Answer
$z_0\lt z_α$: reject the null hypothesis.
There is enough evidence to conclude that the proportion of adult Americans who are current smokers has declined.
Work Step by Step
$H_0:~p=0.21$ versus $H_1:~p\lt0.21$
$np_0(1-p_0)=199\times0.21(1-0.21)=33.0141\gt10$
$p̂ =\frac{x}{n}=\frac{25}{199}=0.1256$
$z_0=\frac{p̂ -p_0}{\sqrt {\frac{p_0(1-p_0)}{n}}}=\frac{0.1256-0.21}{\sqrt {\frac{0.21(1-0.21)}{199}}}=-2.92$
Using the classical method:
$z_α=z_{0.05}$
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, $-z_α=-1.645$
Since $z_0\lt z_α$, we reject the null hypothesis.
