Calculate the z-score first.
[tex]z = \frac{|\mu-\hat{\mu}|}{\frac{s}{\sqrt{n}}}[/tex]
with "s" being the standard deviation within the sample of n=4 and [tex]\hat\mu[/tex] being the sample mean.
So
[tex]z=\frac{|21-22|}{\frac{4}{2}}=\frac{1}{2}[/tex]
Note the absolute value - it takes care of a difference in both directions (sample mean being lower and higher) but we need to keep it mind when looking up the z values in tables - we must choose the "two-tailed test" table.
Now, look up the required z value for alpha=0.05: z>=1.96 (using a two-tailed table).
So our z would have be >=1.96 for the evidence to be significant at the required level of 0.05. The calculated value falls far short at 0.5. Therefore there is no evidence for a significant treatment effect, given the data/sample.