Statistics / Unit 8: Sampling Distributions and Confidence Intervals
8.3 Sampling Distribution of a Sample Mean
Notes
For a sample mean $\bar{x}$ from a population with mean $\mu$ and standard deviation $\sigma$:
- Mean: $\mu_{\bar{x}} = \mu$. Standard deviation: $\displaystyle \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$.
- If the population is normal, $\bar{x}$ is normal for any $n$.
- Central Limit Theorem: for large $n$ (about 30 or more), the sampling distribution of $\bar{x}$ is approximately normal, even if the population is skewed.
- Example: $\sigma = 20$ and $n = 25$ give $\displaystyle \sigma_{\bar{x}} = \frac{20}{5} = 4$.
Practice
A population has $\sigma = 20$. Samples of size $n = 25$ are taken. The standard deviation of $\bar{x}$ is:
$\displaystyle \frac{20}{\sqrt{25}} = \frac{20}{5} = 4$.
A population is strongly skewed right. Why can we still use a normal model for $\bar{x}$ with $n = 100$?
The Central Limit Theorem applies to large samples from almost any population.
A population has $\mu = 50$ and $\sigma = 10$. For samples of size 25, about $95\%$ of sample means fall between:
$\displaystyle \sigma_{\bar{x}} = \frac{10}{5} = 2$. Within 2 standard deviations of 50 is 46 to 54.
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