Answer
$z_0\lt z_α$: null hypothesis is rejected.
There is enough evidence to conclude that $M_D\lt0$.
Work Step by Step
Left-tailed test.
Test statistic: $T=T_+=210$
Large-sample case:
$z_0=\frac{T-\frac{n(n+1)}{4}}{\sqrt {\frac{n(n+1)(2n+1)}{24}}}=\frac{210-\frac{35(35+1)}{4}}{\sqrt {\frac{35(35+1)(2\times35+1)}{24}}}=-1.72$
$z_α=z_{0.05}$
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, $-z_α=-z_{0.05}=-1.645$
Since $z_0\lt z_α$, we reject the null hypothesis.
