Erro Padrão da Média (SEM)

Standard error of the mean: precision of an average

The standard error of the mean (SEM) tells you how much a sample mean wobbles around the true population mean if you were to repeat the sampling over and over. When you know the population σ, it's SE = σ/√n; when you only have the sample standard deviation, you use SE = s/√n. Bigger samples mean a smaller SE, but the shrinkage follows 1/√n, which is the catch. Cutting the SE in half costs you four times as much data. Keep in mind that SE < σ for any n above 1, because SE is about how precise the average is, not how scattered the individual readings are. The Central Limit Theorem is what makes this work: the sample mean lands on a normal distribution with mean μ and standard deviation SE = σ/√n no matter what shape the original data took, as long as the variance is finite. So with s = 10 and n = 25 you get SE = 10/5 = 2.

Applications

You'll reach for it when building confidence intervals (CI ≈ x̄ ± 1.96·SE for 95%), planning sample sizes for research and surveys, running quality control, designing clinical trials, or doing A/B testing in product analytics. If you know the SE up front, you can pick an n that's large enough to catch the effect you care about.

FAQ

SE or SD? SD (σ or s) is about how spread out the individual observations are. SE is about how precise the sample mean is. For any n > 1, the SE comes out smaller.

Why √n and not n? The variance of the sample mean works out to σ²/n. Take the square root to get back to a standard deviation and you land on σ/√n. The upshot is that precision only grows with the square root of how much effort you put in.

How do I get a 95% confidence interval? Roughly x̄ ± 1.96·SE. If your sample is small (n < 30) and you don't know σ, swap the 1.96 for the matching value from Student's t-distribution.

How big must n be? That comes down to σ and how precise you need the answer. Want to cut the SE in half? Multiply n by 4. Want it ten times smaller? You're looking at 100 times the sample.