Use the distribution of the sample mean and the central limit theorem to build confidence intervals for a population mean

What this dot point is asking

SCSA Unit 4 statistical inference is about reasoning from a sample to a population. You must know how the sample mean behaves across repeated samples, state and use the central limit theorem, and construct and interpret confidence intervals for a population mean.

The sampling distribution of the sample mean

Take repeated random samples of size $n$ from a population with mean $\mu$ and standard deviation $\sigma$. Each sample gives a sample mean $\bar{x}$, and these vary from sample to sample. The random variable $\bar{X}$ has:

The quantity $\dfrac{\sigma}{\sqrt{n}}$ is the standard error. It shrinks as $n$ grows, so larger samples give sample means clustered more tightly around the true mean $\mu$.

The central limit theorem

This is what makes inference possible: even if the underlying population is skewed, the sample mean behaves normally for large $n$, so we can use normal-distribution probabilities and $z$-values.

Confidence intervals for a population mean

An approximate confidence interval estimates $\mu$ from one sample. With known (or large-sample estimated) population standard deviation $\sigma$, the interval is

where $z$ is the critical value for the chosen confidence level: $z \approx 1.645$ for $90\%$, $z \approx 1.96$ for $95\%$, and $z \approx 2.576$ for $99\%$. The term $z\dfrac{\sigma}{\sqrt{n}}$ is the margin of error.

The correct interpretation of a $95\%$ confidence interval: if we repeated the sampling many times and built an interval each time, about $95\%$ of those intervals would contain the true mean $\mu$. It is not a $95\%$ probability that $\mu$ lies in this one interval.

Effect of sample size and confidence level

A wider confidence level (say $99\%$ instead of $95\%$) gives a larger $z$ and a wider interval: more confidence costs precision. A larger sample reduces the standard error and narrows the interval. Because the standard error has $\sqrt{n}$ in the denominator, increasing the sample size by a factor of $4$ halves the width.