Question

USA Today (Feb 15. 2007) reported on the results of an opinion poll in which adults were asked what one thing they are most likely to do when they are home sick with a cold or the flu. In the survey, 63% said that they are most likely to sleep and 18% said they would watch television. Although the sample size was not reported, typically opinion polls include approximately 1,000 randomly selected respondents. (a) Assuming a sample size of 1,000 for this poll, construct a 95% confidence interval for the true percentage of all adults who would choose to sleep when they are at home sick. (b) If the true percentage of adults who would choose to sleep when they are home sick is 70%, would you be surprised? Explain.

Answer 1

A) The sample size is 1000

63 % of adult people like to sleep out of 1000 people

probability is 63 % / 1000 = .63

construct 95 % confidence interval: p $\pm$ m

p is probability

m- is a marginal error at 95 % confidence interval

m = Z * SE

Z is z-score value at 95 % confidence interval = 1.96

SE is standard error of probability

SE =   n is sample size = 1000

p(1-p)

p is probability = 0.63

therefore, SE = 0.01527

M =1.96*0.01527 = 0.0299292

so, confidence interval = 0.63$\pm$0.0299292

b) I would be surprised because in 1000 sample survey 63 % people like to sleep and it varies 60.01 to 65.99 % at 95 % confidence interval, if the true percentage of choose to sleep is 70% so, there is a big gap between sample size percentage and true percentage so, I would be surprised.