Answer
$z_0\lt -z_α$: null hypothesis is rejected.
There is enough evidence to conclude that more than half of U.S. adults tend to not trust the press.
Work Step by Step
$H_0:~p=0.5$ versus $H_1:~p\gt0.5$
$n=2302$ (large sample)
$number~of~plus~signs=1243$ (tend to not trust the press)
$number~of~minus~signs=2302-1243=1059$ (tend to trust the press)
Right-tailed test:
$k=number~of~minus~signs=1059$
$z_0=\frac{(k+0.5)-\frac{n}{2}}{\frac{\sqrt n}{2}}=\frac{(1059+0.5)-\frac{2302}{2}}{\frac{\sqrt {2302}}{2}}=-3.81$
$z_α=z_{0.05}$
If the area of the standard normal curve to the right of $z_{0.05}$ is 0.05, then the area of the standard normal curve to the left of $z_{0.05}$ is $1−0.05=0.95$
According to Table V, there are 2 z-scores which give the closest value to 0.95: 1.64 and 1.65. So, let's find the mean of these z-scores: $\frac{1.64+1.65}{2}=1.645$
So, $-z_α=-z_{0.05}=-1.645$
Since $z_0\lt -z_α$, we reject the null hypothesis.
