Sampling Distribution

Study guides: Exam VEE-MS, Exam MAS-I, Exam MAS-II

A Sampling Distribution is the probability distribution of a statistic (e.g., the sample mean $\bar{X}$ or sample variance $S^2$ ) computed from all possible random samples of size $n$ drawn from a population.

$\bar{X} \sim N\!\left(\mu,\, \frac{\sigma^2}{n}\right) \quad \text{if } X_i \stackrel{\text{iid}}{\sim} N(\mu, \sigma^2)$

$\frac{(n-1)S^2}{\sigma^2} \sim \chi^2_{n-1} \quad \text{(normal population)}$

Key sampling distributions used in hypothesis testing — each exact for a random sample from a Normal population; the $F$ row compares two independent Normal samples, of sizes $n_1$ and $n_2$ :

Statistic	Distribution	Use
$Z = \dfrac{\bar{X} - \mu}{\sigma/\sqrt{n}}$	$N(0,1)$	Mean test, known $\sigma$
$T = \dfrac{\bar{X} - \mu}{S/\sqrt{n}}$	$t_{n-1}$	Mean test, unknown $\sigma$
$\chi^2 = \dfrac{(n-1)S^2}{\sigma^2}$	$\chi^2_{n-1}$	Variance test
$F = \dfrac{S_1^2/\sigma_1^2}{S_2^2/\sigma_2^2}$	$F_{n_1-1,\,n_2-1}$	Two-variance test

The standard error is $\text{SE}(\bar{X}) = \sigma/\sqrt{n}$ , the standard deviation of the sampling distribution of $\bar{X}$
By the Central Limit Theorem (CLT), $\bar{X}$ is approximately normal for large $n$ even when the population is not normal

Probability That a Sample Mean Falls in a Range Claims $X_i$ are i.i.d. with $\mu = 2{,}000$ and $\sigma = 400$ . For a random sample of $n = 64$ , find $P(1{,}900 < \bar{X} < 2{,}100)$ .

Answer $\text{SE} = 400/\sqrt{64} = 50$ . $P(1{,}900 < \bar{X} < 2{,}100) = P\!\left(\frac{1{,}900-2{,}000}{50} < Z < \frac{2{,}100-2{,}000}{50}\right) = P(-2 < Z < 2) \approx 0.9545$