Consider data from a sample of statistics students that is s

Consider data from a sample of statistics students that is stored in StudentSurvey. One of the survey questions asked which award students would most like to win from among an Academy Award, Nobel Prize, and Olympic gold medal. Among the 193 male students who responded, 109 chose the Olympic gold medal, while 73 of the 169 females also picked Olympic gold.

Consider data from a sample of statistics students that is stored in StudentSurvey. One of the survey questions asked which award students would most like to win from among an Academy Award, Nobel Prize, and Olympic gold medal. Among the 193 male students who responded, 109 chose the Olympic gold medal, while 73 of the 169 females also picked Olympic gold. Use this information to find a 95% confidence interval for the difference in the two proportions, represent the proportion of males and females who would most like to win an Olympic gold medal, respectively. and , where

Solution

p1=109/193=0.5647668

p2=73/169 =0.4319527

Given a=1-0.95=0.05, Z(0.025) = 1.96 (from standard normal table)

So the lower bound is

(p1-p2)- Z*sqrt(p1*(1-p1)/n1+p2*(1-p2)/n2)

= (0.5647668-0.4319527) - 1.96*sqrt(0.5647668*(1-0.5647668)/193+0.4319527*(1-0.4319527)/169) =0.03048984

So the upper bound is

(p1-p2)+ Z*sqrt(p1*(1-p1)/n1+p2*(1-p2)/n2)

= (0.5647668-0.4319527) + 1.96*sqrt(0.5647668*(1-0.5647668)/193+0.4319527*(1-0.4319527)/169) =0.2351384