Question 1149413
95% interval is sample proportion +/- z*sqrt (p*(1-p)/n), where z=1.96
268/400 is 0.67.  The 8000 just shows that the population is a lot larger than the sample (although it could be factored in to this using the finite population correction, which takes into part the fact that the sample is a significant part of the population.

CI is 0.67+/- 1.96 sqrt (.67*.33/400)=0.67+/- 0.0461

this interval is (0.6239, 0.7161) which can be rounded.