Question 975169
This is a two sample proportion test.

The z value +/- the standard error will give the answer.

z= (phatw-phatm)/sqrt {[ (pw)(1-pw)/nw] + (pm)(1-pm)/pm]}

can do this by hand, but calculators are fine, so long as one knows what is going on.
We expect the difference to be 4.2%, and the confidence interval will contain that value should there be no statistical significance at the 5% level.  Any result in the CI is considered equal to any other result.

It's worth looking at the point estimates:  For women, it is 0.286 and men 0.288.

This is a difference of -0.002.

z=-0.053 p=0.478

standard error is (.286)(.714)/350=0.0005834
and (.288)(.712)/400 = 0.0005126
sum is 000110; sqrt(sum) is 0.0331=SE
multiply by z-value of 1.96, and we get 0.0649

CI (-0.0651,+0.0647)
The CI contains the difference of 0 for the estimation of the true difference in proportions; therefore, the null hypothesis of no difference cannot be rejected.  Given the sample size, a percentage as large as shown in the Almanac could be found in a sample if there were no true difference in the parameters.