Solutions — Margin of Error and Level of Confidence

Fluency A 95% CI for p is based on n = 400 and p̂ = 0.6. Calculate the margin of error.

ME = z* × √(p̂(1−p̂)/n)
ME = 1.96 × √(0.6 × 0.4 / 400)
ME = 1.96 × √(0.0006)
ME = 1.96 × 0.02449
ME ≈ 0.0480

The margin of error is approximately 4.8%.

Fluency A researcher wants a margin of error of at most 0.05 with 95% confidence. No prior estimate of p is available. Find the minimum required sample size.

No prior estimate → use p̂ = 0.5 (conservative)

n ≥ (z*)² × p̂(1−p̂) / ME²
n ≥ (1.96)² × 0.5 × 0.5 / (0.05)²
n ≥ 3.8416 × 0.25 / 0.0025
n ≥ 0.9604 / 0.0025
n ≥ 384.16

Minimum sample size: n = 385

Fluency Calculate the margin of error for a 99% CI with p̂ = 0.35 and n = 500.

z* = 2.576 for 99% CI

SE = √(0.35 × 0.65 / 500) = √(0.000455) ≈ 0.02133
ME = 2.576 × 0.02133 ≈ 0.05495

The margin of error is approximately 5.5%.

Fluency A 95% CI is reported as (0.44, 0.56). What is the margin of error and the sample proportion?

The CI is centred at the sample proportion:
p̂ = (0.44 + 0.56) / 2 = 0.50

The margin of error is the half-width:
ME = (0.56 − 0.44) / 2 = 0.06

The interval extends 6 percentage points either side of the sample proportion.

Understanding A pollster currently uses n = 400 for 95% CIs. How large a sample would they need to halve the current margin of error?

ME ∝ 1/√n, so halving ME requires multiplying n by 4.

Formal derivation: If ME_new = ME_old/2, then:
z* × √(p̂(1−p̂)/n_new) = (1/2) × z* × √(p̂(1−p̂)/n_old)
√(1/n_new) = (1/2) × √(1/n_old)
1/n_new = 1/(4n_old)
n_new = 4 × n_old = 4 × 400 = 1600

To halve the margin of error, the pollster needs to quadruple the sample size to n = 1600.

Understanding A pilot study finds p̂ ≈ 0.25. A researcher wants ME ≤ 0.03 at the 95% confidence level. Find the required n and compare it to the conservative estimate using p̂ = 0.5.

Using p̂ = 0.25:
n ≥ (1.96)² × 0.25 × 0.75 / (0.03)²
n ≥ 3.8416 × 0.1875 / 0.0009
n ≥ 0.7203 / 0.0009
n ≥ 800.3 → n = 801

Conservative (p̂ = 0.5):
n ≥ (1.96)² × 0.25 / 0.0009
n ≥ 0.9604 / 0.0009
n ≥ 1067.1 → n = 1068

The pilot estimate saves 1068 − 801 = 267 participants. Using prior information makes the study considerably more efficient.

Understanding Two surveys have identical n = 200 and p̂ = 0.5. Survey A reports a 90% CI and Survey B reports a 99% CI. Which has a larger margin of error? Calculate both and find the ratio of the margins.

SE = √(0.5 × 0.5 / 200) = √(0.00125) ≈ 0.03536

Survey A (90% CI): z* = 1.645
ME_A = 1.645 × 0.03536 ≈ 0.05817

Survey B (99% CI): z* = 2.576
ME_B = 2.576 × 0.03536 ≈ 0.09108

Survey B has the larger margin of error.

Ratio: ME_B / ME_A = 2.576 / 1.645 ≈ 1.566

The 99% CI has a margin of error about 56.6% larger than the 90% CI for the same data.

Understanding A TV network claims its news program is watched by 30% of households ± 3% (margin of error). If the 95% CI formula was used, back-calculate the approximate sample size used.

Given: p̂ = 0.30, ME = 0.03, z* = 1.96

ME = z* × √(p̂(1−p̂)/n)
0.03 = 1.96 × √(0.30 × 0.70 / n)
0.03 / 1.96 = √(0.21/n)
0.015306 = √(0.21/n)
(0.015306)² = 0.21/n
0.0002343 = 0.21/n
n = 0.21 / 0.0002343 ≈ 896

The network surveyed approximately 896 households.

Problem Solving A health researcher wants to estimate the proportion of adults with high blood pressure. From previous studies, p is thought to be near 0.22. The researcher wants a 99% CI with ME ≤ 0.025. Find the required sample size using (a) the prior estimate and (b) the conservative approach. How many extra participants does the conservative approach require?

z* = 2.576 (99% CI), ME = 0.025

(a) Using p̂ = 0.22:
n ≥ (2.576)² × 0.22 × 0.78 / (0.025)²
n ≥ 6.6355 × 0.1716 / 0.000625
n ≥ 1.1387 / 0.000625
n ≥ 1822.0 → n = 1822

(b) Conservative (p̂ = 0.5):
n ≥ (2.576)² × 0.25 / 0.000625
n ≥ 6.6355 × 0.25 / 0.000625
n ≥ 1.6589 / 0.000625
n ≥ 2654.2 → n = 2655

Extra participants: 2655 − 1822 = 833 extra

Using the prior estimate from previous studies saves 833 participants — a significant saving for an expensive clinical study.

Problem Solving An election analyst wants to be 95% confident that their estimate of a candidate’s true proportion of support is within 2 percentage points of the actual value.

(a) Find the minimum sample size needed (conservative).
(b) If the analyst can only afford n = 600, what is the actual margin of error?
(c) The analyst suggests increasing the confidence level to 99% while keeping n = 600. What is the new margin of error?

(a) ME = 0.02, z* = 1.96, conservative p̂ = 0.5:
n ≥ (1.96)² × 0.25 / (0.02)²
n ≥ 3.8416 × 0.25 / 0.0004
n ≥ 0.9604 / 0.0004
n ≥ 2401
Minimum sample size: n = 2401

(b) n = 600, p̂ = 0.5 (conservative), z* = 1.96:
ME = 1.96 × √(0.5 × 0.5 / 600)
ME = 1.96 × √(0.000417)
ME = 1.96 × 0.02041
ME ≈ 0.0400 (4.0%)

With n = 600, the analyst can only achieve ±4.0% accuracy at 95% confidence — twice the desired margin of error.

(c) n = 600, p̂ = 0.5, z* = 2.576 (99%):
ME = 2.576 × 0.02041 ≈ 0.0526 (5.26%)

Increasing confidence to 99% widens the margin to ±5.26%. The analyst faces a trade-off: with budget for n = 600, they can have either 95% confidence with ±4% accuracy or 99% confidence with ±5.26% accuracy — neither meets the original ±2% target.