Question

A newsgroup is interested in constructing a 99% confidence interval for the proportion of all Americans who are in favor of a new Green initiative. Of the 513 randomly selected Americans surveyed, 393 were in favor of the initiative. Round answers to 4 decimal places where possible. a. With 99% confidence the proportion of all Americans who favor the new Green initiative is between and b. If many groups of 513 randomly selected Americans were surveyed, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population proportion of Americans who favor the Green initiative and about percent will not contain the true population proportion. Hint: Hints -

Answer 1

for 99%. CI P(-2 <p Lz) - 0.99 = 2.576 o feror of the inticative = 393 EL n. 513 393 ols - = 0.7661 513 q= å = 1-P - | -0.7661 = 0,2339 margin of edoon E -'zx ê ain = 2.576 * N0.7661 X0.2339/513 - 0.0481 Ô +E) confirclence in-enal: C - E , (0.7661-0,0481, 0.7661 +0.0481) 0,8142) ) 6 0,8142 0,7179 (0.7179, About the 99°/- percent of these CI will conteen tove population Proportion of Americans who fever the about 1'l. will percent Green Intiative and not contain the true po.pro postion.