Answer:
The confidence interval for the difference of proportions of the population p2003-p2008 is
.
Explanation:
In this question we have to construct a 90% confidence interval (CI) for a difference of proportions.
The proportion for 2003 is
.
The proportion for 2008 is
.
The difference in proportions is
.
The standard deviation of the difference of proportions can be estimated as:

For a 90% CI, the z-value is 1.64.
Then we can construct the CI as
