Answer
(a) $H_0:~p_A=p_B$ versus $H_1:~p_A\ne p_B$
(b) $z_0=0.70$
(c) $z_{\frac{α}{2}}=z_{0.025}=1.96$
(d) $P$-value $=0.484$
The null hypothesis is not rejected.
Work Step by Step
(a) $H_0:~p_A=p_B$ versus $H_1:~p_A\ne p_B$
(b) $z_0=\frac{|f_{12}-f_{21}|-1}{\sqrt {f_{12}+f_{21}}}=\frac{|19-14|-1}{\sqrt {19+14}}=\frac{4}{\sqrt {33}}=0.70$
(c) $z_{\frac{α}{2}}=z_{0.025}$
If the area of the standard normal curve to the right of $z_{0.025}$ is 0.025, then the area of the standard normal curve to the left of $z_{0.025}$ is $1−0.025=0.975$
According to Table V, the z-score which gives the closest value to 0.975 is 1.96.
(d) $P$-value $=2P(z\gt z_0)=2P(z\gt0.70)=2[1-P(z\lt0.70)]=2(1-0.7580)=0.484$
Since $z_0\lt z_{\frac{α}{2}}$ we do not reject the null hypothesis.
