The Journal of the American Medical Association (April 1993) reported on the results of a National Health
Interview Survey designed to determine the prevalence of smoking among US adults. Of the 43,732 survey
respondents, 11,239 indicated that they were current smokers and 10,539 indicated that they were former
smokers. Construct a 90% confidence interval for the percentage of US adults who currently (circa 1993) smoke
cigarettes. Construct a 95% confidence interval for the percentage of US adults who former (circa 1993)cigarette smokers.

Question

The Journal of the American Medical Association (April 1993) reported on the results of a National Health
Interview Survey designed to determine the prevalence of smoking among US adults. Of the 43,732 survey
respondents, 11,239 indicated that they were current smokers and 10,539 indicated that they were former
smokers. Construct a 90% confidence interval for the percentage of US adults who currently (circa 1993) smoke
cigarettes. Construct a 95% confidence interval for the percentage of US adults who former (circa 1993)cigarette smokers.

karleighangelica · Answer

A. find point estimate
b. find z or t critical value
c. margin of error
d. confidence interval
e. statement of confidence

 I am confused on how to get the mean and standard deviation

agent0smith · Answer

You don't need mean or standard deviation since it's a proportion. See the other question that I replied to.

agent0smith · Answer

http://openstudy.com/study#/updates/58081872e4b06a18dc449b76