Answer
$z_0\lt z_α$: null hypothesis is not rejected.
There is not enough evidence to conclude that a majority of pregnant women nap at least twice a week.
Work Step by Step
$H_0:~p=0.5$ versus $H_1:~p\gt0.5$
Requirement:
$np_0(1-p_0)=150\times0.5(1-0.5)=37.5\gt10$
$p̂ =\frac{x}{n}=\frac{81}{150}=0.54$
$z_0=\frac{p̂ -p_0}{\sqrt {\frac{p_0(1-p_0)}{n}}}=\frac{0.54-0.5}{\sqrt {\frac{0.5(1-0.5)}{150}}}=0.98$
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: $\frac{1.64+1.65}{2}=1.645$
Since $z_0\lt z_α$, we do not reject the null hypothesis.
