Answer
$z_0\lt -z_α$: reject the null hypothesis.
There is enough evidence to conclude that less than 25% of adults prefer mint chocolate chip ice cream.
Work Step by Step
$H_0:~p=0.25$ versus $H_1:~p\lt0.25$
Requirement:
$np_0(1-p_0)=320\times0.25(1-0.25)=60\gt10$
$p̂ =\frac{x}{n}=\frac{58}{320}=0.18125$
$z_0=\frac{p̂ -p_0}{\sqrt {\frac{p_0(1-p_0)}{n}}}=\frac{0.18125-0.25}{\sqrt {\frac{0.25(1-0.25)}{320}}}=-2.84$
Using the Classical Method:
$z_α=z_{0.01}$
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.
