Find a 90% confidence interval for the true proportion in the sampled population of women between the ages of?

In a survey of women between the ages of 20 and 40, it was found that 88 out of 200 women take prenatal vitamins. Find a 90% confidence interval for the true proportion in the sampled population of women between the ages of 20 and 40 who take prenatal vitamins.





Please help (show work)!!!Find a 90% confidence interval for the true proportion in the sampled population of women between the ages of?
p=X/n=88/200=.44





the Zscore for a 90% CI is 1.645 (use a normal distribution table and find the Z Score associate with .05, because 1-90%=.1 and half of .1 is in one tail of the distribution and .05 is in the other)





p +or- Z * Square root of ((p(1-p))/n)





So the upper limit is


=.44 + 1.645 * SQRT((.44*.66)/200)


=.44 + 1.645 * .38105118


=.44 + .06268


=.50268





The lower limit


=.44 - 1.645 * SQRT((.44*.66)/200)


=.44 - 1.645 * .38105118


=.44 - .06268


=.37732





Therefore, you can be 90% certain the percent of women between 20 and 40 years of age that take prenatal vitamins lies between 37.73% and 50.27%