Question

A recent study of 3000 children randomly selected found 23​% of them deficient in vitamin D. ​a) Construct the​ 98% confidence interval for the true proportion of children who are deficient in vitamin D. left parenthesis nothing comma nothing right parenthesis ​(Round to three decimal places as​ needed.) ​b) Explain carefully what the interval means. A. We are​ 98% confident that the interval contains the true proportion of children deficient in vitamin D. B. We are​ 98% confident that the percent of people deficient in vitamin D is 23 %. C. We are​ 98% confident that the percent of children deficient in vitamin D is 23 %. D. We are​ 98% confident that the interval contains the true proportion of people deficient in vitamin D. ​c) Explain what​ "98% confidence" means. About​ 98% of random samples of size 3000 will produce ▼ a true proportion a population proportion confidence intervals that contain the ▼ true proportion population proportion confidence interval of children that are deficient in vitamin D.

Answer 1

sample proportion, = 0.23
sample size, n = 3000
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.23 * (1 - 0.23)/3000) = 0.0077

Given CI level is 98%, hence α = 1 - 0.98 = 0.02
α/2 = 0.02/2 = 0.01, Zc = Z(α/2) = 2.33

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.23 - 2.33 * 0.0077 , 0.23 + 2.33 * 0.0077)
CI = (0.212 , 0.248)

We are​ 98% confident that the interval contains the true proportion of children deficient in vitamin D.