Confidence interval for the difference between the

population proportions: (large samples)

As it was discussed earlier, for a large sample the sample proportion is approximately normally distributed with mean p and standard deviation

.

Suppose that a random sample of size observations from a population with proportion of “success” has a sample proportion of success , and that an independent random sample of size observations from a population with proportion of “success” yields sample proportion .

Since and both are large, their sample proportions and are approximately normally distributed with means and , and standard deviations

and

respectively.

Then the random variable , and the variance

.

The standardized random variable

is approximately standard normal.

In order to find confidence interval for , we must either know or estimate the quantity of

.

We can estimate the population proportion by the sample proportion ; and we can estimate the population proportion by the sample proportion . Then

Then an approximate confidence interval is given by

where

.

Example:

Mike and Tom like to throw darts. Mike throws 100 times and hits the target 54 times; Tom throws 100 times and hits the target 49 times. Find a 95 % confidence interval for , where represents the true proportion of hits in Mike’s tosses, and represents the true proportion of hits in Tom’s tosses.

Solution:

and

Thus, with 95 % confidence we can state that the difference between the proportions of Mike’s and Tom’s tosses is between -0.088 and 0.188.

Дата добавления: 2015-08-05