As your testing H0:p=0.79, then you have a two-tailed test.
The p-value at two-tailed test is given by:
[tex]\begin{gathered} p=P(Z\leq-z)+P(Z\ge z) \\ p=P(Z\leq-z)+(1-P(Z\leq z) \\ p=2P(Z\leq-z) \\ 0.79=2P(Z\leq-z) \\ P(Z\leq-z)=\frac{0.79}{2} \\ P(Z\leq-z)=0.395 \end{gathered}[/tex]
Then, you need to check into the standard normal cumulative table to find at which -z you do have a probability of 0.395. Thus:
As you can see in the picture at z=-0.26 you have a P(Z<=-z)=0.3974
And at z=-0.27 you have a P(Z<=-z)=0.3936.
The average of these values is:
[tex]\frac{0.3974+0.3936}{2}=0.3955[/tex]
Then, you have a probability of 0.395 at a z of:
[tex]\frac{-0.26+(-0.27)}{2}=-0.265[/tex]
Then -z=-0.27, thus z=0.27
