Answer
$z_0\lt -z_α$: null hypothesis is rejected.
There is enough evidence to conclude that a lower proportion of women in the experimental group experienced a bone fracture than the women in the control group.
Work Step by Step
$N_1,n_1~and~p_1$ refer to experimental group and $N_2,n_2~and~p_2$ refer to control group.
$H_0:~p̂ _1=p̂ _2$ versus $H_1:~p̂ _1\lt p̂ _2$
$p̂ _1=\frac{x_1}{n_1}=\frac{27}{696}=0.0388$ and $p̂ _2=\frac{x_2}{n_2}=\frac{49}{678}=0.0723$
Requirements:
$n_1p̂ _1(1-p̂ _1)=696\times0.0388(1-0.0388)=25.96\geq10$
$n_2p̂ _2(1-p̂ _2)=678\times0.0723(1-0.0723)=45.48\geq10$
$n_1\leq0.05N_1$
$n_2\leq0.05N_2$
$p̂ =\frac{x_1+x_2}{n_1+n_2}=\frac{27+49}{696+678}=0.0553$
$z_0=\frac{p̂_1-p̂ _2}{\sqrt {p̂ (1-p̂ )}\sqrt {\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{0.0388-0.0723}{\sqrt {0.0553(1-0.0553)}\sqrt {\frac{1}{696}+\frac{1}{678}}}=-2.72$
Left-tailed test:
$z_α=z_{0.01}$
If the area of the standard normal curve to the right of $z_{0.01}$ is 0.01, then the area of the standard normal curve to the left of $z_{0.01}$ is $1−0.01=0.99$
According to Table V, the z-score which gives the closest value to 0.99 is 2.33
So, $-z_α=-2.33$
Since $z_0\lt -z_α$, we reject the null hypothesis.
