Answer
There is enough evidence to conclude that the sentence structure makes a difference in deciding whether the politician would be re-elected.
Work Step by Step
$N_1,n_1~and~p_1$ refer to sentence A and $N_2,n_2~and~p_2$ refer to sentence B.
$H_0:~p̂ _1=p̂ _2$ versus $H_1:~p̂ _1\ne p̂ _2$
$p̂ _1=\frac{x_1}{n_1}=\frac{71}{98}=0.7245$ and $p̂ _2=\frac{x_2}{n_2}=\frac{49}{98}=0.5$
Requirements:
$n_1p̂ _1(1-p̂ _1)=98\times0.7245(1-0.7245)=19.56\geq10$
$n_2p̂ _2(1-p̂ _2)=98\times0.5(1-0.5)=24.5\geq10$
$n_1\leq0.05N_1$
$n_2\leq0.05N_2$
$p̂ =\frac{x_1+x_2}{n_1+n_2}=\frac{71+49}{98+98}=0.6122$
$z_0=\frac{p̂_1-p̂ _2}{\sqrt {p̂ (1-p̂ )}\sqrt {\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{0.7245-0.5}{\sqrt {0.6122(1-0.6122)}\sqrt {\frac{1}{98}+\frac{1}{98}}}=3.23$
Two-tailed test:
Let's use $α=0.01$ level of significance.
$z_\frac{α}{2}=z_{0.005}$
If the area of the standard normal curve to the right of $z_{0.005}$ is 0.005, then the area of the standard normal curve to the left of $z_{0.005}$ is $1−0.005=0.995$
According to Table V, there are 2 z-scores which give the closest value to 0.995: 2.57 and 2.58. So, let's find the mean of these z-scores: $\frac{2.57+2.58}{2}=2.575$
Since $z_0\gt z_{\frac{α}{2}}$, we reject the null hypothesis.
Notice that the null hypothesis would be rejected even for the $α=0.05$ or $α=0.10$ level of significance. They would provide a lower value of $z_\frac{α}{2}$.
