Large Samples with Known Population Variance

When the sample is sufficiently large, population normality is unnecessary for a mean test justified by CLT. If the population variance is supplied, use it directly.

For an independent random sample with known finite variance \(\sigma^2\), under \(H_0:\mu=\mu_0\), \[\overline{X}\approx N\left(\mu_0,\frac{\sigma^2}{n}\right),\qquad Z=\frac{\overline{X}-\mu_0}{\frac{\sigma}{\sqrt{n}}}\approx N(0,1).\] If the population is stated to be normal, the distribution is exact instead.

Bridge the idea

A random sample of 100 independent measurements is taken from a population of unknown shape with known variance 64. For testing \(\mu=20\), the standard error is \(\dfrac{8}{\sqrt{100}}=0.8\). Population normality need not be assumed; the approximate normal mean distribution is justified by the large sample.

Exam wording: “By the Central Limit Theorem, the sample mean is approximately normally distributed because the independent random sample is sufficiently large. No population-normality assumption is required.”

Keep the sampling conditions: the observations must concern the target population and be appropriately independent. If a process is changed, an old stated variance does not automatically become the known variance of the new process; the question must supply or justify its continued use.