Answer
$-z_{\frac{α}{2}}\lt z_0\lt z_{\frac{α}{2}}$: null hypothesis is not rejected.
There is not enough significant evidence to conclude that the proportion of American university undergraduate students who have at least one tattoo is different from 0.23.
Work Step by Step
$H_0:~p=0.23$ versus $H_1:~p\ne0.23$
Requirement:
$np_0(1-p_0)=1026\times0.23(1-0.23)=181.7046\gt10$
$p̂ =\frac{x}{n}=\frac{254}{1026}=0.2476$
$z_0=\frac{p̂ -p_0}{\sqrt {\frac{p_0(1-p_0)}{n}}}=\frac{0.2476-0.23}{\sqrt {\frac{0.23(1-0.23)}{1026}}}=1.34$
Using the classical method:
$z_{\frac{α}{2}}=z_{0.05}$
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: $\frac{1.64+1.65}{2}=1.645$
Also, $-z_{\frac{α}{2}}=-z_{0.05}=-1.645$
Since $-z_{\frac{α}{2}}\lt z_0\lt z_{\frac{α}{2}}$, we do not reject the null hypothesis.
