# SOLUTION: 66% of students as a university live on campus. A random sample found that 20 of 40 male students and 40 of 50 of female students lived on campus. At the .05 level of significanc

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 Question 321545: 66% of students as a university live on campus. A random sample found that 20 of 40 male students and 40 of 50 of female students lived on campus. At the .05 level of significance, is there sufficient evidence to conclude that a difference exists between the proportion of male students who live on campus and the proportion of female students who live on campus? what are the steps to figure this problem and what formula(s) should be used.Answer by stanbon(57323)   (Show Source): You can put this solution on YOUR website!66% of students as a university live on campus. A random sample found that 20 of 40 male students and 40 of 50 of female students lived on campus. At the .05 level of significance, is there sufficient evidence to conclude that a difference exists between the proportion of male students who live on campus and the proportion of female students who live on campus? what are the steps to figure this problem and what formula(s) should be used. --------------------- Ho: p(men)-p(women) = 0 Ha: p(m) - p(w) is not 0 -------------------------- Sample proportion: (20/40)-(40/50) = 0.5-0.8 = -0.3 --- test statistic: z(difference) = -0.3/sqrt[(0.5)^2/40 + (0.8*0.2)/50] = -3.0861 ---- p-value: since this is a 2-tail test = 2*P(z< -3.0861) = 0.002 --- Conclusion: Since the p-value is less than 5%, reject Ho. The test results do not support equality of the proportions. =============== Cheers, Stan H.