Answer:
The 95% confidence interval for the difference between the proportion of women who drink alcohol and the proportion of men who drink alcohol is (-0.102, -0.014) or (-10.2%, -1.4%).
Step-by-step explanation:
We want to calculate the bounds of a 95% confidence interval of the difference between proportions.
For a 95% CI, the critical value for z is z=1.96.
The sample 1 (women), of size n1=1000 has a proportion of p1=0.494.

The sample 2 (men), of size n2=1000 has a proportion of p2=0.552.

The difference between proportions is (p1-p2)=-0.058.
The pooled proportion, needed to calculate the standard error, is:

The estimated standard error of the difference between means is computed using the formula:

Then, the margin of error is:

Then, the lower and upper bounds of the confidence interval are:

The 95% confidence interval for the difference between proportions is (-0.102, -0.014).