Sampling Distributions

Sampling Distribution of $\bar{x}$ (Sample Mean)

• Mean: $\mu_{\bar{x}} = \mu$
• Standard deviation (Standard Error): $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$
• Shape: If the population is normal, $\bar{x}$ is normal. If not, the CLT applies for large $n$.

Central Limit Theorem (CLT)

For large sample size ($n \geq 30$ as a rule of thumb), the sampling distribution of $\bar{x}$ is approximately normal regardless of the population shape.

Sampling Distribution of $\hat{p}$ (Sample Proportion)

• Mean: $\mu_{\hat{p}} = p$
• Standard deviation: $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$
• Shape: Approximately normal if $np \geq 10$ and $n(1-p) \geq 10$.

Key Ideas

• Larger $n$ → smaller standard error → more precise estimates.
• The sampling distribution is centered at the parameter.
• Standard error measures the typical deviation of the statistic from the parameter.

Sampling Distribution of $\hat{p}_1 - \hat{p}_2$

$\mu = p_1 - p_2, \quad \sigma = \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}$

Sampling Distribution of $\bar{x}_1 - \bar{x}_2$

$\mu = \mu_1 - \mu_2, \quad \sigma = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$

Problem: A population has $\mu = 50$ and $\sigma = 12$. For samples of $n = 36$, describe the sampling distribution of $\bar{x}$.

Solution:

• Mean: $\mu_{\bar{x}} = 50$.
• Standard error: $\sigma_{\bar{x}} = 12/\sqrt{36} = 2$.
• Shape: By CLT ($n = 36 \geq 30$), approximately normal.
• $\bar{x} \sim N(50, 2)$.

$P(\bar{x} > 53) = P(z > 1.5) \approx 0.067$.

• Sampling distributions describe statistic variability across samples.
• $SE_{\bar{x}} = \sigma/\sqrt{n}$; $SE_{\hat{p}} = \sqrt{p(1-p)/n}$.
• CLT: for large $n$, sampling distributions are approximately normal.
• Larger samples → smaller standard error → better estimates.