p2 = 71/411 = 0.1727
p1 = 64/253 = 0.2530
(p1 - p2) = 0.2530 - 0.1727 = 0.0803
SE = sqrt{[p1(1 - p1)/n1] + [p2(1 - p2)/n2]} = sqrt{[0.2530(1 - 0.2530)/253] + [0.1727(1 - 0.1727)/411]} = sqrt(0.000747 + 0.0003476) = sqrt(0.001095) = 0.0331
Therefore, 95%
confidence interval for the difference in the proportions of males and
females at this university who would respond less than one week = (p1 - p2) + or - 1.96(SE) = 0.0803 + or - 1.96(0.0331) = 0.0803 - 0.0648 to 0.0803 + 0.0648 = 0.0155 to 0.1451
b.) Yes.