Answer
$-z_{\frac{α}{2}}\lt z_0\lt z_{\frac{α}{2}}$: null hypothesis is not rejected.
There is not enough evidence to conclude that $M\ne100$
Work Step by Step
$H_0:~M=100$ versus $H_1:~M\ne100$
$n=21+28=49$ (large sample)
$number~of~minus~signs=21$
$number~of~plus~signs=28$
$k=21$
$z_0=\frac{(k+0.5)-\frac{n}{2}}{\frac{\sqrt n}{2}}=\frac{(21+0.5)-\frac{49}{2}}{\frac{\sqrt {49}}{2}}=-0.86$
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 $-z_{\frac{α}{2}}\lt z_0\lt z_{\frac{α}{2}}$, we do not reject the null hypothesis.
