Normally distributed: large sample size

Let be a random sample of n observation from a normal population with unknown and unknown variance . Let be the sample mean. If the sample size n is large , then according to the central limit theorem, for a large sample the sampling distribution of the sample mean is (approximately) normal irrespective of the shape of the population from which the sample is drawn. Therefore, when the sample size is 30 or larger, we will use the normal distribution to construct a confidence interval for .

If the variance of the population is unknown, it should be estimated by the sample variance, , and replaced by in confidence interval formula for the case when population variance is known.

Remark:

For large sample sizes, is usually close to the true value of .

An approximate confidence interval for the population mean with unknown variance is given by

where is based on a sample of at least thirty observations, is the number for which and the random variable Z has a standard normal distribution.

Example:

A sample of 64 observations from a large population yielded the sample values, and . Find an approximate 99 % confidence interval for .

Solution:

First we find the standard deviation of . Because is not known, we will use as an estimator of .

.

Then

and

Substituting all the values in the formula, the 99 % confidence interval

for is

An approximate 99 % confidence interval for is (166.4, 177.6).

Example:

Radiation measurements on a sample of 69 microwave ovens produced and . Determine a 94 % confidence interval for the mean radiation.

Solution:

Again since we can use as an estimator of .

Then confidence interval for the population mean with unknown variance is given by

and

Substituting all the values in the formula we obtain

.

Thus, we can state with 94 % confidence that average radiation measure of microwave ovens is between (0.21, 0.139).

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