Answer
$z_0\lt -z_{\frac{α}{2}}$: null hypothesis is rejected.
There is enough evidence to conclude that $M\ne68$
Work Step by Step
$H_0:~M=68$ versus $H_1:~M\ne68$
$n=45+27=72$ (large sample)
$number~of~minus~signs=45$
$number~of~plus~signs=27$
$k=27$
$z_0=\frac{(k+0.5)-\frac{n}{2}}{\frac{\sqrt n}{2}}=\frac{(27+0.5)-\frac{72}{2}}{\frac{\sqrt {72}}{2}}=-2.00$
Two-tailed test:
$z_{\frac{α}{2}}=z_{0.025}$
If the area of the standard normal curve to the right of $z_{0.025}$ is 0.025, then the area of the standard normal curve to the left of $z_{0.025}$ is $1−0.025=0.975$
According to Table V, the z-score which gives the closest value to 0.975 is 1.96.
Since $z_0\lt -z_{\frac{α}{2}}$, we reject the null hypothesis.
