If Zrb is the true population proportion of boys presenting, H0: Zrb = zr and Hi: Zrb r n. Then, Pb = 0.68. Using the normal approximation, a = 0.0707. From Eq. (15.8), z = (IPb -- nl)/a -- 0.18/0.0707 = 2.55. From Table I, z = 2.55 yields two-tailed p = 0.010. We reject H0 and conclude that more boys than girls present with limb fractures.