We have 
and s = 5000, for a sample with n =
200. The figure of 28000 is our point estimate of the population mean,
. What is our 95 per cent confidence interval? Well, we figure that
with probability .95 our sample mean was drawn from the central 95 per cent
of its sampling distribution, which is centered on the unknown population
mean 
, and has standard error 
. If this sampling
distribution is normal (and it will tend to be, for large n, even if the
parent population is non-normal), then the central 95 per cent is given by
(approximately) 
plus or minus two standard errors. Therefore, we can
be 95 per cent confident that our sample mean of 28000 is no more than two
standard errors away from 
(or in other words the maximum error for this
degree of confidence is two standard errors). Now in this case, strictly
speaking, the standard error is unknown since the population standard
deviation, 
, is unknown. But we can take s as an estimate of
, which than gives us an estimated standard error of
Our interval is then:
(Note: in this sort
of case, where 
is unknown, and is replaced by its estimator, s, we
should strictly speaking move from the assumption of a normal sampling
distribution for 
to the t-distribution, with degrees of freedom
. On the other hand, when the sample size is `large',
i.e. greater than 30, the difference in results is negligible.)
Here you may return to the
question.