formula)

4. If no prior estimate of p were available, which value is used, and what does it do to n?
   A. p = 0.5, giving the largest (safe) n
   B. p = 0.5, giving the smallest n
   C. p = 1, giving the largest n
   D. p = 0, so n cannot be found
   Answer: A. pq is maximum at p = 0.5, so it guarantees enough sample; here it gives n = 2654 instead of 2383. (Key point: Conservative p = 0.5)

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