Try using a binomial proportion CI formula:
CI90 = p + or - 1.645(√pq/n)
...where p = proportion in the problem, q = 1 - p, n = sample size, √ = square root, and + or - 1.645 represents 90% CI using a z-table.
Your data:
p = 200/500
q = 1 - p
n = 500
Convert all fractions to decimals. Substitute values into the formula and determine the confidence interval.
I hope this will help get you started.
In a certain city, there are 10,000 persons age 18 to 24. A simple random sample of
500 such persons is drawn, of whom 200 turn out to be currently enrolled in college. Find a 90%
con¯dence interval for the percentage of persons age 18 to 24 in the city who are attending college.
2 answers
500.12