6.4 - Confidence intervals for a mean

The confidence procedure

Does a $95\%$ confidence interval contain $95\%$ of the individual measurements? No: it estimates the population mean. Its width comes from the variation of sample means, which is smaller than the variation of individual observations.

Suppose a normal population has known standard deviation $\sigma$, and an independent sample of size $n$ is taken. The sample mean has standard error $\sigma/\sqrt n$. The standard normal tables give approximately

Before sampling, therefore, there is probability $0.95$ that the random sample mean lies within $1.96$ standard errors of the fixed population mean. The same inequality can be read in reverse:

After observing $\bar x$, use its value for the interval's centre.

[DIAGRAM: asset_name: Repeated samples and confidence intervals; asset_slug: mean-confidence-procedure; caption: The population mean stays fixed while the sample interval changes; description: Five horizontal confidence intervals from different possible samples are plotted against one fixed vertical population-mean line. Four intervals cross the line and one misses it. Each interval is centred on its own sample mean. A note says that the realised proportion covering the mean in a small collection need not equal the confidence level.]

The confidence level describes the procedure's long-run coverage, not a guaranteed proportion within a small collection of intervals. Once you have calculated one interval, its endpoints and $\mu$ are fixed: it either contains $\mu$ or it does not. The repeated-sampling interpretation does not assign a new probability distribution to $\mu$.

A $95\%$ mean interval also does not say that $95\%$ of future sample means lie in this particular observed interval. Its purpose is to estimate the fixed population mean from one sample. Individual-observation prediction is a different task.

Identify the correct interpretation in context.

Known variance and normal population

For a normal population with known variance, the sample mean's normal distribution gives a confidence interval at any sample size. Choose a central normal probability equal to the requested confidence level, leaving equal areas in the two tails.

For $90\%$ confidence, each tail is $0.05$, so $\Phi(z)=0.95$ and $z\approx1.645$. For $95\%$, use $1.960$; for $99\%$, use $2.576$. A $98\%$ interval leaves $0.01$ in each tail, so uses $2.326$, not the critical value for a left area of $0.98$.

Keep the sample and population spread unchanged, but increase only the confidence level. Greater long-run coverage requires a wider interval.