Three Points from a Thousand People, and the Two Roundings That Cancel

A thousand people are asked which candidate they prefer and 600 of them say A. What is the margin of error? The answer is about three points, and getting there needs no calculator, no table of normal quantiles and no memory of any constant. It needs one square root of 1000.

The shortcut is this. For a proportion measured on a sample of size $N$, the usual 95 percent margin of error is almost exactly $1/\sqrt{N}$. Since $\sqrt{1000} \approx 31.6$, the margin is about 3.16 percent, and the exact figure is 3.04. That is close enough for any purpose a poll is put to.

What the exact number is

Each answer is one draw from a two-outcome distribution with unknown success probability $p$, and the estimator is the sample proportion $\hat p$. Its variance is $p(1-p)/N$, so the standard error is

A 95 percent interval reaches 1.96 standard errors on each side, and 1.96 times that is 0.0303642. Three points and a bit, and the whole computation took a square root of 15 that nobody wants to do in their head.

Two roundings, chosen because they are cheap

Both awkward numbers in equation (1) can be replaced by round ones, and the size of each error can be bounded before the swap rather than hoped for afterwards.

The first is $p(1-p)$. As a function of $p$ it is a downward parabola with maximum $1/4$ at $p = 1/2$, and a parabola is flat at its vertex. At $p = 0.6$ the value is 0.24, which sits 4 percent below the peak. Replacing it by a quarter therefore costs at most 2 percent in the standard error, because the square root halves relative errors.

The second is 1.96, which is the 97.5th percentile of the standard normal distribution. Calling it 2 costs 2.04 percent of itself. Neither approximation is subtle, and their product accounts for the 4.1 percent gap between 3.16 and 3.04.

The two errors cancel algebraically

Here is the part worth remembering. With $p(1-p)$ replaced by $1/4$ and 1.96 replaced by 2, the margin becomes

The factor of two from the confidence level and the factor of a half from the square root of a quarter are reciprocal, so they annihilate. Equation (2) is an identity in $N$, not a coincidence at 1000, and the result carries no constants at all. That is the reason the shortcut survives being remembered badly: there is nothing left in it to misremember.

The consequence is a small table anyone can carry. A sample of 100 gives 10 points, 400 gives 5, 1000 gives roughly 3, 2500 gives 2 and 10,000 gives 1.

The shortcut errs on the safe side, and that is not luck

A cheap approximation is worth much more when you know which way it leans. $1/\sqrt{1000} = 0.031623$ against an exact 0.030364, so the shortcut overstates the margin by 3.98 percent of itself. That direction is guaranteed whenever the reported proportion is not exactly a half, since $1/4$ is the maximum of $p(1-p)$ and can only be too large. The remaining question is whether the 1.96-to-2 rounding can outweigh it. It cannot: the quantile rounding contributes a fixed factor of $2/1.96 = 1.0204$ upward, so both swaps push the same way and the product is above 1 for every $p$ and every $N$.

A simulation of 200,000 synthetic polls at $p = 0.6$ and $N = 1000$ confirms the story from the other end. The exact interval caught the true value 95.06 percent of the time, the shortcut interval 95.76 percent, and the interval of one standard error 68.37 percent.

Reporting one standard error is a different answer

The common failure here is not arithmetic. Equation (1) gives 1.55 percent, and 1.55 is a real number about a real poll, so it feels like an answer. It is the answer to a question nobody asked. An interval of one standard error is a 68 percent interval, and quoting it as the margin of error understates the uncertainty by nearly half.

Working out sensible-looking numbers for a neighbouring question is a more dangerous kind of mistake than dropping a factor, because re-checking the arithmetic never catches it. The check that does catch it is dimensional in spirit: ask what fraction of repeated polls the interval is supposed to contain, and see whether your constant matches.

Where the rule stops being usable

Away from a half, the quarter becomes badly generous. At $p = 0.05$ the true $p(1-p)$ is 0.0475, so $1/4$ overstates the variance by a factor of 5.3 and the margin by a factor of 2.3. The shortcut still gives a valid upper bound, and it stops being informative. For rare categories, compute equation (1) properly.

At small $N$, or when $Np$ is small, the normal approximation itself fails and the interval can extend below zero or above one. The usual patch is a continuity-corrected or score-based interval, which no longer has a one-line form.

The margin never depends on how large the population is, which surprises people who expect a national survey to need a bigger sample than a municipal one. Re-derive the variance from the two-outcome distribution and no population term appears. Sampling without replacement does introduce a factor $\sqrt{(M-N)/(M-1)}$ for a population of size $M$, and at $N = 1000$ out of millions that factor is 0.9998.

One thing equation (2) cannot see at all: it measures only the randomness of who was sampled. Whether the sample resembles the population, whether the question was clear, whether refusals correlate with the answer, none of that appears anywhere in the algebra. The margin of error is the smallest of the errors in most real surveys, and it is the only one with a formula.

Sources and further reading

• The quantity being estimated — Margin of error
• Equation (1) and its better-behaved replacements — Binomial proportion confidence interval
• Why 1.96 and not something else — Central limit theorem
• The distinction the trap loses — Standard error

Every figure above was checked twice: by exact rational arithmetic on $\sqrt{15}/250$ and its multiples, and by measuring the actual coverage of the three intervals over 200,000 simulated polls.