America\'s young people are heavy Internet users; 87% of Americans ages 12 to 17

Question

America's young people are heavy Internet users; 87% of Americans ages 12 to 17 are Internet users (The Cincinnati Enquirer, February 7, 2006). MySpace was voted the most popular website by 9% in a sample survey of Internet users in this age group. Suppose 1,450 youths participated in the survey. What is the margin of error, and what is the interval estimate of the population proportion for which MySpace is the most popular website? Use a 95% confidence level and round your answers to 3 decimals. For interval estimate Enter your answer using parentheses and a comma, in the form (n1,n2). Do not use commas in your numerical answer (i.e. use 1200 instead of 1,200, etc.)

Margin error = Interval estimate =

Explanation / Answer

from above margin of error E =0.015

Interval estimate = 0.075 to 0.105

sample size n                    = 1450 sample proportion     p x/n= 0.0900 std error       =Se            =(p*(1-p)/n) = 0.0075 for 95 % CI value of z= 1.9600 margin of error E=z*std error                            = 0.015 lower confidence bound=sample proportion-margin of error 0.075 Upper confidence bound=sample proportion+margin of error 0.105

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