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Chapter 11: Confidence Intervals – Large Samples
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Provided that
n
is large (
30, as a rule-of-thumb), the sampling
distribution of the sample mean
̅
will be approximately normally
distributed with mean of
̅
, and a standard deviation
of
̅
√
..
If the sampling distribution is normal, the sampling distribution of the
sample means will be an exact normal distribution for any sample size.
Now, in finding confidence interval estimates for the unknown parameter
,
we would need to compute
̅
√
.. The general equation used in
constructing a (1 -
)
100 percent confidence interval for the population
mean, is given next.
The margin of error is given by
Example 11-6:
A random sample of 100 public school teachers in a
particular state has a mean salary of $37,000. It is known from past history
that the standard deviation of the salaries for all the teachers in the state is
$2,122. Construct a 95% confidence interval estimate for the true mean
salary for public school teachers for that given state if it is assumed that the
population of high school teach salaries in the given state is normally
distributed.
Solution:
Given
= 0.05,
⁄
= 1.95996,
̅
= 37,000,
= 100,
= 2,122,
and
̅
√
= 212.2. Thus the 95% confidence interval estimate for the
mean salary, using the formula, is 37,000
1.95996
212.2 = 37,000
415.9035. That is, we are 95% confident that the average salary for public
school teachers for the given state will be between $36,584.10 and
$37,415.90 to two decimal places.