Margin of Error Math Example 4

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Example 4

hard
Two polls: Poll A (n=100n=100): margin of error ±10%. Poll B (n=1600n=1600): margin of error ±2.5%. By what factor was n increased, and by what factor did E decrease? Verify using E=z∗0.5×0.5nE = z^* \sqrt{\frac{0.5 \times 0.5}{n}}.

Solution

  1. 1
    Factor increase in n: 1600100=16\frac{1600}{100} = 16
  2. 2
    Factor decrease in E: 10%2.5%=4\frac{10\%}{2.5\%} = 4
  3. 3
    Relationship: E∝1nE \propto \frac{1}{\sqrt{n}}; 16=4\sqrt{16} = 4 — E decreased by factor 4 when n increased by factor 16 ✓
  4. 4
    Verify: EA=1.960.25100=1.96×0.05=0.098≈10%E_A = 1.96\sqrt{\frac{0.25}{100}} = 1.96 \times 0.05 = 0.098 \approx 10\% ✓; EB=1.960.251600=1.96×0.0125=0.0245≈2.5%E_B = 1.96\sqrt{\frac{0.25}{1600}} = 1.96 \times 0.0125 = 0.0245 \approx 2.5\% ✓

Answer

n increased 16×, E decreased 4×. Margin of error decreases as 1n\frac{1}{\sqrt{n}}.
The relationship E∝1/nE \propto 1/\sqrt{n} is fundamental: to halve the margin of error, quadruple the sample size. To reduce E by 4, multiply n by 16. This explains why major polls use n=1000-2000 to achieve ±3% margins.

About Margin of Error

The maximum expected difference between the sample statistic and the true population parameter; it is half the width of a confidence interval.

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Example 2 medium

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Example 3 easy

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