As discussed in Exercise 6.10, the General Social Survey reported a sample where about 61% of US residents thought marijuana should be made legal. If we wanted to limit the margin of error of a 95% confidence interval to 2%, about how many Americans would we need to survey?

Answer: 2285Step-by-step explanation:Formula to find the sample size is given by :-[tex]n=p(1-p)(\dfrac{z_c}{E})^2[/tex], where p = prior estimate of population proportion.E= Margin of error.[tex]z_c[/tex] = z-value for confidence interval of c.Given : Confidence interval : 95%From the z-value table , the z-value for 95% confidence interval = [tex]z_c=1.96[/tex]The General Social Survey reported a sample where about 61% of US residents thought marijuana should be made legal. i.e. the prior estimate of population proportion of US residents thought marijuana should be made legal : p=0.61    [∵ sample proportion is the best estimate for population proportion.]Margin of error : E= 2%=0.02Now, the required minimum sample size would be :-[tex]n=0.61(1-0.61)(\dfrac{1.96}{0.02})^2[/tex]Simplify ,[tex]n=0.2379\times9604=2284.7916\approx2285[/tex]Thus , we need to survey 2285 Americans .