Testing whether a scientists finds a cure or not A scientists claims that he found a drug that in a short time cures depression with a probability of $70\%$. It was tested over a sample of $100$ people and $80$ of them were cured. 
How can we determine whether the scientist afirmation is correct?
First for the null-hypothesis we can set $H_0: p=\frac{7}{10}$ and for the alternative-hypothesis $H_1: p<\frac7{10}$ so this is an one-tailed situation.
I've been given a hint to use the z-score as: 
$$z=\frac{x-np}{\sqrt{np(1-p)}}$$
Where $x$ denotes the number of favourable results and $n$ is the sample's volume.
 $$z=\frac{80-100\cdot \frac{7}{10}}{\sqrt{100\cdot \frac{7}{10}\cdot \frac{3}{10}}}=\frac{10}{\sqrt{21}}=2.182$$
But I don't quite understand why to use exactly this z-score, also what can be concluded if $z=2.182$?
 A: Comment: Perhaps this is a trick question or perhaps it your incorrect version of what is being asked. 
If drug is claimed to be 70% effective in the population, and you want to know whether it meets or exceeds this claim of effectiveness, then $H_0: p \ge 0.7,$ which some authors would write as $H_0: p = 0.7.$ The alternative would be $H_a: p < 0.7.$ 
The null distribution is that the number X cured out of $n = 100$ randomly selected depressed subjects would be $X \sim \mathsf{Binom}(n = 100, p=.7).$ 
If you observe only $X = 67,$ you might wonder if that is sufficiently far below the mean 80 to show that the drug does not meet
expectations. The probability $P(X\le 65 \,|\, p=.7)= 0.1629$ is the P-value of the test. 
Results are a little disappointing, but not bad enough to reject $H_0.$ Ordinarily, rejection would require a P-value $< 0.05.$
(That's testing at the 5% level of significance.) Computation using
R statistical software:
pbinom(65, 100, .7)
[1] 0.1628583

By contrast, if $X = 80$ as you say, then there is clearly no evidence that the
drug is worse than claimed. In that case the P-value is nearly 1.
pbinom(90, 100, .7)
[1] 0.9999996

In practice, whenever you see a P-value very near 1, you should question whether you have the wrong formulation of the problem.
Note: In your question you are using the normal approximation to the binomial distribution. It is always better to use exact probability computations (perhaps using statistical software or calculator) than a normal approximation. Nevertheless, in the situations considered above, you will find that normal approximations are very nearly correct.
