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The reason leading to estimation of a mean also applies to the problem of estimation of a population proportion.
Suppose that n elements are randomly selected from the large population. And let n consists of X elements possessing some characteristic.
Common sense suggests the sample proportion
as an estimator of p.
When the sample size n is only small fraction of the population size, the sample count X has the binomial distribution with mean and standard deviation .
When n is large, the binomial variable X is well approximated by a normal distribution with mean and standard deviation .
That is
is approximately standard normal.
If we divide both numerator and denominator by n we will get a statement about proportions:
As we see the denominator of Z contains p and q. If sample size is large that , then a good approximation is obtained if p replaces the point estimator in the denominator.
Hence, for large sample size, the distribution of
is approximately standard normal.
We can use this result to obtain confidence interval for the population proportion. (Fig.6.12)
Using Fig.6.12 we obtain
Definition:
If sample of n observations selected from the population is large enough that
, then a confidence interval for the population proportion is given by
where is the sample proportion and is the number for which
.
Example:
From a country labor force a random sample of 800 persons was selected and 75 people were found unemployed out of random sample of 800 persons. Compute 90 % confidence interval for the rate of unemployment in the country.
Solution:
The confidence interval for the population proportion is obtained from
The observed and .
Since , we say that sample size is large and a normal approximation to the sample proportion is justified. Then
and
After substituting, we obtain
Therefore, a 90 % confidence interval for the rate of unemployment in the country is (0.0768; 0.1107), or (7.68 %; 11.07 %).
Because our procedure will produce true statements 90 % of the time, we can be 90 % confident that the rate of unemployment is between
7.68 and 11.07.
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