Question

​Sleep apnea is a disorder in which you have one or more pauses in breathing or shallow breaths while you sleep. In a​ cross-sectional study of

420420

adults who suffer from sleep​ apnea, it was found that

210210

had gum disease.​ Note: In the general​ population, about​ 17.5% of individuals have gum disease.

(c) Estimate the proportion of individuals who suffer from sleep apnea who have gum disease with 99​% confidence. Interpret your result. Select the correct choice below and fill in the answer boxes to complete your choice. ​(Use ascending order. Round to three decimal places as​ needed.) A. There is 99​% confidence that the population proportion of individuals who suffer from sleep apnea who have gum disease is between nothing and nothing. B. 99​% of the population lies in the interval between nothing and nothing. C. There is a 99​% chance that the true population proportion of individuals who suffer from sleep apnea who have gum disease is between nothing and nothing.

Answer 1

sample proportion, = 0.5
sample size, n = 420
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.5 * (1 - 0.5)/420) = 0.0244

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.58

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.5 - 2.58 * 0.0244 , 0.5 + 2.58 * 0.0244)
CI = (0.437 , 0.563)

There is 99​% confidence that the population proportion of individuals who suffer from sleep apnea who have gum disease is between 0.437 and 0.563