Sample Size & Margin of Error Planner

plan survey sample size and precision

Worked example — Sample Size & Margin of Error Planner

Confidence	95% (z* = 1.96)
Margin of error	5.0%

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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A national poll needs no more people than a town poll 🖖

How many people you need barely depends on how many people there are. At 95% confidence and ±5%, this tool asks for 385 respondents whether the population is one million or entirely unbounded — the two give the identical answer. Only when the population shrinks to something like 5,000 does the finite-population correction bite at all, and even then it merely trims 385 to 357. The intuition it overturns is the idea that a sample should be a percentage of the whole, which would make polling a country hopeless; what the formula actually cares about is how much random variation you are willing to tolerate, and a population of millions contributes none of its own.

Why precision quadruples the cost 🖖

This planner answers one question: how many people must you ask to pin a percentage down to a chosen margin of error? Because the margin shrinks with the square root of the sample size, buying twice the precision costs four times the respondents — moving from ±3% to ±1.5% roughly quadruples n. When the expected proportion is unknown, set p = 0.5: that maximizes p(1−p) and gives the largest, safest sample size.

When 2.4 million answers got it wrong 🖖

The margin-of-error formula measures only random sampling error — it assumes every respondent was chosen at random. In 1936 the Literary Digest polled about 2.4 million people and confidently predicted Landon would beat Roosevelt; George Gallup called it correctly with barely 50,000. The Digest's list over-sampled car and telephone owners during the Depression, so a huge but biased sample lost to a small representative one. No sample size can fix a skewed sampling frame.

Problem solved in full

The sample size for a poll accurate to ±5% at 95% confidence 5 steps
You want a poll accurate to ±5% at 95% confidence. Derive the sample size — then find what it costs to halve the error, and why the population size never appears.
1. The margin of error is the half-width of the confidence interval for a proportion. It shrinks as the square root of the sample size, which is where everything surprising comes from.
  
  $ME = z^{*} \frac{p ( 1 - p )}{n}$
2. Rearrange for n. The 1.96 is the standard normal value cutting off 2.5% in each tail — it is a property of the confidence level, not of your data.
  
  $n = \frac{( z ^{*} ) ^{2} p ( 1 - p )}{ME ^{2}}, z^{*} = 1.96 at 95%$
3. Substitute. The planner above prints this n, and the answer of 385 is the reason so many polls quote around a thousand respondents.
  
  $n = \frac{1.9 6 ^{2} \times 0.25}{0.0 5 ^{2}} = \frac{0.9604}{0.0025} = 384.2 \to 385$
4. Notice p = 0.5 was used without knowing the true proportion. That is deliberate: p(1−p) peaks there, so it is the worst case and the sample size is safe whatever the answer turns out to be.
  
  $p (1 - p) \leq 0.25 always, maximised at p = 0.5$
5. Now halve the target error. The square in the denominator does the damage.
  
  $ME = 2.5% : n = \frac{0.9604}{0.000625} = 1537$
Answer

385 people for ±5%, and 1537 for ±2.5%. Four times the cost to double the precision — the square-root law running backwards, and the reason polls stall at a few thousand rather than pushing to ±1%, which would need 9604. The other surprise is what is missing: the population never enters. A sample of 385 is as accurate for a city of 100 000 as for a country of 100 million, because the estimate depends on how many you asked, not on how many you did not.

$n = \frac{( z ^{*} ) ^{2} p ( 1 - p )}{ME ^{2}} = 385 ME halved \Rightarrow n quadrupled$

Learning path

When is a difference real?

References (1)

The 1936 Literary Digest poll, and why 2.4 million responses lost to 50,000: P. Squire, "Why the 1936 Literary Digest Poll Failed." Public Opinion Quarterly 52(1), 125–133, 1988.

By the 42Math team · Updated 2026-07-29

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

quick poll 95/5 - 95% confidence and ±5% margin -> classic quick poll planning
tight 95/3 - Tighter ±3% precision requires substantially larger sample size
finite population - Finite population correction lowers needed n for small populations
given n=1000 - Given n=1000, compute resulting margin of error at 95% confidence