Confidence intervals and tests for one mean

What you’ll learn

• Construct and interpret a Normal confidence interval.
• Test a population mean using a standardised sample mean.
• Apply large-sample approximations with appropriate caveats.

Understand the idea

For a Normal population with known variance, standardise the sample mean using its standard error. A two-sided confidence interval extends the observed mean by a critical Normal quantile times the standard error. Higher confidence widens the interval; a larger sample narrows it when other quantities are fixed.

A 95% procedure produces intervals that contain the fixed population mean in about 95% of repeated samples under the model. It does not say that 95% of individual observations lie inside the interval. The uncertainty is about the estimation procedure, not about a randomly changing population mean.

For testing, evaluate the standardised statistic under the hypothesised mean and compare it with the appropriate one- or two-tailed critical value. With a large independent sample, the central limit theorem supports approximate intervals and tests even for a non-Normal population; an unknown standard deviation can then be estimated by $s$. Small non-Normal samples need more care than this approximation allows.

Key formulae

$$\bar x\pm z_{1-\alpha/2}\frac{\sigma}{\sqrt n}$$

$$Z=\frac{\bar X-\mu_0}{\sigma/\sqrt n}$$

$$Z\approx\frac{\bar X-\mu_0}{S/\sqrt n}\quad\text{for large }n$$

Worked example

An independent sample of 64 from a Normal population has mean 51.2. The population standard deviation is known to be 4. Find a 95% confidence interval and test $H_0:\mu=50$ against $H_1:\mu\ne50$ at 5%. Use 1.96 as the critical value.

1. The standard error is $4/\sqrt{64}=0.5$.
2. The confidence margin is $1.96(0.5)=0.98$, giving $(50.22,52.18)$.
3. The test statistic is $z=(51.2-50)/0.5=2.4$.
4. Since $|2.4|>1.96$, reject the null; the interval also excludes 50.

Answer: 95% interval $(50.22,52.18)$; evidence at 5% that the population mean differs from 50.