Question

A major metropolitan newspaper selected a simple random sample of 1,600 readers from their list of 100,000 subscribers. They asked whether the paper should increase its coverage of local news. Forty percent of the sample wanted more local news. What is the 99% confidence interval for the proportion of readers who would like more coverage of local news? 1. What is p hat 2. What is the critical value you should use?(round to 2 decimal places) 3. What is the margin of error?(round to 4 decimal places) 4. We are % confident that the true proportion of readers that want more local news is between % and

Answer 1

Solutions Given that, n= 100000 x = 1600 2² X 1600 0.016 n 100000 = 2. At 99% Confidence interval forp - a - 1-0.gg a 0.01 q 0.01 0.005 za 20.005 – 2.58 — Usług z 3. Maagin of eases E. Z f(1-P) n . 2.58 0.016(1-0.016) 100000 2 0.0010 The ggf. Confidence interval pis. êt f 0.016 - 0.0010 Luc 0.016 + 0.0010 (0.015, 0.017) 4. We are 99% Confidence that the thue pro- postion of readers that want more Local news in between 0.615 % and 0.017%