The test you are running is:
[tex]\begin{gathered} H_0\colon p=0.55 \\ H_a\colon p>0.55 \end{gathered}[/tex]
Then, you have a right-tailed p-value, then:
[tex]\begin{gathered} p-value=P(Z>z_0) \\ p-value=1-P(Z\le z_0) \end{gathered}[/tex]
By replacing p-value =0.55:
[tex]\begin{gathered} 0.55=1-P(Z\le z_0) \\ P(Z\le z_0)=1-0.55 \\ P(Z\le z_0)=0.45 \end{gathered}[/tex]
Then, in a standard normal table look for the corresponding z:
As you can see, z=-0.12 has a p-value=0.4522 and z=-0.13 has a p-value=0.4483.
Then, the average will be:
[tex]\begin{gathered} \frac{0.4522+0.4483}{2}=0.4502 \\ \text{Then the z-score is:} \\ \frac{-0.12+(-0.13)}{2}=-0.125 \end{gathered}[/tex]
Answer: the test statistic z=-0.125
