week 5 5

There is no prior information about the proportion of Americans who support gun control in 2018. If we want to estimate 95% confidence interval for the true proportion of Americans who support gun control in 2018 with a 0.36 margin of error, how many randomly selected Americans must be surveyed? Answer: (Round up your answer to nearest whole number) There is no prior information about the proportion of Americans who support gun control in 2018. If we want to estimate 95% confidence interval for the true proportion of Americans who support gun control in 2018 with a 0.36 margin of error, how many randomly selected Americans must be surveyed? Answer: (Round up your answer to nearest whole number) 8 v v+ Hide question 5 feedback Z-Critical Value =NORM.S.INV(.975) = 1.96 pxqxZ> ME? 5% .5%1.962 .362 Z-Critical Value =NORM.S.INV(.975) = 1.96 p*q*Z2 ME? 5x .5 % 1.962 .362