TheoremProved

A confidence interval for the mean, via the Central Limit Theorem

Statement

If $X_1,\dots,X_n$ are independent, identically distributed with mean $\mu$ and finite variance $\sigma^2$ , then for large $n$ the confidence interval $\bar X - z_{\alpha/2}\frac{\sigma}{\sqrt n} \le \mu \le \bar X + z_{\alpha/2}\frac{\sigma}{\sqrt n}$ contains $\mu$ with probability approximately $1-\alpha$ .

Why is it true?

The Central Limit Theorem says the standardized sample mean behaves like a standard normal variable once $n$ is large, so we can use normal quantiles $z_{\alpha/2}$ to bracket $\mu$ with a known long-run success rate, regardless of the shape of the original population.

Proof sketch

By the Central Limit Theorem, $\frac{\bar X - \mu}{\sigma/\sqrt n}$ converges in distribution to $N(0,1)$ as $n$ $\to\infty$ . So for large $n$ , $-z_{\alpha/2} \le \frac{\bar X - \mu}{\sigma/\sqrt n} \le z_{\alpha/2}$ holds with probability approximately $1-\alpha$ , where $z_{\alpha/2}$ is the value cutting probability $\alpha/2$ from each tail of $N(0,1)$ .

Multiply all three sides of the inequality by $\frac{\sigma}{\sqrt n}$ $>0$ (this preserves the inequality direction): $-z_{\alpha/2}\frac{\sigma}{\sqrt n} \le \bar X - \mu \le z_{\alpha/2}\frac{\sigma}{\sqrt n}$ .

$\bar X$ is subtracted from all sides, then all sides are multiplied by $-1$ (flipping the inequalities): $\bar X - z_{\alpha/2}\frac{\sigma}{\sqrt n} \le \mu \le \bar X + z_{\alpha/2}\frac{\sigma}{\sqrt n}$ . This is exactly the confidence-interval formula: since the underlying probability statement held with approximate probability $1-\alpha$ , so does this rearranged interval containing $\mu$ .

Topics that use this theorem

Estimation

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

NIST/SEMATECH (2013). Confidence Limits for the Mean
Diez, D.; Cetinkaya-Rundel, M.; Barr, C. (2019). OpenIntro Statistics