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.) A in % of samples of teens in the country, the percent who go online several times a day will be within % of the estimated % B. 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

Answer:(a) margin error = 2.4%

(b) The margin error gives the measure in percentage of how the population parameter determined differ from the real population statistics or value.

(c) in 90% of the samples of teens in the country, the percent who go online several times a day will be within 50.6% and 55.4%. of the estimated 100%

Explanation:

Using the proportion formulae

Margin error = z √p(1-p)/n

n= 1170, p = 53% = 0.53, 1-p = 0.47

and the z value at 90% C.I = 1.645

M error= 1.645 √0.53×0.47/1170

Margin error = 0.024 = 0.024 ×100

Margin error = 2.4%

53 - 2.4 = 50.6% and 53 + 2.4 = 55.4%

In other words 90% of the time: the number of teens who go online several time a day will be between 50.6 and 55.4%.