A confidence interval (CI) is a range of values, calculated from sample data, that is likely to contain the true population parameter. The confidence level (usually 95%) expresses how often such intervals capture the true value if the process were repeated many times.
$$\bar{x} \pm z \cdot \frac{\sigma}{\sqrt{n}}$$
For a 95% CI, $z \approx 1.96$ (from the standard normal distribution).
$$\text{CI} = \left(\bar{x} - 1.96 \cdot \frac{\sigma}{\sqrt{n}},\; \bar{x} + 1.96 \cdot \frac{\sigma}{\sqrt{n}}\right)$$
Margin of error: $E = 1.96 \cdot \dfrac{\sigma}{\sqrt{n}}$
$$\hat{p} \pm 1.96\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$
where $\hat{p}$ is the sample proportion and $n$ is the sample size.
A random sample of 64 daily temperature readings has $\bar{x} = 18.5°\text{C}$ and population $\sigma = 3.2°\text{C}$.
$$SE = \frac{3.2}{\sqrt{64}} = \frac{3.2}{8} = 0.4$$
$$\text{95\% CI} = 18.5 \pm 1.96 \times 0.4 = 18.5 \pm 0.784$$
$$\text{CI} = (17.716,\; 19.284)°\text{C}$$
We are 95% confident the true mean temperature lies between 17.7°C and 19.3°C.
A survey of 200 voters finds 112 support a policy. $\hat{p} = 112/200 = 0.56$.
$$SE = \sqrt{\frac{0.56 \times 0.44}{200}} = \sqrt{\frac{0.2464}{200}} = \sqrt{0.001232} \approx 0.03511$$
$$\text{95\% CI} = 0.56 \pm 1.96 \times 0.03511 = 0.56 \pm 0.0688$$
$$\text{CI} = (0.491,\; 0.629)$$
We are 95% confident the true proportion of supporters is between 49.1% and 62.9%.
Larger $n$ $\Rightarrow$ smaller $SE$ $\Rightarrow$ narrower CI $\Rightarrow$ more precise estimate.
To halve the margin of error, you need four times the sample size (since $SE \propto 1/\sqrt{n}$).
REMEMBER: “95% confident” means: if this sampling procedure were repeated many times, 95% of such intervals would contain the true population parameter.
EXAM TIP: Show all four steps: (1) calculate $\hat{p}$ or state $\bar{x}$, (2) calculate $SE$, (3) compute margin of error, (4) state the interval and interpret it in context.
