FS2 5.2, 5.3 - Confidence intervals for a normal mean

The interval varies from sample to sample

How can the same data give a wider interval when you ask for greater confidence? An interval that catches the unknown mean more often must allow more room for sampling error. Confidence and precision compete.

A useful surprise is that halving an interval's width, while keeping confidence and population spread fixed, requires four times as many independent observations. You will calculate that square-root effect rather than just quote it.

Imagine repeating the entire sampling procedure. Each new sample gives a new mean, so its confidence limits move. The population mean stays fixed.

Once one sample has been observed, its interval is fixed and either contains $\mu$ or does not. The $95\%$ refers to the procedure's repeated-sampling performance. It does not mean that $95\%$ of the population's individual observations lie in the interval.

[DIAGRAM: asset_name: Repeated confidence intervals and a fixed population mean; asset_slug: 9fm0-fs2-129-confidence-coverage; caption: The intervals move between samples while the true population mean remains fixed; description: Twelve horizontal confidence intervals of equal width are plotted against a measurement axis. A fixed vertical grey line marks the true mean fifty-two. Ten intervals crossing this line have pale sage backing and two that miss have pale clay backing. Grey point markers show the different sample means, and grey end caps show each interval. The legend identifies covering and missing intervals. A footer states that one run need not have exactly ninety-five percent coverage because confidence is a repeated-sampling property.]

A particular run of twelve or twenty samples need not achieve exactly the nominal percentage. The confidence level is a long-run property, rather than a quota imposed on each small collection of samples.

Keep that meaning separate from a prediction for one future measurement.

Use the correct central normal probability

For a two-sided interval, split the excluded probability equally between the two tails. A $95\%$ interval excludes $0.025$ in each tail, rather than $0.05$.

The booklet's normal percentage-point table defines $z_p$ by $\mathrm P(Z>z_p)=p$. It gives $z_{0.025}=1.9600$, $z_{0.005}=2.5758$ and $z_{0.05}=1.6449$. Quote tabulated values in full.

This interval is exact under the stated normal and known-variance model, apart from numerical rounding. Replacing an unknown population standard deviation by a small-sample estimate does not preserve that exact claim; the present method uses a genuinely known $\sigma$.

Worked example 2 changes only the confidence level, from $95\%$ to $99\%$.

Use enough figures in the limits to make the intended accuracy clear; keep the full critical value until the final calculation.

Change only the confidence level again.

Read the centre and half-width backwards

A symmetric normal interval records two pieces of information: its midpoint is the observed sample mean, and its half-width is the critical value times the standard error. The midpoint is not proof that the population mean equals that number.

For endpoints $L$ and $U$, let $h=(U-L)/2$. Rearranging $h=z_{\alpha/2}\sigma/\sqrt n$ gives

These are derived rearrangements, not separately printed booklet formulas. If the question asks for a variance, square the recovered standard deviation.