Answer

Where does the 1/sqrt(n) margin of error come from?

A frequently asked statistics question · September 21, 2026

confidence intervalsmargin of errorsampling

The formulas do have derivations

Statistics formulas are derived in the same spirit as those in calculus or physics, but introductory courses often skip the derivations because they lean on probability theory that comes later. The margin of error shortcut is a good example: it looks arbitrary, yet it follows from three ideas that you can check one at a time.

Step 1: the variability of a proportion

Suppose each person in a poll answers yes with probability p, independently of the others. One answer is a Bernoulli variable, and its variance is p(1−p). When you average n independent answers to get the sample proportion, the variance of that average is p(1−p)/n, because variances of independent quantities add and dividing by n squares into the denominator.

Take the square root and you get the standard error: sqrt(p(1−p)/n). It says how far a typical poll result lands from the true proportion. Notice that it shrinks with the square root of n, so quadrupling the sample only halves the error.

Step 2: turning that into a 95% margin

With a reasonably large sample, the central limit theorem says the sample proportion is close to normally distributed around p. A normal distribution puts about 95% of its probability within 1.96 standard deviations of its center, and 1.96 is where the familiar multiplier comes from. So the 95% margin of error is:

1.96 × sqrt(p(1−p)/n)

Step 3: remove the unknown p

You do not know p before running the poll, so you ask what the largest possible margin could be. The product p(1−p) is biggest when p = 0.5, where it equals 0.25. Its square root is 0.5, so the worst-case margin is 1.96 × 0.5 / sqrt(n) = 0.98 / sqrt(n). Round 0.98 up to 1 and you have the shortcut.

So 1/sqrt(n) is a slightly generous rule of thumb rather than an arbitrary choice: it holds for any true proportion, at 95% confidence.

Check it by simulation

The script below builds 10,000 imaginary polls of 1,000 people where the truth is exactly 50%, and compares the results with the formula. The figures quoted here come from one seeded run.

set.seed(2026)
n <- 1000
p <- 0.5

# Formula: 95% margin of error for a proportion, and the 1/sqrt(n) shortcut
moe <- 1.96 * sqrt(p * (1 - p) / n)
round(c(exact_formula = moe, shortcut = 1 / sqrt(n)), 4)

# Simulation: draw 10,000 polls of n = 1000 people from a population with p = 0.5
phat <- rbinom(10000, size = n, prob = p) / n
round(sd(phat), 4)                       # spread of poll results (standard error)
round(quantile(phat - p, c(0.025, 0.975)), 4)   # the middle 95% of polling error
round(mean(abs(phat - p) <= 1 / sqrt(n)), 3)    # share of polls within 1/sqrt(n)

# The worst case is p = 0.5; other true proportions give a smaller margin
p_grid <- c(0.1, 0.3, 0.5)
round(1.96 * sqrt(p_grid * (1 - p_grid) / n), 4)

Output from one seeded run (same numbers on a second run):
exact_formula      shortcut
       0.0310        0.0316
[1] 0.0158
  2.5%  97.5%
-0.031  0.030
[1] 0.957
[1] 0.0186 0.0284 0.0310

The formula gives 0.031 and the shortcut gives 0.0316. In the simulation, the middle 95% of polling errors ran from about −0.031 to +0.030, and 95.7% of the polls landed within 1/sqrt(n) of the truth. The last line shows that a true proportion of 10% or 30% gives a smaller margin (0.0186 and 0.0284), which is why 50% is called the worst case.

Limits of the shortcut

• It assumes a simple random sample. Real surveys with clustering or weighting usually have a larger margin of error than this formula suggests.
• It covers sampling error only. Bad question wording, non-response and coverage problems add error that no margin of error captures.
• It is for proportions. For a mean, the standard error is the standard deviation divided by sqrt(n), which follows the same logic as step 1.
• Small samples and extreme proportions need better intervals. When n is small or p is close to 0 or 1, the normal approximation is weak, and methods such as the Wilson interval behave better.

Where to look for more derivations

Most standard results have similar short proofs: least squares estimates come from setting derivatives to zero, and many estimators come from maximizing a likelihood. A mathematical statistics text shows these in full. Once you know the pieces (variance rules, the central limit theorem, normal quantiles), formulas that looked like magic become routine to rebuild.

Related tools and guides

• Which statistical test should I use?
• Binomial proportion confidence interval (Wikipedia)
• Central limit theorem (Wikipedia)

Written by AskStats with AI assistance. This is general information, not advice for your specific data or study. When the results matter, check your approach with a qualified statistician.