Question

A poll found that 53% of a sample of 1170 teens in a certain large country go online several times a day. (Treat this sample as a simple random sample.) Complete parts a through e below. a) Find the margin of error for this poll if we want 90% confidence in our estimate of the percent of teens in the country who go online several times a day. 24% (Round to one decimal place as needed.) b) Explain what the margin of error means Select the correct choice below and fill in the answer boxes to complete your choice. (Round to one decimal place as needed.) O A in % of samples of teens in the country, the percent who go online several times a day will be within % of the estimated % OB. There is an) % chance that the percent of toons in the country who go online several times a day is within % of the estimated % C. The percent of teens in the country who go online several times a day is within % of the estimated D . The pollsters are confident that the true percent of teens in the country who go online several times a day is within O % of the estimated %.

Answer 1

Solution: Sample size = n = 1170 Sample proportion of teens that go online several times a day p 53% = 0.53 Part-A: Standard Error of the Point estimate: P(1- p) 0.53 x (1 0.53) = 0.0146 1170 n For 90% confidence interval, a 1- 0.90 0.10 Hence, a/2=0.10/2=0.05 From normal distribution table; zo.05 = 1.645 Margin of error: E = Zo.05 X op E 1.645 x 0.0146 E 0.0240 2.4% Part-B: The pollsters are 90% confident that the true percent of teens in the country who go online several times a day is within 2.4% of the estimated 53%