Answer
$z_0\lt z_α$: null hypothesis is not rejected.
There is not enough evidence to conclude that the percentage of American adults who eat salad at least once a week is greater than 85%.
Work Step by Step
$H_0:~p=0.85$ versus $H_1:~p\gt0.85$
Requirement:
$np_0(1-p_0)=200\times0.85(1-0.85)=25.5\gt10$
$p̂ =\frac{x}{n}=\frac{171}{200}=0.855$
$z_0=\frac{p̂ -p_0}{\sqrt {\frac{p_0(1-p_0)}{n}}}=\frac{0.855-0.85}{\sqrt {\frac{0.85(1-0.85)}{200}}}=0.20$
Using the classical method:
$z_α=z_{0.1}$
If the area of the standard normal curve to the right of $z_{0.1}$ is 0.1, then the area of the standard normal curve to the left of $z_{0.1}$ is $1−0.1=0.9$
According to Table V, the z-score which gives the closest value to 0.9 is 1.28.
Since $z_0\lt z_α$, we do not reject the null hypothesis.
